Category: CBSE Sample Papers

Students must start practicing the questions from CBSE Sample Papers for Class 10 Science with Solutions Set 1 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Science Set 1 with Solutions

Time : 3 Hr.
Max. Marks : 80

General Instructions:

• This question paper consists of 39 questions in 5 sections.
• Alt questions are compulsory. However, an internal choice is provided in some questions. A student is expected to attempt only one of these questions.
• Section A consists of 20 objective type questions carrying 1 mark each.
• Section B consists of 6 Very Short questions carrying 02 marks each. Answers to these questions should in the range of 30 to 50 words.
• Section C consists of 7 Short Answer type questions carrying 03 marks each. Answers to these questions should in the range of 50 to 80 words.
• Section D consists of 3 Long Answer type questions carrying 05 marks each. Answer to these questions should be in the range of 80 to 120 words.
• Section E consists of 3 source-based/case-based units of assessment of 04 marks each with sub-parts.

SECTION – A (20 Marks)
(Select and write one most appropriate option out of the four options given for each of the questions 1-20)

Question 1.
The change in colour of the moist litmus paper in the given set up is due to (1)

(I) presence of acid
(II) presence of base
(III) presence of H (ag) in the solution
(IV) presence of Litmus which acts as an indicator
(a) (I) and (II)
(b) Only (II)
(c) Only (III)
(d) Only (IV)
Answer:
(c) Only (III)

Explanation: The colour of litmus paper changes only in the presence of ions like hydrogen (H+) or hydronium (H30+) ions. H₂SO₂ can produce these ions only in an aqueous solution. Hence dry H₂SG₄ gas does not change the colour of dry litmus paper.

Question 2.
In the redox reaction (1)
MnO₂ + 4HCI -> MnCl₂ + 2H₂O + Cl₂
(a) MnO₂ is reduced to MnCl₂ & HCl is oxidized to H₂O
(b) MnO₂ is reduced to MnCl₂ & HCl is oxidized to Cl₂
(c) MnO₂ is oxidized to MnCl₂ & HCl is reduced to Cl₂
(d) MnO₂ is oxidized to MnCl₂ & HCl is reduced to H₂O
Answer:
(b) MnO₂ is reduced to MnCl₂ & HCL is oxidized to Cl₂

Explanation: In the reaction described, Mn is reduced from +4 to +2, and CL is oxidised from -1 to 0.

The substance reduced is MnO₂ and the substance oxidised is HCl. Hence MnO₂ is reduced to MnCl₂ and HCl is oxidised to Cl₂.

Caution
Students should know that as MnO₂ is being reduced and oxidized HCl should be acting as an oxidising agent in the reaction. While reduced MnO₂ acts as a reducing agent as HCl is being oxidised. It is crucial to correctly assign the oxidation number in order to identify the species being oxidised and reduced.

Question 3.

Which of the following is the correct observation of the reaction shown in the above set up? (1)
(a) Brown powder of Magnesium oxide is formed.
(b) Colourless gas which turns lime water milky is evolved.
(c) Magnesium ribbon burns with brilliant white light.
(d) Reddish brown gas with a smell of burning Sulphur has evolved.
Ansnswer:
(c) Magnesium ribbon burns with brilliant white light

Explanation: Magnesium ribbon burns with a dazzling white Light and produces a white povVder which is magnesium oxide as shown in the figure. The reaction can be represented as:
2Mg + O₂ → 2MgO
(White powder)

Question 4.
With the reference to four gases CO₂, CO, Cl₂ and O₂, which one of the options in the table is correct? (1)

Option	Acidic oxide	Used in treat ment of water	Product of respiration	Product of incom plete combustion
(a)	CO	Cl2	O2	CO
(b)	CO2	Cl2	CO2	CO
(C)	CO2	O2	O2	CO2
(d)	CO	O2	CO2	CO2

Answer:
(b) CO₂, Cl₂, CO₂, CO

Explanation: Since carbon dioxide (CO₂) dissolves in water to produce carbonic acid, which is acid, it is an acidic oxide.

Chlorine (Cl₂) is most frequently used to disinfect water during water treatment. It has disadvantages as a disinfectant, but it also has benefits.

The biochemical process known as respiration is where an organism’s cells obtain energy by combining oxygen and glucose, which releases carbon dioxide (CO₂), water, and ATP.

Incomplete combustion produces soot or carbon monoxide (CO) when some of the carbon is not fully oxidised.

Question 5.
On placing a copper coin in a test tube containing green ferrous sulphate solution, it will be observed that the ferrous sulphate solution. (1)
(a) turns blue, and a grey substance is deposited on the copper coin.
(b) turns colourless and a grey substance is deposited on the copper coin.
(c) turns colourless and a reddish-brown substance is deposited on the copper coin.
(d) remains green with no change in the copper coin.
Answer
(d) Ferrous sulphate solution remains green with no change in the copper coin

Explanation: Since Cu is less reactive than Fe, it cannot replaced from the solution, so there won’t be any reaction. Due to the lack of a reaction, the given solution will continue to be yellowish-green in colour, whose chemical reaction can be described as

Question 6.
Anita added a drop each of diluted acetic acid and diluted hydrochloric acid on pH paper and compared the colors. Which of the following is the correct conclusion? (1)
(a) pH of acetic acid is more than that of hydrochloric acid.
(b) pH of acetic acid is less than that of hydrochloric acid.
(c) Acetic acid dissociates completely in aqueous solution.
(d) Acetic acid is a strong acid
Answer:
(a) pH of acetic acid is more than that of hydrochloric acid

Explanation: A solution of HCl has a lower pH than an acetic acid solution of the same concentration. Strong and fully dissociated, HCl is an acid. Acetic acid is a dissociated and weak acid. As a result, HCl has a higher hydrogen ion concentration than acetic acid. Lower is the pH, greater is the hydrogen ion concentration and greater is the acid strength.

Question 7.
The formulae of four organic compounds are shown below. Choose the correct option (1)

(a) A and B are unsaturated hydrocarbons
(b) C and D are saturated hydrocarbons
(c) Addition of hydrogen in presence of catalyst changes A to C
(d) Addition of potassium permanganate changes B to D
Answer:
(c) Addition of hydrogen in presence of catalyst changes A to C

Explanation: Ethene, an unsaturated hydrocarbon, undergoes hydrogenation in the presence of catalysts made of nickel, platinum, or palladium to produce ethane, a saturated hydrocarbon. Adding hydrogen to a double bond causes it to become a single bond, a process known as hydrogenation.

Question 8.
In the given transverse section of the leaf identify the layer of cells where maximum photosynthesis occurs.

(a) (I), (II)
(b) (II), (III)
(c) (III), (IV)
(d) (I), (IV)
Explanation: The given figure represents the transverse section of a leaf:

There are two distinct types of mesophyll tissue in some plant species (dicots and some monocots). While spongy mesophyll cells are arranged more loosely to allow gases to pass through them, palisade mesophyll cells are closely clustered together. Additionally, compared to spongy mesophyll cells, palisade mesophyll cells have more chloroplasts.

Question 9.
Observe the experimental setup shown below. Name the chemical indicated as ‘X’ that can absorb the gas which is evolved as a byproduct of respiration. (1)

(a) NaOH
(b) KOH
(c) Ca(OH)₂
(d) K₂CO₃
Ansnswer:
(b) KOH

Explanation: In the experiment, CO₂ released by the plant during respiration is absorbed using potassium hydroxide. KOH absorbs CO₂, which causes the flask to become vacuum- sealed. The conical flask receives the air that is in the bent glass tube. This raises the water level in the curved tube.

Question 10.
If a tall pea plant is crossed with a pure dwarf pea plant then, what percentage of F₁ and F₂ generation respectively will be tall? (1)
(a) 25%, 25%
(b) 50%, 50%
(c) 75%,100%
(d) 100%, 75%
Answer:
(d) 100%, 75%

Explanation: Mendel bred a pure dwarf plant (tt) with a pure tall plant (TT). 100% of the tall plants (Tt) in the F₁ generation are produced. Self-pollination on an F₁ plant results in the formation of 3 tall plants and 1 dwarf plant in the F₂ generation. Therefore, 75% of F₂ generation plants are tall.

Question 11.
Observe the three figures given below. Which of the following depicts tropic movements appropriately? (1)

(a) (II) and (III)
(b) (I) and (III)
(c) (II) only
(d) (III) only
Answer:
(d) (III) only

Explanation: Figure III is correct. Roots are positive to the geotropism while stems are negative to it. The primary roots and a few other areas of the root system, among other plant parts, exhibit positive geotropism by expanding directly in the direction of the centre of gravity. Because they extend away from the centre of gravity, the stems are known as negatively geotropic.

Question 12.
The diagram shown below depicts pollination. Choose the options that will show a maximum variation in the offspring. (1)

(a) A, B and
(b) B and D
(c) B, C and D
(d) A and C
Answer:
(b) B and D

Explanation: Male and female gametes are formed from the same parent plant in self-pollination (shown by “A” and “C”), and the progeny produced almost exactly resemble the parent plant. As they grow on the same plant, this indicates that the genetic makeup of the male and female flowers is the same. Cross-pollination, also known as xenogamy (represented by the letters “B” and “D”), allows two genetically different plant characteristics from the same species to fuse. Due to the mixing of genetically diverse gametes, it causes genetic recombination and variability in plants.

Question 13.
A complete circuit is left on for several minutes, causing the connecting copper wire to become hot. As the temperature of the wire increases, the electrical resistance of the wire (1)
(a) decreases.
(b) remains the same.
(c) increases.
(d) increases for some time and then decreases.
Answer:
(c) increases

Explanation: The thermal velocity of the free electrons increases as the temperature rises, which causes the resistance of a conductor to rise. The number of collisions between the free electrons increases as a result of this. The resistance will rise as the temperature of the metallic conductor rises.

Related Theory
When a conductor is involved, the ions inside the metal conductor gather energy and start to oscillate around their mean positions as the temperature rises. All of these oscillating ions hit electrons and increase resistance.

Question 14.
A copper wire is held between the poles of a magnet.

The current in the wire can be reversed. The pole of the magnet can also be changed over. In how many of the four directions shown can the force act on the wire?
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2 (Either North or South)

Explanation: By applying Fleming’s left-hand rule, we can determine that the force acting on the wire is parallel to both the magnetic field and the wire’s current. As a result, there are only two options for the force’s direction: either upward or downward.

Question 15.

Plastic insulation surrounds a wire having diameter d and length l as shown above. A decrease in the resistance of the wire would be produced by an increase in the
(a) length l of the wire
(b) diameter d of the wire
(c) temperature of the wire
(d) thickness of the plastic insulation
Answer:
(b) diameter d of the wire

Explanation: Resistance of a wire is directly proportional to the length of the conductor and inversely proportional to the area of the cross-section.
That is R ∝ \(\frac { L }{ A }\)
From the formula, we can observe that the resistance is inversely proportional to the cross – section area. So, a decrease in the resistance of the wire would be produced by an increase in the diameter of the wire.

Question 16.
Which of the following pattern correctly describes the magnetic field around a long straight wire carrying current?
(a) straight lines perpendicular to the wire.
(b) straight lines parallel to the wire.
(c) radial lines originating from the wire.
(d) concentric circles centred around the wire.
Answer:
(d) The field consists of concentric circles centred around the wire.

Explanation: A long, straight wire has a magnetic field that consists of concentric circles. They are centred on the wire. Concentric circles centred on the wire make up the magnetic field lines of a long, straight wire. The right- hand thumb rule provides the magnetic field Lines’ direction.

Q. no 17 to 20
are Assertion – Reasoning based questions.
These consist of two statements – Assertion
(A) and Reason (R). Answer these questions selecting the appropriate option given below:
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true and R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true

17. Assertion (A): Silver bromide decomposition is used in black and white photography.
Reason (R): Light provides energy for this exothermic reaction. (1)
Answer:
(c) A is true but R is false

Explanation: When exposed to sunlight, a substance called silver bromine breaks down into sodium metal and bromine gas. This reaction is an endothermic photolytic decomposition reaction because energy is being absorbed and used.

Question 18.
Assertion (A): Height in pea plants is controlled by efficiency of enzymes and is thus genetically controlled.
Reason (R): Cellular DNA is the information source for making proteins in the cell. (1)
Answer:
(a) Both A and R are true and R is the correct explanation of A

Explanation: Height in pea plants is controlled by effciency of enzymes and is thus genetically the structure of DNA. A genome sequence is another name for this particular sequence. The genome sequence is transcriptionally transcribed for each unique protein. A ribosome translates this mRNA into an amino acid sequence, in order to create a particular protein, this amino acid sequence is then altered.

Question 19.
Assertion (A): Amphibians can tolerate mixing of oxygenated and deoxygenated blood.
Reason (R): Amphibians are animals with two chambered heart (1)
Answer:
(c) A is true but R is false

Explanation: The hearts of amphibians have three chambers: two atria and one ventricle. Because of how the withdrawals occur between the atria, the mixing of oxygenated and deoxygenated blood is kept to a minimum. Since amphibians don’t require a lot of energy, they cam withstand some amounts of blood that is both oxygenated and deoxygenated mixing.

Question 20.
Assertion (A): On freely suspending a current – carrying solenoid, it comes to rest in Geographical N-S direction.
Reason (R): One end of current carrying straight solenoid behaves as a North pole and the other end as a South pole, just like a bar magnet. (1)
Answer:
(a) Both A and R are true and R is the correct explanation of A

Explanation: A bar magnet with fixed orientations at both ends and current carrying resembles a freely hanging solenoid. The current-carrying linear electromagnet has two ends, one of which functions as the North Pole and the other as the South Pole. As a result, it is oriented similarly to a bar magnet in the North-South direction.

SECTION – B (12 Marks)
(Q. no. 21 to 26 are very short answer questions)

Question 21.
A clear solution of slaked lime is made by dissolving Ca(OH)₂ in an excess of water. This solution is left exposed to air. The solution slowly goes milky as a faint white precipitate forms. Explain why a faint white precipitate forms, support your response with the help of a chemical equation. (2)
OR
Keerti added dilute Hydrochloric acid to four metals and recorded her observations as shown in the table given below:

Metal	Gas Evolved
Copper	Yes
Iron	Yes
Magnesium	No
Zinc	Yes

Select the correct observation(s) and give chemical equation(s) of the reaction involved.

Total Marks	Breakdown (As per CBSE Marking Scheme)
2m (VSA)	Give Reason (1m) Write Chemical Equation (1m)

Answer:
Calcium hydroxide reacts with carbon dioxide present in the atmosphere to form calcium carbonate which results in milkiness/ white ppt / formation of calcium carbonate
Ca(OH)₂ + CO₂ -> CaCO₃ + H₂O

Total Marks	Breakdown (As per CBSE Marking Scheme)
2m (VSA)	Write Metal(s) that evolve gas (1m) Write Chemical Equation(s) of those metals (1m) No mark will be deducted for unbalanced equations and states of

Fe + HCL FeCl₂/ FeCl₃ + H₂ (No deduction for balancing/ states)
Zn + HCl ZnCl₂ + H₂

Question 22.
How is the mode of action in beating of the heart different from reflex actions? Give four examples. (2)

Total Marks	Breakdown (As per CBSE Marking Scheme)
2m (VSA)	Write any four differences (0.5 m+0.5 m+0.5 m+0.5 m)

Answer:

Beating of heart	Reflex actions
Involuntary actions are the actions which are not controlled by our will.	Reflex actions are the sudden action in response to something
They do not need any kind of stimulus to work.	They required stimulus for its action.
These actions are regulated by the brain.	These actions are regulated by the spinal cord.
They do not involve skeletal muscle.	They do involve skeletal muscle.
These actions are performed throughout one’s life.	These actions area produced in response to an event of an emergency.
This action may be quick or slow.	Reflex actions are always quick.

Question 23.
Patients whose gall bladder are removed are recommended to eat less oily food. Why? (2)

Total Marks	Breakdown (As per CBSE Marking Scheme)
2m (VSA)	Function of Gall bladder (1m) Impact in absence of Gall bladder (1m) Keywords: bile, emulsification of fats

Answer:
Gall bladder stores bile which helps in emulsification of lipids (lmark). In the absence of stored bile, emulsification of fats will be negligible/ affected/ less (1 mark) and thus fat digestion will be slow. Hence there are such diet restrictions.

Question 24.
Name the substances other than water, that are reabsorbed during urine formation. What are the two parameters that decide the amount of water that is reabsorbed in the kidney?

Total Marks	Breakdown (As per CBSE Marking Scheme)
2m (VSA)	Write two substances (0.5 m+0.5 m) Two parameters (0.5 m+0.5 m)

Answer:
Glucose, amino acids, salts (any 2, 1 mark each) and a major amount of water are selectively re-absorbed as the urine flows along the tube. The amount of water reabsorbed depends on how much excess water there is in the body (0.5 marks), and on how much of dissolved waste there is to be excreted (0.5marks)

Question 25.

State the phenomena observed in the above diagram. Explain with reference to the diagram, which of the two lights mentioned above will have the higher wavelength?
OR
How will you use two identical prisms so that a narrow beam of white lightincident on one prism emerges out of the second prism as white light? Draw the diagram. (2)

Total Marks	Breakdown (As per CBSE Marking Scheme)
2m (VSA)	Name of phenomena +1 line defintion (1m) Higher wavelength – Yellow or Blue? (0.5m) Why? (0.5 m)

Answer:
Dispersion – The splitting of white light into seven colours on passing through a prism. Velocity is directly proportional to wavelength given constant frequency. So yellow will have greater wavelength than blue as the velocity of yellow light is greater than blue.

Total Marks	Breakdown (As per CBSE Marking Scheme)
2m (VSA)	Explain the method (1 m) Diagram with labellings (1m) Key focus: angle of deflection

OR
Angle of deflections of the two prisms need to be equal and opposite. While the first prism splits the light in the seven colours due to different angles of deflection, the second prism combines the spectrum along a single ray and the colours again combine to give white light as the emergent light.

Question 26.
A lot of waste is generated in neighbourhood. However, almost all of it is biodegradable. What impact will it have on the environment or human health? (2)

Total Marks	Breakdown (As per CBSE Marking Scheme)
1m (VSA)	harmful impact on environment (1m) harmful impact on human health (1m)

Answer:
Excess generation of biodegradable wastes can be harmful as – Its decomposition is a slow process leading to production of foul smell and gases. It can be the breeding ground for germs that create unhygienic conditions.

SECTION – C (21 Marks)
(Q. no. 27 to 33 are short answer questions.)

Question 27.

Identify the types of reaction mentioned above in (A) and (B). Give one example for each type in the form of a balanced chemical equation. (3)

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (VSA)	Type of Reaction in (A) and (B) (0.5 m+0.5 m) Balanced chemical equation as example in each (1m + 1m) Try to write extra balanced chemical equations (examples) just to be safe with getting balancing right

Answer:
(A) Displacement – 1 / 2 M

Fe_(s) + CuSO_4(aq) —> FeSO_4(aq) + Cu_(s)
Zn_(s) + CuSO_4(aq) —> ZnSO_4(aq) + Cu_(s)
Pb_(s) + CuCl_2(aq) —> PbCl_2(aq) + Cu_(s)

(Any one of the reaction or other displacement reaction.)

(B) Double displacement
Na₂SO_4(aq) + BaCl_2(aq) -> BaSO_4(s) + 2NaCl_(aq)
(Any one of the reaction or other double displacement reaction.)

Question 28.

(A) Identify the gases evolved at the anode and cathode in the above experimental set up.
(B) Name the process that occurs. Why is it called so?
(C) Illustrate the reaction of the process with the help of a chemical equation. (3)

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (VSA)	(a) Gas released at Anode and Cathode (0.5 m+0.5 m) (b) Name and Reason (0.5 m+0.5 m) (c) Balanced chemical equation (1m)

Answer:
(A) Anode: Chlorine; Cathode: Hydrogen
(B) Chlor alkali process as the products obtained are alkali, chlorine gas and hydrogen gas Electric current

Question 29.
The leaves of a plant were covered with aluminium foil, how would it affect the physiology of the plant?
OR
How is lymph an important fluid involved in transportation? If lymphatic vessels get blocked, how would it affect the human body? Elaborate. (3)

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (VSA)	3 points – impact on physiology (1 m + 1 m + 1 m) Key focus: Photosynthesis, Transpiration and Temperature regulation

Answer:
No photosynthesis will occur so no glucose will be made. Also no respiration will take place as no Oxygen will be taken in. No transpiration will occur so there would be no upward movement of water or minerals from the soil as there will be no transpirational pull. Temperature regulation of leaf surface will be affected.

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (VSA)	Two points on importance of Lymph (1 m + 1 m) Two impacts of blockage (0.5 m+0.5 m)

Lymph carries digested and absorbed fat from the intestine and drains excess fluid from extracellular space back into the blood. Blockage of lymphatic system will lead to water retention and poor fat absorption in the body.

Question 30.
Rohit wants to have an erect image of an object using a converging mirror of focal length 40 cm.
(A) Specify the range of distance where the object can be placed in front of the mirror. Justify.
(B) Draw a ray diagram to show image formation in this case.
(C) State one use of the mirror based on the above kind of image formation. (3)

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (VSA)	(A) Range in and reason (0.5 m + 0.5 m) (B) Ray diagram (1m) (C) 1 practical use of this concept (1m) Ray diagram should include complete labelling (A’ B’) of inverted image

Answer:
(A) The object has to be placed at a distance between 0-40 cm. This is because image is virtual, erect and magnified when the object is placed between F and P.
(B)

(C) Used as shaving mirror or used by dentists to get enlarged image of teeth.
(any one use)

Question 31.
(A) A lens of focal length 5 cm is being used by Debashree in the laboratory as a magnifying glass. Her least distance of distinct vision is 25 cm. What is the magnification obtained by using the glass?
(B) Ravi kept a book at a distance of 10 cm from the eyes of his friend Hari. Hari is not able to read anything written in the book. Give reasons for this? (3)

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (VSA)	(A) Magnification formula with value input (1m) Correct answer (1m) (B) Reason (1 point only) (1m)

Answer:
(A) Given, image distance = v = – 25 cm, focal
length = f= 5 cm, magnification = m = ?
From lens formula,
\(\frac { 1 }{ f }\) = \(\frac { 1 }{ v }\) – \(\frac { 1 }{ u }\) = \(\frac { 1 }{ u }\) = \(\frac { 1 }{ v }\) – \(\frac { 1 }{ f }\)
\(\frac { 1 }{ u }\) = \(\frac { 1 }{ -25 }\) – \(\frac { 1 }{ 5 }\)
= \(\frac { -1-5 }{ 25 }\) = \(\frac { -6 }{ 25 }\)
Object distance = u = \(\frac { -25 }{ 6 }\) cm
We know that, m = \(\frac { v }{ u }\) = \(\frac { 25 x 6 }{ -25 }\) = 6

(B) This is because the least distance of distinct vision is 25 cm.

Question 32.
A student fixes a white sheet of paper on a drawing board. He places a bar magnet in the centre and sprinkles some iron filings uniformly around the bar magnet. Then he taps gently and observes that iron filings arrange themselves in a certain pattern.
(A) Why do iron filings arrange themselves in a particular pattern?
(B) Which physical quantity is indicated by the pattern of field lines around the bar magnet?
(C) State any two properties of magnetic field lines. (3)
OR
A compass needle is placed near a current carrying wire. State your observations for the following cases and give reasons for the same in each case:
(A) Magnitude of electric current in wire is increased.
(B) The compass needle is displaced away from the wire.

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (VSA)	(a) Explain via property of magnets (b) Name the quantity (c) Point out two properties (0.5 m + 0.5 m)

Answer:
(A) When iron filings are placed in a magnetic field around a bar magnet, they behave like tiny magnets. The magnetic force experienced by these tiny magnets make them rotate and align themselves along the direction of field lines.

(B) The physical property indicated by this arrangement is the magnetic field produced by the bar magnet.

(C) Magnetic field Lines never intersect, magnetic field lines are closed curves.

OR

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (VSA)	(A) Write observation and reason (0.75 m + 0.75 m) (B) Write observation and reason (0.75 m + 0.75 m)

(A) The deflection in the compass needle increases as Magnetic field of the current carrying conductor is directly proportional to current flowing through it

(B) The deflection in the needle decreases as the magnetic field is inversely proportional to the perpendicular distance from the wire.

Question 33.
Why is damage to the ozone layer a cause for concern? What are its causes and what steps are being taken to limit this damage? (3)

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (VSA)	Explain wr.t harm it causes (1m) What element(s) causes this (1m) Any 1 step taken (1m) Key focus: UV Radiations, CFC and UNEP

Answer:
Damage to the ozone layer is a cause for concern because the ozone layer shields the surface of earth from harmful UV radiations from the sun which cause skin cancer in human beings.

Synthetic chemicals like chlorofluorocarbons (CFCs) which are used as refrigerants and in the fire – extinguishers are the main reason for the depletion of the ozone layer.

Steps taken to limit this damage – Many developing and developed countries have signed and are obeying the directions of UNEP (United Nations Environment Programme) to freeze or limit the production and usage of CFCs at 1986 levels.

SECTION – D (15 Marks)
(Q. no. 34 to 36 are Long answer questions.)

Question 34.
Shristi heated Ethanol with a compound A in presence of a few drops of concentrated sulphuric acid and observed a sweet smelling compound B is formed. When B is treated with sodium hydroxide it gives back Ethanol and a compound C.
(A) Identify A and C.
(B) Give one use each of compounds A and B.
(C) Write the chemical reactions involved and name the reactions.
OR
(A) What is the role of concentrated Sulphuric acid when it is heated with Ethanol at 443 K? Give the reaction involved.
(B) Reshu by mistake forgot to label the two test tubes containing Ethanol and Ethanoic acid. Suggest an experiment to identify the substances correctly? Illustrate the reactions with the help of chemical equations.(5)

Total Marks	Breakdown (As per CBSE Marking Scheme)
5m (LA)	(a) Name of A and C (0.5 m + 0.5 m) (b) One application of each A and B (1 m + 1 m) (c) Write balanced chemical equations and the type of reactions (1 m + 1 m)

Answer:
(A) A – Ethanoic acid/ Or any other carboxylic acid, C- Sodium salt of ethanoic acid/ any other carboxylic acid/ sodium ethanoate
(B) Use of A- dil solution used as vinegar in cooking/ preservative in pickles Use of B – making perfumes, flavoring agent

OR

Total Marks	Breakdown (As per CBSE Marking Scheme)
5m (LA)	(A) Function of acid and chemical equation (1 m+1 m) (B) Suggest an experiment, its result and chemical equation involved (1 m + 1 m + 1 m)

Answer:
(A) Sulphuric acid acts as dehydrating agent

(B) By reaction with sodium carbonate/ bi carbonate 1M with the samples, ethanol will not react whereas ethanoic acid gives brisk effervescence
2CH₃COOH + Na₂CO₃ x 2CH₃COONa + H₂O + CO₂
OR
CH₃COOH + NaHCO₃ x CH₃COONa + H₂O + CO₂

Question 35.
(A) Why is it not possible to reconstruct the whole organism from a fragment in complex multicellular organisms?
(B) Sexual maturation of reproductive tissues and organs are necessary link for reproduction. Elucidate.
OR
(A) How are variations useful for species if there is drastic alteration in the niches?
(B) Explain how the uterus and placenta provide necessary conditions for proper growth and development of the embryo after implantation? (5)

Total Marks	Breakdown (As per CBSE Marking Scheme)
5m (LA)	(A) Write reason in two points (1 m + 1 m) (B) Explain within two points (1.5 m + 1.5 m) Keywords: Specialised cells, cell division

(A) The reason is that many multi-cellular organisms are not simply a random collection of cells. Specialised cells are organised as tissues, and tissues are organised into organs, which then have to be placed at definite positions in the body. Therefore, cell-by-cell division would be impractical.

(B) Sexual maturation of reproductive tissues is a necessary link for reproduction because of the need for specialised cell called germ-cells to participate in sexual reproduction. The body of the individual organism has to grow to its adult size, the rate of general body growth begins to slow down, reproductive tissues begin to mature.

A whole new set of changes in the appearance of the body takes place like change in body proportions, new features appear. This period during adolescence is called puberty.

There are also changes taking place that are different between boys and girls. In girts, breast size begins to increase, with darkening of the skin of the nipples at the tips of the breasts. Also, girls begin to menstruate at around this time. Boys begin to have new thick hair growth on the face and their voices begin to crack

OR

Total Marks	Breakdown (As per CBSE Marking Scheme)
5m (LA)	(A) Importance of variations (2m) (B) Write four points to explain and role of placenta (0.5 m + 0.5 m + 0.5 m + 0.5 m + 1 m) Key focus: Lining of uterus, villi, placenta, rhythmic contractions

(A) If the niche were drastically altered, the population could be wiped out. However, if some variations were to be present in a few individuals in these populations, there would be some chance for them to survive. Variation is thus useful for the survival of species over time.

(B)

• The lining of the uterus thickens and is richly supplied with blood to nourish the growing embryo.
• The embryo gets nutrition from the mother’s blood with the help of placenta. It is embedded in the uterine wall.
• It contains villi on the embryo’s side of the tissue. On the mother’s side are blood spaces, which surround the villi.
• This provides a large surface area for glucose and oxygen to pass from the mother to the embryo. The developing embryo will also generate waste substances which can be removed by transferring them into the mother’s blood through the placenta.
• The child is born as a result of rhythmic contractions of the muscles in the uterus.

The diagram above is a schematic diagram of a household circuit. The house shown in the above diagram has 5 usable spaces where electrical connections are made. For this house, the mains have a voltage of 220 V and the net current coming from the mains is 22A.

(A) What is the mode of connection to all the spaces in the house from the mains?
(B) The spaces 5 and 4 have the same resistance and spaces 3 and 2 have respective resistances of 20Ω and 30Ω. Space 1 has a resistance double that of space 5. What is the net resistance for space 5.
(C) What is the current in space 3?
(D) What should be placed between the main
connection and the rest of the house’s electrical appliances to save them from accidental high electric current?

Total Marks	Breakdown (As per CBSE Marking Scheme)
5m (LA)	(A) Name the type of connection (1m) (B) Write formula, given values, calculate and final answer (0.5 m + 0.5 m + 0.5 m + 0.5 m) (C) Write formula, calculate and final answer (0.5 m + 0.5 m) (D) Name the device (1)

Answer:
(A) All spaces are connected in parallel

(B) Let Resistance of Space 5 and 4 be R ohms respectively
Resistance of Space 1 = 2 R ohms
Resistance of Space 2 = 30 ohms
Resistance of Space 3 = 20 ohms
Current = 22 A V = 220 V
Total Resistance = \(\frac { V }{ I }\)


60R = 10(150 + 5R)
60R =1500 + 50R
10R = 1500
R = 150Ω.

(C) V = IR
I in space – 3 = \(\frac { V }{ R }\)
= \(\frac { 220 }{ 20 }\)
= 11 A

(D) A fuse wire.

SECTION – E (12 Marks)
(Q.no, 37 to 39 are case – based/data -based questions with 2 to 3 short sub – parts. Internal choice is provided in one of these sub-parts)

Question 37.
Two students decided to investigate the effect of water and air on iron object under identical experimental conditions. They measured the mass of each object before placing it partially immersed in 10 mi of water. After a few days, the object were removed, dried and their masses were measured. The table shows their results. (4)

Student	Object	Mass of Object before Rusting in g	Mass of the coated object in g
A	Nail	3.0	3.15
B	Thin Plate	6.0	6.33

(A) What might be the reason for the varied observations of the two students?

(B) In another set up the students coated iron nails with zinc metal and noted that, iron nails coated with zinc prevents rusting. They also observed that zinc initially acts as a physical barrier, but an extra advantage of using zinc is that it continues to prevent rusting even if the layer of zinc is damaged. Name this process of rust prevention and give any two other methods to prevent rusting.

OR

(B) In which of the following applications of Iron, rusting will occur most? Support your answer with valid reason.

(i) iron Bucket electroplated with Zinc
(ii) Electricity cables having iron wires covered with aluminium
(iii) Iron hinges on a gate
(iv) Painted iron fence

Total Marks	Breakdown (As per CBSE Marking Scheme)
4m (CBQ)	(A) Reason in two points (1m + 1m) (B) Identify the phenomenon preventive measures (0.5 m + 0.5 m + 1 m)

Answer:
(A) Rusting occurs in both (i) and (ii) so there is an increase in mass.
As the surface area of (ii) is more, extent of rusting is more

(B) Galvanization
Oiling/ greasing/ painting/ alloying/ chromium plating or any other

Total Marks	Breakdown (As per CBSE Marking Scheme)
4m (CBQ)	(b) Identifly and give reason (1m + 1m)

(B) (iii) – Iron hinges on a gate – Iron is in contact with both atmospheric oxygen and moisture/ water vapour.

Question 38.
Pooja has green eyes white her parents and brother have black eyes. Pooja’s husband Ravi has black eyes while his mother has green eyes and father has black eyes. (4)
(A) On the basis of the above given information, is the green eye colour a dominant or recessive trait? Justify your answer.

(B) What is the possible genetic makeup of Pooja’s brother’s eye colour?

(C) What is the probability that the offspring of Pooja and Ravi will have green eyes? Also, show the inheritance of eye colour in the offspring with the help of a suitable cross.

OR

(C) 50% of the offspring of Pooja’s brother are green eyed. With help of cross show how this is possible.

Total Marks	Breakdown (As per CBSE Marking Scheme)
4m (CBQ)	(A) Identify the trait and give reason (0.5 m + 0.5 m) (B) Genotypes (0.5 m + 0.5 m) (C) Find probability by drawing Genetic cross (1 m + 1 m)

Answer:
(A) Yes, green eye colour is recessive (jh mark) as it will express only in homozygous condition.
(B) BB, Bb
(C) bb x Bb Genetic cross
50% of the offsprings can have green eye colour (0.5)

Total Marks	Breakdown (As per CBSE Marking Scheme)
4m (CBQ)	(C) Genotypes and show Genetic cross (1m + 1m)

Brother is heterozygous (Bb) and wife is homozygous (bb)

50°h of the off springs can have green eye colour as per the cross shown.

Question 39.

The above images are that of a specialized slide projector. Slides are small transparencies mounted in sturdy frames ideally suited to magnification and projection, since they have a very high resolution and a high image quality. There is a tray where the slides are to be put into a particular orientation so that the viewers can see the enlarged erect images of the transparent slides. This means that the slides will have to be inserted upside down in the projector tray.

To show her students the images of insects that she investigated in the lab, Mrs. Iyer brought a slide projector. Her slide projector produced a 500 times enlarged and inverted image of a slide on a screen 10 m away.  (4)

(A) Based on the text and data given in the above paragraph, what kind of lens must the slide projector have?

(B) If v is the symbol used for image distance and u for object distance then with one reason state what will be the sign for \(\frac { v }{ u }\) in the given case?

(C) A slide projector has a convex lens with a focal length of 20 cm. The slide is placed upside down 21 cm from the lens. How far away should the screen be placed from the slide projector’s lens so that the slide is in focus?

OR

(C) When a slide is placed 15 cm behind the lens in the projector, an image is formed 3 m in front of the lens. If the focal length of the lens is 14 cm, draw a ray diagram to show image formation, (not to scale) (4)

Total Marks	Breakdown (As per CBSE Marking Scheme)
4m (CBQ)	(A) Type of Lens (1m) (B) Sign (+/-) and Reason (0.5 m + 0.5 ) (C) Write formula, calculate and final answer (0.5 m + 0.5 m + 1 m)

Answer:
(A) Convex lens
(B) Negative as the image is real and inverted.
(C) \(\frac { 1 }{ f }\) = \(\frac { 1 }{ v }\) – \(\frac { 1 }{ u }\)
\(\frac { 1 }{ 20 }\) = \(\frac { 1 }{ v }\) – \(\frac { 1 }{ (-20) }\)
\(\frac { 1 }{ v }\) = \(\frac { 1 }{ 20 }\) – \(\frac { 1 }{ 21 }\)
= \(\frac { (21-30) }{ 420 }\)
= \(\frac { 1 }{ 420 }\)
v = 420 cm

OR

Total Marks	Breakdown (As per CBSE Marking Scheme)
2m (CBQ)	(C) Draw Ray diagram with labellings (1 m + 1 m) Labellings should be correct and as per the question

Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths with Solutions Set 8 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Maths Standard Set 8 with Solutions

Time Allowed: 3 Hours
Maximum Marks: 80

General Instructions:

1. This Question Paper has 5 Sections A-E.
2. Section A has 20 MCQs carrying 1 mark each
3. Section B has 5 questions carrying 02 marks each.
4. Section C has 6 questions carrying 03 marks each.
5. Section D has 4 questions carrying 05 marks each.
6. Section E has 3 case based integrated units of assessment (04 marks each) with sub-parts of the values of 1, 1 and 2 marks each respectively.
7. All Questions are compulsory. However, an internal choice in 2 Qs of 5 marks, 2 Qs of 3 marks and 2 Questions of 2 marks has been provided. An internal choice has been provided in the 2 marks questions of Section E
8. Draw neat figures wherever required. Take π =22/7 wherever required if not stated.

Section – A
(Section A consists of 20 questions of 1 mark each.)

Question 1.
A prime number greater than 91 but less than 100 is: (1)
(a) 94
(b) 97
(c) 96
(d) 95
Answer:
(b) 97

Explanation: prime number greater than 91 but less than 100 is 97.

Question 2.
Find a zero of the polynomial x³ – 8. (1)
(a) \( \sqrt{2} \)
(b) –\( \sqrt{2} \)
(c) 3
(d) 2
Answer:
(d) 2

Explanation: Here, x³ – 8 = 0 gives x³ = 8 i.e. x= \(\sqrt[3]{8}\) = 2

Question 3.
Write a quadratic polynomial whose sum of zeroes is \(\frac{-1}{4}\) and product of zeroes is \(\frac{1}{4}\) . (1)
(a) 4x² + x + 1
(b) x² + 4x – 1
(c) 2x² + 3x – 1
(d) x² – 2x + 1
Answer:
(a) 4x² + x + 1

Explanation:
Sum of zeroes = \(\frac{-1}{4}\)
Product of zeroes = \(\frac{1}{4}\)
∴ Quadratic Polynomial is
x² – (sum of zeroes) x + product of zeroes = 0
∴ x² – \(\frac{-1}{4}\) x + \(\frac{1}{4}\) = 0
⇒ 4x² + x + 1 = 0

Question 4.
Determine the roots of the equation S x² – 2x – 73 =0 (1)
(a) \(\sqrt{3},-\frac{1}{\sqrt{3}}\)
(b) 1, \(\frac{-1}{2}\)
(c) \(\sqrt{3}\), -1
(d) \(\frac{1}{\sqrt{3}}\), -1
Answer:
(a) \(\sqrt{3},-\frac{1}{\sqrt{3}}\)

Explanation:
The given equation has roots, whose sum is \(\frac{2}{\sqrt{3}}\) and product is -1. This is possible only with \(\sqrt{3}\) and \(-\frac{1}{\sqrt{3}}\).

Question 5.
Find the 15th term of the AP: x – 7, x – 2, x + 3….. (1)
(a) x + 67
(b) x + 4
(c) x + 36
(d) x + 63
Answer:
(d) x + 63

Explanation:
Here, a = x – 7, d = 5
So, 15th term = a + 14 d = (x – 7) + 14 (5)
= x + 63

Question 6.
Find The discriminant of the equation (x + 1)³ = 4 – 1 + x³ (1)
(a) 52
(b) 53
(c) 64
(d) 72
Answer:
(a) 52

Explanation:
Given, equation is :
(x + 1)³ = 4 – x + x³
⇒ x³ + 1 + 3x(x + 1) = 4 – x + x³
⇒ 1 + 3x² + 3x = 4 – x
⇒ 3x² + 4x – 3 = 0
On comparing this equation with ax² + bx + c = 0, we get a = 3, b = 4, c = – 3
∴ Discriminant, D = b² – 4ac
= 4² – 4 ×3 × (-3)
= 16 + 36
= 52

Question 7.
The next term of the A.P.: 3, 3 + \(\sqrt{2}\), 3 + 2 \(\sqrt{2}\), 3 + 3\(\sqrt{2}\) ,…. is: (1)
(a) 0
(b) 3 + 4\(\sqrt{2}\)
(c) 3 – 4\(\sqrt{2}\)
(d) 1
Answer:
(b) 3 + 4\(\sqrt{2}\)

Explanation:
Here,
d= \(\sqrt{2}\)
So, next term is (3+ 3\(\sqrt{2}\)) + \(\sqrt{2}\) i.e. 3 + 4\(\sqrt{2}\).

Question 8.
The value of x and y : x + 2y = 9, 2x – y = 8 is: (1)
(a) 0, 0
(b) 3, 5
(c) 5, 2
(d) 0,1
Answer:
(c) 5, 2

Explanation:
Here, x + 2y = 9 …(i)
2x – y = 8 …(ii)
Multiply equation (ii) by 2 and add both the equations.
x + 2y = 9
4x – 2 y = 16
5x = 25
x = \(\frac{25}{5}\) = 5
Then,
y = \(\frac{9 – x}{2}\) = \(\frac{9 – 5}{2}\) = 2

Question 9.
The condition for the points (a, 0), (0, b) and (1,1) to be collinear, is:
(a) \(\frac{1}{a} – \frac{1}{b}\) = 1
(b) \(\frac{1}{ab}\) = 1
(c) 0
(d) \(\frac{1}{a} + \frac{1}{b}\) = 1
Answer:
(d) \(\frac{1}{a} + \frac{1}{b}\) = 1

Explanation:
As the given points are collinear, a(b – 1) + 0 (1 – 0) + 1 (0 – b) = 0
⇒ ab – a – b = 0
⇒ ab = a + b
⇒ \(\frac{1}{a} + \frac{1}{b}\) = 1

Question 10.
The length of the altitude of an isosceles triangle of sides 6 cm, 6 cm and 4 cm is:
(a) 4\(\sqrt{2}\) cm
(b) 5\(\sqrt{2}\) cm
(c) 6\(\sqrt{2}\) cm
(d) 2\(\sqrt{2}\) cm
Answer:
(a) 4\(\sqrt{2}\) cm

Explanation:
Since, AD is an altitude in isosceles ∆ABC.

Question 11.
Find the coordinates of a point on y-axis which is equidistant from the points (6,5)and (-4, 3).
(a) (4, 2)
(b) (3, 2)
(c) (0,9)
(d) (9,2)
Answer:
(c) (0,9)

Explanation:
Let, the coordinate on y axis be P(0, y)
The given points are A(6, 5) and B(-4, 3)
∴ PA = PB
⇒ PA² = PB²
⇒ A (0 – 6)² + (y – 5)² = (0 + 4)² + (y – 3)²
⇒ 36 + y² + 25 – 10y = 16 + y² + 9 – 6y
⇒ – 4y = – 36
⇒ y = 9
Then, coordinate on y-axis is (0, 9).

Question 12.
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is: (1)
(a) 7 cm
(b) 4 cm
(c) 9 cm
(d) 10 cm
Answer:
(a) 7 cm

Explanation:
Here, OP is a tangent to a circle with radius ‘r’ of length 24 cm.

And OQ = 25 cm
Now, ∠QPO = 90°, as tangent at any point to a circle is ⊥r to the radius.
∴ In AOPQ by pythagoras theorem
OQ² = OP² + PQ²
25² = r² + 24²
⇒ r² = 625 – 576
= 49
⇒ r = 7 cm

Question 13.
If 3cos A = 1, then find the value of cosec A. (1)
(a) \(\frac{1}{\sqrt{2}}\)
(b) \(\frac{5}{\sqrt{2}}\)
(c) \(\frac{2 \sqrt{2}}{3}\)
(d) \(\frac{3}{2 \sqrt{2}}\)
Answer:
(d) \(\frac{3}{2 \sqrt{2}}\)

Explanation:
Here, 3 cos A = 1
cos A = \(\frac{1}{3}\)
Then cos² A = \(\frac{1}{9}\)
Then, sin² A = 1 – cos² A
= 1 – \(\frac{1}{9}\) = \(\frac{8}{9}\)
Sin A = \(\frac{2 \sqrt{2}}{3}\)
Then cosec A = \(\frac{1}{sin A}\) = \(\frac{3}{2 \sqrt{2}}\)

Question 14.
The perimeter of a quadrant of a circle of radius V is: (1)
(a) \(\frac{r^{2}}{2}\)
(b) π+ 4
(c) \(\frac{r}{2}\) (π + 4)
(d) \(\frac{r}{2}\)
Answer:
(c) \(\frac{r}{2}\) (π + 4)

Explanation:
Perimeter of quadrant,

Perimeter = OB + OA + \(\overparen{\mathrm{AB}}\)
= r + r + \(\frac{2πr}{4}\)
= 2r + \(\frac{π}{2}\) r
= \(\frac{r}{2}\)(π + 4)

Question 15.
The total surface area of a quadrant of a sphere of radius V is:
(a) πr²
(b) 2πr²
(c) 2πr
(d) \(\frac{πr^{2}}{2}\)
Answer:
(b) 2πr²

Explanation:
Total surface area of quadrant
= \(\frac{4 \pi r^2}{4}+\frac{\pi r^2}{2}+\frac{\pi r^2}{2}\)
= πr² + πr²
= 2πr²

Question 16.
The probability of drawing at random a green coloured ball from a bag containing 6 red and 5 black balls is: (1)
(a) 0
(b) \(\frac{1}{2}\)
(c) \(\frac{3}{5}\)
(d) 1
Answer:
(a) 0

Explanation:
Explanation: From the 11(6+5) balls in the bag, no ball is of green colour.
P(a green ball)=\(\frac{0}{11}\) i.e. 0

Question 17.
The median of the first 50 even natural numbers is: (1)
(a) 48
(b) 49
(c) 50
(d) 51
Answer:
(d) 51

Explanation:
First 50 even natural number are:
2, 4, 6, …………, 98,100.
As median is the middle-most value,
median = \(\frac{50 + 52}{2}\) = 51

Question 18.
The value of \(\frac{1+\tan ^2 \theta}{1+\cot ^2 \theta}\) is:
(a) tan θ
(b) cos² θ
(c) 1 + sin θ
(d) 0
Answer:
(a) tan θ

Explanation:
\(\frac{1+\tan ^2 \theta}{1+\cot ^2 \theta}\)

DIRECTION: In the question number 19 and 20, a statement of assertion (A) is followed by a statement of reason (R).
Choose the correct option as:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)
(b) Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A)
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.

Question 19.
Statement A (Assertion): There are infinite number of tines which passes through (1, 13).
Statement R (Reason): A Linear equation in two variabLes has infiniteLy many soLutions. (1)
Answer:
(b) Both assertion (A) and reason (R.) are true and reason (T) is not the correct explanation of assertion (A)

Explanation:
Through o point infinite tines can be drown. Through (1,13), infinite number of tines can be drawn.
Also o Line has infinite points on it, hence a linear equation representing a tine has infinite solutions.

Question 20.
Statement A (Assertion): Two simiLar triangLes are always congruent.
Statement R (Reason): Two similar triangles are said to be congruent if their areas are equaL (1)
Answer:
(d) Assertion (A) is false but reason(R) is true.

Explanation: Two similar triangles are not congruent generally.
If the area of two similar triangles are equal then the triangles ore congruent.

SECTION – B
(Section B consists of 5 questions of 2 marks each.)

Question 21.
Write the prime factorisation of 8190.
OR
Find the HCFof2205, 5145 and 4410. (2)
Answer:
The prime factorisation of 8190 is:
8190 = 2 ×3 × 3 × 5 × 7 × 13.

OR
HCF of 2205, 5145 and 4410

2205 = 3 × 3 × 5 × 7 × 7
5145 = 3 × 5 × 7 × 7 × 7
4410 = 2 × 3 × 3 × 5 × 7 × 7
HCF = 3 × 5 × 7 × 7 = 735

Question 22.
In an A.P., a = 5, d = 2 and n = 50, find an. (2)
Answer:
In given A.P.
Where a is 1st term, d is common difference and n is number of terms.
a_n = a + (n – 1 )d
= 5 + (50 – 1) × 2
= 5 + 49 × 2
= 5 + 98
= 103

Question 23.
If Q (0, 1) is equidistant from P(5, -3) and R (x, 6), find the values of ‘x’. Also, find the distances of QR and PR.
OR
Find the ratio in which P(4, p) divides the line segment joining the points A(2, 3) and B(6, 3). Hence find the value of p. (2)
Answer:
Since, Q(0, 1) is equidistant from P(5, -3) and R(x, 6),
PQ = QR
⇒ PQ² = QR²
i.e (5 – 0)² + (-3 – 1)² = (x – 0)² + (6 – l)²
i.e 25 + 16 = x² + 25
i.e x² = 16
or x = ±4

Thus, R(4, 6) or R (-4, 6)
For R (4, 6),
QR = \(\sqrt{(4-0)^2+(6-1)^2}\)
= \(\sqrt{16+25}\) = \(\sqrt{41}\)
and PR = \(\sqrt{(4-5)^2+(6+3)^2}\)
= \(\sqrt{1+81}\) = \(\sqrt{82}\)

For R(-4,6),
QR = \(\sqrt{(-4-0)^2+(6-1)^2}\)
= \(\sqrt{16+25}\) = \(\sqrt{41}\)
and PR = \(\sqrt{(-4-5)^2+(6+3)^2}\)
= \(\sqrt{81+81}\) = \(\sqrt{162}\)
= \(9 \sqrt{2}\)
OR
Given, point P(4, p) and line segment A(2,3) and B(6, 3)
Let the point P divide given line segment AB in the ratio of k: 1.
Then, P(4,p) = [\left(\frac{k \times 6+2}{k+1}\right) \cdot\left(\frac{k \times 3+3}{k+1}\right)]
∴ On comparing, we get
\(\frac{6k + 2}{k+1}\) = 4
⇒ 6k + 2 = 4k + 4
⇒ k = 1
Then point P divides line AB in the ratio of 1:
Now, p = \(\frac{1×3 + 3}{1+1}\) = \(\frac{6}{2}\)
⇒ p = 3
Hence, value of p is 3.

Question 24.
A tree casts a shadow of 4\(\sqrt{3}\) m on ground. If the sun’s elevation is 60°. then find the height of the tree. (2)
Answer;

Let PQ be the tree and RQ be its shadow.
∴ In ∆PQR
tan 60° = \(\frac{PQ}{QR}\)
\(\sqrt{3}\) = \(4 \sqrt{3}\)
PQ = 12 m.
Hence, the height of the tree is 12 m.

Question 25.
Find the mode of the following frequency distribution: (2)

Class	15-20	20-25	25-30	30-35	35-40	40-45
Frequency	3	8	9	10	3	2

Answer:
In the given frequency distribution, modal class
Is 30-35, with maximum frequency 10.
Here tower Limit of modal class. I = 30
frequency of class preeceding modal class, f_o = 9
frequency of modal doss, f₁ = 10
frequency of cLass succeeding modal class, f₂ = 3
\(M_0=l+\frac{f_1-f_0}{2 f_1-f_0-f_2} \times h\)
\(=30+\frac{10-9}{20-9-3} \times 5\)
= 30 + \(\frac{5}{8}\) = 30 + 0.625
= 30.625

SECTION – C
(Section C consists of 6 questions of 3 marks each.)

Question 26.
Prove that an irrational number, using the fact that 2\(\sqrt{3}\) – 4 is an irrational number. (3)
Answer:
Let us assume on the contrary, that 2\(\sqrt{3}\) – 4 be a rational number.
Then,
2\(\sqrt{3}\) – 4 = \(\frac{p}{q}\). where p and q are co-primes and q ≠ 0.
⇒ \(\sqrt{3}=\frac{1}{2}\left(\frac{p}{q}+4\right)\)
Since p and q are integers, \(\frac{1}{2}\left(\frac{p}{q}+4\right)\) is rational and s0 \(\sqrt{3}\) is rationaL But this contradicts the fact that \(\sqrt{3}\) is irrational.
Hence, 2\(\sqrt{3}\) – 4 is an irrational number.

Question 27.
The sum of two numbers, as well as, the difference of their squares is 9. Find the numbers.
OR
Find the values of k for which the following equations have an infinite number of solutions: 2x + 3y = 7; (k – 1)x + (k + 2) y = 3k (3)

Answer:
Let the two numbers be x and y. (x > y).
Then, x + y = 9 and x² – y² = 9
⇒ x + y = 9 and (x – y) (x + y) = 9
⇒ x + y = 9 and x – y = 1
Solving the two equations, we get x = 5 and y = 4
Thus, the two numbers are 5 and 4.

OR

Given, equations are
2x + 3y = 7
(k – 1)x + (k + 2)y = 3k
Here, a₁ = 2, b₁ = 3, c₁ = 7
a₂ = k – 1, b₂ = k + 2 c₂ = 3k
Since, given system of equation has infinite solution.
∴ \(\begin{gathered}
\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} \\
\frac{2}{k-1}=\frac{3}{k+2}=\frac{7}{3 k}
\end{gathered}\)
On comparing the two ratios
⇒ 2(k+2) = 3(k-1)
⇒ 2k + 4 = 3k – 3
⇒ -k = -7
⇒ k = 7
and, \(\frac{3}{k+2}=\frac{7}{3 k}\)
⇒ 9k = 7k + 14
⇒ 2k = 14
⇒ k = 7
Hence, the value of k is 7.

Question 28.
Show that ∆ABC with vertices A(-2, 0), B(2, 0) and C(0, 2) is similar to A PQR with vertices P(-4, 0), Q(4, 0) and R(0,4). (3)
Answer:
Here, vertices of ∆ABC are A(-2, 0), B(2, 0) and C(0, 2)
Another ∆PQR, with vertices P(-4, 0), Q(4, 0) and R(0, 4).
In ∆ABC,

Then, lenth of AB = \(\sqrt{(2+2)^2+(0-0)^2}\)
= 4 units

lenth of BC = \(\sqrt{(0-2)^2+(2-0)^2}\)
= \(\sqrt{4+4}\)
= \(\sqrt{8}\)
= \(2 \sqrt{2}\)

lenth of AC = \(\sqrt{(-2-0)^2+(0-2)^2}\)
= \(\sqrt{4+4}\)
= \(2 \sqrt{2}\)

In ∆PQR,

lenth of PQ = \(\sqrt{(4+4)^2+(0-0)^2}\)
= \(\sqrt{8^2}\) = 8 units

lenth of QR = \(\sqrt{(4-o)^2+(o-4)^2}\)
= \(\sqrt{4^2+4^2}\)
= \(\sqrt{32}\)
= \(4 \sqrt{2}\)

lenth of AC = \(\sqrt{(-4)^2+(4)^2}\)
= \(4 \sqrt{2}\) units

∴ \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{4}{8}=\frac{1}{2}\)
\(\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{2 \sqrt{2}}{4 \sqrt{2}}=\frac{1}{2}\)
and \(\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{2 \sqrt{2}}{4 \sqrt{2}}=\frac{1}{2}\)
Now, \(\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}=\frac{1}{2}\)
Hence, ∆ABC ~ ∆PQR (by SSS – Similarity)

Question 29.
What is the ratio between the areas of the circle and the square when a square is inscribed in a circle? (3)
Answer:
Let,
radius of the circle be r units.
Then, diagonaL of the square = 2r
⇒ Side of the square = \(\frac{2 r}{\sqrt{2}}=\sqrt{2} r=\frac{\pi}{2}\)
∴ \(\frac{\text { Area of the circle }}{\text { Area of the square }}=\frac{\pi r^2}{(\sqrt{2} r)^2}=\frac{\pi r^2}{2 r^2}\)
= π : 2

Question 30.
If sum and product of quadratic polynomial are 2 and -8 respectively, then find a quadratic polynomial and zeroes of the polynomial so obtained.
OR
The area of a sector of a circle of radius 36 cm is 54 n. sq cm. Find the length of the corresponding arc of the sector. (3)
Answer:
Let α and β be the zeroes of the polynomial.
Given: Sum of zeroes, α + β = 2
Product of zeroes, α × β = -8
Equation of polyomial:
x² – (sum of zeroes)x + product of zeroes
x² – (2)x + (- 8)
⇒ x² – 2x – 8
∠eores of polynomial
Let P(x) = x² – 2x – 8
= x² – 4x + 2x – 8
= x(x – 4) + 2(x – 4)
= (x – 4)(x + 2)
Hence, the zeroes of the polynomial are 4 and -2.

OR

Let 9 be the angle of the sector. Then,

Question 31.
Find the mean of the following frequency distribution: (3)

Marks	Below 10	Below 20	Below 30	Below 40	Below 60	Below 70	Below 80	Below 90	Below 100
Number of students	12	22	35	50	86	97	104	109	115

Answer:

SECTION – D
(Section D consists of 4 questions of 5 marks each.)

Question 32.
if tan q + sin q = m and tan q – sin q = n, show that: m² – n² = \(4 \sqrt{mn}\)
OR
The angle of elevation of the top of a building from the foot of a tower is 30°, and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building. (5)
Answer:
We have,

OR

Question 33.
A circle is inscribed in a DABC having sides 8 cm, 10 cm and 12 cm as shown in the figure. Find AD, BE and CF.

OR
The 6th term of an AP is five times the 1st term and the 11th term exceeds twice the 5th term by 3. Find the 8th term of the AP. (5)
Answer:
We know that the tangents drawn from an external point to a circle are equal. Therefore,

gives y = 5
Hence, AD = x = 7 cm, BE = y = 5 cm and CF = z = 3 cm.

OR

Let a and d be the first term and the common difference of the AP respectively. Then,
a₆ = a + 5d = 5a …(i)
and a₁₁ = 2 a₅ + 3
⇒ a + 10d = 2 (a + 4d) + 3 …..(ii)
Simplifying (i) and (ii), we get
4a = 5d and a – 2d + 3 = 0
Solving these equations, we get d = 4 and a = 5
Thus, a₈ = a + 7d = 5 + 7(4) = 33

Question 34.
On the ground, a tree with a length of 6 m creates a shadow that is 4 m long, while another tree creates a shadow that is 28 m long. Find the tree’s height.

Answer:
Given,
Length of the vertical pole = 6 m Shadow of the pole = 4 m Let the height of the tower be h m.
Length of the shadow of the tower = 28 m
In ∆ABC and ∆DFE,
∠C = ∠E (angle of elevation)
∠B = ∠F = 90°
By AA similarity criterion,
∆ABC ~ ∆DFE
We know that the corresponding sides of two similar triangles are proportional.
\(\frac{AB}{DF}\) = \(\frac{BC}{EF}\)
\(\frac{6}{h}\) = \(\frac{4}{28}\)
h = \(\frac{6 \times 28}{4}\)
h = 6 × 7
h = 42
Hence, the height of the tower = 42 m.

Question 35.
Two dice have the following numbers: 1, 2, 3, 4, 5, 6 and 1,1, 2, 2, 3, 3. The total of the numbers on them is calculated once they are thrown. Calculate the probability of getting each sum from 2 to 6 individually. (5)
Answer:
Number of total outcome = n(S) = 36
(i) Let E₁ be the event ‘getting sum 2’
Favourable outcomes for the event E₁ = {(1,1),(1,1)}
n(E₁) = 2
\(P\left(E_1\right)=\frac{n\left(E_1\right)}{n(S)}=\frac{2}{36}=\frac{1}{18}\)

(ii) Let E₂ be the event ‘getting sum 3’
Favourable outcomes for the event E₂ = {(1,2),(1,2),(2,1),(2,1)}
n(E₂) = 4
\(P\left(E_2\right)=\frac{n\left(E_2\right)}{n(S)}=\frac{4}{36}=\frac{1}{9}\)

(iii) Let E₃ be the event ‘getting sum 4’
Favourable outcomes for the event E₃ = {(2,2),(2,2),(3,1),(3,1)(1,3),(1,3)}
n(E₃) = 6
\(P\left(E_3\right)=\frac{n\left(E_3\right)}{n(S)}=\frac{6}{36}=\frac{1}{6}\)

(iv) Let E₄ be the event ‘getting sum 5’
Favourable outcomes for the event E₄ = {(2,3),(2,3),(4,1),(4,1)(3,2),(3,2)}
n(E₄) = 6
\(P\left(E_4\right)=\frac{n\left(E_4\right)}{n(S)}=\frac{6}{36}=\frac{1}{6}\)

(v) Let E₅ be the event ‘getting sum 6’
Favourable outcomes for the event E₅ = {(3,3),(3,3),(4,2),(4,2)(5,1),(5,1)}
n(E₅) = 6
\(P\left(E_5\right)=\frac{n\left(E_5\right)}{n(S)}=\frac{6}{36}=\frac{1}{6}\)

SECTION – E
(Case Study Based Questions)
(Section E consists of 3 questions. All are compulsory.)

Question 36.
On the roadway, Points A and B, which stand in for Chandigarh and Kurukshetra, respectively, are located nearly 90 kilometres apart. At the same time, a car departs from Kurukshetra and one from Chandigarh. These cars will collide in 9 hours if they are travelling in the same direction, and in 9/7 hours if they are travelling in the other direction. Let X and Y be two cars that are travelling at x and y kilometres per hour from places A and B, respectively.

On the basis of the above information, answer the following questions:
(A) When both cars move in the same direction, then find the situation can be represented algebraically.
OR
When both cars move in the opposite direction, then find the situation can be represented algebraically. (2)
(B) Find the speed of car x. (1)
(C) Find the speed of car y. (1)
Answer:
(A) Suppose two cars meet at point Q.
Then, Distance travelled by car X = AQ,
Distance travelled by car Y = BQ.
It is given that two cars meet in 9 hours.
Distance travelled by car X in 9 hours
= 9x km = AQ = 9x
Distance travelled by car Y in 9 hours
= 9y km = BQ = 9y

Clearly, AQ – BQ = AB
= 9x – 9y = 90
= x – y = 10

OR

Suppose two cars meet at point P.
Then Distance travelled by car X = AP
and Distance travelled by car Y = BP.
In this case, two cars meet in 9/7 hours.
Distance travelled by car X in \(\frac{9}{7}\) hours
= \(\frac{9}{7}\) X km
⇒ AP = \(\frac{9}{7}\) X

Distance travelled by car Y in \(\frac{9}{7}\) hours
= \(\frac{9}{7}\) y km

clearly AP + BP = AB
⇒ \(\frac{9}{7}\) x + \(\frac{9}{7}\) y = 90
⇒ \(\frac{9}{7}\) (x + y) = 90
⇒ x + y = 70

(B) We have, x – y = 10
⇒ x + y = 70
Adding equations (i) and (ii), we get
2x = 80
⇒ x = 40
Hence, speed of car X is 40 km/hr.

(C) We have x-y = 10
⇒ 40 – y = 10
⇒ y = 30
Hence, speed of car y is 30 km/hr

Question 37.
Eshan purchased a new building for her business. Being in the prime location, she decided to make some more money by putting up an advertisement sign for a rental ad income on the roof of the building.


From a point P on the ground level, the angle of elevation of the roof of the building is 30° and the angle of elevation of the top of the sign board is 45°. The point P is at a distance of 24 m from the base of the building.
On the basis of the above information, answer the following questions:
(A) Find the height of the building (without the sign board).
OR
The height of the building (with the sign board) (2)
(B) Find the height of the sign board. (1)
(C) Find the distance of the point P from the top of the sign board. (1)
Answer:
(A) Without the sign board, the height of the shop is AB.
In ∆PAB,

OR
Considering, the diagram in the above question, AC as the new height of the shop including the sign-baard.
In ∆APC,
tan 45° = \(\frac{AC}{AP}\)
1 = \(\frac{AC}{24}\)
⇒ AC = 24 m

(B) From Q (i) and Q (ii).
Length of sign board, BC = AC – AB
= 24 – 14
= 10 m

(C) In ∆APC,
cos 45° = \(\frac{AP}{AC}\)
\(\frac{1}{\sqrt{2}}\) = \(\frac{24}{AC}\)
⇒ PC = \(24 \sqrt{2}\) m

Question 38.
In a toys manufacturing company, wooden parts are assembled and painted to prepare a toy. One specific toy is in the shape of a cone mounted on a cylinder.
For the wood processing activity center, the wood is taken out of storage to be sawed, after which it undergoes rough polishing, then is cut, drilled and has holes punched in it. It is then fine polished using sandpaper.

For the retail packaging and delivery activity center, the polished wood sub-parts are assembled together, then decorated using paint.
The total height of the toy is 26 cm and the height of its conical part is 6 cm. The diameters of the base of the conical part is 5 cm and that of the cylindrical part is 4 cm.
On the basis of the above information, answer the following questions:
(A) If its cylindrical part is to be painted yellow, find the surface area need to be painted. (1)
(B) if its conical part is to be painted green, find the surface area need to be painted.
OR
Rnd the volume of the wood used in making this toy. (2)
(C) If the cost of painting the toy is 3 paise per sq cm, then find the cost of painting the toy. (Use it = 3.14) (1)
Answer:
(A) C.S.A. of cylinder = 2nrH + πr2 = nr(2H + r)
= 2π(2 × 20 + 2)
= 84TI

(B) C.S.A. of cone = πrl + π(R² – r²)
= \(\begin{aligned}
&\pi\left[r \sqrt{r^2+h^2}+\left(R^2-r^2\right)\right] \\
&\pi\left[2.5 \sqrt{2.5^2+6^2}+\left(2.5^2-2^2\right)\right]
\end{aligned}\)
= π [2.5 × 6.5 + 0.5 × 4.5]
= π [16.25 + 2.25]
= 18.571 sq units

OR

Volume of toy = Volume of cone + Volume of cylinder
= \(\frac{1}{3}\) πr²h + πR²H
= π[\(\frac{1}{3}\) × 2.5 × 2.5 × 6 + 2 × 2 × 20]
= π [12.5 + 80]
= 92.5TC cm³

(C) Surface area = SA of cone + S. A of cylinder
= 84π + 18.5π = 102.571
∴ Cost of painting
= 0.03 × 102.571
= ₹ 9.65

Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths with Solutions Set 7 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Maths Set 7 with Solutions

Time Allowed: 3 Hours
Maximum Marks: 80

General Instructions:

• This Question Paper has 5 Sections A, B C D. and E
• Section A has 20 Multiple Choice Questions (MCQs) carrying 1 mark each.
• Section B has 5 Short Answer-I (SA-I) type questions carrying 2 marks each.
• Section Chas 6 Short Answer-II (SA-II) type questions carrying 3 marks each.
• Section D has 4 Long Answer (LA) type questions carrying S marks each.
• Section E has 3 Case Based integrated units of assessment (4 marks each) with sub-parts of the values of 1, 1 and 2 marks each respectively.
• AU Questions are compulsory. However an internal choice in 2 Qs of 2 marks, 2 Qs of 3 marks and 2
  Questions of 5 marks has been provided. An internal choice has been provided in the 2 marks questions
  of Section E
• Draw neat figures wherever required. Take π =22/7 wherever required if not stated.

SECTION – A (20 marks)
(Section – A consists of 20 questions of 1 mark each)

Question 1.
The standard form of the equation y(2y + 15) = 3(y² + y + 8) is: [1]
(a) y² + 25y + 24 = 0
(b) y² – 27y – 23 = 0
(c) y² – 27y + 25 = 0
(d) y² – 12y + 24 = 0
Answer:
(d) y² – 12y + 24 = 0

Explanation:
y(2y + 15) = 3(y² + y + 8)
⇒ 2y² + 15 y = 3y² + 3y + 24
⇒ y² – 12y + 24 = 0

Question 2.
The value of c for which pair of linear equations cx – y = 2 and 6x – 2y = 4 will have infinitely many solutions is: [1]
(a) 3
(b) 5
(c) -1
(d) 0
Answer:
(a) 3

Explanation:
We have cx – y = 2
and 6x – 2y = 4

For infinitely many solutions
\(\frac{c}{6}=\frac{-1}{-2}=\frac{2}{4}\)
\(\frac{c}{6}=\frac{1}{2}\)
⇒ c = 3

Question 3.
The distance between (7, 0) and (1, – 8) is: [1]
(a) 7 units
(b) 10 units
(c) 8 units
(d) 9 units
Answer:
(b) 10 units

Explanation:
Distance between (7, 0) and (1.-8)
[By distance formula]

Question 4.
School divides its students into 5 houses A, B, C, D and E. Class X A has 23 students, 4 from house A, 8 from house B, 5 from house C, 2 from house D and rest from house E. A single student is chosen at random to become the monitor of the class. The probability that the chosen student is not from houses A, B and C is: [1]
(a) \(\frac{4}{23}\)
(b) \(\frac{6}{23}\)
(c) \(\frac{8}{23}\)
(d) \(\frac{17}{23}\)
Answer:
(b) \(\frac{6}{23}\)

Explanation:
Total students in class XA = 23
students from house A = 4
students from house B = 8
students from house C = 5
students from house D = 2
students from house E = 4
Total. students in house D and E = 2 + 4
= 6
Required probability = \(\frac{6}{23}\)

Question 5.
A line of length 10 units has one end at the point (-3, 2). If the ordinate of the other end is 10, then the abscissa will be: [1]
(a) 9 or 3
(b) 4 or -3
(c) -3 or 5
(d) -9 or 3
Answer:
(d) -9 or 3

Explanation:
Let A and B denote the points (-3, 2) and (a, 10), where a is to be determined.
Here, AB = 10 units.

So \(\sqrt{(a+3)^2+(10-2)^2}\) = 10
⇒ (a + 3)² + 64 = 100
or (a + 3)² = 36
⇒ a + 3 = ±6
⇒ a = -9 or 3.

Question 6.
If ΔABC – ΔDEF, such that ∠A = 47° and ∠E = 83°, then the value of ∠C is: [1]

(a) 50°
(b) 60°
(c) 45°
(d) 90°
Answer:
(a) 50°

Explanation:
Since ΔABC – ΔDEF.
∠A = ∠D, ∠B = ∠E and ∠C = ∠F
⇒ ∠A = 47° and ∠B = 83°
So, in ΔABC, ∠C = 180° – (∠A + ∠B)
= 180° – (47° + 83°) = 50°

Question 7.
The zeros of 2x² – x – 45 are respectively: [1]
(a) 5, \(\frac{1}{2}\)
(b) 2, –\(\frac{9}{2}\)
(c) 5, –\(\frac{9}{2}\)
(d) 5, -9
Answer:
(c) 5, –\(\frac{9}{2}\)

Explanation:
2x² – x – 45 = 2x² – 10x + 9x – 45
= 2x(x – 5) + 9(x – 5)
= (2x + 9)(x – 5)
So, its zeros are 5 and

Question 8.
If cot A + \(\frac{1}{\cot A}\) = 2, then the value of cot²A + \(\frac{1}{\cot ^2 A}\) is: [1]
(a) 2
(b) 4
(c) 1
(d) 5
Answer:
(a) 2

Explanation:
Given, cot A + \(\frac{1}{\cot A}\) = 2, we have
on squaring on both sides, we get

Question 9.
The mean of twenty observations is 15. If the observations 3 and 14 are replaced by 8 and 5 respectively, then the new mean will be: [1]
(a) 15.65
(b) 15.2
(c) 15
(d) 14.8
Answer:
(d) 14.8

Explanation:
We know,
Mean = \(\frac{\text { Sum of observations }}{\text { Total number of observations }}\)
15 = \(\frac{\text { Sum of observations }}{20}\)
⇒ Sum of observations
= 15 × 20 = 300
Since, 3 and 14 are replaced by 8 and 5 New sum = 300 – (3 + 14) + (8 + 5) = 296
∴ New mean = \(\frac{296}{20}\) = 14.8

Question 10.
If sinA = \(\frac{1}{2}\) then the value of (cot A – cos A) is: [1]
(a) √3
(b) \(\frac{1}{\sqrt{3}}\)
(c) \(\frac{3}{2}\)
(d) \(\frac{\sqrt{3}}{2}\)
Answer:
(d) \(\frac{\sqrt{3}}{2}\)

Explanation:
sin A = \(\frac{1}{2}\) gives A = 30°
So, cot A – cos A = cot 30° – cos 30°
= √3 – \(\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}\)

Question 11.
From a group of 4 girls and 6 boys, a child is selected. The probability that the selected child is a girl is: [1]
(a) \(\frac{2}{5}\)
(b) \(\frac{3}{5}\)
(c) \(\frac{4}{5}\)
(d) \(\frac{1}{5}\)
Answer:
(a) \(\frac{2}{5}\)

Explanation:
Total no.of children = 4 + 6 = 10
∴ P(a girl) = \(\frac{4}{10}\) i.e \(\frac{2}{5}\)

Question 12.
The perimeter of a quadrant of a circle of radius ‘r’ is: [1]
(a) π + 4
(b) \(\frac{r^2}{2}\) (π + 2)
(c) 0
(d) \(\frac{r}{2}\)(π + 4)
Answer:
(d) \(\frac{r}{2}\)(π + 4)

Explanation:
Perimeter of a Quadrant

Question 13.
The total surface area of the hemisphere of radius V is: [1]
(a) 3πr²
(b) 5πr²
(c) 2πr²
(d) 4πr²
Answer:
(a) 3πr²

Explanation:
Total Surface area of hemisphere

= \(\frac{4 \pi r^2}{2}\) + πr²
= 2πr² + πr² = 3πr²

Question 14.
What is the upper limit of the median class for the given below distribution? [1]

(a) 14
(b) 10
(c) 15
(d) 20
Answer:
(c) 15

Explanation:

Here, N = 57. So, \(\frac{N}{2}\) = 28.5
Cumulative frequency just greater than 28.5 is 38. which belongs to 10 – 15.
So, the median class is 10 – 15
Thus, its upper Limit is 15.

Question 15.
Two coins are tossed simultaneously. The probability of getting at least one head is: [1]
(a) \(\frac{3}{4}\)
(b) \(\frac{1}{4}\)
(c) \(\frac{2}{5}\)
(d) \(\frac{5}{4}\)
Answer:
(a) \(\frac{3}{4}\)

Explanatlon:
Possibte outcomes are {HH, HT. TH, TT}
Favourable outcomes are {HH, HT, TH}
So, required probability = \(\frac{3}{4}\)

Question 16.
The common difference of the A.P., \(\sqrt{3}, \sqrt{12}, \sqrt{27}\) …………. is: [1]
(a) \(\sqrt{12}\)
(b) √4
(c) √3
(d) 2√3
Answer:
(c) √3

Explanation:
Here the terms of the A.P are √3, 2√3, 3√3, ……………
So, the common difference = (2√3 – √3),
= √3.

Question 17.
The nature of roots of the quadratic equation ax² – 3bx – 4a = 0 (a ≠ 0) is: [1]
(a) equal
(b) real and distinct
(c) unreal and equal
(d) unreal
Answer:
(b) real and distinct

Explanation:
Here, given quadratic equation is:
ax² – 3 bx -4a = 0
Discriminant D,
b² – 4ac = (-3b)² – 4(a)(-4a)
= 9b² + 16a² > 0
So, the roots are real and distinct.

Question 18.
From a point Q, the length of the tangent to a circle is 12 cm and distance of d from the centre is 13 cm. The radius of the circle is: [1]
(a) 4 cm
(b) 6 cm
(c) 5 cm
(d) 9 cm
Answer:
(c) 5 cm

Explanation: Let the radius of circle be V.
And, tangent is perpendicular to radius at the point of contact.
∴ ∠OTQ = 90°

In ΔOQT, by Pythagoras thereom,

OQ² = OT² + QT²
⇒ 13² = r² + 12²
⇒ r² = 13² – 12²
= (13 – 12)(13 + 12) = 25
∴ r = 5 cm

Direction for questions 19 and 20: In question number 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R).
Choose the correct option:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A)
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.

Question 19.
Assertion (A): From a pack of 52 cards, the probability of drawing a red queen is \(\frac{1}{20}\)
Reason (R) : Probability of occurring of an event P(A) = \(\frac{\text { Favourable outcomes }}{\text { Total outcomes }}\) [1]
Answer:
(d) Assertion (A) is false but reason (R) is true.

Explanation:
Red queens in pack of 52 cards = 2
Total number of cards = 52
P(red queen) = \(\frac{2}{52}=\frac{1}{26}\)

Question 20.
Assertion (A): The simplest form of \(\frac{1095}{1168}\) is \(\frac{15}{16}\)
Reason (R) : For finding the simplest form of a fraction the numerator and denominator are divided by their HCF. [1]
Answer:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)

Explanation:
Here, HCFof 1095 and 1168 is 73.
\(\frac{1095 \div 73}{1168 \div 73}=\frac{15}{16}\) simplest form of fraction.

SECTION – B (10 marks)
(Section – B consists of 5 questions of 2 marks each.)

Question 21.
Write the prime factorisation of 8190.
OR
Find the HCF of (2³ × 3² × 5¹), (2² × 3³ × 5²) and (2⁴ × 3¹ × 5² × 7). [2]
Answer:
The prime factorisation of 8190 is:
8190 = 2 × 3 × 3 × 5 × 7 × 13.

OR
The given factors are (2³ × 3² × 5¹), (2² × 3³ × 5²) and (2⁴ × 3¹ × 5² × 7)
Now, HCF = Product of each prime factors with smallest power.
HCF = 2² x 3¹ x 5¹ i.e. 60.

Question 22.
Form a quadratic polynomial whose zeros are 3 + √2 and 3 – √2. [2]
Answer:
Sum of zeros = (3 + √2) + (3 – √2) = 6
Product of roots = (3 + √2)(3 – √2) = 9 – 2 = 7
A quadratic polynomial with sum and product of zeros is given as,
x² – (sum of zeros) x + (Product of zeros)
i.e, x² – 6x + 7.

Question 23.
In a right angle triangle ABC, right¬angled at B, if sin (A – C) = \(\frac{1}{2}\) find the measures of angles A and C. [2]
OR
If sin θ = \(\frac{2 m n}{m^2+n^2}\). find the value of \(\frac{\sin \theta \cot \theta}{\cos \theta}\)
Answer:
Since sin(A – C) = \(\frac{1}{2}\)

A – C = 30°
But A + C = 90° (as A + B + C = 180°)
So, C = 30° and A = 60°
OR

Question 24.
A path of width 7 m runs around outside a circular park whose radius is 18 m. Find the area of the path. [2]
Answer:
Area of the path = Area of the outer circle –
Area of the inner circle = [π(18 + 7)2 – π(18)2] sq. m

= (625π – 324π) sq. m
= 301π sq. m, or 946 sq. m.

Question 25.
A die is thrown once. Find the probability of getting:
(A) a prime number greater than 3. [2]
Answer:
P (prime number greater than 3) = \(\frac{1}{6}\)
(∵ Only 5 is the prime number greater than 3)

(B) an even prime number greater than 3.
Answer:
P (even prime number greater than 3)
= \(\frac{0}{6}\) i.e, 0.
(∵ Only even prime number is 2, which is not greater than 3)

SECTION – C (18 marks)
(Section – C consists of 6 questions of 3 marks each.)

Question 26.
Prove that: 2√3 – 4 is an irrational number, using the fact that √3 is an irrational number.
OR
The figure shows a rectangle with its length and breadth as indicated. [3]

Given that the perimeter of the rectangle is 120 cm, find:
(A) the values of x and y;
(B) the length and the breadth;
(C) the area of the rectangle. 3
Answer:
Let us assume on the contrary, that 2√3 – 4 be a rational number.
Then, 2√3 – 4 = \(\frac{p}{q}\). where p and q are co-primes and q ≠ 0.
⇒ √3 = \(\frac{1}{2}\left(\frac{p}{q}+4\right)\)
Since p and q are integers, \frac{1}{2}\left(\frac{p}{q}+4\right) is rational and so √3 is rational But, this contradicts the fact that √3 is irrational
Hence, 2√3 – 4 is an irrational number.
OR
(A) Perimeter of the rectangle ABCD = AD + AB + BC + CD
= 3x-y + 2x-3 + 2x + y + 2x – 3
= 9x – 6
⇒ 9x – 6 = 120
⇒ x = \(\frac{126}{9}\) = 14
Also, opposite sides of a rectangle are equal. So,
AD = BC
3x – y = 2x + y
⇒ x = 2y
Thus, y = 7 (as x = 14)

(B) Length = 3x – y = 2x + y = 35 cm
Breadth = 2x – 3 = 25 cm

(C) Area of rectangle = (35 × 25) sq. cm, i.e. 875 sq. cm.

Question 27.
If Q (0,1) is equidistant from P (5, -3) and R (x, 6); find the values ofx. Also, find the distances QR and PR. [3]

Answer:
Since Q (0, 1) is equidistant from P (5, -3) and R (x, 6),
PQ = QR
⇒ PQ² = QR²
i.e. (5 – 0)² + (-3 – 1)² = (x – 0)² + (6 – 1)²
i.e. 25 + 16 = x² + 25
i.e. x² = 16 or x = ± 4
Thus, R is R (4, 6) or R (-4, 6)
For R (4, 6),

Question 28.
Prove that the area of the semi-circle drawn on the hypotenuse of a right-angled triangle is equal to the sum of the area of the semi-circles drawn on the other two sides of the triangle. [3]
Answer:
We need to prove that
ar (semi-circle III) = ar (semi-circle I) + ar (semi-circle II)

Here, ΔABC is a right triangle, right-angled at B.
So, using Pythagoras Theorm,
AC² = AB² + BC² ……(1)

Adding (2) and (3), and using (1), we have ar (semi-circle I) + ar (semi-circle II)
= \(\frac{\pi}{8}\)(BC² + AB²)
= \(\frac{\pi}{8}\)AC²
= ar (semi-circle III)
Hence, proved.

Question 29.
If the median of the distribution given below is 28.5, find the values of x and y. [3]

Answer:
With the given frequency distribution table, we first prepare a cumulative frequency distribution table as given below:

Class interval	Frequency	Cum. frequency
0-10	5	5
10-20	X	5 + x
20-30	20	25+ x
30-40	15	40 + x
40-50	y	40 + x + y
50-60	5	45 + x + y = 60

Since, median given is 28.5, the median class is 20-30.
For this class,
l = 20, h = 10, \(\frac{n}{2}\) = 30, f = 20

According to the formula,
⇒ median = l + \(\left(\frac{\frac{n}{2}-c f}{f}\right)\) × h
28.5 = 20 + \(\frac{30-(5+x)}{20}\) × 10
⇒ \(\frac{25-x}{2}\) = 8.5
⇒ x = 8
Since, 45 + x + y = 60
⇒ 45 + 8 + y = 60
⇒ y = 60 – 53 = 7
Thus, required values of x and y are 8 and 7 respectively.

Question 30.
If sin A = m sin B and tan A = n tan B then show that (n² – 1) cos²A = m² – 1. [3]
OR
Find the mass of a 3.5 m long lead pipe if the external diameter of the pipe is 2.4 cm, thickness of the metal is 2 mm; and the mass of 1 cm3 of lead is 12 grams.
Answer:
Given: tan A = n tan B
and sin A = m sin B
To Prove: (n² – 1) cos² A = m² – 1
Proof: sin A = m sin B (given)
tan A = n tan B
\(\frac{\sin A}{\cos A}=n \frac{\sin B}{\cos B}\)
on substituting sin B from eq. (i) We get
⇒ 1 – cos²A = m²(1 – \(\frac{n^2}{m^2}\)cos²A)
⇒ 1 – cos²A = m²\(\left(\frac{m^2-n^2 \cos ^2 \mathrm{~A}}{m^2}\right)\)
1 – cos²A = m² – n² cos²A
n² cos²A – cos²A = m² – 1
cos²A (n² – 1) = m² – 1
Hence, Proved
OR
Length of pipe (h) = 3.5 m = 350 cm 24
External radius (R) = \(\frac{2.4}{2}\) = 1.2 cm
Thickness (t) = 2 mm = 0.2 cm
Internal radius (r) = 1.2 cm – 0.2 cm = 1.0 cm

Volume of lead in the pipe
= π[(1.2)² – (1)²] × 350 cu.cm
= \(\frac{22}{7}\) × (1.44 – 1) × 350 cu.cm
= 484 cu.cm
∴ total mass of lead = (484 × 12) grams
= 5808 grams
= 5.808 kg

Question 31.
ABCD is a trapezium with AB ∥ DC. If ΔAED ~ ΔBEC, then prove that AD = BC. [3]

Answer:
For triangles AEB and CED, we have:
∠EAB = ∠ECD and ∠EBA = ∠EDC [alternate angles as AB ∥ DC]
∴ By AA similarly criterion, we have:
ΔAEB ~ ΔCED

SECTION – D (20 marks)
(Section – D consists of 4 questions of 5 marks each.)

Question 32.
Solve the pair of equations graphically: 4x – y = 5; x + y = 5. [5]
OR
Which term of the A.P. 3, 8, 13, 18, … is 78?
Answer:
4x – y = 5

x	1	2
y	-1	3

x + y = 5

X	3	2
y	2	3

The solution is x = 2, y = 3.
OR
Let nth term of the A.P. be 78.
Then, a_n = a + (n – 1 )d = 78
⇒ 3 + (n – 1)(5) = 78 (Here, a = 3, d = 5)
⇒ 5(n – 1) = 75
⇒ n – 1 = 15
⇒ n = 16
So, 16th term of the A.P. is 78.

Question 33.
Rahul got a total of 28 in math and science on a class quiz. The product of his scores would have been 180 if he had received 3 more in mathematics and 4 less in science. Find out what he scored in the two subjects. [5]
Answer:
Let Rahul has obtained V marks in mathematics and ‘y marks in science.
Then by the problem,
x + y = 28 …(i)
If Rahul would have got 3 marks more in mathematics, then Rahul would have got (x + 3) in mathematics and if Rahul would have got marks less in science then Rahul would have got (y – 4) marks in science.
A.T.Q.
⇒ (x + 3) (y – 4) = 180
⇒ (x + 3) (28 – x – 4) = 180
⇒ (x + 3) (24 – x) = 180
⇒ 72 + 21x – x² = 180
⇒ x² -21x + 108 = 0
⇒ (x – 12)(x – 9) = 0
x = 12, 9
Then, y = 16, 19
Hence, Rahul obtained 12 marks in mathematics and 16 marks in science.
Or Rahul obtained 9 marks in mathematics and 19 marks in science.

Question 34.
A parallelogram, PQRS is inside a DABC in which AB ∥ PS. Prove that OC ∥ SR. [5]

Answer:
Given: PQRS is Parallelogram in which AB ∥ PS
To Prove: OC ∥ SR
Proof: In ΔOAB and ΔOPS
PS ∥ AB (given ……..(i)
∠1 = ∠2
(Corresponding Pair of angles)
∠3 = ∠4
(Corresponding Pair of angles)
∴ ΔOPS – ΔOAB (by AA-similarity criterion)
\(\frac{O P}{O A}=\frac{O S}{O B}=\frac{P S}{A B}\) ……..(ii)

PQRS is a parallelogram, so PS ∥ QR …(iii)
⇒ QR ∥ AB (from (i) & (iii)
In ΔCQR and ΔCAB
∠5 = ∠CAB (Coresponding angle)
∠6 = ∠CBA (Coresponding angle)
∴ ΔCQR – ΔCAB (by AA similarity)
\(\frac{C Q}{C R}=\frac{C R}{C B}=\frac{Q R}{A B}\)
(Corresponding sides of similar triangles)
∴ PS = QR
\(\frac{P S}{A B}=\frac{C R}{C B}=\frac{C Q}{C A}\)
⇒ \(\frac{C R}{C B}=\frac{O S}{O B}\) (from (ii) and (v))
These are the rates of two sides of ABOC and are equal.
So, by converse of BPT, SR ∥ OC.
Hence, Proved.

Question 35.
A Ferris wheel, often known as a big wheel in the UK, is a type of amusement ride that consists of a rotating, upright wheel with several passenger-carrying units or passenger cars linked to the rim, so that they remain upright as the wheel rotates. AB is a chord of the outer wheel which touches the inner wheel at P. The radius of the inner wheel = 8 m and radius of outer wheel = 10 m. [5]

(A) Find the length of the chord AB of the outer circle.
(B) The chord AB of the inner wheel is extended to a point C. If BC = 9 m, then find distance of the point C from the centre of the wheel.
OR
The angle of elevation of the top of a tower from two points 8 m and 32 m from its base and in the same straight line with it, are complementary. Find the height of the tower.
Answer:
(A) Triangle OPB is right-angled at P since OP ⊥ AB.

OP = 8 m (radius of inner wheel) and
OB = 10
m (radius of outer wheel).
Therefore by Pythagoras theorem in ΔOPB,
OB² = OP² + BP²
⇒ BP² = OB² – OP²
= 10² – 8² = 36 = 62
⇒ BP = 6 m
AP = BP = \(\frac{1}{2}\)AB
So, AB = 2PB
= 2 × 6
= 12 m

(B) As BC = 9 m
∴ PC = PB + BC
= 6 + 9 = 15 m
Triangle OPC is right angled at P.
Therefore, applying Pythagoras theorem, we get
OC² = OP² + PC²
= 8² + 15²
= 64 + 225 = 289
⇒ OC = 17 m.
OR
Let AB represents the tower and points P and Q be two points at a distances of 32 m and 8 m from the base A of the tower, respectively.
∴ AQ = 8 and AP = 32 m

Let ‘h’ metres be the height of the tower.
Also, let ∠AQB = θ
Then, ∠APB = 90° – θ
[∵ ∠P and ∠Q are complementary i.e., their sum is 90°]
In ΔQAB,
\(\frac{A B}{A Q}\) = tan θ, or AB = 8 tan θ …(i)
In APAB,
\(\frac{A B}{A P}\) = tan (90° – θ) = cot θ
or AB = 32 cot θ …(ii)
From eqn. (i) and (ii) [∵ tan θ. cot θ = 1]
AB. AB = 8 tan θ × 32 cot θ
⇒ (AB)² = 256
⇒ AB = 16
Thus, the height of the tower is 16 metres.

SECTION – E (12 marks)
(Case Study Based Questions)
(Section – E consists of 3 questions. All are compulsory.)

Question 36.
Satellite TV manufacturing businesses “economies of scale.” When economies of tend to have what economists call scale exist, bigness can be its own reward.

The more TV’s you manufacture in a single run, lower the costs per unit, which in turn increases your bottom-line margins.
Keeping that in mind, a T.V. manufacturing company increases its production uniformly by fixed number every year: The company produces 8000 sets in the 6th year and 11,300 sets in the 9th year.
On the basis of the above information, answer the following questions:
(A) Find the company’s production of the first year.
OR
In which year the company’s production is 9100 sets ? [2]
(B) Find the company’s production of the 8th year. [1]
(C) Find the company’s total production of the first 6 years. [1]
Answer:
(A) Given a₆ = 8000 a₉ = 11,300
Let, the first term be ‘o’ and common difference be ‘d.
Then, a + (6 – 1)d = 8000
⇒ a + 5d = 8,000 …(i)
and a + 8d = 11,300 …(ii)
On solving (i) and (ii) we get
d = 1,100
⇒ a = 8000 – 5 × 1100
= 2500
OR
Let, the year in which production is 9100 be ‘n’
Then, an = a + (n – 1 )d
9100 = 2500 + (n – 1) × 1100
⇒ (n – 1) × 100 = 6600
⇒ n – 1 = 6
⇒ n = 7

(B) Since, a = 2500 and d = 1100
∴ a_n = a + (8 – 1)d
= 2500 + 7 × 1100
= 2500 + 7700 = 10,200

(C) Production in 6th year = 8,000
∴ S_n = \(\frac{n}{2}\)[a + l] = \(\frac{6}{2}\)[2500 + 8000]
= 3 × 10500
= 31,500

Question 37.
Eshan purchased a new building for her business. Being in the prime location, she decided to make some more money by putting up an advertisement sign for a rental ad income on the roof of the building.

From a point P on the ground level, the angle of elevation of the roof of the ouilding is 30° and the angle of elevation of the top of the sign board is 45°. The point P is at a distance of 24 m from the base of the building.
On the basis of the above information, answer the following questions:
(A) Find The height of the building (without the sign board).
OR
The height of the building ( with the sign board). [2]
(B) Find The height of the sign board. [1]
(C) Find the distance of the point P from the top of the sign board. [1]
Answer:
(A) Without the sign board, the height of the shop is AB.
In ΔPAB,

OR
Considering, the diagram in the above question, AC as the new height of the shop including the sign-baard.
∴ In ΔAPC,
tan 45° = \(\frac{\mathrm{AC}}{\mathrm{AP}}\)
1 = \(\frac{\mathrm{AC}}{24}\)
⇒ AC = 24 m

(B) Length of sign board, BC = AC – AB
= 24 – 14
= 10 m

(C) In ΔAPC
cos 45° = \(\frac{\mathrm{AP}}{\mathrm{AC}} \Rightarrow \frac{1}{\sqrt{2}}=\frac{24}{\mathrm{PC}}\)
⇒ PC = 24√2

Question 38.
in a toys manufacturing company, wooden parts are assembled and painted to prepare a toy. One specific toy is in the shape of a cone mounted on a cylinder.
For the wood processing activity center, the wood is taken out of storage to be sawed, after which it undergoes rough polishing, then is cut, drilled and has holes punched in it. It is then fine polished using sandpaper

For the retail packaging and delivery activity center, the polished wood sub¬parts are assembled together, then decorated using paint.
The total height of the toy is 26 cm and the height of its conical part is 6 cm. The diameters of the base of the conical part
is 5 cm and that of the cylindrical part is 4 cm.
On the basis of the above information, answer the following questions:
(A) If its cylindrical part is to be painted yellow, then find the surface area need to be painted. [1]
(B) If its conical part is to be painted green, then find the surface area need to be painted and also find the volume of the wood used in making this toy.
OR
If the cost of painting the toy is 3 paise per sq cm, then find the cost of painting the toy. ( Use π = 3.14) [2]
(C) The paint company gives a discount of 5% if the number of toys to be painted is 100 or above, then find the cost of painting 200 toys. [1]
Answer:
(A) C.S.A. of cylinder = 2πrH + πr²
= πr(2H + r)
= 2π(2 × 20 + 2)
[∵ H = 26 – 6 = 20] = 84 π

(B) C.S.A. of cone = πrl + π(R² – r²)
= π[r\(\sqrt{h^2+r^2}\) +(R² – r²)]
= π[2.5\(\sqrt{2.5^2+6^2}\) +(2.5² – 2²)]
= π[2.5 × 6.5 + 0.5 × 4.5]
= π[16.25 + 2.23]
= 18.5π sq units

Volume of toy
= Volume of cone + Volume of cylinder
= – πr2h + πR2H
= π[\(\frac{1}{3}\) × 2.5 × 2.5 × 6 + 2 × 2 × 20]
= π [12.5 + 80]
= 92.5π cm3
OR
Surface area
= S.A. of cone + S. A. of cylinder
= 84π + 18.5π
= 102.5π
∴ cost of painting = 0.03 × 102.5π
= ₹ 9.65

(C) Cost of painting 200 toys
= 9.65 × 200
= ₹ 1930
∴ Discount = \(\frac{5}{100}\) × 1930
= 96.50
∴ cost = 1930 – 96.50
= ₹ 1833.5

Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths with Solutions Set 1 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Maths Set 1 with Solutions

Time Allowed: 3 Hours
Maximum Marks: 80

General Instructions:

• This Question Paper has 5 Sections A, B, C, D, and E.
• Section A has 20 Multiple Choice Questions (MCQs) carrying 1 mark each.
• Section B has 5 Short Answer-I (SA-I) type questions carrying 2 marks each.
• Section C has 6 Short Answer-ll (SA-II) type questions carrying 3 marks each.
• Section D has 4 Long Answer (LA) type questions carrying 5 marks each.
• Section E has 3 Case Based integrated units of assessment (4 marks each) with sub-parts of the values of 1, 1 and 2 marks each respectively.
• All Questions are compulsory. However, an internal choice in 2 Qs of 2 marks, 2 Qs of 3 marks and 2 Questions of 5 marks has been provided. An internal choice has been provided in the 2 marks questions of Section
• Draw neat figures wherever required. Take π = 22/7 wherever required if not stated.

Section – A (20 marks)
(Section – A consists of 20 questions of 1 mark each.)

Question 1.
If two positive integers p and q can be expressed as p = ab² and q = a³b; a, b being prime numbers, then LCM (p, q) is [1]
(a) ab
(b) a²b²
(c) a³b²
(d) a³b³
Answer:
(c) a³b²

Explanation:
Given that, p = ab2 = a × b × b
And q = a³b = a × a × a × b
LCM of p and q = LCM (ab², a³b)
= a × b × b × a × a
= a³b²
[Since, LCM is the product of the greatest power of each prime factor involved in the numbers]

Question 2.
What is the greatest possible speed at which a man can walk 52 km and 91 km in an exact number of hours? [1]
(a) 17 km/hours
(b) 7 km/hours
(c) 13 km/hours
(d) 26 km/hours
Answer:
(c) 13 km/hours

Explanation: 52 = 13 × 4
91 = 13 × 7
So, greatest no is 13

Question 3.
If one zero of the quadratic polynomial x² + 3x + k is 2, then the value of k is: [1]
(a) 10
(b) – 10
(c) 5
(d) – 5
Answer:
(b) -10

Explanation: Let the given quadratic polynomial be f(x) = x² + 3x + k
Given one of the zero of the quadratic polynomial is 2. .
Hence f(2) = 0
Put x = 2 in f(x), we get
f(2) = 2² + 3(2) + k
0 = 4 + 6 + k
0 = 10 + k
k = – 10
Therefore, The value of k is -10.

Question 4.
Graphically, the pair of equations given by 6x – 3y + 10 = 0 2x – y + 9 = 0 represents two lines which are [1]
(a) intersecting at exactly one point
(b) parallel
(c) coincident
(d) intersecting at exactly two points
Answer:
(b) Parallel

Explanation:
The given equations are,
6x – 3y + 10 = 0
On dividing by 3, we get
⇒ 2x – y + \(\frac{10}{3}\) = 0 ……. (i)
And 2x – y + 9 = 0 ……… (ii)
The table for 2x – y + \(\frac{10}{3}\) = 0, is given below

x	0	-5/3
y	10/3	0

The table for 2x – y + 9 = 0 is given below

x	0	9/2
y	9	0

Hence, the pair of equations represents two parallel lines.

Question 5.
If the quadratic equation x² + 4x + k = 0 has real and equal roots, then
(a) k < 4 (b) k > 4
(c) k = 4
(d) k > 4 1
Answer:
(c) k = 4

Explanation:
Here, a = 1, b = 4 and c = k
As we know that D = b² – 4ac
Putting the value of a = 1, b = 4 and c = k
(4)² – 4 × 1 × k = 16 – 4k
The given equation will have real and distinct roots, if D = 0
16 – 4k = 0
4k = 16
k = \(\frac{16}{4}\)
= 4
Therefore, the value of k = 4.

Question 6.
The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is [1]
(a) 5 units
(b) 12 units
(c) 11 units
(d) (7 + √5) units 1
Answer:
(b) 12 units

Explanation:
Let A(0, 4), 0(0, 0) and B(3, 0) be the vertices of ∆AOB.
Using distance formula, we get
OA = \(\sqrt{(0-0)^2+(4-0)^2}\) = √16 = 4 units
OB = \(\sqrt{(3-0)^2+(0-0)^2}\) = √9 = 3 units
AB = \(\sqrt{(3-0)^2+(0-4)^2}\) = √9 + 16 = √25 = 5 units
∴ Perimeter of ∆AOB = OA + OB + AB
= 4 + 3 + 5 = 12 units
Thus, the required perimeter of the triangle is 12 units.

Question 7.
If in triangles ABC and DEF, \(\frac{A B}{D E}=\frac{B C}{F D}\) , then they will be similar, when [1]
(a) ∠B = ∠E
(b) ∠A = ∠D
(c) ∠B = ∠D
(d) ∠A = ∠F
Answer:
(c) ∠B = ∠D

Explanation:
\(\frac{A B}{D E}=\frac{B C}{F D}\)

Angle formed by AB and BC is ∠AB.
Angle formed by DE and FD is ∠AD.
So, ∠B = ∠D
∴ ∆ABC ~ ∆EDF [By SAS similarity criterion]
Therefore, the given triangles will be similar when ∠B = ∠D.

Question 8.
In which ratio the y-axis divides the line segment joining the points (5, – 6) and (-1, -4)? [1]
(a) 1:5
(b) 5:1
(c) 1:1
(d) 1:2
Answer:
(b) 5:1

Explanation:
Using the section formula, if a point (x, y) divides the line joining the points (x₁, y₁) and (x₂, y₂) in the ratio m: n, then
(x, y) = \(\left(\frac{m x_2+n x_1}{m+n}, \frac{m y_2+n y_1}{m+n}\right)\)
Let the point on y-axis be P(0, y) and
AP : PB = k : 1
Therefore, using section formula:
\(\frac{5 \times 1+(-1) \times k}{k+1}\) = 0
5 – k = 0
⇒ k = 5
Hence, required ratio is 5:1.

Question 9.
In the figure, if PA and PB are tangents to the circle with centre O such that ∠APB = 50°, then ∠OAB is equal to ________ [1]

(a) 25°
(b) 30°
(c) 40°
(d) 50°
Answer:
(a) 25°

Explanation: We know that the radius and tangent are perpendicular at their point of contact
∵ ∠OBP = ∠OAP = 90°
Now, In quadrilateral AOBP
∠AOB + ∠OBP + ∠APB + ∠OAP = 360°
[Angle sum property of a quadrilateral]
⇒ ∠AOB + 90° + 50° + 90° = 360°
⇒ 230° + ∠AOB = 360°
⇒ ∠AOB = 130°
Now, In isosceles triangle AOB
∠AOB + ∠OAB + ∠OBA = 180°
[Angle sum property of a triangle]
⇒ 130° + 2∠OAB = 180°
[∵ ∠OAB = ∠OBA]
⇒ ∠OAB = 25°

Question 10.
If sin A = \(\frac{1}{2}\), then the value of sec A is: [1]
(a) \(\frac{2}{\sqrt{3}}\)
(b) \(\frac{1}{\sqrt{3}}\)
(c) √3
(d) 1
Answer:
(a) \(\frac{2}{\sqrt{3}}\)

Question 11.
√3 cos² A + √3 sin² A is equal to:
(a) 1
(b) \(\frac{1}{\sqrt{3}}\)
(c) √3
(d) 0
Answer:
(c) √3

Explanation:
√3 cos²A + √3 sin²A = √3 (cos²A + sin²A)
= √3 × 1
[∵ cos²A + sin²A = 1]
= √3

Question 12.
The value of cos 1° cos 2° cos 3° cos 4° ……. cos 90° is: [1]
(a) 1
(b) 0
(c) -1
(d) 2
Answer:
(b) 0

Explanation: cos 1° cos2° cos 3°….. cos 90°
∴ (cos 1°) × (cos 2°) × (cos 3°) × … × (cos 89°) × (cos 90°)
cos 90° = 0
∴ (cos 1°) × (cos 2°) × (cos 3°) × …. × (cos 89°) × (cos 90°) = 0

Question 13.
If the perimeter of a circle is equal to that of a square, then the ratio of their areas is [1]
(a) 22 : 7
(b) 14 : 11
(c) 7 : 22
(d) 11 : 14
Answer:
(b) 14 : 11

Explanation:
Let radius of circle be r and side of a square be a.
According to the given condition,
Perimeter of a circle = Perimeter of a square
∴ 2πr = 4 a

Question 14.
If the radii of two circles are in the ratio of 4:3, then their areas are in the ratio of ____. [1]
(a) 4 : 3
(b) 8 : 3
(c) 16 : 9
(d) 9 : 16
Answer:
(c) 16 : 9
Explanation:
Let the radii of the two circles be r and R, the circumferences of the circles be c and C and the areas of the two circles be a and A.

Question 15.
The total surface area of a solid hemisphere of radius 7 cm is: [1]
(a) 447 π cm²
(b) 239 π cm²
(c) 174 π cm²
(d) 147 π cm²
Answer:
(d) 147 π cm²

Explanation: Total surface of hemisphere of radius r = 3πr² [1]
= 3π × 7 × 7
= 147π cm²

Question 16.
For the following distribution :

the upper limit of the modal class is: [1]
(a) 10
(b) 15
(c) 20
(d) 25
Answer:
(c) 20

Explanation:

class	Frequency	Cumulative Frequency
0-5	10	10
5-10	15	25
10-15	12	37
15-20	20	57
20-25	9	66

Here, N = 66.
The highest frequency is 20, which lies in the interval 15 – 20.
So, the upper limit of modal class is 20.

Question 17.
If the mean of the following distribution is 2.6, then the value of y is _____. [1]

(a) 3
(b) 8
(c) 13
(d) 24
Answer:
(b) 8

Explanation:

xi	fi	fixi
1	4	4
2	5	10
3	y	3y
4	1	4
5	2	10
	Σfi = 12+ y	Σfixi = 28 + 3y

Mean = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\)
⇒ 2.6 = \(\frac{28+3 y}{12+y}\)
⇒ 2.6 (12 + y) = 28 + 3y
⇒ 31.2 + 2.6 y = 28 + 3 y
⇒ 3y – 2.6y = 31.2 – 28
⇒ 0.4y = 3.2
Hence, y = 8

Question 18.
A card is selected at random from a well shuffled deck of 52 cards. The probability of its being a red face card is: [1]
(a) \(\frac{3}{26}\)
(b) \(\frac{3}{13}\)
(c) \(\frac{2}{13}\)
(d) \(\frac{1}{2}\)
Answer:
(a) \(\frac{3}{26}\)

Explanation:
There are 52 cards in a deck out of which half are red cards.
In a deck of cards, there are 6 red-faced cards.
Probability = \(\frac{\text { No. of favourable events }}{\text { Total no.of events }}\)
So, the probability of drawing a red-faced card is 6 out of 52 cards.
Thus, the probability of drawing a red face card
from a deck of cards = \(\frac{6}{52}\)
= \(\frac{3}{26}\)

Direction for question 19 8c 20: In question number 19 and 20, a statement of Assertion
(A) is followed by a statement of Reason (R). Choose the correct option as:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (IQ is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.

Question 19.
Assertion (A): If HCF of 510 and 92 is 2, then the LCM of 510 & 92 is 32460
Reason (R): as HCF(a, b) × LCM(a, b) = a × (R) b. [1]
Answer:
(d) Assertion (A) is false but reason (R) is true

Explanation:
510 and 92
510 = 2 × 3 × 5 × 17
92 = 2 × 2 × 23
HCF = 2
LCM = 2 × 2 × 3 × 5 × 17 × 23
= 23460
Hence, Product of the two numbers = 510 x 92
= 46920
HCF × LCM = 2 × 23460 = 46920
Hence, the product of two numbers = HCF × LCM

Question 20.
Assertion (A): The ratio in which the line segment joining (2, -3) and (5, 6) internally divided by x axis is 1: 2.
Reason (R): as formula for the internal division is
\(\left(\frac{m x_2+n x_1}{m+n}, \frac{m y_2+n y_1}{m+n}\right)\) [1]
Answer:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

Explanation: Using the section formula, if a point (x, y) divides the line joining the points (x₁, y₁) and (x₂, y₂) in the ratio m: n, then
(x, y) = \(\left(\frac{m x_2+n x_1}{m+n}, \frac{m y_2+n y_1}{m+n}\right)\)
Substituting (x₁, y₁) = (2, – 3) and (x₂, y₂) = (5, 6) in the section formula,
we get the point = \(\left(\frac{m(5)+n(2)}{m+n}, \frac{m(6)+n(-3)}{m+n}\right)\)
Since the point of intersection lies on the x-axis, y-coordinate = 0
= \(\frac{6 m-3 n}{m+n}\) = 0
⇒ 6m – 3n = 0
⇒ 6m = 3m
2m = n
⇒ m : n = 1 : 2

Section – B (10 marks)
(Section – B consists of 5 questions of 2 marks each.)

Question 21.
For what values of k will the following pair of linear equations have infinitely many solutions? [2]
kx + 3y – (k – 3) = 0
12x + ky – k = 0

Total Marks	Breakdown (As per CBSE Marking Scheme)
2m (SA-I)	Recall the condition for the pair of linear equations to have infinitely many solutions. Write the values of a1, b1, c1 and a2, b2, c2 from the given equations and substitute them to the condition, to get the value of k. (lm)

Answer:
For a pair of linear equations to have infinitely many solutions:
\(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\) ⇒ \(\frac{k}{12}=\frac{3}{k}=\frac{k-3}{k}\)
\(\frac{k}{12}=\frac{3}{k}\)
⇒ k² = 36
⇒ k = ± 6
Also, \(\frac{3}{k}=\frac{k-3}{k}\)
⇒ k² – 6k = 0
⇒ k = 0, 6.
Therefore, the value of k, that satisfies both the conditions, is k = 6.

Question 22.
In the figure, altitudes AD and CE of ∆ ABC intersect each other at the point P. Show that: [2]

(A) ∆ADB ~ ∆CEB
(B) ∆PDC ~ ∆BEC

Total Marks	Breakdown (As per CBSE Marking Scheme)
2m (SA-I)	Use AA criterion to prove ∆ABD ~ ∆CBE (1 m) Use AA criterion to prove ∆PDC ~∆BEC (1 m)

OR

In the figure, DE || AC and DF || AE. Prove that \(\frac{B F}{F E}=\frac{B E}{E C}\)

Total Marks	Breakdown (As per CBSE Marking Scheme)
2m (SA-I)	Apply basic proportionality theorem to prove BD/AD=BE/EC=BE/FE (1 m) prove BF/FE=BE/EC (1 m)

Answer:
(A) In ∆ABD and ∆CBE
∠ADB = ∠CEB = 90°

∠ABD = ∠CBE (Common angle)
⇒ ∆ABD ~ ∆CBE (AA criterion)

(B) In ∆PDC and ∆BEC
∠PDC = ∠BEC = 90°
∠PCD = ∠BCE (Common angle)
⇒ ∆PDC ~ ∆BEC (AA criterion)

In ∆ABE, DF || AE
\(\frac{\mathrm{BD}}{\mathrm{AD}}=\frac{\mathrm{BE}}{\mathrm{EC}}\) ….. (ii) (Using BPT)

From (i) and (ii)
\(\frac{B D}{A D}=\frac{B E}{E C}=\frac{B F}{F E}\)
Thus, \(\frac{B F}{F E}=\frac{B E}{E C}\)

Question 23.
Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle. [2]

Total Marks	Breakdown (As per CBSE Marking Scheme)
2m (SA-I)	Draw a correct figure with radius 5 cm and 3 cm (0.5 m) Apply the properties of tangent (0.5 m) Apply pythagoras theorem and find the length of chord AB (1 m)

Answer:
Let O be the centre of the concentric circle of radii 5 cm and 3 cm respectively. Let AB be a chord of the larger circle touching the smaller circle at P

Then AP = PB and OP ⊥ AB
Applying Pythagoras theorem in ∆OPA, we have
OA² = OP² + AP²
⇒ 25 = 9 + AP²
⇒ Ap² = 16
⇒ AP = 4 cm
∴ AB = 2AP = 8 cm

Question 24.
If cot θ = \(\frac{7}{8}\), evaluate \(\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}\) [2]

Total Marks	Breakdown (As per CBSE Marking Scheme)
2m (SA-I)	Use trigonometric identities to start simplifying the given expression (1 m) Evaluate the value of the given expression (1 m)

Answer:
Now,

Question 25.
Find the perimeter of a quadrant of a circle of radius 14 cm. [2]

Total Marks	Breakdown (As per CBSE Marking Scheme)
2m (SA-I)	Write the formula  for perimeter of a Quadrant (0.5 m) Substitute the values (1 m) Calculate the perimeter of a quadrant (0.5 m)

OR
Find the diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm.

Total Marks	Breakdown (As per CBSE Marking Scheme)
2m (SA-I)	Write the formula for area of the circle (0.5 m) Substitute the values (0.5 m) Calculate radius (0.5 m) Find diameter (0.5 m)

Answer:
Perimeter of quadrant = 2r + \(\frac{1}{4}\) × 2 πr
⇒ Perimeter = 2 × 14+ \(\frac{1}{2}\) × \(\frac{22}{7}\) × 14
⇒ Perimeter = 28 + 22 = 50 cm
OR
Area of the circle = Area of first circle + Area of second circle
⇒ πR² = π (r₁)² + π (r₂)²
⇒ πR² = π (24)² + π (7)²
⇒ πR² = 576π + 49π
⇒ πR² = 625π
⇒ R² = 625
⇒ R = 25
Thus, diameter of the circle = 2R = 50 cm.

Section – C (18 marks)
(Section – C consists of 6 questions of 3 marks each.)

Question 26.
Prove that √5 is an irrational number. [3]

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (SA-II)	Suppose √5 to be a rational number (0.5 m) Write it in fraction form i.e., p/q, where p, q are co-primes. (1 m) Find prime factors of p and q (0.5 m) p and q will have common prime factors which contradict our assumption that p, q are co-primes and √5 is an irrational number (1 m)

Answer:
Let us assume to the contrary, that √5 is rational Then we can find a and b (≠ 0) such that √5 = \(\frac{a}{b}\) (assuming that a and b are co-primes).
So, a = √5 b ⇒ a² = 5b²
Here 5 is a prime number that divides a² then 5 divides a also
(Using the theorem, if a is a prime number and if a divides p², then a divides p, where a is a positive integer)
Thus 5 is a factor of a
Since 5 is a factor of a, we can write a = 5c
(where c is a constant). Substituting a = 5c
We get (5c)² = 5b² ⇒ 5c² = b²
This means 5 divides b² so 5 divides b also
(Using the theorem, if a is a prime number and if a divides p², then a divides p, where a is a positive integer).
Hence a and b have at least 5 as a common factor.
But this contradicts the fact that a and b are coprime. This is the contradiction to our assumption that p and q are co-primes.
So, √5 is not a rational number. Therefore, the -√5 is irrational.

Question 27.
Find the zeroes of the quadratic polynomial 6x² – 3 – 7x and verify the relationship between the zeroes and the coeffcients. [3]

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (SA-II)	Equate the given polynomial to zero and factorise (1 m) Equate each factor to zero, to get the zeros of x (0.5 m) Write the formulae for sum and product of zeros (0.5 m) Substitute the values and verify the desired result (1 m)

Answer:
6x² – 7x – 3 = 0
⇒ 6x² – 9x + 2x – 3 =0
⇒ 3x(2x – 3) + 1(2x – 3) = 0
⇒ (2x – 3)(3x + 1) = 0
⇒ 2x – 3 = 0 and 3x + 1 = 0
x = \(\frac{3}{2}\) and x = \(\frac{-1}{3}\)
Hence, the zeros of the quadratic polynomials are 3/2 and -1/3.
For verification

Therefore, the relationship between zeros and their coefficients is verified.

Question 28.
A shopkeeper gives books on rent for reading. She takes a fixed charge for the first two days, and an additional charge for each day there after. Latika paid ₹ 22 for a book kept for six days, while Anand paid ? 16 for the book kept for four days. Find the fixed charges and the charge for each extra day. [3]

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (SA-II)	Assume the fixed charge and additional charge as x and y (0.5 m) Form the pair of linear equations from the given situation (1 m) Solve the two equations simultaneously to get the values of x and y (1 m) Calculate the fixed charge (0.5 m)

OR
Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (SA-II)	Assume the speed of both cars in form of x and y (0.5 m) Form the pair of linear equations from the given situation Solve the two equations (1 m) simultaneously to get the values of x and y (1 m) Calculate the speed of both cars (0.5 m)

Answer:
Let the fixed charge by ₹ x and additional charge by ₹ y per day.
Number of days for Latika = 6 = 2 + 4
Hence, Charge x + 4y = 22
x = 22 – 4y ……….(1)
Number od days for Anand = 4 = 2 + 2
Hence, Charge x + 2y = 16
x = 16 – 2 y ………. (2)
On comparing equation (1) and (2), we get,
22 – 4y = 16 – 2y ⇒ 2y = 6 ⇒ y = 3
Substituting y = 3 in equation (1), we get.
x = 22 – 4 (3) ⇒ x = 22 – 12 ⇒ x = 10
Therefore, fixed charge = ₹ 10 and additional charge = ₹ 3 per day

Detailed Answer:
₹ x is the fixed charge for the first two days
₹ y is the additional charge for each.
From the first condition,
Latika paid ₹ 22 for a book kept for six days
x + 4y = 22 ……… (1)
According to the second condition,
Anand paid ₹ 16 for a book kept for four days
x + 2y = 16 ………. (2)
Now solve equations (1) and (2)
Subtracting (2) from (1), we get,
2y = 6
y = 3.
Substitute the value of y in (2), we get,
x + 2 × 3 = 16
x + 16 – 6 = 10
x = 10.
x = 10.
y = 3.
Therefore, the fixed charge = ₹ 10 and the charge for each extra day = ₹ 3.

OR

AB = 100 km. We know that, Distance = Speed × Time.
AP – BP = 100
⇒ 5x – 5x = 100
⇒ x – y = 20 ………. (i)
AQ + BQ = 100
⇒ x + y = 100 ………. (ii)
Adding equations (i) and (ii), we get,
x – y + x + y = 20 + 100
⇒ 2x = 120
⇒ x = 60
Substituting x = 60 in equation (ii), we get
60 + y = 100
⇒ y = 40
Therefore, the speed of the first car is 60 km/hr
and the speed of the second car is 40 km/hr.

Question 29.
In the figure, PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T. Find the length TP. [3]

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (SA-II)	Apply the property of tangent (0.5 m) Use pythagoras theorem to find the value of OR (1 m) Use AA similarity criteria to prove TP/PO = RP/RG (0.5 m) Calculate the length TP (1 m)

Answer:
Since OT is perpendicular bisector of PQ.
Therefore, PR = RQ = 4 cm
Now,OR = \(\sqrt{O P^2-P R^2}\) = \(\sqrt{5^2-4^2}\) = 3 cm

Now, ∠TRP + ∠RPO = 90° (∵ ∠TPO = 90°)
and ∠TPR + ∠PTR = 90° (∵ ∠TRP = 90°)
So, ∠RPO = ∠PTR
So, ∆TRP – ∆PRO
[By A-A Rule of similar triangles]
So, \(\frac{T P}{P O}=\frac{R P}{R G}\)
⇒ \(\frac{\mathrm{TP}}{5}=\frac{4}{3}\)
⇒ TP = \(\frac{20}{3}\) cm

Question 30.
Prove that
\(\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}\) = 1 + sec θ cosec θ. [3]

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (SA-I)	Use trigonometric identities to start simplifying LHS (1 m) Eventually reach LHS = RHS stage (2 m)

OR
If sin θ + cos θ = √3, then prove that tan θ + cot θ = 1.

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (SA-II)	Use trigonometric identities to start simplifying LHS (1 m) Eventually reach LHS = RHS stage (2 m)

Answer:

OR

sin θ + cos θ = √3
⇒ (sin θ + cos θ)² = √3
⇒ sin²θ + cos²θ + 2 sin θ cos θ = 3
⇒ 1 + 2 sin θ cos θ = 3
⇒ 1 sin θ cos θ = 1
Now, tan θ + cot θ = \(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\)
= \(\frac{\sin ^2 \theta+\cos ^2 \theta}{\sin \theta \cos \theta}\)
= \(\frac{1}{\sin \theta \cos \theta}\) = \(\frac{1}{1}\) = 1

Question 31.
Two dice are thrown at the same time. What is the probability that the sum of the two numbers appearing on the top of the dice is : [3]
(A) 8?
(B) 13?
(C) less than or equal to 12?

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (SA-II)	(A) Determine the probability of getting 8 (1 m) (B) Determine the probability of getting 13 (1 m) (C) Determine the probability of getting less than or equal to 12 (1 m)

Answer:
(A) P(8) = \(\frac{5}{36}\)
(B) P(13) = \(\frac{0}{0}\)
(C) P (less than or equal to 12) = 1

Section – D (20 marks)
(Section – D consists of 4 questions of 5 marks each.)

Question 32.
An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11km/h more than that of the passenger train, find the average speed of the two trains. [5]

Total Marks	Breakdown (As per CBSE Marking Scheme)
5m (LA)	Assuming the average speed of both trains as x km/hr and (x+11) km/hr hours (0.5m) Form a quadratic equation in x (2 m) Solve it to find the values of x (1 m) Pick the positive/realistic value of x (0.5 m) Calculate the average speed of two trains (1 m)

OR

A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.

Total Marks	Breakdown (As per CBSE Marking Scheme)
5m (LA)	Assume the average of stream in x(0.5m) Form a quadratic equation in x (2 m) Solve it to find the values of x (1 m) Pick the positive/realistic value of x (0.5 m) Calculate the average speed of two trains (1 m)

Answer:
Let the average speed of passenger train = x km/h.
and the average speed of express train = (x + 11) km/h
As per given data, time taken by the express train to cover 132 km is 1 hour less than the passenger train to cover the same distance. Therefore,
\(\frac{132}{x}-\frac{132}{x+11}\) = 1
⇒ \(\frac{132(x+11-x)}{x(x+11)}\) = 1
⇒ \(\frac{132 \times 11}{x(x+11)}\) = 1
⇒ 132 × 11 = x(x + 11)
⇒ x² + 11x – 1452 = 0
⇒ x² + 44x – 33x – 1452 = 0
⇒ x (x + 44) – 33(x + 44) = 0
⇒ (x + 44)(x – 33) = 0
⇒ x = – 44, 33
As the speed cannot be negative, the speed of the passenger train will be 33 km/h and the speed of the express train will be 33 + 11 = 44 km/h.
OR
Let the speed of the stream be x km/hr
So, the speed of the boat in upstream
= (18 – x) km/hr
and the speed of the boat in downstream
= (18 + x) km/hr

⇒ 48x = 324 – x²
⇒ x² + 48x – 324 = 0
⇒ (x + 54) (x – 6) = 0
⇒ x = – 54 or 6
As speed to stream con never be negative, the speed of the stream is 6 km/hr.

Question 33.
Prove that If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. In the figure, find EC if \(\frac{A D}{D B}=\frac{A E}{E C}\) using the above theorem. [5]

Total Marks	Breakdown (As per CBSE Marking Scheme)
5m (LA)	Draw a figure with necessary construction (1 m) Apply basic proportionality theorem to prove ar(ADE) = ar(ABC) (1 m) Prove AD/DB = AE/EC (1.5m) Apply basic proportionality theorem and calculate the value of EC (1.5m)

Answer:
We are given a triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively.

We need to prove that \(\frac{A D}{A E}=\frac{D B}{E C}\)
Let us join BE and CD and then draw DM ⊥ AC and EN ⊥ AB.
Now, area of ∆ADE (\(\frac{1}{2}\) base × height) = \(\frac{1}{2}\) AD × EN. area of AADE is denoted as ar(ADE).
So, ar (ADE) = \(\frac{1}{2}\) AD × EN
Similarly, ar(BDE) = \(\frac{1}{2}\) DB × EN, ar (ADE = \(\frac{1}{2}\)
AE × DM and ar(DEC) = \(\frac{1}{2}\) EC × DM.

Note that ABDE and ADEC are on the same base DE and between the same parallels BC and DE.
So, ar(BDE) = ar(DEC) …(iii)
Therefore, from (i), (ii) and (iii), we have:
\(\frac{A D}{D B}=\frac{A E}{E C}\)
\(\frac{2}{6}=\frac{3}{E C}\)
EC = \(\frac{6 \times 3}{2}\)
EC = 9 cm

Question 34.
A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand. [5]

Total Marks	Breakdown (As per CBSE Marking Scheme)
5m (LA)	Write values and calculate the volume of one conical depression (1 m) Calculate the volume of cuboidal box (1 m) Deduct volume of cuboidal box and find value  (2 m) Calculate the volume of wood in the entire stand (1 m)

OR
Ramesh made a bird-bath for his garden in the shape of a cylinder with a hemispherical depression at one end. The height of the cylinder is 1.45 m and its radius is 30 cm. Find the total surface area of the bird-bath.

Total Marks	Breakdown (As per CBSE Marking Scheme)
5m (LA)	Calculate the radius and height of bird-bath (0.5m) Write the formula of surface area of combination of solids (0.5m) Calculate the CSA of cylinder, hemisphere (2m) Add the CSA of hemisphere and cylinder (1 m) Find the total surface area of bird- bath (1 m)

Answer:
Volume of one conical depression
= \(\frac{1}{3}\) × πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 0.52 × 1.4 cm³
= 0.366 cm³
Volume of 4 conical depression
= 4 × 0.366 cm³
= 1.464 cm³
Volume of cuboidal box
= L × B × H
= 15 × 10 × 3.5 cm³ = 525 cm³

Remaining volume of box
= Volume of cuboidal box – Volume of 4 conical depressions
= 525 cm³ – 1.464 cm³
= 523.5 cm³

OR

Let h be height of the cylinder, and r the common radius of the cylinder arid hemisphere.

Then, the total surface area = CSA of cylinder + CSA of hemisphere
= 2 πrh + 2πr² = 2πr(h + r)
= \(\frac{22}{7}\) × 390(145 + 30) cm²
= \(\frac{22}{7}\) × 30 × 175 cm²
= 33000 cm² = 3.3 m²

Detailed Answer:
Given: The height of the cylinder is 1.45 m and its radius is 30 cm.
Let h be height of the cylinder, and r the common radius of the cylinder and hemisphere. Then,
The total surface area of the bird-bath = CSA of cylinder + CSA of hemisphere + Area of the circle in bottom
= 2πrh + 2 πr²
= (2 × \(\frac{2 \dot{2}}{7}\) × 30 × 145) + (2 × \(\frac{2 \dot{2}}{7}\) × 302) [∵ 1.45 m = 145 cm]
= \(\frac{191400}{7}+\frac{39600}{7}\)
= \(\frac{231000}{7}\)
= 33000 cm²
= 3.3 m²
Hence, the total surface area of the birdbath is 3.3 m².

Question 35.
A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years. [5]

Age (in years)	Number of policy holders
Below 20	2
20-25	4
25-30	18
30-35	21
35-40	33
40-45	11
45-50	3
50-55	6
55-60	2

 

Total Marks	Breakdown (As per CBSE Marking Scheme)
5m (LA)	Make frequency distribution table with all given data (2m) Write the formula of median (0.5m) Find the median class (1.5m) Calculate the median age (1 m)

Answer:

Class Interval	Number of policy holders (f)	Cumulative Frequency (cf)
Below 20	2	2
20-25	4	6
25-30	18	24
30-35	21	45
35-40	33	78
40-45	11	89
45-50	3	92
50-55	6	98
55-60	2	100

n = 100 ⇒ \(\frac{n}{2}\) = 50, Therefore, median class = 35 – 40,
Class size, h = 5, Lower limit of median class, l = 35,
frequency f = 33, cumulative frequency cf = 45 ‘
⇒ Median = l + \(\left[\frac{\frac{n}{2}-c f}{f}\right]\)
⇒ Median = 35 + \(\left[\frac{50-45}{33}\right]\) × 5
= 35 + \(\frac{25}{33}\) = 35 + 0.76
= 35.76
Therefore, median age is 35.76 years

SECTION – E (12 marks)
(Case Study Based Questions)
(Section E consists of 3 questions. All are compulsory.)

Question 36.
Case Study -1
In the month of April to June 2022, the exports of passenger cars from India increased by 26% in the corresponding quarter of 2021-22, as per a report. A car manufacturing company planned to produce 1800 cars in 4th year and 2600 cars in 8th year. Assuming that the production increases uniformly by a fixed number every year.

Based on the above information answer the following questions.
(A) Find the production in the 1st year. [1]
(B) Find the production in the 12th year. [1]
(C) Find the total production in first 10 years.

OR

In which year the total production will reach to 15000 cars? [2]

Total Marks

Breakdown (As per CBSE Marking Scheme)

4m (CBQ)

Apply the distance formula (0.5m) Calculate the distance between LB (0.5m) Apply the mid-point formula (0.5m) Find the coordinates of K  (0.5m) Equate the distances of the point (x, y) from two points (0.5m) Apply the distance formula (0.5m) Get the type of triangle (1 m) OR Equate the distances of the point (x,y) from two points (0.5m) Apply the distance formula and find ther value of b (0.5m) Get the coordinates (1 m)

Answer:
(A) Since the production increases uniformly by a fixed number every year, the number of Cars manufactured in 1st, 2nd, 3rd, ….years will form an AP.
So, a + 3d = 1800 and a + 7d = 2600
So d = 200 and a = 1200

(B) t₁₂ = a + 11d
⇒ t₃₀ = 1200 + 11 × 200
⇒ t₁₂ = 3400

(C) S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
⇒ S₁₀ = \(\frac{10}{2}\) [2 × 1200+ (10 – 1)200]
⇒ S₁₀ = \(\frac{13}{2}\) [2 × 1200 + 9 × 200]
⇒ S₁₀ = 5 × [2400 + 1800]
⇒ S₁₀ = 5 × 4200 = 21000

OR

Let in n years the production will reach to 31200
S_n = \(\frac{n}{2}\) [2a + (n – 1)d] = 31200
⇒ \(\frac{n}{2}\) [2 × 1200 + (n – 1)200] = 31200
⇒ \(\frac{n}{2}\) [2 × 1200 + (n – 1)200] = 31200
⇒ \(\frac{n}{2}\) [2 × 1200 + (n – 1)200] = 31200
⇒ n² + 11n – 312 = 0
⇒ n² + 24n – 13n – 312 = 0
⇒ (n + 24) (n – 13) = 0
⇒ n = 13 or – 24. As n can’t be negative
So, n = 13

Question 37.
Case Study – 2
In a GPS, The lines that run east-west are known as lines of latitude, and the lines running north-south are known as lines of longitude. The latitude and the longitude of a place are its coordinates and the distance formula is used to find the distance between two places. The distance between two parallel lines is approximately 150 km. A family from Uttar Pradesh planned a round trip from Lucknow (L) to Puri (P) via Bhuj (B) and Nashik (N) as shown in the given figure below.

Total Marks

Breakdown (As per CBSE Marking Scheme)

4m (CBQ)

Apply the distance formula (0.5m) Calculate the distance between LB (0.5m) Apply the mid-point formula (0.5m) Find the coordinates of K (0.5m) Equate the distances of the point (x.y) from two points (0.5m) Apply the distance formula (0.5m) Get the type of triangle (1 m) OR Equate the distances of the point (x,y) from two points (0.5m) Apply the distance formula and find the value of b (0.5m) Get the coordinates (1 m)

Based on the above information answer the following questions using the coordinate geometry.
(A) Find the distance between Lucknow (L) to Bhuj (B). [1]
(B) If Kota (K), internally divide the line segment joining Lucknow (L) to Bhuj (B) into 3 : 2 then find the coordinate of Kota (K). [1]
(C) Name the type of triangle formed by the places Lucknow (L), Nashik (N) and Puri (P)
OR
Find a place (point) on the longitude (y-axis) which is equidistant from the points Lucknow (L) and Puri (P). [2]
Answer:

(A) LB = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
⇒ LB = \(\sqrt{(0-5)^2+(7-10)^2}\)
LB = \(\sqrt{(5)^2+(3)^2}\)
⇒ LB = √25 + 9
LB = √34

(B) Coordinate of kota(K) is

(C) L(5, 10), N(2, 6), P(8, 6)

as LN = PL ≠ NP, so D LNP is an isosceles triangle

OR

Let A (O, b) be a point on the y – axis then AL = AP
⇒ \(\sqrt{(5-0)^2+(10-b)^2}\) = \(\sqrt{(8-0)^2+(6-b)^2}\)
⇒ (5)² + (10 – b)² = (8)² + (6 – b)²
⇒ 25 + 100 – 20b + b² = 64 + 36 – 12b + b²
⇒ 8b = 25 ⇒ b = \(frac{25}{8}\)
So, the coordinate on y axis is (0, \(frac{25}{8}\))

Question 38.
Case Study – 3
Lakshaman Jhula is located 5 kilometers north-east of the city of Rishikesh in the Indian state of Uttarakhand. The bridge connects the villages of Tapovan to Jonk. Tapovan is in Tehri Carhwal district, on the west bank of the river, while Jonk is in Pauri Garhwal district, on the east bank. Lakshman Jhula is a pedestrian bridge also used by motorbikes. It is a landmark of Rishikesh. A group of Class X students visited Rishikesh in Uttarakhand on a trip. They observed from a point (P) on a river bridge that the angles of depression of opposite banks of the river are 60° and 30° respectively. The height of the bridge is about 18 meters from the river. [2]

Based on the above information answer the following questions.
(A) Find the distance PA. [1]
(B) Find the distance PB. [1]
(C) Find the width AB of the river.
OR
Find the height BQ if the angle of the elevation from P to Q be 30°. [2]

Total Marks	Breakdown (As per CBSE Marking Scheme)
4m (CBQ)	(A) Apply trigonometric ratios (0.5m) Calculate the distance PA (0.5 m) (B) Apply trigonometric ratios (0.5m) Calculate the distance PB (0.5 m) (C) Apply trigonometric ratios to find the length AC and CB (1 m) Calculate the width of the river(1 m) OR Apply trigonometric ratios to find the length RB and QR (1 m) Calculate the height (1 m)

Answer:

(A) sin 60° = \(\frac{P C}{P A}\)
⇒ \(\frac{\sqrt{3}}{2}=\frac{18}{\mathrm{PA}}\)
⇒ PA = 12√3 m

(B) sin 30° = \(\frac{\mathrm{PC}}{\mathrm{PB}}\)
⇒ \(\frac{1}{2}=\frac{18}{P B}\)
⇒ PB = 36 m

(C) tan 60° = \(\frac{P C}{A C}\)
⇒ √3 = \(\frac{18}{\mathrm{AC}}\)
⇒ AC = 6√3 m
tan 30° = 6√3 m
⇒ \(\frac{1}{\sqrt{3}}=\frac{18}{C B}\)
⇒ CB = 18√3 m
Width AB = AC + CB = 6√3 + 18√3 = 24√3 m

OR

RB = PC = 18 m and PR = CB = 18√3 m
tan 30° = \(\frac{\mathrm{QR}}{\mathrm{PR}}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{Q R}{18 \sqrt{3}}\)
⇒ OR = 18m
QB = QR + RB = 18 + 18 = 36 m.
Hence height BQ is 36 m.

Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths with Solutions Set 2 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Maths Set 2 with Solutions

Time Allowed: 3 Hours
Maximum Marks: 80

General Instructions:

• This Question Paper has 5 Sections A, B C D. and E
• Section A has 20 Multiple Choice Questions (MCQs) carrying 1 mark each.
• Section B has 5 Short Answer-I (SA-I) type questions carrying 2 marks each.
• Section Chas 6 Short Answer-II (SA-II) type questions carrying 3 marks each.
• Section D has 4 Long Answer (LA) type questions carrying S marks each.
• Section E has 3 Case Based integrated units of assessment (4 marks each) with sub-parts of the values of 1, 1 and 2 marks each respectively.
• AU Questions are compulsory. However an internal choice in 2 Qs of 2 marks, 2 Qs of 3 marks and 2
  Questions of 5 marks has been provided. An internal choice has been provided in the 2 marks questions
  of Section E
• Draw neat figures wherever required. Take π =22/7 wherever required if not stated.

SECTION – A (20 marks)
(Section – A consists of 20 questions of 1 mark each)

Question 1.
The common zero of the polynomial x³ + 1, x² – 1 and x² + 2x + 1 is: [1]
(a) 1
(b) -1
(c) 3
(d) -2
Answer:
(b) -1

Explanation:
x³ + 1 = (x + 1)(x² – x + 1)
x² + 1 = (x + 1)(x – 1)
x² + 2x + 1 = (x + 1)²
The common factor is (x + 1), i.e. the common zero is -1.

Question 2.
If p, 2p – 1, 2p + 1 are three consecutive terms of an A.P., then the value of ‘p’ is: [1]
(a) 3
(b) 4
(c) 5
(d) 6
Answer:
(a) 3

Explanation:
Since p, 2p – 1 and 2p + 1 are consecutive terms of an A.P.,
2(2p – 1) = p + (2p + 1)
⇒ 4p – 2 = 3p + 1
⇒ P = 3

Question 3.
In which quadrant does the point (-1, -2) lie? [1]
(a) I
(b) II
(c) III
(d) IV
Answer:
(c) III

Explanation:
Third quadrant since, both ‘x’ and ‘y’ are negative.

Question 4.
If radii of two concentric circles are 4 cm and 5 cm, then the length of the chord to one circle which is tangent to the other circle is: [1]
(a) 6 cm
(b) 5 cm
(c) 8 cm
(d) 9 cm
Answer:
(a) 6 cm

Explanation:
Here, OP ⊥ AB and
AP = PB = \(\frac{1}{2}\)AB
We know tangent is perpendicular to radious at the point of contact.
∴ OP ⊥ AB
Also, AB is a chord to smaller circle and perpendicular from centre to the chord bisects it

Question 5.
The tangent of a circle makes angle with radius at point of contact is: [1]
(a) 60°
(b) 30°
(c) 90°
(d) none of these
Answer:
(c) 90°

Explanation:
Tangent at any point of a circle is perpendicular to the radius at the point of contact So, the tangent makes a right angle with the radius at the point of contact

Question 6.
The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is: [1]
(a) 10 units
(b) 14 units
(c) 15 units
(d) 12 units
Answer:
(d) 12 units

Explanation:
Perimeter of ΔABC.
= OA + AB + OB

= 3 + \(\sqrt{(3-0)^2+(0-4)^2}\) + 4
= 3 + 5 + 4
= 12 units

Question 7.
What is the area of a circle inscribed in a square of side ‘a’ units ? [1]
(a) \(\frac{\pi a^2}{4} \)
(b) πa²
(c) \(\frac{\pi}{2}\)
(d) \(\frac{\pi a}{2}\)
Answer:
(a) \(\frac{\pi a^2}{4} \)

Explanation:
The radius of the inscribed circle is \(\frac{a}{2}\).

So, its area is π\left(\frac{a}{2}\right)^2, i.e \frac{\pi a^2}{4}

Question 8.
A number from 11 to 30 was chosen at random. The probability of this chosen number being a multiple of 2 is: [1]

(a) 50-60
(b) 10-20
(c) 40-50
(d) 30-40
Answer:
(d) 30-40

Explanation:

Here, maximum frequency is 30, which belongs to class 30 – 40.
∴ Modal class = 30 – 40

Question 9.
A number from 11 to 30 was chosen at random. The probability of this chosen number being a multiple of 2 is: [1]
(a) \(\frac{1}{2}\)
(b) \(\frac{2}{3}\)
(c) \(\frac{1}{4}\)
(d) \(\frac{3}{2}\)
Answer:
(a) \(\frac{1}{2}\)

Explanation:
Total numbers from 11 to 30 = 20
∴ Total number of outcomes = 20
Multiples of 2 from 11 to 30 = {12, 14, 16, … 28, 30}
⇒ Number of favourable outcomes = 10
∴ P (Multiple of 2) = \frac{10}{20}=\frac{1}{2}

Question 10.
A car travels 0.99 km in which each wheel makes 450 complete revolutions. The radius of the wheel is: [1]
(a) 30 cm
(b) 32 cm
(c) 35 cm
(d) 40 cm
Answer:
(c) 35 cm

Explanation:
The wheel covers a distance of 2πr, in one revolutions.
So, 450 × 2πr = 0.99 × 1000 × 100
450 × 2 × \(\frac{22}{7}\) × r = 99 × 1000
r = 35
Thus, the radius of the wheel is 35 cm.

Question 11.
If α and β are the zeros of the polynomial p(x) = x² – px + q, then the value of \frac{1}{\alpha}+\frac{1}{\beta} is: [1]
(a) p
(b) \(\frac{-p}{q}\)
(c) pq
(d) \(\frac{p}{q}\)
Answer:
(d) \(\frac{p}{q}\)

Explanation:
Since a and P are the zeros of p(x) = x² – px + q,
∴ α + β = p and αβ = q
Now, \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha \beta}=\frac{p}{q}\)

Question 12.
The sum of first 20 even numbers is: [1]
(a) 420
(b) 320
(c) 140
(d) 450
Answer:
(a) 420

Explanation:
The list of first 20 even numbers is: 2, 4, 6, 8, ………., 40
It is an A.P. with a = 2, d = 2 20
So, the required sum = \(\frac{20}{2}\) [2 × 2 + (20 – 1)(2)]
= 10(4 + 38)
= 420

Question 13.
The discriminant of the quadratic equation (p + 3)x² – (5 – p)x + 1 = 0 is: [1]
(a) (p + 13)
(b) (p + 1)
(c) (P + 2)
(d) (p – 13) (p – 1)
Answer:
(d) (p – 13) (p – 1)

Explanation:
Discriminant = b² – 4ac
= [-(5 – p)]² – 4(p + 3)(1)
= 25 + p² – 10p – 4p – 12
= p² – 14p + 13
= p²– 13p – p+ 13
= p (p – 13) – 1 (p – 13)
= (p – 13) (p – 1)

Question 14.
In a ΔABC, DE ∥ BC with D on AB and E on AC. If \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{2}{3}\), then \(\frac{B C}{D E}\) is: [1]
(a) \(\frac{5}{2}\)
(b) \(\frac{3}{2}\)
(c) \(\frac{5}{2}\)
(d) \(\frac{2}{3}\)
Answer:
(a) \(\frac{5}{2}\)

Explanation: Since DE is parallel to BC,
∴ ΔADE ~ ΔABC
⇒ \(\frac{A D}{A B}=\frac{D E}{B C}\) ………….(i)

Question 15.
The median of the following data is: [1]
5, 2, 7, 9, 3, 2, 4, 8.
(a) 5.5
(b) 4
(c) 4.5
(d) 5
Answer:
(c) 4.5

Explanation:
Arranging the data in ascending order,
2, 2, 3. 4, 5, 7, 8, 9

Here, number of terms (n) = 8 i.e., even
∴ median =

Question 16.
The discriminant 3x² + 2x = 0 is: of the equation [1]
(a) 4
(b) 2
(c) 1
(d) -5
Answer:
(b) 2

Explanation:
Here, equation is 3x² + 2x = 0 Then comparing it with ax² + bx + c = 0, we get
a = 3, b = 2, c = 0

Then, discriminant,
D = \(\sqrt{b^2-4 a c}\)
= \(\sqrt{(2)^2-4 \times 3 \times 0}\)
= √4 = 2

Question 17.
The 8th term from the end of A.P.: – 12, -7,-2, ………., 68 is: [1]
(a) 33
(b) 35
(c) 30
(d) 36
Answer:
(a) 33

Explanation:
Write the given A.P in reverse order, then series is,
68, 63, ……. – 2, – 7, – 12
Then, a = – 68, d = – 5
a8 = a + (8 – 1) d = 68 + 7 (- 5) = 33

Question 18.
In the given figure, ΔABC – ΔPQR. The value of x is: [1]

(a) 4 cm
(b) 6 cm
(c) 3 cm
(d) 5 cm

Direction for questions 19 and 20: In question number 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R).
Choose the correct option:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A)
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer:
(c) 3 cm

Explanation:
Since, ΔABC ~ ΔPQR

Question 19.
Assertion (A): The length of the tangent will be 7 cm in a circle with a radius of 3 cm and a point’s distance from the centre of the circle is 5 cm. [1]
Reason (R) : (hypotenuse)² = (base)² + (height)²
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A)
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer:
(d) Assertion (A) is false but reason (R) is true.

Explanation:

A tangent at any point of a circle is perpendicular to the radius through the point of contact
Therefore, DOBA = 90° and AOBA is a right-angled triangle.
By Pythagoras theorem,
OA² = OB² + AB²
5² = 3² + AB²
AB²= 25 – 9
AB² = 16
AB = ±4

Question 20.
Assertion (A): If the number of runs scored by 11 players of a cricket team of India are 5, 19, 42, 11, 50, 30, 21, 0, 52, then median is 21. [1]
Reason (R): Median \(\left(\frac{n+1}{2}\right)^{\text {th }}\) value, if n is odd.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A)
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Answer:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)

Explanation:
Arranging the terms in ascending order, 0, 5, 11, 19, 21, 30, 42, 50, 52.
Median = Value of (n + 1)/2^th observation
Median score = Value of (9 + 1)/2^th term
= Value of 5^th term
= 21

SECTION – B (10 marks)
(Section – B consists of 5 questions of 2 marks each)

Question 21.
Show that 3 + √5 is an irrational number, assume that √5 is an irrational number.
OR
Solve algebraically: 4x + 3y = 14 and 3x – 4y = 23. [2]
Answer:
If possible, let us assume that 3 + √5 be a rational number. So, there exists positive integers a and b such that 3 + √5 = \(\frac{a}{b}\), where a and b are integers having no common foctor other than 1 and b ≠ 0.
⇒ √5 = \(\frac{a}{b}\) – 3
⇒ √5 = \(\frac{a-3 b}{b}\)
Since, \(\frac{a-3 b}{b}\) is a rational number, √5 is a rational number which is a contradiction to the fact that ”√5 is irrational”.
Hence, 3 + √5 is irrational
OR
Given, 4x + 3y = 14 …(i)
and 3x – 4 y = 23 ……(ii)
Adding (i) and (ii). we get
7x – y = 37
y = 7x – 37 …(iii)
Substituting the value of y in equation (i), we get
4x + 3 (7x – 37) = 14
⇒ 4x + 21x – 111 = 14
⇒ 25x = 125
⇒ x = 5
Putting x = 5 in eq. (iii).we get
y = 7 (5) – 37
= 35 – 37
= – 2
The solution is x = 5 and y = – 2.

Question 22.
Evaluate: (sin⁴ 60° + sec⁴ 30°) – 2 (cos² 45° – sin² 90°)
OR
Prove that \(\frac{\sin \theta-2 \sin ^3 \theta}{2 \cos ^3 \theta-\cos \theta}\) = tan θ. [2]
Answer:
(sin⁴ 60° + sec⁴ 30°) – 2 (cos² 45° – sin² 90°)

Question 23.
A dice is thrown twice. Find the probability that 5 will not come up either time. [2]
Answer:
Total number of possible outcomes = 36
Number of favourable outcomes = 36 – 11 = 25
[Excluding (1, 5), (5, 1), (2, 5). (5, 2), (3, 5), (5, 3), (4, 5), (5, 4), (5, 5), (5, 6), (6, 5)]
So, required probability = \(\frac{25}{36}\)

Question 24.
Find a point on x-axis which is equidistant from A (-3, 4) and B (7, 6).
Answer:
Let P (x, 0) be a point on x-axis, equidistant from A (-3, 4) and B (7, 6), i.e.,
AP = BP or AP² = BP²
i.e. (-3 – x)² + (4 – 0)² = (7 – x)² + (6 – 0)²
i.e. 9 + x² + 6x + 16 = 49 + x² – 14x + 36
⇒ 6x + 25 = – 14x + 85
⇒ 20x = 60
⇒ x = 3
Thus, the required point is (3, 0).

Question 25.
If the circumfrence of a circle increases from 4π to 8π, then find the percentage increase in the area of the circle. [2]
Answer:
Let r and R be the radii of the initial circle and the increased circle, respectively. Then,
2πr = 4π and 2πR = 8π
⇒ r = 2 and R = 4
∴ Area of initial circle
= π(2)² i.e. 4π
and Area of increased circle
= π(4)² i.e. 16π

Thus, % increase = \(\frac{16 \pi-4 \pi}{4 \pi}\) × 100
= \(\frac{12 \pi}{4 \pi}\) × 100
= 300%

SECTION – C (18 marks)
(Section – C consists of 6 questions of 3 marks each.)

Question 26.
The sum of the squares of two consecutive multiples of 7 is 637. Find the two multiples.
OR
Find the last term of an AP having 9 terms whose last term is 28 and sum of all the terms is 144. [3]
Answer:
Let 7x and 7(x + 1) be two consecutive multiples of 7.
Then,
(7x)² + [7(x + 1)]² = 637
⇒ 49x² + 49 (x² + 2x + 1) = 637
⇒ 98x² + 98x + 49 = 637
⇒ 98x² + 98x – 588 =0
⇒ x² + x – 6 = 0
⇒ (x + 3)(x – 2) = 0
⇒ x + 3 = 0 or x-2 = 0
⇒ x = -3 or x = 2
Thus, two consecutive multiples of 7 are 14 and 21, or – 21 and – 14.
OR
Let ‘a’ be the first term of A.P. and‘d be the common difference.
Here, total number of terms of A.P. is 9, i.e. n = 9
nth term = last term = = an = a + 8d = 28 …(i)
Also, S_n = S₉ = \(\frac{9}{2}\)[2a + (9 – 1)d] = 144
⇒ 9 (a + 4 d) = 144
or 9a+ 36d= 144 …(ii)

Solving (i) and (ii) simultaneously, we get a = 4; d = 3
Thus, the required first term is 4.

Question 27.
The figure drawn on the graph paper shows a AABC with vertices A (-4, 1), B (-4, 6) and C (8, 1). [3]

(A) Find the length of BC;
Answer:
Length of BC = \(\sqrt{A B^2+A C^2}\)
= \(\sqrt{(1-6)^2+(8-(4))^2}\)
= \(\sqrt{5^2+12^2}\)
= \(\sqrt{25+144}=\sqrt{169}\) i.e 13

(B) Find sin ∠ABQ
Answer:
Here, AB = 5cm, AC = 12 cm and BC = 13 cm
So, sin ∠ABQ = \(\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{12}{13}\)

(C) Find cos ∠BCA.
Answer:
cos ∠BCA = \frac{\mathrm{AC}}{\mathrm{BC}}=\frac{12}{13}\(\)

Question 28.
Prove that the length of two tangents drawn from an external point to a circle are equal. [3]
Answer:
Here, AP and AQ are two tangents drawn from an external point A, to the circle with centre O.
We need to prove that, AP = AQ
Join OP, OQ and OA.

Proof:
Since a tangent at any point of a circle is perpendicular to the radius through the point of contact.
∴ OQ ⊥ QA and OP⊥ PA
So, ∠OQA = 90° = OPA
Now, in two Δs OQA and CPA. we have:
(j) QQ = OP(Rodii of the same cirde)
(ii) ∠OPA = ∠OQA (each 90°)
(iii) OA = OA (common)
So, by RHS congruence criterion,
ΔOPA = ΔOQA
This concludes that, AP = AQ
Thus, the lengths of tangents drawn from an external point to a circle are equal.

Question 29.
The minute hand of a clockis 2 cm long. Find the area of the face of the clock described by the minute hand between 7 am and 7:15 am.
OR
A cone of height 24 cm and radius of base 6 cm is made up from modelling clay. A child reshapes it in the form of a sphere. Find the diameter of the sphere. [3]
Answer:
We know that a minute hand sweeps by 6° angle in one minute.

So, it will sweep by an angle of 90° in 15 minutes.
So, area swept by the minute-hand in 15 minutes
= \(\frac{90}{360}\) × π(2)² sq cm
= π sq cm, or \(\frac{22}{7}\) sq cm.
OR
Height of the cone, h = 24 cm
Radius of base of the cone, r = 6 cm
Let, the radius of the sphere formed by reshaping cone be R.
As the cone is reshaped into sphere:
∴ Volume of cone = Volume of sphere
\(\frac{1}{3}\)πr²h = \(\frac{4}{3}\)πR³
r²h = 4R³
6 × 6 × 24 = 4 × R³
R³ = 6 × 6 × 6
R = 6 cm
∴ Radius (R) of the sphere formed = 6 cm
Hence, diameter of sphere = 6 × 2 = 12 cm

Question 30.
Determine the median for the following frequency distribution: [3]

Answer:

Class	Frequency	Cum. frequency
100-120	12	12
120-140	14	26
140-160	8	34
160-180	6	40
180-200	10	50

Here, N = 50, \(\frac{N}{2}\) = 25
Cumulative frequency just greater than 25 is 26, which belongs to class 120 – 140.
Hence, the median class is 120-140
For this class,
l = 120, f = 14, cf = 12, h = 20
The Median = l + \(\frac{\frac{N}{2}-c f}{f}\) × h
= 120 + \(\frac{25-12}{14}\) × 20
= 120 + 18.57
= 138.57
Thus, median of the given data is 138.57.

Question 31.
A bag contains 15 white balls and some black balls. If the probability of drawing a black ball from the bag is thrice that of a white ball, find the number of black balls in the bag. [3]
Answer:
Let the bag contains “x ’ number of black balls. Then, the total number of balls in the bag is (x+15).
It is given that,
P(a black ball) = 3 × P(a white ball)
⇒ \(\frac{x}{x+15}\) = 3 × \(\frac{15}{x+15}\)
⇒ x = 45
Thus, there are 45 black balls in the bag.

SECTION – D (20 marks)
(Section – D consists of 4 questions of 5 marks each)

Question 32.
Prove the foLlowing identities, where the angles involved are acute angles for which the expressions are defined:
(A) \(\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}\) = 1 + sec θ cosec θ
(B) (sinA+ cosecA)² + (cosA + secA)² = 7 + tan²A + cot²A
OR
Two men on either side of a 75 m high building and in line with base of the building observe the angles of elevation of the top of the building as 30° and 60°. Find the distance between the two men (use √3 = 1.73).
Answer:

OR
Let AB be the building and, M₁ and M₂ be the positions of two men.

∴ Given, AB = 75 m,
∠AM₁B = 30°
and ∠AM₂B = 60°
Let AM₁ = x and AM₂ = y
From right triangle. M₁ AB.
\(\frac{A B}{M_1 A}\) = tan 30°
x = \(\frac{75}{x}=\frac{1}{\sqrt{3}}\)
x = 75√3

From right triangle. M₂AB.
\(\frac{A B}{M_2 A}\) = ton 60°
\(\frac{75}{y}\) = √3
y = \(\frac{75}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)
y = 25√3
Thus. M₁M₂ = x + y
= (1√3 + 2√3)
= 100√3
= 100 × 1.73
= 173 metres.

Question 33.
Solve for x and y graphically: x – y + 1 = 0; 3x + 2y – 12 = 0. [5]
OR
The shadow of a vertical tower on level ground increases by 16 m, when the altitude of the Sun changes from angles of elevation 60° to 45°. Find the height of the tower, correct to one pLace of decimal (Use √3 = 1.73).
Answer:
Table for the value of x – y + 1 = 0

AS₁, AS₂ be is shadows when sun’s attitude is 60° and 45° respectively.
∴ S₁S₂ = 16 m
From S₁AB
\(\frac{\mathrm{AB}}{\mathrm{S}_1 \mathrm{~A}}\) = tan 60°

Question 34.
In the given figure, PA, QB and RC are perpendicuLars to AC such that PA = RC = y, QB = z, AB = a and BC = b. Prove that \(\frac{1}{x}+\frac{1}{y}=\frac{1}{z}\). [5]

Answer:
Given PA, QB and RC are perpendiculars to AC
To Prove: \(\frac{1}{x}+\frac{1}{y}=\frac{1}{z}\)
Proof:
Since PA, QB and RC are perpendicuLars, drawn on same tine segment AC.
∴ PA ∥ QB ∥ RC
In ΔARC, QB ∥ RC.
By Basic proportional Theorem,

Question 35.
If a + b = c + d where a, b, c, d are rational numbers, them prove that either a=c and b = d or b and d are squares of rational numbers. [5]
Answer:
Let a = c, then
a + √b = c + √d
⇒ √b = √d
⇒ b = d

So let a ≠ c, then, there exist a postive rational number x, such that a = c + x.
Now, a + √b = c + √d
⇒ c + x + √b = c + √d [∵ a = c + x]
⇒ x + √b = √d ……(i)
⇒ (x + √b)² = (√d)²
⇒ x² + 2 √b x + b = d
⇒ √b = \(\frac{d-x^2-b}{2 x}\)
⇒ √b is rational [∵ d, x, b are rationals, \(\frac{d-x^2-b}{2 x}\) is rational
⇒ b is a square of a rational number
From (i), we have
⇒ √d is rational
⇒ d is the square of a rational number
Hence, either a = c and b = d or b and d are the squares of rationats number.

SECTION – E (12 marks)
(Case Study-Based Questions)
(Section E consists of 3 questions. All are compulsory.)

Question 36.
Mr. Naik is a paramilitary Intelligence Corps officer who is tasked with planning a coup on the enemy at a certain date. Currently he is inspecting the area standing on top of the cliff. Agent Vinod is on a hot air balloon in the sky. When Mr. Naik looks down below the cliff towards the sea, he has Ajay and Maran in boats positioned to get a good vantage point.
The main goal is to scope out the range and angles at which they should train their soldiers.

On the basis of the above information, answer the following questions:
(A) Write a pair of ‘angles of elevation’ and ‘angle of depression’. [1]
(B) If the vertical height of the balloon from the top of the cliff is 12 m and ∠b = 30°, then find the distance between the Naik and vinod. [1]
(C) Ajay’s boat is 25 m away from the base of the cliff. If ∠d = 30°. What is the height of the cliff? (use √3 = 1.73).
OR
If the height of the cliff is 30 m , ∠c = 45° and ∠d = 30°, then find the horizontal distance between the two boats (use √3 = 1.73). [2]
Answer:
(A) A pair of ‘angle of elevation’ is ∠b°, ∠e and one pair of angle of depression is ∠c° and ∠d°

(B) Then, sin 30°
= \(\frac{\text { Vertical height }}{\text { Distance between Naik and Vinod }}\)
⇒ \(\frac{1}{2}=\frac{12}{D_{\text {Nand } V}}\)
⇒ Distance = 24 m

(C) Here. ∠d° = ∠f° = 30°
Then. \(\frac{\text { Height of cliff. }}{\text { Distance of Ajay s boat from the base of cliff }}\) = tan 30°
⇒ \(\frac{1}{\sqrt{3}}=\frac{h}{25}\)
⇒ h = \(\frac{25}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{25}{3} \times \sqrt{3}\)
= 14.45 m
OR
Here, height of cliff = 30 m
Then, ∠c = ∠e = 45°
⇒ tan 45° = \(\frac{\text { height of cliff }}{\text { Distance of Maran’s boat }}\)
⇒ 1 = \(\frac{30}{\text { Distance of Maran’s boat }}\)
⇒ Distance of morones boat = 30 m
And ∠d =∠f = 30°
tan 30° = \(\frac{\text { height of cliff }}{\text { Distance of Ajay’s boat }}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{30}{D_A}\)
⇒ D_A = 30√3

Distance between boats
= 30√3 – 30
= 30(√3 – 1)
= 30(173.-1)
= 30 × 0.73
= 21.9 m

Question 37.
Nimmi, a 10th class student makes a project on corona virus in Science for an exhibition in her schooL For this project, she picks a sphere which has volume 38808 cu. cm and 11 cylindrical shapes rods each of volume 1540 cu. cm and length 10 cm.

Based on the above, answer the following questions :
(A) What is the diameter of the base of a cylindrical rod?
OR
What is the diameter of the sphere? [2]
(B) What is the curved surface area of a cylindrical rod? [1]
(C) How much curved surface area of the sphere is covered by 11 cylindrical rods ? [1]
Answer:
(A) Volume of cylinderical shape = 1540 cu. Cm
Height of cylinderical shape h = 10 cm
Let r be the radius of the cylinderical shape
∴ πr²h = V
⇒ \(\frac{22}{7}\) × r² × 10 = 1540
⇒ r² = \(\frac{1540 \times 7}{10 \times 22}\)
⇒ r = 7 cm
∴ Diameter = 2r = 14 cm
OR
Volume of sphere = 38808 cm³
Let R be the radius of the sphere
∴ \(\frac{4}{3}\)πR³ = 38808
⇒ R³ = \(\frac{38808 \times 3}{4 \pi}\)
= \(\frac{38808 \times 3}{4 \times 22}\) × 7
= 9261
⇒ R = 21cm
∴ Diameter = 2R = 42 cm

(B) For cylindericol rod,
CSA = 2πrh
= 72 × \(\frac{22}{7}\) × 7 × 7
= 440 sq cm

(C) Base Area of each cylindricaL rod = it,2
= \(\frac{22}{7}\) × 7 × 7
= 154 sq. cm
Total area covered = 11 × 154
= 1694 sq.cm

Question 38.
Prime Minister’s National Relief Fund (PMNRF) was established to help the families of earthquake affected village. The allotment officer is trying to come up with a method to calculate fair division of funds across various affected families so that the fund amount and amount received per family can be easily adjusted based on daily revised numbers.

The total fund allotted is formulated by the officer as:
x³ – 5x² – 2x – 6
The officer has also divided the fund equally among families of the village and each family receives an amount of x² + 2x + 1. After distribution, an amount of 11x + 1 should be left to have some buffer for future disbursements.
On the basis of the above information, answer the following questions:
(A) If an amount of ? 540 is left after distribution, what is value of x? [1]
(B) How much amount (In rupees) does each family receive? [1]
(C) What is the amount of fund (In rupees) allocated ?
OR
Find the value of k, if x – 1 is a factor of p(x) = 2x² + kx + √2. [2]
Answer:
(A) Amount lift = 11x + 1
∴ 11x + 1 = 540
x = \(\frac{539}{11}\) = 49

(B) Since, x = 49
∴ Amount received by each family is
x² + 2x + 1 = (49)² + 2(49) + 1
= 2401 + 98 + 1
= 2500

(C) Since, x = 49
∴ Fund alloted is
x³ – 5x² – 2x – 6
= (49)³ – 5(49)² – 2(49) – 6
= 117649 – 12005 – 98 – 6
= 1,05,540

OR

Let g(x) = x – 1
g(x) = 0 = x – 1 ⇒ x = 1
so, p(x) = 2x² + kx + √2
P( 1) = 2(1)² + k(1) + S.
= 2 + k + √2 ………. (i)
As (x – 1) is a factor of p(x)
∴ P(x) = 0
2 + k+ √2 = 0
k = -[2 + √2]

Students must start practicing the questions from CBSE Sample Papers for Class 10 Social Science with Solutions Set 6 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Social Science Set 6 with Solutions

Time : 3 Hours
Maximum Marks: 80

General Instructions:

1. Question paper comprises five Sections – A, B, C, D and E. There are 37 questions in the question paper. All questions are compulsory.
2. Section A – From question 1 to 20 are MCQs of 1 mark each.
3. Section B – Question no. 21 to 24 are Very Short Answer Type Questions, carrying 2 marks each. Answer to each question should not exceed 40 words.
4. Section C contains Q.25to Q.29 are Short Answer Type Questions, carrying 3 marks each. Answer to each question should not exceed 60 words.
5. Section D – Question no. 30 to 33 are long answer type questions, carrying 5 marks each. Answer to each question should not exceed 120 words.
6. Section-E – Questions no from 34 to 36 are case based questions with three sub questions and are of 4 marks each.
7. Section F – Question no. 37 is map based, carrying 5 marks with two parts, 37a from History (2 marks) and 37b from Geography (3 marks).
8. There is no overall choice in the question paper. However, an internal choice has been provided in few questions. Only one of the choices in such questions have to be attempted.
9. In addition to this, separate instructions are given with each section and question, wherever necessary.

Section – A
MCQs (1 x 20 = 20)

Question 1.
Which of the following changes was a part of the 1992 amendment to the Indian Constitution which was a major step towards decentralisation of power?
(a) Telangana was created.
(b) It became constitutionally mandatory to hold regular elections for the local government bodies.
(c) Central Election Commission was institutionalised.
(d) It was made mandatory for the leaders to declare their physical assets before their election. [1]
Answer:
(b) It became constitutionally mandatory to hold regular elections to the local government bodies.
Explanation: A major step towards decentralisation was taken in 1992. The indian Constitution was amended to make the third- tier of democracy more powerful and effective. In this pursuit, holding regular elections was made compulsory.

Related Theory:
An independent institution called the State Election Commission has been created in each state to conduct panchayat and municipal elections.

Question 2.
How do banks mainly use the deposits of their customers?
(a) to pay salaries of its employees
(b) to pay interests to its customers
(c) to return the money which the Bank has lent from entrepreneurs to balance its books
(d) to extend loans to people [1]
Answer:
(d) to extend loans to people Explanation: Banks make use of the deposits to meet the loan requirements of the people. Banks pays the salaries of its employees through the interest it earns by lending money to people.

Related Theory:
Banks mediate between those who have surplus funds (the depositors) and those who are in need of these funds (the borrowers).

Question 3.
Read the given graph and answer the following question:

Which is the most important sector in terms of its contribution to the GDP in 1973?
(a) Primary Sector
(b) Secondary Sector
(c) Tertiary Sector
(d) Both Primary and Tertiary Sector
Answer:
(a) Primary Sector
Explanation: The most important sector in terms of its contribution to the GDP in 1973 is the primary or the agricultural sector.

Question 4.
What is credit?
(I) Credit is the money lent to someone.
(II) Credit is the money borrowed from someone.
Codes:
(a) (I) is true.
(b) (II) is true.
(c) Both (I) and (II) are true.
(d) Both (I) and (II) are false. [1]
Answer:
(c) Both (I) and (II) are true.
Explanation: Credit is the money that a person borrows from another. The lent amount is known as credit. Credit (loan) refers to an agreement in which the Lender supplies the borrower with money, goods or services in return for the promise of future payment.

Question 5.
Identify the political party based on the given hints:
(I) Its symbol is an elephant.
(II) It was formed under the leadership of Kanshi Ram.
(III) Its main base is in Uttar Pradesh.
(a) NCP
(b) BSP
(c) BJP
(d) INC [1]
Answer:
(b) BSP
Explanation: BSP (Bahujan Samaj Party) was formed in 1984 under the leadership of Kanshi Ram. It seeks to represent and secure power for the bahujan samaj which includes the dalits, adivasis, OBCs and religious minorities.

Question 6.
How did the policy of the Indian government regarding Indian producers and companies change after 1991?
(a) The government decided that the Indian producers could now compete with foreign companies and hence opened Indian market to the world.
(b) The government decided that the Indian producers required more protection and increased tax barriers.
(c) The government asked the Indian producers to learn foreign methods of production.
(d) They decided that the foreign companies should not be allowed to sell their products in India because they are inferior. [1]
Answer:
(a) The government decided that the Indian producers could now compete with the foreign companies and hence opened the Indian market to the world.
Explanation: Starting around 1991, the government decided that the time had come for the Indian producers to compete with the producers around the globe. It felt that competition would improve the performance of producers within the country since they would have to improve their quality.

Related Theory:
Barriers on foreign trade and foreign investment were removed to a large extent as a result of this liberalisation.

Question 7.
Match the items in both the columns and choose the code which reflects the correctly matched pairs:

	Symbol		Political Parties
(A)	Hand	(I)	Indian National Lok Dal
(B)	Bulb	(II)	INC
(C)	Reading Glasses or Spectacles	(III)	CPI
(D)	Sickle and Wheat	(IV)	Mizoram People’s Conference

(a) (A)- (IV), B-(III), C-(I), D-(II)
(b) (A)- (I), B-(II), C-(IV), D-(III)
(c) (A)- (II), B-(III), C-(I), D-(IV)
(d) (A)- (II), B-(IV), C-(I), D-(III) [1]
Answer:
(d) A- (II), B- (IV), C-(I), D-(III)
Explanation: INC and CPI are both national parties while the other two, namely, Indian National Lok Dal and Mizoram People’s Conference are state parties. They have been recognised based on the criterion of the Election Commission.

Related Theory:
Every party in the country has to register with the Election Commission. While the Commission treats all parties equally, it offers some special facilities to large and established parties. These parties are given a unique symbol – only the official candidates of that party can use that election symbol.

Question 8.
The East India Company established political power, it could assert a monopoly on right to trade. Which of the following steps was not taken to assert a monopoly on the right to trade?
(a) Company tried to eliminate the existing traders.
(b) Company appointed Gomasthas to supervise weavers.
(c) Weavers were controlled by system of advances.
(d) Company tried to conquer new provinces. [1]
Answer:
(d) Company tried to conquer new provinces. Explanation: The EIC proceeded to develop a system of management and control that would eliminate competition, control costs, and ensure regular supplies of cotton and silk goods. Conquering new lands was to primarily establish political power and take over the administration of new lands.

Related Theory:
The company appointed a paid servant called the gomastha to supervise weavers, collect supplies, and examine the quality of cloth.

Question 9.
What is the other name given to the great Indian Bustard?
(a) Chinkara
(b) Hangul
(c) Gharial
(d) Godawan [1]
Answer:
(d) Godawan
Explanation: The Great Indian Bustard or Godawan is the state bird of Rajasthan and is considered India’s most critically endangered bird. It is considered the flagship grassland species, representing the health of the grassland ecology. Its population is confined mostly to Rajasthan and Gujarat.

Related Theory:
The bird is under constant threats due to collision/ electrocution with power transmission lines, hunting (still prevalent in Pakistan), habitat loss and alteration as a result of widespread agricultural expansion, etc.

Question 10.
In a democracy, the citizens must have freedom to choose among various contenders for power which takes the form of competition among different parties. How is this competition beneficial for the country?
(a) It introduces more candidates in a democracy.
(b) It ensures that power and responsibility is distributed among various parties and groups.
(c) It brings out new methods of administration.
(d) The competition leaves behind only the best administrators in power. [1]
Answer:
(b) It ensures that power and responsibility is distributed among various parties and groups.
Explanation: In contemporary democracies, competition among different parties ensures that power does not remain in one hand. In the long run, power is shared among different political parties that represent different ideologies and social groups. This makes the administration more representative.

Related Theory:
When two or more parties form an alliance to contest elections and if their alliance gets elected, they form a coalition government and thus share power.

Question 11.
Choose the correct matched pair.
(a) Farmer from Punjab- insurance for his car
(b) An aspirant of government job-facilities to visit religious places
(c) An employee in an IT firm-Peaceful work atmosphere
(d) A poor man in a water scarce region-ways to dig deeper wells [1]
Answer:
(c) An employee in an IT firm-Peaceful work atmosphere
Explanation: Ideal goals of development for the given people are Farmers from Punjab- insurance for his crops and equipment. An aspirant of government job- coaching facilities. A poor man in a water scarce region-free water supply through pipelines etc.

Related Theory:
Besides seeking more income, one way or the other, people also seek things like equal treatment, freedom, security, and respect of others. They resent discrimination. All these are important goals.

Question 12.
There are two statements marked as Assertion (A) and Reason (R). Mark your answer as per the codes provided below:
Assertion (A): Consequences of environ-mental degradation do not respect national or state boundaries.
Reason (R): Destruction due to exploitation of nature takes place everywhere.
(a) Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but the Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true. [1]
Answer:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Explanation: Destruction due to natural disasters doesn’t restrict itself to a specific nation or region. The actions of any country can affect the entire world at the same time.

Related Theory:
Sustainability of development is essential for the upcoming generations to enjoy the fruits of the development achieved by our generation.

Question 13.
How can we suffer from water scarcity if planet Earth is three-fourth covered by water?
(a) Most water is polluted.
(b) Most water is in the form of ice and vapour and cannot be utilised.
(c) Over-exploitation of fresh water resources causes water scarcity.
(d) Drinking water is not accessible to all the countries in the world. [1]
Answer:
(c) Over-exploitation of freshwater resources causes water scarcity.
Explanation: The availability of water resources varies over space and time, mainly due to the variations in seasonal and annual precipitation, but water scarcity is mostly caused due to over-exploitation and excessive usage of resources.

Related Theory: Freshwater can be obtained directly from precipitation, surface run-off and groundwater.

Question 14.
Which of the three states is the most developed based on all three indicators?

State	Infant Mortality Rate per 1,000 (2003)	Literacy Rate (%) (2001)	Net Attendance Ratio for Class l-V (1995-96)
Punjab	49	70	81
Kerala	11	91	91
Bihar	60	47	41

(a) Kerala
(b) Bihar
(c) Punjab
(d) All three of them are equally developed. [1]
Answer:
(a) Kerala
Explanation: All the indicators show that Kerala has better literacy rates and net attendance ratio which means Kerala provides better educational facilities. Infant mortality rates are low which show that the state has excellent medical and health care facilities.

Related Theory:
Bihar has high infant mortality and low literacy and net attendance ratio. This means it is not developed in comparison to other states.

Question 15.
Read the following data:

	% share of national Income
Name of the Countries	Top 20%	Bottom 20%
South Africa	64.8	2.9
Brazil	63.0	2.6
Russia	53.7	4.4
USA	50.0	4.0
United Kingdom	45.0	6.0
Denmark	34.5	9.6
Hungary	34.4	10.0

Which of the following nations has an equitable distribution of income?
(a) Brazil
(b) South Africa
(c) Denmark
(d) Hungary [1]
Answer:
(d) Hungary
Explanation: This is because maximum share of income in Hungary is not entrusted to the rich classes unlike in the other countries. In Hungary, the poor-most class shares 10% of the total wealth of the country. The richest class has 34.4 %. The difference between the two is not huge. Hence, it has an equitable distribution of economic resources and opportunities.

Question 16.
Find the odd one out from the following options:
(a) Solar energy, Hydel energy, Geothermal energy
(b) Wind energy, Tidal energy
(c) Nuclear energy, Natural gas, Petroleum and Coal
(d) Biomass, Atomic energy [1]
Answer:
(c) Nuclear energy, Natural gas, Petroleum and Coal
Explanation: All the other options include only clean form of energy. These forms of energy are also examples of renewable resources. They can be renewed and rejuvenated periodically. While the odd option out includes Petroleum and Coal which are not clean forms of energy. They pollute the environment and cause climate change by emitting carbon di oxide upon burning.

Question 17.
Fill in the blank by choosing the most appropriate option.
The Constitution declares India a of ……………. states.
(a) federation
(b) group
(c) confederation
(d) union [1]
Answer:
(d) union
Explanation: The Constitution declares India as a Union of States. It does not use the word federation even though the Indian union is based on the principles of federalism. It is an indestructible union because the states do not have the right to secede from this union.

Related Theory:
America, on the other hand, is a federation of states.

Question 18.
Which of the following workers can be classified as an employee of the primary sector?
(a) A Cricketer
(b) A Fisherman
(c) A industry worker
(d) A Salesman [1]
Answer:
(b) A Fisherman
Explanation: The primary sector includes activities like agriculture, dairy, fishing, forestry. This sector is also called agriculture and related sector. A fisherman can be thus understood as an employee of the primary sector.

Related Theory:
A salesman and a cricketer are both performing services for other consumers and hence belong to the tertiary sector.

Question 19.
Which of the following minerals is NOT found in abundance in Meghalaya?
(a) Coal
(b) Iron Ore
(c) Limestone
(d) Copper [1]
Answer:
(d) Copper
Explanation: In most of the tribal areas of the north-east India, minerals are owned by individuals or communities. In Meghalaya, there are large deposits of coal, iron ore, limestone and dolomite etc.

Related Theory:
Coal mining in Jowai and Cherapunjee is done by family members in the form of a long narrow tunnel, known as ‘Rat hole’ mining.

Caution:
Boxes given in the chapter also contain important information. Students must pay attention to them while reading the chapter.

Question 20.
Why is collateral necessary to borrow money?
(I) It is a physical asset.
(II) It is a guarantee against the borrowed money.
(III) It can be sold if the borrower defaults on the repayment even once.
(IV) lt makes borrowing quicker.
Codes:
(a) I and II (b) II only
(c) II and III (d) I, II, III and IV [1]
Answer:
(a) I and II
Explanation: Collateral is a physical asset that the borrower owns (such as land, building, vehicle, livestocks, deposits with banks) and uses this as a guarantee to a lender until the loan is repaid.

Related Theory:
If the borrower fails to repay the loan, the lender has the right to sell the asset or collateral to obtain payment.

Section – B
Very Short Answer Type Questions (2 x 4 = 8)

Question 21.
Besides the violence at Chauri Chaura, what were the other reasons why Gandhiji called off the Non-cooperation movement? [2]
Answer:
Other reasons are:
(1) Within the Congress, leaders wanted to participate in elections to the provincial councils that had been set up by the Government of India Act of 1919 to try and bring changes in the administration from the inside.

(2) People were getting tired of the mass struggles and protests everyday. There were not many alternate Indian institutions that they could be a part of after boycotting the british institutions.

Question 22.
Mention the major iron-ore belts found in India.
OR
Why is Aluminium an important metal? [2]
Answer:
Two major Iron-ore belts found in India are:

• Durg Bastar Chandarpur Belt in Chhattisgarh and Maharashtra – It contains super high grade hematite iron ore. It has the best physical properties needed for steel-making.
• Balia ri-Chitradurga-Chikkamagaluru- Tumakuru belt lies in Karnataka- Kudremukh has the largest iron-ore reserves in the world. The Kudremukh mines located in the Western Ghats of Karnataka are a 100 per cent export unit.

OR
Aluminium is an important metal because:

• It is very strong, resistant to corrosion and light. It is used to make equipment, airplanes and utensils.
• Aluminium shows good conductivity and great malleability. It is a great conductor of heat.

Question 23.
Has the tertiary sector grown equally in terms of employment generation as its contribution to the GDP? Explain in two points. [2]
Answer:
The Tertiary sector has not grown equally in terms of employment generation from 1973 to 2014. From 1973. to 2014, the primary sector continues to be the largest employer even now.

Caution:
Such questions not only require thorough understanding of the given graph but also the other graphs mentioned in the chapter. Students should focus on each one of them.

Question 24.
How does the Equal Remuneration Act of 1956 help in overcoming the discrimination faced by women at work places? [2]
Answer:
Women face immense discrimination at work place. They are not paid equally despite doing the same amount of work. The ERA of 1956 helped in the following ways:

• The ERA 1956 made it mandatory for offices to pay a man and woman equally if they are both employed at the same position and are performing equally well.
• This will reduce the wage gap and promote equality at work places. Women will feel appreciated and valued.

Section – C
Short Answer Type Questions (3 x 5 = 15)

Question 25.
How are state parties beneficial for Indian democracy? [3]
Answer:
State parties are beneficial for the Indian democracy because:

• India is a diverse country with numerous communities residing in different parts. State or regional parties help accomodate each of their aspirations.
• State parties strengthen the concept of federalism in the Indian democracy because often national parties have to form coalition governments with state parties to govern a state/the country.
• Common people and talented administrators can reach and participate in state parties easily. It is harder for them to approach national political parties.

Question 26.
Why is the road transport becoming more important for the development of a nation with time? [3]
Answer:
Road transport is becoming more important with time because:

• It is the most commonly used form of transportation for shorter distances. Goods are transported from factories to markets and vice versa using roads.
• The construction and maintenance costs of roads is less than that of any other form of transportation. They are quicker to mend. Hence, they are the cheapest mode of transportation too.
• They connect different modes of transportation to people’s residences and offices. For passangers, roadways is an intermediary form of transportation that connects their destination to their homes and one form of transportation to other.

Question 27.
Why did the developing nations organise themselves into G-77? Mention three points. [3]
Answer:
Most developing countries did not benefit from the fast growth and economic institutions which were made by the developed countries.

• The development that the Western economies experienced in the 1950s and 1960s was completely different from that experienced by developing and least developed countries.
• To aid their development, the developing countries organised themselves as a Group of 77 (or G-77) – to demand a new international economic order (NIEO).
• They wanted to gain real control over their natural resources, more development assistance, fairer prices for raw materials, and better access for their manufactured goods in developed countries’ markets which they weren’t able to get unless the financial institutions were neutrally administered.

Question 28.
Mention three reasons why Self Help Groups are necessary for people living in rural areas.
OR
Give three reasons why banks should increase their lending in rural areas. [3]
Answer:
Three reasons why Self Help Groups are necessary for the Indian economy are as follows:

• Self help group members residing in rural areas can take small loans from the group itself to meet their financial needs on low interests. This prevents them from borrowing credit from moneylenders who charge exorbitant interests.
• Banks are also willing to lend to the poor women when organised in SHGs, even though they have no collateral as such because SHGs bring accountability to them.
• SHGs also help its members to inculcate saving habits. Hence the SHGs are beneficial for rural people.

OR
Three reasons why banks should increase their lending in rural areas are:

• Absence of bank branches in rural areas push them towards lending from moneylenders who charge exorbitant rates of interest.
• Most people are unable to get loans from these lenders too because moneylenders are limited in number. Farmers require credit regularly but are unable to get the same because of this dearth.
• Formal credit needs to be distributed more equally among lenders from all classes and communities. Opening branches in rural areas will help in that.

Question 29.
Printing created an appetite for new kinds of writing in the nineteenth century. Lives and feelings of women began to be written in particularly vivid and intense ways. How did this change things for the middle class women of the time? [3]
Answer:
Printing created an appetite for new kinds of writing in the nineteenth century. This created favourable conditions for middle class women.

• Liberal husbands and fathers started sending their women to women’s schools to read and learn about the world.
• Many journals began carrying writings by women. They explained why women should be educated. They also carried a syllabus and attached suitable reading matter to be used for home-based schooling.
• Few Bengali women like Kailashbashini Debi wrote books highlighting the experiences of women. Conservative Hindus and Muslim families were still opposed the education of their women.

Section – D
Long Answer Type Questions (5 x 4 = 20)

Question 30.
Differentiate between Food and Non Food crops by giving examples. Which of them is more important for the development of a nation?
OR
What is resource planning? Why is resource planning essential? Explain. [5]
Answer:
The difference can be listed on the basis:

Basis of Distinction	Food Crops	Non-Food Crops/Cash Crops
Definition	Crops which are grown to be directly consumed are food crops.	Crops grown to produce goods
Sowing purpose	They are grown for sustenance.	They are industrial crops-grown for sales.
Constituents	They supply nutrition-vitamins.	They do not necessarily supply nutrition.
Examples	Paddy, Wheat Rice.	Tea, Coffee, Jute, Cotton.

For a nation, both of these types are equally important because one of them is used for sustenance of the population while the other helps them earn their livelihood. Either one of them cannot be completely depended upon for the development of a nation.
OR
Resource planning is a technique of proper utilisation of resources which aimed at sustainable development. Resource planning is essential because of the following reasons:

• Most of resources available on earth are limited in supply.
• The resources available to us are distributed unevenly all over the country.
• Overutilization of the resources may lead to environmental pollution and depletion of resources as well. Therefore, planning of resources can reduce pollution and overutilization of resources as well.
• Planning of resources can lead to having a balanced development at national, state, regional and local levels.

Question 31.
Highlight the various functions of political parties in India.
OR
How can the relationship between politics and religion be beneficial and problematic at the same time? Explain. [5]
Answer:
The functions are:

• A political party contests elections through candidates.
• In a democratic country, parties form a link between the government and the people. They convey their aspirations to the government at the centre.
• Those parties which lose elections form the opposition. They voice different view points than the government and also mobilize opposition against them.
• Political parties shape public opinion.
• Political parties represent the views of various communities and mould it to achieve satisfaction for the majority.

OR
The relationship between politics and religion can be beneficial:

• Influence of religion can make politics ethical.
• Religious communities can politically express their needs and interests. Politics can help ensure their freedom and equality.
• Political authorities can monitor and control religious oppression.

The relationship can be problematic:

• Religion can lead to development of nationalist sentiments leading to birth of conflicts.
• Political parties use religion to make political gains by appeasing one and ignoring others.
• State power may be used to establish the domination of one religious group over another.

Question 32.
Following the defeat of Napoleon in 1815, European governments were driven by a spirit of conservatism. Do you agree? Elaborate upon your stand.
OR
Explain how print culture developed in India. [5]
Answer:
Following the defeat of Napoleon in 1815, European governments were driven by a spirit of conservatism. However this conservatism was coloured
in a new hue.

• Conservatives believed that established, traditional institutions of state and society-like the monarchy, the Church, social hierarchies etc., should be preserved.
• They did not however propose a return to the society of pre-revolutionary days.
• They agreed that the modernisation introduced by Napoleon could in fact strengthen traditional institutions like the monarchy.
• Conservatives believed that modern army, an efficient bureaucracy, a dynamic economy, the abolition of feudalism and serfdom could strengthen the autocratic monarchies of Europe.
• They still believed in the orthodox institutions but they were prepared to colour them in the hues of modernism.

OR
The print culture developed in India by the following means:

• India had a very rich and old tradition of handwritten manuscripts in Sanskrit, Arabic, Persian, as well as in various vernacular languages.
• The printing press first came to Goa with Portuguese missionaries in the mid 16th century.
• By the 1870s, besides Urdu, Marathi, Bengali, Hindi printing also began giving voice to local opinions.
• Print in the 18th century : By the close of the 18th century, a number of newspapers and journals appeared in print. The first Indian weekly i.e. Bengal Gazette also came into picture in the late 18th century.
• Print in the 19th century : By the end of the 19th century, a visual culture started taking place. By 1870s, caricatures and cartoons were being published in journals and newspapers. (Mention any 3 points)

Question 33.
Mention five factors which have made the phenomenon of Globalisation possible.
OR
Money in your pocket cannot buy all the goods and services that you may need to live well Do you agree? Elaborate. [5]
Answer:
Five such factors are:

• Rapid improvement in transportation has helped in facilitating Globalisation. Improved transportation has made transfer of goods and services simpler and quicker.
• Liberalisation of trade restrictions and foreign investment have enabled the globalisation process.
• The drastic improvement in communication and Information technology has played a vital role in faster delivery of ideas and information across continents at very low costs. This has helped in intermingling of cultures.
• Internet has facilitated strong contact between people from different nations. At one click, people can now virtually visit other nations. They can talk to people across the world through video calls, phone calls, emails and by using social media.
• Barriers on foreign investment were also removed to a large extent enabling many MNCs to set up their units in the country.

OR

• Money or Income by itself is not a completely adequate indicator of development, goods and services that citizens are able to use.
• Money cannot buy a pollution-free environment or clean drinking water.
• It cannot buy citizens rights and freedom to express their opinions.
• Money cannot also ensure holistic national development. It might ensure economic growth but not development.
• Money cannot ensure proper education. Money can afford to pay school fees but it cannot afford values and literacy which is more than just learning addition, subtraction etc.

Section – E
Case Based Questions (4 x 3 = 12)

Question 34.
Read the given source and answer the following questions:
The IMF and the World Bank were designed to meet the financial needs of the industrial countries. They were not equipped to cope with the challenge of poverty and lack of development in the former colonies. But as Europe and Japan rapidly rebuilt their economies, they grew less dependent on the IMF and the World Bant Thus from the late 1950s the institutions began to shift their attention more towards developing countries.

As colonies, many of the less developed regions of the world had been part of Western empires. Now, ironically, as newly independent countries facing urgent pressures to lift their populations out of poverty, they came under the guidance of international agencies dominated by the former colonial powers.
(A) Why was the World Bank established? [1]
(B) How did the developing countries come under the guidance of international agencies? [1]
(C) Mention two characteristics of the post¬war international economic system. [2]
Answer:
(A) The International Bank for Reconstruction and Development or the World Bank was set up to finance postwar reconstruction. This institution gives loan to least developed countries at low rates.
Related Theory:
The framework was agreed upon at the United Nations Monetary and Financial Conference held in July 1944 at Bretton Woods in New Hampshire, USA.

(B) The institutions were build to preserve the economic stability of the new world and to provide more employment. The developed countries outgrew the institution and did not need any more help.

This caused them to focus on the developing and least developed countries. They began solving their economic issues by granting them cheap loans and grants.

(C) Two characteristics of post-war international economic system or the Bretton Woods system are:

• It was based on fixed exchange rates.
• In this system, national currencies were fixed against the price of the dollar and the price of the dollar was fixed against gold.

Question 35.
Read the given source and answer the following questions:
Air pollution is caused by the presence of high proportion of undesirable gases, such as Sulphur dioxide and Carbon monoxide. Airborne particulate materials contain both solid and liquid particles like dust, sprays mist and smoke.

Smoke is emitted by chemical and paper factories, brick kilns, refineries and smelting plants, and burning of fossil fuels in big and small factories that ignore pollution norms. Toxic gas leaks can be very hazardous with long-term effects. Are you aware of the Bhopal Gas tragedy that occurred? Air pollution adversely affects human health, animals, plants, buildings and the atmosphere as a whole.
(A) What is meant by the term “Airborne”? [1]
(B) How was the Bhopal Gas Tragedy caused? [1]
(C) Mention two ways to reduce Air pollution. [2]
Answer:
(A) The term “Airborne” means transported by the air or carried by the air from one place to another.

(B) Bhopal Gas Tragedy was a major Industrial accident which occured in 1984. A dangerous gas, methyl isocyanate, escaped from an insecticide plant that was owned by the Indian subsidiary of the American firm Union Carbide Corporation. The gas killed thousands of people. Survivors continue to suffer from various diseases.

Caution:
Such answers are based on real life incidents which have been mentioned in the book. It is always wise to make a detailed list of such incidents mentioned in a chapter and read about them separately.

(C) Two ways in which Air pollution can be reduced are:

• Reducing the dependence upon fossil fuels for energy.
• Fixing smoke filters, electric scrubbers and chimneys in industries to reduce and treat the disposed gas before releasing it in the atmosphere.

Related Theory:
Afforestation can also help in improving the air quality of a region. Trees help clear the quality by exhaling oxygen and inhaling carbon dioxide.

Question 36.
Read the given source and answer the following questions:
Do we prefer democracy only for moral reasons? Or are there some prudential reasons to support democracy too? Over a hundred countries of the world today claim and practice some kind of democratic politics: they have formal constitutions, they hold elections, they have parties and they guarantee rights of citizens.

While these features are common to most of them, these democracies are very much different from each other in terms of their social situations, their economic achievements and their cultures. Clearly, what may be achieved or not, achieved under each of these democracies will be very different. But is there something that we can expect from every democracy, just because it is democracy?
(A) What is a prudential reason for supporting democracy? [1]
(B) Mention two features of a democratic nation. [1]
(C) Mention two legitimate expectations from a democratic government. [2]
Answer:
(A) A prudential reason for supporting democracy is that it helps to avoid social and political conflicts between communities with different interests and practises.

(B) A democratic nation has following features:

• It ensures free and fair elections.
• It has formally adopted constitutions.
• It has a universal adult franchise and right to contest elections.
• It has political parties who compete for power. (Mention any two points)

(C) Two such legitimate expectations are:

• One can expect an accountable and responsible government from a democratic government.
• A democracy should also facilitate transparent decision-making processes and results.

Section – F
Map Based Questions (2 + 3 = 5)

Question 37.
(a) On the given outline Political Map of India, identify the places marked as A and B with the help of following information and write their correct name on che lines marked near it.
(A) The place where Indian National Congress session was held in 1927.
(B) Mahatma Gandhi organized a Satyagraha Movement at this place for indigo planters. [2]

(b) On the same given map of India, locate and label any three of the following with suitable symbols:
(a) Ramagudam thermal power plant
(b) Thiruvananthapuram Software Technology Park
(c) Chhatrapati Shivaji Airport
(d) Haldia port [3]

Answer:
(a) (A) Madras (B) Champaran

Students must start practicing the questions from CBSE Sample Papers for Class 10 Social Science with Solutions Set 5 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Social Science Set 5 with Solutions

Time : 3 Hours
Maximum Marks: 80

General Instructions:

1. Question paper comprises five Sections – A, B, C, D and E. There are 37 questions in the question paper. All questions are compulsory.
2. Section A – From question 1 to 20 are MCQs of 1 mark each.
3. Section B – Question no. 21 to 24 are Very Short Answer Type Questions, carrying 2 marks each. Answer to each question should not exceed 40 words.
4. Section C contains Q.25to Q.29 are Short Answer Type Questions, carrying 3 marks each. Answer to each question should not exceed 60 words.
5. Section D – Question no. 30 to 33 are long answer type questions, carrying 5 marks each. Answer to each question should not exceed 120 words.
6. Section-E – Questions no from 34 to 36 are case based questions with three sub questions and are of 4 marks each.
7. Section F – Question no. 37 is map based, carrying 5 marks with two parts, 37a from History (2 marks) and 37b from Geography (3 marks).
8. There is no overall choice in the question paper. However, an internal choice has been provided in few questions. Only one of the choices in such questions have to be attempted.
9. In addition to this, separate instructions are given with each section and question, wherever necessary.

Section – A
MCQs (1 x 20 = 20)

Question 1.
Which of the following describes the issues faced by local governments in India today?
(a) Elections are not held regularly.
(b) Candidates do not want to contest elections.
(c) Gram Sabhas are held regularly but they do not reach at any joint decision.
(d) Most state governments have not transferred significant powers to the local governments. [1]
Answer:
(d) Most state governments have not transferred significant powers to the local governments.
Explanation: India has introduced the concept of local self government but still struggles with achieving it. Elections are held regularly and candidates want to contest elections because of the power they are promised but state governments have not transferred enough powers to these governments. They are still controlled financially by state governments.

Related Theory:
Constitutional status of local state governments have helped to deepen Democracy in India.

Question 2.
Dictatorships are based on political and social equality. Choose the incorrect part of the given sentence.
(a) Dictatorships
(b) Political
(c) Social
(d) Equality [1]
Answer:
(a) Dictatorships
Explanation: Democracy are based on political and social equality. Democracy considers the will of the people above everything. All individuals have equal weight in electing representatives. They have the right to vote or contest elections directly or indirectly through their representatives. They can participate in the decision-making process and can change the government in the next election if the government is not taking care of them.

Related Theory:
Non-democratic governments such as monarchy, military rule, religious leaders or dictatorship do not consider the will of its people. People do not have any right to choose their representatives or change them, nor are these governments accountable to its people.

Question 3.
Identify which of the following statements can appropriately describe the picture.

(a) The Election Commission has passed several orders to reform political parties but parties are reluctant to follow these regulations.
(b) The political parties are preparing to protest against the dictatorial behaviour of the Election Commission.
(c) Candidates of the political parties go to the Election Commission for their training and allotment of duties.
(d) Political Parties have to request Election Commission for funds to fulfill their promises to the public. [1]
Answer:
(a) The Election Commission has passed several orders to reform political parties but parties are reluctant to follow these regulations.
Explanation: Election Commission was formed to monitor and hold free and fair elections with regularity all across the country. They also monitor the working of Political Parties. Political Parties often indulge in corruption and indiscipline. The Election Committee takes care of these by passing rules and regulations (code of conduct), however political parties are reluctant to fulfill them.

Caution:
For picture questions, it is important to read the captions or dialogues if any. Doing that helps to answer the questions in a better way.

Question 4.
Choose the correctly matched pair.
(a) Panchayat Samiti- Group of Gram Panchayats
(b) Gram Panchayat – Supervises Gram Sabha
(c) Sarpanch- Head of Zilla Parishad
(d) Zilla Parishad- Group of Gram Sabhas [1]
Answer:
(a) Panchayat Samiti- Group of Gram Panchayats
Explanation: Sarpanch is the president or head of the Gram Panchayat. All the panchayat samitis or mandals in a district together constitute the zilla (district) parishad.
Gram Sabha supervises the Gram Panchayats. It is the decision-making body for the entire village.

Related Theory:
The local government structure goes right up to the district level. A few gram panchayats are grouped together to form what is usually called a panchayat samiti or block or mandal.

Question 5.
Identify the industry with the help of given features.
(I) This industry requires chemicals as raw materials.
(II) After green revolution, the industry expanded exponentially.
(III) Industry has a huge market in Gujarat.
(a) Automobile industry
(b) Aluminium Smelting industry
(c) Textile industry
(d) Fertiliser industry [1]
Answer:
(d) Fertiliser Industry Explanation: The Fertilisers industry requires chemicals like potash, urea, ammonium phosphate etc., as raw materials.
Related Theory:
After the Green Revolution, the industry expanded to several other parts of the country. Gujarat, Tamil Nadu, Uttar Pradesh, Punjab and Kerala contribute towards half of the fertilizer production.

Question 6.
Identify which of the following statements is false about money borrowed from Banks and Cooperative societies.
(a) The money borrowed can be returned in the form of regular monthly instalments, the amount of which is pre-decided.
(b) This type of loan helps to increase the income of the borrowers.
(c) This type of loan can be borrowed without any legal documents.
(d) The Reserve Bank of India supervises and monitors this type of loan. [1]
Answer:
(c) This type of loan can be borrowed without any legal documents.
Explanation: Loan borrowed from Banks and Cooperative sectors is considered as the credit borrowed from Formal Sector and can only be borrowed by fulfilling terms of credit- collateral, legal documents, rate of interest and time period of borrowing.

Question 7.
Match the following items given in Column A with those in Column B and choose the most appropriate code which reflects the correctly matched pairs.

	Column A		Column B
(A)	Dipa	(I)	Himalayan Belt
(B)	Milpa	(II)	Chhattisgarh
(C)	Valre	(III)	Mexico
(D)	Khil	(IV)	Rajasthan

Codes:
(a) A-(I), B-(III), C-(IV), D-(II)
(b) A-(III), B-(IV), C-(II), D-(I)
(c) A-(II), B-(III), C-(IV), D-(I)
(d) A-(IV), B-(III), C-(II), D-(I) [1]
Answer:
(c) A-(II), B-(III), C-(IV), D-(I)
Explanation: Jhumming is a kind of Slash and Burn agriculture. In this practice, farmers clear a patch of land and produce cereals and other food crops to sustain their family. When the soil fertility decreases, the farmers shift and clear a fresh patch of land for cultivation. The shifting allows regeneration of fertility in the land.

Related Theory:
The ‘slash and burn’ agriculture is known as ‘Conuco’ in Venezuela, ‘Roca’ in Brazil, ‘Masole’ in Central Africa, tadang’ in Indonesia, ‘Ray’ in Vietnam.
In India, this primitive form of cultivation is called ‘Bewar’ or ‘Dahiya’ in Madhya Pradesh, ‘Podu’ or ‘Penda’ in Andhra Pradesh, ‘Pama Dabi’ or ‘Koman’ or Bringa’ in Odisha, Xumari’ in Western Ghats.

Question 8.
Why is the tertiary sector becoming so important in India? Which of the following statements cannot answer the given statement?
(a) Service sector is responsible for providing basic services like educational, medical, banking and other market related facilities to people.
(b) The development of agriculture and industry leads to the development of services.
(c) The Service sector supplies raw materials to other sectors.
(d) As income levels rise, certain sections of people start demanding better and more services. [1]
Answer:
(c) The Service sector supplies raw materials to other sectors.
Explanation: The Service sector doesn’t produce any goods at all for them to act like raw materials for other sectors. Thus, this sentence cannot be used as an answer to the question mentioned above. Services develop because other sectors are growing. This sector also grows because of greater demand. Demand is generated when people change lifestyles and want greater comfort and luxury.

Related Theory:
Over the past decade, certain new services such as those based on information and communication technology have become important and essential. The production of these services has been rising rapidly

Question 9.
Which of the following statements about Middle classes in Europe is True?
(a) They owned estates in the countryside and also town-houses.
(b) They were a small group, mostly including peasantry.
(c) The ideas of national unity following the abolition of aristocratic privileges gained popularity in this class first.
(d) They spoke French for purposes of diplomacy and in high society. [1]
Answer:
(c) The ideas of national unity following the abolition of aristocratic privileges gained popularity in this class first.
Explanation: In the wake of Industrial Revolution in nineteenth century, a working- class population and middle classes made up of industrialists, businessmen, professionals rose.

It was among the educated, liberal middle classes that the ideas of national unity rose.

Question 10.
Which of the following institutions has been entrusted with the responsibility to conduct Panchayat and Municipal elections?
(a) Zilla Parishad
(b) District Courts
(c) State Election Commission
(d) State Municipal Commission [1]
Answer:
(c) State Election Commission
Explanation: State Election Commission was constituted in 1992 to monitor and hold Panchayat and Municipal Elections.

Question 11.
Which of the following indicators is used by the UNDP to compare development in different countries across the world through its Human Development Report?
(a) Sanitation facilities
(b) Life expectancy at birth
(c) Age at which the child began schooling
(d) Availability of nutritional food [1]
Answer:
(b) Life expectancy at Birth
Explanation: Human Development Report published by UNDP compares countries based on the educational levels of the people, their health status and per capita income. This includes GNI, Life expectancy at birth and mean years of schooling.

Related Theory:
According to the Human Development Report, 2020, United Nations Development Programme, New York, India stands at rank 130.

Question 12.
There are two statements marked as Assertion (A) and Reaspm (R). Mark your answer as per the codes provided below:
Assertion (A): Caste system was based on exclusion of and discrimination against the ‘outcaste’ groups.
Reason (R): Lower or Downtrodden communities were prevented from fraternising, meeting and even walking with Upper caste individuals.
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true. [1]
Answer:
(a) Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of Assertion (A).

Explanation: Members of the same caste group were supposed to form a social community that practiced the same or similar occupation, married within the caste group and did not eat with members from other caste groups. This caused exclusion.

Question 13.
What is meant by ‘ethnic’?
(a) A social division based on shared culture
(b) A division of society based on language spoken by people
(c) A cultural division based on social conditions
(d) A socio-cultural division based on the linguistic lines [1]
Answer:
(a) A social division based on shared culture
Explanation: People belonging to the same ethnic group believe in their common descent because of similarities of physical type or of culture or both. They need not always have the same religion or nationality.

Question 14.
Fill in the blank marked A in the given table. [1]

Name of the Book	Subject/Genre
1. Jam-i-Jahan Nama	Newspaper carrying Nationalist information
2. Alamnacs	Calenders
3. Biliotheque Bleue	A
4. Shilling series	Cheap versions of popular works

(a) Psychological thrillers of the Victorian English society
(b) Romances and Historical narratives
(c) Religious books
(d) Republished folk tales
Answer:
(b) Romances and Historical narratives
Explanation: In France, Biliotheque Bleue were low-priced small books printed on poor quality paper, and bound in cheap blue covers. Then there were the romances, printed on four to six pages, and the more substantial ‘histories’ which were stories about the past.

Related Theory:
In England, penny chapbooks were carried by petty pedlars known as chapmen, and sold for a penny, so that even the poor could buy them.

Question 15.
Which of the following connects countries across the world with each other? [1]
(I) Technology
(II) Inter-personal interactions
(III) Movement of goods and services
(IV) Geography
(V) Investment Codes:
(a) I and IV
(b) I, II, III and IV
(c) I, II, III and V
(d) I, II and III 1
Answer:
(c) I, II, III and I
Explanation: Geographical boundaries divide countries and regions. Apart from that, other four options connect one country to another. People usually move from one country to another in search of better income, better jobs or better education. Technology and investments also move from nation to nation.

Related Theory:
Globalisation is this process of rapid integration or interconnection between countries. MNCs are playing a major role in the globalisation process.

Question 16.
Lending libraries had been in existence from the seventeenth century onwards. In the nineteenth century, lending libraries in England became instrument for educating …………….
Fill in the given blank by choosing the most appropriate option.
(a) Women from rich households
(b) Artisans and lower middle-class people
(c) Political leaders and revolutionaries
(d) The Upper middle class people [1]
Answer:
(b) Artisans and lower middle-class people
Explanation: In the nineteenth century, lending libraries in England became instrument for educating white-collar workers, artisans and lower-middle-class people. Self- educated working class people also wrote for themselves.

Workers also wrote political tracts and autobiographies in large numbers.

Related Theory:
Earlier in the nineteenth century, women became important as readers as well as writers. Penny magazines were especially meant for women. There were manuals, teaching proper behaviour and housekeeping, published especially for them.

Question 17.
Choose the odd one out –
(a) House, Paintings, Vehicles
(b) Buildings, Gold, Weather report for the duration of loan
(c) Interest rates, Ornaments, Land documents
(d) Duration of loan, Interest rate, Loan agreement [1]
Answer:
(b) Buildings, Gold, Weather report for the duration of loan
Explanation: Every loan agreement specifies an interest rate which the borrower must pay to the lender along with the repayment of the principal. In addition, lenders may demand collateral (security) against loans. The rest of the options have different terms of credit. Weather report is not a term of credit.

Related Theory:
Interest rate, collateral and.documentation requirement, and the mode of repayment together comprise what is called the terms of credit. The terms of credit vary substantially from one credit arrangement to another.

Question 18.
” ……………. It is said of “passive resistance” that it is the weapon of the weak, but the power which is the subject of this article can be used only by the strong. This power is not passive resistance; indeed it calls for intense activity. The movement in South Africa was not passive but active ……………..
Which of the following is the given source related to?
(a) Violence in Protest and Clashes
(b) Satyagraha Mass Agitation
(c) Guerrilla Warfare techniques
(d) Hunger Strikes [1]
Answer:
(b) Satyagraha Mass Agitation
Explanation: Satyagraha was a new method which Gandhi ji used to fight against the racist regime in South Africa and later in India.
Gandhi ji’s idea of Satyagraha was based on two principles-truth and non-violence.

Related Theory:
Gandhi ji launched Satyagraha successfully in Champaran in 1916 for indigo planters, in 1917 in Kheda district of Gujarat to support the peasants, and in 1918 in Ahmedabad for cotton milt workers.

Question 19.
The party that loses the election plays the role of the opposition in the Parliament or Legislative Assembly. Find the incorrect statement about the opposition party from those given below:
(a) They act as a check on the actions of the ruling party.
(b) People who have lost the election form a new party with winning MLAs or MPs to remain in power.
(c) The opposition is smaller in number than the ruling party.
(d) The members of the opposition do not necessarily oppose the ruling party on aU matters. [1]
Answer:
(b) People who have lost the election form a new party with winning MLAs or MPs to remain in power.

Question 20.
Which of the following statements determine(s) that the ethnic composition of Belgium is very complex?
(I) 59 per cent lives fn the Flemish region and speaks Dutch language
(II) 40 per cent lives in the Wallonia region and speaks French
(III) 1 per cent of the Belgians speak German.
(IV) In Brussels, 80 percent people speak French while 20 percent are Dutch¬speaking
Codes:
(a) (I), 01) and (IV)
(b) (I), (II) and (III)
(c) Both (I) and (II)
(d) (I), (II), (III) and (IV) [1]
Answer:
(d) (l), (II), (III) and (IV)
Explanation: Ethnic Composition of Belgium is complex because there are multiple linguistic communities living in the country. They are scattered throughout the country and form majorities in some regions and minorities in others. Hence, no blanket welfare scheme has proven to be beneficial for them.

Section – B
Very Short Answer Type Questions (2 x 4 = 8)

Question 21.
Analys two reasons why people dislike political parties. [2]
Answer:
Political parties are unpopular among people because:

• People tend to blame parties for associated issues with democracy and political life.
• Parties are corrupt and indisciplined. They do not practise democratic internal working.

Question 22.
Within the wide swathe of territory that came under his control, Napoleon set about introducing many of the reforms that he had already introduced in France. Highlight two such reforms introduced by him. [2]
Answer:
Two such reforms are:

• He did away with all the privileges based on birth.
• He secured the right to property.

Question 23.
Demand deposits share the essential features of money. Discuss two situations where this feature can be used for one’s benefit.
OR
Mention two reasons which prevent poor people from borrowing credit from banks or other registered institutions? [2]
Answer:
Demand deposits share the essential features of money. This can be used in the following ways:

• Demand deposits can be used to settle payments without the use of cash.
• Demand deposits can be used to withdraw money in times of need. One can withdraw money from the bank using demand deposits at any time.

OR
Two reasons which present poor people from borrowing credit from the bank are-

• Absence of collateral to pledge to borrow money
• Illiteracy and lack of physical bank branches in the villages

Question 24.
Mention two ways of rainwater harvesting practised in ancient India.
Answer:
Two ways of Rainwater Harvesting are:

• Kuls or Guls: They were traditionally built as diversion channels to channelise the flow of surface run-off into water bodies or reservoir tanks in Himalayas. This was also used for irrigation.
• Tankas: In the semi-arid and arid regions of Rajasthan, particularly in Bikaner, Phalodi and Barmer, houses traditionally had underground tanks or tankas almost of the size of a room, for storing drinking water.

Section – C
Short Answer Type Questions (3 x 5 = 15)

Question 25.
State any three characteristics of commercial farming in India.
OR
Mention three forms in which minerals occur in India. [3]
Answer:
The main characteristics of commercial farming in India are as follows:

• Crops are grown on a large scale in large farms or plantations for commercial purposes and for export to other countries.
• It is a capital intensive farming and requires high application of modern inputs.
• Modern inputs like high yielding variety (HYVjseeds, chemical fertilisers, insecticides and pesticides are used to obtain higher productivity. Irrigation is also used to meet water requirements.

OR

• They were sent to foreign lands under false promises of sending back to their native lands after they had worked five years on their employer’s plantation.
• On arrival at plantations, labourers found different conditions from what they had imagined or were informed about.
• They had absolutely no legal rights and were underpaid. The living and working conditions were tough and they were oppressed. They were their master’s underpaid slaves who he could exploit in any way.

Question 26.
The Supreme Court’s order to bring down the influence of money and criminals in politics is still fraught with defects. Examine the statement. [3]
Answer:
(1) In order to bring down the influence of money and criminals on politics, the Supreme Court in India has made it mandatory for every election contesting candidate to file an affidavit where he/she shall declare details of his property and criminal cases pending against him/her.

(2) This system of filing an affidavit has made a lot of information available to the public. However, there is no system of checking if the information submitted by candidates is true and correct.

(3) This order of the Supreme Court does not mention the punishment to be given to the leaders /candidates in case of non-compliance. There is no mechanism to check if a candidate has declared all his assets legally and honestly.

Question 27.
19th century Indentured labour system has been described as a ‘New system of slavery’. Explain. [3]
Answer:
In the 19th century, many Indians and Chinese labourers went to work on plantations in mines and construction projects around the world to sustain in times of economic depression. This system of indenture has been described as a ‘new system of slavery’ because of the following reasons:

• They were sent to foreign lans under false promises of sending back to their native lands after they had worked five years on their employer’s plantation.
• On arrival at plantations, labourers found different conditions from what they had imagined or were informed about.
• They had absolutely no legal rights and were underpaid. The living and working conditions were tough and they were oppressed. They were their master’s underpaid slaves who he could exploit in any way.

Question 28.
Define National Development. How can a country achieve it? [3]
Answer:

• National Development entails equitable economic, social and political development of all the communities residing in the nation regardless of their gender, caste, class, creed or domicile.
• It means a holistic growth of the nation in terms of their intellectual, emotional, physical and any other capabilities.
• National development can be achieved if the government ensures the provision of proper housing, sanitation, educational and health care facilities to all the citizens of the country equally.

Question 29.
Conservation in the background of rapid decline in wildlife population and forestry has become essential. Do you agree? Support your stand using three points. [3]
Answer:
Conservation in the background of rapid decline in wildlife population and forestry has become extremely essential. This is because:

• Due to rapid deforestation and decline in wildlife population, the ecological balance of the planet gets disturbed causing destruction of certain species and imbalance in natural food chains.
• Deforestation has also added to the impact of Global warming and Climate change. It has increased the incidence of diseases across all the countries.
• Destruction of forest resources has led to a steep fall in the revenue generated by it. Tribals who are directly dependent on such resources have been severely affected.

Section – D
Long Answer Type Questions (5 x 4 = 20)

Question 30.
Can Tourism be considered as a trade? Elaborate your stand using five points.
OR
Explain the pro-active approach adopted by the National Thermal Power Corporation (NTPC) for preserving the natural environment and resources. (5)
Answer:
Yes, Tourism can and should be considered as a trade because of its growing scope and its contribution to the country’s economy.

• Besides boosting the economy, tourism promotes national integration and provides support to various cultural pursuits.
• Tourist arrival and expenditure in the country witnessed an increase of 11.8 percent during the year 2010 as against the year 2009, contributing ? 64,889 crore of foreiqn exchange to the Indian economy in 2010.
• Tourism also promotes local handicrafts of various remote parts of the country and provides the artisans employment.
• More than 15 million people are engaged in this industry directly.
• This proves that tourism has a potential of developing into a very prosperous industry if proper attention was paid to it.
• More attention needs to be paid to diversify and deepen the roots of this industry to branches Like heritage tourism, eco tourism, adventure tourism, cultural tourism, medical tourism and business tourism industry. (Mention any five points)

OR
The proactive approach adopted by the National Thermal Power Corporation (NTPC) for preserving the natural environment and resources like water, oil and gas and fuels in places where it is settling up its power plants entails the following steps:

• Equipment: Optimum utilisation of equipment and adoption of latest techniques and upgrading existing equipment to avoid energy loss or inferior quality production.
• Waste Regeneration: Minimising waste generation by maximising ash utilisation.
• Maintain Ecological Imbalance: Creating green belts for nurturing ecological balance and addressing the question of special purpose vehicles for afforestation.
• Wags to reduce environmental pollution: Reducing environmental pollution through activities like ash pond management, ash water recycling system and liquid waste management.
• Monitoring: Ecological monitoring, reviews and online database management for all its power stations.

Question 31.
Differentiate between organised and unorganised sectors.
OR
Suggest any five measures to reduce unemployment. (5)
Answer:
The differences between the organised and the unorganised sectors are as follows:

Basis of distinction	Organised Sector	Unorganised Sector
1. Meaning	All organisations registered under the government or some other regulatory agency belong to the organised sector.	These are generally small and scattered units or businesses which are largely outside the control of the government.
2. Followed Status of rules and regulations	Most of their codes of conduct have been framed by the government. They are strictly monitored.	There are rules and regulations but these are not followed. They are not menitored regularly.
3. Nature of job and work availability	Regular work is ensured and the employees cannot be dismissed from their job without valid reasons.	Employees can be asked to leave immediately without any reason. There’s no job security. Work available is also seasonal.
4. Examples	A worker in a large factory, a clerk in an office, a doctor in a hospital. a teacher in a school, etc.	A daily wage labourer/worker in a shop, a handloom weaver. an agricultural labourer, a shop assistant, street vendors etc.
5. Benefits	Workers get several benefits from their employers like paid leaves, provident fund, incentives, pensions, gratuity, medical benefits etc.	There is no provision for extra payment for overtime, paid leaves, holidays, medical leaves or even pension after retirement etc.

OR
Five measures to reduce unemployment are as follows:

• Unemployment can be reduced by capacity building. Capacity must be built to acquire jobs and new skills through education and training.
• More investments in employment creation schemes can be another way.
• The National Rural Employment Programme scheme is an effective way of generating employment too. This programme aims at creating community assets for strengthening rural infrastructure like drinking water wells, community irrigation wells, village tanks, rural roads and schools.
• Credit should be made easily available for young men and women from all classes and regions to push them towards establishing startups and native industries.
• Farmers should be given credit easily to encourage them to set up food processing units which would generate more employment.

Question 32.
How did the handloom industry collapse in India under the British rule? Explain.
OR
Explain the attitude of the Indian merchants and the industrialists towards the ‘Civil Disobedience’ Movement. (5)
Answer:
The handloom industry collapsed in India under the British rule:

• Factories in Manchester began producing cotton textiles for the domestic market.
• The government put more import duties on the textile coming from India to encourage the local industries and the Indian weavers Lost their overseas market.
• Manchester goods started flooding the Indian markets. Indian weavers could not compete with low cost Manchester cloth.
• The British government in India also levied more taxes on the handloom units. This made Indian textiles costlier.
• The civil war in the USA forced the British to buy more raw cotton from India for their Manchester textile industries. This created an acute shortage of raw material for the weavers and the Indian handloom industry collapsed.

OR
The attitude of the Indian merchants and the industrialists towards the Civil Disobedience Movement can be explained through the following points:

• During the First World War, Indian merchants and industrialists wanted to protect their companies and their products against cheap foreign imports goods. They also wanted a Rupee-Sterling foreign exchange ratio to discourage import.
• To organise their business interests, they formed the Indian Industrial and Commercial Congress (in 1920) and the Federation of Indian Chamber of Commerce and Industries-(FICCI) in 1972.
• Further, they gave financial assistance to the movement and refused to buy or sell imported goods.
• Most businessmen came to see ‘Swaraj’ as a time when colonial restrictions on business would no longer exist and trade would flourish without constraints.
• After the failure of the Round Table Conference, business groups were no longer enthusiastic. They were apprehensive of the spread of militant activities. They were worried that they would gain nothing out of protesting.

Question 33.
Why are Democratic governments slow at making decisions? How are Non-democratic countries different from them?
OR
What are the main.forms of power sharing in modern democracies? (5)
Answer:
Democratic governments are slow at making decisions because:
(1) Democracy is based on the idea of deliberation and negotiation. Many leaders, ministers and officers have to be convinced to arrive at a conclusion regarding every issue.

(2) Such deliberations, debates and negotiations take time. The opinion of every person is considered while deciding for the country. Hence, for democratic government, decision making takes time and has to be achieved through a fixed process.

(3) despite the delay, their decisions are consensual and popular. They are widely accepted.

Non-democratic nations are different from Democratic governments in the following ways:
(1) Non-democratic nations do not have to negotiate and deliberate before taking a decision. Usually, only one person or one group has all the authority to take decisions.

(2) This process takes less time and is not as rigorous. The opinion of the people is not considered while making decisions. Hence their decisions are taken faster but are unpopular and imposed.
OR
Power can be shared or distributed among various organisations or institution in two ways, Horizontal and Vertical
(1) Horizontal Distribution of Power Sharing: In this system power is shared among different organs of the governments in which each organ of government, i.e. Judiciary, Legislature and Executive are placed at the some level to exercise different powers in such a way that none of the organ exercise unlimited power. Besides, each organ checks the others which makes a balance of power among various institutions.
Judiciary → Legislature → Executive
The Horizontal Distribution of Power Sharing

(2) Vertical Distribution of Power Sharing: In this system, power is shared among governments at different levels. For example: Central/Union Government for the entire country at national level States Governments/Provincial or Regional Governments at state level while Municipality and Panchayats at the local level share the power in our country. This division of powers involving higher and lower levels of government is called vertical distribution of power sharing.

Section – E
Case Based Questions (4 x 4 = 16)

Question 34.
Read the source given below and answer the following questions:
In the German regions a large number of political associations whose members were middle-class professionals, businessmen and prosperous artisans came together in the city of Frankfurt and decided to vote for an all-German National Assembly.

On 18 May 1848, 831 elected representatives marched in a festive procession to take their places in the Frankfurt parliament convened in the Church of St. Paul. They drafted a constitution for a German nation to be headed by a monarchy subject to a parliament.
(A) What can we conclude about the treatment of Women in Europe in this century through their participation in this assembly? [1]
(B) Why was this Parliament disbanded? [1]
(C) What led to the unification of Germany? [2]
Answer:
(A) Even though women made their own associations, published news and supported men in their struggles, they were treated as inferiors and admitted as only observers in visitor’s gallery while men made laws and were entitled suffrage.

(B) It was disbanded because:

• The parliament was dominated by the middle classes who resisted the demands of workers and artisans and lost their support.
• The Parliament was also opposed by the military and aristocracy due to the large number of participants from the middle class.

(C)

• Germany was unified by the efforts of Otto von Bismarck who carried this out with the help of the Prussian army and bureaucracy.
• Three wars over seven years – with Austria, Denmark and France – ended in Prussian victory and completed the process of unification.

Question 35.
Read the source given below and answer the following questions:
S0ubsequently, in 1974, Gandhian philosophy was once again presented by Schumacher in his book ‘Small is Beautiful’. The seminal contribution with respect to resource conservation at the global level was made by the Brundtland Commission Report, 1987. This report introduced the concept of ‘Sustainable Development’ and advocated it as a means for resource conservation, which was subsequently published in a book entitled ‘Our Common Future’. Another significant contribution was made at the Earth Summit at Rio de Janeiro, Brazil in 1992.
(A) Which Gandhian philosophy does the source talk about? [1]
(B) What was the subject and purpose of this report? [1]
(C) Which significant contribution was made at Earth Summit, 1992? [2]
Answer:
(A) Gandhiji believed that there was enough for everybody’s need and not for anybody’s greed on Earth and advocated that the greed for development has caused immense destruction to the environment. If greed was kept below one’s most important needs to grow and develop in the list of priorities, the environment could be saved.

Related Theory:
Other philosophies of Gandhiji as mentioned were used by him to bring India its much deserved Independence from its oppressors- the British.

(B) The concept of sustainable development was introduced through the Brundtland commission report. Thus, the report was based on the subject of Resource conservation.

Its objective was to give a reality check about the frail health of the environment to the world.
(C) (1) Agenda 21 aims at achieving global sustainable development and was declared at the Earth Summit 1992.
(2) It is an agenda to combat environmental damage, poverty, disease through global cooperation.

Related Theory:
United Nations Environment Programme was founded in 1972 at Stockholm Conference.

Question 36.
Read the source given below and answer the following questions:
Large MNCs in the garment industry in Europe and America order their products from Indian exporters. These large MNCs with worldwide networks look for the cheapest goods in order to maximize their profits. To get these large orders, Indian garment exporters try hard to cut their own costs.

As the cost of raw materials cannot be reduced, exporters try to cut labour costs. Where earlier a factory used to employ workers on a permanent basis, now they employ workers only on a temporary basis so that they do not have to pay workers for the whole year. Workers also have to put in very long working hours and work night shifts on a regular basis during the peak season. Wages are low and workers are forced to work overtime to make both ends meet.
(A) Why do large MNCs with worldwide network look for the cheapest raw materials? [1]
(B) How has Globalisation changed the market? mention two points. [1]
(C) Can you consider the impact of Globalisation on the factory owners and employees equal and fair in the given source? Support your answer using examples. [2]
Answer:
(A) MNCs with large capital in their hands look for countries where they get cheaper raw materials and other resources to earn maximum profits to establish their factories.

(B) Globalisation has brought greater employment opportunities. It has also improved the living standards of the people as their choices have
widened. Globalisation has aLso posed major challenges for a large number of small producers and workers.

Related Theory:
The small industries inIndia which employ the largest number of workers (20 million) in the country have been shut down due to globalisation as these industries were unable to compete with large MNCs.
OR
(C) No, the impact of GLobalisation has not been equally beneficial for the workers and the factory owners. This can be evidenced by the following arguments:
(1) Factory owners have been given an opportunity to maximize their profits through international trade but it is the workers who actually bear the brunt. Their wages are reduced to reduce the cost of production.

(2) Factory owners recieve and utilise these profits, for selfish motives while the workers who put in long working hours, rigorous schedules and low wages receive no such benefits.

Caution:
To answer questions like this, it is important that the student read each word in the given source and understand it thoroughly.
The answer should be balanced but clearly inclined towards an agreement or disagreement with what is asked in the question.

Section – F
Map Based Questions (5)

Question 37.
(a) On the given political outline map of India identify the places marked as A and B with the help of the following information and write their correct names on the lines marked near them.
(A) The place known for peasant Satyagraha in Gujarat.
(B) The place where the session of Indian National Congress was held in December 1920. [2]

(b) On the same given map of India, locate and label any three of the following with suitable symbols.
(a) Singrauli – Thermal Power Plant
(b) Tuticorin – Seaport
(c) Bhakra Nangal – Dam/Multi-purpose
(d) Kalapakkam- Nuclear power plant [3]

Answer:
(1) (A) Kheda
(B) Nagpur

(2) Located and labeled on the map

Students must start practicing the questions from CBSE Sample Papers for Class 10 Social Science with Solutions Set 4 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Social Science Set 4 with Solutions

Time : 3 Hours
Maximum Marks: 80

General Instructions:

1. Question paper comprises five Sections – A, B, C, D and E. There are 37 questions in the question paper. All questions are compulsory.
2. Section A – From question 1 to 20 are MCQs of 1 mark each.
3. Section B – Question no. 21 to 24 are Very Short Answer Type Questions, carrying 2 marks each. Answer to each question should not exceed 40 words.
4. Section C contains Q.25to Q.29 are Short Answer Type Questions, carrying 3 marks each. Answer to each question should not exceed 60 words.
5. Section D – Question no. 30 to 33 are long answer type questions, carrying 5 marks each. Answer to each question should not exceed 120 words.
6. Section-E – Questions no from 34 to 36 are case based questions with three sub questions and are of 4 marks each.
7. Section F – Question no. 37 is map based, carrying 5 marks with two parts, 37a from History (2 marks) and 37b from Geography (3 marks).
8. There is no overall choice in the question paper. However, an internal choice has been provided in few questions. Only one of the choices in such questions have to be attempted.
9. In addition to this, separate instructions are given with each section and question, wherever necessary.

Section – A
MCQs (1 x 20 = 20)

Question 1.
Which of the following companies can be called a multinational company?
(a) Company X, which sells goods in two countries.
(b) Company Y which produces goods in multiple countries and then sells the same in multiple countries too.
(c) Company Z which has a huge market in its home country,
(d) Company A which has a huge online presence and foreigners buy goods and products from this website. [1]
Answer:
(b) Company Y which produces goods in multiple countries and then sells the same in multiple countries too.
Explanation: A MNC is a company that owns or controls production in more than one nation. MNCs set up offices and factories for production in regions where they can get cheap labour and other resources.

Related Theory:
Trade was the main channel connecting distant countries.

Caution:
Such questions are asked based on real life situations. They are application of the concepts taught in the chapter. To answer these, it is important that students understand the concept rather than to rote-learn. Students might underline important statements associated with the topic and make topic notes separately.

Question 2.
Which Italian revolutionary was born in Genoa in 1807?
(a) Giuseppe Garibaldi
(b) Jacob Grimn
(c) Wihelm Grimn
(d) Giuseppe Mazzini [1]
Answer:
(d) Giuseppe Mazzini
Explanation: Mazzini was born in Genoa in 1807. He went on to become a member of the secret society of the Carbonari. As a young man of 24, he was sent into exile in 1831 for attempting a revolution in Liguria. He subsequently founded two more underground societies, first, Young Italy in Marseilles, and then, Young Europe in Berne, whose members were like-minded young men from Poland, France, Italy and the German states. Mazzini believed that God had intended nations to be the natural units of mankind.

Question 3.
Which of the following principles of a democratic government does the action in the given picture violate?

(a) Accountability
(b) Legitimacy
(c) Popularity
(d) Transparency [1]
Answer:
(d) Transparency
Explanation: Democracy ensures that
decision making will be based on norms and procedures. A citizen, who wants to know if a decision was taken through the correct procedures, has the right and the means to examine the process of decision making. This is known as transparency. The picture shows that the government is able to view and observe everything about the public but the vice versa is not being facilitated. Hence, it is transparency of a democratic government that is being violated her.

Question 4.
Choose the correctly matched pairs:
(a) Flemish Region of Belgium-French
(b) Northern Sri Lanka-Tamil
(c) Wallonia Region in Belgium-Dutch
(d) Eastern Sri Lanka-Sinhalese [1]
Answer:
(b) Northern Sri Lanka- Tamil
Explanation: In Belgium, 59 per cent lives in the Flemish region and speaks Dutch language. Another 40 per cent people live in the Wallonia region and speak French. Remaining 1 per cent of the Belgians speak German.

Sri Lankan Tamils who speak Tamil are concentrated in Northern and Eastern parts of Sri Lanka.

Question 5.
Identify the type of soil by using the given hints:
(I) The soil develops under tropical and subtropical climate with alternate wet and dry season.
(II) It is the product of intense leaching due to heavy rain.
(III) They are deep to very deep, acidic (pH<6.0), generally deficient in plant nutrients.
(a) Laterite Soil
(b) Arid soil
(c) Alluvial Soil
(d) Black Soil [1]
Answer:
(a) Laterite Soil
Explanation: Laterite has been derived from the Latin word ‘later’ which means brick. It develops under tropical and subtropical climate with alternate wet and dry season. They are prone to erosion and degradation due to their position on the landscape. The soil is suitable for cashew nuts, tea and coffee.

Related Theory:
The soils occur mostly in southern states, Western Ghats region of Maharashtra, Odisha, some parts of lA/est Bengal and North-east regions.

Question 6.
What is meant by the term ‘scheduled’ in the word scheduled languages?
(a) The languages are scheduled to be made official.
(b) The languages are spoken by scheduled tribes.
(c) The languages are spoken by a lot of people.
(d) The languages are included in a schedule of the Indian constitution. [1]
Answer:
(d) The languages are included in a schedule of the Indian constitution.
Explanation: There are 22 languages recognised as ‘Scheduled’ in the eighth schedule of the Indian constitution.

Related Theory:
Assamese, Bengali, Bodo, Dogri, Gujarati, Hindi, Kannada, Kashmiri, Konkani, Maithili, Malayalam, Manipuri, Marathi, Nepali, Oriya. Punjabi, Tamil, Telugu, Urdu included in the list of 22 Scheduled Languages of India.

Question 7.
Match the people in Column A with the sectors of economic activity, they belong to in Column B and choose the most appropriate code which reflects the correctly matched pairs.

	Column A		Column B
(A)	Auto rickshaw driver	(I)	Primary Sector
(B)	A Businessman who produces bread	(II)	Tertiary Sector
(C)	A Small trader	(III)	Secondary Sector
(D)	A Miner	(IV)	Tertiary Sector

Codes:
(a) (A)-(I), (B)-(III), (C)-(IV), (D)-(II)
(b) (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
(c) (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
(d) (A)-(IV), (B)-(III), (C)-(II), (D)-(I) [1]
Answer:
(d) A-IV, B-lll, C-ll, D-l
Explanation: Agriculture, fishing and mining are classified as primary sector activities a driver and trader provide services. A businessman who produces processed goods (bread) from natural products (wheat) belongs to the secondary sector.

Question 8.
Choose the correct statement about collateral.
(a) Collateral is what pushes the lender into a painful debt trap situation.
(b) Collateral is the amount that RBI gets from other banks as security deposit for emergency times.
(c) Collateral is an asset that the borrower owns and uses as a guarantee against the amount borrowed.
(d) Collateral is not involved during borrowing of loans from the Formal Sector. [1]
Answer:
(c) Collateral is an asset that the borrower owns and uses as a guarantee to a lender against the amount borrowed.
Explanation: Collateral acts like a guarantee to a lender until loan is repaid. If borrower fails to repay the loan, the lender can reimburse his loan by selling the collateral or confiscating it.

Related Theory:
Property such as land titles, deposits with banks (Jewellery, bonds), livestock are some common examples of collateral used for borrowing. RBI takes a part of the minimum cash balance out of the deposits received by the bank for emergency and other purposes. Collateral is a safety valve for lender’s benefit. It does not put the lender in a debt trap because lender tends and not borrows.

Question 9.
Why were dams built traditionally?
(a) to generate electricity
(b) to stop landslides
(c) for generating hydroelectric energy
(d) to impound rivers and rainwater for irrigation [1]
Answer:
(d) to impound rivers and rainwater for irrigation
Explanation: Dams were traditionally built to impound rivers and rainwater that could be used later to irrigate agricultural fields.

Related Theory:
Dams are now built not just for irrigation but also for electricity generation, water supply for domestic and industrial uses, flood control, recreation, inland navigation and fish breeding.

Question 10.
What did the Corn laws allow the government to do?
(a) to tax corn producers
(b) to fix a quantity of corn that can be harvested everyday
(c) to restrict the import of Corn
(d) to restrict the export of Corn [1]
Answer:
(c) to restrict the import of Corn
Explanation: Under pressure from landed groups, the government also restricted the import of corn as per the ‘Corn Laws’. Unhappy with high food prices, industrialists and urban dwellers forced the abolition of the Corn Laws.

Related Theory:
After the Corn Laws were scrapped, food could be imported into Britain more cheaply. British agriculture was unable to compete with imports.

Question 11.
Which of the following pairs is matched correctly?

	Column A		Column B
(a)	Gram Panchayat	(I)	Panchayat Samitis
(b)	Municipal Corporation	(II)	Sarpanch
(c)	Zila Parishad	(III)	Mayor
(d)	State	(IV)	Governor

Answer:
(d) State-(iv) Governor
Explanation: In India, the governor is the head of the state appointed by the Central Government of India.

Related Theory:
Alt the panchayat samitis or mandals in a district together constitute the zila parishad. Sarpanch is the head of the gram panchayat in a village in India while the Mayor is the political head of the municipal corporation in towns.

Question 12.
There are two statements marked as Assertion (A) and a Reason (R). Mark your answer as per the codes provided below:
Assertion (A): The textile industry occupies unique position in the Indian economy.
Reason (R): It is the only industry in the country which is self-reliant.
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true. [1]
Answer:
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
Explanation: The textile industry occupies unique position in the Indian economy, because it contributes significantly to industrial production, employment generation and foreign exchange earnings.

Related Theory:
The industry produces all its goods from raw materials to finished goods all by itself.

Question 13.
There are two statements marked as Assertion (A) and a Reason (R). Mark your answer as per the codes provided below:
Assertion (A): Health and nutritional status of people with PDS implemented in their states are certainly likely to be better.
Reason (R): Adequate provision of basic health and educational facilities ensures physical and mental development of the citizens of the state.
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true. [1]
Answer:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
Explanation: Money does not buy everything. Some developmental indicators are not bought by money. Health and nutritional status of people in states with Public Distribution System is very developed because affordable food and healthcare facilities ensure physical and mental well-being of the people.

Question 14.
Which continent are the Scandinavian countries located in?
(a) Asia
(b) America
(c) Africa
(d) Europe [1]
Answer:
(d) Europe
Explanation: Scandinavian countries such as Sweden, Norway and Finland are located in Northern Europe. The participation of women in public life in these countries is very high.

Related Theory:
These countries are extremely developed and evolved in terms of education, financial and political life.

Caution:
Such questions are asked based on some general knowledge and awareness of the current affairs. It is important for the students to take initiative to research a little bit outside of the chapter about important topics.

Question 15.
The table below shows the estimated number of employees working in the organised and unorganised sectors in India. Read the table carefully and answer the following question:

Sector	Organised	Unorganised	Total
Primary	20	242	262
Secondary	9	54	73
Tertiary	25	76	101
Total	64	372	436
Total in Percentage	14.7%	85.3%	100%

What percentage of primary sector workers in India are employed in the Unorganised Sector according to the table?
(a) 61.7%
(b) 92.3%
(c) 55.5%
(d) 88.0% [1]
Answer:
(b) 92.3%
Explanation: To calculate percentage of primary sector workers employed in an unorganised sector:
Percentage = \(\frac{x}{\text { Total } \times 100}\)
Total number of employees in primary sector = Number of employees in organised + number of employees in unorganised sector
Total number of workers in primary sector: 20 + 242 = 262
Percentage of primary sector employees employed in unorganised sector:
\(\frac { 242 }{ 262 }\) ×100 = 92.3%

Question 16.
Choose the odd one out:
(a) Sweden, USA, China, Saudi Arabia
(b) Norway, France, India and UAE
(c) Italy, Finland, Sri Lanka
(d) Denmark, Canada, Congo and Yemen [1]
Answer:
(c) Italy, Finland, Sri Lanka
Explanation: Each group has countries
arranged in a certain fashion. The first country on the left has the largest women representation in its political institutions (out of the group), followed by countries with representation decreasing as one goes towards the right. This is not true for the chosen option. Finland, as a Scandinavian country has larger representation of women in its political life than Italy. Hence, this option is the odd one out.

Related Theory:
Women’s political representation can be best represented through the following graph. The Nordic countries are also known as Scandinavian countries.

Question 17.
Complete the following table with correct information with regard to cultivation of Sugarcane by identifying A:

	Rainfall required annually	Type of climate required	Second largest producing country
Sugarcane	75 cm to 100 cm	(A)	India

(a) Cool and dry
(b) Hot and humid
(c) Cold and wet
(d) Warm and dry [1]
Answer:
(b) Hot and Humid
Explanation: Sugarcan needs hot and humid climate with a temperature of about 21°-27°C. Brazil is the largest producer of sugarcane in the world.

Question 18.
By the late nineteenth century, manu-facturers were printing calendars to popularise their products. Why were the manufacturers using calendars for promotion?
(a) Calenders were used by rich people.
(b) Calenders had no language restriction.
(c) Calendars were used even by people who could not read.
(d) Calenders were used by only middle class people. [1]
Answer:
(c) Calendars were used even by people who could not read.
Explanation: By the late nineteenth century, manufacturers were printing calendars to popularise their products. Unlike newspapers and magazines, calendars were used even by people who could not read. They were hung in tea shops and in poor people’s homes just as much as in offices and middle-class apartments.

Related Theory
Like the images of gods, figures of important personages, of emperors and nawabs, adorned advertisement and calendars.

Caution:
It is important for the students to understand that every line in the chapter is not to be memorised. Students should underline assertions mentioned in the chapter and their reasons separately to relate and understand a concept completely.

Question 19.
Fill in the blank by choosing the most appropriate option.
The practise of buying technology from foreign nations for use is called ………………. the technology.
(a) exporting
(b) sharing
(c) granting
(d) importing [1]
Answer:
(d) importing
Explanation: Importing entails buying foreign goods and services while exporting entails selling goods to foreign consumers. Granting involves donation and not purchase of a technology.

Related Theory:
Foreign trade connects the markets or integrates markets in different countries.

Question 20.
Indian Industrialists began shifting from yarn to cloth production in early 1900s. Which of the following statements is possibly a reason?
(I) Early cotton mills in India produced coarse cotton yarn because British yarn was not imported in India. This yarn was exported to China.
(II) The export of Indian yarn to China declined in 1906.
(III) Industrialists had to shift to cloth production from yarn production.
(IV) Indian markets had no demand for coarse yarn.
Codes:
(a) I, II and IV
(b) I and III
(c) I and IV
(d) I, II and III 1
Answer:
(d) I, II and III
Explanation: Lack of British yarn exports forced Indian industrialists to produce coarse yarn which was exported to Chinese markets. From 1906, the export of Indian yarn to China declined since produce from Chinese and Japanese mills flooded the Chinese market. So industrialists in India began shifting from yarn to cloth production. Cotton goods production in India doubled between 1900 and 1912.

Section – B
Very Short Answer Type Questions (2 x 4 = 8)

Question 21.
When and why was the young Italy movement organised?
OR
Mention two consequences of the great depression of 1929. [2]
Answer:
The Young Italy movement was designed by Mazzini through his secret society caLled the Young Italy created to work for the unification of Italy. It was joined by Garibaldi later. The movement can be understood as a struggle to evoke collective belongingness among ItaLians living in divided regions.
OR
The great depression led to:

• Catastrophic declines in production, employment, income and trade among countries of the world.
• Fall in agricultural and industrial goods prices.

Question 22.
Why are Political Parties such significant institutions in Democracy? [2]
Answer:

• Political Parties are Linked to the emergence of representative democracies particularly in large societies like India as an agency to gather different views on various issues and to present these to the government.
• They also help to spread the ideas of democracy across the nation by forming closer links to people.
• They help to educate them about their political, economics and social rights and duties.

Question 23.
What is tidal energy? Mention one use of the same. [2]
Answer:
Oceanic tides are used to generate energy by trapping running water in floodgate dams and thereafter utilising the potential energy. Tidal Energy can be used as a clean source to generate electricity. In India, the Gulf of Khambhat, the Gulf of Kuchchh in Gujarat on the western coast is a great source of tidal energy.

Question 24.
How is the competition in the market directly related to uncertain employment opportunities? [2]
Answer:
Cutthroat competition in the market leads the producers to strive to offer their goods for the lowest cost possible without risking their profit margin. To achieve that, producers try to employ workers onLy seasonally so that their expenditure on wages is contained. This makes their employment uncertain.

Section – C
Short Answer Type Questions (3 x 5 = 15)

Question 25.
Distinguish between open and disguised unemployment. [3]
Answer:
The differences between open unemployment and disguised unemployment are as follows:
(1) Open unemployment: Agriculture sector offers large opportunities in terms of employment but a large number of landless labourers still face unemployment. Since they are visibly unemployed, it is called open unemployment.

(2) Disguised unemployment: It is a situation in which more people than actually required are employed in an activity. This causes partial unemployment because each worker performs less and earns below his capacity. This is known as disguised unemployment because on the face of it, no worker is unemployed.

(3) Casual workers such as electricians, plumbers etc. in urban areas are an example of disguised unemployment as these people do not find gainful work every day.

Question 26.
Describe any three types of soils available in India.
OR
Explain any three steps that can be taken to solve the problem of land degradation? [3]
Answer:
Three types of soils found in India are:
(1) Alluvial Soil: It is the most widespread soil and makes up the whole northern plains. Alluvial soil is deposited by three rivers: Ganga, Brahmaputra and Indus. It consists of various proportions of sand, silt and clay.

(2) Black Soil: They are also called Regur Soils. This type of soil is found in the Deccan plateau. They are highly moisture retentive. They are rich in calcium carbonate, magnesium, potash and lime.

(3) Laterite Soil: Laterite soil develops in areas of high temperature and heavy rainfall. It is formed due to intense leaching. These are found in Karnataka, Kerala, Tamil Nadu, Madhya Pradesh.
OR
Steps to tackle land degradation are:
(1) Contour Ploughing: Ploughing along the contour lines can decelerate the flow of water down the slopes, also reducing soil erosion.

(2) Terrace Cultivation: Making terraces by cutting steps out on slope and cultivating them can help stop sheet erosion. This can stop land from losing its fertility.

(3) Strip Cropping: Large fields can be divided into strips. Strips of grass are left to grow between the crops which breaks up the force of the wind reducing its effect.

Question 27.
Has democracy been successful in removing inequalities among various classes in the Indian society? [3]
Answer:

• Democracies Gre expected to remove economical and social inequalities but they have not been very successful.
• Inequalities have been widening. A small number of ultra rich people are entrusted with a great share of wealth and resources to live life luxuriously while a large number of people have almost no resources to thrive upon.
• The lower classes of Indian society are getting poorer and poorer- with their falling incomes- while the top most are getting richer and richer.

Question 28.
Describe how the Non-Cooperation Movement lost its meaning in the countryside using the example of peasant’s revolution in Awadh. [3]
Answer:
Non-Cooperation Movement was widespread across the countryside but it slowly lost its significance. This can be explained through the following points:

• In Awadh, Baba Ramchandra led peasants against talukdars and landlords who demanded high rents and unnecessary cesses.
• The peasant movement demanded reduction of revenue, abolition of Begar and social boycott of oppressive landlords. Awadh Kisan Sabha was set up under Nehru. Nai-dhobi Bandhs were organised.
• However, soon the movement lost its direction. Houses of talukdars and merchants were attacked: bazaar looted and grain hoards were taken over. Tribal peasants misinterpreted Gandhi’s idea of Swaraj and grew violent.

Question 29.
Both Belgium and Sri Lanka are democracies but they follow different systems of power sharing. Support the statement by giving three points of difference. [3]
Answer:
Difference in power sharing arrangements of Belgium and Sri Lanka can be understood through the following points:
(1) In Belgium, the government does not follow preferential policies in matters of jobs and education unlike Sri Lanka, In Sri Lanka, the majoritarian government prefers only the Sinhala language speaking community.

(2) In Belgium, there is a unique set up calLed the ‘community government’ to look after the cultural, educational and language related issues of minorities and majority. In Sri Lanka, issues of minorities are neglected.

(3) In Belgium, there is no discrimination between different religions. In Sri Lanka, Buddhism is the official religion. People practising other religions are isolated and disrespected.

Section – D
Long Answer Type Questions (5 x 4 = 20)

Question 30.
How was the Print culture responsible for the Protestant reformation?
OR
Dalit participation was limited in the Civil Disobedience Movement. Examine the statement. [5]
Answer:
Print created the possibility of wide circulation of ideas. It led to a new world of debate and discussion.

• People were able to print messages which agreed and disagreed with established authorities and influence people with printed messages.
• Such books and pamphlets were severely criticised. In the wake of this criticism, in 1517, the religious reformer Martin Luther wrote Ninety Five Theses criticising many of the practices and rituals of the Roman Catholic Church.
• Luther’s writings which challenged the Church to debate his ideas, were immediately reproduced in vast numbers and read widely.
• This caused a rift within the Church and lead to the beginning of the Protestant Reformation.
• The print thus helped spread the new ideas that led to the Reformation.

OR
Dalit participation was limited in the Civil Disobedience Movement. This was because:

• Dalit had long been ignored by the Congress before the movement started due to the fear of ostracization of the party by higher Hindu castes. They were underrepresented and neglected.
• In Spite of Gandhiji’s efforts for immediate reconciliation, people were hesitant.
• Dalits expressed only lukewarm support to Civil Disobedience due to their insecurities of being dominated and ignored for decades.
• Many Dalit leaders were keen on a different political solution. They wanted the upliftment of their community first.
• Dalits wanted political empowerment before Swaraj. Gandhiji was against separate electorates. This clash made Dalits even more reluctant.

Related Theory:
Dalit leaders Dr . B. R. Ambedkar, who organised the Dalits into the Depressed Classes association in 1930, clashed with Mahatma Gandhi at the second Round Table Conference by demanding separate electorates for Dalits to secure their interests.

Question 31.
The creation of Linguistic States was the first and a major test for democratic politics in our country. Do you agree? Explain.
OR
“Dynastic succession is one of the most serious challenges before the political parties.” Analyse the statement. [5]
Answer:
Yes, I agree that the reformation and creation of linguistic states was the first obstruction for the newly set up democratic system. This is because:

• In 1947, states were rearranged to ensure that people who spoke the same language, shared the same culture, geography and descent were living in the same state.
• Most intellectuals and experts stated that rearrangement of state could lead to disintegration of the country, further risking its security, internal harmony and sovereignty. This was a test of democracy.
• However, later, it was realised that the reformation of states only made the country more united and made the administration easier. It brought harmony among the people.
• Creation of Linguistic states helped to fulfill people’s will and aspirations from a democratic system.

OR
Dynastic succession is one of the most serious challenges before the political parties. This can be asserted by the following points.

• Most political parties do not practice open and transparent election procedures for their functioning and ticket distribution. Only certain members receive positions of influence due to their contacts and influence.
• There are few ways for an ordinary worker to rise to the top in a party.
• In many parties the top positions are always controlled by members of one family or kin. They hold superior positions in the party and are reluctant to give the same up even to more talented candidates in power thirst.
• This practice is unfair to other members of that party and is also bad for a democracy.
• People who do not have adequate experience or popular support come to occupy positions of power. This tends to disrupt the working and leads to mistrust and splitting of parties.

Question 32.
Explain any two features each of formal sector loons and informal sector loans.
OR
What should India do or achieve to become a developed country? Explain. [5]
Answer:
Formal Sector Loans: Three features of formal sector loans are:

• Formal sectors provide cheap and affordable loans and their rates of interest are decided and monitored by the RBI.
• Formal sector strictly follows the terms of credit which includes fixed interest rate, collateral, documentation and the mode of repayment
• These loans are given by banks and cooperatives. There is no exploitation of borrowers as lending activities of all the banks are supervised and monitored by the government.

Informal Sector Loans: Loans from local moneylenders, traders, employers, relatives, friends are called informal sector loans. Features are:

• Their credit activities are not governed by any organisation, therefore they charge exorbitant rates which keep changing as per their profit.
• Informal sector loan providers know the borrowers personally, and hence they provide loans without collateral and documentation. These loans are not backed by the government.

OR
India should focus on the following points to become a developed country:

• It should reduce the gap between rich and the poor.
• It should make provision for accommo-dating all its existing able people in jobs who are willing and able to do but unable to get a suitable job.
• Primary health and education must access to all even in the remote parts of the country.
• Government should make provisions for making the country self-reliant by providing skill education to all.
• Government should encourage small-scale and cottage industries by giving cheaper credits and training to those who are willing to be the entrepreneurs.

Question 33.
Why is a dense and efficient network of transport and communication a prerequisite for the development of local, national and global trade of today? Give your opinion.
OR
Assess the need for the conservation of forests and wildlife in India. [5]
Answer:
Dense and efficient network of transport and communication are required for development and local and global trade because:
(1) We use different goods and services in our daily life. Some of these are available in our immediate surroundings while other requirements are met by transporting and exporting. Movement of these goods and services thus becomes necessary for us to sustain.

(2) Today, the world has converted into a large village with the help of efficient and fast communication. Ideas and information can be easily transferred from market to market, integrating and linking their requirements and trends.

(3) Transport has been able to achieve the integration with the help of equally developed transportation and communication systems.

(4) Easy availability of goods and services added to the multitude of products present in every market due to well developed transportation and communication systems boosts the demand and the trade ultimately. Trade helps boosting the country’s economy.

(5) Therefore transport, communication and trade are complementary to each other.
OR
The conservation of forests and wildlife in India
is extremely essential.
(1) Living organisms on Earth form a complex web or ecological system which humans are just a minute part of.

(2) We are all interdependent on each other. For example, Humans depend upon living organisms for hasic necessities such as pure air- trees purify the air- take in C02 and release oxygen.

(3) The loss of forest and wildlife is not just a biological problem but it also raises a cultural question on the existence of the people who attach their identity with the forests such as the tribals and the aboriginals.

(4) From 1951 and 1980, according to the forest survey of India, over 26,200 sq km of forest areas was converted into agricultural lands all over India.

(5) Destruction of forests has caused extreme ecological and social disbalance. It has led to a disastrous impact on environment in form of climate change.

Section – E
Case Based Questions (4 x 3 = 12)

Question 34.
Read the source given below and answer the following questions:
After arriving in India, Mahatma Gandhi successfully organised Satyagraha movements in various places. In 1916, he travelled to Champaran in Bihar to inspire the peasants to struggle against the oppressive plantation system. Then in 1917, he organised a Satyagraha to support the peasants of the Kheda district of Gujarat. Affected by crop failure and a plague epidemic, the peasants of Kheda could not pay the revenue, and were demanding that revenue collection be relaxed.
(A) Why did Gandhi organize a movement in South Africa before coming to India? [1]
(B) When and why was the Kheda Satyagraha organised? [1]
(C) What does the concept of Satyagraha emphasize upon? [2]
Answer:
(A) In South Africa, Mahatma Gandhi had begun a revolution to help their society get rid of racism. He fought against their contemporary racist regime in favour of Blacks, Non-whites and native residents through Satyagraha.

Related Theory:
Nelson Mandela fought against Apartheid decades later and was able to get South Africa its Independence in 1994.

(B) Peasants of Kheda were forced to pay tax and rent even when their crops failed due to adverse climate or some natural disaster like epidemic, pest attack etc. Gandhi encouraged these peasants to fight for relaxation of revenues through a Satyagraha movement.

Related Theory:
Champaran Satyagraha was organised against the oppressive Plantation system and Ahmedabad Satyagraha was organised against Cotton Mill owners.

(C) (1) The idea of Satyagraha emphasised upon the force/power of truth and the dire need to search for truth.

(2) It suggested that if the cause was true and one was struggling against injustice, then physical force was not necessary to fight the oppressor, truth was powerful enough.

Question 35.
Read the source given below and answer the following questions:
When power is taken away from Central and State governments and given local governments, it is called decentralisation. The basic idea behind decentralisation is that there are a large number of problems and issues which are best settled at the local level. People have better knowledge of problems in their localities.

They also have better ideas on where to spend money and how to manage things more efficiently. Besides, at the local level it is possible for the people to directly participate in decision making. This helps to inculcate a habit of democratic participation. Local government is the best way to realise one important principle of democracy, namely local self-governance.

The need for decentralisation was recognised in our Constitution. Since then, there have been several attempts to decentralize power to the level of villages and towns. Panchayats in villages’ and municipalities in urban areas were set up in all the States.

But these were directly under the control of state governments. Elections to these local governments were not held regularly. Local governments did not have any powers or resources of their own. Thus, there was very little decentralisation in effective terms.
(A) What is a Gram Sabha? What is it responsible for? [1]
(B) What is decentralisation? [1]
(C) Mention two features of local government in India. [2]
Answer:
(A) Gram Sabha is a local self government institution which has to meet at least twice or thrice in a year to approve the annuaL budget of the gram panchayat and to review the performance of the gram panchayat.

Related Theory:
Each village, or a group of villages in some states has a gram panchayat. This is a council consisting of ward members called Ranch or Sarpanch. They are directly elected by all the adult population living in that ward or village. It is the decision-making body for the entire village.

It works under the supervision of the Gram Sabha. Gram Panchayats are grouped together to form a Panchayat Samiti or Block or Mandal Samiti. Its members are elected by all the panchayat members in that area.

(B) Decentralisation is the process of sharing power and responsibility by taking some away from the central government and distributing them further to local and state governments. Generally, the power distributed to the local and state governments include issues that can be settled better at local level.

(C) Two such features are:
(1) India has a three tier system of governments, the central, state and local governments. Each of them have been given their set of powers and responsibilities by the Indian Constitution.

(2) Local self governments have seats reserved in decision making bodies for various ill-represented communities like women, scheduled caste and tribes and others.

Related Theory:
At least one-third of all positions in local self government institutions are reserved for women.

Question 36.
Read the source given below and answer the following questions:
When we compare different things, they could have similarities as well as differences. Which aspects do we use to compare them? Let us look at students in the class itself. How do we compare different students? They differ in their height, health, talents and interests. The healthiest student may not be the most studious one.

The most intelligent student may not be the friendliest one. So, how do we compare students? The criterion we may use depends on the purpose of comparison. We use different criterion to choose a sports team, a debate team, a music team or a team to organise a picnic. Still, if for some purpose, we have to choose the criterion for the all-round progress of children in the class, shall we do it?

Usually we take one or more important characteristics of persons and compare them based on these characteristics. Of course, there can be differences about what are important characteristics that should form the basis of comparison: friendliness and spirit of cooperation, creativity or marks secured?
(A) According to the source given above, why does comparing the total income of a country not tell us what an average person is likely to earn? [1]
(B) In the given source, how does the author drive home the point that being rich might not always mean being developed? [1]
(C) According to you, what are the indicators that should be measured while reviewing the development of a country? [2]
Answer:
(A) Population in each country is variable and so total income cannot be a good criterion to compare the countries. Average income or total income is notan adequate indicator to measure development also because monetary wealth does not mean holistic equal development.

(B) (1) In the given source, the author has tried to drive home the point that each individual is different and hence indicators which are not customised cannot measure development properly.

(2) For example, the author says that the students have different heights, weights, health, talent and interests. They cannot be measured by one single standard. A healthy person might not be the brightest and vice versa. For measuring holistic development, checking multiple indicators is important.

(C) According to me, the indicators that should be measured are:

• Income of the household
• Education received by various classes and communities across different regions
• Sanitation and Housing facilities available to them.
• Value systems- political, social and technological awareness.

Section – F
Map Based Questions (2 + 3 = 5)

Question 37.
(a) On the given political outline map of India, identify the places with the help of the following information and write their correct names on the lines drawn on them.
(A) The city where the Jallianwala Bagh incident occurred.
(B) The place where peasant Satyagraha was started. [2]

(b) On the same outline map of India, locate and label any three of the following with suitable symbols.
(a) Tarapur – Nuclear Thermal Power Plant
(b) Kochi – Major Seaport
(c) Tungabhadra – Dam
(d) Chennai – Meenam Bakkam international Airport [3]

Answer:
(a) (A) Amritsar
(B) Kheda

Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths with Solutions Set 1 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Maths Standard Set 1 with Solutions

Time Allowed: 3 Hours
Maximum Marks: 80

General Instructions:

1. This Question Paper has 5 Sections A-E.
2. Section A has 20 MCQs carrying 1 mark each
3. Section B has 5 questions carrying 02 marks each.
4. Section C has 6 questions carrying 03 marks each.
5. Section D has 4 questions carrying 05 marks each.
6. Section E has 3 case based integrated units of assessment (04 marks each) with sub-parts of the values of 1, 1 and 2 marks each respectively.
7. All Questions are compulsory. However, an internal choice in 2 Qs of 5 marks, 2 Qs of 3 marks and 2 Questions of 2 marks has been provided. An internal choice has been provided in the 2 marks questions of Section E
8. Draw neat figures wherever required. Take π =22/7 wherever required if not stated.

Section – A
(Section A consists of 20 questions of 1 mark each.)

Question 1.
Let a and b be two positive integers such that a = p³q⁴ and b = p²q³ , where p and q are prime numbers. If HCF(a, b) = p^mqⁿ and LCM(a, b) = pr(f, then (m + n)(r + s) =        (1)
(a) 15
(b) 30
(c) 35
(d) 72 1
Answer:
(c) 35

Explanation:
Given,
a = p³q⁴
and b = p²q³
HCF of (ab) = p²q³
p^mqⁿ = p²q³ .
⇒ m = 2
⇒ n = 3

LCM of (a, b) = p³q⁴
p^rq^s = p³q⁴
⇒ r= 3
⇒ s = 4
(m + n)(r + s) = (2 + 3)(3 + 4)
⇒ 5 × 7
= 35

Question 2.
Let p be a prime number. The quadratic equation having its roots as factors of p is: (1)
(a) x² – px + p = 0
(b) x² – (p + 1)x + p = 0
(c) x² + (p + 1)x + p = 0
(d) x² – px + p + 1 = 0
Answer:
(b) x² – (p + 1)x + p = 0

Explanation:
If p is prime number, then it’s factors are 1 and p itself.
And it’s roots, a and p are it’s factors
Hence α = 1, β = p
x = α or β
0 = (x – α)(x – β)
Which means = (x – 1)(x – p)
= x²+ (-1- p)x + (-1 × -p)
⇒ x² – (p + 1)x + p is the required quadratic equation

Question 3.
If a and p are the zeros of a polynomial f(x) = px² – 2x + 3p and α + p = αp, then p is (1)
(a) \(\frac{-2}{3}\)
(b) \(\frac{2}{3}\)
(c) \(\frac{1}{3}\)
(d) \(\frac{-1}{3}\)
Answer:
(b) \(\frac{2}{3}\)

Explanation:
Given, f(x) = px² – 2x + 3p
Compare it with standard equation ax² + bx + c = 0, we get
a = p, b = -2, c = 3p
Sum of zeros = α + β = –\(\frac{b}{3}\)
α + β = \(\frac{2}{p}\)
Product of zeros = αβ = \(\frac{c}{a}\)
αβ = \(\frac{c}{a}\)
αβ = \(\frac{3p}{p}\) = 3

It is given that, α + β = αβ
\(\frac{2}{p}\) = 3
p = \(\frac{2}{3}\)

Question 4.
If the system of equations 3x + y = 1 and (2k – 1)x + (k – 1)y = 2k + 1 is inconsistent, then k =      (1)
(a) -1
(b) 0
(c) 1
(d) 2
Answer:
(d) 2

Explanation:
Given,
3x + y = 1 ……..(i)
(2k – 1)x + (k – 1)y = 2k + 1 …….(ii)

Both the equation are in the from of ax + by + c = 0
a₁ = 3, b₁ = 1, c₁ = 1
a₂ = (2k – 1), b₂ = (k – 1), c₂ = 2k + 1

The system of equation is inconsistent, if
\(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
∴ \(\frac{3}{2k – 1}=\frac{1}{k – 1} \neq \frac{1}{2k + 1}\)
\(\frac{3}{2k – 1}=\frac{1}{k – 1}\)
3k – 3 = 2k – 1
k = 2

Caution: While comparing the given equations with standard equation, we should also consider the signs of constant

Question 5.
If the vertices of a parallelogram PQRS taken in order are P(3, 4), Q(-2, 3) and R(-3, -2), then the coordinates of its fourth vertex S are:  (1)
(a) (-2, -1)
(b) (-2, -3)
(c) (2, -1)
(d) (1,2)
Answer:
(c) (2, -1)

Explanation:
Since, the diagonals of a parallelogram bisect each other.
Let the coordinate of S is (x, y)
∴ Coordinates of the mid point of PR = coordinates of the mid point of QS
⇒ \(\left(\frac{3+(-3)}{2}, \frac{4-2}{2}\right)=\left(\frac{-2+x}{2}, \frac{3+y}{2}\right)\)
(0, 1) = \(\left(\frac{x-2}{2}, \frac{3+y}{2}\right)\)
\(\frac{x-2}{2}\) = 0 and \(\frac{3+y}{2}\) = 1
⇒ x – 2 = 0
x= 2
and 3 + y = 2
y = -1
Hence, the fourth vertex of the parallelogram is (2, -1)

Question 6.
∆ABC – ∆PQR. If AM and PN are altitudes of ∆ABC and ∆PQR respectively and AB²: PQ² = 4: 9, then AM : PN = (1)
(a) 3 : 2
(b) 16 : 81
(c) 4: 9
(d) 2 : 3
Answer:
(d) 2 : 3

Explanation:
We Know that the ratio the sides of the two similar triangles is equal to the ratio of their corresponding sides.
Therefore, (1)
\(\frac{AB}{PN}\) = \(\frac{AB}{PQ}\)
\(\frac{AB}{PN}\) = \(\sqrt{\frac{4}{9}}\)
\(\frac{AB}{PN}\) = \(\frac{2}{3}\)

Find AB: PQ,
So that,
AM : PN = 2 : 3
Hence, the ratio of AM to PN is 2 : 3.

Question 7.
If x tan 60°cos 60° = sin60°cot 60°, then x =    (1)
(a) cos 30°
(b) tan 30°
(c) sin 30°
(d) cot 30°
Answer:
(b) tan30°

Explanation:
x tan 60° cos 60° = sin 60° cot 60°
\((x) \times(\sqrt{3}) \times \frac{1}{2}=\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{3}}\)
\(\frac{\sqrt{3} x}{2}=\frac{1}{2}\)
\(x=\frac{1}{\sqrt{3}}\) = tan30°

Question 8.
If sin θ + cos θ = \(\sqrt{2}\) , then tan θ + cot θ =        (1)
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

Explanation:
Given, sin θ + cos θ = \(\sqrt{2}\)
On squaring both the sides, we get

From (i) and (ii) we get,
tan θ + cot θ = 2

Question 9.
In the given figure, DE || BC, AE = a units, EC = b units, DE = x unit and BC = y units. Which of the following is true? (1)

(a) \(x=\frac{a + b}{ay}\)
(b) \(y=\frac{ax}{a+b}\)
(c) \(x=\frac{ay}{a+b}\)
(d) \(\frac{x}{y}=\frac{a}{b}\)
Answer:
(c) \(x=\frac{ay}{a+b}\)

Explanation:
Given AE = a, EC = b, DE = x and BC = y.
As DE || BC
∠ADE = ∠ABC
∠AED = ∠ACB
So by AAA property,
∆ADE ~ ∆ABC
\(\frac{DE}{BC} = \frac{AE}{AC}\)
\(\frac{x}{y} = \frac{a}{a+b}\)
\(x = \frac{ay}{a+b}\)

Question 10.
ABCD is a trapezium with AD || BC and AD = 4cm. If the diagonals AC and BD intersect each other at O such that = \(\frac{AO}{OC} = \frac{DO}{OB} = \frac{1}{2}\) then BC =  (1)
(a) 6 cm
(b) 7 cm
(c) 8 cm
(d) 9 cm
Answer:
(c) 8 cm

Explanation:
Given ABCD is a trapezium in which BC || AD and AD = 4 cm.
Also, the diagonals AC and BD intersect at O such that \(\frac{AO}{OC} = \frac{DO}{OB} = \frac{1}{2}\)

In ∆AOD and ∆COB,
∠OAD = ∠OCB [alternate angles]
∠ODA = ∠OBC [alternate angles]
∠AOD = ∠BOC [vertically opposite angles]
We know that AAA similarity criterion states that in two triangles, if corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar.
∴ ∆AOD ~ ∆COB
We know that two triangles are similar if their corresponding sides are proportional.
\(\frac{AO}{OC} = \frac{DO}{OB} = \frac{AD}{BC}\)
⇒ \(\frac{1}{2} = \frac{AD}{BC}\)
⇒ \(\frac{1}{2} = \frac{4}{BC}\)
∴ BC = 8 cm

Question 11.
If two tangents inclined at an angle of 60° are drawn to a circle of radius 3 cm, then the length of each tangent is equal to (1)
(a) \(\frac{3 \sqrt{3}}{2}\) cm
(b) 3 cm
(c) 6 cm
(d) \(3 \sqrt{3} \) cm
Answer:
(d) \(3 \sqrt{3} \) cm

Explanation:
Let P be an external point and a pair of tangents is drawn from point P and angle between these two tangents is 60°

Radius of the circle = 3 cm
Join OA and OP.
Also, OP is bisector line of ∠APC
∴ ∠APO = ∠CPO = 30°
OA ⊥ AP
Also, tangents at any point of a circle is perpendicular to the radius through the point of contact.
In right angled ∆OAP, we have
tan 30° = \(\frac{OA}{AP}\) = \(\frac{3}{AP}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{3}{AP}\)
⇒ AP = \(3 \sqrt{3} \)
AP =CP
= \(3 \sqrt{3} \) cm
[Tangents drawn from an external point are equal]
Hence, the length of each tangent is \(3 \sqrt{3} \) cm.

Caution: Remember that the point where tangent touches the circle is perpendicular to the radius.

Question 12.
The area of the circle that can be inscribed in a square of 6cm is: (1)
(a) 36π cm²
(b) 18π cm²
(c) 12π cm²
(d) 9π cm²
Answer:
(d) 9π cm²

Explanation:
Diameter of circle = side of square = 6 cm
Radius = \(\frac{6}{2}\)
Radius = 3 cm
Area of circle = πr²
π(3)²
= 9π
Therefore, the area of the circle is 9π square cm.

Question 13.
The sum of the length, breadth and height of a cuboid is \(6 \sqrt{3} \) cm and the length of its diagonal is \(2 \sqrt{3} \) cm. The total surface area Qf the cuboid is: (1)
(a) 48 cm²
(b) 72 cm²
(c) 96 cm²
(d) 108 cm²
Answer:
(c) 96 cm²

Explanation:
According to condition,
\(\sqrt{\left(l^2+b^2+h^2\right)}=2 \sqrt{3}\)
⇒ l² + b² + h² = 12
Again, 1 + b + h = 6V3
⇒ (l + b + h)² = 108
⇒ l² + b² + h² + 2 (lb + bh + h[) = 108
⇒ 2 (lb + bh + hl)= 108
⇒ 2 (lb + bh + hl) = 96
∴ The total surface area of cuboid = 96 cm2

Question 14.
If the difference of Mode and Median of a data is 24, then the difference of median and mean is: (1)
(a) 8
(b) 12
(c) 24
(d) 36 1
Answer:
(b) 12

Explanation:
Since, Mode = 3 Median – 2 Mean
⇒ Mode = 2 Median + Median – 2 Mean
⇒ Mode – Median = 2 Median – 2 Mean
⇒ Mode – Median = 2 (Median – Mean)
Given, Difference of mode and median of a data is 24
24 = 2 (Median – Mean)
⇒ Median – Mean = 12

Question 15.
The number of revolutions made by a circular wheel of radius 0.25m in rolling a distance of 11km is:
(a) 2800
(b) 4000
(c) 5500
(d) 7000
Answer:
(d) 7000

Explanation:
Given radius = 0.25m.
Distance covered in 1 revolution = 2πr
= 2 × \(\frac{22}{7}\) × 0.25
= \(\frac{11}{7}\)
Given total distance = 11km = 11000m.
Then the number of revolutions = 11000 × \(\frac{11}{7}\)
= \(\frac{77000}{11}\)
= 7000
Therefore the number of revolutions = 7000.

Caution: Remember that circumference of a wheel is the distance covered by the wheel in one round.

Question 16.
For the following distributions, (1)

Class	Frequency
0-5	10
5-10	15
10-15	12
15-20	20
20-25	9

the sum of the lower limits of the median and model class is:
(a) 15
(b) 25
(c) 30
(d) 35
Answer:
(b) 25
Explanation:

Class	Frequency	Cumulative Frequency
0-5	10	10
5-10	15	25
10-15	12	37
15-20	20	57
20-25	9	66

Here N = 66
∴ \(\frac{N}{2}\) = 33, which lies in the interval 10 – 15.
So, the lower limit of the median class is 10. The highest frequency is 20, which lies in the interval 15 – 20.
Therefore, the lower limit of modal class is 15. So, the required sum is 10 + 15 = 25.

Question 17.
Two dice are rolled simultaneously. What is the probability that 6 will come up at least once? (1)
(a) \(\frac{1}{6}\)
(b) \(\frac{7}{36}\)
(c) \(\frac{11}{36}\)
(d) \(\frac{13}{36}\)
Answer:
(c) \(\frac{11}{36}\)
Explanation: Outcomes when 5 will come up at least once are:
(1,6), (2,6), (3,6), (4,6), (5,6), (6,6), (6,1), (6,2), (6,3), (6,4) and (6,5)
Probability that 5 will come up at least once = \(\frac{11}{36}\)

Question 18.
If 5 tan β = 4, then \(\frac{5 \sin \beta-2 \cos \beta}{5 \sin \beta+2 \cos \beta}\) =  (1)
(a) \(\frac{1}{3}\)
(b) \(\frac{2}{5}\)
(c) \(\frac{3}{5}\)
(d) 6
Answer:
(a) \(\frac{1}{3}\)

Explanation:
Given 5 tan β = 4

Direction: In the question number 19 and 20, a statement of assertion (A) is followed by a statement of reason (R).
Choose the correct option as:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)
(b) Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A)
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.

Question 19.
Statement A (Assertion): If product of two numbers is 5780 and their HCF is 17, then their LCM is 340
Statement R (Reason): HCF is always a factor of LCM (1)
Answer:
(b) Both assertion (A) and reason (R) are true and reason (R) is not the correct explanation of assertion (A)

Explanation:
We know that,
HCF × LCM = Product of the numbers
⇒ 17 × LCM = 5780
⇒ LCM = \(\frac{5780}{17}\)
⇒ LCM = 340
The LCM of the two numbers = 340

Question 20.
Statement A (Assertion): If the co-ordinates of the mid-points of the sides AB and AC of ∆ABC are D(3, 5) and E(-3, -3) respectively, then BC = 20 units
Statement R (Reason): The line joining the mid points of two sides of a triangle is parallel to the third side and equal to half of it. (1)
Answer:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)

Explanation:
M and N are the mid-point of AB and AC.
∴ MN || BC (mid-point theorem)
∴ In ∆AMN and ∆ABC
∠A = ∠A (common)
∠AMN = ∠ABC (corresponding angles)
∠ANM = ∠ACB (corresponding angles)
∴ ∆AMN ~ ∆ABC (AAA)
∴ \(\frac{AM}{AB} = \frac{MN}{BC} = \frac{AN}{NC} = \frac{1}{2}\)
∴ \(\frac{MN}{BC}\) = \(\frac{1}{2}\)
BC = 2MN
∴ MN = \(\sqrt{(-3-3)^2+(-3-5)^2}\)
= \( \sqrt{36 + 64} \) = 10
∴ BC = 2 × 10 = 20

SECTION – B
(Section B consists of 5 questions of 2 marks each.)

Question 21.
If 49x + 51y = 499, 51x + 49y = 501, then find the value of x and y (2)

Total Marks	Breakdown (As per CBSE Marking Scheme)
2m (SA-I)	Add and subtract equations to simplify (1m) Calculate x (0.5m) Calculate y (0.5m)”

Answer:
Adding the two equations and dividing by 10,
we get: x + y = 10
Subtracting the two equations and dividing by -2,
we get: x – y =1
Solving these two new equations, we get.
x = \(\frac{11}{2}\)
y = \(\frac{9}{2}\)

Detailed Answer:
The given equations are:
49x + 51y = 499 …(i)
51x + 49y = 501 …(ii)
On adding the equations, we get,
100x + 100y = 1000
Dividing by 100
⇒ x + y = 10 ….(iii)
On subtracting the equations, we get,
-2x + 2y = -2
Dividing by 2
⇒ -x + y = -1 …(iv)
Now, solving equation (iii) and equation (iv),
x + y = 10
-x + y = -1
On adding,
2y = 9
⇒ y = \(\frac{9}{2}\)
Substituting y = \(\frac{9}{2}\) in equation (iii)
⇒ x + \(\frac{9}{2}\) = 10
⇒ x = 10 – \(\frac{9}{2}\)
⇒ x = \(\frac{11}{2}\)
Therefore, the value of x is \(\frac{11}{2}\) and that of y is \(\frac{9}{2}\).

Caution: When we make the coefficients equal then check their signs carefully to decide whether to add or subtract the two equations.

Question 22.
In the given figure below, \(\frac{AD}{AE} = \frac{AC}{BD}\) and M a AE BD ∠1 = ∠2. Show that ∆BAE ~ ∆CAD. (2)

Total Marks	Breakdown (As per CBSE Marking Scheme)
2m (SA-I)	Use AD/AE=AC/BD  to prove AC/AB = AD/AE     (1m) Use SAS similarity criterion to prove ∆BAE ~ ∆CAD (1m)

Answer:
In ∆ABC,
∠1 = ∠2
∴ AB = BD ……(i)
Given \(\frac{AD}{AE} = \frac{AC}{BD}\)
Using equation 0, we get \(\frac{AD}{AE} = \frac{AC}{AB}\) ……(ii)
AE AB „(ii)
In ∆BAE and ∆CAD, by equation (ii),
\(\frac{AC}{AB} = \frac{AD}{AE}\)
∠A = ∠A (common)
∴ ∆BAE ~ ∆CAD [By SAS similarity criterion]

Question 23.
In the given figure, O is the centre of circle. Find ∠AQB, given that PA and PB are tangents to the circle and ∠APB= 75°. (2)

Total Marks	Breakdown (As per CBSE Marking Scheme)
2m (SA-I)	Apply angle sum property to find ∠AOB (1m) Use property of tangent to find ∆AQB (1m)

Answer:
∠PAO = ∠PBO = 90°
(angle b/w radius and tangent)
∠AOB = 105°
(By angle sum property of a triangle)
∠AQB = \(\frac{1}{2}\) × 105° = 52.5°
(Angle at the remaining part of the circle is half the angle subtended by the arc at the centre)

Question 24.
The length of the minute hand of a clock is 6cm. Find the area swept by it when it moves from 7:05 p.m. to 7:40 p.m. (2)

Total Marks	Breakdown (As per CBSE Marking Scheme)
2m (SA-I)	Find the angle swept by the minute hand of the clock in 1 min (0.5m) Find the angle swept by the minute hand of the clock in 35 min (0.5m) Calculate the area swept by the minute hand of the clock (lm)

OR
In the given figure, arcs have been drawn of radius 7cm each with vertices A, B, C and D of quadrilateral ABCD as centres. Find the area of the shaded region.

Total Marks	Breakdown (As per CBSE Marking Scheme)
2m (SA-I)	Write the formula for required area with centre A, B, C and D (0.5m) Substitute the values (0.5m) Apply angle sum property of triangle (0.5m) Calculate the required area and get the desired results (0.5m)

Answer:
We know that, in 60 minutes, the tip of minute hand moves 360°

In 1 minute, it will move = \(\frac{360}{60}\) = 6°
∴ From 7 : 05 pm to 7: 40 pm i.e. 35 min, it will move through = 35 × 6° = 210°
∴ Area of swept by the minute hand in 35 min = Area of sector with sectorial angle 0 of 210° and radius of 6 cm
= \(\frac{210}{320}\) × π × 6²
= \(\frac{7}{12}\) × \(\frac{22}{7}\) × 6 × 6
= 66 cm²
OR
Let the measure of ∠A, ∠B, ∠C and ∠D be θ₁, θ₂, θ₃ and θ₄ respectively Required area = Area of sector with centre A + Area of sector with centre B + Area of sector with centre C + Area of sector with centre D

Question 25.
If sin(A + B) = 1 and cos(A – B) = \(\frac{\sqrt{3}}{2}\), 0°< A + B < 90° and A > B, then find the measures of angles A and B. (2)

Total Marks	Breakdown (As per CBSE Marking Scheme)
2m (SA-I)	Form a pair of linear equations in two variables, namely A and B, from the given relations (0.5m) Add equations to simplify (0.5m) Calculate A (0.5m) Calculate B (0.5m)

OR
Find an acute angle 0 when
\(\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}=\frac{1-\sqrt{3}}{1+\sqrt{3}}\)

Total Marks	Breakdown (As per CBSE Marking Scheme)
2m (SA-I)	Divide numerator and denominator by COS θ (1m) Simplify it and find an acute angle (1m)

Answer:
sin(A + B) = 1 = sin 90, so A + B = 90 ……. (i)
cos(A – B) = \(\frac{\sqrt{3}}{2}\) = cos 30,
so A – B= 30 …(ii)
From (i) & (ii) ∠A = 60°
And ∠B = 30°
Detailed Answer:
We have sin(A + B) = 1
(A + B) = sin^-11
We know, sin^-1 1 = 90°
A + B = 90° ….(i)
cos(A – B) = \(\frac{\sqrt{3}}{2}\)
A – B = cos^-1 \(\frac{\sqrt{3}}{2}\)
we know cos^-1 \(\frac{\sqrt{3}}{2}\) = 30°
solve (i) and (ii)
A + B = 90°
A – B = 30°
add
2A = 120°
A = 60° and ∠B = 30°
OR
\(\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}=\frac{1-\sqrt{3}}{1+\sqrt{3}}\)
Dividing the numerator and denominator of LHS by cos θ, we get
\(\frac{1-\tan \theta}{1+\tan \theta}=\frac{1-\sqrt{3}}{1+\sqrt{3}}\)
Which on simplification (or comparison) gives
tan θ = \( \sqrt{3} \)
Or θ = 60°

SECTION – C
(Section C consists of 6 questions of 3 marks each.)

Question 26.
Given that \( \sqrt{3} \) is irrational, prove that 5 + \(2 \sqrt{3} \) is irrational. (3)

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (SA-II)	Suppose 5 + \(2 \sqrt{3} \) to be a rational number (0.5m) Write it in fraction form i.e., p/q, where p, q are co-primes. (1m) Find prime factors of p and q (0.5m) p and q will have common prime factors which contradict our assumption that p, q are co-primes and 5 + \(2 \sqrt{3} \) is an irrational number (1m)

Answer:
Let us assume 5 + \(2 \sqrt{3} \) is rational, then it must be in the form of \(\frac{p}{q}\) where p and q are coprime
integers and q ≠ 0
i.e 5 + \(2 \sqrt{3} \) = \(\frac{p}{q}\)
So \( \sqrt{3} \) = \(\frac{p-5q}{2q}\) ….(i)
Since p, q, 5 and 2 are integers and q ≠ oR,
HS of equation (i) is rational. But LHS of (i) is \( \sqrt{3} \) which is irrational.
This is not possible. This contradiction has arisen due to our wrong assumption that 5 + 2\( \sqrt{3} \)is rationaL So, 5+2\( \sqrt{3} \) is irrational.

Question 27.
If the zeroes of the polynomial x² + px + q are double in value to the zeroes of the polynomial 2x² – 5x – 3, then find the values of p and q. (3)

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (SA-II)	Find the sum of zeroes of the polynomial (0.5m) Find the product of zeroes of the polynomial (0.5m) Find the sum of zeroes of the polynomial (0.5m) Calculate p (0.5m) Find the product of zeroes of the polynomial (0.5m) Calculate q (0.5m)

Answer:
Let a and p be the zeros of the polynomial 2x² – 5x – 3
Then α + β = \(\frac{5}{2}\)
And αβ = \(– \frac{3}{2}\)
Let 2α and 2β be the zeroes x² + px + q
Then 2α + 2β = -p
2(α + β) = -p
2 × \(\frac{5}{2}\) = -p
So P = -5
And 2α × 2β = q
4αβ = q
So q = 4 × \(– \frac{3}{2}\)
= -6

Question 28.
A train covered a certain distance at a uniform speed. If the train would have been 6 km/h faster, it would have taken 4 hours less than the scheduled time. And, if the train were slower by 6 km/hr ; it would have taken 6 hours more than the scheduled time. Find the length of the journey.

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (SA-II)	Assume the actual speed of the train and actual time taken as x km/h and y hr (0.5m) Form the pair of linear equations from the given situation (1m) Solve   the two equations simultaneously to get the values of x and y (1m) Calculate the length of journey (0.5m)

OR
Anuj had some chocolates, and he divided them into two lots A and B. He sold the first lot at the rate of ₹2 for 3 chocolates and the second lot at the rate off per chocolate, and got a total of ₹400. If he had sold the first lot at the rate of per chocolate, and the second lot at the rate of 4 for 5 chocolates, his total collection would have been ₹460. Find the total number of chocolates he had.

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (SA-II)	Assume the number of chocolates in both lots in form of x and y (0.5m) Form the pair of linear equations from the given situation (1m) Solve the two equations simultaneously to get the values of x and y (1m) Calculate the total number of chocolates Anuj had (0.5m)

Answer:
Let the actual speed of the train be x km/hr and let the actual time taken be y hours Distance covered is xy km If the speed is increased by 6 km/hr, then time of journey is reduced by 4 hours i.e., when speed is (x + 6) km/hr, time of journey is (y – 4) hours.
∴ Distance covered = (x + 6) (y – 4)
⇒ xy = (x + 6)(y – 4)
⇒ -4x + 6y – 24 = 0
⇒ -2x + 3y – 12 = 0 ……(i)
Similarly xy = (x – 6)(y + 6)
⇒ 6x – 6y – 36 = 0
⇒ x – y – 6 = 0
Solving (i) and (ii) we get x = 30 and y = 24
Putting the values of x and y in equation (i), we obtain
Distance = (30 × 24) km = 720 km.
Hence, the length of the journey is 720 km.

OR

Let the number of chocolates in lot A be x
And let the number of chocolates in lot B be y total number of chocolates = x + y
Price of 1 chocolate = = ₹ \(\frac{2}{3}\), so for x chocolates = \(\frac{2}{3}\) x and price of y chocolates at the rate of ₹ 1 per chocolate = y.
∴ by the given condition \(\frac{2}{3}\)x + y = 400
⇒ 2x + 3y = 1200
Similarly
x + \(\frac{4}{3}\) y = 460
⇒ 5x + 4y = 2300 …(ii)
Solving (i) and (ii) we get
x =300 and y = 200
∴ x + y = 300 + 200 = 500
So, Anuj had 500 chocolates.

Question 29.
Prove the following that (3)

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (SA-II)	Use trigonometric identities to start simplifying LHS (1m) Eventually reach LHS = RHS stage (2m)

Answer:

Question 30.
Prove that a parallelogram circumscribing a circle is a rhombus (3)

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (SA-II)	Form equations using properies of tangents with diagram (1m) Use parallelogram property with above equations to achieve ‘equal sides’ (1.5m) Mention the rhombus property and conclude (0.5m)

OR
In the figure XY and XT’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C interesting XY at A and X’Y’ at B, what is the measure of ∠AOB.

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (SA-II)	Do necessary construction in the figure (0.5m) Apply the properties of tangent (0.5m) Use SSS congruency criterion to prove ∆OPA ≅ ∆OCA (lm) Calculate the angle betweem the tangents (0.5m) Use angle sum property, find ∆AOB (0.5m)

Answer:
Let ABCD be the rhombus circumscribing the circle with centre O, such that AB, BC, CD and DA touch the circle at points P, Q, R and S respectively.

We know that the tangents drawn to a circle from an exterior point are equal in length
∴ AP = AS ……..(i)
BP = BQ …(ii)
CR = CQ …(iii)
DR = DS …(iv)
Adding (i), (ii), (iii) and (iv) we get
AP + BP + CR + DR = AS + BQ + CQ + DS (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
AB + CD = AD + BC …(v)
Since AB = DC and AD = BC (opposite sides of parallelogram ABCD)
putting in (v) we get, 2AB = 2AD
or AB = AD.
∴ AB = BC = DC = AD
Since a parallelogram with equal adjacent sides is a rhombus, so ABCD is a rhombus
OR

Join OC
In ∆OPA and ∆OCA
OP = OC (radii of same circle)
PA = CA
(length of two tangents from an external point)
AO = AO (Common)
Therefore, ∆OPA ≅ ∆OCA
(By SSS congruency criterion)
Hence, ∠1 = ∠2 (CPCT)
Similarly
∠3 = ∠4
∠PAB + ∠QBA = 180°
(co interior angles are supplementary as XY || X’Y)
2∠2 + 2∠4 = 180°
∠2 + ∠4 = 90° …(1)
∠2 + ∠4 + ∠AOB = 180° (Angle sum property)
Using (i), we get, ∠AOB = 90°

Question 31.
Two coins are tossed simultaneously. What is the probability of getting
(A) At least one head?
(B) At most one tail?
(C) A head and a tail? (3)
Answer:
(A) P (At least one head) = \(\frac{3}{4}\)
(C) P(A head and a tail) = \(\frac{3}{4}\)
(B) P(At most one tail) =\(\frac{2}{4}\) = \(\frac{1}{2}\)

SECTION – D
(Section D consists of 4 questions of 5 marks each.)

Question 32.
To fill a swimming pool two pipes are used. If the pipe of larger diameter used for 4 hours and the pipe of smaller diameter for 9 hours, only half of the pool can be filled. Find, how long it would take for each pipe to fill the pool separately, if the pipe of smaller diameter takes 10 hours more than the pipe of larger diameter to fill the pool?

Total Marks	Breakdown (As per CBSE Marking Scheme)
5m (LA)	Identify co-relation in time taken by both pipes (0.5m) Form a quadratic equation in x (2m) Solve it to find the values of x (1m) Pick the positive/realistic value of x (0.5m) Calculate the time taken by larger pipe and smaller pipe (1m)

OR
In a flight of 600km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr from its usual speed and the time of the flight increased by 30 min. Find the scheduled duration of the flight. (5)

Total Marks	Breakdown (As per CBSE Marking Scheme)
5m (LA)	Assuming the usual speed of plane and the reduced speed of the plane as x km/hr and (x-200) km/hr hours (0.5m) Form a quadratic equation in x (2m) Solve it to find the values of  x (1m) Pick the positive/realistic value of x (0.5m) Calculate scheduled duration of the flight (1m)

Answer:
Let the time taken by larger pipe alone to fill the tank = x hours
Therefore, the time taken by the smaller pipe = x + 10 hours
Water filled by larger pipe running for 4 hours = \(\frac{4}{x}\) litres
Water filled by smaller pipe running for 9 hours = \(\frac{9}{x+10}\) litres
We know that, \(\frac{4}{x}\) + \(\frac{9}{x+10}\) = \(\frac{1}{2}\)

Which on simplification gives:
x² – 16x – 80 = 0
x² – 20x + 4x – 80 =0
x(x – 20) + 4(x – 20) = 0
(x + 4)(x – 20) = 0
x = -4, 20
x cannot be negative.
Thus, x = 20
x + 10 = 30
Larger pipe would alone fill the tank in 20 hours and smaller pipe would fill the tank alone in 30 hours.

OR

Let the usual speed of plane be x km/hr and the reduced speed of the plane be (x – 200) km/hr
Distance = 600 km [Given]
According to the question,
(time taken at reduced speed) – (Schedule time) = 30 minutes = 0.5 hours
\(\frac{600}{x-200}\) + \(\frac{600}{x}\) = \(\frac{1}{2}\)

Which on simplification gives:
x² – 200x – 240000 = 0
x² – 600x + 400x – 240000 = 0
x(x – 600) + 400(x – 600) = 0
(x – 600)(x + 400) = 0
x= 600 or x= -400
But speed cannot be negative.
∴ The usual speed is 600 km/hr and the scheduled duration of the flight is \(\frac{600}{600}\) = 1 hour

Question 33.
Prove that if a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.
Using the above theorem prove that a line through the point of intersection of the diagonals and parallel to the base of the trapezium divides the non parallel sides in the same ratio. (5)

Total Marks	Breakdown (As per CBSE Marking Scheme)
5m (LA)	Draw a figure with necessary construction (1m) Use AA similarity criterion to prove ∆ADE ~ ∆ABC (1m) Apply basic proportionality theorem to prove AD/ DB = AE/ EC (1.5m) Apply basic proportionality theorem to prove DE/EA= CF/FB (1.5m)

Answer:
For the Theorem :
Given, To prove, Construction and figure
Proof

Let ABCD be a trapezium DC || AB and EF is a line parallel to AB and hence to DC.
To prove : \(\frac{DE}{EA}\) = \(\frac{CF}{FB}\)
Construction : Join AC, meeting EF in G.
Proof: In ∆ABC, we have
GF || AB
\(\frac{CG}{GA}\) = \(\frac{CF}{FB}\) [By BPT] …(i)
In ∆ADC, we have
EG || DC (EF || AB & AB || DC)
\(\frac{DE}{EA}\) = \(\frac{CG}{GA}\) [By BPT] …(ii)
From (i) & (ii), we get,
\(\frac{DE}{EA}\) = \(\frac{CF}{FB}\)

Detailed Answer:
We are given a triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively.

We need to prove that \(\frac{AD}{AE}\) = \(\frac{DB}{EC}\)
Let us join BE and CD and then draw DM ⊥ AC and EN ⊥ AB.
Now, area of ∆ADE (\(\frac{1}{2}\) base × height) = \(\frac{1}{2}\) AD × EN. area of AADE is denoted as ar(ADE).
So, ar(ADE) = \(\frac{1}{2}\) AD × EN

Note that ABDE and ADEC are on the same base DE and between the same base DE and between the same parallels BC and DE.
So, ar(BDE) = ar(DEC) …(iii)
Therefore, form (i), (ii) and (iii), we have:
\(\frac{AD}{DB}\) = \(\frac{AE}{EC}\)
Let ABCD be a trapezium with AB||DC
AD and BC are non-parallel sides.
Let, EF be a line parallel to AB, DC.
Join AC such that ∆ACD and ∆ACB are triangles and AC meets EF at G.

Apply Thales theorem on ∆ACD and ∆ACB Basic Proportionality Theorem: In a triangle, a line drawn parallel to one side to intersect the other sides distinct points divides two sides in same ratio.
In ∆ACD
Here, EG||DC. According to Basic Proportionality Theorem
\(\frac{AE}{ED}\) = \(\frac{AG}{GC}\) ……….(i)

In ∆ACB
Here, GF||AB. According to Basic Proportionality Theorem
\(\frac{EC}{BF}\) = \(\frac{AG}{GC}\)
⇒ \(\frac{BF}{FC}\) = \(\frac{AG}{GC}\)
Equating equation (i) and equation (ii)
\(\frac{AE}{ED}\) = \(\frac{BF}{FC}\)
Hence proved that a line drawn parallel to the parallel sides of a trapezium divides the non-parallel sides proportionally.

Question 34.
Due to heavy floods in a state, thousands were rendered homeless. 50 schools collectively decided to provide place and the canvas for 1500 tents and share the whole expenditure equally. The lower part of each tent is cylindrical with base radius 2.8 m and height 3.5 m and the upper part is conical with the same base radius, but of height 2.1 m. If the canvas used to make the tents costs ?120 per m², find the amount shared by each school to set up the tents. (5)

Total Marks	Breakdown (As per CBSE Marking Scheme)
5m (LA)	Write values and slant height of conical part formula (1m) Calculate height (1m) Deduct Area of Canvas formula and find value (2m) Use unit cost to calculate final value (1m)

OR
There are two identical solid cubical boxes of side 7cm. From the top face of the first cube a hemisphere of diameter equal to the side of the cube is scooped out. This hemisphere is inverted and placed on the top of the second cube’s surface to form a dome. and
(A) the ratio of the total surface area of the two new solid formed
(B) volume of each new solid formed. (5)

Total Marks	Breakdown (As per CBSE Marking Scheme)
5m (LA)	(A) Calculate the surface area new solid of first (1m) Calculate the surface area of second new solid (1m) Find the ratio of the total surface area of two new solids (1m) (B) Calculate the volume for first new solid  (1m) Calculate the volume for second new solid (1m)

Answer:
Radius of the base of cylinder (r) = 2.8 m = Radius of the base of the cone (r)
Height of the cylinder (h) = 3.5 m
Height of the cone (H) = 2.1 m.
Slant height of conical part (l) = \( \sqrt{r^{2} + H^{2}} \)
= \( \sqrt{2.8^{2} + 2.1^{2}} \)
= \( \sqrt{7.84 + 4.41} \)
= \( \sqrt{12.25} \) = 3.5 m

Area of canvas used to make tent = CSA of cylinder + CSA of cone
2 × π × 2.8 × 3.5 + π × 2.8 × 3.5
= 61.6 + 30.8 = 92.4 m²
Cost of 1500 tents at 120 per sq.m = 1500 × 120 × 92.4 = 16,632,000

Share of each school to set up the tents \(\frac{16632000}{50}\)
= ₹ 332,640

OR

(A) SA for first new solid (S₁):
6 × 7 × 7 + 2π × 3.52 – π × 3.5²
= 294 + 77 – 38.5 = 332.5 cm²

SA for second new solid (S₂):
6 × 7 × 7 + 2 π × 3.5² – π × 3.5²
= 294 + 77 – 38.5 = 332.5 cm²
So S₁ : S₂ = 1 : 1

(B) Volume for first new solid (V₁)
= 7 × 7 × 7 – \(\frac{2}{3}\) π × 3.5³
= 343 – \(\frac{539}{6}\)
= \(\frac{1519}{6}\) cm³

Volume for second new solid (V₂)
= 7 × 7 × 7 + \(\frac{2}{3}\) π × 3.5³
= 343 + \(\frac{539}{6}\) = \(\frac{2597}{6}\) cm³

Question 35.
The median of the following data is 525. Rnd the values ofx and y, if the total frequency is 100 (5)

Class interval	Frequency
0-100	2
100-200	5
200-300	x
300-400	12
400-500	17
500-600	20
600-700	y
700-800	9
800-900	7
900-1000	4

Total Marks	Breakdown (As per CBSE Marking Scheme)
5m (LA)	Make frequency distribution table with all given data (2m) Using the sum of frequency, make an equation with variables x and y (0.5m) Apply the formula of median to form another equation with variables x and y (1.5m) Solve the two equations to find the values of x and y (1m)

Median = 525, so Median Class = 500 – 600

Total Marks	Breakdown (As per CBSE Marking Scheme)
5m (LA)	(A) Apply the distance formula (0.5m) (B) Apply the mid-point formula (0.5m) Find the coordinates of Z (0.5m) (C) Equate the distances of the point (x,y) from two points (0.5m) Apply the distance formula (0.5m) Get the coordinates of Z on y-axis (1m) OR Find the length of GH and AF (1m) Calculate the area of trapezium AFGH (1m)

76 + x + y = 100
x + y = 24
Median = \(l+\frac{\frac{n}{2}-c f}{f} \times h\)
Since, l = 500, h = 100, f = 20, cf = 36 + x and n = 100
Therefore, putting the value in the Median formula, we get; 525 = \(\frac{\frac{50-(36+x)}{20}} \times 100\)
so x = 9, y = 24 – x [from eq. (i)]
y = 24 – 9 = 15
Therefore, the value ofx = 9 and y = 15.

SECTION – E
(Case Study Based Questions)
(Section E consists of 3 questions. All are compulsory.)

Question 36.
A tiling or tessellation of a flat surface is the covering of a plane using one or more geometric shapes, called tiles, with no overlaps and no gaps. Historically, tessellations were used in ancient Rome and in Islamic art. You may find tessellation patterns on floors, walls, paintings etc. Shown below is a tiled floor in the archaeological Museum of Seville, made using squares, triangles and hexagons.

A craftsman thought of making a floor pattern after being inspired by the above design. To ensure accuracy in his work, he made the pattern on the Cartesian plane. He used regular octagons, squares and triangles for his floor tessellation pattern

Use the above figure to answer the questions that follow:
(A) What is the length of the line segment joining points B and F? (1)
(B) The centre ‘Z’of the figure will be the point of intersection of the diagonals of quadrilateral WXOP. Then what are the coordinates of Z? (1)
(C) What are the coordinates of the point on y axis equidistant from A and G?

OR

What is the area of Trapezium AFGH?

Total Marks	Breakdown (As per CBSE Marking Scheme)
5m (LA)	(A) Find ‘a’ and ‘d’ (0.5m) Calculate nth term (0.5m) (B) Apply the formula for sum of terms and substitute the values (1m) Reject negative value and pick answer (1m) OR Write the formula for sum of n terms of A.P. (1m) For n=10, find the sum (1m) (C) Find the middle row (0.5m) Calculate the nth term (0.5m)

Answer:
(A)
B(1,2), F(-2,9)
BF² = (-2 – 1)² + (9 – 2)²
= (-3)² + (7)²
= 9 + 49
= 58

So, BF = \( \sqrt{58} \) units

(B) W(-6, 2), X(-4, 0), 0(5, 9), P(3, 11)
Clearly WXOP is a rectangle

Point of intersection of diagonals of a rectangle is the mid point of the diagonals. So the required point is mid point of WO or XP
= \(\left(\frac{-6+5}{2}, \frac{2+9}{2}\right)\)
= \(\left(\frac{-1}{2}, \frac{11}{2}\right)\)

(C) A(-2, 2), G(-4, 7)
Let the point on y-axis be Z(0, y)
AZ² = GZ²
(0 + 2)² + (y – 2)² = (0 + 4)² + (y – 7)²
(2)² + y² + 4 – 4y = (4)² + y² + 49 – 14y
8 – 4y = 65 – 14y
10y = 57
So, y = 5.7
i.e. the required point is (0, 5.7)

OR

A(-2, 2), F(-2, 9), G(-4, 7), H(-4, 4)
Clearly GH = 7 – 4 = 3units
AF = 9 – 2 = 7 units
So, height of the trapezium AFGH = 2 units
So, area of AFGH = \(\frac{1}{2}\)(AF + GH) × height
= \(\frac{1}{2}\)(7+3) × 2
= 10 sq. units

Question 37.
The school auditorium was to be constructed to accommodate at least 1500 people. The chairs are to be placed in concentric circular arrangement in such a way that each succeeding circular row has 10 seats more than the previous one.

(A) If the first circular row has 30 seats, how many seats will be there in the 10th row? (1)
(B) For 1500 seats in the auditorium, how many rows need to be there?
OR
If 1500 seats are to be arranged in the auditorium, how many seats are still left to be put after 10th row? (2)
(C) If there were 17 rows in the auditorium, how many seats will be there in the middle row? (1)

Total Marks	Breakdown (As per CBSE Marking Scheme)
4m (CBQ)	(A) Find ‘a’ and’d’ (0.5m) Calculate nth term (0.5m) (B) Apply the formula for sum of n terms and substitute the values (lm) Reject negative value and pick answer (lm) OR Write the formula for sum of n terms of A.P. (lm) For n=10, find the sum (lm) (C) Find the middle row (0.5m) Calculate the nth term (0.5m)

Answer:
(A) Since each row is increasing by 10 seats, so it is an AP with first term a = 30, and common difference d = 10.
So number of seats in 10th row
= a₁₀= a + 9d
= 30+ 9 × 10 = 120

(B) S_n = \(\frac{n}{2}\)[2a+(n-1)d]
1500 = \(\frac{n}{2}\)[2×30+(n-1)10]
3000 = 50n + 10n²
n² + 5n – 300 = 0
n² + 20n – 15n – 300 = 0
(n + 20) (n – 15) = 0
Rejecting the negative value, n = 15

OR

No. of seats already put up to the 10th row = S₁₀
S₁₀ = \(\frac{10}{2}\)[2×30 + (10-1)10]
= 5(60 + 90) = 750
So, the number of seats still required to be put are 1500 – 750 = 750

(C) If no. of rows =17
then the middLe row is the 9th row
a₈ = a + 8d
= 30 + 80
= 110 seats

Question 38.
We all have seen the airplanes flying in the sky but might have not thought of how they actually reach the correct destination. Air Traffic Control (ATC) is a service provided by ground-based air traffic controllers who direct aircraft on the ground and through a given section of controlled airspace, and can provide advisory services to aircraft in non- controlled airspace. Actually, all this air traffic is managed and regulated by using various concepts based on coordinate geometry and trigonometry.

At a given instance, ATC finds that the angle of elevation of an airplane from a point on the ground is 60°. After a flight of 30 seconds, it is observed that the angle of elevation changes to 30°. The height of the
plane remains constantly as 3000√2 m. Use the above information to answer the questions that follow:
(A) Draw a neat labelled figure to show the above situation diagrammatically. (1)
(B) What is the distance travelled by the plane in 30 seconds?
OR
Keeping the height constant, during the above flight, it was observed that after 15( √3 -1) seconds, the angle of elevation changed to 45°. How much is the distance travelled in that duration. (2)
(C) What is the speed of the plane in km/hr. (1 )

Total Marks	Breakdown (As per CBSE Marking Scheme)
4m (CBQ)	(A) Draw a correct figure (1m) (B) Apply trigonometric ratios to find the length AB (1m) Calculate the distance travelled by the plane (1m) OR Apply trigonometric ratios to find the length AB (1m) Calculate the distance travelled by the plane (1m) (C) Calculate the speed in km/hr (1m)

Answer:

OR

OR