CBSE Sample Papers for Class 10 Maths Basic Set 3 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths with Solutions Set 3 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Maths Set 3 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.
The value of \(\frac{\cot A+\tan B}{\cot B+\tan A}\) is: [1]
(a) cot A tan B
(b) cot B tan A
(c) cot A
(d) tan B
Answer:
(a) cot A tan B

Explanation:

Question 2.
For any two positive integers ‘a’ and ‘b’, what is the value of HCF (a, b) x LCM (a, b)? [1]
(a) ab²
(b) a²b
(c) a × b
(d) a + b
Answer:
(c) a × b

Explanation:
We know the relation,
HCF (a, b) × LCM (a, b) = Product of a and b.

Question 3.
The mean of first 10 odd natural numbers is: [1]
(a) 20
(b) 40
(c) 30
(d) 10
Answer:
(d) 10

Explanation:
First ten odd numbers are: 1, 3, 5, ………. 19.
Their sum = \(\frac{10}{2}\) (2 × 1 + 9 × 2) = 100
Mean = \(\frac{100}{10}\) = 10

Question 4.
Using prime factorisation method, the HCF and LCM of 210 and 175 is: [1]
(a) 35, 1000
(b) 34, 1050
(c) 35, 1050
(d) 35, 1010
Answer:
(c) 35, 1050

Explanation: The prime factorisations of 210 and 175 are:
210 = 2 × 3 × 5 × 7
175 = 5 × 5 × 7
So, HCF (210, 175) = 5 × 7 = 35; and
LCM (210, 175) = 2 × 3 × 5 × 7 × 5 = 1050

Question 5.
If 2x, x + 10, 3x + 2 are in A.P., the value of x is: [1]
(a) 6
(b) 7
(c) 9
(d) 10
Answer:
(a) 6

Explanation:
Since 2x, x + 10, 3x + 2 are in A.P.,
∴ 2(x + 10) = 2x + (3x + 2)
⇒ 2x + 20 = 5x + 2
⇒ 3x = 18
⇒ x = 6

Question 6.
By taking A = 90° and B = 30°, sin A cos B – cos A sin B is: [1]
(a) 0
(b) \(\frac{\sqrt{3}}{2}\)
(c) \(\frac{1}{2}\)
(d) \(\frac{1}{\sqrt{2}}\)
Answer:
(b) \(\frac{\sqrt{3}}{2}\)

Explanation:
sin A cos B – cos A sin B
= sin 90° cos 30° – cos 90° sin 30°
= 1 × \(\frac{\sqrt{3}}{2}\) – 0 × \(\frac{1}{2}\) = \(\frac{\sqrt{3}}{2}\)

Question 7.
The degree of the polynomial (x + 1) (x² – x – x⁴ + 1) is: [1]
(a) 4
(b) 2
(c) 5
(d) 3
Answer:
(c) 5

Explanation:
(x + 1)(x² – x – x⁴ + 1) = (x³ – x² – x⁵ + x) + (x² – x – x⁴ + 1)
= – x⁵ – x⁴ + x³ + 1
So, the degree of the polynomial is 5.

Question 8.
The ratio of the volume of a right circular cone to that of the volume of right circular cylinder, of equal diameter and height is: [1]
(a) 4 : 3
(b) 1: 4
(c) 2 : 3
(d) 1: 3
Answer:
(d) 1 : 3

Explanation:
Volume of a right circular cone = \(\frac{1}{3}\) ≠ r²h
Volume of a right circular cylinder = πr²h
So, required ratio is 1: 3.

Question 9.
The value of (cos 0° + sin 45° + sin 30°) (sin 90° – cos 45° + cos 60°) is: [1]
(a) \(\frac{1}{4}\)
(b) \(\frac{3}{4}\)
(c) \(\frac{5}{4}\)
(d) \(\frac{7}{4}\)
Answer:
(d) \(\frac{7}{4}\)

Explanation:
(cos 0° + sin 45° + sin 30°)
(sin 90° – cos 45° + cos 60°)

Question 10.
A quadratic polynomial sum of whose zeros is 3 and product is – 6 is: [1]
(a) x² – 3x – 6
(b) 2x²2 + 3x + 6
(c) x² + 3x + 6
(d) x² – 6x – 3
Answer:
(a) x² – 3x – 6

Explanation:
Here, sum of zeros = 3
Product of zeros = – 6
∴ Quadratic polynomial is
x² – (sum of roots) x + product of roots
⇒ x² – 3x – 6, is the required quadratic polynomial

Question 11.
The value of ‘k’ for which the pair of linear equations kx – y = 2 and 6x – 2y = 3 will have infinitely many solutions is: [1]
(a) 4
(b) 3
(c) 2
(d) None of these
Answer:
(d) None of these

Explanation:
The given pair of equations will have infinitely many solutions when
\(\frac{k}{6}=\frac{-1}{-2}=\frac{2}{3}\)
No value of k satisfy the above relation.
So, no value of k exist.

Question 12.
The probability of getting a number which is neither prime nor composite in a single throw of a fair dice is: [1]
(a) \(\frac{1}{5}\)
(b) \(\frac{2}{5}\)
(c) \(\frac{1}{6}\)
(d) \(\frac{1}{4}\)
Answer:
(c) \(\frac{1}{6}\)

Explanation:
Total number of outcomes on rolling a fair dice = 6
∵ 1 is only the number which is neither prime nor composite.
So, required probability is \(\frac{1}{6}\).

Question 13.
The mean of the following data is: [1]
1, 7, 9, 3, 4, 5, 6
(a) 4
(b) 2
(c) 3
(d) 5
Answer:
(d) 5

Explanation:
We know,
Mean = \(\frac{\text { Sum of observation }}{\text { Total no. of observation }}\)
= \(\frac{1+7+9+3+4+5+6}{7}\)
= \(\frac{35}{7}\) = 5

Question 14.
One card is drawn at random from a well shuffled deck of 52 cards. What is the probability to get a face card ? [1]
(a) \(\frac{1}{13}\)
(b) \(\frac{3}{13}\)
(c) \(\frac{2}{13}\)
(d) \(\frac{4}{13}\)
Answer:
(b) \(\frac{3}{13}\)

Explanation:
In a deck of 52 cards, number of face cards is 12.
So, the required probability = \(\frac{12}{52}\), i.e. \(\frac{3}{13}\)

Question 15.
A solid cube is cut into two cuboids of equal volumes. The ratio of surface areas of the given cube and one of the resulting cuboid is: [1]
(a) 2 : 3
(b) 1: 3
(c) 3 : 2
(d) 3 : 1
Answer:
(c) 3 : 2

Explanation:
Since the cube is cut into two cuboids of equal volumes, so the two cuboids are equal.

Let the edge of the cube be 2a units and the cube be cut along its length.
∴ Dimensions of each cuboid formed are a × 2a × 2a.
So, Surface area of cuboid
= 2 (a × 2a + 2a × 2a + 2a × a)
= 16a²
Also, Surface area of cube
= 6(2a)² = 24a²
∴ Required ratio = 24a² : 16a² = 3 : 2.

Question 16.
An unbiased dice is rolled once. What is the probability of getting a prime number? [1]
(a) \(\frac{3}{2}\)
(b) \(\frac{5}{2}\)
(c) \(\frac{1}{3}\)
(d) \(\frac{1}{2}\)
Answer:
(d) \(\frac{1}{2}\)

Explanation:
When a dice is rolled, the total outcomes are 1, 2, 3, 4, 5 and 6.
∴ Total outcomes = 6
Favourable outcomes = 2, 3, 5 i.e. 3
∴ P (getting a prime number) = \(\frac{3}{6}=\frac{1}{2}\)
Hence, required probability is \(\frac{1}{2}\)

Question 17.
If the ratio of the Length of a rod to its shadow ¡s 1 : √3, then, angle of elevation of the sun is: [1]
(a) 30
(b) 60°
(c) 45°
(d) 900
Answer:
(a) 30°

Explanation:
Let θ be the angle of elevation of the sun.

Then, tan θ = \(\frac{BC}{AB}\)
= \(\frac{1}{\sqrt{3}}\)
θ = 30°

Question 18.
A rolling pin is made by joining three cylindrical pieces of wood, as shown in the figure: [1]

Assuming that there is no wastage of wood, the volume of wood used in making the rolling pin is:
(a) 64π cm³
(b) 280π cm³
(c) 480π cm³
(d) 544π cm³
Answer:
(d) 544 π cm³

Explanation:
For the bigger cylinder:
r₁ = 4 cm
h₁ = 30 cm
For the two smaller cylinders:
r₂ = \(\frac{4}{2}\) = 2 cm,
h₂ = 8 cm
Now, Volume of wood used = Volume of the pin
= Volume of bigger cylinder + 2x volume of smaller cylinder
= πr₁²h₁ + 2 × πr₂²h₂
= π × (4)² × 30 + 2 × π × (2)² × 8
= 480π + 64π
= 544 π cm³

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): The base radii of two right circular cylinders of the same height are in the ratio 3 : 5. The ratio of their curved surface area is 3 : 5.
Reason (R): CSA of right circular cylinder is 2πr²h. [1]
Answer:
(c) Assertion (A) is true but reason (R) is false.

Explanation:
Let r₁ h₁ and r₂, h₂ be the radii and the heights of the first and second cylinders respectively.

Question 20.
Assertion (A): If the probability of the occurence of an event is \(\frac{3}{7}\), then the probability of its non-occurence is \(\frac{5}{7}\).
Reason (R): P(A) + P(Ā) = 1, when P(Ā) is a complement. [1]
Answer:
(d) Assertion (A) is false but reason (R) is true.

Explanation:
For an event A, the P(A) + P(Ā) = 1
Here, P(A) = \(\frac{3}{7}\)
Then, P(Ā) = 1 – \(\frac{3}{7}\)
= \(\frac{4}{7}\)

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

Question 21.
Determine the A.P. whose 3^rd term is 5 and the 7^th term is 9. [2]
Answer:
Here, a₃ = a + 2d = 5
and a₇ = a + 6d = 9
Solving the two equations, we get:
a = 3, d = 1
So, required A.P. is 3, 4, 5, 6, …………

Question 22.
Find the zeros of 3x² – x – 4.
Answer:
Let p(x) = 3x² – x – 4
= 3x² – 4x + 3x – 4
= x(3x – 4) + 1 (3x – 4)
= (3x – 4) (x + 1)
So, the two zeros of 3x² – x – 4 are \(\frac{4}{3}\) and – 1.

Question 23.
Find the coordinates of the point which divides the line joining (1, -2) and (4, 7) internally in the ratio 1: 2.
OR
Find the third vertex of a triangle, if two of its vertices are at (-3, 1) and (0, -2) and the centroid is at the origin. [2]
Answer:
Let P(x, y) divide AB in the ratio 1 : 2. Then,

Let the third vertex be (x, y). Then,
\(\left(\frac{x-3+0}{3}, \frac{y+1-2}{3}\right)\) = (0, 0)
⇒ \(\frac{x-3}{3}\) = 0 and \(\frac{y-1}{3}\) = 0
⇒ x = 3 and y = 1
Thus, the third vertex is (3, 1).

Question 24.
Find the area of the shaded region:

OR
Two dice are thrown simultaneously and the outcomes are noted. Find the probability that: [2]
(A) doublets are obtained
(B) sum of numbers on the two dice is 5.
Answer:
Area of the shaded region = Area of semi-circle of radius 14 cm + 2 × Area of semi-circle of radius 7 cm.
= (\(\frac{\pi}{2}\)(14)² + 2 × \(\frac{\pi}{2}\)(7)²) sq.cm
= (\(\frac{1}{2} \times \frac{22}{7}\) × 14 × 14 + \(\frac{22}{7}\) × 7 × 7) sq.cm
= (308 + 154) sq.cm
= 462 sq.cm

OR

On throwing two dice together, Total number of outcomes = 36
(A) Favourable outcomes = {(1, 1), (2, 2), (3, 3), (4, 4). (5, 5), (6, 6)}.
⇒ Number of favourable outcomes = 6 6 1
∴ P (a doublet) = \(\frac{6}{36}\) i.e., \(\frac{1}{6}\)

(B) Favourable outcomes = {(1, 4), (2, 3), (3, 2), (4,1)}.
⇒ Number of favourable outcomes = 4
∴ P (sum of 5 on two numbers)
= \(\frac{4}{36}\) i.e., \(\frac{1}{9}\)

Question 25.
Amrish wakes up in the morning and notices that his digital clock reads 07: 25 am. After noon, he looks at the clock again. [2]

What is the probability that:
(A) the number in column A is 4?
Answer:
The number in column A can be 0,1, 2, 3, 4 and 5.
So, P(4) = \(\frac{1}{6}\)

(B) the number in column B is 8?
Answer:
The number in column B can be 0, 1, 2, 3, …….. 9.
So, P(8) = \(\frac{1}{10}\)

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

Question 26.
On a morning walk, three girls step off together and their steps measure 40 cm, 42 cm and 45 cm respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps? [3]
Answer:
Required minimum distance
= LCM (40, 42, 45)
∵ 40 = 2 × 2 × 2 × 5 i.e., 2³ × 5
42 = 2 × 3 × 7
∴ 45 = 3 × 3 × 5 i.e., 3² × 5
LCM (40, 42, 45) = 2³ × 3² × 5 × 7
= 2520

Question 27.
There is a circular park of radius 24 m and there is a pole at a distance of 26 m from the centre of the park as shown in the figure. It is planned to enclose the park by planting trees along line segments PQ and PR tangential to the park. [3]

(A) Find the length of PQ and PR;
Answer:
In right triangle PRO,
We have PR = \(\sqrt{P O^2-R O^2}\)

= \(\sqrt{26^2-24^2}\)
= \(\sqrt{676-576}\)
= √100
= 10 m
⇒ PR + PQ =10 m

(B) If six trees are to be planted along each tangential line segments at equal distances, find the distance between any two consecutive trees.
Answer:
As six trees are to be planted along PQ and PR each.

Let’s assume the distance between consecutive trees is x.
Total trees are at 5 equal distances.
Hence, 5x = 10
x = 2

Question 28.
In the figure, sectors of two concentric circles of radii 7 cm and 3.5 cm are shown. [3]

Find the area of the shaded region.
Answer:
Area of the shaded region
= Area of the sector of radius 7 cm – Area of the sector of radius 3.5 cm
= (\(\frac{30^{\circ}}{360^{\circ}}\)π(7)² – \(\frac{30^{\circ}}{360^{\circ}}\)π(3.5)²)sq.cm
= \(\frac{30}{360} \times \frac{22}{7}\) × (49 – 12.25) sq.cm
= 9.625 sq.cm

Question 29.
The sum of reciprocals of a child’s age (in years) 3 years ago and 5 years from now is \(\frac{1}{3}\). Find his present age. [3]
OR
Solve for x and y:
7x – 4y = 49; 5x – 6y = 57
Answer:
Let the present age of the child (in years) be x. Then,
\(\frac{1}{x-3}+\frac{1}{x+5}\) = \(\frac{1}{3}\)
⇒ \(\frac{(x+5)+(x-3)}{(x-3)(x+5)}\) = \(\frac{1}{3}\)
⇒ 3(2x + 2) + (x – 3) (x + 5)
⇒ 6x + 6 = x² + 2 x + 15
⇒ x² – 4x – 21 = 0
⇒ x² – 7x + 3x – 21 = 0
⇒ x(x – 7) + 3(x – 7) = 0
⇒ (x – 7) (x + 3) = 0
⇒ x – 7 = 0 or x + 3 = 0
⇒ x = 7 or x = – 3
(x ≠ – 3, as age cannot be negative)
Thus, present age of child is 7 years.

OR

Given equations are:
7x – 4y = 49 ….. (i)
5x – 6y = 57 ….. (ii)
Multiplying eq. (i) by 5 and eq. (ii) by 7, we get:
35x – 20y = 245 …..(iii)
35x – 42y = 399 …… (iv)
Subtracting equation (iv) from equation (iii), we get
⇒ 22y = – 154
⇒ y = – 7
Substituting y = – 7 in eq. (i), we get
7x + 28 = 49
⇒ 7x = 21
⇒ x = 3
Thus, x = 3, y = – 7 is the required solution.

Question 30.
ABP and ACQ are two tangents to the circle with O as its centre in the given figure. If ∠TCQ = 60° and ∠TBP = 50°, then find the measure of ∠BTC. [3]

Answer:

Join OB, OT and OC
We know, tangent is perpendicular to radius at the point of contact.
∴ OB ⊥ AP and OC ⊥ AQ
∴ ∠OBP = 90°
⇒ ∠OBT + ∠TBP = 90°
⇒ ∠OBT + 50° = 90°
⇒ ∠OBT = 90° – 50° = 40°
Similarly, ∠OCQ = 90° and ∠TCQ = 60°
∴ ∠OCT = 30°
Now, in ∆OBT
OB = OT (Radii)
∠OTB = ∠OBT
[Equal angles opposite to equal sides]
⇒ ∠OTB = 40°
Similarly in AOTC
OT = OC
⇒ ∠OCT = ∠OTC = 30°
So, ∠BTC = ∠OTB + ∠OTC
= 40° + 30° = 70°

Question 31.
A natural number, when increased by 12, equals 160 times its reciprocal. Find the number.
OR
Find two numbers whose sum is 27 and product is 182. [3]
Answer:
Let the natural number be x. Then,
x + 12 = 160 × \(\frac{1}{x}\)
⇒ x² + 12x – 160 = 0
⇒ x² + 20x – 8x – 160 = 0
⇒ x(x + 20) – 8 (x + 20) = 0
⇒ (x + 20) (x – 8) = 0
⇒ x + 20 = 0
⇒ x = – 20
(Rejected, as x is a natural number)
or x – 8 = 0
⇒ x = 8
Thus, the required natural number is 8.

OR

Let first number be x and let second number be (27 – x)
According to given condition, the product of two numbers is 182.
Therefore,
x(27 – x) = 180
⇒ 27x – x² = 182
⇒ x² – 27x + 182 = 0
⇒ x² – 14x – 13x + 182 = 0
⇒ x(x – 14) – 13(x – 14)= 0
⇒ (x – 14) (x – 13)= 0
⇒ x = 14, 13
Therefore, the first number is equal to 14 or 13
And, Second number is = 27 – x = 27 – 14 = 13 or Second number = 27 – 13 = 14
Therefore, two numbers are 13 and 14.

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

Question 32.
A number consists of two digits. When it is divided by the sum of the digits, the quotient is 6 with no remainder. When the number is diminished by 9, the digits are reversed. Find the number. [5]
Answer:
Let the number be ab, i.e., 10a + b
As per the question,
\(\frac{10 a+b}{a+b}\) = 6 or 10a + b = 6a + 6b
⇒ 4a = 5b ….. (1)
(10a + b) – 9 = 10b + a
or 9a = 9b + 9
or a = b + 1 ….. (2)
Solving Eq. (1) and Eq. (2), we get
b = 4 and a = 5.
So, the required number is 54.

Question 33.
Amit and Prem were very good cricketers and also represented their school team at district and even state level. One day, after their match, they measured the height of the wickets and found it to be 28 inches. They marked a point P on the ground as shown in the figure below:

(A) If cot P = \(\frac{3}{4}\), then find the length of P
(B) Find the value of cosec P.
(C) Find the value of \(\frac{1+\sin \mathrm{P}}{1+\cos \mathrm{P}}\).
(D) Find the value of sec R.
(E) Find The value of coses² R – cot² R.
OR
Two poles of equal heights are standing opposite to each other on either side of the road which is 80 m wide. From a point between them on the road , the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distance of the point from the two poles.
Answer:
(A) It is given that QR = 28 inches and
cot P = \(\frac{3}{4}\)
We know that cot P = \(\frac{\text { Base }}{\text { Perpendicular }}\)
= \(\frac{P Q}{Q R}=\frac{P Q}{28}\)

Therefore, \(\frac{P Q}{28}=\frac{3}{4}\)
⇒ PQ = \(\frac{28 \times 3}{4}\) = 21 inches

(B) To evaluate cosec P, we will first find PR
by applying Pythagoras theorem ∆PQR.
PR² = PQ² + QR²
= 21² + 28²
= 441 + 784
= 1225
⇒ PR = 35 inches
∴ cosec P = \(\frac{\text { Hypotenuse }}{\text { Perpendicular }}\) = \(\frac{\mathrm{PR}}{\mathrm{QR}}=\frac{35}{28}=\frac{5}{4}\)

(C)

(D) sec R = \(\frac{\text { Hypotenuse }}{\text { Base }}\) = \(\frac{\mathrm{PR}}{\mathrm{QR}}\)
= \(\frac{35}{28}=\frac{5}{4}\)

(E)

In the figure. AB and XY are two poles of equal height, say ‘h’ metres.

Here, AX represents the width of the road and P is a point on the road.
In ∆ BAP,

⇒ h = 20√3 metres
Thus, the height of each pole is 20√3 metres.
From eqn. (i) and (ii), we aLso have,
AP = \(\frac{20 \sqrt{3}}{\sqrt{3}}\) i.e. 20 metres
and XP = √3 × 20√3, i.e. 60 metres.
Thus, distance, of the point from the two poles are 20 metres and 60 metres.

Question 34.
State and prove Basic Proportionality Theorem. [5]
Answer:
Statement:
If a line is drawn parallel to one-side of a triangle intersecting the other sides, the other two sides are divided in the same ratio.

Proof:
Consider a ∆ABC is which DE || BC
To prove: \(\frac{A D}{B D}=\frac{A E}{E C}\)

Construction: Draw EM ⊥ AB, DN ⊥ AC and join BE and CD.
Proof: We know,
Area of triangle = \(\frac{1}{2}\) × AD × EM
Also, (∆ADE) = \(\frac{1}{2}\) × AE × DN
Similarly, ar (ABDE) = \(\frac{1}{2}\) × BD × EM
And, ar (ADEC) = \(\frac{1}{2}\) × EC × DN
Now, triangles BDE and DEC are on same base DE and lying between some parallels DE and BC.
∴ ar (∆BDE) = ar (∆DEC)

Question 35.
The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the missing frequencies fi and

OR
A test tube has a hemisphere-shaped lower portion and a cylindrical upper portion with the same radius. The test tube fills up to the point of being exactly full when \(\frac{4554}{7}\) cu cm of water is poured, but when only 396 cu cm is added, 9 cm of the tube is left empty. Calcualte the test tube’s radius and the cylindrical part’s height. [5]
Answer:
The frequency distribution for calculating the mean, for the given data is:

We know that
30 + f₁ + f₂ = 50
⇒ f₁ + f₂ = 20
⇒ f₂ = 20 – f₁
Now, mean = A + \(\frac{\Sigma f_i d_i}{\Sigma f_i}\)
⇒ 62.8 = 50 + \(\frac{560+20 f_2-20 f_1}{50}\)
⇒ 12.8 = 50 + \(\frac{560+20\left(20-f_1\right)-20 f_1}{50}\) [Using (i)]
⇒ 640 = 960 – 40 f₁
⇒ 40 f₁ = 320
f₁ = 8
∴ f₂ = 20 – 8 = 12
∴ f₁ = 8, f₂ = 12.

OR

Volume of water that can fill the test tube = \(\frac{4554}{7}\) cm³

Let r’ be the radius of cylinder and hemispherical part.
And let ‘h’ be height of cylinderical part

= 21 cm
Hence, radius of the test tube is 3 cm and height of cylinderical part is 21 cm.

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

Question 36.
Rishu is riding in a hot air balloon. After reaching a point P. he spots a car parked at B on the ground at an angle of depression of 30°. The balloon rises further by 50 metres and now he spots the same car at an angle of depression of 45° and a lorry parked at S’ at an angle of depression of 30°. (Use √3 = 1.73)

The measurement of Rishu facing vertically is the height. Distance is defined as the measurement of car/lorry from a point in a horizontal direction. If an imaginary line is drawn from the observation point to the top edge of the car/lorry, a triangle is formed by the vertical, horizontal and imaginary line.
On the basis of the above information, answer the following questions:
(A) If the height of the balloon at point P is ‘h’m and distance AB is’x’ m, then find the relation between ‘x’ and ‘h’: [1]
(B) What is the relation between the height of the balloon at point P and distance AB. [1]
(C) Find the height of the balloon at point P and the distance AB on the ground.
OR
Find the distance B’B on the ground. [2]
Answer:

(A) In ∆APB,
tan 30° = \(\frac{\mathrm{AP}}{\mathrm{AB}}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{h}{x}\)
⇒ x = √3h

(B) In ∆AP’B,
tan 45° = \(\frac{A P^{\prime}}{A B}\)
⇒ AB = AP’
⇒ x = h + 50

(C) On solving equation obtained in (i) and (ii), we get

= 68.25 × 1.732
= 118 m

OR

In ∆AP’B’
tan 30° = \(\frac{A P^{\prime}}{\mathrm{AB}^{\prime}}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{68.25+50}{\mathrm{AB}^{\prime}}\)
⇒ AB’ = 118.25 × 1.732
= 204.809
BB’ = AB’ – AB
= 204.809 – 118
= 86.80 = 87 m

Question 37.
A factory is using an inclined conveyor belt to transport its product from level 1 to level 2 which is 3 m above level 1 as shown in the figure below. The inclined conveyor is supported from one end to level 1 and from the other end to a post located 8 m away from level 1 supporting point.

The factory wants to extend the conveyor belt to reach at a new level 3 which is 9 m above level 1 while maintaining the inclination angle.
On the basis of the above information, answer the following questions:
(A) Find the distance at which a new post is to be installed to support the conveyor belt at level 3. [1]
(B) How much distance is extended from D to B? [1]
(C) Find the length of the conveyor belt up to level 3.
OR
Find the length of the conveyor belt up to level 2. [2]
Answer:
(A) In ∆ADE and ∆ABC,
Since, both triangles are similar, then, their corresponding sides will be proportional
Then \(\frac{A D}{A B}=\frac{D E}{B C}\)
⇒ \(\frac{8}{A B}=\frac{3}{9}\)
⇒ AB = 24 cm

(B) Distance extended = AB – AD
= 24 – 8 = 16 m

(C) Since, ∆ABC is a right-angled at B.
∴ AC² = AB² + BC² (by Pythagoras theorem)
AC² = (24)² + 9²
AC = \(\sqrt{676+81}\) = 25.63
≃ 26 m
Then, distance need to be travelled to reach new level is 26 m

OR

In ∆ADE, by pythagoras theorem
AD² + DE² = AE²
⇒ 8² + 3² = AE²
⇒ AE² = 64 + 9 = 73
⇒ AE = √73 = 8.5 m

Question 38.
To conduct sport day activities in the rectangular school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each along DB.
100 flower pots have been placed at a distance of 1 m from each other along DA as shown in the figure below.

Radha runs \(\frac{1}{4}\)th of the distance DA on 2nd line and post a green flag at X . Preeti runs \(\frac{1}{5}\)th of the distance DA on other line post a red flag at Y.
on the basis of the above information, answer the following questions:
(A) Treating DB as x-axis and DA as y-axis, find the position of green flag. [1]
(B) Treating DB as x-axis and DA as y-axis, find the position of red flag. [1]
(C) Find the distance (in complete metres) between the two flags.
OR
Find the perimeter (in complete metres) of the triangular region OXY. [2]
Answer:
(A) Radha’s distance on x-axis is 2 and on y-axis she is at \(\frac{1}{4}\) × 100 = 25 4
Green flag coordinates are (2, 25)

(B) X-coordinate = 8
Y-coordinate = \(\frac{1}{5}\) × 100 = 20
∴ Coordinates of red flag (8, 20)

(C) Coordinates of green flag is (2, 25)
∴ Coordinates of red flag is (8, 20)
∴ By distance formula

= √61 = 7.8 m
≃ 8 m (approx)

The distance formula \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\) = \(\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\). It gives the same answer.

OR

Perimeter = OX + OY + YY
= 25.07 + 21.54 + 7.81
= 54.42
= 55 m