CBSE Sample Papers for Class 10 Maths Basic Set 7 with Solutions

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

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

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