CBSE Sample Papers for Class 10 Maths Basic Set 11 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths with Solutions Set 11 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Maths Set 11 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.
In the following frequency distribution, what is the upper limit of the median class? [1]

(a) 18.5
(b) 20.5
(c) 25.5
(d) 17.5
Answer:
(d) 17.5

Explanation:

Class	Frequency	Cumulative
– 0.5 – 5.5	13	13
5.5 – 11.5	10	23
11.5 – 17.5	15	38
17.5 – 23.5	8	46
23.5 – 29.5	11	57

Here, N = 57 So, \(\frac{N}{2}\) = 28.5
The cumulative frequency, just greater than 28.5, is 38 which belongs to class 11.5 – 17.5.
So, the median class is 11.5 – 17.5 Its upper limit is 17.5

Question 2.
If the perimeter of o circle is equal to that of a square, then the ratio of the area of circle to the area of the square [1]
(a) 14: 11
(b) 12: 13
(c) 11:14
(d) 13:12
Answer:
(a) 14: 11

Explanation:
Here, it is given that
4s = 2πr

So, the ratio of the area of the circle to the area of square is 14 : 11.

Question 3.
A, B, and C alt begin running in the same direction on a circular track at the same moment. A fin ¡shes a round in 252 seconds, B in 308 seconds, and C In 198 seconds. When will they meet at the beginning point? [1]
(a) 46 min 12 sec
(b) 42 min 6 sec
(c) 52 min 12 sec
(d) 56 min 10 sec
Answer:
(a) 46 min 12 sec

Explanation:
∴ 252 = 22 × 32 × 7
308 = 22 × 7 × 11 198 = 22 × 32 × 11
⇒ Required time = LCM (252, 308, 198)
= 22 × 32 × 7 × 11
= 2772 s
Now, 1 min = 60 s
1 s = \(\frac{1}{60}\)min
∴ 2772 s = \(\frac{2772}{60}\)min = 46 min 12 s

Question 4.
In the figure, ABCÐ is a square of side, 14 cm. If APD and BPC are semi-circles, then the area of the shaded region is [1]

(a) 40 sq. cm
(b) 46 sq. cm
(c) 42 sq. cm
(d) 50 sq. cm
Answer:
(c) 42 sq. cm

Explanation:
Area of the shaded region = Area of the square – 2 × Area of semi-circular regions n
= 14 × 14 – 2 × \(\frac{π}{2}\)(7)²
= 196 – 154
= 42 sq. cm

Question 5.
The coordinates of the points which trisect the line segment joining (1, – 2) and (- 3, 4) is: [1]
(a) (0, 0) (-1, 0)
(b) (-2, 3) (0, -2)
(c) \(\left(-\frac{1}{4}, 0\right)\left(-\frac{4}{3}, 2\right)\)
(d) \(\left(-\frac{1}{3}, 0\right)\left(-\frac{5}{3}, 2\right)\)
Answer:
(d) \(\left(-\frac{1}{3}, 0\right)\left(-\frac{5}{3}, 2\right)\)

Explanation:
Let A (1, – 2) and B (- 3, 4) be the given two points.
Let P and Q be the points of trisection of \(\overline{A B}\). P trisects \(\overline{A B}\) in the ratio 1 : 2; and Q trisects \(\overline{A B}\) in the ratio 2 : 1.

Question 6.
In a ΔABC if BD ⊥ CA and CE ⊥ BA. Then ΔABD congruent to: [1]
(a) ΔACB
(b) ΔACE
(c) ΔAED
(d) ΔABD
Answer:
(b) ΔACE

Explanation:
In Δs ABD and ACE, we have

∠D = ∠E (each is of 90°)
∠A = ∠A (common)
So, by AA similarity criteria,
ΔABD ~ ΔACE
Hence Proved.

Question 7.
The ratio in which x-axis divides the join of (2, -3) and (5, 6) is: [1]
(a) 1: 2
(b) 3 : 4
(c) 1: 3
(d) 1: 5
Answer:
(a) 1 : 2

Explanation:
Let P(x, 0) be the point on x-axis which divides the join of (2, -3) and (5, 6) in the ratio m : n.
∴ By section formula,
P(x, 0) = \(\left(\frac{5 m+2 n}{m+n}, \frac{6 m-3 n}{m+n}\right)\)
\(\frac{6 m-3 n}{m+n}\) = 0
6m = 3n
⇒ \(\frac{m}{n}=\frac{1}{2}\) is the required ratio.

Question 8.
If cosec A = \(\frac{13}{12}\), then the value of \(\frac{2 \sin A-3 \cos A}{4 \sin A-9 \cos A}\) [1]
(a) 4
(b) 5
(c) 6
(d) 3
Answer:
(d) 3

Explanation:
Given cosec A = \(\frac{13}{12}\)

Question 9.
If the angle of elevation of the top of a tower from a point of observation at a distance of 100 m from its base is 45°, then the height of the tower is: [1]
(a) 160 m
(b) 100 m
(c) 200 m
(d) 150 m
Answer:
(b) 100 m

Explanation:
Here, PQ is the tower and A is a point of observation at a distance of 100 m from PQ.
From right ΔAPQ,

Thus, the height of tower is 100 metre.

Question 10.
If the 6th term and 11th term of an A.P. are 12 and 22 respectively, then its 2nd term is: [1]
(a) 8
(b) -5
(c) -8
(d) 16
Answer:
(c) -8

Explanation:
Here, 6th term = a₆ = a + 5d = 12 . . .(i)
and 8th term = a₈ = a + 7d = 22 . . .(ii)
Subtracting equation (i) from equation (ii), we get
2d = 10, i.e. d = 5
a = – 13
So, 2nd term = a₂ = a + d
= (- 13) + 5
= – 8

Question 11.
The value of x and y is: x – 2y = 4 and 6x – 4y = 8 [1]
(a) x = 2, y = 0
(b) x = 1, y = 3
(c) x = 0, y = -2
(d) x = -1, y = -31
Answer:
(c) x = 0, y = -2

Explanation:
Given equations are
x – 2y = 4 …(i)
And 6x – 4y = 8 …(ii)
Multiply the eq. (i) x 2 then subtract eq. (ii) from eq. (i), we get
x = 0, y = -2

Question 12.
How many tangents can be drawn to a circle from a point lying inside the circle? [1]
(a) 0
(b) 2
(c) 1
(d) 3
Answer:
(a) 0

Explanation:
Zero, as tangent can be drawn to a circle from a point Lying inside a circle.

Question 13.
The degree of the polynomial : (x + 1) (x – x + x – 1) is: [1]
(a) 4
(b) 3
(c) 5
(d) 2
Answer:
(c) 5

Explanation:
The given ponynomial is (x + 1) (x² – x + x⁴ – 1)
⇒ x³ + x² – x² – x + x⁵ + x⁴ – x – 1
So, the degree of this polynomial is 5.

Question 14.
If r = 3 is a root of quadratic equation kr2 -kr- 3 = 0, then the value of k is: [1]
(a) \(\frac{1}{2}\)
(b) 3
(c) \(\frac{1}{3}\)
(d) \(\frac{1}{4}\)
Answer:
(a) \(\frac{1}{2}\)

Explanation:
Given, equation is kr² – kr – 3 = 0
If, r = 3
Then, k(3)² – k(3) – 3 = 0
⇒ 9k – 3k – 3 = 0
⇒ 6k =3
⇒ k = \(\frac{1}{2}\)

Question 15.
Two different dice are thrown together. The probability of getting the sum of the two numbers less than 7 is: [1]
(a) \(\frac{5}{12}\)
(b) \(\frac{7}{12}\)
(c) \(\frac{12}{5}\)
(d) \(\frac{3}{11}\)
Answer:
(a) \(\frac{5}{12}\)

Explanation:
Total outcomes = 36
Outcomes in which sum of two numbers is less than 7 = (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (2, 1) (2, 2) (2, 3) (2, 4) (3, 1) (3, 2) (3, 3) (4, 1) (4, 2) (5, 1) i.e., Total number of possible outcomes = 15
∴ Required probability = \(\frac{15}{36}\)
= \(\frac{5}{12}\)

Question 16.
If x = 1 is a root of quadratic equation x² – ax – 1 = 0, then the value of ‘o’ is: [1]
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(a) 0

Explanation:
Put x = 1, in the given equation x² – ax – 1 = 0.
We have
(1)² – a × 1 – 1 = 0
⇒ 1 – a – 1 = 0
⇒ a = 0

Question 17.
Convert the following statement into a pair of linear equations in x and y (x > y)
“The sum of 2 numbers is 58. The greater number exceeds twice the smaller number by 1.” [1]
(a) x + 2y = 1, x + y
(b) x – 2y = 3, x + y = 58
(c) x = 3y + 2, x – y = 58
(d) x = 2y + 1, x + y = 58
Answer:
(d) x = 2y + 1, x + y = 58

Explanation:
x + y = 58
x = 2y + 1

Question 18.
A bag contains 5 red, 8 green and 7 white balls. One ball is drawn at random from the bag. The probability of getting a red or white ball is: [1]
(a) \(\frac{2}{5}\)
(b) \(\frac{3}{5}\)
(c) \(\frac{4}{5}\)
(d) \(\frac{1}{5}\)
Answer:
(b) \(\frac{3}{5}\)

Explanation:
Total balls = 5 + 8 + 7 = 20
Number of Red or White balls = 5 + 7 = 12
∴ Required probability = \(\frac{12}{20}\)
= \(\frac{3}{5}\)

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): If a and (3 are the zeroes of 2x² + 5x + 9, then the 9 value of αβ is \(\frac{9}{2}\)
Reason (R): If a and p are the zeroes of p(x) = ax² + bx + c, then αβ = \(\frac{c}{a}\). [1]
Answer:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)

Explanation:
Here, quadratic equations 2x² + 5x + 9. Then α.β = \(\frac{c}{a}=\frac{9}{2}\).

Question 20.
Assertion (A) :The value of y is 3, if the distance between the points P(2, -3) and Q(10, y) is 10.
Reason (R) : Distance between two points is given by \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\) [1]
Answer:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A)

Explanation:
For y = 3
Distance PQ = \(\sqrt{(10-2)^2+(y+3)^2}\)
= \(\sqrt{8^2+6^2}\)
= \(\sqrt{64+36}\)
= \(\sqrt{100}\)
= 10 units

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

Question 21.
Prove that the number 4ⁿ, n being a natural number, can never end with the digit 0.
OR
Draw a factor tree for the number 546. [2]
Answer:
If 4ⁿ ends with 0, then it must have 2, 5 as its factors.
But, (4)ⁿ = (2²)ⁿ = 2²ⁿ, i.e, the only prime factor of 4ⁿ is 2.

Also, we know from the Fundamental Theorem of Arithmetic that the prime factorisation of each number is unique.
∴ 4ⁿ can never end with digit 0.

Question 22.
The angles of a triangle are in A.P., the least being half the greatest. Find the angles. [2]
Answer:
Let, the angles be a, a + d, a + 2d.
∴ a = \(\frac{1}{2}\)(a + 2d)
⇒ 2a = a + 2d
⇒ a = 2d …(i)
Now, a + a + d + a + 2d = 180° (Angles sum property of a triangle)
⇒ 3a + 3d = 180°
⇒ 3 (2d) + 3d = 180° (Using (i))
⇒ 9d = 180°
⇒ d = 20°
⇒ a = 40°
Thus, the measure of the angles of the triangle are 40°, 60° and 80°.

Question 23.
What number should be added to the polynomial x² – 5x + 4 so that 3 is the zero of the polynomial? [2]
Answer:
Let k be the number to be added to the given polynomial.
Then the polynomial becomes x² – 5x + (4 + k)
As 3 is the zero of this polynomial, we get:
(3)² – 5(3) + (4 + k) = 0
⇒ (4 + k) = 15 – 9
⇒ 4 + k = 6
⇒ k = 2
Thus, 2 is to be added to the given polynomial.

Question 24.
Evaluate: 3 cos² 60° sec² 30° – 2 sin² 30° tan² 60°. [2]
OR
A coin is tossed twice. Find the probability of getting at-most 2 heads.
Answer:
3 cos²60° sec²30° – 2 sin²30° tan²60°
= 3\(\left(\frac{1}{2}\right)^2\left(\frac{2}{\sqrt{3}}\right)^2\) – 2\(\left(\frac{1}{2}\right)^2(\sqrt{3})^2\)
= 1 – \(\frac{3}{2}=-\frac{1}{2}\)
OR
Sample space = (H, H) (H, T) (T, H) (T, T)
∴ Possible outcomes = 4
Favourable outcomes = (H, H) (H, T) (T, H) (T,T)
All are favourable cases;
Required probability = \(\frac{4}{4}\) = 1

Question 25.
A heap of rice is in the form of a cone of base diameter 24 m and height 3.5 m. How much canvas cloth is required to just cover the heap. [2]
Answer:
Slant height of cone

∴ Cloth required
= πrl = \(\frac{22}{7}\) × 12 × 12.5 = 471.42 sq.m

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

Question 26.
Knowing that √5 is an irrational number, show that 3 + 2√5 is an irrational number. [3]
Answer:
Let, 3 + 2√5 be a rational number. Then, it can be written in the form , where p, q are co-primes and q ≠ 0.
Now, 3 + 2√5 = \(\frac{p}{q}\),
2√5 = \(\frac{p}{q}\) – 3
√5 = \(\frac{p-3 q}{2 q}\)
Since, \(\frac{p-3 q}{2 q}\) is a rational number, √5 is an irrational number (as, L.H.S. = R.H.S.) which is a contraction to the given fact 45 is an irrational number.
This concludes that 3 + 2√5 is an irrational number.

Question 27.
Taking A as the origin, find the coordinates of the vertices P, Q and R of the triangle PQR. The class X students of a secondary school have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 metre from each other. There is a triangular grassy lawn in the plot as shown in the figure. [3]

The students have to sow seeds of flowering plants on the remaining area of the plot.
(A) Taking A as the origin, find the coordinates of the vertices P, Q and R of the triangle PQR;
(B) What will be the coordinates of the vertices of ΔPQR if C is the origin?
Answer:
(A) From the figure, the coordinates of R Q, R are:
P = (1, 4)
Q = (4, 3)
R = (3, 1)

(B) Taking C as origin, CB as X-axis and CD as Y-axis, the coordinates of the vertices P, Q and R are (8, 4) (5, 5) and (6, 7) respectively.

Question 28.
A 90 cm tall female is standing next to a lamppost. She now moves 1.2 metres per second away from a lamp post’s base. What is the length of her shadow after 4 seconds, if the lamp is 3.6 m above the ground? [3]
OR
A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use π = 3.14 and √3 = 1.73)

Answer:
Speed of girl = 1.2 m/s
∴ In 4 seconds, travels

distance = 1.2 x 4 = 4.8 m
∴ After 4 seconds, she reaches at D.
∴ BD = 4.8 m
Let CD be the length of her shadow.
Now, ∠ABD = ∠EDC = 90°
∴ AB ∥ ED
Hence, by BPT
\(\frac{A B}{E D}=\frac{B C}{D C}\)
\(\frac{A B}{E D}=\frac{B C}{D C}\)
⇒ 4x = 4.8 + x
⇒ x = 1.6 m
OR
We draw OX ⊥ AB.
Since, AOB is an isosceles triangle, X bisects AB, i.e. AX = XB
Also, ∠AOX = ∠BOX = 60°

From right-angled triangle AXO, we have:
\(\frac{\mathrm{AX}}{\mathrm{AO}}\) = sin 60 and \(\frac{O X}{A O}\) = cos 60
AX = AO × \(\frac{\sqrt{3}}{2}\)
and OX = AO × \(\frac{1}{2}\)
⇒ AX = 6√3 cm and OX = 6 cm
AB = 12√3 cm
Now,
Area of the shaded segment
= Area of sector AOB – Area of AAOB
= \(\frac{120^{\circ}}{360^{\circ}}\)π(12)² – \(\frac{1}{2}\) × AB × OX
= (48π – 36√3) sq. cm
= [48 (3.14) – 36(1.73)] sq. cm
= (150.72 – 62.28) sq. cm
= 88.44 sq. cm

Question 29.
As seen in the figure, a medicine capsule is shaped like a cylinder with two hemispherical ends. Find the capsule’s capacity if its length is 14 mm and its diameter is 6 mm. [3]

Answer:
Diameter of capsule = 6 mm
⇒ Radius of capsule (r) = = 3 mm
So, Length of cylindrical part (h)
= Length of capsule – 2 × Radius of hemispherical ends
= 14 – 2 × 3
= 14 – 6 = 8 mm
So, capacity of the capsule
= 2 × Volume of hemispherical ends + Volume of cylinderical part
= 2 × \(\frac{2}{3}\)πr³ + πr²h
= \(\frac{4}{3}\)πr³ + πr²h
= \(\frac{4}{3}\) × π × (3)³ + π × (3)² × 8
= 36π + 72π
= 108π mm³

Question 30.
The distribution below gives the makes of 100 students of a class, if the median makes are 24, find the frequencies f₁ and f₂ 

All the black face cards are removed from a pack of 52 playing cards. The reaming cards are well shuffled and then a card is drawn at random. Find the probability of getting o: [3]
(A) face card
(B) red card
(C) black card.
Answer:

Class	Frequency	Cumulative
0 – 5	4	4
5 – 10	6	10
10 – 15	10	20
15 – 20	f1	20 + f1
20 – 25	25	45 + f1
25 – 30	f2	45 + f1 + f2
30 – 35	18	63 + f1 + f2
35 – 40	5	68 + f1 + f2

Now, Median = 24 (Given)
So,median class = 20 – 25
For this class,
I = 20. h = 5, \(\frac{N}{2}\) = 50, cf = 20 + f₁, f= 25
We know, Median = l + \(\frac{\frac{N}{2}-c f}{f}\) × h
⇒ 24 = 20 + \(\frac{50-\left(20+f_1\right)}{25}\) × 5
⇒ 4 = \(\frac{\frac{N}{2}-c f}{f}\)
⇒ 20 = 30 – f₁

Also, sum of frequencies = 100
⇒ 68 + f₁ + f₂ = 100
⇒ f₁ + f₂ = 32
⇒ 10 + f₂ = 32
⇒ f₂ = 22
∴ f₁ = 10, f₂ = 22.
OR
When all the three black face cards are removed,
Remaining number of cards = 52 – 3 = 49
(A) Number of face cards in the remaining deck = 9
∴ P(getting a face card) = \(\frac{9}{49}\)

(B) Number of red cardsin the remaining deck = 26
∴ P (getting a red card) = \(\frac{26}{49}\)

(C) Number of black cards in the remaining deck = 23
∴ P (getting a black card) = \(\frac{23}{49}\)

Question 31.
In the figure, two chords AB and CD intersect each other at the point P. [3]
Prove that
(A) ΔAPC ~ ΔDPB
(B) AP. PB = CP. DP

Answer:
(A) Consider Δs APC and DPB
∠APC = ∠DPB
(Vertically opopsite angles)
Also,
∠CAP = ∠PDB

(Angles made by the same arc CB)
So, by AA similarity criteria,
ΔAPC ~ ΔDPB.
Hence Proved.

(B) This further concludes that
\(\frac{\mathrm{AP}}{\mathrm{PC}}=\frac{\mathrm{DP}}{\mathrm{PB}}\)
or AP.PB = CP.DP
Hence Proved.

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

Question 32.
A man sold a chair and a table for f 1520, thereby making a profit of 25% on the chair and 10% on the table. By selling together for ₹ 1535, he would have made a profit of 10% on the chair and 25% on the table. Find the cost price of each. [4]
OR
If Zeba was younger by 5 years than what she really is, then the square of her are (in years) would have been 11 more than five times her actual age. What is her age now?
Answer:
Let the cost of 1 chair be ₹ p and of cost 1 table is ₹ q.
Then,
(p + 25% p) + (q + 10% q) = 1520;
(p + 10% p) + (q + 25% q) = 1535
⇒ \(\frac{5 p}{4}+\frac{11}{10}\)q = 1520
\(\frac{11 p}{10}+\frac{5}{4}\)q = 1535
⇒ 25p + 22q = 30400; …(i)
22p + 25q = 30700 …(ii)

Adding the two equations, we get 47 p + 47 q = 61100
⇒ 47p + 47q = 61100
⇒ p + q = 1300 …(iii)
From Eq. (i) and Eq. (ii), we get
25p + 22 (1300 – p) = 30400
⇒ 3p = 1800
or p = 600
From equation (iii), we have q = 700.
Thus, the cost of 1 chair is ₹ 600 and of 1 table is ₹ 700.
OR
Let Zeba’s present age (in years) be x.
Then (x – 5)² = 11 + 5x
⇒ x² – 10x + 25 = 11 + 5x
⇒ x² – 15x + 14 = 0
⇒ (x – 14) (x – 1) = 0
⇒ x – 14 = 0 [∵ (x – 1) ≠ 0)]
⇒ x = 14
Thus, her present age is 14 years.

Question 33.
In the figure, a circle is inscribed in a quadrilateral ABCD in which ∠B = 90° If AD = 23 cm, AB = 29 cm and DS = 5 cm, find the radius (r) of the circle. [4]

Answer:
Join OP and OQ.
We know, Lengths of tangents drawn from an external point to a circle are equal.
So,
Here, AR = AQ, BP = BQ,
CP = CS, DR = DS
Here, ∠B = 90° and OP = OQ = r (say)
OPBQ is a square and QB = r.

Now, AD = AR + RD
⇒ AQ + SD
⇒ 23 = AQ + 5
⇒ AQ = 18 cm
Also 29 = AB = AQ + BQ
⇒ 18 + r
⇒ r = 11
Thus, radius of the circle is 11 cm.

Question 34.
If sin θ + cos θ = √3, then prove that tan θ + cot θ = 1. [4]
Answer:
Given, sin θ + cos θ = √3
(sin θ + cos θ)² = (√3)²
⇒ sin²θ + cos²θ + 2 sin θ cos θ = 3
⇒ 1 + 2 sin θ cos θ = 3 (∵ sin²θ + cos²θ = 1) …(i)

Question 35.
Three uniformly flowing pipes fill a paddling pool. While the third pipe is the only one filling the paddling pool, the first two pipes are simultaneously filling the paddling pool. Five hours faster than the first pipe and four hours later than the third pipe, the second pipe fills the paddling pooL Calculate how long it takes for each pipe to fill the paddling pool independently. [4]
OR
A cylindrical hole with a diameter of 7 cm is drilled out of a cuboid solid metallic block with dimensions of 15 cm 10 cm 5 cm. Find the block’s remaining surface area.
Answer:
Let V be the volume of the pool and x the number of hours required by the second pipe alone to fill the pool. Then, the first pipe takes (x + 5) hours, while the third pipe takes (x – 4) hours to fill the pool. So, the parts of the pool filled by the first, second and third pipes in one hour are respectively.
\(\frac{V}{x+5}, \frac{V}{x}, \frac{V}{x-4}\)
Let the time taken by the first and second pipes to fill the pool simultaneously be t hours.
Then, the third pipe also takes the same time to fill the pool

⇒ (2x + 5) (x – 4) = x2 + 5x
⇒ x² – 8x – 20 = 0
⇒ x² – 10x + 2x – 20 = 0
⇒ (x – 10) (x + 2) = 0
⇒ x = 10 or x = -2
But x cannot be negative. so x = 10
Hence the timings required by first, second and third pipes to fill the pool individually are 15 hours, 10 hours and 6 hours respectively.
OR
Length of solid mettalic block l = 15 cm
Breadth of solid mettalic block, b = 10 cm
Height of solid mettalic block, h = 5 cm
Diameter of cylinderical block (d) = 7 cm

Radius (r) = \(\frac{d}{2}=\frac{7}{2}\) = 3.5 cm
Height of cyLindericaL block, h = 5 cm
Surface area of cuboid = 2(lb + bh + hl)
= 2(15 × 10 + 10 × 5 + 5 × 15)
= 2(150 + 50 + 75)
= 2 × 375 = 550 cm2
Curved surface area of cylinder k = 2prh
= 2 × \(\frac{22}{7}×\frac{7}{2}\) × 5 = 11 × 7
= 77 cm²
Surface area of the reamining block
= Surface area of cuboidal block + CSA of cylinder – Area of 2 cylinderal holes
= 550 + 110 – 77
= 660 – 77
= 583 cm²
Hence the surface area of remaining block = 583 cm².

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

Question 36.
The diagram shows a grid of squares. A button is placed on one of the squares. A fair dice is thrown. If 1,2,3 or 4 is thrown, the button is moved one square to the left. If 5 or 6 is thrown, the button is moved one square to the right.

On the basis of the above information, answer the following questions:
(A) The button is placed on square X.
the dice is thrown once. Then find the probability that the button is moved to the right. [1]
(B) On the other occasion, the button is placed on square Y. The dice is thrown once and the button is moved. The die is thrown a second time and the button is moved again. Find the probability that the button is moved at P. [1]
(C) Find the probability that the button is moved at Q and also find the probability that the botton is moved at Y.
OR
Find the probability that the button is moved at P, Q or Y. [2]
Answer:
(A) P(right) = \(\frac{2}{6}=\frac{1}{3}\)

(B) P(gatting 1, 2, 3 or 4 twice) = \(\frac{2}{6}=\frac{1}{3}\)
= \(\frac{1}{3} \times \frac{1}{3}\)
= \(\frac{1}{9}\)

(C) As per the situation given in part (B), if the button is at Y and it is moved twice, then there is no possibility that it reach to Q.
As per the situation given in part (B), if the button is at Y and it is moved twice, then it will reach to Y again if it is first move to left then to right or vice-versa.
P(Y) = \(\frac{2}{6} \times \frac{4}{6}+\frac{4}{6} \times \frac{2}{6}=\frac{16}{36}\)
= \(\frac{4}{9}\)
OR
The probability that the button is moved at P = \(\frac{1}{9}\)
The probability that the button is moved at Q = 0
The probability that the button is moved at y = \(\frac{4}{9}\)

Question 37.
Earth is excavated to make a railway tunnel. The tunnel is a cylinder of radius 5 m and length 450 m. A level surface is laid inside the tunnel to carry the railway lines. The Diagram 1 shows the circular cross – section of the tunnel. The level surface is represented by AB, the centre of the circle is 0 and ∠AOB = 90°. The space below AB is filled with rubble (debris from the demolition buildings).

Steel girders are erected above the tracks to strengthen the tunnel. Some of these are shown in Diagram 2. The girders are erected at 6 m intervals along the length of the tunnel, with one at each end.
On the basis of the above information, answer the following questions:
(A) Find the volume of earth removed to make the tunnel. [1]
(B) Find the area of ΔAOB shown in Diagram 1. [1]
(C) Find the length of each girder.
OR
How many girder are erreced and find total length of steel required in the 450 m length of tunnel. [2]
Answer:
(A) Volume of tunnel = πr²h
= π × 5 × 5 × 450
= 11250π cu cm

(B) Area of ΔAOB = \(\frac{1}{2}\) × AO × OB
= \(\frac{1}{2}\) × 5 × 5
= 12.5 sq.m

(C) As steel girders are in the shape of a circle or a sector of the circle with centra/angle 270°.
So, circumference of the circle with central angle 270°
= \(\frac{270^{\circ}}{360^{\circ}}\) × 2πR
= \(\frac{3}{4}\) × 2πR
= 1.5π × 5
= 7.5π cm
OR
No. of girders = \(\frac{450}{6}\)
= 75
∴ Toatl grider = 75 + 1 (one at least)
= 76
Toatl length = 7.5π × 76
= 570 π m

Question 38.
A radio mast PQ , of height ‘h’ metres, is standing vertically on the horizontal ground. From A, the angle of elevation of the top of the mast is found to be 45°. On moving 50 m up a slope of 15°, the angle of elevation ofP is found to be 75° from B, The horizontal through B is BC.

On the basis of the above information, answer the following questions:
(A) What is the measure of ∠APC? [1]
(B) Find the measure of ∠BPC. [1]
(C) Find the length BP.
OR
Find the length AP and the approaximat value of h. [2]
Answer:
(A) In ΔAPQ
∠A + ∠P + ∠Q = 180°
⇒ 45° + ∠P + 90° = 180°
∠P = 180° – 90° – 45°
= 180° – 135°
= 45°

(B) In ΔBPC
∠B + ∠BPC + ∠C = 180°
⇒ 75° + ∠BPC + 90° = 180°
⇒ ∠BPC = 180° – 165°
= 15°

(C) In ΔAPQ,
tan 45° = \(\frac{PQ}{AQ}\)
PQ = AQ = h

In ΔABQ’,
sin 15° = \(\frac{\mathrm{BQ’}}{\mathrm{AB}}\)
BQ’ = 50 sin 15°
= 13 m
(Taking sin 15° = 0.26)
Then, BQ’ = CQ = 13 m
PC = h – 13

In ΔABQ’
cos 15° = \(\frac{AQ’}{AB}\)
AQ’ = 50 cos 15°
= 48.25 ≈ 49
(Taking cos 15° = 0.965)
QQ’ = AQ – AQ’
= h – 49
BC = QQ’ = h – 49

Now, in ΔPBC
tan 75° = \(\frac{P C}{B C}\)
3.73 = \(\frac{h-13}{h-49}\)
3.73h – 182.77 = h – 13
2.73h = 169.77
h = 62.186
sin 75° = \(\frac{\mathrm{PC}}{\mathrm{PB}}\)
1 = \(\frac{62.2-13}{P B}\)
PB = 49.2 = 50 m

OR

In ΔAPQ
sin45° = \(\frac{h}{\mathrm{AP}}\)
AP = \(\frac{h}{\frac{1}{\sqrt{2}}}\) = √2h

In ΔPBC
tan 75 = \(\frac{\mathrm{PC}}{\mathrm{BC}}\)
3.73 = \(\frac{\mathrm{PC}}{\mathrm{BC}}\)
3.73h – 182.77 = h – 13
h = 62.186
h ~ 60 m