Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths with Solutions Set 4 are designed as per the revised syllabus.

CBSE Sample Papers for Class 10 Maths Set 4 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 HCF of 40 and 54 is: [1]
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a) 2

Explanation:
The prime factorisations of 40 and 54 are:
40 = 2 × 2 × 2 × 5, or 23 × 51
54 = 2 × 3 × 3 × 3, or 21 × 33
So, HCF (40, 54) = 21, i.e. 2.

CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions

Question 2.
The value of k for which the polynomial 2kx2 – 3kx + 7 has real roots is: [1]
(a) 35
(b) 34
(c) 19
(d) None of these
Answer:
(d) None of these

Explanation:
For real roots,
Discriminant > 0
⇒ b2 – 4ac ≥ 0
Here, a = 2k, b = – 3k and c = 7.
∴ (- 3k)2 – 4 × 2k × 7 ≥ 0
⇒ 9k2 – 56 k ≥ 0
⇒ k (9k – 56) ≥ 0
But k ≠ 0
⇒ k > 0 or k ≥\(\frac{56}{9}\)

Question 3.
If the value of ‘x’ in the equation 2x + 3y = 13 is 2, then the corresponding value of y is: [1]
(a) 1
(b) 2
(c) 4
(d) 3
Answer:
(d) 3

Explanation:
Putting the value of x in the equation 2x + 3y = 13, we have
2(2) + 3y = 13
⇒ 4 + 3y = 13
⇒ 3y = 13 – 4 = 9
⇒ y = 3

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

Explanation:
Let P (x, 0) be a point on x-axis which divides A (2, – 3) and B (5, 6) in the ratio k : 1.
Then, by section formula
⇒ P(x, 0) = \(\left(\frac{5 k+2}{k+1}, \frac{6 k-3}{k+1}\right)\)
⇒ \(\frac{6 k-3}{k+1}\) = 0
⇒ 6k – 3 = 0
⇒ k = \(\frac{1}{2}\)
⇒ Required ratio = k : 1 = \(\frac{1}{2}\) : 1 = 1 : 2.

CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions

Question 5.
The ratio of the height of a tower and the length of its shadow is Vi: 1. The angle of elevation of the Sun is: [1]
(a) 60°
(b) 30°
(c) 45°
(d) 90°
Answer:
(a) 60°

Explanation:
Let the angle of elevation be θ.
Then, tan θ = \(\frac{B C}{A B}\)
CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions 1
⇒ θ = 60°
So, the angle of elevation of the sun is 60°.

Question 6.
The value of sec2 60° cos 45° – cosec2 30° tan 45° is: [1]
(a) 2√2 – 4
(b) 3√2 + 2
(c) 2√2
(d) 7√2 – 2
Answer:
(a) 2√2 – 4

Explanation:
sec2 60° cos45° – cosec2 30° tan 45°
= (2)2 × \(\frac{1}{\sqrt{2}}\) – (2)2 × 1
= 2√2 – 4

Question 7.
The 10th term of the A.P.: 2, 7, 12, …………. is: [1]
(a) 47
(b) 36
(c) 45
(d) 30
Answer:
(a) 47

Explanation:
Here, a = 2, d = 5
So, a10 = a + 9d
= 2 + 45 = 47

CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions

Question 8.
The sum of the first 10 multiples of 2 is: [1]
(a) 110
(b) 210
(c) 150
(d) 140
Answer:
(a) 110

Explanation:
First 10 multiple of 2 are 2, 4, 6, 8, ….. 20.
This is an A.P. with a = 2 and d = 2.
So, their sum = \(\frac{10}{2}\) [2 × 2 + 9 × 2] = 110

Question 9.
A quadratic polynomial whose zeros are 2 and -5 is: [1]
(a) x2 + 10x – 3
(b) x2 + 3x – 10
(c) x2 – 3x – 10
(d) x2 + 5x + 4
Answer:
(b) x2 + 3x – 10

Explanation:
A quardratic polynomial with sum and product of zeroes as S and P, respectively is given as,
x2 – Sx + P
∴ Required polynomial is x2 + 3x – 10.
[∵ sum of zeroes = -3 and product of zeroes = -10]

Question 10.
The sum of the digits of a 2-digit number is 10. A number is selected at random. The probability of the chosen number to be divisible by 3 is: [1]
(a) 1
(b) 3
(c) 4
(d) 0
Answer:
(d) 0

Explanation:
2-digit numbers, where the sum of the digits is 10, are 19, 28, 37, 46, 55, 64, 73, 82 and 91.
Of these nine numbers, no number is divisible by 3.
So, the required probability = \(\frac{0}{9}\) i.e. 0.

CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions

Question 11.
The median of the given data is: [1]
2, 4, 6, 12, 3, 5, 10, 8, 2, 4, 9, 2, 10
(a) 4
(b) 9
(c) 5
(d) 10
Answer:
(c) 5

Explanation:
Rearranging the given data in ascending order, we have,
2, 2, 2, 3, 4, 4, 5, 6, 8, 9,10, 10, 12
Here, number of terms (n) = 13
∴ Median = \(\left(\frac{n+1}{2}\right)^{\text {th }}\) term
= 7th term = 5.

Question 12.
In a single throw of an unbiased die with 6 faces, what is the probability of getting a prime number ? [1]
(a) \(\frac{1}{2}\)
(b) \(\frac{1}{3}\)
(c) \(\frac{2}{3}\)
(d) \(\frac{1}{5}\)
Answer:
(a) \(\frac{1}{2}\)

Explanation:
Out of the six numbers 1, 2, 3, 4, 5, 6, three are prime numbers, namely 2, 3 and 5.
So, P (a prime number) = \(\frac{3}{6}\), i.e. \(\frac{1}{2}\)

Question 13.
If θ is the angle (in degrees) of a sector of a circle of radius ‘r’, then what is the area of the sector ? [1]
(a) \(\frac{\theta}{360^{\circ}}\) × πr2
(b) \(\frac{\theta}{360^{\circ}}\) × 2πr
(c) \(\frac{360^{\circ}}{\theta}\) × πr2
(d) \(\frac{360^{\circ}}{\theta}\) × 2πr
Answer:
(a) \(\frac{\theta}{360^{\circ}}\) × πr2

Explanation:
Area of sector with angle θ and radius r is \(\frac{\theta}{360^{\circ}}\) × πr2.

CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions

Question 14.
A cylindrical pencil sharpened at one edge is the combination of which two solid figures ? [1]
(a) Sphere and cone
(b) Cone and triangle
(c) Cuboid and cone
(d) Cylinder and cone
Answer:
(d) Cylinder and cone

Explanation:
CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions 2
So, it a combination of cylinder and a cone.

Question 15.
If one zero of P(y) = 4y2 – 8ky – 9 is negative of other, then the value of ‘k’ is: [1]
(a) 1
(b) 4
(c) 0
(d) 7
Answer:
(c) 0

Explanation:
Since one zero is the negative of the other zero, so the sum of two zeroes is 0.
⇒ \(\frac{8 k}{4}\) = 0 ⇒ k = 0

Question 16.
The value of ‘k’ if the given system of equations 5x + ky = – 7 and x + 2y = 3 is inconsistent, is: [1]
(a) 10
(b) 20
(c) 15
(d) 7
Answer:
(a) 10

Explanation:
A pair of system of equations is inconsistent, it
CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions 3

Question 17.
AT is a tangent to circle with centres such that OT = 4 cm and ∠OTA = 30°. The length of AT is: [1]
CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions 4
(a) 2 cm
(b) √3 cm
(c) 4√3 cm
(d) 2√3 cm
Answer:
(d) 2√3 cm

Explanation:
Join OA,
Then ∠OAT = 90° (Since, OA ⊥ AT)
Now, cos 30° = \(\frac{\mathrm{AT}}{\mathrm{OT}}\)
⇒ \(\frac{\sqrt{3}}{2}=\frac{A T}{4}\)
⇒ AT = 2√3

CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions

Question 18.
If ∆ABC ~ ∆PQR, perimeter of ∆ABC = 32 cm, perimeter of ∆PQR = 48 cm and PR = 6 cm, then the length of AC is: [1]
(a) 3 cm
(b) 6 cm
(c) 5 cm
(d) 4 cm
Answer:
(d) 4 cm

Explanation:
∴ ∆ABC ~ ∆PQR
CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions 5

Direction for questions 19 & 20: In the 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 polynomial 2x2 + 14x + 20 have two zeroes.
Reason (R): A quadratic polynomial can have two or more than two zeroes. [1]
Answer:
(c) Assertion (A) is true but reason (R) is false.

Explanation:
Here,
2x2 + 14x + 20 = 0
⇒ x2 + 7x + 10 = 0
⇒ x2 + 5x + 2x – 10 = 0
⇒ (x + 5) (x + 2) = 0
⇒ x = – 5, – 2

Question 20.
Assertion (A): The two tangents are drown to a circle from an external point, than they subtend equal angles at the centre.
Reason (R): A parallelogram circumscribing a circle is a rhombus. [1]
Answer:
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A)

Explanation:
From an external point the two tangents drawn subtend equal angles at the centre. So A is true. Also, a parallelogram circumscribing a circle is a rhombus.

CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions

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

Question 21.
If HCF (150, 210) = 30, then find LCM (150, 210). [2]
Answer:
We know that
LCM (150, 210) = \(\frac{150 \times 210}{{HCF}(150,210)}\)
= \(\frac{150 \times 210}{30}\) = 1050

Question 22.
Find the value of x for which 2x, (x + 10) and (3x + 2) are three consecutive terms of an A.P.
OR
If the first term of on A.P. is P and its common difference is q. then find its 6th term. [2]
Answer:
Since, 2x, (x + 10) and (3x + 2) are three consecutive terms of A.P.
∴ (x + 10) – 2x = (3x + 2) – (x + 10)
⇒ 10 – x = 2x – 8
⇒ 3x = 18
⇒ x = 6
OR
Let a be the first term and d be the common difference of the A.P.
Then, a = p and d = q
∴ 6th term = a + 5d
= p + 5q

Question 23.
Find a relationship between x and y such that the point (x, y) is equidistant from the points (3, 6) and (- 3, 4). [2]
Answer:
Let P (x, y), A (3, 6) and B (-3, 4).
As P is equidistant from A and B,
So PA = PB or PA2 = PB2
i.e., (x – 3)2 + (y – 6)2 = (x + 3)2 + (y – 4)2
⇒ (x2 – 6x + 9) + (y2 – 12y + 36)
= (x2 + 6x + 9) + (y2 – 8y + 16)
⇒ 12x + 4y – 20 = 0.
or 3x + y – 5 = 0
which is the required relationship between x and y.

CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions

Question 24.
The shadow of a 5 m long stick is 2 m long. At the same time, find the length of the shadow of a 12.5 m high tree. [2]
Answer:
Let the length of a shadow of 12.5 m high tree x m
Now, ratio of lengths of objects = Ratio of lengths of their shadows
\(\frac{5}{12.5}\) = \(\frac{2}{x}\)
x = \(\frac{2 \times 12.5}{5}\) = \(\frac{25}{5}\) = 5 m

Question 25.
The area of a circle is 154 sq. cm. Find its circumference.
OR
A bag contains 3 red and 5 blue balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is:
(A) red?
(B) yellow?
Answer:
Let ‘r’ cm be the radius of the circle
Then, πr2 = 154
⇒ r2 = \(\frac{154 \times 7}{22}\) = 49 ⇒ r = 7 cm
So, circumference
= 2πr = 2 × \(\frac{22}{7}\) × 7 = 44 cm.

OR

Total balls = 3 + 5 = 8
(A) P (red ball) = \(\frac{3}{8}\)
(B) P (yellow ball) = \(\frac{0}{8}\), i.e. 0. (As there is no yellow ball in the bag.)

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

Question 26.
Find the greatest 4-digit number which is divisible by 15, 24 and 36. [3]
OR
Solve for x and y:
3x + 2y = 11, 2x + 3y = 4
Answer:
LCM (15, 24, 36) is the smallest number which is divisible by 15, 24, 36.
Also, every multiple of the LCM (15, 24, 36) is also divisible by 15, 24 and 36.
Now,
15 = 3 × 5
24 = 23 × 3
36 = 22 × 32
∴ LCM (15, 24, 36) = 23 × 32 × 5 i.e. 360
The greatest 4-digit number which is the multiple of 360, is 9720 (27 × 360), which is the required number.

OR

3x + 2y = 11 …….(i)
2x + 3y = 4 …… (ii)
Adding (i) and (ii), we get
5x+ 5 y = 15
⇒ x + y = 3 …… (iii)
Also, subtracting (ii) from (i), we get
x – y = 7 ……. (iv)
Adding (iii) and (iv), we get
2x = 10
⇒ x = 5
Putting the value of x in equation (iii), we get
5 + y = 3
⇒ y = – 2
∴ x = 5 and y = – 2.

CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions

Question 27.
Find the coordinates of the points of trisection of the line segment joining the points (2, – 2) and (- 7, 4). [3]
Answer:
Let, the given points be A and B. Then, A (2, -2) and B (-7, 4).
Let, P and Q be the two points of trisection of \(\overline{\mathrm{AB}}\) such that, P divides \(\overline{\mathrm{AB}}\) in the ratio 1 : 2; and Q divides \(\overline{\mathrm{AB}}\) in the ratio 2 : 1.
CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions 6

Question 28.
In the figure, DEFG is a square and ∠BAC = 90°. Prove that [3]
CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions 7
(A) ∆AGF ~ ∆DBG
(B) ∆AGF ~ ∆EFC
(C) ∆DBG ~ ∆EFC
OR
Determine the A.P. whose 3rd term is 5 and the 7th term is 9.
Answer:
(A) Consider As AGF and DBG.
Here, GF || BD and AB is a transversal
So, ∠AGF = ∠DBG
(corresponding angles are equal)
Also, ∠GAF = ∠GDB (each is 90°)
So, by AA similarity criteria,
∆AGF ~ ∆DBG

(B) Consider A’s AGF and EFC.
Here FG || CE and AC is a transversal
So, ∠AFG = ∠FCE
(corresponding angles are equal)
Also, ∠FAG = ∠FEC (each is 90°)
So, by AA similarity criteria,
∆AGF ~ ∆EFC

(C) Since ∆DBG ~ ∆AGF [by (i)]
and ∆EFC ~ ∆AGF, [by (ii)]
we get ∆DBG ~ ∆EFC
OR
Let a be the first term and d be the common difference of A.P.
Then, a3 = a + 2d = 5
and a7 = a + 6d = 9
Solving these simultaneously, we get:
a = 3 and d = 1
Thus, the required A.P. is 3, 4, 5, 6, ………….

CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions

Question 29.
A quadrilateral ABCD circumscribe a circle. Prove that AB + CD = BC + DA. [3]
Answer:
We know, length of tangents, drawn from a same external points are equal.
So here,
AP = AS; BP = BQ; CQ = CR and DS = DR. ……….. (i)
CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions 8
Now, AB + CD = (AP + BP) + (CR + DR)
= (AS + BQ) + (CQ + DS) [By eqn. (i)]
= (AS + DS) + (BQ + CQ)
= AD + BC, or BC + DA.

Question 30.
A automobile has wheels that are each 80 cm in diameter. When the car is moving at a speed of 66 km/h, how many complete rotations do each wheel complete in 10 minutes? [3]
Answer:
Diameter of wheel = 80 cm
⇒ Radius of wheel (r) = 40 cm
Distance covered by wheel in one revolution
= 2πr = 2 × \(\frac{22}{7}\) × 40 = \(\frac{1760}{7}\) cm
∵ Distance covered by wheel in 1 hour = 66 km = 66000 m = 6600000 cm
Distance covered by wheel in 10 minutes
= \(\frac{6600000}{60}\) × 10 = 1100000 cm
No. of revolutions
= \(\frac{\text { Total distance }}{\text { distance of one revolution }}\) = \(\frac{1100000 \times 7}{1760}\)
= 4375

Question 31.
A girl of height 90 cm is standing near a lamp-post. Now, she starts walking away from the base of a lamp post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, then what is the length of her shadow after 4 seconds? [3]
Answer:
Speed of girl = 1.2 m/s
∴ In 4 seconds, travels
CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions 9
distance = 12 × 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
\(\frac{A B}{E D}=\frac{B C}{D C}\)
\(\frac{3.6}{0.9}=\frac{4.8+x}{x}\)
⇒ 4x = 4.8 + x
⇒ x = 1.6 m

CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions

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

Question 32.
Solve for x:
\(\frac{1}{x+4}-\frac{1}{x-7}\) = \(\frac{11}{30}\),
OR
Using quadratic formula, solve for x:
3x2 + 2√5x – 5 = 0. [5]
Answer:
We have
CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions 10
⇒ (x + 4) (x – 7) = – 30
⇒ x2 – 3x – 28 = – 30
⇒ x2 – 3x + 2 = 0
⇒ x2 – 2x – x + 2 = 0
⇒ x(x – 2) – 1 (x – 2) = 0
⇒ (x – 1) (x – 2) = 0
⇒ x = 1 or 2

OR

Using the quadratic formula, we have:
CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions 11

Question 33.
Izmir Clock Tower is a historic clock tower in Konak Square in the center of Izmir, Turkey. The French architect Raymond Charles Pere designed the Izmir Clock Tower.
Let us assume that the height of the tower AB = 14 m, height of tree CD = 5 m and BD – BC = 1 m. As the tower is vertical ∠ABC = 90°. Further, let us denote ∠CBD by ‘a’ and ∠BAD by ‘b’.
CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions 14
Find the value of sin a and tan b.
OR
The angle of elevation of the top of a tower from o point on the ground, which is 30 m away from the foot of the tower is 30°. Find the height of the tower. [5]
Answer:
To find sin a, we will first find BD. It is given that BD – BC = 1 m and CD = 5 m. Therefore, applying Pythagoras theorem in triangle BCD, we get:
BD2 = BC2 + CD2
⇒ BD2 = (BD – 1)2 + 52
⇒ BD2 = BD2 – 2BD + 1 + 25
Solving further, 2BD = 26, or BD = 13 m
Therefore, BC = 12 m.
In ∆BCD, sin a = \(\frac{\text { Perpendicular }}{\text { Hypotenuse }}\)
= \(\frac{C D}{B D}=\frac{5}{13}\)
To find tan b, we will find AE and DE (drawn parallel to BC).
CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions 12
We construct DE || BC and we get a rectangle.
∴ AB = BE + AE
or 14 = 5 + AE
Therefore, AE = 9 m
and DE = BC = 12 m
Therefore, tan b = \(\frac{\text { Perpendicular }}{\text { Hypotenuse }}\)
\(\frac{\mathrm{DE}}{\mathrm{AE}}=\frac{12}{9}\)

OR

Let P be the point of observation on the ground and TQ be the tower.
So, PQ = 30 m and ∠TPQ = 30°.
Let TQ = h metres
From rt. ∠d ∆PQT,
\(\frac{\mathrm{TQ}}{\mathrm{PQ}}\) = tan 30°
CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions 13
⇒ h = \(\frac{30}{\sqrt{3}}\)
= 10√3 metres

CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions

Question 34.
As shown in the figure, two hemispheres of a solid metal ball ore divided in half and joined. The solid is positioned in a water-filled, cylindrical tub such that it is completely submerged. The cylindrical tub has a radius of 4 cm and a height of 11 cm, respectively. Additionally, a spherical ball has a 3 cm radius. Calculate how much water is left in the cylindrical tub. [5]
CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions 15
Answer:
Radius of spherical ball r = 3 cm
∴ Radius of each hemisphere, r = 3cm
Radius of cylinderical tub, R = 4 cm
Height of cylinderical tub, H = 11 cm
Volume of water left in tub = volume of water in cylinderical tub – Volume of the solid
= πR2H – 2 × \(\frac{2}{3}\) πr3
= \(\frac{22}{7}\) × 4 × 4 × 11 – \(\frac{4}{3} \times \frac{22}{7}\) × 3 × 3 × 3
= \(\frac{22}{7}\) (176 – 36)
= \(\frac{22}{7}\) × 140
= 22 × 20
= 440 cm2

Question 35.
Find the mean, median and mode of the following frequency distribution: [5]
CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions 16
Answer:
Given, distribution is:
CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions 17
Mean X̄ = \(\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{4225}{25}\)
For median, n = 25
\(\frac{n}{2}\) = 12.5
Then, median class is 150 – 200.
we have, l = 150, c.f. = 10, f = 6, h = 50
CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions 18
For mode
modal class = 150 – 200
f0 = 5, fi = 6, f2 = 5, h = 50, l = 150
Mode = l + \(\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right)\) × h
= 150 + \(\left(\frac{6-5}{12-5-5}\right)\) × 50
= 150 + \(\frac{50}{2}\)
= 150 + 25
= 175

CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions

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

Question 36.
Due to ongoing COVID-19 crises, Surbhi Medical store has started stocking up and sell masks of decent quality as sourced from a disposable medical device manufacturer. The owner of Surbhi Medical store is selling two types of masks currently – A and B. The cost of one type A mask is ₹ 10 and of one type B mask is ₹ 12. In the month of April, 2020, the store sold 100 masks for total sales of ₹ 1082.

Due to great demand and short supply, the store has increased the price of each type by ₹ 1 from May 1, 2020 . In the month of May, 2020, the store sold 250 masks for total sales of ₹ 2920.
CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions 19
On the basis of the above information, answer the following questions:
(A) How many masks of each type were sold in the month of April? [1]
OR
How many masks of each type were sold in the month of May?
(B) If the store had sold 125 masks of each type, what would be its sale in month of May? [1]
(C) What percent of masks of each type sale was increased in the month of May, compared with the sale of month April? [2]
Answer:
(A) Let, the mask of type A sold in April be x and type of mask B sold in April be y.
Then, x + y = 100 …(i)
and 10x + 12y = 1082 …(ii)
Multiply equation (i) by 10 and subtract (ii) from (i).
10x + 10y = 1000
10x + 12y = 1082
-2y = – 82
y = 41
Then, x = 100 – 41 = 59

OR

For May, Let, the mask of type A sold be x and type B be y.
Then, x + y = 250 …(i)
and 11x + 13y = 2920 …(ii)
Multiply equation (i) by 11 and subtract it from equation (ii), we get
11x + 11y = 2750
11x + 13y = 2920
– 2 y = – 170
y = 85
and x = 250 – 85 = 165

(B) 11 × 125 + 13 × 125 = 1375 + 1625
= ₹ 3000

(C) Increase in type A = \(\frac{165-59}{59}\) × 100
= 179.66%
= 180%

Increase in type B = \(\frac{85-41}{41}\) × 100
= 107.31%
= 110%

Question 37.
A game at a stall in Diwali fare involves using a spinner first as a pre-cursor to complete the game with certain rules. If the spinner stops at a particular number, then the player is allowed to roll a 6-faced unbiased die.
CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions 20
Rules:
(1) If the spinner stops at a particular number, then the player is allowed to roll a 6- faced unbiased dice.
(2) If the spinner stops at any other number, you get to try again and only two tries allowed maximum.
(3) If you reach the next stage and roll a dice, the shopkeeper will open a chit to disclose the number if it matches, the player gets a prize.

On the basis of the above information, answer the following questions:
(A) What is the probability of getting an odd number on the spinner? [1]
(B) If getting an even number on the spinner allows a player to roll the die, then find the probability of his rolling the die. [1]
(C) If the player is allowed to roll the die and getting a prime number entitles him to get prize, then find the probability of his winning the prize and if getting a square number on the spinner allows a player to roil the die, then find the probability of his rolling the die.
OR
If the player is allowed to roll the die and getting a number greater than 5 entitles him to get prize, then find the probability of his winning the prize. [2]
Answer:
(A) Total number of cases = 6
Favourable outcomes = (1, 9, 15) i.e., 3
P (dice will be thrown) = \(\frac{3}{6} = \frac{1}{2}\)

(B) Even number = 4, 8, 12
∴ P(getting in even number) = \(\frac{3}{6} = \frac{1}{2}\)

(C) Prime no. on dice = 2, 3, 5, i.e., (3) outcomes
∴ Total outcomes = 6
∴ P(getting a prime no.) = \(\frac{3}{6} = \frac{1}{2}\)
Total outcomes = 6
Favourable outcomes = {1, 4, 9} i.e., 3
P (dice will be thrown) = \(\frac{3}{6} = \frac{1}{2}\)
OR
Total outcomes = 6
Favorable outcomes is 6 i.e., 1
P(getting a no. greater than 5) = \(\frac{1}{6}\)

CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions

Question 38.
Radio towers are typically tall structures designed to support antennas for telecommunications and broadcasting, including television. There are 2 main types: guyed and self-supporting structures.
They are among the tallest human-made structures. Masts are often named after the broadcasting organizations that originally built them or currently use them.
CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions 21
On a similar concept, a radio-station tower was built in two sections A and B. From a point 24 m from the base of the tower, the angle of elevation of the top of section A is 30° and the angle of elevation of the top of section B is 45°.
CBSE Sample Papers for Class 10 Maths Basic Set 4 with Solutions 22
On the basis of the above information, answer the following questions:
(A) Find the height of the section A. [1]
(B) Find the height of the section B. [1]
(C) Find the length of the wire structure from the point O to the top of section A.
OR
Find the length of the wire structure from the point O to the top of section B. [2]
Answer:
(A) In ∆AOC,
tan 30° = \(\frac{A C}{O C}\)
⇒ AC = 24 × \(\frac{1}{\sqrt{3}}\)
= 8√3
= 13.84 m

(B) In ∆BOC,
tan 45° = \(\frac{B C}{O C}\)
⇒ BC = OC
⇒ BC = 24 m
Now, AB = 24 – 13.84
= 10.16 m

(C) In ∆OAC,
cos 30° = \(\frac{O C}{O A}\)
⇒ \(\frac{\sqrt{3}}{2}=\frac{24}{\mathrm{OA}}\)
⇒ OA = \(\frac{48}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)
= 16√3
= 27.68 ≃ 27.7 m

OR
In ∆OBC,
cos 45° = \(\frac{O C}{O B}\)
⇒ \(\frac{1}{\sqrt{2}}=\frac{24}{\mathrm{OB}}\)
⇒ OB = 24√2
= 33.84 m