HBSE Class 10 Maths Question Paper 2022 Answer Key

Q1. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red (ii) not red ? 

Sol. 

No. of Red balls = 3 

No. of Black balls = 5 

so, Total balls = 3 + 5 = 8 

(i) P(red) = 3/8 

(ii) P(not red) = P(black) = 5/8 

 

Q2. Find two positive numbers whose sum is 27 and product is 182. 

Sol. 

Let two numbers are x and y. 

x + y = 27 …….(i) 

xy = 182 …….(ii) 

From eqn.(i),  

y = 27 – x ……(iii) 

Putting value of y = 27 – x in equation (ii), 

x(27 – x) = 182 

27x – x² = 182 

x² – 27x + 182 = 0 

x² – 14x – 13x + 182 = 0 

x(x – 14) – 13(x – 14) = 0 

(x – 14)(x – 13) = 0

Either x – 14 = 0 or x – 13 = 0

So, x = 14 or 13 

y = 27 – x = 27 – 14 = 13  or  y = 27 – 13 = 14 

Two positive integers are 13 and 14. 

 

Q3. Find the co-ordinates of a point A where AB is the diameter of a circle whose centre is O(2,-3) and co-ordinates of B is (1, 4). 

 

O is Mid point of AB,

A(x, y) and O(2, -3) and B(1, 4) 

Using Mid Point Formula,

O(2, -3) = (x+1)/2 , (y+4)/2 

(x+1)/2 = 2 and (y+4)/2 = -3

x+1 = 4 and y+4 = -6 

x = 3 and y = -10 

Co-ordinates of A(x, y) are (3, -10). 

 

Q4. Evaluate: 

[5cos²60° + 4sec²30° – tan²45°] ÷ [sin²30° + cos²30°] 

Sol. 

= [5(1/2)² + 4(2/√3)² – (1)²] ÷ [(1/2)² + (3/4)²]

= [5/4 + 16/3 – 1] ÷ [4/4] 

= [67/12] ÷ 1 = 67/12 

 

Q5. An underground water tank is in the form of a cuboid of edges 48 m, 36 m and 28 m. Find the volume of the tank. 

Sol. 

Volume of cuboid = length × breadth × height = 48 × 36 × 28 = 48384 m³ 

 

Q6. Find the median of the following data : 

Sol. 

Here  cf = 36,  n = 100,  l = 7,  f = 40,  h = 10-7 = 3 

Median = l + (n/2 – cf)/f × h

              = 7 + (50 – 36)/40 × 3 

              = 7 + 14/40 × 3 

              = 7 + 1.05 = 8.05 

 

Q7. A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator, it becomes 5/6. Find the fraction. 

Sol. 

Let Numerator = x

and Denominator = y

so, Fraction = x/y

Given that, 

(Numerator+2) ÷ (Denominator+2) = 9÷11 

(x+2)/(y+2) = 9/11

11x + 22 = 9y + 18

11x – 9y = – 4  ……..(i) 

Multiply eqn.(i) by 5,

55x – 45y = -20 ………(iii) 

Given that, 

(Numerator+3) ÷ (Denominator+3) = 5÷6 

(x+3)/(y+3) = 5/6 

6x + 18 = 5y + 15

6x – 5y = -3 ……….(ii) 

Multiply eqn.(ii) by 9, 

54x – 45y = -27 ………(iv) 

Subtract eqn.(iv) from eqn.(iii), 

x = 7 

Putting value of x=7 in eqn.(i),

11(7) – 9y = -4 

77 – 9y = -4 

-9y = -81 

y = 9 

So, Fraction = x/y = 7/9 

 

Q8. Find the 31st term of an A. P. whose 11th term is 38 and 16th term is 73. 

Sol. 

A.P. is a, a + d, a + 2d, …….., a + (n-1)d 

a11 = a + 10d = 38 …….(i) 

a16 = a + 15d = 73 …….(ii) 

Subtract eqn.(i) from eqn.(ii), 

5d = 35 

d = 7 

Putting value of d = 7 in eqn.(i), 

a + 10(7) = 38 

a = 38 – 70 = -32 

So, a31 = a + 30d = -32 + 30(7) = -32 + 210 = 178 

 

Q9. A ladder is placed against a wall such that its foot is at a distance of 2.5m from the wall and its top reaches a window 6m above the ground. Find the length of the ladder. 

 

Height of window = AC = 6 m

Ladder from base of wall = BC = 2.5 m

Since the wall will be perpendicular to ground ∠ACB = 90° 

∆ACB is a right angle triangle

So, Using Pythagoras theorem

(Hypotenuse)² = (Perpendicular)² + (Base)²

AB² = AC² + BC²

AB² = 6² + (2.5)²

AB² = 36 + 6.25 = 42.25 

AB = √42.25 = 6.5 m 

Length of the ladder = AB = 6.5 m 

 

Q10. Find the area of the shaded region as shown in figure where ABCD is a square of side 14 cm. 

 

Area of shaded region = Area of square ABCD – Area of 4 circles

Area of square ABCD = (side)² = 14² = 196 cm² 

Diameter of each circle = AB/2 = 14/2 = 7 cm 

Radius of each circle = Diameter/2 = 7/2 = 3.5 cm

Area of one circle = πr² = 22/7 x 3.5 x 3.5 = 38.5 cm²

So, Area of 4 circle = 4 x Area of one circle 

                                    = 4 x 38.5 = 154 cm²

Area of shaded region = Area of square ABCD – Area of 4 circles = 196 – 154 = 42 cm² 

Area of shaded region = 42 cm² 

 

Q11. The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower.

 

Given tower be AB

When Sun’s altitude is 60°, ∠ACB = 60°

Length of shadow = BC

When Sun’s altitude is 30°, ∠ADB = 30° 

Length of shadow= DB

Shadow is 40 m when angle changes from 60° to 30°,  CD = 40 m

Since tower is vertical to ground, ∠ABC = 90° 

In right angle triangle ∆ABC, 

tanC = AB/CB

tan60° = AB/CB 

√3 = AB/CB 

CB = AB/√3 …….(i) 

In right angle triangle ∆ABD, 

tanD = AB/DB 

tan30° = AB/DB 

1/√3 = AB/DB 

DB = √3AB

DC + CB = √3AB 

40 + CB = √3AB 

CB = √3AB – 40 ……..(ii) 

From equation (i) & (ii), 

AB/√3 = √3AB – 40 

AB = √3(√3AB) – 40√3 

AB = 3AB – 40√3 

3AB – AB = 40√3

2AB = 40√3 

AB = 20√3 

Hence, Height of the tower = 20√3 m 

 

                                             OR 

The angle of elevation of the top of a tower from a point on the ground, which is 30m away from the foot of the tower, is 30° . Find the height of the tower. 

 

Let tower be AB

Distance of point C from foot of tower = BC = 30 m

∠ACB = Angle of elevation = 30°

Tower is vertical, ∠ABC = 90°

In right triangle ∆ABC,

tanC = AB/BC 

tan30° = AB/30