**Haryana (HBSE)** Class 10 Maths Question Paper 2022 Answer Key. HBSE Class 10 Maths Solved Question Paper 2022. HBSE Class 10 Maths Previous Year Question Paper with Answer. HBSE Board Solved Question Paper Class 10 Maths 2022. HBSE 10th Question Paper Download 2022 Answer Key. HBSE Class 10 Maths Paper Solution 2022. HBSE Class 10th Maths Question Paper 2022 PDF Download with Answer Key.

**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