HBSE Class 10th Maths Solved Question Paper 2022

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 

1/√3 = AB/30 

AB = 30/√3 

Multiply numerator and denominator by √3, 

AB = (30/√3) × √3/√3 = 10√3 m

Hence, Height of tower h = AB = 10√3 m 

Q12. Solve the following pair of equations by reducing them to a pair of linear equations : 

5/(x-1) + 1/(y-2) = 2 and 6/(x-1) – 3/(y-2) = 1 

Sol. 

Let 1/(x-1) = u and 1/(y-2) = v 

So, 

5u + v = 2 …….(i) 

6u – 3v = 1 ……..(ii) 

Multiply eqn.(i) by 3 and add in eqn.(ii), we get 

(15u + 3v) – (6u – 3v) = 6 + 1 

21u = 7 

u = 1/3 

Putting value of u=1/3 in eqn.(i), 

5(1/3) + v = 2 

5/3 + v = 2 

v = 1/3 

But 1/(x-1) = u and 1/(y-2) = v 

So, 1/(x-1) = 1/3 and 1/(y-2) = 1/3 

x – 1 = 3 and y – 2 = 3 

x = 4, y = 5 

                                         OR 

Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu ? 

Sol. 

Let Present age of Nuri = x  years

& Present age of Sonu = y  years

Five years ago,

Nuri’s Age = x – 5

Sonu’s Age = y – 5

Nuri was thrice as old as Sonu, 

x – 5 = 3(y – 5) 

x – 5 = 3y – 15 

x – 3y = – 15 + 5

x – 3y = – 10 ……..(i) 

Also,

Ten years later, 

Nuri’s Age = x + 10

Sonu’s Age = y + 10 

Nuri will be twice as old as Sonu, 

x + 10 = 2(y + 10)

x + 10 = 2y + 20 

x – 2y = 20 – 10 

x – 2y = 10 ……..(ii) 

Subtract eqn.(ii) from eqn.(i), 

-y = -20 

y = 20  

Putting y = 20 in eqn.(ii), 

x – 2(20) = 10

x – 40 = 10

x = 10 + 40 = 50 

So, Present age of Nuri (x) = 50 years

& Present age of Sonu (y) = 20 years

Q13. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60° and write steps of construction.

Steps of construction :

Step: I- Draw a circle of radius 5 cm

Step: II- Draw horizontal radius OQ

Step: III- Draw angle 120° from point O 

Let the ray of angle intersect the circle at point R 

Step: IV- Now draw 90° from point Q 

Step: V- Draw 90° from point R 

Step: VI- Where the two arcs intersect, mark it as point P

So, PQ and PR are the tangents at an angle of 60°. 

                                             OR 

Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2∠OPQ.

We know that, the lengths of tangents drawn from an external point to a circle are equal.

∴ TP = TQ

In ΔTPQ,

TP = TQ

∠TQP = ∠TPQ ……(i) (In a triangle, equal sides have equal angles opposite to them)

∠TQP + ∠TPQ + ∠PTQ = 180º (Angle sum property)

∴ 2∠TPQ + ∠PTQ = 180º (using eqn.i)

∠PTQ = 180º – 2 ∠TPQ …….(ii)

We know that, a tangent to a circle is perpendicular to the radius through the point of contact.

OP ⊥ PT,

∴ ∠OPT = 90º

∠OPQ + ∠TPQ = 90º

∠OPQ = 90º – ∠TPQ

2∠OPQ = 2(90º – ∠TPQ) = 180º – 2 ∠TPQ ..

…..(iii)

From (ii) and (iii), we get

∠PTQ = 2∠OPQ

SET-A (Objective Questions) 

Q1. Express 5005 as a product of prime factors.

Sol. 5005 = 5 × 7 × 11 × 13

Q2. 3 + 2√5 is a rational number or irrational number. 

Ans. irrational number

Q3. Find HCF of 510 and 92. 

Sol. 

510 = 2 × 3 × 5 × 17

92 = 2 × 2 × 23 

HCF = 2 

Q4. The rational number 17/8 is a terminating or non-terminating decimal expansion. 

Ans. Terminating decimal expansion (denominator 8 = 2 × 2 × 2, if denomination is 2 or 5 or 2&5) 

Q5. Product of roots of the quadratic polynomial 6x²  – 7x – 3 is ……….. 

Sol. 

Here  a = 6, b = -7, c = -3 

αβ = constant/coefficient of x² = c/a 

αβ = -3/6 = -1/2 

Q6. If sum of roots of the quadratic polynomial is -1/4 and product is 1/4 then the quadratic polynomial is :

(A) 4x² + x + 1

(B) 4x² – x – 1

(C) 4x² – x + 1

(D) 4x² + x – 1 

Ans. (A) 4x² + x + 1 

α + β = -b/a = -1/4 ,  αβ = c/a = 1/4 

x² – (α+β)x + αβ

x² – (-1/4)x + 1/4 

x² + 1/4 x + 1/4 

Multiple by 4, then 

4x² + x + 1 

Quadratic polynomial is 4x² + x + 1. 

Q7. Find out whether the following pair of linear equations is consistent or inconsistent :

3x + 2y = 5 and 2x – 3y = 7 

Sol. 

a1/a2 = 3/2 ,  b1/b2 = 2/-3 

a1/a2 ≠ b1/b2 

3/2 ≠ 2/-3 

So, it is consistent. 

Q8. The solution of linear equations 3x + 4y = 10 and x – y = 1  is : 

(A) x = 1, y = 2 

(B) x = 3, y = 1

(C) x = 2, y = 1

(D) x = 4, y = 3 

Ans. (C) x = 2, y = 1

3x + 4y = 10 ………(i) 

x – y = 1 ………(ii) 

Multiply eqn.(ii) by 4 and add in eqn.(i), 

7x = 14 

x = 14/7 = 2 

Put x = 2 in equation (ii), 

2 – y = 1 

– y = 1 – 2 = – 1 

y = 1

So, x = 2, y = 1 

Q9. Factorise the quadratic equation 2x² + x – 6=0 into linear factors. 

Sol. 

2x² + x – 6=0 

2x² + 4x – 3x – 6 = 0

2x(x+2) – 3(x+2) = 0

(x + 2)(2x – 3) = 0 

Q10. The roots of the quadratic equation 6x² – x – 2 = 0 are ……….. 

Sol. 

6x² – x – 2 = 0 

6x² – 4x + 3x – 2 = 0 

2x(3x-2) + 1(3x-2) = 0 

(3x – 2)(2x + 1) = 0 

Either  3x – 2 = 0  or  2x + 1 = 0

x = 2/3  or  x = -1/2 

The roots of the quadratic equation are 2/3 and -1/2 . 

Q11. If the roots of the quadratic equation 2x² + kx + 2 = 0 are equal, then the value of k is : 

(A) ± 2 

(B) + 5

(C) ± 3

(D) ± 4 

Ans. (D) ± 4 

2x² + kx + 2 = 0 , here a = 2, b = k, c = 2 

Discriminant (D) = b² – 4ac = 0 

k² – 4(2)(2) = 0 

k² = 16 

k = √16 = ± 4 

Q12. For which value of p does the pair of equations 4x + py + 8 = 0 and 2x + 2y + 2 = 0 has a unique solution ? 

Sol. 

For unique solution, 

a1/a2 ≠ b1/b2 

4/2 ≠ p/2 

p ≠ 4 

All values of p except 4. 

Q13. Write the common difference of A. P.  3, 1, -1, -3, ………. 

Sol. d = a2 – a1 = 1 – 3 = -2 

Q14. 17th term of A. P. 7, 13, 19, …… is …………. 

Here a = 7,  d = 13 – 7 = 6,  n = 17 

an = a + (n-1)d 

a17 = 7 + (17-1)6 = 7 + (16)6 = 7 + 96 = 103 

Q15. The sum of first n natural numbers is ……….. 

Ans. n(n+1)/2 

Q16. Write the next four terms of the A. P. 1, -1, -3, -5, ……….. 

Sol. 

d = – 1 – 1 = -2 

– 5 – 2 = -7 

– 7 – 2 = -9

– 9 – 2 = -11 

– 11 – 2 = -13 

Next four terms are -7, -9, -11, -13

Q17. All ……….. triangles are similar. (isosceles, equilateral) 

Ans. Equilateral 

Q18. ∆MNL and ∆QPR are similar triangles. In given figure which similarity criterion is used? 

(A) S. S. S.

(B) A. A. A.

(C) S. A. S.

(D) None of these 

Ans. (C) S. A. S.

Q19. Sides of triangles are given below. Determine which of them is a right triangle?

(i) 3 cm, 8 cm, 6 cm

(ii) 13 cm, 12 cm, 5 cm 

Ans. (ii) 13 cm, 12 cm, 5 cm 

For right angle triangle, H² = P² + B² 

13² = 12² + 5² 

169 = 144 + 25

Q20. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q, so that OQ = 12 cm. Then the length PQ is :

      (A) 12 cm

      (B) 13 cm

      (C) 8.5 cm

      (D) √119 cm 

Ans. (D) √119 cm 

Using Pythagoras theorem, H² = P² + B² 

PQ² = QO² – OP² = 12² – 5² = 144 – 25 = 119 

PQ = √119 cm 

Q21. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80°, then ∠POA is equal to :

(A) 50°

(B) 60°

(C) 70°

(D) 80° 

Ans. (A) 50° 

∠APO = ∠APB/2 = 80/2 = 40° 

∠POA + ∠PAO + ∠APO = 180°  (in ∆APO) 

∠POA + 90° + 40° = 180° 

∠POA + 130° = 180° 

∠POA = 180 – 130 = 50° 

Q22. A circle can have ……….. parallel tangents at the most. 

Ans. Two

Q23. In figure DE || BC. Find EC. 

Sol. 

AD/DB = AE/EC 

1.5/3 = 1/EC 

EC = 3/1.5 = 2 cm 

Q24. The distance between the points (-5, 7) and (-1, 3) is …….. 

Sol. 

A(-5, 7) and B(-1, 3) 

x1 = -5, x2 = -1, y1 = 7, y2 = 3 

Using Distance Formula, 

AB = √x2-x1)²+(y2-y1)² 

      = √(-1-(-5))²+(3-7)²

      = √(-1+5)²+(-4)² 

      = √16+16 = √32 = 4√2 unit 

      

Q25. The ratio in which the y-axis divides the line-segment joining the points (5, -6) and (-1, -4) is :

(A) 1 : 5

(B) 5 : 1

(C) 3 : 2

(D) 2 : 3 

Ans. (B) 5 : 1 

Using Section Formula, 

x = (m1x2 + m2x1) ÷ (m1 + m2) 

0 = [k(-1) + 1(5)] ÷ (k + 1) 

0 × (k+1) = -k + 5

0 = -k + 5

k = 5 

k : 1 = 5 : 1 

Q26. The co-ordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2 : 3 is :  

(A) (3, 1)

(B) (5, 2)

(C) (2, 5)

(D) (1, 3) 

Ans. (D) (1, 3)

Using Section Formula, 

x = (m1x2 + m2x1) ÷ (m1 + m2) 

   = [2(4) + 3(-1)] ÷ (2 + 3) 

   = (8 – 3) ÷ 5 = 5 ÷ 5 = 1 

y = (m1y2 + m2y1) ÷ (m1 + m2) 

   = [2(-3) + 3(7)] ÷ (2 + 3) 

   = (-6 + 21) ÷ 5 = 15 ÷ 5 = 3

So, P(x,y) = (1,3) 

Q27. If sinA = 3/4, find the value of tanA. 

sinA= Perpendicular/Hypotenuse = P/H = 3/4 

P = 3, H = 4

Using Pythagoras theorem, H² = P² + B² 

B² = H² – P² = 4² – 3² = 16 – 9 = 7 

Base, B = √7 

tanA = Perpendicular/Base = P/B = 3/√7 

Q28. (1-tan²45°)÷(1+tan²45°) is equal to : 

(A) tan 90°

(B) 1

(C) sin 45°

(D) 0 

Ans. (D) 0 

(1-tan²45°) ÷ (1+tan²45°) = (1 – 1²) ÷ (1 + 1²)

= (1 – 1) ÷ (1 + 1) = 0 ÷ 2 = 0 

Q29. sin 60° cos 30° + sin 30° cos 60° is equal to ………….. 

Sol. 

sin 60° cos 30° + sin 30° cos 60°

= √3/2 × √3/2 + 1/2 × 1/2 

= 3/4 + 1/4 = 4/4 = 1 

Q30. The point on the x-axis which is equidistant from (2, -5) and (-2, 9) is :

(A) (-7, 0)

(B) (0, -7)

(C) (-5, 0)

(D) None of these

Ans. (A) (-7, 0)

A(x,0) , B(2,-5) , C(-2,9) 

Using distance formula = √(x2-x1)²+(y2-y1)²

AB = AC 

√(x-2)²+(-5)² = √(x+2)²+(9)² 

Cancel square root from both side, 

x² – 4x + 4 + 25 = x² + 4x + 4 + 81 

– 8x = 56 

x = -7 

So, A(-7, 0) 

Q31. Area of the sector of a circle with radius 6 cm and angle of sector 30° is :

(A) 7π cm²

(B) 9π cm²

(C) 3π cm² 

(D) 6π cm² 

Ans. (C) 3π cm²

Area = θ/360 × πr² = 30/360 × π(6)² 

= 1/12 × π(36) = 3π cm² 

Q32. The length of the minute hand of a clock is 14 cm. The area swept by the minute hand in 5 minutes is :

(A) 162 cm²

(B) 154/3 cm² 

(C) 205/3 cm²

(D) None of these 

Ans. (B) 154/3 cm² 

Minute hand of a clock, 

Angle in 60 minutes = 360° 

Angle in 1 minutes = 360/60 = 6°

Angle in 5 minutes = 6 × 5 = 30° 

Area = θ/360 × πr² 

= 30/360 × 22/7 × (14)² = 154/3 cm² 

Q33. Find the volume of the cone of height 24 cm and radius of base 6 cm. 

Sol.

Volume = 1/3 × πr²h = 1/3 × 22/7 × (6)² × 24 

= 288π cm³  or  6336/7 cm³

Q34. Write the formula for finding out the curved surface area of the cylinder. 

Ans. Curved Surface Area (CSA) = 2πrh 

Q35. A metallic sphere of radius 3 cm is melted and recast into the shape of a cylinder of radius 2 cm. The height of the cylinder is :

(A) 10cm

(B) 8cm

(C) 9 cm

(D) 7 cm 

Ans. (C) 9 cm

Volume of sphere = Volume of cylinder

4/3 × πR³ = πr²h 

4/3 × 22/7 × 3 × 3 × 3 = 22/7 × 2 × 2 × h 

h = 9 cm 

Q36. The probability of an event is greater than or equal to …….. and less than or equal to …….. 

Ans. 0, 1 (0 ≤ P ≤ 1)

Q37. Which of the following can not be the probability of an event ?

(A) 2/3 

(B) 15%

(C) 0.7

(D) -1.5 

Ans. (D) -1.5 

0 ≤ P ≤ 1  (probability never be negative) 

Q38. A die is thrown once. The probability of getting an odd number is …………. 

Sol. 

Total numbers= 6 (1,2,3,4,5,6) 

Odd numbers = 3 (1,3,5) 

Probability (Odd) = 3/6 = 1/2 = 0.5 

Q39. The mean of the following data is : 

(A) 10.5

(B) 8.1 

(C) 11.5 

(D) 9.5 

Ans. (B) 8.1 

Mean = ∑fX/∑f = (1×1 + 2×3 + 1×5 + 5×7 + 6×9 + 2×11 + 3×11) ÷ (20) 

          = 162 ÷ 20 = 8.1  

Q40. The wickets taken by a bowler in 10 cricket matches are as follows : 

          2, 6, 4, 5, 0, 2, 1, 3, 2, 3  

Find the mode of the data. 

Ans. Mode = 2 (most repeated number) 

SET-B (Objective Questions) 

Q1. Express 3825 as a product of its prime factors. 

Sol. 3825 = 5 × 5 × 3 × 3 × 17 = 5² × 3² × 17 

Q2. What is HCF of 64 and 96 ? 

Sol. 

96 = 2 × 2 × 2 × 2 × 2 × 3 

64 = 2 × 2 × 2 × 2 × 2 × 2 

HCF = 2 × 2 × 2 × 2 × 2 = 32 

Q3. Find a quadratic polynomial, the sum and product of whose zerocs arc -3 and 2 respectively. 

Sol. 

α + β = -3 

αβ = 2 

x² – (α+β)x + αβ 

x² – (-3)x + 2

Quadratic polynomial is x² + 3x + 2. 

Q4. For what values of K does the pair of linear equations 4x + Ky + 8 = 0 and 2x + 2y + 2 = 0 has unique solution ? 

Sol. 

For unique solution, 

a1/a2 ≠ b1/b2 

4/2 ≠ K/2 

K ≠ 4 

all values of K except 4. 

Q5. For what values of K, quadratic equation Kx² – 4x + 2 = 0 has equal roots ? 

Sol. 

Here a = K, b = -4, c = 2 

Discriminant, D = b² – 4ac = 0 

b² – 4ac = 0 

(-4)² – 4(K)(2) = 0 

16 – 8K = 0 

8K = 16 

K = 16/8 = 2 

Q6. Find 5th term of A. P. 10, 7, 4 ……… 

Sol. 

Here a = 10,  d = 7 – 10 = -3,  n = 5 

an = a + (n-1)d 

a5 = 10 + (5-1)(-3) = 10 + (4)(-3) = 10 -12 = -2 

Q7. In figure DE||BC. Find length of EC.

Ans. 

AD/DB = AE/EC 

1.5/3 = 1/EC 

EC = 3/1.5 = 2 

Q8. If sinA = 3/4 , find value of cosA. 

Sol. 

sin²θ + cos²θ = 1  

cos²θ = 1 – sin²θ 

cos²θ = 1 – (3/4)² = 1 – 9/16 = 7/16 

cosθ = √7/4 

Q9. Find the value of sin45° + cos45°. 

Sol. 

sin45° + cos45° = 1/√2 + 1/√2 = 2/√2

Rationalise the denominator, 

2/√2 × √2/√2 = √2 

Q10. Find the area of a Sector of a circle with radius 7 cm if angle of the Sector is 60°. 

Sol. Area = θ/360 × πr² = 60/360 × 22/7 × 7 × 7 = 77/3 cm² 

Q11. Whether the Rational number 17/8 will have a terminating decimal expansion or non terminating repeating decimal expansion. 

Ans. Terminating decimal expansion (denominator 8 = 2×2×2, if denomination is 2 or 5 or 2&5) 

Q12. Find discriminant of quadratic equation 3x² – 5x + 2 = 0. 

Sol. 

Here a = 3, b = -5, c = 2 

Discriminant (D) = b² – 4ac = (-5)² – 4(3)(2) = 25 -24 = 1 

D = 1 

Q13. Find common difference of A. P. 7, 5, 3, 1 ………. 

Sol. Common difference (d) = a2 – a1 = 5 – 7 = -2 

Q14. Find the sum of first 6 terms of A.P. 2, 7, 12 ………… 

Sol. 

Here a = 2,  d = 7 – 2 = 5,  n = 6 

Sn = n/2 [2a + (n-1)d] 

S6 = 6/2 [2×2 + (6-1)5] = 3 [4 + 25] = 3(29) = 87 

Q15. All circles are ………. (Similar / Congruent) 

Ans. Similar 

Q16. A tangent to a circle intersects it in ……… points. 

Ans. One Point (Point of contact or Point of tangency) 

Q17. The lengths of tangents drawn from an external point to a circle are ……….. 

Ans. Equal Tangents 

Q18. Find distance between the points (-5, 7) and (-1, 3). 

Sol. 

Here A(-5, 7) and B(-1, 3) 

Using Distance Formula, 

AB = √(x2-x1)²+(y2-y1)² 

      = √(-1-(-5))²+(3-7)²

      = √(-1+5)²+(-4)² = √(4)²+(-4)² 

      = √16+16 = √32 = 4√2 unit  

Q19. The value of cos²θ + sin²θ = ……… 

Ans. 1 

Q20. If P(E) = 0.15, what is the probability of event ‘not E’ ? 

Sol. 

P(E) + P(not E) = 1 

P(not E) = 1 – P(E) = 1 – 0.15 = 0.85 

Q21. Which of these is an irrational number?

(A) √4

(B) √5

(C) √9

(D) √16 

Ans. (B) √5

√4 = 2, √9 = 3, √16 = 4

Q22. Sum of zeroes of quadratic Polynomial 2x² + 5x – 3 is : 

(A) -3/2 

(B) -2/5 

(C) -5/2

(D) -3/5 

Ans. (C) -5/2

Here a = 2, b = 5, c = -3 

α + β = – coefficient of x/coefficient of x² = -b/a = -5/2 

Q23. The pair of linear equations x – 3y – 3 = 0 and 3x – 9y – 2 = 0 has solution:

(A) Unique Solution 

(B) No Solution

(C) Infinitely Many Solutions

(D) None of these 

Ans. (B) No Solution

x – 3y – 3 = 0 and 3x – 9y – 2 = 0 

Here a1 = 1, a2 = 3, b1 = -3, b2 = -9, c1 = -3, c2 = -2 

a1/a2 = 1/3, b1/b2 = -3/-9 = 1/3, c1/c2 = -3/-2 = 3/2 

a1/a2 = b1/b2 ≠ c1/c2 

1/2 = 1/2 ≠ 3/2 

No Solution 

Q24. Which of these is a quadratic equation ?

(A) (x + 2)(x + 1) = (x – 1)(x + 3)

(B) x² + 3x + 1 = (x – 2)²

(C) (x + 1)² = 2(x – 3)

(D) (x + 2)³ = 2x(x² – 1) 

Ans. (C) (x + 1)² = 2(x-3)

(x + 1)² = 2(x-3) 

x² + 2x + 1 = 2x – 6 

x² + 7= 0

Q25. Roots of the quadratic equation x² – 3x – 10 = 0 are : 

(A) 5, 2

(B) 5,-2

(C) -5, 2

(D) 5, 3 

Ans. (B) 5,-2

x² – 3x – 10 = 0 

x² – 5x + 2x – 10 = 0 

x(x-5) + 2(x-5) = 0 

(x-5)(x+2) = 0 

Either x-5 = 0  or  x+2 = 0 

So,  x = 5 or -2 

Roots of quadratic equation are 5 and -2. 

Q26. Which one is A.P. series ? 

(A) 2, 4, 8, 12 ………..

(B) 0.2, 0.22, 0.222 ……….

(C) 3, 5, 7, 9 ……….

(D) 1, 2, 4, 8 ……….  

Ans. (C) 3, 5, 7, 9 ……….

Common difference (d) = 5 – 3 = 7 – 5 = 9 – 7 = 2 

Q27. ∆ABC and ∆QRP are similar. Which similarity criterion is used? 

(A) S.A.S. 

(B) A.A.A.

(C) S.S.S. 

(D) None of these 

Ans. (C) S.S.S. 

Q28. From a point Q, the length of the tangent to a circle is 24 cm and distance of Q from the centre is 25 cm. The radius of the circle is : 

(A) 7 cm

(B) 12 cm

(C) 15 cm

(D) 24.5 cm 

Ans. (A) 7 cm

Using Pythagoras Theorem, 

H² = P² + B² 

25² = r² + 24² 

r² = 25² – 24² = 625 – 576 = 49 

r = √49 = 7 cm 

Q29. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to :

(A) 80°

(B) 70°

(C) 60°

(D) 50° 

Ans. (D) 50° 

∠APO = ∠APB/2 = 80/2 = 40° 

∠POA + ∠PAO + ∠APO = 180°  (in ∆APO) 

∠POA + 90° + 40° = 180° 

∠POA + 130° = 180° 

∠POA = 180° – 130° = 50° 

Q30. Co-ordinates mid point of line joining two points (-1, 7) and (4,-3) is :

(A) (1/2, 2)

(B) (2, 1)

(C) (3/2, 2)

(D) (-3/2, 1/2) 

Ans. (C) (3/2, 2)