# HBSE Class 10th Maths Solved Question Paper 2021

HBSE Class 10th Maths Solved Question Paper 2021

HBSE Class 10 Maths Previous Year Question Paper with Answer. HBSE Board Solved Question Paper Class 10 Maths 2021. HBSE 10th Question Paper Download 2021. HBSE Class 10 Maths Paper Solution 2021. Haryana Board Class 10th Maths Question Paper 2021 Pdf Download with Answer.

Subjective Questions

Q1. Find two consecutive positive integers, the sum of whose squares is 365.

Sol.

Let First integer = x

So, Second integer = x+1

Also given that

Sum of squares = 365

(First number)² + (Second number)² = 365

x² + (x+1)² = 365

x² + x² + 1² + 2x = 365

2x² + 2x +1 – 365 = 0

2x² + 2x – 364 = 0

divide both side by 2,

x² + x – 182 = 0

x² + 14x – 13x – 182 = 0

x(x+14) – 13(x+14) = 0

(x+14)(x-13) = 0

so, x = -14, 13

x is positive, so use x=13

First integer = x = 13

Second integer = x+1 = 1+13 = 14

Q2. If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a parallelogram, taken in order, find the value of p.

Sol.

We know that diagonals of parallelogram bisect each other

So, O is the mid-pint of AC & BD

We find x co-ordinate of O from both AC & BD

Finding mid-point of AC,

x-coordinate of O = (x1+x2)/2 = (6+9)/2 = 15/2

Finding mid-point of BD,

x-coordinate of O = (x1+x2)/2 = (8+p)/2

Comparing x-coordinate of both,

(8+p)/2 =15/2

8+p = 15

p = 15-8 = 7

Q3. Evaluate :  [7sin²30° + 6cosec²60° – cot²45°] ÷ [sin²60° + cos²60°]

Sol.

= [7(1/2)² + 2(2/√3)² – (1)²] ÷ [(√3/2)² + (1/2)²]

= [7/4 + 8 – 1] ÷ [3/4 + 1/4]

= [(7+32-4)/4] ÷ [(3+1)/4]

= [35/4] ÷ [4/4]

= 35/4

Q4. Two cubes each of volume 64cm³ are joined end to end. Find the surface area of the resulting cuboid.

Sol.

It is given that

Volume of 1 cube = 64 cm³

(side)³ = 64 cm³

a³ = 64 = 4³

side = a = 4 cm

Length of cuboid = l = a + a = 4 cm + 4 cm = 8 cm

Breadth = b = a = 4 cm

Height = h = a = 4 cm

Surface Area of cuboid = 2(lb + bh + hl)

= 2(8×4 + 4×4 + 4×8)

= 2(32 + 16 + 32) = 2(80) = 160 cm²

Hence, surface area of cuboid 160 cm².

Q5. A bag contains 7 red balls and 3 green balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is (i) red (ii) green.

Sol.

No. of Red balls = 7

No. of green balls = 6

so, Total balls = 7 + 6 = 13

(i) P(red) = 7/13

(ii) P(green) = 6/13

Q6. If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes fraction 1/2 if we only add 1 to the denominator. What is the fraction?

Sol.

Let Numerator = x

and Denominator = y

so, Fraction = x/y

Given that,

(Numerator+1) ÷ (Denominator-1) =1

(x+1) ÷ (y-1) = 1

x+1 = y-1

x – y = -2 ………(i)

Given that,

(Numerator) ÷ (Denominator+1) = 1/2

(x) ÷ (y+1) = 1/2

2x = y+1

2x – y = 1 ………(ii)

Subtract equation (i) from equation (ii),

x = 3

use x = 3 in equation (i)

3 – y = -2

y = 3+2 = 5

Hence, Original Fraction = x/y = 3/5

Q7. Find the 20th term from the last of the A.P. 3, 8, 13, …….., 253.

Sol.

Given AP 3, 8, 13, …….., 253

Now, 20th term from last of AP 3, 8, 13, ….., 253 = 20th term of AP 253, 248, 243, …….., 8, 3

Thus, our AP is 253, 248, 243, ….., 8, 3

Here a = 253, d = 248 – 253 = – 5, n = 20

an = a + (n-1)d

an = 253 + (20-1)(- 5)

an = 253 + 19(- 5)

an = 253 – 95 = 158

So the 20th term from the Last is 158.

Q8. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.

Sol.

Height of window = AC = 8m

Length of ladder = AB = 10m

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

10² = 8² + BC²

100 = 64 + BC²

100 – 64 = BC²

36 = BC²

BC = √36 = 6 m

Hence, distance of the foot of the ladder from base of the wall (BC) = 6m

Q9. Find the area of the shaded region as shown in figure, where ABCD is a square of side 16 cm.

Sol.

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

Area of square ABCD = (side)² = 16² = 256 cm²

Diameter of each circle = AB/2 = 16/2 = 8 cm

Radius of each circle = Diameter/2 = 8/2 = 4 cm

Area of one circle = πr² = 3.14 x 4 x 4 = 50.24 cm²

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

= 4 x 50.24 = 200.96 cm²

Area of shaded region = Area of square ABCD – Area of 4 circles = 256 – 200.96 = 55.04 cm²

Area of shaded region = 55.04 cm²

Q10. Find the median of the following data :

Sol.

Here  cf = 13,  n = 60,  l = 20,  f = 20,  h = 10

Median = l + (n/2 – cf)/f × h

= 20 + (30 – 13)/20 × 10

= 20 + 17/2 = 20 + 8.5 = 28.5

Q11. Solve the following pair of equations by reducing them to a pair of linear equations :

6x + 3y = 6xy and 2x + 4y = 5xy

Sol.

Divide both equations by xy, then

6/y + 3/x = 6 ……..(i)

2/y + 4/x = 5 ……..(ii)

Let  1/x = u,  1/y = v

So, our equations become

6v + 3u = 6 ……..(iii)

2v + 4u = 5 ……..(iv)

Multiple equation (iv) by 3, then it subtract from equation (iv),

(6v-6v) + (3u-12u) = 6-15

-9u = -9

u = 1

Put u = 1 in equation (iv)

2v + 4(1) = 5

2v = 5 – 4 = 1

v = 1/2

Hence,  u = 1, v = 1/2

Now u = 1/x , x = 1/u = 1/1 = 1

and  v = 1/y, y = 1/v = 2

Hence x = 1, y = 2 is the solution of the given equation.

OR

Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages ?

Sol.

Let Present age of Jacob = x  years

& Present age of Jacob’s son = y  years

Five years hence (later),

Jacob’s Age = x + 5

Jacob son’s Age = y + 5

Age of Jacob will be three times of his son,

x + 5 = 3(y + 5)

x + 5 = 3y + 15

x – 3y = 15 – 5

x – 3y = 10 ……..(i)

Also,

Five years ago,

Jacob’s Age = x – 5

Jacob son’s Age = y – 5

Age of Jacob was seven times of his son,

x – 5 = 7(y – 5)

x – 5 = 7y – 35

x – 7y = -35 + 5

x – 7y = -30 ……..(ii)

Subtract equation (ii) from (i),

4y = 40

y = 10

Putting y = 10 in equation (i),

x – 3(10) = 10

x = 10 + 30 = 40

So, Present age of Jacob = x = 40 years

& Present age of Jacob’s son = y = 10 years

Q12. Draw a pair of tangents to a circle of radius 6 cm which are inclined to each other at an angle of 60° and write steps of construction.

Sol.

Steps of construction :

Step: I- Draw a circle of radius 6 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 concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Sol.

Connecting OP, OA and OB

OP = Radius of smaller circle = 3 cm

OA = OB = Radius of larger circle = 5 cm

Since AB is tangent to circle C₁

OP⊥AB (Tangent at any point of circle is perpendicular to the radius through point of contact)

:. ∠OPA = ∠OPB = 90°

Using Pythagoras theorem

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

In right triangle ∆OAP,

OA² = OP² + AP²

5² = 3² + AP²

25 = 9+ AP2

25 – 9 = AP²

AP² = 16

AP = √16 = 4 cm

In right triangle ∆OPB,

OB² = OP² + PB²

5² = 3² + PB²

25 = 9 + PB²

25 – 9 = PB²

PB² = 25 – 9

PB² = 16

PB = √16 = 4 cm

Hence, AB = AP + PB = 4 + 4 = 8 cm

Q13. The angle of elevation of the top of a tower from a point on the ground, which is 20 m away from the foot of the tower, is 30° . Find the height of the tower.

Sol.

Let tower be AB

Distance of point C from foot of tower = BC = 20 m

∠ACB = Angle of elevation = 30°

Tower is vertical, ∠ABC = 90°

In right triangle ∆ABC,

tanC = AB/BC

tan30° = AB/20

1/√3 = AB/20

AB = 20/√3

Multiply numerator and denominator by √3,

AB = (20/3)√3 = 6.6√3 m

Hence, Height of tower = AB = (20/3)√3 = 6.6√3 m

OR

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.

Sol.

Given tower be AB

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

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

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

Objective Questions

Q1. Express 3825 as a product of prime factors.

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

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

Ans. irrational number

Q3. Find HCF of 336 and 64.

Sol.

336 = 2 × 2 × 2 × 2 × 3 × 7

54 = 2 × 3 × 3 × 3

HCF = 2 × 3 = 6

Q4. The rational number 64/455 is a terminating or non-terminating decimal expansion.

Ans. Non-Terminating decimal expansion (denominator 455 = 5 × 7 × 13)

Q5. Sum of the roots of the quadratic polynomial 7x² – 3x + 1  is ……………

Sol.

Here, a = 7, b = -3, c = 1

α + β = -coefficient of x/coefficient of x² = -b/a = -(-3)/7 = 3/7

Q6. If the sum and products of the roots of the quadratic polynomial are 1/3 and -1/3 respectively, then the quadratic polynomial is :

(A)  3x² + x + 1

(B)  3x² – x – 1

(C)  3x² + x – 1

(D)  3x² – x + 1

Ans. (B)  3x² – x – 1

α + β = 1/3,  αβ = -1/3

According to question,

x² – (α+β)x + αβ

x² – (1/3)x – 1/3

3x² – x -1 = 0

Q7. The pair of linear equations (4/3)x + 2y = 8 and 2x + 3y = 12  is consistent or inconsistent.

Sol.

(4/3)x + 2y – 8 = 0,  2x + 3y – 12 = 0

Here, a= 4/3, a2 = 2, b= 2, b2 = 3, c= -8, c= -12

a1/a2 = (4/3)÷2 = 2/3

b1/b2 = 2/3

c1/c2 = -8/-12 = 2/3

a1/a2 = b1/b2 = c1/c2

so it is  Consistent

Q8. The solution of linear equations s − t = 3 and s/3 + t/2 = 6 is  :

(A)  s = 3, t = 5

(B)  s = 6, t = 9

(C)  s = 9, t = 6

(D)  s = 5, t = 3

Ans. (C)  s = 9, t = 6

s – t = 3   ……(i)

s/3 + t/2 = 6  ……(ii)

2s + 3t = 36  ……(iii) (multiplied by 6 eqn.ii)

2s – 2t = 6  ……(iv) (multiplied by 2 eqn.i)

Subtract eqn.(iv) from eqn.(iii),

5t = 30,  t = 6

put t = 6 in eqn.i,

s – 6 = 3,  s = 9

Q9. Factories the quadratic equation 6x² – x – 2 = 0.

Sol.

Using Splitting middle term method,

6x² – 4x + 3x – 2 = 0

2x(3x – 2) + 1(3x – 2) = 0

(2x + 1)(3x – 2) = 0

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

Sol.

Using Splitting middle term method,

2x² + 4x – 3x – 6 = 0

2x(x + 2) – 3(x + 2) = 0

(2x – 3)(x + 2) = 0

Either 2x – 3 = 0  or  x + 2 = 0

x = -2, 3/2

Q11. If the roots of the quadratic equation 3x² + kx + 3 = 0 are equal, then the value of k is :

(A)  ±6

(B)  ±3

(C)  ±9

(D) None of these

Ans. (A)  ±6

Here,  a = 3, b = k, c = 3

Discriminant, D = b² – 4ac = 0

k² – 4 × 3 × 3 = 0

k² – 36 = 0

k² = 36

k = √36 = ± 6

Q12. For which value of p does the pair of equations x + y + 1 = 0 and 4x + py + 8 = 0 has a unique solution.

Sol.

Here,  a1 = 1, a2 = 4, b1 = 1, b2 = p, c1 = 1, c2 = 8

so, for unique solution

a1/a2 ≠ b1/b2

1/4 ≠ 1/p

p ≠ 4

Hence, the given pair of equation has unique solution for all values of p except 4.

Q13. Write the common difference of A. P. 4, 1, –2, –5, …………..

Sol.

d = a2 – a1 = 1 – 4 = – 3

common difference, d = – 3

Q14. 14th term of A. P. 5, 12, 19, ……. is …………..

Sol.

Here, a = 5,  b = 7

an = a + (n -1)d

a14 = 5 + (14 – 1)7 = 5 + (13)7 = 5 + 91

a14 = 96

Q15. The  sum  of  first  50  natural  numbers  is  …………..

Sol.

1 + 2 + 3 + 4 + …….. + 50

Here,  a = 1,  l = 50

Sn = n/2 (a + l)   or  n/2 [2a + (n-1)d]

S50 = 50/2 (1 + 50)

S50 = 25 × 51 = 1275

Q16. Write  the  next  three  terms  of  A.  P.  2, -1,  -4,  ……………

Sol.

d = a2 – a1 = – 1 – 2 = -3

Next Three terms,

a4 = a3 + d = – 4 – 3 = – 7

a5 = a4 + d = – 7 – 3 = – 10

a6 = a5 + d = – 10 – 3 = – 13

Q17. All  squares  are  …………..  (similar,  congruent)

Ans. similar

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.

(A)  7 cm,  24 cm,  25 cm

(B)  8 cm,  6 cm,  3 cm

(C)  3 cm,  6cm,  5cm

(D)  4 cm,  5 cm,  6cm

Ans. (A)  7 cm,  24 cm,  25 cm

Using Pythagoras Theorem,

For Right Angle Triangle,

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

H² = P² + B²

25² = 7² + 24²

625 = 49 + 576 = 625

Q20. If  tangent  PA  and  PB  from  a  point  P  to  a  circle  with  centre  O  are  inclined  to each  other  at  an  angle  100°,  then angle POA is equal to……….

(A)  40°

(B)  80⁰

(C)  50°

(D)  None of these

Ans. (A)  40°

∠APO = ∠APB/2 = 100/2 = 50°

∠POA + ∠PAO + ∠APO = 180°  (in ∆APO)

∠POA + 90° + 50° = 180°

∠POA + 140° = 180°

∠POA = 180 – 140 = 40°

Q21. The  common  point  of  a  tangent  to  a  circle  and  the  circle  is  called  …………..

Ans. Point of contact

Q22. From  a  point  Q,  the  length  of  the  tangent  to  a  circle  is  24  cm,  and  the  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

Q23. In figure DE || BC. Find the value of EC :

(A)  1 cm

(B)  2 cm

(C)  3 cm

(D)  4 cm

Ans. (B)  2 cm

By BPT Theorem,

2/4 = 1/EC

EC = 2 cm

Q24. The distance between the points (–1, −1) and (–4, 4) is ………….

Sol.

A(-1, -1) and B(-4, 4)

Here,  x1 = – 1,  x2 = – 4,  y1 = – 1,  y2 = 4

Distance Formula,

AB = √(x2x1)²+(y2-y1)²

= √(-4-(-1))²+(4-(-1))²

= √(-4+1)²+(4+1)²
= √(-3)²+(5)² = √9+25 = √34 unit

Q25. The ratio in which the x-axis divides the line-segment joining the points A(1, –5) and B(–4, 5) is :

(A)  2 : 1

(B)  1 : 1

(C)  1 : 2

(D)  3 : 2

Ans. (B)  1 : 1

Using Section Formula,

y = (m1y2 + m2y1) ÷ (m1 + m2

0 = [k(5) + 1(-5)] ÷ (k + 1)

0 × (k+1) = 5k – 5

0 = 5k – 5

k = 1

k : 1 = 1 : 1

Q26. The  co-ordinates  of  the  point  which  divides  the  join  of  (4,  −3)  and  (8,  5)  in  the ratio  3  :  1  internally,  are  :

(A)  (4,  5)

(B)  (−3,  5)

(C)  (7,  3)

(D)  (3,  7)

Ans. (C)  (7,  3)

R(x,y) = [(m1x2 + m2x1)/(m1 + m2)], [(m1y2 + m2y1)/(m1 + m2)]

Here  m1 = 3,  m2 = 1,  x1 = 4,  y1 = – 3,  x2 = 8,  y2 = 5

R(x,y) = [(3×8 + 1×4)/(3+1) , (3×5 + 1× -3)/(3+1)]

R(x,y) = (28/4, 12/4)

R(x,y) = (7, 3)

Q27. The  point  on  the  x-axis  which  is  equidistant  from  (2,  −5)  and  (−2,  9)  is  :

(A)  (0, −7)

(B)  (-7, 0)

(C)  (−5, 0)

(D)  (0, -5)

Ans. (B)  (-7, 0)

Coordinates of Q(2, -5) and R(-2, 9)

Coordinates of point on x-axis = P(x, 0)

Using Distance Formula,

D = √(x2x1)²+(y2-y1)²

Distance PQ = Distance PR

√(x-2)² + (0+5)² = √(-2-x)² + (9-0)²

Cancelled square root both side,

(x-2)² + 5² = (2+x)² + 9²

x² + 4 – 4x + 25 = 4 + x² + 4x + 81

– 4x – 4x = 81 – 25

– 8x = 56

x = – 7

Coordinates of point on x-axis = P(-7, 0)

Q28. If secθ = 13/12 , then  find  sinθ.

Sol.

secθ = Hypotenuse/Base = 13/12

so,  Hypotenuse = 13,  Base = 12

Using Pythagoras Theorem,

H² = P² + B²

P² = H² – B² = 13² – 12² = 169 – 144 = 25

Perpendicular = P = √25 = 5

sinθ = Perpendicular/Hypotenuse = 5/13

Q29. 2tan30⁰/(1+tan²30⁰) is equal to :

(A)  sin 60⁰

(B)  cos 60⁰

(C)  tan 60⁰

(D)  sin 30⁰

Ans. (A)  sin 60⁰

2tan30⁰/(1+tan²30⁰) =

= (2×1/√3)/(1+(1/√3)²) = (2/√3) ÷ (4/3)

= √3/2 = sin60⁰

Q30. The value of 2tan²45⁰ + cos²30⁰ – sin²60⁰  is …………..

Sol.

2tan²45⁰ + cos²30⁰ – sin²60⁰ =

= 2(1)² + (√3/2)² – (√3/2)² = 2 ×1 = 2

Q31. Area of the sector of a circle with radius 14 cm and angle of sector 60° is :

(A)  154 cm²

(B)  208/3 cm²

(C)  308/3 cm²

(D)  196 cm²

Ans. (C)  308/3 cm²

Area of Sector of circle = θ/360⁰ × πr² = 60/360 × 22/7 × 14 × 14 = 308/3 cm²

Q32. The length of the minute hand of a clock is 7 cm. The area swept by the minute hand in 15 minutes is :

(A)  77/2 cm²

(B)  154/3 cm²

(C)  49/2 cm²

(D)  170/3 cm²

Ans. (A)  77/2 cm²

Minute hand of clock,

Angle made in 60 minutes = 360⁰

Angle made in 1 minute = 360/60 = 6⁰

Angle made in 15 minutes = 6 × 15 = 90⁰

Area of Sector of circle = θ/360⁰ × πr² = 90/360 × 22/7 × 7 × 7 = 77/2 cm²

Q33. The volume of the right circular cone of height 21 cm and radius of its base 5 cm, is ………

Sol. Volume of Cone = 1/3 πr²h = 1/3 × 22/7 × 5 × 5 × 21 = 22 × 25 = 550 cm³

Q34. The volume of the sphere of radius 8 cm is ……………

Sol. Volume of sphere = 4/3 πr³ = 4/3 × 22/7 × 8 × 8 × 8 = 2145.5 cm³

Q35. A metallic sphere of radius 4 cm is melted and recast into the shape of a cylinder of radius 2 cm. The height of the cylinder is :

(A)  20/3 cm

(B)  64/3 cm

(C)  16/3 cm

(D)  None of these

Ans. (B)  64/3 cm

Radius of sphere = 4 cm

Volume of sphere = 4/3 πr³ = 4/3 π(4)³

Radius of cylinder = 2 cm

Volume of cylinder = πr²h = π(2)²h

Volume of cylinder = Volume of sphere

π(2)²h = 4/3 π(4)³

4 × h = 4/3 × 64

h = 64/3 cm

Q36. The probability of an event that cannot happen is ………….. such an event is called …………..

Ans. 0,  impossible event

Q37. Which  of  the  following  cannot  be  the  probability  of  an  event  ?

(A)  3/4

(B)  -1.3

(C)  17%

(D)  0.5

Ans. (B)  -1.3

0 ≤ P(E) ≤ 1  (probability never be negative)

0% ≤ P(E) ≤ 100%

Here, 0 means impossible,  1 means Certain

Q38. A  die  is  thrown  once.  The  probability  of  getting  a  prime  number  is  …………..

Sol.

Number on die or dice = (1, 2, 3, 4, 5, 6)

Total Possible Outcomes = 6

Prime Number = 2, 3, 5

Total Favorable Outcomes = 3

P(Prime Number) = 3/6 = 1/2 = 0.5

Q39. The  mean  of  the  following  data  is  :

(A)  9.2

(B)  8.5

(C)  10.2

(D)  7.6

Ans. (A)  9.2

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

3,  5,  2,  1,  2,  0,  5,  1,  2,  4

Find  the  mode  of  the  data.

Ans. Mode = 2 (most repeated number)

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