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RBSE 10th Maths Board Model Paper 2022 with Answers in English 

March 21, 2022 by Veer Leave a Comment

Students must start practicing the questions from RBSE 10th Maths Model Papers Board Model Paper 2022 with Answers in English Medium provided here.

RBSE Class 10 Maths Board Model Paper 2022 with Answers in English

Time: 2:45 Hours
Maximum Marks: 80

General Instructions:

  • All the questions are compulsory.
  • Write the answer of each question in the given answer book only.
  • For questions having more than one part the answers to their parts are to be written together in continuity.
  • Candidate must write first his/her Roll No. on the question paper compulsorily.
  • Question numbers 17 to 23 have internal choices.
  • The marks weightage of the questions are as follows :
Section Number of Questions Total Weightage Marks for each question
Section A 1 (i to xii), 2(i to vi), 3(i to xii) = 30 30 1
Section B 4 to 16 =13 26 2
Section C 17 to 20 = 4 12 3
Section D 21 to 23 = 3 12 4

Section – A

Question 1.
Multiple Choice Questions :
(i) Decimal expansion of rational number \(\frac{129}{2^{5} \times 5^{7} \times 5^{7}}\) is: (1)
(a) terminating
(b) non-terminating
(c) non-terminating repeating
(d) non-terminating non-repeating
Answer:
(c) non-terminating repeating

(ii) Which of these is not a quadratic polynomial? (1)
(a) x2
(b) x2 – 4
(c) x2 – 4x + 4
(d) x3 – 3x2 + 3x – 1
Answer:
(d) x3 – 3x2 + 3x – 1

(iii) The solution of linear pair equation x + y = 4, x – y = 2 is : (1)
(a) x = 3, y = 1
(b) x = 1, y = 3
(c) x = y = 2
(d) x = 4, y = 0
Answer:
(a) x = 3, y = 1

(iv) Which one of the following methods is used to solve the quadratic equation: (1)
(a) factorization
(b) perfect square
(c) quadratic formula
(d) all of these
Answer:
(d) all of these

(v) The first term and common difference for the given AP; 4. 10, 16, 22 will be respectively: (1)
(a) 10, 5
(b) 4, 6
(cl 4, 10
(d) 6, 10
Answer:
(b) 4, 6

(vi) if the radius of a circle is 15 cm, then diameter of circle is : (1)
(a) 6 cm
(b) 7 cm
(c) 8.5 cm
(d) none of these
Answer:
(b) 7 cm

(vii) Point (-3. 5) belongs to which quadrant? (1)
(a) first
(b) second
(c) third
(d) fourth
Answer:
(b) second

RBSE 10th Maths Board Model Paper 2022 with Answers in English 

(viii) Value of tan 45° + cot 45°: (1)
(a) 1
(b) 2
(c) 3
(d) 0
Answer:
(b) 2

(ix) Value of cos (90° – 48°) is : (1)
(a) sec 48°
(b) tan 48°
(c) sin 48°
(d) cot 48°
Answer:
(c) sin 48°

(x) One of these is not a formula to find mean: (1)
(a) \(\frac{\sum x}{n}\)
(b) \(\frac{\Sigma x+a h}{n}\)
(c) \(\frac{\Sigma f x}{N}\)
(d) \(\frac{\Sigma f d}{N}\)
Answer:
(b) \(\frac{\Sigma x+a h}{n}\)

(xi) The mode of data is: (1)
RBSE 10th Maths Board Model Paper 2022 with Answers in English 1
(a) 9
(b) 5
(c) 7
(d) 11
Answer:
(c) 7

(xii) Which of the following can not be the probability of an event: (1)
(a) 2/3
(b) -1.5
(c) 15%
(d) 0.7
Answer:
(b) -1.5

Question 2.
Fill in the blanks :
(i) If \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\) , then solution of linear pair of equation is ________________ (1)
Answer:
infinite

(ii) 7th term of AP; 10, 8, 6, 4 is ________________ (1)
Answer:
-2

(iii) How many tangents are drawn to a circle from the point of outsite the circle? (1)
Answer:
Two

(iv) The formula to find the distance between two points in the co-ordinate geometry is ________________ (1)
Answer:
\(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)

(v) sin (90° – θ) = ________________ (1)
Answer:
cos θ

(vi) Median of distribution 7, 4, 6, 3, 8, 5, 9 is ________________ (1)
Answer:
6

Question 3.
Very Short Answer Type Questions:
(i) If the product of two numbers is 180 and their HCF is 3. then find their LCM. (1)
Answer:
Product of two numbers = Product of LCM and HCF
∴ 180 = LCM × 3
⇒ LCM = 180 + 3
LCM = 60

(ii) Find the remainder, if x2 + 3x + 1 is divided by x – 2? (1)
Answer:
RBSE 10th Maths Board Model Paper 2022 with Answers in English 2
So, remainder is 11.

RBSE 10th Maths Board Model Paper 2022 with Answers in English 

(iii) Write the sum and product of zeroes of the polynomial x2 – 6x + 7? (1)
Answer:
Let the zeroes of polynomial x2 – 6x + 7 are α and β then,
Sum of zeroes, α + β = –\(\frac{b}{a}\)
⇒ α + β = \(\left(\frac{-6}{1}\right)\) ⇒ α + β = 6
and (product of zeroes) αβ = \(\frac{c}{a}\)

(iv) Find the condition of unique solution for pair of linear equation in two variables. (1)
Answer:
If two pairs of linear equations are a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, then the condition for unique solution is \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\).

(v) Find the sum of roots of quadratic equation 2x2 + 3x + 4 = 0 ? (1)
Answer:
In equation in 2x2 + 3x + 4 = 0.
Sum of roots = \(-\frac{b}{a}=-\frac{3}{2}\)

(vi) Write Shree Dharacharya formula for finding roots of quadratic equation. (1)
Answer:
For quadratic equation ax2 + bx + c; where a ≠ 0, Shree Dharacharya formula is
x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

(vii) Find the angle between radius of a circle and tangent of circle. (1)
Answer:
90°

(viii) Find the abscissa and ordinate in co-ordinate (4,3). (1)
Answer:
Abscissa = 4,
Ordinate = 3

(ix) Find the value : cos 30° – sin 60° (1)
Answer:
cos 30° – sin 60° = \(\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}\) = 0

(x) If sin 3x = 1, then find the value of x. (1)
Answer:
If sin 3x = 1, then
⇒ sin 3x = sin \(\frac{\pi}{2}\)

(xi) Find the mean of distribution 10, 12, 8, 7, 13. (1)
Answer:
Mean of distribution 10, 12, 8, 7, 13
= \(\frac{10+12+8+7+13}{5}=\frac{50}{5}\) = 10

(xii) Find the probability of getting a head and a tail when a coin is tossed once. (1)
Answer:
If a coin is tossed once, then total outcomes S = [H, T]
⇒ n (S) = 2
n(H) 1
Probability of getting head P(H) = \(\frac{n(H)}{n(S)}=\frac{1}{2}\)
Probability of getting tail P(T) = \(\frac{n(T)}{n(S)}=\frac{1}{2}\)

Section – B

Question 4.
Find the HCF of 90 and 144 by using Euclid’s division method. (2)
Answer:
Step 1. The numbers 90 and 144 are such that 144 > 90
So, by Euclid’s division lemma for 144 and 90,
144 = 90 × 1 + 54

Step 2. ∵ Remainder is hot 0.
∴ For 54 and 90, again by Euclid’s division lemma,
90 = 54 × 1 + 36

Step 3. ∵ Remainder is not 0.
∴ For 36 and 54, again by Euclid’s division lemma,
54 = 36 × 1 + 18

Step 4. ∵ Remainder is not 0.
∴ For 18 and 36, again by Euclid’s division lemma,
36 = 18 × 2 + 0 ∵ Remainder is 0.

∴ The divisor of Step 4 is 18.
So, HCF of 90 and 144 is 18.

Question 5.
Find the zeroes of a quadratic polynomial x2 – 2x – 15. (2)
Answer:
Given, x2 – 2x – 15
⇒ x2 – 2x – 15 = x2 – 5x + 3x — 15
= x (x – 5) + 3 (x – 5)
= (x – 5) (x + 3)
For zeroes
(x – 5) (x + 3) = 0
⇒ x – 5 = 0 or x + 3 = 0
⇒ x = 5 or x = -3
So, the zeroes are 5, -3.

RBSE 10th Maths Board Model Paper 2022 with Answers in English 

Question 6.
Solve the following pair of linear equations by elimination method: 2x + y = 6 and 2x-y = 2 (2)
Answer:
Given equations are
2x + y = 6 …(i)
and 2x-y = 2 …(ii)
Adding equations (i) and (ii),
⇒ 4x = 8 ⇒ x = 2

Subtracting equations (ii) and (i),
2y = 4 ⇒ y = 2
So, x = 2 and y = 2 are the solution of given linear equations.

Question 7.
Find the roots of the following quadratic equation by factorisation:
2x2 – 2x – 6 = 0 (2)
Answer:
Given, quadratic equation is 2x2 + x – 6 = 0
⇒ 2x2 + 4x – 3x – 6 = 0
⇒ 2x (x + 2) – 3 (x + 2) = 0
⇒ (x + 2) (2x – 3) = 0
⇒ x + 2 = 0 or 2x – 3 = 0
⇒ x = -2 or x = \(\frac{3}{2}\)
So, the roots are – 2 and \(\frac{3}{2}\) .

Question 8.
Find the number of terms of the following AP : 7, 13, 19, ………… 205
Answer:
Given, AP series is
7, 13, 19, ………… 205
First term a = 7
Common difference d = 13 – 7 = 6
Last term l = 205

Let the number of term be n.
By l = a + (n – 1) d
⇒ 205 = 7+ (n – 1)6
⇒ (n – 1)6 = 205 – 7
⇒ (n – 1) = \(\frac{198}{6}\)
⇒ n = 33 + 1 = 34
So, the total number of terms is 34.

Question 9.
How many three digit numbers are divisible by 7 ? (2)
Answer:
The list of three-digit numbers is :
100, 101, 102, ……… 999,
The first number of three digits divisible by 7 = 105 and last number of three digits divisble by 7 = 994
Then list of three digit numbers divisble by 7 is 105, (105 + 7), (105 + 7 + 7), … 994
= 105, 112, 119, …, 994

Let the total number be n.
First term a = 105, common difference d = .7.
∴ nth term an = 994
⇒ a + (n – 1 )d = 994
⇒ 105 + (n – 1) × 7 = 994
⇒ (n – 1) × 7 = 994 – 105
⇒ (n – 1) = \(\frac{889}{7}\) =127.
∴ n = 127 + 1 = 128
So, the total three-digit numbers divisble by 7 is 128.

Question 10.
Draw a line segment of length 5 cm and divide in the ratio 2:3. (2)
Answer:
AP : PB = 2:3
RBSE 10th Maths Board Model Paper 2022 with Answers in English 3

Question 11.
Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°. (2)
Answer:
Steps of Construction:

  • Draw a circle with centre O and radius 5 cm.
  • Construct radii OA and OB such that ∠AOB = 360° – (90° + 90° + 60°) = 120°.
  • Draw AX ⊥ OA at A and BY ⊥ OB at B.

They intersect at T.
Then TA and TB are the required tangents inclined to each other at 60°.
Justification : TA ⊥ OA and TB ⊥ OB and ∠AOB = 120°.
RBSE 10th Maths Board Model Paper 2022 with Answers in English 4
∠ATB = 3600 (90° + 90° + 120°)
∠ATB = 360° – 300°
∠ATB = 60°.
Therefore, tangents TA and TB are inclined at 60°.

RBSE 10th Maths Board Model Paper 2022 with Answers in English 

Question 12.
If sin A = \(\frac{3}{4}\),calculate cos A and tan A. (2)
Answer:
Given,
sin A = \(\frac{3}{4}=\frac{B C}{A C}\)
RBSE 10th Maths Board Model Paper 2022 with Answers in English 5
From figure
AB2 = AC2 – BC2
= 42– 32 = 16 – 9
= 7
AB = \(\sqrt{7}\)
So, cos A = \(\frac{A B}{A C}=\frac{\sqrt{7}}{4}\)
and tan A = \(\frac{B C}{A C}=\frac{3}{\sqrt{7}}\)

Question 13.
Evaluate the following:
2tan2 45° + cos2 30° – sin2 60°
Answer:
2tan2 45° + cos2 30° – sin2 60°
= 2(1)2 + \(\left(\frac{\sqrt{3}}{2}\right)^{2}-\left(\frac{\sqrt{3}}{2}\right)^{2}\)
= 2 × 1 + \(\frac{3}{4}-\frac{3}{4}\)
= 2

Question 14.
Find the mean of the following distribution: (2)
RBSE 10th Maths Board Model Paper 2022 with Answers in English 6
Answer:
RBSE 10th Maths Board Model Paper 2022 with Answers in English 7
So, mean (x̄) = \(\frac{\sum f_{i} x_{i}}{\sum f_{i}}\)
= \(\frac{4460}{100}\) = 44.60

Question 15.
Find the mean òf the following distribution: (2)
Answer:

Class-interval No. of students
0-20 10
20-40 35
40 – 60 52
60 – 80 61
80-100 38
100-120 20

The class interval 60—80 has maximum
frequency. So, it is the modal class.
∴ l = 60, f1 = 61, f0 = 52, f2 = 38 h = 20
RBSE 10th Maths Board Model Paper 2022 with Answers in English 8

Question 16.
A bag contains 5 white and 2 red balls. From this bag one ball is the drawn randomly. What is the probability that drawn ball is : (i) white (ii) red. 2
Answer:
Number of all possible out comes = 5 + 2 = 7
(i) Number of white balls = 5
Number of favourable outcomes = 5
∴ P (white ball) = \(\frac{5}{7}\)
So, the probability of drawing white ball is \(\frac{5}{7}\)

(ii) Number of red balls = 2
Number of favourable outcomes = 2
P (Red ball) = \(\frac{2}{7}\)
So, the probability of red ball outcomes is \(\frac{2}{7}\)

RBSE 10th Maths Board Model Paper 2022 with Answers in English 

Section – C

Question 17.
If second and third terms of an AP are 24 and 28 respectively, then find the sum of first 61 terms. (3)
OR
Find the sum of all natural numbers divisible by 5 between 2 and 101. (3)
Answer:
According to the question, a2 = 24
and a3 = 28
then, an = a + (n – 1) d (formula)
a2 = a + d = 24 …(i)
and = a + 2d = 28 …(ii)
Subtracting equation (i) from (ii)
RBSE 10th Maths Board Model Paper 2022 with Answers in English 9
Substituting d = 4 in equation (i)
a + 4 = 24
⇒ a = 20
So, a – 20 and d = 4
Sum of first 61 terms of AP is

S61 = \(\frac{61}{2}\) [2 × 20 +(61 – 1)4] [∵ Sn = \(\frac{n}{2}\)(2a + (n – 1 )d)]
= \(\frac{61}{2}\) × (40 + 240) = \(\frac{61}{2}\) × 280
= 61 × 140 = 8540
So, the sum of first 61 terms is 8540.
OR
The list of numbers divisible by 5 between 2 to 101 is
5, 10, 15, 20, 25, 30, …………… , 100

The following numbers are in AP,
then a = 5, d = 5 , an = 100

Here, an = 100
a + (n – 1)d= 100
⇒ 5 + (n – 1) 5 = 100
⇒ (n – 1) = \(\frac{100-5}{5}\)
⇒ n = 19 + 1
⇒ n = 20
∵ We know that,
Sn = \(\frac{n}{2}\)[a + l]
⇒ Sn = \(\frac{20}{2}\) [5 + 100]= 10 × 105
So, S20 = 1050

Question 18.
Find the point on x-axis which is equidistant from the point (-2, 3) and (-3, 4). (3)
OR
In which ratio, point (7, 6) divides the line joining the points (11,8) and (1, 3). (3)
Answer:
Let the required point on x-axis be (x, 0), then
Distance between points (x, 0) and (-2, 3)
= Distance between points (x, 0) and (-3, 4)
RBSE 10th Maths Board Model Paper 2022 with Answers in English 10
Squaring both sides
x2 + 4x + 13 = x2 + 6x + 25
⇒ 6x – 4x = 13 – 25
⇒ 2x = -12
⇒ x = -6
So, the required point is (-6, 0)

OR

Let the point (7, 6) divides the line segment joining the given points in the ratio of λ : 1.
RBSE 10th Maths Board Model Paper 2022 with Answers in English 11
By Internal division formula
x = \(\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}\)
7 = \(\frac{\lambda \times 1+1 \times 11}{\lambda+1}\)
⇒ 7λ + 7 = λ + 11
⇒ 6λ = 11 – 7
⇒ λ = \(\frac{4}{6}=\frac{2}{3}\)
∴ λ : 1 = 2 : 3
So the required ratio is 2 : 3

Question 19.
Find the value of:
\(5 \frac{\sin 17^{\circ}}{\cos 73^{\circ}}+2 \frac{\cos 67^{\circ}}{\sin 23^{\circ}}-6 \frac{\sin 15^{\circ}}{\cos 75^{\circ}}\)
OR
cos4θ + sin4θ = 1 – 2cos2θsin2θ
Answer:
RBSE 10th Maths Board Model Paper 2022 with Answers in English 12
= 5 + 2 – 6 = 7 – 6 = 1

OR

L.H.S = cos4θ + sin4θ
{cos2θ)2 + (sin2θ)2 + 2sin2θcos2θ} – 2sin2θcos2θ
= (sin2θ + cos2θ)2 – 2sin2θ cos2θ
= (1)2 – 2sin2θ cos2θ
Hence Proved.

RBSE 10th Maths Board Model Paper 2022 with Answers in English 

Question 20.
Find the mean of the following distributions: (3)
RBSE 10th Maths Board Model Paper 2022 with Answers in English 13
OR
Find the mode of the following frequency distribution: (3)
RBSE 10th Maths Board Model Paper 2022 with Answers in English 14
Answer:
RBSE 10th Maths Board Model Paper 2022 with Answers in English 15
Mean (x̄) = \(\frac{\Sigma f_{i} x_{i}}{\sum f_{i}}=\frac{930}{40}\) = 23.50
So, the required mean is 23.50
OR
The class interval 40 – 55 has maximum frequency, so it is the modal class
∴ l = 40, f1 = 44, f0 = 20, f2 = 26, h = 15
RBSE 10th Maths Board Model Paper 2022 with Answers in English 16

Section – D

Question 21.
Solve the following pair of linear equations by graphical method : (4)
3x – 5y = 1 and 2x – y = 3
OR
Solve the following pair of linear equations by graphical method :
4x – 5y = 20 and 3x + 5y = 15 4
Answer:
Given, pair of linear equations is :
3x – 5y = 1
and 2x – y = 3
From equation (i),
3x – 5y = 1
3x = 1 + 5y
⇒ x = \(\frac{1+5 y}{3}\)
Putting y = 1,
x = \(\frac{1+5(1)}{3}=\frac{1+5}{3}=\frac{6}{3}\) = 2

Putting y = 4,
x = \(\frac{1+5(4)}{3}=\frac{1+20}{3}=\frac{21}{3}\) = 7

Putting y = – 2,
x = \(\frac{1+5(-2)}{3}=\frac{1-10}{3}=\frac{-9}{3}\) = -3

Table-I
RBSE 10th Maths Board Model Paper 2022 with Answers in English 17
From equation (ii),
2x – y = 3
⇒ y = 2x – 3

Putting x = 0,
y = 2(0) – 3

Putting x = 2,
y = 2(2) – 3 = 4 – 3 = 1

Putting x = 3
y = 2(3) – 3 = 6 – 3 = 3

Table – II
RBSE 10th Maths Board Model Paper 2022 with Answers in English 18

Now plotting the points (2, 1), (7, 4) and (-3, -2) and joining the graph of the equation 3x – 5y = 1 is a straight line AB and plotting the points (0, – 3), (2,1) and (3,3) and joining the graph of the equation 2x – y = 3 is straight line CD.
RBSE 10th Maths Board Model Paper 2022 with Answers in English 19
So. the solution of given equations is
x = 2 and y = 1.
OR
Given pair of linear equation are
4x – 5y = 20 ..(i)
3x + 5y = 15 …(ii)

From equation (i),
4x – 5y = 20
⇒ 4x = 5y + 20
⇒ x = \(\frac{5 y+20}{4}\)

Putting, y = 0,
For, x = \(\frac{5(0)+20}{4}=\frac{0+20}{4}\) = 5

Putting, y = 3,
x = \(\frac{5(-4)+20}{4}=\frac{-20+20}{4}\) = 0

Putting, y = -3,
x = \(\frac{5(-4)+20}{4}=\frac{-20+20}{4}\) = 10

Table – I
RBSE 10th Maths Board Model Paper 2022 with Answers in English 20
From equation (ii),
3x – 5y = 15
⇒ 3x = 15 – 5y
⇒ x = \(\frac{15-5 y}{3}\)

Putting, y = 0,
x = \(\frac{15-5(0)}{3}=\frac{15-0}{3}\) = 5

Putting, y = 3,
x = \(\frac{15-5(3)}{3}=\frac{15-15}{3}\) = 0

Putting. y = – 3,
x = \(\frac{15-5(-3)}{3}=\frac{15+15}{3}\) = 10

Table – II
RBSE 10th Maths Board Model Paper 2022 with Answers in English 21
Now plotting the pints (5, 0), (0, – 4) and (10, 4) and joining the graph of the equation 4x – 5y = 20 is a straight line AB and plotting the points (5,0), (0,3) and (10, -3) and joining the graph of the equation 3x + 5y = 15 is a straight line CD.
So, the solution of given equations is x = 5, y= 0.

RBSE 10th Maths Board Model Paper 2022 with Answers in English 

Question 22.
Draw a line-segment of 10 cm and divide it into the ratio of 2 : 3 (4)
OR
Draw a line segment AB of 5 cm and then draw a circle of 2.2 cm taking B as centre and draw tangents from point A on this circle. (40
Answer:
Steps of Construction :

  • Draw a ray AX of 10 cm, making an acute angle with AB.
  • Along AX mark (2 + 3) = 5 points A1; A2, A3, A4 and A5 such that AA1 = A1A2 = A2A3 = A3A4 = A4A5.
  • Join A5B.
  • From the point A2 draw A2 C ∥ A5B, meeting AB at C. Then AC : BC = 2:3.

RBSE 10th Maths Board Model Paper 2022 with Answers in English 22
Justification: In ∆ACA2 and ∆ABA5, we have CA2 ∥ BA5
\(\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\mathrm{AA}_{2}}{\mathrm{~A}_{2} \mathrm{~A}_{5}}\)
[By Basic proportionality theorem]
⇒ \(\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{2}{3}\)
Hence, AC : BC = 2 : 3.

OR

Steps of Construction
(1) Draw the line segment ‘AB = 5 cm.
(2) Draw a circle of radius 2.2 cm from the point B of AB.

RBSE 10th Maths Board Model Paper 2022 with Answers in English 23
(3) Draw the bisector of AB and mark the bisector as M.
(4) Taking Mas center, draw a circle of radius MB which intersects the circle with center B at T1 and T2.
(5) Joint AT1 and AT2. AT1 and AT2 are the required tangents.

Justification:
Joint BT1 and BT2.
We know that the tangent is perpendicular to the radius passing through the point of contact.
So, ∠AT1B = ∠AT2B = 90°.

AB is diameter of the circle with center M.
∠AT1B = ∠AT2B = 90°.
(The angle in semi-circle is a right angle) So, AT1 and AT2 are tangents.

Question 23.
Find the median of the following distribution: (4)

Class interval Frequency (f)
10-25 6
25-40 20
40-55 44
55-70 26
70-85 3
85- 100 1

OR
Find the made of the following distribution. (4)

Marks No. of Students
0-20 5
20-40 10
40-60 12
60-80 6
80-100 3

Answer:
RBSE 10th Maths Board Model Paper 2022 with Answers in English 24

where N = 100 ⇒ \(\frac{\mathrm{N}}{2}=\frac{100}{2}\) = 50
So, the comutative frequency more than 50 comes in the class interval is in 70 which is 40 – 55
Median class = 40 – 55
l = 40, c = 26, f = 44, h = 15

Median = l + \(\left(\frac{\frac{N}{2}-c}{f}\right)\) × h
= 40 + \(\left(\frac{50-26}{44}\right)\) × 15
= 40 + \(\frac{24}{44}\) × 15
= 40 + \(\frac{6}{11}\) × 5
= 40 + 8.18 = 48.18

OR

The class interval 40 – 60 has maximum frequency, 12, it is modal class.
l = 40, f1 = 12,f0 = 10, f2 = 6, h = 20
Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h
= 80 + \(\left(\frac{12-10}{2 \times 12-10-6}\right)\) × 20
= 80 + \(\left(\frac{2}{8}\right)\) × 20 = 85
So, mode is 85.

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