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RBSE 10th Maths Model Paper Set 2 with Answers in English

March 22, 2022 by Fazal Leave a Comment

Students must start practicing the questions from RBSE 10th Maths Model Papers Set 2 with Answers in English Medium provided here.

RBSE Class 10 Maths Model Paper Set 2 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) The HCF and LCM of 12, 21, 15 respectvely are : (1)
(a) 3, 140
(b) 12, 420
(c) 3, 420
(d) 420, 3
Answer:
(c) 3, 420

(ii) If 2 is factor of polynomial f(x) = x3 – 3x2 – 10x + k, then the value of k is : 1
(a) 12
(b) – 12
(c) 24
(d) – 24
Answer:
(b) – 12

RBSE 10th Maths Model Paper Set 2 with Answers in English

(iii) The value of k for which equations 2x + 4y – 5 = 0 and 6x + ky – 9 = 0 has no solution is:
(a) 10
(b) 12
(c) 13
(d) 11
Answer:
(b) 12

(iv) The roots of the equation 3x2 – 4x + 3 = 0 are:
(a) real and unequal
(b) real and equal
(c) imaginary
(d) none of these.
Answer:
(c) imaginary

(v) The first term of an AP is p and common difference is q, then its 10th term is :1
(a) q + qp
(b) p – 9q
(c) p + 9q
(d) 2p + 9q.
Answer:
(c) p + 9q

(vi) In the given figure, if PA1 = A1A2 = A2A3 = A3A4 = A4Q and A1B1 || A2B2 || A3B3 || A4, B4|| QR, then A3, A3 divides the line segment PR in the ratio : 1
RBSE 10th Maths Model Paper Set 2 with Answers in English 1
(a) 3 : 5
(b) 2 : 3
(c) 3 : 2
(d) 3 : 6
Answer:
(c) 3 : 2

(vii) The distance between the points (a cos θ + 6 sin θ, θ) and (θ, a sin θ – b cos θ), is :
(a) a2 + b2
(b) a2 – b2
(c) \(\sqrt{a^{2}+b^{2}}\)
(d) \(\sqrt{a^{2}-b^{2}}\)
Answer:
(c) \(\sqrt{a^{2}+b^{2}}\)

(viii) If sin (A + 2B) = \(\frac{\sqrt{3}}{2}\) and cos (A + 4B) = 0, where A > B and A + 4B ≤ 90°, then A and B is:
(a) 30°, 45°
(b) 15°, 45°
(e) 60°, 15°
(d) 30°, 15°
Answer:
(d) 30°, 15°

RBSE 10th Maths Model Paper Set 2 with Answers in English

(ix) sin 38° cos 52° is equal to :
(a) 1
(b) 0
(c) 14
(d) None of these
Answer:
(b) 0

(x) For the following distribution:
RBSE 10th Maths Model Paper Set 2 with Answers in English 2
The modal class is :
(a) 10 – 20
(b) 20 – 30
(c) 30 – 40
(d) 50 – 60
Answer:
(c) 30 – 40

(xi) In the formula x̄ = a + h\(\left(\frac{\sum f_{i} u_{i}}{\sum f_{i}}\right)\), for finding the mean
of grouped frequency distribution, ui is equal :
(a) \(\frac{x_{i}+a}{h}\)
(b) h(xi – a)
(c) \(\frac{x_{i}-a}{h}\)
(d) \(\frac{a-x_{i}}{h}\)
Answer:
(c) \(\frac{x_{i}-a}{h}\)

(xii) The probability of an impossible event is :
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(a) 0

Question 2.
Fill in the blanks:
(i) If the two line coincide, the pair of equations has ……………………… many solutions. (1)
Answer:
infinitely

(ii) If a, b, c are in AP, then b = \(\frac{a+c}{2}\) and b is called the arithmetic ………………….. (1)
Answer:
mean

(iii) There can be …………………. tangent drawn to the circlé from point inside it. (1)
Answer:
no

(iv) To locate the position of a point on a plane, a pair of coordinate is required. …………………..(1)
Answer:
axis

(v) sin θ.cos(90° – θ) + cos θ.sin(90°- θ) is equal to …………………. .
Answer:
1

(vi) If Pis equal to …………………., then mode of data 9, 5, 4, 7, 4, 5, 7, 2, 7, P, 7 is 7. (1)
Answer:
7

RBSE 10th Maths Model Paper Set 2 with Answers in English

Question 3.
Very Short Answer Type Questions:
(i) Find the HCF of Integers 375 and 975 by the prime factorization method. (1)
Answer:
375 = 3 × 5 × 5 × 5
= 3 × 53
And 675 = 3 × 3 × 3 × 5 × 5
= 33 × 52
RBSE 10th Maths Model Paper Set 2 with Answers in English 3
HCF (375, 675) = 3 × 52 = 75

(ii) Divide the polynomial 3x3 + x2 + 2x + 5 by the polynomial 1 + 2x + x2. (1)
Answer:
RBSE 10th Maths Model Paper Set 2 with Answers in English 4

(iii) Write the degree of the following polynomial : 2x + \(\frac{7}{3}\)x + \(\frac{9}{5}\)x – \(\frac{1}{7}\).
Answer:
Degree = 3

(iv) Show that x = 2 and y – 1 is solution of equation 3x-2y = 4.
Answer:
Put x = 2 and y = 1 in the equation.
LHS = 3(2) – 2(1) = 6 – 2 = 4 = RHS
So, x = 2 and y = 1, is a solution of the given equation.

(v) Find the roots of equation 4x2 + 3x + 5 = 0 by the method of completing the square.
Answer:
The given equation is:
4x2 + 3x + 5 = 0
RBSE 10th Maths Model Paper Set 2 with Answers in English 5
But (x + \(\frac{3}{8}\))2 cannot be negative for any real value of x.
Therefore, the given equation has no real roots.

RBSE 10th Maths Model Paper Set 2 with Answers in English

(vi) Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find its length and breadth.
Answer:
According to question,
24 × x = 800
[∵ Area of rectangle = Length × Breadth]
⇒ 2x2 = 800
⇒ x2 = \(\frac{800}{2}\) = 400
⇒ x = √400 = ±20
Reject x = – 20
[∵ Breadth cannot be negative.]
∴ x = 20
Hence, the rectangular grove is possible and its breadth = 20 m and length 2 × 20 = 40 m.

(vii) Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm (only figure). 1
Answer:
RBSE 10th Maths Model Paper Set 2 with Answers in English 6

(viii) Find the area of a triangle whose vertices are (1, – 1), (- 4, 6) and (- 3, – 5).
Answer:
Let A(x1, y1) = (1, – 1), B(x2, y1)
= (- 4, 6) and C(x3, y1) = (- 3, – 5)
∴ Area of ∆ABC = \(\frac{1}{2}\) [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
= \(\frac{1}{2}\) [1(6 + 5) + (- 4)(- 5 + 1) + (- 3) (- 1 – 6)]
= \(\frac{1}{2}\) [11 – 4 × (- 4) – 3 × (- 7)]
= \(\frac{1}{2}\) (11 + 16 + 21)
= \(\frac{1}{2}\) × 48 = 24 units

(ix) If tan α = \(\frac{5}{12}\), find the value of sec α.
Answer:
∵ 1 + tan2α = sec2α
∴ sec α = \(\sqrt{1+\tan ^{2} \alpha}\)
RBSE 10th Maths Model Paper Set 2 with Answers in English 7

(x) Show that sin 28° cos 62° + cos 28° sin 62° = 1.
Answer:
We have,
sin 28° cos 62° + cos 28° sin 62°
= sin 28° cos (90° – 28°) + cos 28° sin (90° – 28°)
= sin 28° sin 28° + cos 28° cos 28°
= sin2 28° + cos2 28° = 1

RBSE 10th Maths Model Paper Set 2 with Answers in English

(xi) Compute the mode for the following frequency distribution:
RBSE 10th Maths Model Paper Set 2 with Answers in English 8
Answer:
The class interval 12 – 16 has maximum frequency. So, it is the modal class.
∴ l = 12, f1 = 17, f0 = 9, f2 = 12, h = 4
mode
RBSE 10th Maths Model Paper Set 2 with Answers in English 9

(xii) If the probability of lossing a game is 0.065, what is probability of winning it? (1)
Answer:
P (winning a game) = 0.07 (given)
P (lossing a game) + P (winning a game) = 1
0. 065 + P (winning a game) = 1
P (winning a game) = 1 – 0.065 = 0.935

Section – B

Question 4.
Write the smallest number which is divisible by both 306 and 657.
Answer:
The required smallest number is the LCM of 306 and 657.
We have
306 = 2 × 3 × 3 × 17
= 2 × 32 × 17
And 657 = 3 × 3 × 73
= 32 × 73
RBSE 10th Maths Model Paper Set 2 with Answers in English 10

LCM (306, 657)
= 2 × 32 × 17 × 73
= 22338
RBSE 10th Maths Model Paper Set 2 with Answers in English 11
Hence, the required smallest number = 22338

Question 5.
Divide 3x2 – x3 – 3x + 5 by x – 1 – x2.
Answer:
f(x) = 3x2 – x3 – 3x + 5
and g(x)= x – 1 – x2
We write f(x) and g(x) in standard form of decreasing powers of x
so, f(x)= – x3 + 3x2 – 3x + 5
and g(x)= – x2 + x – 1
Now, we divide f(x) by g(x), as follows:
RBSE 10th Maths Model Paper Set 2 with Answers in English 12
Hence, quotient q(x) = x – 2 and remainder r(x) = 3.

RBSE 10th Maths Model Paper Set 2 with Answers in English

Question 6.
If 2x + y = 23 and 4x – y = 19, find the value of (5y – 2x). (2)
Answer:
The given equations are :
2x + y = 23 ….. (i)
4x – y = 19 …… (ii)
Adding equation (i) and (ii), we get
6x = 42
⇒ x = \(\frac{42}{6}\) = 7
Substituting the value ofx in equation (i),
we get 2 × 7 + y = 23 .
⇒ y = 23 – 14 = 9 .
So, x = 7, y = 9
Putting the values of x and y in 5y – 2x, we get 5 × 9 – 2 × 7
= 45 – 14 = 31.

Question 7.
If the roots of the equation (a2 + b2)x2 – 2(ac + bd)x + (c2 + d2) = 0 are equal, prove that \(\frac{a}{b}=\frac{c}{d}\).
Answer:
Here A = (a2 + b2), B = – 2(ac + bd) and C = (c2 + d2)
Since, roots of the given equation are equal, therefore
⇒ B2 – 4AC = 0
⇒ [- 2(ac + bd)]2 – 4 × (a2 + b2) × (c2 + d2) = 0
⇒ [4 (a2c2 + b2d2 + 2abcd)] – 4(a2c2 + b2c2 + a2d2 + b2d2) = 0
⇒ 4a2c2 + 4b2d2 + 8abcd – 4a2c2 4b2c2 – 4a2d2 – 4b2d2 = 0
⇒ – 4(a2d2 + b2c2 – 2abcd) = 0
⇒ (a2d2 + b2c2 – 2abcd) = 0
⇒ (ad – bc)2 = 0
⇒ ad – bc = 0
⇒ ad= be
⇒ \(\frac{a}{b}=\frac{c}{d}\)
Hence proved

Question 8.
Write the nth terms of the AP : \(\frac{1}{m}, \frac{1+m}{m}, \frac{1+2 m}{m}\)
Answer:
Here, a = \(\frac{1}{m}\) and
d = \(\frac{1+m}{m}-\frac{1}{m}\) = \(\frac{1+m-1}{m}\) = 1
nth term = an = a + (n – 1)d
= \(\frac{1}{m}\) + (n – 1) × 1
= \(\frac{1}{m}\) + (n – 1)
= \(\frac{1+(n-1) m}{m}\)

Question 9.
If the sum of first n terms of an AP is \(\frac{1}{2}\)[3n2 + 7n], then find the nth term.
Answer:
We have
Sn = \(\frac{1}{2}\)[3n2 + 7n]
S1 = \(\frac{1}{2}\)[3 × 12 + 7 × 1]
= \(\frac{1}{2}\) × 10 = 5
S2 = \(\frac{1}{2}\)[3 × 22 + 7 × 2]
= \(\frac{1}{2}\) [12 + 14]
= \(\frac{1}{2}\) × 26 = 13
∴ a1 = S1 = 5, a2 = S2 – S1 = 13 – 5 = 8
d = 8 – 5 = 3
So, AP is 5, 8, 11
nthterm(an) = a + (n – 1)d
= 5 + (n – 1) × 3
= 5 + 3n – 3
= 3n + 2

RBSE 10th Maths Model Paper Set 2 with Answers in English

Question 10.
Construct a triangle ABC with sides 3 cm, 4 cm and 5 cm. Now, construct another triangle whose sides \(\frac{4}{5}\)times the corresponding sides of ∆ABC. (2)
Answer:
Given a triangle ABC, we required to construct a triangle whose sides are \(\frac{4}{5}\) times the corresponding sides of ∆ABC.
RBSE 10th Maths Model Paper Set 2 with Answers in English 13
Steps of Construction:

  • Draw a line segment BC = 4 cm.
  • With B as centre and radius 3 cm draw and arc.
  • With C as centre and radius 5 cm draw another are to intersect the previous arc at A.
  • Join AB and AC to get AABC.
  • Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
  • Along BX mark 5 points B1, B2, B3, B4 and B5 such that BB1 = B1B2 = B2B3 = B3B4 = B4B5
  • Join B5C.
  • From B4 draw B4C’||B5C meeting BC at C.
  • From C’ draw C’ A’ || CA meeting BA at A’. Then A’ BC’ is required triangle, each of whose side is \(\frac{4}{5}\) of corresponding sides of ∆ABC.

Question 11.
Draw a circle of radius 6 cm from a point 10 cm away from its centre. Construct the pair of tangents to the circle. (2)
Answer:

  • Draw a circle with centre O of radius 6 cm.
  • Mark a point A, 10 cm away from the centre.
  • Join AO and bisect it at M.
    RBSE 10th Maths Model Paper Set 2 with Answers in English 14
  • Draw a circle with M as a centre and radius equal to AM intersects the given circle at points P and Q.
  • Join AP and AQ.
    Then AP and AQ are the required tangents. Lengths of AP and AQ are 8.2 cm. (approx.)

Question 12.
Prove that : \(\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}\) = 2 Sec A
Answer:
We have,
RBSE 10th Maths Model Paper Set 2 with Answers in English 15

RBSE 10th Maths Model Paper Set 2 with Answers in English

Question 13.
If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A. (2)
Answer:
We have,
tan 2A = cot (A 18°)
⇒ cot (90° – 2A) = cot (A – 18°) [∵ tan θ = cot (90° – θ)]
⇒ 90° – 2A = A – 18°
⇒ 90° + 18° = A + 2A
⇒ 3A = 108°
⇒ A = \(\frac{108^{\circ}}{3}\) = 36°
Hence, A = 36°.

Question 14.
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate. (2)
RBSE 10th Maths Model Paper Set 2 with Answers in English 17Answer:
RBSE 10th Maths Model Paper Set 2 with Answers in English 16
From the table, we have
Σfi = 35, Σfidi = – 20, a = 70
∴ x̄ = a + \(\frac{\Sigma f_{i} d_{i}}{\Sigma f_{i}}\)
⇒ x̄ = 70 + \(\left(-\frac{20}{35}\right)\)
⇒ x̄ = 70 – 0.57 = 69.43%.
Hence, the mean literacy rate = 69.43%.

Question 15.
The marks distribution of 30 students in a mathematics examination are given below in table. Find the mode of this data. Also compare and interpret the mode and the mean.
RBSE 10th Maths Model Paper Set 2 with Answers in English 18
Answer:
The class 40-55 has maximum frequency, so it is the modal class.
∴ l = 40, f1 = 7, f0 = 3, f2 = 6, and h = 15
Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h
= 40 + \(\left(\frac{7-3}{2 \times 7-3-6}\right)\) × 15
= 40 + \(\left(\frac{4}{14-3-6}\right)\) × 15
= 40 + \(\frac{60}{5}\) = 40 + 12
Mode = 52
So, the mode marks is 52.
Mean marks (x̄) = \(\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}\) = \(\frac{1860}{30}\) = 62
So, the mean marks = 62.
Hence, maximum number of students obtained 52 marks, while on an average a student obtained 62 marks.

RBSE 10th Maths Model Paper Set 2 with Answers in English

Question 16.
A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be (i) red ? (ii) not green ? (2)
Answer:
Total number of possible out comes
= 5 + 8 + 4= 17
(i) Number of red marbles = 5
Number of favourable outcomes = 5
P (a red marble) = \(\frac{5}{17}\)

(ii) Number of marbles which are not green
= 5 + 8 = 13
Number of favourable outcomes = 13
∴ P(a marble not green) = \(\frac{13}{17}\)

Section – C

Question 17.
How many terms of AP : – 6, \(\frac{-11}{2}\), – 5, \(\frac{-9}{2}\), ….. are needed to give their sum zero.
Or
Find the sum of first 100 positive integers.
Answer:
The given sequence of AP is
– 6, \(\frac{-11}{2}\), – 5, \(\frac{-9}{2}\), …………
= \(\frac{-11}{2}+\frac{12}{2}=\frac{1}{2}\) and Sn = 0(given)
Let AP has n terms
Sn = 0
⇒ \(\frac{n}{2}\) [2a + (n – 1)d] = 0
⇒ \(\frac{n}{2}\) [2 × (- 6) + (n – 1) × \(\frac{1}{2}\)] = 0
⇒ \(\frac{n}{2}\) [- 12 + \(\frac{n}{2}\) – \(\frac{1}{2}\)] = 0
⇒ \(\frac{n}{2}\) [- 24 + n – 1] = 0
⇒ n [- 25 + n] = 0
⇒ n = 0 or n – 25 = 0
⇒ n = 0 or = n = 25
Reject n = 0
Hence, terms needed= 25

Question 18.
The distance between the point A(5, – 3) and B(13, m) is 10 units. Calculate the value of m. (3)
Or
If the distances of POc, y) from A(5, 1) and B(- 1, 5) are equal, then prove that 3x = 2y. (3)
Answer:
Here, x1 = 5, y1 = – 3,
x2 = 13, y2 = m
Distance (AB) = 10
⇒ \(\sqrt{(13-5)^{2}+(m+3)^{2}}\) = 10
⇒ \(\sqrt{8^{2}+m^{2}+9+6 m}\) = 10
⇒ \(\sqrt{64+m^{2}+9+6 m}\) = 10
⇒ \(\sqrt{73+m^{2}+6 m}\) = 10
⇒ m2 + 6m + 73 = 102
⇒ m2 + 6m + 73 – 100 = 0
⇒ m2 + 6m – 27 = 0
⇒ m2 + (9m – 3m) – 27 = 0

RBSE 10th Maths Model Paper Set 2 with Answers in English

Question 19.
Prove that: \(\frac{\cot \theta+{cosec} \theta-1}{\cot \theta-{cosec} \theta+1}=\frac{1+\cos \theta}{\sin \theta}\).
Or
If cot θ = \(\frac{7}{8}\), then evaluate: \(\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}\)
Answer:
RBSE 10th Maths Model Paper Set 2 with Answers in English 19

Question 20.
The following table shows the ages of the patients admitted in a hospital during a year : (3)
RBSE 10th Maths Model Paper Set 2 with Answers in English 20
Find the mean of the data.
Or
The following distribution gives the goes of worker of a factory. Calculate the modal age. (3)
RBSE 10th Maths Model Paper Set 2 with Answers in English 21
Answer:
Let us find the mean of the data. Here h = 10
RBSE 10th Maths Model Paper Set 2 with Answers in English 22
From the table, we have
Σfi = 80, Σfiui = – 37, a = 40, and h = 10
Mean = a + \(\left(\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\right)\) × h
= 40 + \(\frac{-37}{80}\) × 10 = 40 – 4.63
= 35.37 years.

RBSE 10th Maths Model Paper Set 2 with Answers in English

Section – D

Question 21.
Check graphically whether the pair of equations x + 3y = 6 and 2x – 3y = 12 is consistent. If so, solve them graphically. (4)
Or
In a parking place, the parking charge of 5 cars and 4 scooters is ₹ 37 and parking charge of 6 cars and 3 scooters is ₹ 39. Represent this situation algebraically and graphically.
Answer:
We draw the graphs of given linear equations as follows :
x + 3y = 6
⇒ 3y = 6 – x
⇒ y = \(\frac{6-x}{3}\)
We put the different values of x in this equation then we get different values of y and we prepare the table of x, y for equation x + 3y = 6
RBSE 10th Maths Model Paper Set 2 with Answers in English 23
and 2x – 3y = 12
⇒ 3y = 2x – 12
⇒ y = \(\frac{2x-12}{3}\)
We put the different values of x in this equation then we get the different values of y and we prepare the table of x, y for equation 2x – 3y = 12.
RBSE 10th Maths Model Paper Set 2 with Answers in English 24
Now, we plot the points (0, 2), (3, 1) and (6, 0) on the graph paper and we draw a graph passing through these points.
We get a graph of linear equation x + 3y = 6, again we plot the points (0, – 4), (3, – 2) and (6, 0) on the graph paper and we draw a graph passing through these points again we get a graph of linear equation 2x – 3y = 12. We observe that graphs of equations x + 3y = 6 and 2x – 3y – 12 intersect each other at point (6, 0). Hence, x = 6, y = 0 is the solution of given equations. Therefore, given pair of equations is consistent.
RBSE 10th Maths Model Paper Set 2 with Answers in English 25

Question 22.
Construct a ∆ABC in which CA = 6 cm, AB = 5 cm and ∠BAC = 450rn Then construct a triangle whose sides are \(\frac{3}{5}\) of the corresponding sides of ∆ABC. (4)
Or
Draw a line segment AB of 8 cm taking A as centre. Draw a circle of radius 3.5 cm and taking B as centre, draw another circle of radius 2.5 cm. Construct tangent to the circle from center of other circle. (4)
Answer:
Steps of Construction:

  • Draw a line segment AB = 5 cm.
  • At A, draw ∠BAY = 45°
  • From A draw an arc AC = 6 cm meeting A at C
  • Join BC. Thus, AABC obtained.
  • Draw any ray AX making an acute angle with AB on the side opposite to the vertex C.
    RBSE 10th Maths Model Paper Set 2 with Answers in English 26
  • Along AX mark 5 points A1( A2, A3, A4 and A5 such that AA1 = A1A2 = A2A3 = A3A4 = A4A6
  • Join A5B
  • From A3 draw A3B’ || A5B meeting AB at B’
  • From B’ draw B’C’ || BC meeting at C’.
    Then AB’C is required triangle, each of whose side is \(\frac{3}{5}\) of corresponding sides of ∆ABC.

Jutification:
Since
B’C || BC ‘
Therefore, AB’C1 ~ ∆ABC
\(\frac{\mathrm{AB}^{\prime}}{\mathrm{AB}}=\frac{\mathrm{B}^{\prime} \mathrm{C}^{\prime}}{\mathrm{BC}}=\frac{A C^{\prime}}{\mathrm{AC}}=\frac{3}{5}\)

RBSE 10th Maths Model Paper Set 2 with Answers in English

Question 23.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the mean and mode of the data and compare them. (4)
RBSE 10th Maths Model Paper Set 2 with Answers in English 27
Or
A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year. (4)
RBSE 10th Maths Model Paper Set 2 with Answers in English 28
Answer:
For calculating mean, we prepare the table as given below:
RBSE 10th Maths Model Paper Set 2 with Answers in English 29
From the table, we have
Σfi = 68, Σfiui =7, a = 135, h = 20
∴ Mean = a + \(\left(\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\right)\) × h
= 135 + \(\frac{7}{68}\) × 20
= 135 + \(\frac{140}{68}\)
= 135 + 205
= 13705.
For calculating mode, the class 125 – 145 has maximum frequency. So, it is the modal class.
∴ l = 125, f0 = 13, f1 = 20, f2 = 14, and h = 20
Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h
= 125 + \(\left(\frac{20-13}{2 \times 20-13-14}\right)\) × 20
= 125 + \(\frac{7 \times 20}{13}\)
= 125 + \(\frac{140}{13}\)
= 125 + 10.76 = 135.76.
Hence, mean and mode are 137.05 and 135.76 respectively. The two measures are approximately same in this case.

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