RBSE 10th Maths Model Paper Set 1 with Answers in English 

Students must start practicing the questions from RBSE 10th Maths Model Papers Set 1 with Answers in English provided here.

RBSE Class 10 Maths Model Paper Set 1 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) Which of the following rational number has terminating decimal expansion? (1)
(a) \(\frac{11}{3000}\)
(b) \(\frac{91}{270}\)
(c) \(\frac{343}{2^{3} \times 5^{2} \times 7^{3}}\)
(d) \(\frac{31}{2^{4} \times 3^{5}}\)
Answer:
(c) \(\frac{343}{2^{3} \times 5^{2} \times 7^{3}}\)

(ii) If one zero of the quadratic polynomial x² + 3x + k is 2, then value of k is : (1)
(a) 10
(b) – 10
(c) – 7
(d) – 2
Answer:
(b) – 10

(iii) The value of k for which the system of linear equations x + 2y = 3, 5x + ky + 7 = 0 is inconsistent is : (1)
(a) \(\frac{-14}{3}\)
(b) \(\frac{2}{3}\)
(c) 5
(d) 2
Answer:
(d) 2

(iv) The roots of the quadratic equation x² – 0.04 = 0 are:
(a) ± 0.2
(b) ± 0.02
(c) 0.4
(d) 2
Answer:
(a) ± 0.2

(v) The common difference of the AP \(\frac{1}{p}, \frac{1-p}{p}, \frac{1-2 p}{p}\) is : (1)
(a) 1
(b) \(\frac{1}{p}\)
(c) -1
(d) –\(\frac{1}{p}\)
Answer:
(c) -1

(vi) To divide a line segment AB in the ratio 3: 4. We draw a ray AX, so that ∠BAX is an acute angle and then marks the points on the ray AX at equal distances such that the minimum number of these points is: (1)
(a) 3
(b) 7
(c) 4
(d) 12
Answer:
(b) 7

(vii) The coordinates of the point which is reflection of point (-8, 6) in x-axis are : (1)
(a) (8, 6)
(b) (8, – 6)
(c) (- 8, – 6)
(d) (- 8, 6)
Answer:
(c) (- 8, – 6)

(viii) The value of sin 32° cos 58° + cos 32° sin 58° is : (1)
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(b) 1

(ix) The value of \(\frac{\cos 80^{\circ}}{\sin 10^{\circ}}\) + cos 59° cosec 31° is : (1)
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(c) 2

(x) If x,s are the mid points of the class intervals of grouped data, f_i are the corresponding frequencies and x is the mean, then Σ(f_ix_i – x̄) is equal : (1)
(a) 0
(b) -1
(c) 1
(d) 2
Answer:
(a) 0

(xi) If the mode of a data is 18 and mean is 24, then median is : (1)
(a) 18
(b) 24
(c) 22
(d) 21
Answer:
(c) 22

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

Question 2.
Fill in the blanks :
(i) If \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\), then the pair of equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 is _____________ (1)
Answer:
inconsistent

(ii) Value of x for which 2x, (x + 10) and (3x + 2) are three consecutive terms of an AP is _____________ (1)
Answer:
6

(iii) The part of a curve between two given points on the curve is known as _____________ (1)
Answer:
curve

(iv) The distance of a point from the y-axis is known its _____________ (1)
Answer:
abscissa

(v) If AABC is right angled triangle at C, then value of cos(A + B) = _____________ (1)
Answer:
0

(vi) The cumulative frequency table is useful in determining the _____________ (1)
Answer:
median

Question 3.
Very Short Answer Type Questions :
(i) What type of decimal expansion does a rational number has ? How can you distinguish it from decimal expansion of irrational numbers? (1)
Answer:
A rational number may has its decimal expansion either terminating decimal expansion or a non-terminating repeating. But an irrational number has its decimal expansion non-repeating and non-terminating.

(ii) Find a quadratic polynomial, the sum and product of whose zeroes are – 3 and 2 respectively. (1)
Answer:
Given: (α + β) = -3 and = (α × β) = 2
∴ Required polynomial
p(x) = [x² – (α + β)x + α × β]
= x² – (-3)x + 2
= x² + 3x + 2

(iii) Find the value of k such that the polynomial x² – (k + 6)x + 2 (2k – 1) has sum of its zeroes equal to half to their products. (1)
Answer:
Here, α + β = \(-\left[\frac{-(k+6)}{1}\right]\)
= k + 6
and α × β = \(\frac{2(2 k-1)}{1}\) = 2(2k – 1)

Given, (k + 6) = \(\frac{1}{2}\) × 2(2k – 1)
⇒ k + 6 = 2k – 1
⇒ k = 7

(iv) Find the values of k so that the pair of equations x + 3y = 6 and 2x + ky + 12 = 0 has a unique solution. (1)
Answer:
From the given equations,
a₁ = 1, b₁ = 3, c₁ = – 6, a₂ = 2, b₂ = k, and c₂ = 12

Condition for unique solution is :
\(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \Rightarrow \frac{1}{2} \neq \frac{3}{k}\) ⇒ k ≠ 6
Hence, k have all real values except 6.

(v) Find the value of k for which the roots of the equation 3x² – 10x + k = 0 are reciprocal of each other. (1)
Answer:
Let one root of the equation be a other 1 root is \(\frac{1}{\alpha}\).
Product of two roots = \(\frac{c}{a}\)
⇒ α × \(\frac{1}{\alpha}=\frac{k}{3}\)
⇒ 1 = \(\frac{k}{2}\) ⇒ k = 3

(vi) For what value of k, the given quadratic equation kx² – 6x – 1 = 0 has no real roots ? (1)
Answer:
The condition for no real roots is:
D < 0
⇒ b² – 4ac < 0
⇒ (- 6)² — 4ac – 1 < 0
⇒ 36 + 4k < 0
⇒ 4k < – 36
⇒ k < –\(\frac{36}{4}\)
⇒ k < – 9
Hence, k should be less than -9

(vii) Construct a triangle with sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are \(\frac{4}{5}\) times of the corresponding sides of the given triangle. (1)
Answer:

(viii) If k(5, 4) is the mid-point of line segment PQ and coordinates of Q are (2, 3) then find the coordinates of point P. (1)
Answer:
Let the coordinates of be (x, y) then, according question
∴ 5 = \(\frac{x+2}{2}\) and 4 = \(\frac{y+3}{2}\)
⇒ 10 = x + 2 and 8 = y + 3
⇒ x = 10 – 2 and y = 8 – 3
⇒ x = 8 and y = 5
∴ Coordinates of point P are (8, 5).

(ix) Find the value of \(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\) where A is equal to 45°. (1)
Answer:

(x) Evaluate : sin 25° cos 65° + cos 25° sin 65° (1)
Answer:
sin 25° cos 65° + cos 25° sin 65° = sin 25° cos (90° – 25°) + cos 25° sin (90° – 25°)
= sin 25° sin 25° + cos 25° cos 25°
= sin² 25° + cos² 25°
= 1 [∵ sin2θ + cos2θ = 1]
Hence, sin 25° cos 65° + cos 25° sin 65° = 1

(xi) Find the class-marks of the classes 15 – 35 and 45 – 60. (1)
Answer:
Class-mark of 15 – 35
= \(\frac{15+35}{2}=\frac{50}{2}\) = 25
And classmark of 45 – 60
= \(\frac{45+60}{2}=\frac{105}{2}\) = 52.5

(xii) If the probability of winning a game is 0.07, what is the probability of lossing it? (1)
Answer:
P (winning a game) = 0.07 (given) P (winning a game) + P (lossing a game) = 1
⇒ 0.07 + P (lossing a game) = 1
⇒ P (lossing a game) = 1 – 0.07 = 0.93

Section – B

Question 4.
Prove that √3 is an irrational number. (2)
Answer:
Let us assume, that √3 is a rational number. It can be expressed in the form of \(\frac{a}{b}\), where a, b are coprime b
positive integers and b ≠ 0.
∴ √3 = \(\frac{a}{b}\)[HCF of a and b is 1]
3 = \(\frac{a^{2}}{b^{2}}\) (Squaring both sides)
⇒ 3b² = a² ….(i)
Therefore, 3 divides a2. It follows that 3 divides a
Let a = 3c, where- c is any integer

Put a = 3c in (i) we get,
⇒ 3b² = (3c)²
⇒ 3b² = 9c²
⇒ b² = 3c² …(ii)
It means b² is divisible by 3 and so b is divisible by 3.
From (i) and (ii) we say that 3 is a common factor of botb a and b. But this contradicts the fact that a and b are coprimes, i.e., they have no common factor. So our assumption that √3 is a rational number is wrong. Therefore, √3 is an irrational number.
Proved

Question 5.
If the α and β are the zeroes of the polynomial f(x) = x² – 6x + k, find the value of k, such that α² + β² = 40. (2)
Answer:
According to the questions α² + β² = \(-\frac{b}{a}=\frac{-(-6)}{1}\) = 6 ……..(i)
and α × β = \(\frac{c}{a}=\frac{k}{1}\) = k …(ii)
Now, α² + β² = 40 (given)
⇒ (α + β)² – 2ap = 40
⇒ 6² – 2 × k = 40
⇒ 36 – 2k = 40
⇒ – 2k = 40 – 36 = 4
⇒ k = \(\frac{4}{-2}\) = -2

Question 6.
Raghav scored 70 marks in a test, getting 4 marks for each right answer and losing 1 mark for each wrong answer. Had 5 marks been awarded for each correct answer and 2 marks been deducted for each wrong answer, then Raghav would have scored 80 marks. How many questions were there in the test? (2)
Answer:
Let the number of right answers be x and number of wrong answers be y.
According to the question,
4x – y = 70 …(i)
and 5x – 2y = 80 …(ii)
Multiplying equation (i) by 2 and substracting equation (ii) from equation (i), we get

Substituting x = 20 in equation (i), we get >
4 × 20 – y = 70
⇒ – y = 70 – 80 = – 10
⇒ y = 10.
Hence, total number of questions = x + y = 20 + 10 = 30.

Question 7.
If – 3 is a root of the quadratic equation 2x² + Px – 15 = 0 while the quadratic equation x² – 4Px + k = 0 has equal roots. Find the value of k. (2)
Answer:
We put x = – 3 in 2x² + Px – 15 = 0 as it is it one root, we get
2(- 3)2 + P(- 3) – 15 = 0
⇒ 18 – 3P – 15 = 0
⇒ 3 – 3P= 0
⇒ P = \(\frac{-3}{-3}\) = 1
Putting the value of P in given equation, we get
x² – 4 x 1 + k = 0
⇒ x2 – 4x + k = 0

The equation x² – 4x + k = 0 has equal roots.
So, D = 0
⇒ b² – 4ac = 0
⇒ (- 4)² – 4 x 1 x k = 0
⇒ 16 – 4k = 0
k = \(\frac{-16}{-4}\) = 4

Question 8.
Find the sum of all 11 terms of an A.P. whose middle terms is 30. (2)
Answer:
Let first term be a and common difference be d of AP consisting of 11 terms, \(\left(\frac{11+1}{2}\right)^{\text {th }}\)i.e. 6th terms is middle term.
But it is given that middle most term is 30.
a₆ = 30
⇒ a + (6 – 1)d =30
⇒ a + 5d = 30
Sum of 11 terms
= \(\frac{11}{2}\)[2a + (11 – 1)d]
= \(\frac{11}{2}\)[2a + 107]
= \(\frac{11}{2}\)[2a + 57] = 11[a + 5d]
= 11 × 30 = 330

Question 9.
How many terms of the AP : 45, 39, 33, must be taken so that their sum is 180? (2)
Answer:
The given sequence of AP is 45, 39, 33, …………..
Let AP has n terms
Here, a – 45, d = 39 – 45 = – 6 and S_n = 180 (given)
S_n = 180
⇒ \(\frac{n}{2}\) [2a + (n – 1)d] = 180
⇒ \(\frac{n}{2}\)[2 × 45 + (n – 1) × (-6)] = 180
⇒ \(\frac{n}{2}\) [90 – 6n + 6] = 180
⇒ \(\frac{n}{2}\)[96 – 6n] = 180
⇒ 96n – 6n² = 360
⇒ 6n² – 96n + 360 = 0
⇒ n² – 16n + 60 = 0
⇒ n² – (10 + 6)n + 60 = 0
⇒ n – 10n – 6n + 60 =0
⇒ n(n – 10) – 6(n – 10) = 0
⇒ (n – 10) (n – 6) =0
⇒ n = 10 or n = 6
Hence, 10 or 6 terms must be taken.

Question 10.
Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are \(\frac{3}{4}\) of the corresponding sides of the triangle ABC. (2)
Answer:
Steps of Construction :

• Draw a line segment BC = 6 cm.
• At B, draw ∠CBY = 60°.
• From B draw an arc AB = 5 cm meeting BY at A. Join AC. Thus, ∆ABC was obtained.
• Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
• Along BX mark four points B₁, B₂, B₃, B₄ such that BB₄ = B₁B₂ = B₂B₃ = B₃B₄.
• Join B₄C.

• From B₃ draw B₃C’ ∥ B₄C meeting BC at C.
• From C’ draw C’A’ ∥ CA, meeting AB at A.

Then A’BC is the required triangle, each of whose side is \(\frac{3}{4}\) of corresponding sides of ∆ABC.

Question 11.
Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 70°. (2)
Answer:
Steps of Construction :

• Draw a circle with centre O and radius 5 cm.
• Construct raddi OA and OB such that ∠AOB (360°- 90° + 90° + 70°) = 110°.
• Draw AL ⊥ OA at A and BM ⊥ OB at B. They intersect at P.
• Then PA and PB are the required tangents inclined to each other at 70°.

Question 12.
Prove that : (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A. (2)
Answer:
We have,
L.H.S.= (sin A + cosec A)² + (cos A + sec A)²
= sin² A + cosec² A + 2 sin A cosec A + cos² A + sec² A + 2 cos A sec A
= (sin² A + cos²A) + cosec² A + sec² A + 2 sin A × \(\frac{1}{\sin A}\) + 2 cos A × \(\frac{1}{\cos A}\)
= 1 + 1 + cot² A + 1 + tan² A + 2 + 2
= 7 + tan² A + cot² A = R.H.S.
∴L.H.S. = R.H.S.

Question 13.
If tan(A + B) = √3 and tan(A – B) = \(\frac{1}{\sqrt{3}}\) <(A + B) ≤ 90°, A > B, find A and B. (2)
Answer:
We have
tan(A + B) = √3
⇒ tan(A + B) = tan 60°
⇒ A + B = 60° …(1)
and tan(A – B) = \(\frac{1}{\sqrt{3}}\)
⇒ tan(A – B) = tan 30°
⇒ A – B = 30° …(2)

Question 14.
Find the mode of the following frequency distribution. (2)

Answer:
The class interval 30 – 35 has maximum frequency. So, it is the modal class.
∴ l = 30, f₁ = 10, f₀ = 9, f₂ = 3, h = 5
mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h
30 + \(\left(\frac{10-9}{2 \times 10-9-3}\right)\) × 5 = 30 + \(\frac{5}{8}\)
= 30 + 0.625 = 30.625
Hence, mode = 30.625

Question 15.
Perday expenses of 25 families of the frequency distribution of a Dhani of a village is given as follows : (2)

Find the mean expense of families by direct method.
Answer:
We prepare the cumulative frequency table as given :

∴ Mean = \(\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}=\frac{1240}{25}\) = ₹ 49.6
Hence, mean expense of families = 49.6

Question 16.
A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is \(\frac{2}{3}\). Find the number of blue marbles in the jar. (2)
Answer:
Total marbles in the jar = 24
Let the number of green marbles in the jar be x.
Then blue marbles in the jar = 24 – x
Total number of possible outcomes = 24
P (getting a green marble) = \(\frac{x}{24}\)
⇒ \(\frac{2}{3}=\frac{x}{24}\)
⇒ x = \(\frac{24 \times 2}{3}\) = 16
Number of blue marbles in jar
= 24 – 16 = 8.
Hence, number of blue marbles in the jar = 8

Section – C

Question 17.
Which term of the A.P. : 20, 19\(\frac{1}{4}\), 18\(\frac{1}{2}\), 17\(\frac{1}{4}\), is the first negative term. (3)
Or
If m times the mth term of an A.P. is equal to n times its rath term and m ≠ n, show that (m + n)th term of A.P. is zero. (3)
Answer:
The given sequence of A.P is 20, 19\(\frac{1}{4}\), 18\(\frac{1}{2}\), 17\(\frac{1}{4}\)……….
Here, a = 20, d = 19\(\frac{1}{4}\) – 20 = \(\frac{-3}{4}\)
Let nth term of an AP be the first negative term, then,
a < 0
⇒ a + (n – 1)d < 0
⇒ 20 + (n – 1) × \(\left(-\frac{3}{4}\right)\) < 0
⇒ 20 – \(\frac{3}{4}\)n + \(\frac{3}{4}\) < 0
⇒ \(\frac{-3}{2}\)n < –\(\frac{83}{4}\) < 0
⇒ \(\frac{-3}{2}\)n < –\(\frac{83}{4}\) ⇒ n > \(\frac{83}{3}\)
⇒ n >27.66
⇒ n > 28
Hence, 28 term of the AP is the first negative term.

Question 18.
The line segment joining the points A(2, 1) and 13(5, – 8) is trisected at the points P and Q such that P is nearer to A. If P also lies on the line given by 2x – y + k = 0, find the value of k. (3)
Or
Find the ratio in which the line x – 3y = 0 divides the line segment joining the points (-1, – 2, – 5) and (6, 3). Find the coordinates of the points of intersection. (3)
Answer:
Points P and Q trisects the line segment joining the points A(2, 1) and B(5, – 8)
∴ AP = PQ = BQ.
PB = PQ + BQ = AP + AP = 2AP

\(\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{\mathrm{AP}}{2 \mathrm{AP}}=\frac{1}{2}\) = 1 : 2
∴ Point P divides the line segment AB in the ratio 1 : 2.

By section formula, coordinates of point P are

So, coordinates of point P is (3, – 2). Since, point P (3, – 2) lies on the given line 2x – y + k = 0
2 × 3 – (- 2) + k = 0
⇒ 6 + 2 + k = 0
⇒ k = – 8

Question 19.
Prove that 2 (sin⁶ θ + cos⁶ θ) – 3 (sin⁴ θ + cos⁴ θ) + 1 = 0 (3)
0r
Prove that : \(\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta}\) using the indentity sec² θ =1 + tan² θ. (3)
Answer:
We have
L.H.S. = 2 [(sin² θ)3 + (cos² θ)3] – 3 (sin4 0 + cos40) +1=0
= 2 [(sin² θ + cos² θ) (sin4 0 + cos4 0) – sin² θ cos² θ] – 3 (sin⁴ θ + cos⁴ θ) + 1 [∴ a³ + b³ = (a + b) (a² + b² – ab)]
= 2 sin⁴ θ + 2 cos⁴ θ – 2 sin² θ cos² θ – 3 cos⁴ θ – 3 sin⁴ θ + 1
= – sin⁴ θ – cos⁴ θ – 2 sin² θ cos² θ + 1
= – (sin⁴ θ + cos⁴ θ + 2 sin² θ cos² θ) + 1
= – (sin² θ + cos² θ)2 + 1
= -1 + 1 = 0 = R.H.S.
Proved

Question 20.
The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and data obtained is represented in the following table : (3)

Find the median length of the leaves.
Or
To find out the concentration of S02 in the air (in parts per million, i.e., ppm), the data was collected for 20 localities in a certain city and is presented below :

Find the mean concentration of S02 in the air. (3)
Answer:
The series is in inclusive form. We convert it into exclusive form and prepare the cumulative frequency table as given below :

n = 40
⇒ \(\frac{n}{2}\) = 20
But 20 comes under the cumulative frequency 29 and the class-interval against the cumulative frequency 29 is 144-5 – 153-5. So, it is the median class.
∴ l = 144.5, cf = 17, f = 12, and h = 9
Median = l + \(\left(\frac{\frac{n}{2}-c f}{f}\right)\) × h
= 144.5 + \(\left(\frac{20-17}{12}\right)\) × 9
= 144.5 + \(\frac{3 \times 9}{12}\) = 144.5 + \(\frac{9}{4}\)
= 144.5 + 2.25 = 146.75
Hence, median length of leaves = 146-75 mm.

Section – D

Question 21.
The cost of 5 apples and 3 organges is ₹ 35 and the cost of 2 apples and 4 oranges is ₹ 28. Formulate the problem algebraically and solve it graphically. (4)
Or
Cost of 2 exercise books and 3 pencils is ₹ 17 and the cost of 3 exercise books and 4 pencils is ₹ 24. Formulate the problem algebrically and solve it graphically. (4)
Answer:
Let the cost of 1 apple be ₹ x and cost of 1 orange be ₹ y.
∴ Cost of 5 apples is ₹ 5x and cost of 3 oranges is ₹ 3y.

According to question.
5x + 3y = 35 …(i)
Cost of’2 apples is ? 2x and cost of 4 oranges is ? 4y

According to question
2x + 4y = 28 …(ii)
The algebraic representation of this situation is :
5x + 3y = 35 …(1)
And 2x + 4y = 28 …(2)

For representation of these equations graphically, we draw the graphs of these equations as follows :
5x + 3y = 35
⇒ y = \(\frac{35-5 x}{3}\)

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 the equation 5x + 3y = 35.
Table 1

And 2x + 4y = 28
⇒ y = \(\frac{28-2 x}{4}=\frac{14-x}{2}\)
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 the equation 2x + 4y = 28.

Table 2

Now, we plot the values of x and y from tables 1 and 2 on the graph paper and we draw the graphs of the equations 1 and 2, those passes through these values.

From the graph, we observe that the two straight lines intersect each other at point (4, 5). Hence, cost of 1 apple is ‘ 4 and cost of 1 orange is 5.

Question 22.
Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are \(\frac{2}{3}\) of the corresponding sides of the first triangle. (4)
Or
Draw a line segment PQ of length 6 cm. Taking P as centre, draw a circle of radius 3 cm and taking Q as centre, draw another circle of radius 2 cm. Construct tangents to each circle from the centre of the other circle. (4)
Answer:

Steps of Construction :

• Draw a line segment BC = 5 cm.
• With B as a centre and radius 4 cm draw an arc.
• With C as a centre and radius 6 cm draw another arc to intersect the first arc. at A.
• Join BA and CA to get ABC.
• Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
• Along BX mark 3 points B₁; B₂ and B₃ such that BB₁ = B₁B₂ = B₂B₃.
• Join B₃C.
• From B₂ draw B₂C’ ∥ B₃C meeting BC at C’.
• From C’ draw C’A’ ∥ CA meeting BA at A’.

Then A’BC’ is the required triangle, each of whose side is \(\frac{2}{3}\) of corresponding sides of ∆ABC.

Justification: Since,
A’C’ ∥ AC
Therefore, ∆A’BC’ ~ ∆ABC
\(\frac{\mathrm{A}^{\prime} \mathrm{B}}{\mathrm{AB}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{A}^{\prime} \mathrm{C}^{\prime}}{\mathrm{AC}}=\frac{2}{3}\)

Question 23.
By changing the folio ving frequency distribution to less than type distribution, draw its ogive. (4)

Or
The distribution below given the weights of 42 students of a class. Find the median weight of the students. (4)

Answer:
We prepare the cumulative frequency table by less than type method as given :

We plots the points (15, 6), (30, 14), (45, 24). (60, 30) and (75, 34) on the graph paper. Joining these points with a free hand to obtain less than type ogive curve as shown in graph.