RBSE 10th Maths Model Paper Set 4 with Answers in English

Students must start practicing the questions from RBSE 10th Maths Model Papers Set 4 with Answers in English Medium provided here.

RBSE Class 10 Maths Model Paper Set 4 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 is the prime factorisation of 60 : 1
(a) 12 × 5
(b) 6 × 10
(c) 20 × 3
(d) 22 × 3 × 5.
Answer:
(d) 22 × 3 × 5.

(ii) Sum of the zeroes of the polynomial ax³ + bx² + cx + d is :
(a) \(\frac{c}{a}\)
(b) – \(\frac{d}{a}\)
(c) \(\frac{-b}{a}\)
(d) \(\frac{b}{a}\)
Answer:
(c) \(\frac{-b}{a}\)

(iii) In a triangle ABC, ∠C = 3 ∠B = 2 (∠A + ∠B), then ∠C = ? 1
(a) 100°
(b) 40°
(c) 20°
(d) 120°
Answer:
(d) 120°

(iv) If x = 3 is one root of the quadratic equation x² – 2kx – 6 = 0, then value of k is :
(a) 1
(b) \(\frac{1}{2}\)
(c) \(\frac{1}{3}\)
(d) 3
Answer:
(c) \(\frac{1}{3}\)

(v) If the first term of an AP is p and last term is q, then sum of n terms is : 1
(a) \(\frac{n}{2}\)(P + q)
(b) \(\frac{n}{2}\)(p – q)
(c) (p + q)
(d) (- p – q).
Answer:
(a) \(\frac{n}{2}\)(P + q)

(vi) To divide a line segment AB in the ratio 5 : 7, first a ray AX is drawn so that ZBAX is an acute angle and then at equal distances points are marked on the ray AX such that the minimum number of these points is : 1
(a) 8
(b) 10
(c) 11
(d) 12
Answer:
(d) 12

(vii) If the points (a, 0) (0, b) and (1, 1) are collinear, then, \(\frac{1}{a}+\frac{1}{b}\) is :
(a) not possible
(b) 1
(c) -1
(d) 0
Answer:
(b) 1

(viii) (1 + tan² θ) cos² θ is equal to :
(a) sin² θ – cos² θ
(b) 1
(c) sec2² θ
(d) sin² θ
Answer:
(b) 1

(ix) \(\frac{\tan ^{2} \theta}{\sec \theta+1}\) = sec θ is equal to
(a) – 1
(b) 1
(c) 0
(d) none of these
Answer:
(a) – 1

(x) The numbers 5, 7, 10, 12, 2P – 8, 2P + 10, 35, 41, 42, 50 are order. If their median is 25, then P is:
(a) 8
(b) 10
(c) 12
(d) 14
Answer:
(c) 12

(xi) If mean = 24, median = 26, mode is :
(a) 23
(b) 26
(c) 25
(d) 30
Answer:
(d) 30

(xii) If P (A) denotes the probability of an event A, then :
(a) P (A) < 0 (b) PA > 1
(c) 0 < PA ≤ 1
(d) – 1 ≤ P ≤ 1
Answer:
(c) 0 < PA ≤ 1

Question 2.
Fill in the blanks :
(i) A pair of linear equations in two variables can be represented and solved by the algebraic and …………………………. method. (1)
Answer:
graphical

(ii) If 2p – 1, 7 and 3p are three consecutive terms of A.P., then p = …………………………. (1)
Answer:
3

(iii) To divide a line segment in 3 : 1 the number of points located in ray AX are …………………………. (1)
Answer:
four

(iv) The distance of a point from the x-axis is called its …………………………. (1)
Answer:
ordinate

(v) If cot A = \(\frac{12}{5}\), then (sin A + cos A) × cosec A = …………………………. (1)
Answer:
\(\frac{17}{5}\)

(vi) If the mean-of first n natural numbers is \(\frac{5 n}{9}\) then n = …………………………. (1)
Answer:
9

Question 3.
Very Short Answer Type Questions : .
(i) HCF of two numbers is 18 and their product is 12960. Find their LCM. (1)
Answer:
∵ HCF = 18
∴ LCM = \(=\frac{\text { Product of two numbers }}{\text { HCF }}\)
= \(\frac{12960}{18}\) = 720

(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
= [x² – (α + β) x + a × β]
= x² – (- 3) x + 2
= x² + 3x + 2

(iii) Divide polynomial f(x) = 4x³ – 3x² + 2x – 4 by other polynomial g(x) = x – 1. Find the quotient and remainder. (1)
Answer:

(iv) How many solutions, do a consistent system of linear equation have? (1)
Answer:
At least one solution

(v) Find the value of k, for which x = 3 is solution of the quadratic equation x² – kx + 6 = 0 (1)
Answer:
∵ x = 3 is a solution.
∴ (3)² – k × 3 + 6 = 0
⇒ 9 – 3k + .6 = 0
⇒ 15 – 3k = 0
⇒ 3k = 15
⇒ k = 5

(vi) Write the standard form of quadratic equation; (1)
Answer:
Standard form of quadratic equation ax² + bx + c = 0, where a, b, c, ∈ R.

(vii) Write the name of point at which the tangent touches the circle. (1)
Answer:
Point of contact.

(viii) What is the distance of point P (2, 3) from the ac-axis? (1)
Answer:
3

(ix) Evaluate : sin 30° cos 60° + cos 30° sin 60°. (1)
Answer:
sin 30° cos 60° + cos 30° sin 60°
= \(\frac{1}{2} \times \frac{1}{2}+\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}\)
= \(\frac{1}{4}+\frac{3}{4}=\frac{1+3}{4}=\frac{4}{4}\) = 1

(x) Evaluate : 2 sin² 60° cos² 45° – 4 tan² 30° cot² 60°. (1)
Answer:
2 sin² 60° cos² 45° – 4 tan² 30° cot² 60°

(xi) Find the median of the following numbers 3, 5, 1, 2, 4, 6, 0, 2, 2, 3. (1)
Answer:
On arranging the given numbers in ascending order, we get
0, 1, 2, 2, 2, 3, 3, 4, 5, 6
Here, number of observations (n) = 10

(xii) Harpreet tosses two different coins simultaneously (say, one is of ₹ 1 and other of ₹ 2). What is the probability that she gets at least one head ? (1)
Answer:
When two coins are tossed simultaneously, all possible out comes are (HH), (HT), (TH), (TT) which are all equally likely.
∴ Number of all possible outcomes = 4
Let E be event of getting at least one head .
Then, the favourable outcomes are (HT), (TH), (HH)
Number of favourable outcomes = 3
∴ P(E) = \(\frac{3}{4}\)
Hence, P(E) = \(\frac{3}{4}\).

Section – B

Question 4.
Find the simplest form of \(\frac{90}{144}\).
Answer:
For the simplest form of a given fraction, first we find the HCF of 90 and 144.

90 = 2 × 3 × 3 × 5
144 = 2 × 2 × 2 × 2 × 3 × 3
HCF (90, 144) = 2 × 3 × 3 = 18
∴ \(\frac{90}{144}\) = \(\frac{90 \div 18}{144 \div 18}=\frac{5}{8}\)
Hence, simplest form of \(\frac{90}{144}\) is \(\frac{5}{8}\).

Question 5.
What must be subtracted from the polynomial 2x³ – 3x² + 6x +7 so that it may be exactly divisible by x² – 4x + 8. – 2
Answer:
f{x) = 2x³ – 3x² + 6x + 7
g(x) = x² – 4x + 8
According to division algorithm of polynomials
f(x) = g(x) × q(x) + r(x)
f(x) – r(x) = g(x) × q(x)
If we subtract r(x) from the f(x) then resulting polynomial is divisible by g(x).
We find r(x) on dividing f(x) by g(x) as follows:

Hence, quotient q(x) – 2x + 5
and remainder r(x) = 10x – 33.
Hence, if we subtract 10x – 33 from f(x) so it may be exactly divisible by g(x).

Question 6.
Use elimination method to find all possible solutions of the following pair of linear equations : (2)
2x + 3y = 8, 4x + 6y = 7
Answer:
The given equations are :
2x + 3y = 8 …(i)
4x + 6y = 7 …(ii)
Multiplying the equation (i) by (ii) then subtract equation (ii), from it we get

which is a false statement.
Therefore, the pair of equations has no solution.

Question 7.
Find the roots of equation 4x² + 3x + 5 = 0 by the method of completing the square. (2)
Answer:
The given equation is

But \(\left(x+\frac{3}{8}\right)^{2}\) cannot be negative for any real value of x.
Therefore, the given equation has no real roots.

Question 8.
Find the 10^th term of the AP : 2, 7, 12, … (2)
Answer:
The given sequence of AP is 2, 7, 12, …
Here, a = 2, d = a₂ – a₁ = 7 – 2 = 5
We know that nth term of AP is an = a + (n – 1 )d
⇒ a₁₀ = 2 + (10 – 1) × 5
⇒ a₁₀ = 2 + 9 × 5
⇒ a₁₀ = 2 + 45
⇒ a₁₀ = 47
Hence, 10^th term of given AP = 47.

Question 9.
How many two-digit numbers are divisible by 3? (2)
Answer:
The list of two-digit numbers divisible by 3 is:
12, 15, 18, 21, …, 99.
It is an AP with a = 12, d = 3 and a_n = 99
We know that nth term of AP is a_n = a + (n – 1) d
⇒ 99= 12 + (n – 1) × 3
⇒ 99 – 12 = 3n – 3
⇒ 87 = 3n – 3
⇒ 3n = 87 + 3
⇒ 3n = 90
⇒ n = \(\frac{90}{3}\) = 30
Hence, there are 30 two-digit numbers divisible by 3.

Question 10.
Construct a triangle of sides 4 cm, 5 cni and 6 cm and then a triangle similar to it whose sides are \(\frac{2}{3}\) of the corresponding sides of the first triangle. (2)
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 three 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
\(\frac{\mathrm{AB}^{\prime}}{\mathrm{AB}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}\) = \(\frac{\mathrm{A}^{\prime} \mathrm{C}^{\prime}}{\mathrm{AC}}=\frac{2}{3}\).

Question 11.
Draw a circle of radius 3 cm with centre 0 and take a point P outside the circle such that OP = 6 cm, from P, draw two tangents to the circle. (2)
Answer:
Steps of Construction:

• Draw a circle with O as the centre and radius 3 cm.
• Mark a point P outside the circle such that OP = 6 cm.
• Join OP and draw perpendicular bisector of PO meeting PO at M.

• Draw a circle with M as the center and radius equal to PM = MO intersecting the given circle at points Q and R.
• Join PQ and PR.
  Then PQ and PR are the required tangents.

Question 12.
If A = 60°, B = 30°, verify that sin(A + B) = sin A cos B + cos A sin B. (2)
Answer:
We have, A = 60°, B = 30° and sin(A + B) = sin A cos B + cos A sin B
LHS = sin(A + B)
= sin(60° + 30°) = sin 90° = 1
RHS = sin A cos B + cos A sin B
= sin 60° cos 30° + cos 60° sin 30°
= \(\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}+\frac{1}{2} \times \frac{1}{2}\) = \(\frac{3}{4}+\frac{1}{4}\) = 1
∴ LHS = RHS

Question 13.
If sin 3A = cos (A – 10°), where 3A is an acute angle, then find the value of A. (2)
Answer:
We have, sin 3A
= cos (A – 10°)
⇒ cos (90° – 3A) = cos (A – 10°) [∵ sin θ = cos (90° – 0)]
⇒ 90° – 3A = A – 10°
⇒ – 3A – A = – 10° – 90°
⇒ – 4A = – 100°
⇒ A = \(\frac{-100^{\circ}}{-4}\) = 25°
Hence, A = 25°.

Question 14.
The daily income of a sample of 50 employees are tabulated as follows:

Find the mean daily income of employees. (2)
Answer:
Here, h = 200

From the table, we have
Σf_i = 50, Σf_iu_i = – 36, a = 500-5 and h = 200
∴ x̄ = a + \(\left(\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\right)\) × h
⇒ 5c = 500.5 + \(\left(\frac{-36}{50}\right)\) × 200
⇒ 5c = 500.5 – 144
⇒ x = 3565.
Hence, mean daily income of employees = ₹ 356.5

Question 15.
A survey conducted on 20 households in a locality by a group of students resulted in the following frequency table for the number of family members in a household:

Find the mode of this data. (2)
Answer:
The class interval 3-5 has maximum frequency. So, it is the modal class.
∴ l = 3, f₁ = 8, f₀ = 7, f₂ = 2, h = 2

= 3 + \(\frac{2}{7}\) = 3 + 0.286 = 3.286
Hence, mode of the data = 3.286.

Question 16.
Savita and Hamida are friends. What is the probability that both will have (i) different birthdays ? (ii) the same birthday ? (ignoring a leap year). (2)
Answer:
Out of the two friends, one girl say, Savita’s birthday can be any day of the year. Now, Hamida’s birthday can also be any day of 365 days in the year.
(i) If Hamida’s birthday is different from Savita’s, the number of favourable outcomes for her birthday
= 365 – 1 = 364.
So, P(Hamida’s birthday is different from Savita’s birthday) = \(\frac{364}{365}\)

(ii) P(both have the same birthday)
= 1 – P (Both have different birthday)
= 1 – \(\frac{364}{365}\) = \(\frac{1}{365}\)
[Using not P (E) = 1 – P (E)]

Section – C

Question 17.
Find the middle term of AP : 10, 7, 4, …….., – 62. (3)
Or
Find the 25th term from the last term (towards the first term) of the AP : 12, 16, 20, 24, … 248. (3)
Answer:
The given sequence is 10, 7, 4, …, – 62
Here, a = 10, d = 7 – 10 = – 3, a_n = – 62,
We know that nth term of AP is
a_n = a + (n – 1 )d
⇒ – 62 = 10 + (n – 1) × (- 3)
⇒ – 62 – 10 = (n – 1) × (- 3)
⇒ – 72 = – 3n + 3
⇒ – 72 – 3 = – 3n
⇒ 3n = 75
⇒ n = \(\frac{75}{3}\) = 25
So, the given AP contains 25 terms and 25 is an odd number. So,
Middle term = \(\left(\frac{25+1}{2}\right)\)^th term
= 13^th term.
∴ a₁₃ = 10 + (13 – 1) × (- 3)
⇒ a₁₃ = 10 + 12 × (-3)
⇒ a₁₃ = 10 – 36
⇒ a₁₃ = – 26
Hence, the middle term is – 26.

Question 18.
Find the co-ordinates of the point which divides the join of (- 1, 7) and (4, – 3) in the ratio 2:3. 3
Or
One end point of diameter of a circle is at (2, 3) and the centre is (- 2, 5), what are the co-ordinates of the other end of this diameter ? (3)
Answer:
Let P(x, y) be the required point


Here, x₁ = – 1, y₁ = 7, x₂ = 4, y₂ = – 3 and m₁ : m₂ = 2 : 3

Hence, the co-ordinates of required point are (1, 3).

Question 19.
If a sin θ + 6 cos θ = c, then prove that a cos θ – b sin θ = \(\sqrt{a^{2}+b^{2}-c^{2}}\). (3)
Or
Prove that : sin θ (1 + tan θ) + cos θ (1 + Cot θ) = (sec θ + cosec θ) (3)
Answer:
We have, a sin θ + b cos θ = c
⇒ (a sin θ + b cos θ)² = c²
⇒ a² sin² θ + b² cos² θ + 2ab sin θ cos θ = c²
⇒ a²(1 – cos² θ) + b²(1 – sin² θ) + 2ab sin θ cos θ = c²
⇒ a² – a² cos² θ + b² – b² sin² θ + 2ab sin θ cos θ = c²
⇒ – a² cos² θ – b² sin² θ + 2ab sin θ cos θ= – a² – b² + c²
⇒ a² cos² θ + b² sin² θ – 2ab sin θ cos θ = a² + b² – c²
⇒ (a cos θ – b sin θ)² = a² + b² – c²
⇒ a cos θ – b sin θ = \(\sqrt{a^{2}+b^{2}-c^{2}}\)

Question 29.
The following distribution gives the state-wise teacher-student ratio in the higher secondary schools of India. Find the mode of this data. (3)

Or
Find the median marks for the following frequency distribution: (3)

Answer:

Section – D

Question 21.
Solve the following system of linear equations graphically :
4x – 5y = 20 and 3a: + 5y = 15. With the help of this find the value of m while 4x + 3y = m. (4)
Or
Solve graphically the system of equations 4x – 5y – 20 = 0, 3a; + 5y – 15 = 0.
Find the co-ordinates of the vertices of the triangle formed by these two lines and the y-axis. (4)
Answer:
The given equations are :
4x – 5y = 20 …(1)
And 3x + 5y = 15 …(2)
For representation of these equations graphically, we draw the graphs of these equations as follows :
4x – 5y = 20
⇒ 4x – 20 = 5y
⇒ y = \(\frac{4 x-20}{5}\)
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 equaiton 4x – 5y = 20.

And 3x + 5y = 15
⇒ y = \(\frac{15-3 x}{5}\)
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 3x + 5y = 15.

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.
Observe that we get two straight lines which intersect each other at point (5, 0). Hence, x = 5, y = 0 is the required solution. Putting these values of x and y in the equation 4x + 3y = m, we get
4 × 5 + 3 × 0 = m
⇒ 20 + 0 = m
⇒ m = 20

Question 22.
Draw a triangle ABC with BC = 7 cm, ∠B = 45° and ∠A = 105°, then construct a triangle whose sides are \(\frac{4}{3}\) times the corresponding sides of ∆ABC.
Or
Draw a triangle ABC with side BC = 6 cm, ∠C = 30° and ∠A = 105°. Then construct another triangle whose sides are \(\frac{4}{3}\) times the corresponding sides of ∆ABC. (4)
Answer:
Steps of Construction :

• Draw a line segment BC = 7 cm. In ∆ABC, ∠B = 45° and ∠A = 105°, ∠C = 180° – (45° + 105°) = 30°
• At B draw ∠B = 45° and at C draw ∠C = 30° intersecting each other at A to get ∆ ABC.
• Draw a ray BX making an acute angle with BC on the side opposite to the vertex A.
• Along BX mark four points B₁, B₂, B₃ and B₄ such that BB₁ = B₁B₂ = B₃B₄.

• Join B₃C.
• From B₄ draw B₄C’ || B₃C intersecting the produced line segment BC at C’.
• From C’ draw C’A’ ]| CA intersecting the produced line segment BA at A’. Then A’BC’ is the required triangle, each of whose side is \(\frac{4}{3}\) times the corresponding sides of ∆ABC.
  Justification: Since,
  AC || A’C’
  Therefore, ∆A’BC’ ~ ∆ABC
  ⇒ \(\frac{\mathrm{A}^{\prime} \mathrm{B}}{\mathrm{AB}}=\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{A}^{\prime} \mathrm{C}^{\prime}}{\mathrm{AC}}\) = \(\frac{4}{3}\).

Question 23.
The following distribution gives the daily income of 50 workers of a factory.

Convert the distribution above to less than type cumulative frequency distribution, and draw its ogive. (4)
Or
Draw an ogive and the cumulative frequency polygon for the following frequency distribution by less than method. (4)

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

We plot the points (120,12), (140, 26), (160, 34), (180,40), (200, 50) on the graph paper. Joining these points with a free hand to get less than curve as shown in graph.