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

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

(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

(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°

(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:

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, u_i is equal :
(a) \(\frac{x_{i}+a}{h}\)
(b) h(x_i – 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

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

HCF (375, 675) = 3 × 5² = 75

(ii) Divide the polynomial 3x³ + x² + 2x + 5 by the polynomial 1 + 2x + x². (1)
Answer:

(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

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

(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:

(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 = \(\sqrt{1+\tan ^{2} \alpha}\)

(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

(xi) Compute the mode for the following frequency distribution:

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

(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

LCM (306, 657)
= 2 × 32 × 17 × 73
= 22338

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:

Hence, quotient q(x) = x – 2 and remainder r(x) = 3.

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 n^th 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
n^th term = a_n = 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 n^th term.
Answer:
We have
S_n = \(\frac{1}{2}\)[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
n^thterm(a_n) = a + (n – 1)d
= 5 + (n – 1) × 3
= 5 + 3n – 3
= 3n + 2

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.

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.
  
• 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,

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)
Answer:

From the table, we have
Σf_i = 35, Σf_id_i = – 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.

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.

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 S_n = 0(given)
Let AP has n terms
S_n = 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

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:

Question 20.
The following table shows the ages of the patients admitted in a hospital during a year : (3)

Find the mean of the data.
Or
The following distribution gives the goes of worker of a factory. Calculate the modal age. (3)

Answer:
Let us find the mean of the data. Here h = 10

From the table, we have
Σf_i = 80, Σf_iu_i = – 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.

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

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.

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.

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.
  
• 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
\(\frac{\mathrm{AB}^{\prime}}{\mathrm{AB}}=\frac{\mathrm{B}^{\prime} \mathrm{C}^{\prime}}{\mathrm{BC}}=\frac{A C^{\prime}}{\mathrm{AC}}=\frac{3}{5}\)

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)

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)

Answer:
For calculating mean, we prepare the table as given below:

From the table, we have
Σf_i = 68, Σf_iu_i =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.