RBSE 10th Maths Model Paper Set 7 with Answers in English

Students must start practicing the questions from RBSE 10th Maths Model Papers Set 7 with Answers in English Medium provided here.

RBSE Class 10 Maths Model Paper Set 7 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) Euclid’s division lemma states that for two positive integers a and b, there exists unique integers q and r satisfying a = bq + r, and: (1)
(a) 0 < r < b
(b) 0 > r ≤ b
(c) 0 ≤ r < b
(d) 0 ≤ r ≤ b
Answer:
(c) 0 ≤ r < b

(ii) The zeroes of the polynomial x² – 3x – m (m + 3) are: (1)
(a) m, (m + 3)
(b) – m, (m + 3)
(c) m, – (m + 3)
(d) – m, – (m + 3)
Answer:
(b) – m, (m + 3)

(iii) The values of a and b for which the following system of equations: (1)
2x – 3y = 7, (a + b)x – (a + b – 3)y = 4a + b has infinitely many solutions, are :
(a) a = – 5, b = 2
(b) a = – 2, b = – 5
(c) a = – 5, b = – 1
(d) a = – 1, b = – 5.
Answer:
(c) a = – 5, b = – 1

(iv) Which of the following equations has two distinct real roots ? (1)
(a) 2x² – 3√2x + \(\frac{9}{4}\)
(b) x² + x – 5 = 0
(c) x² + 3x + 2√2 = 0
(d) 5x² – 3x + 1 = 0.
Answer:
(b) x² + x – 5 = 0

(v) The sum of first 20 odd natural numbers is : (1)
(a) 400
(b) 200
(c) 500
(d) 800.
Answer:
(a) 400

(vi) To construct a triangle similar to a given ∆ABC with its sides \(\frac{3}{7}\) of the corresponding sides of ∆ABC, first draw a ray BX such that ∠CBX is an acute angle and X lies on the opposite side of A with respect to BC. Then locate points B₁, B₂, B₃, …. on BX at equal distance and next step is to join : (1)
(a) B₁₀ to C
(b) B₃ to C
(c) B₇ to C
(d) B₄ to C
Answer:
(c) B₇ to C

(vii) The coordinates of the point which is reflection of point (-3, 5) in x-axis are: (1)
(a) (3, 5)
(b) (3, – 5)
(c) (- 3, – 5)
(d) (- 3, 5)
Answer:
(c) (- 3, – 5)

(viii) The value of sin² 29° + sin² 61° is : (1)
(a) 0
(b) 1
(c) 2 sin² 29°
(d) 2 cos² 61°
Answer:
(b) 1

(ix) If tan A = cot B, and A, B are acute angles, then (A + B) is equal to : (1)
(a) 90°
(b) 60°
(c) 45°
(d) 120°
Answer:
(a) 90°

(x) In an arranged discrete series in which total number of observations n is even, median is: (1)
(a) \(\left(\frac{n}{2}\right)^{t h}\) terms
(b) \(\left(\frac{n}{2}+1\right)^{\mathrm{th}}\)
(c) The mean of \(\left(\frac{n}{2}\right)^{t h}\) term and \(\left(\frac{n}{2}+1\right)^{\mathrm{th}}\) term
(d) none of these
Answer:
(c) The mean of \(\left(\frac{n}{2}\right)^{t h}\) term and \(\left(\frac{n}{2}+1\right)^{\mathrm{th}}\) term

(xi) The mean of ungrouped data is given by: (1)
(a) Mean = \(\frac{\sum x_{i}}{\sum f}\)
(b) Mean = \(\frac{\sum x}{n}\)
(c) Mean = \(\frac{\sum f x}{\sum n}\)
(d) Mean = α + \(\frac{\sum f x}{n}\)
Answer:
(b) Mean = \(\frac{\sum x}{n}\)

(xii) Cards bearing numbers 2, 3, 4, …………, 11 are kept in a bag. A card is drawn at random from the bag. The probability of getting a card with a prime number is : (1)
(a) \(\frac{1}{2}\)
(b) \(\frac{2}{5}\)
(c) \(\frac{3}{10}\)
(d) \(\frac{5}{9}\)
Answer:
(a) \(\frac{1}{2}\)

Question 2.
Fill in the blanks :
(i) Every solution of a linear equation is two variables is a _______________ on the line representing it. (1)
Answer:
point

(ii) The common difference of an AP can be positive, negative or _______________ (1)
Answer:
zero

(iii) The angle between diameter and tangent is a _______________ angle. (1)
Answer:
right

(iv) The distance of a point P (x, y) from _______________ is \(\sqrt{x^{2}+y^{2}}\). (1)
Answer:
origin

(v) If cot A = \(\frac{12}{5}\), then the value of (sinA + cosA) × cosecA is _______________. (1)
Answer:
17/5

(vi) The data having only one mode is called _______________ data. (1)
Answer:
unimodal

Question 3.
Very Short Answer Type Questions:
(i) Find LCM and HCF of 6 and 20 by the prime factorization method.
Answer:
6 = 2 × 3
20 = 2 × 2 × 5 = 2² × 5
HCF (6, 20) = 2
(Product of smallest power of each common prime factor in the numbers).
LCM (6, 20) = 2² × 3 × 5 = 60 (Product of the greatest power of each prime factor in the numbers.)

(ii) If the zeroes of the polynomial x² + 4x + 2a are α and \(\frac{2}{\alpha}\) then find the value of a.
Answer:
Given : f(x) = x² = 4x + 2a
∴ α × \(\frac{2}{\alpha}=\frac{C}{A}\) = \(\frac{2 a}{1}\)
⇒ 2 = 2a
⇒ α = \(\frac{2}{2}\) = 1

(iii) Find a cubic polynomial whose zeroes are a, p and y such that α + β + γ, αβ + βγ + γα = – 7 and αβγ = 14.
Answer:
∴ Required polynomial
= x³ – (α + β + γ)x² + (αβ + βγ + γα)x – αβγ
= x³ – 2x² – 7x – 14

(iv) Find the value of k so that the pair of equations x + 2y = 5 and 3x + ky + 15 = 0 has a unique solution. (1)
Answer:
Here,
a₁ = 1, b₁ = 2, c₁ = – 5
a₂ = 3, b₂ = k, and c₂ = 15
Condition for unique solution is:
⇒ \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\) ⇒ \(\frac{1}{3} \neq \frac{2}{k}\) ⇒ k ≠ 6
Hence, k have all real values except 6.

(v) Find the roots of the equation 2x² – 5x + 3 = 0, by factorisation method. (1)
Answer:
The given equation is :
2x² – 5x + 3= 0
⇒ 2x² – (2 + 3)x + 3 = 0
⇒ 2x² – 2x – 3x + 3= 0
⇒ 2x (x – 1) – 3(x – 1) = 0
⇒ (x – 1) (2x – 3) = 0
⇒ x – 1 = 0 or 2x – 3 = 0
⇒ x = 1 or x = – 2
Hence, x = 1 and x = \(\frac{3}{2}\) are the roots of the given quadratic equation.

(vi) If one root of the quadratic equation 6x² – x – k = 0 is \(\frac{2}{3}\), then find the value of k. (1)
Answer:
Since, \(\frac{2}{3}\) is one root of the quadratic equation 6x² – x – k = 0
6 × \(\left(\frac{2}{3}\right)^{2}-\frac{2}{3}\) – k = 0
⇒ 6 × \(\frac{4}{9}-\frac{2}{3}\) – k = 0
⇒ \(\frac{8}{3}-\frac{2}{3}\) – k = 0
⇒ \(\frac{6}{3}\) – k = 0
⇒ 2 – k = 0
⇒ k = 2

(vii) How many tangents can be constructed to any point on the circle of radius 4 cm?
Answer:
only one

(viii) If the points A(6, 1), B(8, 2), C(9, 4) and D(P, 3) are the vertices of a parallelogram, taken in order. Find the value of P. (1)
Answer:
We know that diagonals of a parallelogram bisect each other. Therefore, coordinates of the mid point of AC is equal to the coordinates of the mid point of BD. i.e.

⇒ 8 + P = 15
⇒ P = 15 – 8 = 7
Hence, P = 7.

(ix) Find the value of: \(\frac{\tan 67^{\circ}}{\cot 23^{\circ}}\). (1)
Answer:
\(\frac{\tan 67^{\circ}}{\cot 23^{\circ}}\) = \(\frac{\tan \left(90^{\circ}-23^{\circ}\right)}{\cot 23^{\circ}}\)
= \(\frac{\cot 23^{\circ}}{\cot 23^{\circ}}\) [∵ tan (90° – θ) = cot θ]
= 1

(x) If sin 3A = cos (A – 26°), where 3A is an acute angle, find the value of A. (1)
Answer:
We have, sin 3A = cos (A – 26°)
cos (90° – 3A) = cos (A – 26°)
[∵ sin θ = cos (90° – θ)]
⇒ 90° – 3A = A – 26°
⇒ 4A = 90° + 26° = 116°
⇒ A = \(\frac{116^{\circ}}{4}\) = 29°

(xi) Find the medn of the following data: (1)

Answer:

∴ Mean = \(\frac{\sum f_{i} x_{i}}{\sum f_{i}}\)
= \(\frac{326}{40}\)
= 8.15

(xii) If probability of ‘not E’ = 0.95, then find P (E).
Answer:
∵ P (E) + P (not E) = 1
∴ P (E) + 0.95 = 1
P (E) = 1 – 0.95
⇒ P (E) = 0.05

Section – B

Question 4.
Prove that one of three consecutive positive integer is divisible by 3. (2)
Answer:
Let x be any positive integer. By Euclid’s division lemma x = 3q + r, where 0 < r ≤ 3 [∴ r = 0, 1, 2]
Putting r = 0, we get x = 3q + 0 = 3q
which is divisible by 3.
Putting r = 1, we get, x = 3q + 1
which is not divisible by 3.
Putting r = 2, we get x = 3q + 2,
which is not divisible by 3.
So, one of every three consecutive positive integers is divisible by 3.

Question 5.
Divide polynomial f(x) = 4x² – 3x² + 2x – 4 by other polynomial g(x) = x – 1. Find the quotient and remainder. (2)
Answer:
Given f(x) = 4x² – 3x² + 2x – 4
and g(x) = x – 2
Now, we divide f(x) by g(x) as follows:

Hence, quotient q(x) = 4x² + x + 3 and remainder r(x) = – 1

Question 6.
If x = a and y = b is the solution of the pair of equation x – y = 2 and x + y = 4, then find the values of a and b. (2)
Answer:
x – y = 2 …….. (i)
x + y = 4 …….. (ii)
Adding equations (i) and (ii), we get 2x = 6
⇒ x = \(\frac{6}{2}\) = 3
Substituting the value of x in equation (i) we get
3 – y = 2
⇒ y – 1
∴ x – 3 and x = a (given)
⇒ a = 3
And y = 1 and y – b (given)
⇒ 6 = 1
Hence, a = 3, 6 = 1

Question 7.
Is the following situation possible? If so, determine their present ages.
“The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.” (2)
Answer:
Let the present age of first friend be x years.
Then present age of second friend = (20 – x) years
Four years ago age of .first friend = (x – 4) years
Four years ago age of second friend = (20 – x – 4) years = (16 – x) years
According to question,
(x – 4)(16 – x) = 48
⇒ 16x – x² – 64 + 4x = 48
⇒ – x² + 20x – 64 – 48 = 0
⇒ – x² + 20x – 112 = 0
⇒ x² – 20x + 112= 0
Here, a = 1, b = -20, c = 112
∴ D = b² – 4ac
⇒ D= (- 20)² – 4 × 1 × 112
⇒ D = 400 – 448
⇒ D = – 48 ∵ D < 0
∴ The equation has no real value of x. Hence, the situation is not possible.

Question 8.
Find the sum of the first 40 positive integers divisible by 6. (2)
Answer:
The first 40 positive integers divisible by 6 are: 6, 12, 18, 24, …, 240.
Since, each number is divisible by 6.
So it is an AP in which
a = 6, d = a₂ – a₁ = 12 – 6 = 6, n = – 40 and l = 240
Now, S₄₀ = \(\frac{40}{2}\) [6 + 240] [∵ S_n = \(\frac{n}{2}\) (a + l)]
⇒ S₄₀ = 20 × 246
⇒ S₄₀ = 4920
Hence, the sum of the first 40 positive integers divisible by 6 is 4920.

Question 9.
Find the sum of the first 15 multiples of 8. (2)
Answer:
The first 15 multiples of 8 are 8, 16, 24, 32, …, 120.
Since each number is multiple of 8, So, it is an AP in which a = 8, d = a₂ – a₁ = 16 – 8 = 8, n = 15 and l = 120
Now, S₁₅ = \(\frac{15}{2}\) [8 + 120]
[∵ S_n = \(\frac{n}{2}\) (a + l)]
⇒ S₁₅ = \(\frac{15}{2}\) × 128
⇒ S₁₅ = 15 × 64 = 960
Hence, the sum of the first 15 multiples of 8 is 960.

Question 10.
Draw two tangents to a circle of radius 4 cm, which are included to each other at an angle of 60°. (2)
Answer:
Steps of Construction:

• Draw a circle with centre O and radius 4 cm.
• Construct raddi OA and OB such that ∠AOB = 360° – (90° + 90° + 60°) = 120°
• 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 60°

Question 11.
Divide a line segment in the ratio of 2 : 3. (2)
Answer:
Steps of Construction :

• Draw a line segment AB.
• Draw a ray AX making an acute angle with AB.
• Draw a ray BY (On opposite side of AX) parallel to AX making ∠ABY = ∠BAX.
• Along AX mark the points A₁, A₂ and along BY mark the points B₁; B₂, B₃ such that AA₁ = A₁A₂ = BB₁ = B₁B₂ = B₂B₃
  
• Join A₂B₃ intersecting AB at point C. Then
  AC : BC = 2 : 3

Question 12.
Show that: \(\frac{\cos ^{2}\left(45^{\circ}+\theta\right)+\cos ^{2}\left(45^{\circ}-\theta\right)}{\tan \left(60^{\circ}+\theta\right) \tan \left(30^{\circ}-\theta\right)}\) = 1
Answer:

Question 13.
Prove that : sec A (1 – sin A) (sec A + tan A) = 1. (2)
Answer:
LHS= sec A (1 – sin A) (sec A + tan A)

Question 14.
The following table shows the marks obtained by 50 students in mathematics of class 10 in a school. 2

Find the median marks.
Answer:

n = 50 ⇒ \(\frac{n}{2}\) = 25
But 25 comes under the cumulative frequency 34 and class interval against the cumulative frequency 34 is 50 – 60.
So, it is the median class.
∴ l = 50, cf = 22, f = 12 and h = 10
∴ Median = \(l+\left(\frac{\frac{n}{2}-c f}{f}\right) \times h\)
= 50 + \(\left(\frac{25-22}{12}\right)\) × 10
= 50 + \(\frac{3}{12}\) × 10
= 50 + 2.5 = 52.5
Hence, median marks = 52.5

Question 15.
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data. (2)

Answer:
The class 40 – 50 has maximum frequency.
So, it is the modal class.
∴ l = 40, f₁ = 20, f₀ = 12, f₂ = 11 and h = 10
Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h
= 40 + \(\left(\frac{20-12}{2 \times 20-12-11}\right)\) × 10
= 40 + \(\left(\frac{8}{40-12-11}\right)\) × 10
= 40 + \(\frac{80}{17}\)
= 40 + 4.7 = 44.7
Hence, mode = 44.7

Question 16.
A number x is chosen at random from the numbers -3, -2, -1, 0, 1, 2, 3. What is the probability that x² ≤ 4 ? (2)
Answer:
A number x is chosen at random from the numbers -3, -2, -1, 0, 1, 2, 3
x² = 9, 4, 1, 0, 1, 4, 9
The number all possible outcomes = 7
The numbers x² ≤ 4 are 4, 1, 0, 1, 4, 9 i.e. 5 numbers
Let E be event of chosen at random x² ≤ 4
Number of outcomes to favourable to event
E = 5
∴ P(E) = \(\frac{5}{7}\)

Section – C

Question 17.
Show that the sum of an AP whose first term is a, the second term 6 and the last term c, is equal to \(\frac{(a+c)(b+c-2 a)}{2(b-a)}\)
Or
In an-AP, the first term is 22, n^th term is – 11 and sum of first n^th term is 66. Find n and the common difference (d). (3)
Answer:
We have, first term = a
second term = b
Then, common difference (d) = b – a
last term (l) = c
Let n be the number of terms in the AP
∴ a_n = l = c
⇒ a + (n – 1) × (b – a) = c
⇒ (n – 1) (b – a) = c – a

Question 18.
Find the ratio in which the line segment joining the points (- 3, 10) and (6, – 8) is divided by (- 1, 6). (3)
Or
If (x, y) be on the line joining the two points (-1, 3) and (4, -2), then prove that : x + y – 2 = 0. (3)
Answer:
Let the required ratio be m₁ : m₂ By section formula, we have

⇒ 6m₁ – 3m₂ = – m₁ – m₂
⇒ 6m₁ + m₁ = 3 m₂ – m₁
⇒ 7m₁ = 2m₂
⇒ \(\frac{m_{1}}{m_{2}}=\frac{2}{7}\)
⇒ m₁ : m₂ = 2 : 7
Hence, the required ratio is 2 : 7

Question 19.
Prove that: \(\frac{\tan A-\sin A}{\tan A+\tan A}=\frac{\sec A-1}{\sec A+1}\)
Or
Prove that : \(\frac{\sin \theta+\tan \theta}{\sec \theta-\tan \theta}\) = (sec θ + tan θ)² = 1 + 2 tan² θ + 2 tan θ sec θ (3)
Answer:
We, have = \(\frac{\tan A-\sin A}{\tan A+\tan A}\)

Question 20.
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes. (3)

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose ?
Or
Calculate the median from the following data :

Answer:
The given series is an inclusive series. It is not necessary to change it in an exclusive series because the class interval includes both 50 and 52. So we take h = 3 (not 2).

From the table, we have,
Σf_i = 400, Σf_iU_i = 25,
a = 57, h = 3
∴ x̄ = a + \(\left(\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\right)\) × h
⇒ x̄ = 57 + \(\left(\frac{25}{400}\right)\) × 3
⇒ x̄ = 57 + 0.91×
⇒ x̄ = 57.19
Hence, the mean number of mangoes = 57.19.
We used step-deviation method.

Section – D

Question 21.
Two rails are represented by the equations x + 2y – 4 = 0 and 2x + 4y – 12 = 0. Represent this situation geometrically. 4
Or
Solve graphically the following system of equations 4x – 5y + 16 = 0 and 2x + y – 6 = 0. Also find the points, where these lines meet the x-axis. (4)
Answer:
Two solutions of each of the equations are.
According to question,
x + 2y – 4 = 0 …(i)
and 2x + 4y – 12 =0 …(ii)
For representation of these equations geometrically, we draw the graphs of these equation as follows:
x + 2y – 4 = 0
⇒ 2y = 4 – x
⇒ y = \(\frac{4-x}{2}\)
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 the equation x + 2y – 4 = 0.

and 2x + 4y – 12 = 0
⇒ 4y = 12 – 2x
⇒ y = \(\frac{12-2 x}{4}\)
⇒ y = \(\frac{6-x}{2}\)
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 – 12 = 0.

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

Question 22.
Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle. Give justification too. (4)
Or
Construct a triangle similar to a given triangle ABC with its side equal to 4/3 of the corresponding side of the triangle ABC. (4)
Answer:
Steps of Construction :

• Draw a line segment AB = 8 cm.
• A as centre and radius equal to 4 cm draw a circle.
• B as centre and radius equal to 3 cm draw another circle.
• Bisect line segment AB at M.
  
• M is the centre and radius equal to AM draw the circle which intersects the previous circles at P, Q (A as centre) and R, S (B as centre).
• Join BP, BQ, AR and AS.
• Then, BP, BQ, AR and AS are the required tangents.

Justification : Join AP, RB
∠APB = 90°
[Angle in a semicircle is 90°]
PB ⊥ AP.
Therefore, AP is radius and BP is tangent.
Similarly, AR, AS and BQ are tangents.

Question 23.
On the annual day of school age wise participation of students is given in the following distribution table. (4)

Find the median of students.
Or
Find the mean, and of the following data : (4)

Answer:
We convert the cumulative frequency table into depecting class intervals with their respective frequencies as given below:

\(\frac{n}{2}\) = \(\frac{76}{2}\) = 38
But 38 comes under the cumulative frequency 42 and the class interval against cumulative frequency 42 is 12 – 14. So, it is the median class.
Here l = 12, f = 20, cf = 22, h = 2
∴ Median = l + \(\frac{\left(\frac{\mathrm{n}}{2}-c f\right)}{f}\) × h
= 12 + \(\left(\frac{38-22}{20}\right)\) × 2
= 12 + \(\frac{16}{10}\)
= 12 + 1.6
= 13.6