RBSE 10th Maths Board Model Paper 2022 with Answers in English 

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RBSE Class 10 Maths Board Model Paper 2022 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) Decimal expansion of rational number \(\frac{129}{2^{5} \times 5^{7} \times 5^{7}}\) is: (1)
(a) terminating
(b) non-terminating
(c) non-terminating repeating
(d) non-terminating non-repeating
Answer:
(c) non-terminating repeating

(ii) Which of these is not a quadratic polynomial? (1)
(a) x²
(b) x² – 4
(c) x² – 4x + 4
(d) x³ – 3x² + 3x – 1
Answer:
(d) x³ – 3x² + 3x – 1

(iii) The solution of linear pair equation x + y = 4, x – y = 2 is : (1)
(a) x = 3, y = 1
(b) x = 1, y = 3
(c) x = y = 2
(d) x = 4, y = 0
Answer:
(a) x = 3, y = 1

(iv) Which one of the following methods is used to solve the quadratic equation: (1)
(a) factorization
(b) perfect square
(c) quadratic formula
(d) all of these
Answer:
(d) all of these

(v) The first term and common difference for the given AP; 4. 10, 16, 22 will be respectively: (1)
(a) 10, 5
(b) 4, 6
(cl 4, 10
(d) 6, 10
Answer:
(b) 4, 6

(vi) if the radius of a circle is 15 cm, then diameter of circle is : (1)
(a) 6 cm
(b) 7 cm
(c) 8.5 cm
(d) none of these
Answer:
(b) 7 cm

(vii) Point (-3. 5) belongs to which quadrant? (1)
(a) first
(b) second
(c) third
(d) fourth
Answer:
(b) second

(viii) Value of tan 45° + cot 45°: (1)
(a) 1
(b) 2
(c) 3
(d) 0
Answer:
(b) 2

(ix) Value of cos (90° – 48°) is : (1)
(a) sec 48°
(b) tan 48°
(c) sin 48°
(d) cot 48°
Answer:
(c) sin 48°

(x) One of these is not a formula to find mean: (1)
(a) \(\frac{\sum x}{n}\)
(b) \(\frac{\Sigma x+a h}{n}\)
(c) \(\frac{\Sigma f x}{N}\)
(d) \(\frac{\Sigma f d}{N}\)
Answer:
(b) \(\frac{\Sigma x+a h}{n}\)

(xi) The mode of data is: (1)

(a) 9
(b) 5
(c) 7
(d) 11
Answer:
(c) 7

(xii) Which of the following can not be the probability of an event: (1)
(a) 2/3
(b) -1.5
(c) 15%
(d) 0.7
Answer:
(b) -1.5

Question 2.
Fill in the blanks :
(i) If \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\) , then solution of linear pair of equation is ________________ (1)
Answer:
infinite

(ii) 7th term of AP; 10, 8, 6, 4 is ________________ (1)
Answer:
-2

(iii) How many tangents are drawn to a circle from the point of outsite the circle? (1)
Answer:
Two

(iv) The formula to find the distance between two points in the co-ordinate geometry is ________________ (1)
Answer:
\(\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\)

(v) sin (90° – θ) = ________________ (1)
Answer:
cos θ

(vi) Median of distribution 7, 4, 6, 3, 8, 5, 9 is ________________ (1)
Answer:
6

Question 3.
Very Short Answer Type Questions:
(i) If the product of two numbers is 180 and their HCF is 3. then find their LCM. (1)
Answer:
Product of two numbers = Product of LCM and HCF
∴ 180 = LCM × 3
⇒ LCM = 180 + 3
LCM = 60

(ii) Find the remainder, if x² + 3x + 1 is divided by x – 2? (1)
Answer:

So, remainder is 11.

(iii) Write the sum and product of zeroes of the polynomial x² – 6x + 7? (1)
Answer:
Let the zeroes of polynomial x² – 6x + 7 are α and β then,
Sum of zeroes, α + β = –\(\frac{b}{a}\)
⇒ α + β = \(\left(\frac{-6}{1}\right)\) ⇒ α + β = 6
and (product of zeroes) αβ = \(\frac{c}{a}\)

(iv) Find the condition of unique solution for pair of linear equation in two variables. (1)
Answer:
If two pairs of linear equations are a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, then the condition for unique solution is \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\).

(v) Find the sum of roots of quadratic equation 2x² + 3x + 4 = 0 ? (1)
Answer:
In equation in 2x² + 3x + 4 = 0.
Sum of roots = \(-\frac{b}{a}=-\frac{3}{2}\)

(vi) Write Shree Dharacharya formula for finding roots of quadratic equation. (1)
Answer:
For quadratic equation ax² + bx + c; where a ≠ 0, Shree Dharacharya formula is
x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

(vii) Find the angle between radius of a circle and tangent of circle. (1)
Answer:
90°

(viii) Find the abscissa and ordinate in co-ordinate (4,3). (1)
Answer:
Abscissa = 4,
Ordinate = 3

(ix) Find the value : cos 30° – sin 60° (1)
Answer:
cos 30° – sin 60° = \(\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}\) = 0

(x) If sin 3x = 1, then find the value of x. (1)
Answer:
If sin 3x = 1, then
⇒ sin 3x = sin \(\frac{\pi}{2}\)

(xi) Find the mean of distribution 10, 12, 8, 7, 13. (1)
Answer:
Mean of distribution 10, 12, 8, 7, 13
= \(\frac{10+12+8+7+13}{5}=\frac{50}{5}\) = 10

(xii) Find the probability of getting a head and a tail when a coin is tossed once. (1)
Answer:
If a coin is tossed once, then total outcomes S = [H, T]
⇒ n (S) = 2
n(H) 1
Probability of getting head P(H) = \(\frac{n(H)}{n(S)}=\frac{1}{2}\)
Probability of getting tail P(T) = \(\frac{n(T)}{n(S)}=\frac{1}{2}\)

Section – B

Question 4.
Find the HCF of 90 and 144 by using Euclid’s division method. (2)
Answer:
Step 1. The numbers 90 and 144 are such that 144 > 90
So, by Euclid’s division lemma for 144 and 90,
144 = 90 × 1 + 54

Step 2. ∵ Remainder is hot 0.
∴ For 54 and 90, again by Euclid’s division lemma,
90 = 54 × 1 + 36

Step 3. ∵ Remainder is not 0.
∴ For 36 and 54, again by Euclid’s division lemma,
54 = 36 × 1 + 18

Step 4. ∵ Remainder is not 0.
∴ For 18 and 36, again by Euclid’s division lemma,
36 = 18 × 2 + 0 ∵ Remainder is 0.

∴ The divisor of Step 4 is 18.
So, HCF of 90 and 144 is 18.

Question 5.
Find the zeroes of a quadratic polynomial x² – 2x – 15. (2)
Answer:
Given, x² – 2x – 15
⇒ x² – 2x – 15 = x² – 5x + 3x — 15
= x (x – 5) + 3 (x – 5)
= (x – 5) (x + 3)
For zeroes
(x – 5) (x + 3) = 0
⇒ x – 5 = 0 or x + 3 = 0
⇒ x = 5 or x = -3
So, the zeroes are 5, -3.

Question 6.
Solve the following pair of linear equations by elimination method: 2x + y = 6 and 2x-y = 2 (2)
Answer:
Given equations are
2x + y = 6 …(i)
and 2x-y = 2 …(ii)
Adding equations (i) and (ii),
⇒ 4x = 8 ⇒ x = 2

Subtracting equations (ii) and (i),
2y = 4 ⇒ y = 2
So, x = 2 and y = 2 are the solution of given linear equations.

Question 7.
Find the roots of the following quadratic equation by factorisation:
2x² – 2x – 6 = 0 (2)
Answer:
Given, quadratic equation is 2x² + x – 6 = 0
⇒ 2x² + 4x – 3x – 6 = 0
⇒ 2x (x + 2) – 3 (x + 2) = 0
⇒ (x + 2) (2x – 3) = 0
⇒ x + 2 = 0 or 2x – 3 = 0
⇒ x = -2 or x = \(\frac{3}{2}\)
So, the roots are – 2 and \(\frac{3}{2}\) .

Question 8.
Find the number of terms of the following AP : 7, 13, 19, ………… 205
Answer:
Given, AP series is
7, 13, 19, ………… 205
First term a = 7
Common difference d = 13 – 7 = 6
Last term l = 205

Let the number of term be n.
By l = a + (n – 1) d
⇒ 205 = 7+ (n – 1)6
⇒ (n – 1)6 = 205 – 7
⇒ (n – 1) = \(\frac{198}{6}\)
⇒ n = 33 + 1 = 34
So, the total number of terms is 34.

Question 9.
How many three digit numbers are divisible by 7 ? (2)
Answer:
The list of three-digit numbers is :
100, 101, 102, ……… 999,
The first number of three digits divisible by 7 = 105 and last number of three digits divisble by 7 = 994
Then list of three digit numbers divisble by 7 is 105, (105 + 7), (105 + 7 + 7), … 994
= 105, 112, 119, …, 994

Let the total number be n.
First term a = 105, common difference d = .7.
∴ nth term a_n = 994
⇒ a + (n – 1 )d = 994
⇒ 105 + (n – 1) × 7 = 994
⇒ (n – 1) × 7 = 994 – 105
⇒ (n – 1) = \(\frac{889}{7}\) =127.
∴ n = 127 + 1 = 128
So, the total three-digit numbers divisble by 7 is 128.

Question 10.
Draw a line segment of length 5 cm and divide in the ratio 2:3. (2)
Answer:
AP : PB = 2:3

Question 11.
Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°. (2)
Answer:
Steps of Construction:

• Draw a circle with centre O and radius 5 cm.
• Construct radii OA and OB such that ∠AOB = 360° – (90° + 90° + 60°) = 120°.
• Draw AX ⊥ OA at A and BY ⊥ OB at B.

They intersect at T.
Then TA and TB are the required tangents inclined to each other at 60°.
Justification : TA ⊥ OA and TB ⊥ OB and ∠AOB = 120°.

∠ATB = 3600 (90° + 90° + 120°)
∠ATB = 360° – 300°
∠ATB = 60°.
Therefore, tangents TA and TB are inclined at 60°.

Question 12.
If sin A = \(\frac{3}{4}\),calculate cos A and tan A. (2)
Answer:
Given,
sin A = \(\frac{3}{4}=\frac{B C}{A C}\)

From figure
AB² = AC² – BC²
= 4²– 3² = 16 – 9
= 7
AB = \(\sqrt{7}\)
So, cos A = \(\frac{A B}{A C}=\frac{\sqrt{7}}{4}\)
and tan A = \(\frac{B C}{A C}=\frac{3}{\sqrt{7}}\)

Question 13.
Evaluate the following:
2tan² 45° + cos² 30° – sin² 60°
Answer:
2tan² 45° + cos² 30° – sin² 60°
= 2(1)² + \(\left(\frac{\sqrt{3}}{2}\right)^{2}-\left(\frac{\sqrt{3}}{2}\right)^{2}\)
= 2 × 1 + \(\frac{3}{4}-\frac{3}{4}\)
= 2

Question 14.
Find the mean of the following distribution: (2)

Answer:

So, mean (x̄) = \(\frac{\sum f_{i} x_{i}}{\sum f_{i}}\)
= \(\frac{4460}{100}\) = 44.60

Question 15.
Find the mean òf the following distribution: (2)
Answer:

Class-interval	No. of students
0-20	10
20-40	35
40 – 60	52
60 – 80	61
80-100	38
100-120	20

The class interval 60—80 has maximum
frequency. So, it is the modal class.
∴ l = 60, f₁ = 61, f₀ = 52, f₂ = 38 h = 20

Question 16.
A bag contains 5 white and 2 red balls. From this bag one ball is the drawn randomly. What is the probability that drawn ball is : (i) white (ii) red. 2
Answer:
Number of all possible out comes = 5 + 2 = 7
(i) Number of white balls = 5
Number of favourable outcomes = 5
∴ P (white ball) = \(\frac{5}{7}\)
So, the probability of drawing white ball is \(\frac{5}{7}\)

(ii) Number of red balls = 2
Number of favourable outcomes = 2
P (Red ball) = \(\frac{2}{7}\)
So, the probability of red ball outcomes is \(\frac{2}{7}\)

Section – C

Question 17.
If second and third terms of an AP are 24 and 28 respectively, then find the sum of first 61 terms. (3)
OR
Find the sum of all natural numbers divisible by 5 between 2 and 101. (3)
Answer:
According to the question, a₂ = 24
and a₃ = 28
then, a_n = a + (n – 1) d (formula)
a₂ = a + d = 24 …(i)
and = a + 2d = 28 …(ii)
Subtracting equation (i) from (ii)

Substituting d = 4 in equation (i)
a + 4 = 24
⇒ a = 20
So, a – 20 and d = 4
Sum of first 61 terms of AP is

S₆₁ = \(\frac{61}{2}\) [2 × 20 +(61 – 1)4] [∵ S_n = \(\frac{n}{2}\)(2a + (n – 1 )d)]
= \(\frac{61}{2}\) × (40 + 240) = \(\frac{61}{2}\) × 280
= 61 × 140 = 8540
So, the sum of first 61 terms is 8540.
OR
The list of numbers divisible by 5 between 2 to 101 is
5, 10, 15, 20, 25, 30, …………… , 100

The following numbers are in AP,
then a = 5, d = 5 , a_n = 100

Here, a_n = 100
a + (n – 1)d= 100
⇒ 5 + (n – 1) 5 = 100
⇒ (n – 1) = \(\frac{100-5}{5}\)
⇒ n = 19 + 1
⇒ n = 20
∵ We know that,
S_n = \(\frac{n}{2}\)[a + l]
⇒ S_n = \(\frac{20}{2}\) [5 + 100]= 10 × 105
So, S₂₀ = 1050

Question 18.
Find the point on x-axis which is equidistant from the point (-2, 3) and (-3, 4). (3)
OR
In which ratio, point (7, 6) divides the line joining the points (11,8) and (1, 3). (3)
Answer:
Let the required point on x-axis be (x, 0), then
Distance between points (x, 0) and (-2, 3)
= Distance between points (x, 0) and (-3, 4)

Squaring both sides
x² + 4x + 13 = x² + 6x + 25
⇒ 6x – 4x = 13 – 25
⇒ 2x = -12
⇒ x = -6
So, the required point is (-6, 0)

OR

Let the point (7, 6) divides the line segment joining the given points in the ratio of λ : 1.

By Internal division formula
x = \(\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}\)
7 = \(\frac{\lambda \times 1+1 \times 11}{\lambda+1}\)
⇒ 7λ + 7 = λ + 11
⇒ 6λ = 11 – 7
⇒ λ = \(\frac{4}{6}=\frac{2}{3}\)
∴ λ : 1 = 2 : 3
So the required ratio is 2 : 3

Question 19.
Find the value of:
\(5 \frac{\sin 17^{\circ}}{\cos 73^{\circ}}+2 \frac{\cos 67^{\circ}}{\sin 23^{\circ}}-6 \frac{\sin 15^{\circ}}{\cos 75^{\circ}}\)
OR
cos⁴θ + sin⁴θ = 1 – 2cos²θsin²θ
Answer:

= 5 + 2 – 6 = 7 – 6 = 1

OR

L.H.S = cos⁴θ + sin⁴θ
{cos²θ)² + (sin²θ)² + 2sin²θcos²θ} – 2sin²θcos²θ
= (sin²θ + cos²θ)² – 2sin²θ cos²θ
= (1)² – 2sin²θ cos²θ
Hence Proved.

Question 20.
Find the mean of the following distributions: (3)

OR
Find the mode of the following frequency distribution: (3)

Answer:

Mean (x̄) = \(\frac{\Sigma f_{i} x_{i}}{\sum f_{i}}=\frac{930}{40}\) = 23.50
So, the required mean is 23.50
OR
The class interval 40 – 55 has maximum frequency, so it is the modal class
∴ l = 40, f₁ = 44, f₀ = 20, f₂ = 26, h = 15

Section – D

Question 21.
Solve the following pair of linear equations by graphical method : (4)
3x – 5y = 1 and 2x – y = 3
OR
Solve the following pair of linear equations by graphical method :
4x – 5y = 20 and 3x + 5y = 15 4
Answer:
Given, pair of linear equations is :
3x – 5y = 1
and 2x – y = 3
From equation (i),
3x – 5y = 1
3x = 1 + 5y
⇒ x = \(\frac{1+5 y}{3}\)
Putting y = 1,
x = \(\frac{1+5(1)}{3}=\frac{1+5}{3}=\frac{6}{3}\) = 2

Putting y = 4,
x = \(\frac{1+5(4)}{3}=\frac{1+20}{3}=\frac{21}{3}\) = 7

Putting y = – 2,
x = \(\frac{1+5(-2)}{3}=\frac{1-10}{3}=\frac{-9}{3}\) = -3

Table-I

From equation (ii),
2x – y = 3
⇒ y = 2x – 3

Putting x = 0,
y = 2(0) – 3

Putting x = 2,
y = 2(2) – 3 = 4 – 3 = 1

Putting x = 3
y = 2(3) – 3 = 6 – 3 = 3

Table – II

Now plotting the points (2, 1), (7, 4) and (-3, -2) and joining the graph of the equation 3x – 5y = 1 is a straight line AB and plotting the points (0, – 3), (2,1) and (3,3) and joining the graph of the equation 2x – y = 3 is straight line CD.

So. the solution of given equations is
x = 2 and y = 1.
OR
Given pair of linear equation are
4x – 5y = 20 ..(i)
3x + 5y = 15 …(ii)

From equation (i),
4x – 5y = 20
⇒ 4x = 5y + 20
⇒ x = \(\frac{5 y+20}{4}\)

Putting, y = 0,
For, x = \(\frac{5(0)+20}{4}=\frac{0+20}{4}\) = 5

Putting, y = 3,
x = \(\frac{5(-4)+20}{4}=\frac{-20+20}{4}\) = 0

Putting, y = -3,
x = \(\frac{5(-4)+20}{4}=\frac{-20+20}{4}\) = 10

Table – I

From equation (ii),
3x – 5y = 15
⇒ 3x = 15 – 5y
⇒ x = \(\frac{15-5 y}{3}\)

Putting, y = 0,
x = \(\frac{15-5(0)}{3}=\frac{15-0}{3}\) = 5

Putting, y = 3,
x = \(\frac{15-5(3)}{3}=\frac{15-15}{3}\) = 0

Putting. y = – 3,
x = \(\frac{15-5(-3)}{3}=\frac{15+15}{3}\) = 10

Table – II

Now plotting the pints (5, 0), (0, – 4) and (10, 4) and joining the graph of the equation 4x – 5y = 20 is a straight line AB and plotting the points (5,0), (0,3) and (10, -3) and joining the graph of the equation 3x + 5y = 15 is a straight line CD.
So, the solution of given equations is x = 5, y= 0.

Question 22.
Draw a line-segment of 10 cm and divide it into the ratio of 2 : 3 (4)
OR
Draw a line segment AB of 5 cm and then draw a circle of 2.2 cm taking B as centre and draw tangents from point A on this circle. (40
Answer:
Steps of Construction :

• Draw a ray AX of 10 cm, making an acute angle with AB.
• Along AX mark (2 + 3) = 5 points A₁; A₂, A₃, A₄ and A₅ such that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅.
• Join A₅B.
• From the point A₂ draw A₂ C ∥ A₅B, meeting AB at C. Then AC : BC = 2:3.

Justification: In ∆ACA₂ and ∆ABA₅, we have CA₂ ∥ BA₅
\(\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\mathrm{AA}_{2}}{\mathrm{~A}_{2} \mathrm{~A}_{5}}\)
[By Basic proportionality theorem]
⇒ \(\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{2}{3}\)
Hence, AC : BC = 2 : 3.

OR

Steps of Construction
(1) Draw the line segment ‘AB = 5 cm.
(2) Draw a circle of radius 2.2 cm from the point B of AB.

(3) Draw the bisector of AB and mark the bisector as M.
(4) Taking Mas center, draw a circle of radius MB which intersects the circle with center B at T₁ and T₂.
(5) Joint AT₁ and AT₂. AT₁ and AT₂ are the required tangents.

Justification:
Joint BT₁ and BT₂.
We know that the tangent is perpendicular to the radius passing through the point of contact.
So, ∠AT₁B = ∠AT₂B = 90°.

AB is diameter of the circle with center M.
∠AT₁B = ∠AT₂B = 90°.
(The angle in semi-circle is a right angle) So, AT₁ and AT₂ are tangents.

Question 23.
Find the median of the following distribution: (4)

Class interval	Frequency (f)
10-25	6
25-40	20
40-55	44
55-70	26
70-85	3
85- 100	1

OR
Find the made of the following distribution. (4)

Marks	No. of Students
0-20	5
20-40	10
40-60	12
60-80	6
80-100	3

Answer:

where N = 100 ⇒ \(\frac{\mathrm{N}}{2}=\frac{100}{2}\) = 50
So, the comutative frequency more than 50 comes in the class interval is in 70 which is 40 – 55
Median class = 40 – 55
l = 40, c = 26, f = 44, h = 15

Median = l + \(\left(\frac{\frac{N}{2}-c}{f}\right)\) × h
= 40 + \(\left(\frac{50-26}{44}\right)\) × 15
= 40 + \(\frac{24}{44}\) × 15
= 40 + \(\frac{6}{11}\) × 5
= 40 + 8.18 = 48.18

OR

The class interval 40 – 60 has maximum frequency, 12, it is modal class.
l = 40, f₁ = 12,f₀ = 10, f₂ = 6, h = 20
Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h
= 80 + \(\left(\frac{12-10}{2 \times 12-10-6}\right)\) × 20
= 80 + \(\left(\frac{2}{8}\right)\) × 20 = 85
So, mode is 85.