RBSE 10th Maths Model Paper Set 9 with Answers in English

Students must start practicing the questions from RBSE 10th Maths Model Papers Set 9 with Answers in English Medium provided here.

RBSE Class 10 Maths Model Paper Set 9 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) If HCF (a, 8) = 4, LCM (a, 8) = 24, then a is (1)
(a) 8
(b) 10
(c) 12
(d) 14
Answer:
(c) 12

(ii) The sum and product respectively of zeroes of the polynomial x² – 4x + 3 are (1)
(a) 3, 3
(b) 4, 3
(c) – 4, 3
(d) \(\frac{4}{3}\), 1
Answer:
(b) 4, 3

(iii) If (6, k) is a solution of the equation 3x + y – 22 = 0, then the value of k is 1
(a) -4
(b) 4
(c) 3
(d) – 3
Answer:
(b) 4

(iv) If one root of the equation ax² + bx + c = 0 is three times of the other, then k is equal to …………………… (1)
(a) 3
(b) lg
(c) 16
(d) 3
Answer:
(b) lg

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

(vi) To divide a line segment AB in the ratio 2 : 5, 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) 2
(b) 5
(c) 4
(d) 7
Answer:
(d) 7

(vii) Which of the following point is equidistant from (3, 2) and (- 5, – 2) ? (1)
(a) (0, -2)
(b) (0, 1)
(c) (0, 2)
(d) (2, 0)
Answer:
(a) (0, -2)

(viii) 9 sec²θ – 9 tan² θ is equal to : 1
(a) 9
(b) 1
(c) 0
(d) 8
Answer:
(a) 9

(ix) If sin θ + sin² θ = 1, then cos² θ + cos⁴ θ is equal to : 1
(a) 0
(b) – 1
(c) 1
(d) 2
Answer:
(c) 1

(x) The statistical data are of two types. These types are : (1)
(a) technical data and presentation data
(b) primary data , and secondary data
(c) primary data and personal data
(d) none of these
Answer:
(b) primary data , and secondary data

(xi) The number of times a particular item occurs in a class interval is called its : (1)
(a) mean
(b) frequency
(c) cumulative frequency
(d) none of these
Answer:
(c) cumulative frequency

(xii) A card is drawn from a packet of 100 cards numbered 1 to 100. The probability of drawing a number which is perfect square is : (1)
(a) \(\frac{1}{10}\)
(b) \(\frac{1}{100}\)
(c) \(\frac{1}{50}\)
(d) \(\frac{9}{100}\)
Answer:
(a) \(\frac{1}{10}\)

Question 2.
Fill in the blanks :
(i) The value of k for which x – 5x = 10 and 3x – 3ky = – 5 have no solution, will be ______________ (1)
Answer:
5

(ii) If a + 1, 2a + 1, 4a – 1 are in AP, then value of a is ______________ . (1)
Answer:
2

(iii) Given a triangle with side AB = 8 cm. To get a line segment AB’ = \(\frac{3}{4}\) of AB. It is required to divide the line segment AB in the ratio ______________ (1)
Answer:
3 : 1

(iv) If the points (- 1, 3), (2, P) and (5, -1) are collinear, then the value of P will be ______________ (1)
Answer:
1

(v) The equation sin²x = \(\frac{a^{2}+b^{2}}{2 a b}\) is possible if ______________ (1)
Answer:
a = b

(vi) The points A (0, 0), B(3, √3) and C(3, -√3) are vertices of a triangle, then triangle will be ______________ (1)
Answer:
equilateral

Question 3.
Very Short Answer Type Questions :
(i) HCF and LCM of two numbers are 9 and 360 respectively. If one number is 45, find the other number. (1)
Answer:
HCM = 9, LCM = 360 and one number = 45
∵ other number = \(\frac{\mathrm{LCM} \times \mathrm{HCF}}{\text { One number }}\)
∴ Required number = \(\frac{360 \times 9}{45}\) = 8 × 9
= 75

(ii) Write division algorithm for polynomials.
Answer:
If p (x) and g (x) are any two polynomials with g (x) ≠ 0, then p (x) = g (x) × q (x) + r (x); where r (x) ≠ 0 or degree of r(x) < degree of g (x).

(iii) Check whether (x – 3) is a factor of polynomial f(x) = x³ + x² – 17x + 15 or not. (1)
Answer:
f (x) = x³ + x² – 17x + 15
f(3) = (3)³ + (3)² – 17 (3) + 15
= 27 + 9 – 51 + 15 = 0

(iv) For which values of P, will the lines represented by the following pair of linear equations be parallel 3x – y – 5 = 0 and 6x – 2y – P = 0. (1)
Answer:
The condition for parallel lines is

So, P can have any real values other than 10.

(v) If x = 3 is one root of quadratic equation x² – 2kx – 6 = 0, then find the value of k. (1)
Answer:
Since, x = 3 is one root of the quadratic equation x² – 2kx – 6 = 0.
∴ (3)² – 2k × 3 – 6 = 0
⇒ 9 – 6k – 6 = 0
⇒ – 6k + 3 = 0
⇒ k = \(\frac{3}{6}\) ⇒ k = \(\frac{1}{2}\)

(vi) Find the roots of the quadratic equation 6x² – x – 2 = 0.
Answer:
The given equation
6x² – x – 2 = 0
⇒ 6x² – (4 – 3)x – 2 = 0
⇒ 6x² – 4x + 3x – 2 = 0
⇒ 2x(3x – 2) + 1(3x – 2) = 0
⇒ (3x – 2) (2x + 1) = 0
⇒ 3x – 2 = 0 or 2x + 1 = 0
⇒ x = \(\frac{2}{3}\) or x = – \(\frac{1}{2}\)
Hence, x = \(\frac{2}{3}\) and x = – \(\frac{1}{2}\) are the roots of given quadratic equation.

(vii) Define scale factor.
Answer:
Scale factor is the ratio of the sides of the triangle to be constructed with corresponding sides of the given triangle.

(viii) Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (- 3, 4). (1)
Answer:
Given P(x, y) be the equidistant from the points A(3, 6) and B(- 3, 4), then
PA = PB
⇒ PA² = PB²
⇒ (3 – x)² + (6 – y)² = (- 3 – x)² + (4 – y)²
⇒ 9 + x² – 6x + 36 + y² – 12y
= 9 + x² + 6x + 16 + y² – 8y
⇒ x² + y² + 6x – 8y + 6x + 12y – x² – y² = 45 – 25
⇒ 12x + 4y = 20
⇒ 3x + y – 5 = 0
Hence, this is the required relation.

(ix) Evaluate: \(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\)
Answer:
\(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\) = \(\frac{\sin \left(90^{\circ}-72^{\circ}\right)}{\cos 72^{\circ}}\)
= \(\frac{\cos 72^{\circ}}{\cos 72^{\circ}}\) = 1

(x) If tan A = cot B, prove that A + B = 90°.
Answer:
We have,
tan A = cot B
⇒ tan A = tan (90° – B)
⇒ A = 90° – B
⇒ A + B = 90°.

(xi) The mean of the following data is 15. Find the missing number P.
12, 15, 32, P, 28, 12, 35
Answer:
According to Question, Sum of data
Mean = \(\frac{\text { Sum of data }}{\text { Total number }}\)
⇒ 21 = \(\frac{12+15+32+p+28+12+35}{7}\)
⇒ 21 × 7 = 134 + p
⇒ p = 147 – 134
⇒ p = 13

(xii) A die is thrown once, what is the probability of getting an even prime number? (1)
Answer:
When a die is thrown once, the number of possible outcomes is 6 [1, 2, 3, 4, 5, 6]. Let E be event of getting a even prime number.
The number of outcomes favourable to E (i.e. getting even prime no. i.e. 2) is 1.
P(E) = \(\frac{\text { No. of favourable outcomes }}{\text { No. of all possible outcomes }}=\frac{1}{6}\)

Section – B

Question 4.
Given that √3 is an irrational number. Prove that (2 + √3) is an irrational number. 2
Answer:
Let us assume, that (2 + √3) is a rational, it can be expressed in the form of \(\frac{a}{b}\), where a and b are coprime positive integers and b ≠ 0.
∴ 2 + √3 = \(\frac{a}{b}\)
(∵ HCF of a and b is 1)
⇒ \(\frac{a}{b}\) = √2
⇒ \(\frac{a-2 b}{b}\) = √3
\(\frac{a-2 b}{b}\) = rational number
(∵ a and b are positive integers) Thus, from eqn. (i) is a rational number.
But this contradicts the fact that is an irrational number. So, our assumption that 2 + √3 is a rational number, is wrong.
Hence, (2 + √3) is an irrational number.

Question 5.
Find a quadratic polynomial whose zeroes are reciprocal of the zeroes of the polynomial f(x) = ax² + bx + c, a ≠ 0, c ≠ 0. (2)
Answer:
f(x) = ax² + bx + c (given)
Let α and β be the zeroes of f(x)
Then, α + β = –\(\frac{b}{a}\) And αβ = \(\frac{c}{a}\)
Zeroes of required polynomial are \(\frac{1}{\alpha}\) and \(\frac{1}{\beta}\).
∴ Sum of zeroes = \(\frac{1}{\alpha}+\frac{1}{\beta}\) = \(\frac{\alpha+\beta}{\alpha \beta}\) = \(\frac{-\frac{b}{a}}{\frac{c}{a}}\)
= \(\frac{-b}{a} \times \frac{a}{c}\) = \(\frac{-b}{c}\)
And product of zeroes

Required polynomial = x² – (sum of zeroes)x + product of zeroes

Question 6.
Solve the equations 3x – 5y = 4 and 9x – 2y = 7 for x and y. (2)
Answer:
The given equation are:
3x – 5y = 4 ….. (i)
9x – 2y = 7 ……. (ii)
From equation (ii), we get
x = \(\frac{7+2 y}{9}\) …… (iii)
Substituting the value of x in equation
(i), we get
\(\frac{3(7+2 y)}{9}\) – 5y = 4
⇒ \(\frac{7+2 y}{3}\) – 5y = 4
⇒ 7 + 2y – 15y = 12
⇒ – 13y = 12 – 7 = 5
⇒ y = \(\frac{5}{-13}\) = \(\frac{-5}{13}\)
Putting the value ofy in equ. (iii), we get

Question 7.
Solve: \(\frac{1}{a+b+x}\) = \(\frac{1}{a}+\frac{1}{b}=\frac{1}{x}\); where a + b ≠ 0
Answer:
The given equation is:

⇒- (a + b) × ab = (a + b)x (a + b + x)
⇒ (a + b)x (a + b + x) + ab (a + b) = 0
⇒ (a + b) [x (a + b + x) + ab] = 0
⇒ x (a + b + x) + ab = 0
⇒ ax + bx + x² + ab = 0
⇒ x² + ax + bx + ab = 0
⇒ x(x + a) + b (x + a) =0
⇒ (x + a) (x + b) = 0
⇒ (x + a) = 0 and (x + b) = 0
⇒ x = – a and x = – b

Question 8.
Find the 11th term of the AP : – 17, – 12, – 7,….
Answer:
The given sequence of AP is – 17, – 12, – 7, ……..
Here, a₁ = – 17, a₂ = – 12, a₃ = – 7 d = a₂ – a₁ = – 12 – (-17) = – 12 + 17 = 5
We know that n^th term of AP is
a_n = a + (n – 1)d
⇒ a₁₁ = – 17 + (11 – 1) × 5
= – 17 + 50 = 33
Hence, 11th term of given AP = 33.

Question 9.
In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 in the third, and so on. There are 5 rose plants in the last row. How many rows are there in the flower bed? (2)
Answer:
The number of rose plants in the 1^st, 2^nd, 3^rd, … rows are 23, 21, 19, …, 5
a₂ – a₁ = 21 – 23 = – 2
a₃ – a₂ = 19 – 21 = – 2
∵ a₂ – a₁ = a₃ – a₂
∴ This sequence forms an AP.
Here a = 23, d = 21 – 23 = – 2, a_n = 5
We know that n^th term of an AP is
a_n = a + (n – 1) d
⇒ 5 = 23 + (n – 1) × (- 2)
⇒ 5 = 23 – 2n + 2
⇒ 2n = 25 – 5
⇒ n = \(\frac{20}{2}\) = 10
Hence, the number of rows are there in the flower bed = 10.

Question 10.
Draw a circle of radius 3.5 cm. From a point P, 6 cm from its centre, draw two tangents to the circle. (2)
Answer:

• Draw a circle with O as the centre and radius 3.5 cm
• Mark a point P outside the circle such that OP = 6 cm.
• Join OP and draw prependicular bisector of PO meeting PO at M.
• Draw a circle with M as the centre and radius equal to PM = OM intersecting the given circle at points Q and R.
  
• Join PQ and PR then PQ and PR are required tangents.

Question 11.
Draw a ∆ABC in which BC = 7 cm, ∠B = 45° and ∠A = 105°. Then construct another triangle whose sides are \(\frac{3}{4}\) times corresponding sides of ∆ABC. (2)
Answer:

• Draw a line segment BC = 7 cm.
• In ∆ABC, ∠B = 45°, 2A = 105°
  ∴ ∠C = 180° – (45° + 105°) = 30°
  At B draw and angle ∠B = 45° and draw ∠C = 30° intersecting each other at A to get ∆ABC
  
• At B draw any ray BX making an acute angle with BC on the side opposite to the vertex A
• Along BX mark 4 points B₁(B₂, B₃ and B₄ such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄
• Join B₄C
• From B₃ draw B<sub3C’ | | B₄C meeting BC at C’
• From C’ draw C’A’ | | AC meeting AB at A
  Then A’BC’ is required triangle, each of whose side is \(\frac{3}{4}\) of corresponding sides of ∆ABC.

Question 12.
Prove that 1 + \(\frac{\cot ^{2} \alpha}{1+{cosec} \alpha}\) = cosec α
Answer:
LHS = 1 + \(\frac{\cot ^{2} \alpha}{1+{cosec} \alpha}\)
= 1 + \(\frac{{cosec}^{2} \alpha-1}{{cosec} \alpha+1}\)
= \(1+\frac{({cosec} \alpha+1)({cosec} \alpha-1)}{({cosec} \alpha+1)}\)
= 1 + cosec α – 1
= cosec α = R.H.S.

Question 13.
If sec θ = x + \(\frac{1}{4 x}\), prove that sec θ + tan θ = 2x or \(\frac{1}{2 x}\)
Answer:

Question 14.
Consider the following distribution of daily wages of 50 workers of a factory. (2)

Find the mean daily wages of the workers of the factory by using an appropriate method.
Answer:
Here, h = 20

From the table, we have
Σf_i = 50, Σf_iu_i = – 12, a = 150, h = 20
∴ x̄ = a + \(\left(\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\right)\) × h
⇒ x̄ = 150 + \(\left(-\frac{12}{50}\right)\) × 20
⇒ x̄ = 150 – 4.80 = 145.20.
Hence, the mean daily wages of the workers = ₹ 145.20.

Question 15.
Find the mode of the following distribution. (2)

Answer:
The class interval 35 – 40 has maximum frequency. So, it is the modal class.
∴ l = 35, f₁ = 50, f₀ = 34, f₂ = 42, h = 5
Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h
= 35 + \(\left(\frac{50-34}{2 \times 50-34-42}\right)\) × 5
= 35 + \(\frac{16 \times 5}{24}\)
= 35 + 3.33
= 38.33

Question 16.
A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanif wins if all the tosses give the same result i.e.,three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game. (2)
Answer:
When a coin tossed three times, the the possible outcomes are
(H, H, H), (H, H, T), (H, T, T), (H, T, H), (T, H, H), (T, H, T), (T, T, H) and (T, T, T)
Total number of possible outcomes = 8
Number of cases when all the tosses give the same results are two i.e., (H, H, H) and (T, T, T)
It is given that Hanif wins when all the tosses give the same results, Number of cases in which Hanif will lose the game = 8 – 2 = 6
Number of favorable outcomes = 6
∴ P (Hanif will lose the game) = \(\frac{6}{8}\) = \(\frac{3}{4}\)

Section – C

Question 17.
A manufactures of TV set produce 720 sets in fourth year and 1080 set in the sixth year, assuming that the production increase uniformly by a fixed number every year, then find total production in first 9 year. (3)
Or
Find the sum of n terms of the series: (3)
(4 – \(\frac{1}{n}\)) + (4 – \(\frac{2}{n}\)) + (4 – \(\frac{3}{n}\)) + ………
Answer:
Production of TV sets in 4th year = 720,
Production of TV sets in 6th year = 1080
Let a and d be first term and common difference respectively then,
a₄ = a + (4 – 1)d
⇒ 720 = a + 3d
⇒ a + 3d = 720 …(1)
And a₆ = a + (6 – 1)d
⇒ 1080 = a + 5 d
⇒ a + 5d = 1080 …(2)
Substracting equ. (2) from equ. (1), We get

Substituting the value of d in equ. (1), We get
a + 3 × 180 = 720
⇒ a + 540 = 720
⇒ a = 720 – 540
⇒ a = 180
Production of TV sets increases uniformly by a fixed numbers. So,
Total productions of TV sets in first 9 year = a₉
⇒ a₉ = a + (9 – 1) 180
= 180 + 8 × 180
= 180 + 1440 = 1620
Hence, production in first 9^th year = 1620.

Question 18.
In what ratio does the point P(\(\frac{24}{11}\), y) divide the line segment joining the points P(2, – 2) and Q(3, 7) ? (3)
Or
Find the value of P, if the points A(2, 3) B(4, P) and C(6, – 3) are collinear. (3)
Answer:
Let the required ratio be k : 1
Here, x₁ = 2, y₁ = -2, x₂ = 3, y₂ = 7, x = \(\frac{24}{11}\) y = y, m₁ = k, m₂ = 1
By section formula, we have

Question 19.
Prove that: \(\frac{1}{1+\sin ^{2} \theta}+\frac{1}{1+\cos ^{2} \theta}+\frac{1}{1+\sec ^{2} \theta}+\frac{1}{1+{cosec}^{2} \theta}\) = 2
Or
If sin θ + cos θ = √3, then prove that tan θ + cot θ = 1
Answer:
We have, LHS

Question 20.
The table below shows the salaries of 280 persons:

Calculate the median salary of the data.
Or
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows :

Find the mean number of letters in the surnames. (3)
Answer:
We prepare the cumulative frequency distribution table as given below:

n = 280
⇒ \(\frac{n}{2}=\frac{280}{2}\) = 140
But 140 comes under the cumulative frequency 182 and class interval against the cumulative frequency 182 is 10 – 15. So, it is the median class.
∴ l = 10, cf = 49, f = 133 and h = 5.
Median = l + \(\left(\frac{\frac{n}{2}-c f}{f}\right)\) × h
= 10 + \(\left(\frac{140-49}{133}\right)\) × 5
= 10 + \(\frac{91 \times 5}{133}\)
= 10 + \(\frac{65}{19}\)
= 10 + 3.42
= 13.42 (thousands)
= ₹ 13.42

Section – D

Question 21.
Akhila went to a fair in her village. She wanted to enjoy rides on the Giant Wheel and play Hoopla (a game in which you throw a ring on the items kept in a stall, and if the ring covers any object completely you get it.) The number of times she played Hoopla is half the number of rides she had on the Giant wheel. If each ride costs ₹ 3, and a game of Hoopla costs ₹ 4, how would you find out the number of rides she had and how many times she played Hoopla, provided she spent ₹ 20 ?
Represent this situation algebraically and graphically (Geometrically). (4)
Or
Cost of 3 balls and 2 bats is ₹ 100 and the cost of 3 balls and 4 bats is ? 150. Formulate the problem algebrically and solve it graphically.(4)
Answer:
Let the number of rides be x and number of times Hoopla played be y. According to question,
y = \(\frac{1}{2}\) x
⇒ x – 2y = 0
and 3x + 4y = 20
The algebraic representation of the given situation is:
x – 2y = 0 …(i)
3x + 4y = 20 ….(ii)
For representation of these equations graphically we draw the graph of these equations as follows:
x – 2y = 0
⇒ y = \(\frac{x}{2}\)
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 x – 2y = 0

and 3x + 4y = 20
⇒ 4y = 20 – 3x ⇒ y = \(\frac{20-3 x}{4}\)
Now, we put the different values of x then different values of y and we prepare the table of x, y for the equation 3x + 4y = 20

Now, we plot the values of x and y from table 1 & 2 on a graph paper and draw the graphs through these values. Observe that we get two straight lines which intersect each other at the point (4, 2).
Hence, number of rides she had taken on giant wheel is 4 and the number of times she played Hoopla is 2.

Question 22.
Construct an Isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1\(\frac{1}{2}\) times the corresponding sides of the isosceles triangle. (4)
Or
Draw a circle of radius 3 cm. From a point 5 cm away from its centre, construct the pair of tangents to the circle and measure their lengths. Also verify the measurement by actual calculation. (4)
Answer:
Steps of Construction :

• Let ABC be the isosceles triangle with base AB = 8 cm and altitude CD = 4 cm.
• Draw a line segment AB = 8 cm.
• Draw the perpendicular bisector of AB intersecting AB at D.
• With D as a centre and radius 4 cm draw an arc intersecting the perpendicular bisector at C.
• Join AC and BC to get the isosceles triangle ABC.
  
• produce AB to B’ such that
  AB’ = 1\(\frac{1}{2}\) of 8 = \(\frac{3}{2}\) × 8 = 12 cm.
• Through B’ draw B’C’ || BC intersecting the produced line segment AC to C’. .
  Then AB’C’ is the required triangle each of whose side is 1\(\frac{1}{2}\) times of the corresponding sides of ∆ABC.

Justification: Since,
BC || B’C’
Therefore, ∆C’AB’ ~ ∆CAB
⇒ \(\frac{A C^{\prime}}{A C}=\frac{A B^{\prime}}{A B}=\frac{B^{\prime} C^{\prime}}{B C}=\frac{3}{2}\)

Question 23.
Change the following distribution to a more than type distribution. Hence, draw the more than type ogive for this distribution. (4)

Or
A survey regarding the heights (in cm) of 51 girls of class X of a school was conducted and the following data was obtained: (4)

Find the median height.
Answer:
We prepare the cumulative frequency table by more than method as given below:

On the graph paper, we plot the points (20, 100), (30, 90), (40, 82), (50, 70), (60, 46), (70, 40), (80, 15). Joining these points with free hand to get more than curve as shown below in graph.