RBSE 10th Maths Model Paper Set 5 with Answers in English

Students must start practicing the questions from RBSE 10th Maths Model Papers Set 5 with Answers in English Medium provided here.

RBSE Class 10 Maths Model Paper Set 5 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 decimal representation of \(\frac{11}{2^{3} \times 5}\) will: (1)
(a) terminate after 1 decimal place
(b) terminate after 2 decimal place
(c) terminate after 3 decimal place
(d) not terminate
Answer:
(c) terminate after 3 decimal place

(ii) If 2 is factor of polynomial f(x) = x⁴ – x³ – 4x² + kx + 10, then k is: (1)
(a) 2
(b) – 2
(c) – 1
(d) 1
Answer:
(c) – 1

(iii) The value of k for which the system of equations : (1)
2x + 3y = 5 and 4x + ky = 10
has infinite number of solutions, is :
(a) 3
(b) 6
(c) 0
(d) 1
Answer:
(b) 6

(iv) If x² + 2kx + 4 = 0 has a root x = 2, then value of k is : (1)
(a) – 1
(b) – 2
(c) 2
(d) – 4
Answer:
(b) – 2

(v) The n^th term of the : AP a, 3a, 5a, ………………. is : (1)
(a) na
(b) (2n – 1)a
(c) 2 (2n + 1)
(d) 2na
Answer:
(b) (2n – 1)a

(vi) To construct, a triangle similar to a given ∆ ABC with its sides \(\frac{8}{5}\) of the corresponding sides of ∆ ABC, draw a ray BX such that ∠CBX is an acute angle and X is on the opposite side pf A with respect to BC. The minimum number of points to be located at equal distances on ray BX is : (1)
(a) 5
(b) 8
(c) 13
(d) 3
Answer:
(a) 5

(vii) The point P on y-axis equidistant from the A (6, 5) and B (- 4, 3) is : (1)
(a) (0, 9)
(b) (0, 2)
(c) (3, 0)
(d) (9, 2)
Answer:
(a) (0, 9)

(viii) If cos A = \(\frac{12}{13}\), then cot A is: (1)
(a) \(\frac{12}{5}\)
(b) \(\frac{13}{12}\)
(c) \(\frac{13}{5}\)
(d) \(\frac{5}{12}\)
Answer:
(a) \(\frac{12}{5}\)

(ix) If sin (A + 2B) = \(\frac{12}{5}\) and cos (A + 4B) = 0, A > B and A + 4B ≤ 90°, then A and B is (1)
(a) 30°, 45°
(b) 15°, 45°
(c) 60°, 15°
(d) 30°, 15°
Answer:
(d) 30°, 15°

(x) Median of data 2, 3, 5, 2, 3, 6, 5, 2, 2, 5, 7, 4, 4, is : (1)
(a) 5
(b) 3
(c) 4
(d) 2
Answer:
(c) 4

(xi) For the following distribution: (1)

The modal class is:
(a) 10 – 20
(b) 20 – 30
(c) 30 – 40
(d) 50 – 60
Answer:
(c) 30 – 40

(xii) If an event cannot occur, then its probability is: (1)
(a) 1
(b) \(\frac{3}{4}\)
(c) \(\frac{1}{2}\)
(d) 0
Answer:
(d) 0

Question 2.
Fill in the blanks:
(i) If \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\), then pair of equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 (1)
Answer:
no solution

(ii) If a, b and c are in AP, then ______________ = \(\frac{a+c}{2}\) and b is called the arithmetic mean. (1)
Answer:
b

(iii) ______________ is a straight line which touches the circle at one point only. (1)
Answer:
Tangent

(iv) The point of intersection of two axes is called ______________ . (1)
Answer:
origin

(v) The value of tan 1° tan 2° ______________ tan 89° is equal to ______________. (1)
Answer:
1

(vi) The difference between the true upper limit and true lower limit of a class is called its ______________. (1)
Answer:
class-size.

Question 3.
Very Short Answer Type Questions :
(i) Write whether rational number \(\frac{7}{75}\) will have terminating decimal expansion or a non-terminating decimal. (1)
Answer:
\(\frac{7}{75}=\frac{7}{3 \times 5^{2}}\)
Since, denominator of given rational number is not of the form 2^m × 5ⁿ. Hence, it is non-terminating decimal expansion.

(ii) Find the quadratic polynomial whose sum and product of zeroes are \(\frac{21}{8}\) and \(\frac{5}{16}\) respectively. (1)
Answer:
Sum of zeroes = α + β = \(\frac{21}{8}\)
And product of zeroes = α × β = \(\frac{5}{16}\)
So, the required quadratic polynomial is:
= x² – (α + β)x + αβ
= x² – \(\frac{21}{8}\)x + \(\frac{5}{16}\)

(iii) If α and β are zeroes of a polynomial x² – 2√3x + 5 then find the value of α + β + αβ
Answer:
Let p(x) = x² – 2√3x + 5
Since, α and β are zeroes of p(x), then α + β = sum of zeroes = \(\frac{-(-2 \sqrt{3})}{1}\) = 2√3
And α × β = product of zeroes = \(\frac{5}{1}\) = 5
Now, α + β + αβ = 2√3 + 5.

(iv) Two rails are represented by the equations x + 2y – 4 = 0 and 2x + 4y – 12 = 0. Will the rails cross each other ? (1)
Answer:

This is the condition for parallelism. So, the given two rails will not cross each other.

(v) Write the discriminant of the quadratic equation (x – 5)² = 2(5x + 3) (1)
Answer:
The given equation is
(x – 5)² = 2(5x + 3)
⇒ x² – 10x + 25 = 10x + 6
⇒ x² – 10x – 10x + 25 – 6 = 0
⇒ x² – 20 + 19 = 0
Here, a = 1, b = – 20 and c = 19
Discriminant = b² – 4ac
= (-20)² – 4 x 1 x 19
= 400 – 76 = 324

(vi) Find the discriminant of the quadratic equation 2x² – 4x + 3 = 0 and hence find the nature of its roots. (1)
Answer:
The given equation is :
2x² – 4x + 3 = 0
Here,
a = 2, b = – 4, c = 3
∴ D = b² – 4ac
⇒ D = (- 4)² – 4 × 2 × 3
⇒ D = 16 – 24
⇒ D = – 8
D < 0.
So, the given equation has no real roots.

(vii) Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. (1)
Answer:

(viii) If the mid-point of two points A(- 2, 5) and B(- 5, y) is \(\left(-\frac{7}{2}, 3\right)\) distance between points A and B.
Answer:
According to question
= \(\frac{5+y}{2}\)
⇒ 6 = 5 + y
⇒ y = 6 – 5 = 1
Coordinates of point is (- 5, 1).
Length of AB = \(\sqrt{(-5+2)^{2}+(1-5)^{2}}\)
= \(\sqrt{(-3)^{2}+(-4)^{2}}\)
= \(\sqrt{9+16}\) = √25 = 5
Hence, length of AB = 5 units

(ix) Solve for x, where 0° < x < 90°: 2 cos²θ = \(\frac{1}{2}\)
Answer:
We have, 2 cos²θ = \(\frac{1}{2}\)
⇒ cos²θ = \(\frac{1}{4}\)
⇒ cos²θ = \(\frac{1}{2}\)
⇒ θ = 60°

(x) Calculate \(\left(\frac{\sin 35^{\circ}}{\cos 55^{\circ}}\right)^{2}+\left(\frac{\cos 43^{\circ}}{\sin 47^{\circ}}\right)^{2}\) – 2 cos 60°
Answer:
we have,

(xi) Find the class mark of the classes 20 – 25 and 35 – 60. (1)
Answer:
class mark of 20-25 = \(\frac{20+25}{2}\) = \(\frac{45}{2}\)
= 22.5
class mark of 35-60 = \(\frac{35+60}{2}\) = \(\frac{95}{2}\) = 47.5

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

Section – B

Question 4.
Show that 5 – √3 is an irrational number.
Answer:
Let us assume, that 5 – √3 is a rational number. It can be expressed in the form of \(\frac{a}{b}\), where a and b are coprime positive integers and b ≠ 0.
∴ 5 – √3 = √3
[HCF of a and b is 1 and b ≠ 0]
⇒ √3 = rational number.
But this contradicts the fact that √3 is an irrational number. So, our assumption that 5 – √3 is a rational number, is wrong.
Hence, 5 – √3 is an irrational number.

Question 5.
If α, β and γ are the zeroes of the cubic polynomial 8x³ + x² – 3x + 1, then find the value of α^-1 + β^-1 + γ^-1. (2)
Answer:

Question 6.
A father’s age is three times the sum of the ages of his two children. After 5 years his age will be two times the sum of their ages. Find the present age of the father. (2)
Answer:
Let the age of father be x years and the sum of ages of his children be y years.
According to questions,
x = 3y ….(1)
After 5 years
Age of father = (x + 5) years And sum of ages of his children = (y + 10) years
According to the question,
x + 5 = 2(y + 10)
⇒ x + 5 = 2y + 20
⇒ x – 2y = 20 – 5
⇒ x – 2y = 15 ….(2)
From equation, (1), substituting the value of x in the equation, we get
3y – 2y = 15
⇒ y – 15 years
Putting the value of y in the equation we get
x = 3 × 15 = 45 years
Hence, present age of father = 45 years.

Question 7.
Find k so that the quadratic equation (k + 1) x² – 2 (k + 1) x + 1 = 0 has equal roots. (2)
Answer:
(k + 1)x² – ²(k + 1) x + r = 0
Since, equation has equa1 roots
∴ D = 0
⇒ b² — 4ac = 0
⇒ [- 2(k + 1)]² – 4 × (k + 1) × 1 = 0
⇒ 4(k² + ²k + 1) – 4k – 4 = 0
⇒ 4k² + 8k + 4 – 4k – 4 = 0
⇒ 4k² + 4k = 0
⇒ 4k(k + 1)= 0 and k + 1 = 0
⇒ k = – 1
k = – 1 (Reject since if we put the k = – 1 in the equation (- 1 + 1)x will be zero
∴ k = 0

Question 8.
If second and third terms of an AP are 3 and 5 respectively, then find the sum of first 20 terms of it. (2)
Answer:
Let a be first term and common difference be d of an AP.
a₂ = 3
⇒ a + d = 3 ….(1)
And a₃ = 5
⇒ a + 2d = 5 ….(2)
Substracting equ. (2) from equ. (1), we get

Substituting the value of d in equ (1), we get
a + 2 = 3
⇒ a = 1
Sum of first 20 terms is
S₂₀ = \(\frac{20}{2}\) [2 × 1 + (20 – 1) × 2]
= 10 [2 + 38]
= 10 × 40 = 400
Hence, S₂₀ = 400

Question 9.
For what value of n are the nth terms of two APs : 63, 65, 67 and 3, 10, 17…….. are equal ? (2)
Answer:
The two APs are 63, 65, 67, ………. and 3, 10, 17,
Let a, d and A, D be I^st term and common difference of two given APs.
Here, a = 63, d = 65 – 63 = 2
And A = 3, D = 10 – 3 = 7
n^th terms of two APs are equal
a_n = A_n
⇒ a + (n – 1 )d = A + (n -1)D
⇒ 63 + (n – 1) × 2 = 3 + (n – 1) × 7
⇒ 63+ 2n- 2 = 3 + 7n – 7
⇒ 2n + 61 = 7n – 4
⇒ 61 + 4 = 7n – 2n
⇒ 65 = 5 n
⇒ n = \(\frac{65}{5}\) = 13
Hence, 13^th term of two given APs are equal.

Question 10.
Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are \(\frac{5}{3}\) times the corresponding sides of the given triangle. (2)
Answer:
Steps of Construction:
Let ABC be right triangle in which ∠B = 90°, BC = 4 cm and AB = 3 cm.
1. Draw BC = 4 cm.
2. Draw ∠CBY = 90° at B.
3. Cut AB = 3 cm from BY. Join AC to get ∆ABC.
4. Draw any ray CX making an acute angle with BC on the side opposite to the vertex A.
5. Along CX mark five points C₁, C₂, C₃, C₄ and C₅ such that CC₁ = C₁C₂ = C₂C₃ = C₃C₄ = C₄C₅. Join C₃B.

6. From C₅ draw C₅B || C₃B intersecting the produced line segment CB at B’.
7. From B’ draw B’A’ || BA intersecting the produced line segment CA at A’. Then, A’CB’ is the required triangle.

Question 11.
Draw a circle of radius 2.0 cm with centre O and take a point P outside the circle such that OP = 6.5 cm. From P, draw two tangents to the circle. 2
Answer:
Steps of Construction:
1. Draw a circle with O as the centre and radius 2.0 cm
2. Mark a point P outside the circle such that OP = 6.5 cm.
3. Join OP and draw prependicular bisector of PO meeting PO at M.

4. Draw a circle with M as the centre and radius equal to PM = OM intersecting the given circle at points Q and R,
5. Join PQ and PR then PQ and PR are required tangents.

Question 12.
If cos 4A = sin (A – 50°), where 4A is an acute angle, find the value of A. (2)
Answer:
We have,
cos 4A = sin (A — 50°)
⇒ cos 4A = cos [90° – (A – 50°)] [∵ sin θ = cos (90° – θ)]
⇒ 4A = 90° – (A – 50°)
⇒ 4A = 90° – A + 50°
⇒ 4A + A = 90° + 50°
⇒ 5A = 140°
⇒ A = \(\frac{140^{\circ}}{5}\) = 28°.
Hence, A = 28°.

Question 13.
If x = a sin θ and y = b tan θ, ithen prove that \(\frac{a^{2}}{x^{2}}-\frac{b^{2}}{b^{2}}\) = 1
Answer:
We have, LHS = \(\frac{a^{2}}{x^{2}}-\frac{b^{2}}{y^{2}}\)
= \(\frac{a^{2}}{a^{2} \sin ^{2} \theta}-\frac{b^{2}}{b^{2} \tan ^{2} \theta}\)
= \(\frac{1}{\sin ^{2} \theta}-\frac{1}{\tan ^{2} \theta}\)
= cosec² θ – cot²θ = 1
= RHS

Question 14.
Find the mode of the following data: (2)

Answer:
The class interval 60 – 80 has maximum frequency. So, it has modal class.
∴ l = 60, f₁ = 24, f₀ = 10, f₂ = 6, h = 20
mode = \(l+\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right) \times h\)
= \(60+\left(\frac{24-10}{2 \times 24-10-6}\right) \times 20\)
= 60 + \(\frac{14 \times 20}{48-16}\)
= 60 + \(\frac{280}{32}\)
= 60 + 8.75
= 68.75

Question 15.
The following distribution shows the daily pocket allowance of children of a locality :2

Find the mean daily pocket allowance by using appropriate method.
Answer:

∴ Mean = \(\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}\)
= \(\frac{955}{25}\) = ₹ 38.20
Hence, mean daily pocket allowance
= ₹ 38.20

Question 16.
A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag. (2)
Answer:
Let the number of blue balls in the bag be x.
Then total number of balls = (5 + x)
∴ Total number of possible outcomes = (5 + x)
∴ P(a blue ball) = \(\frac{x}{(5+x)}\)
∴ P(a red ball) = \(\frac{5}{5+x}\)
According to question P (a blue ball) = 2 × P(a red ball)
⇒ \(\frac{x}{(5+x)}\) × \(\frac{5}{5+x}\)
⇒ x = 10.
Hence, number of blue balls in the bag = 10.

Section – C

Question 17.
The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference. (3)
Or
Find the sum of first 22 terms of an AP in which d -‘1 and 22nd term is 149. (3)
Answer:
We have, a = 5, l = a_n = 45,
S_n = 400
⇒ \(\frac{n}{2}\) [a + l] = 400
⇒ \(\frac{n}{2}\) [5 + 45] = 400
⇒ n × 50 = 400 × 2
⇒ 50n – 800
⇒ n = \(\frac{800}{50}\)
⇒ n = 16
Now, a_n = 45
⇒ a + (n – 1 )d = 45
⇒ 5 + (16 – 1)d = 45
⇒ 5 + 15 d = 45
⇒ 15d = 45 – 5 = 40
⇒ d = \(\frac{40}{15}\) = \(\frac{8}{3}\)
Hence, n = 16, d = \(\frac{8}{3}\).

Question 18.
Find the co-ordinates of the point which divides the line segment joining the points (4, – 3) and (8, 5) in the ratio 3 : 1 internally. 3
Or
In what ratio does the point (- 4, 6) divide the line segment joining the points A(- 6; 10) and B(3, – 8) ? (3)
Answer:
Let P(x, y) be the required point Here (x₁ = 4, y₁ = – 3) and (x₂ = 8, y₂ = 5) and m₁ = 3, m₂ = 1

By section formula, we have

Hence, co-ordinates of required point, are (7, 3).

Question 19.
If sin θ + cos θ = √2 . Prove that : tan θ + cot θ = 2.
Or
Prove that: (1 – sin θ + cos θ)² = 2 (1 + cos θ) (1 – sin θ).
Answer:
We have,
sin θ + cos θ = √2
(sin θ + cos θ)²= (√2)²
Squaring on both sides
⇒ sin² θ + cos²θ + 2 sin θ cos θ = 2
⇒ 1 + 2 sin θ cos θ = 2
⇒ 2 sin θ cos θ = 1
⇒ sin θ cos θ = 2 N
Now, LHS = tan θ + cot θ
= \(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\)
= \(\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cos \theta}\) = \(\frac{1}{\sin \theta \cos \theta}\)
= \(\frac{1}{\frac{1}{2}}\) = 2 = RHS
LHS = RHS (Hence Proved)

Question 20.
The table below shows the daily expenditure on food of 25 households in a locality. (3)

Find the mean daily expenditure on food by a suitable method.
Or
The following table gives the distribution of the life time of 400 neon lamps: (3)

Find the median life time of a lamp.
Answer:
Here h = 50

From the table, we have
Σf_i = 25, Σf_iu_i = – 7, a = 225, h = 50
∴x̄ = a + \(\left(\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}\right)\) × h
⇒ x̄ = \(\left(-\frac{7}{25}\right)\) × h
⇒ x̄ = 225 – 14
⇒ x̄ = 211
Hence, the mean daily expenditure on food = ₹ 211.

Section – D

Question 21.
The cost of 2 kg of apples and 1 kg of grapes on a day was found to be ₹ 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is ₹ 300. Represent the situation algebraically and graphically. (4)
0r
Form the pair of linear equations in the following problems, and find their solution graphically. (4)
“10 Students of of class X took part in a Mathematics quiz. If the number of girls is 4 more than the nubmer of boys, find the number of boys and girls who took part in the quiz.”
Answer:
Let the cost of 1 kg apples be ₹ x and cost of 1 kg of grapes be ₹ y.
According to question,
2x + y = 160 … (1)
and 4x + 2y = 300
⇒ 2x + y = 150 …(2)
For representation of these equations graphically, we draw the graphs of these equations as follows:
2x + y = 160
⇒ y = 160 – 2x
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 + y – 160.

Now, 2 x + y = 150
⇒ y – 150 – 2x
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 + y = 150.

Now, we plot the values of x, y on the graph paper from table 1 and 2 and we draw the graphs of equations (1) and (2) which passes through these values

From the graph, we observe that, there are two straight lines which are parallel to each other.

Question 22.
Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle. (4)
Or
Draw a line segment PQ of length 6.5 cm. Taking P as centre, draw a circle of radius 2.5 cm and taking B as centre, draw another circle of radius 2 cm. Construct tangents to each circle from the centre of the other circle. (4)
Answer:
Take four points A, B, C and D on the given circle and join AB and CD.
1. Draw the perpendicular bisectors of non-parallel chords AB and CD which intersect at O.
2. O is the centre of circle.
3. Take a point P outside the circle.

4. Join OP and bisect it at M.
5. with M as the centre and radius PM = MO, draw another circle which intersect previous circle at R and Q.
6. Join PR and PQ. Then PR and PQ are the required tangents.
Justification: Join OR,
∠PRO= 90°
[Angle in a semi circle is 90°]
PR ⊥ OR
OR is a radius and PR is a tangent.
Similarly, PQ is a tangent.

Question 23.
Given below is a frequency distribution table showing daily income of 100 workers of a factory. (4)

Convert this table to a cumulative frequency distribution table of more than type.
Or
On the annual day of school agewise participation of students is given in the following distribution table.

Age (in years)	Number of students
Less than 4	3
Less than 6	8
Less than 8	15
Less than 10	19
Les^ than 12	46
Lesf than 14	55
Less than 16	68

Find the median of students. (4)
Answer:
We prepare the cumulative frequency table by less than type method as given:

We plots the points (15, 6), (30, 14), (45, 24), (60, 30) and (75, 34) on the graph paper. Joining these points with a free hand to obtain less than type- ogive curve as shown in graph.