CBSE Sample Papers for Class 10 Maths Paper 2 is part of CBSE Sample Papers for Class 10 Maths Here we have given CBSE Sample Papers for Class 10 Maths Paper 2 According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

## CBSE Sample Papers for Class 10 Maths Paper 2

 Board CBSE Class X Subject Maths Sample Paper Set Paper 2 Category CBSE Sample Papers

Students who are going to appear for CBSE Class 10 Examinations are advised to practice the CBSE sample papers given here which is designed as per the latest Syllabus and marking scheme as prescribed by the CBSE is given here. Paper 2 of Solved CBSE Sample Papers for Class 10 Maths is given below with free PDF download solutions.

Time: 3 Hours
Maximum Marks: 80

GENERAL INSTRUCTIONS:

• All questions are compulsory.
• This question paper consists of 30 questions divided into four sections A, B, C and D.
• Section A comprises of 6 questions of 1 mark each, Section B comprises of 6 questions of 2 marks each, Section C comprises of 10 questions of 3 marks each and Section D comprises of 8 questions 1 of 4 marks each.
• There is no overall choice. However, internal choice has been provided in one question of 2 marks, 1 three questions of 3 marks each and two questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
• In question of construction, drawings shall be neat and exactly as per the given measurements.
• Use of calculators is not permitted. However, you may ask for mathematical tables.

SECTION A

Question numbers 1 to 6 carry 1 mark each.

Question 1.
Find the value of k for which the following pair of linear equations has a unique solution:
2x + 3y = 7; (k – 1)x + (k + 2)y = 3k.

Question 2.
Find the nature of the roots of quadratic equation 2x² – √5 x + 1 = 0.

Question 3.
What is the probability that a non-leap year has 53 Mondays?

Question 4.
A die is thrown once. Find the probability of getting a prime number.

Question 5.
Find the mode of the data, whose mean and median are given by 10.5 and 11.5 respectively.

Question 6.
In the adjoining figure, DE || BC. If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, find the value of x.

SECTION B

Question numbers 7 to 12 carry 2 marks each.

Question 7.
Find HCF and LCM of 90 and 144 by method of prime factorisation.

Question 8.
Find the values of a and b for which the following pair of linear equations has infinitely many solutions:
3x – (a + 1)y = 2b – 1; 5x + (1 – 2a)y = 3b.

Question 9.
Without using trigonometric tables, evaluate the following:
(cos² 25° + cos² 65°) + cosec θ . sec (90° – θ) – cot θ tan (90° – θ).

Question 10.
ABC is a triangle and G (4, 3) is the centroid of the triangle. If A, B and C are the points (1, 3), (4, b) and (a, 1) respectively, find the values of a and b. Also find the length of side BC.

Question 11.
In the adjoining figure, DE || AC and $$\frac { BE }{ EC } =\frac { BC }{ CP }$$ . Prove that DC || AP.

Question 12.
In the adjoining figure, a circle is inscribed in a quadrilateral ABCD in which ∠B = 90°. If AD = 23 cm, AB = 29 cm and DS = 5 cm, find the radius (r) of the circle.

SECTION C

Question numbers 13 to 22 carry 3 marks each.

Question 13.
If two zeroes of the polynomial x4 + 3x3 – 20x2 – 6x + 36 are √2 and – √2 , find the other zeroes of the polynomial.

Question 14.
If α and β are zeroes of the polynomial 6x² – 7x – 3, then form a quadratic polynomial whose zeroes are $$\frac { 1 }{ \alpha }$$ and $$\frac { 1 }{ \beta }$$.

Question 15.
How many terms of the A.P. -6, $$\frac { 11 }{ 2 }$$, -5,……. double answer.are needed to give the sum – 25? Explain the
OR
The 19th term of an AP is equal to three times its 6th term. If its 9th term is 19, find the AP.

Question 16.
The father’s present age is six times his son’s ages. Four years hence the age of the father will be four times his son’s age. Find the present ages of the father and son.
OR
The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them manages to save Rs 2000 per month, find their monthly incomes.

Question 17.
ABC is a right triangle, right angled at C. If p is the length of perpendicular from C to AB and a, b, c have usual meanings, then prove that $$\frac { 1 }{ { p }^{ 2 } } =\frac { 1 }{ { a }^{ 2 } } +\frac { 1 }{ { b }^{ 2 } }$$
OR
If the diagonals of a quadrilateral divide each other proportionally, prove that it is a trapezium.

Question 18.
PQ is a tangent to a circle with centre O at the point Q: A chord QA is ‘drawn parallel to PO. If AOB is a diameter of the circle, prove that PB is tangent to the circle at the point B.

Question 19.
The radii of the circular ends of a bucket of height 15 cm are 14 cm and r cm (r < 14 cm). If the volume of bucket is 5390 cm3, then find the value of r.

Question 20.
Find the area of the major segment APB in adjoining figure, of a circle of radius 35 cm and ∠AOB = 90°.

OR
In adjoining figure, a semicircle is drawn with O as centre and AB as diameter. Semicircles are drawn with AO and OB as diameters. If AB = 28 m, find the perimeter of the shaded region.

Question 21.
Prove that :

Question 22.
Prove that :

SECTION D

Question numbers 23 to 30 carry 4 marks each.

Question 23.
Prove that √5 is an irrational number and hence show that 2 + √5 is also an irrational number.

Question 24.
If two vertices of an equilateral triangle are (3, 0) and (6, 0), find the third vertex.
OR
The mid-points D, E and F of the sides AB, BC and CA of a triangle are (3, 4), (8, 9) and (6, 7) respectively. Find the coordinates of the vertices of the triangle.

Question 25.
Water is flowing at the rate of 15 km/hour through a pipe of diameter 14 cm into a cuboidal pond which is 50 m long and 44 m wide. In what time will the level of water in the pond rise by 21 cm?

Question 26.
While boarding an aeroplane, a passenger got hurt. The pilot showing promptness and concern, made arrangements to hospitalise the injured and so the plane started late by 30 minutes. To reach the destination, 1500 km away in time, the pilot increased the speed by 100 km/h. Find the original speed/hour of the plane.
Do you appreciate the values shown by pilot, namely promptness in providing help to the injured and his efforts to reach in time.

Question 27.
Draw a pair of tangents to a circle of radius 3 cm which are inclined at an angle of 60° to each other.

Question 28.
The angle of elevation of the top of a building from the foot of a tower is 30° aid the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.
OR
The angles of depression of two ships from the top of a lighthouse and on the same side of it are found to be 45° and 30°. If the ships are 200 m apart, find the height of the lighthouse.

Question 29.
Three coins are tossed simultaneously, find the probability of getting:

Question 30.
The mean of the following frequency distribution is 62.8 and the sum Of all the frequencies is 50. Compute the missing frequencies f1 and f2:

OR
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mode and mean of the data:

For unique solution, $$\frac { 2 }{ k-1 } \neq \frac { 3 }{ k+2 }$$ ⇒ 2k + 4 ≠ 3k – 3
⇒ k ≠ 7
Hence, the given pair of linear equations will have unique solution for all real values of k except 7.

Given 2x² – √5x + 1 = 0
D = (-√5)² – 4 x 2 x 1 = 5 – 8 = – 3
∵D < 0, therefore, given equation has no real roots.

There are 365 days in a non-leap year.
365 days = 52 weeks + 1 day
∴ One day can be M, T, W, Th, F, S, Su = 7 ways
∴ P(53 Mondays in non-leap year) = $$\frac { 1 }{ 7 }$$

Total number of outcomes = 6(1, 2, 3, 4, 5 or 6)
Favourable number of outcomes = 3(2, 3, 5)
∴ P(prime number) = $$\frac { 3 }{ 6 } =\frac { 1 }{ 2 }$$

Mode = 3 Median – 2 Mean .
= 3 x 11.5 – 2 x 10.5 = 34.5 – 21 – 13.5
Hence, mode = 13.5

∵ DE || BC
∴ By Basic Proportionality Theorem, we have

$$\frac { AD }{ DB } =\frac { AE }{ EC }$$
⇒ $$\frac { x }{ x-2 } =\frac { x+2 }{ x-1 }$$
⇒ x (x – 1) = (x – 2) (x + 2)
⇒ x² – x – x² – 4
⇒ -x = -4
⇒ x = 4.

90 = 2 x 3 x 3 x 5
and 144 = 2 x 2 x 2 x 2 x 3 x 3
∴ HCF = 2 x 3 x 3 = 18
and LCM = 2 x 2 x 2 x 2 x 3 x 3 x 5 = 720
Hence, HCF = 18 and LCM = 720.

Given 3x – (a + 1 )y – (2b – 1) = 0
and 5x + (1 – 2a)y – 3b = 0

Hence, a = 8 and b = 5.

(cos² 25° + cos² 65°) + cosec θ . sec (90° – θ) – cot θ . tan (90° – θ)
= cos² 25° + cos² (90° – 25°) + cosec θ . cosec θ – cot θ . cot θ
= cos² 25° + sin² 25° + cosec² θ – cot² θ
= 1 + 1
= 2.

Since G (4, 3) is the centroid of ∆ABC, we have

In ∆ABC, DE || AC

AR = AQ, DR = DS, BP = BQ (lengths of tangents)
but DS = 5 cm => DR = 5 cm.
AR = AD – DR = 23 cm – 5 cm = 18 cm => AQ = 18 cm.
BQ = AB – AQ = 29 cm – 18 cm = 11 cm.
As AB is tangent to the circle at Q, OQ ⊥ AB
=> ∠OQB = 90°.
Also ∠B = 90° (given) => OQBP is a rectangle.
But BP = BQ => OQBP is a square.
∴ Radius = r = OQ = BQ = 11 cm.

∴ √2 and -√2 are the zero’s of the given polynomial.
∴ (x – √2) (x + √2) i.e. (x² – 2) is a factor of the given polynomial.

To find the other two zero’s, we proceed as follows:
x² + 3x – 18 = 0
⇒ (x + 6) (x – 3) = 0
⇒ x + 6 = 0 or x – 3 = 0
⇒ x = -6 or x = 3
Hence, other zero’s are -6 and 3.

Given α and β are zero’s of quadratic polynomial 6x² – 7x – 3,

Here, a = -6, d = $$-\frac { 11 }{ 2 }- (-6)$$ = $$-\frac { 11 }{ 2 }+6$$ = $$\frac { 1 }{ 2 }$$ ,Sn = -25.
We are required to find n.

Let the father’s present age be x years
and the son’s present age be y years.
According to given, x – 6y …(i)
4 years later,
father’s age = (x + 4) years
and son’s age = (y + 4) years
∴ (x + 4) = 4(y + 4)
=> x – 4y = 12 …(ii)
Putting the value of x from (i) in (ii), we get
6y – 4 y = 12 => 2y = 12 => y = 6
∴ x = 6 x 6 = 36
Hence, father’s present age = 36 years and son’s present age = 6 years
OR
Let the incomes per month of two persons be Rs x and Rs y respectively. As each person saves Rs 2000 per month, so their expenditures are Rs (x – 2000) and Rs (y – 2000) respectively.
According to given, we have
$$\frac { x }{ y } =\frac { 9 }{ 7 }$$ i.e- 7x – 9y = 0 …(i)
and $$\frac { x-2000 }{ y-2000 } =\frac { 4 }{ 3 }$$ i.e. 3x – 4y + 2000 = 0 …(ii)
Multiplying equation (i) by 3 and equation (ii) by 7, we get
21x – 27y = 0 …(iii) and 21x – 28y + 14000 = 0 …(iv)
Subtracting equation (iv) from equation (iii), we get
y – 14000 = 0 => y = 14000.
Substituting this value of y in (i), we get
7x – 9 x 14000 = 0 =>x = 18000.
Hence, the monthly incomes of the two persons are Rs 18000 and Rs 14000 respectively.

In ∆ACB and ∆CDB,
∠ACB = ∠CDB (both 90°)
∠B = ∠B (common)
∴ ∆ACB ~ ∆CDB (by AA similarity criterion)

Given a circle with centre O and PQ is tangent to the circle at the point Q from an external point P. Chord QA is parallel to PO and AOB is a diameter.
We need to prove that PB is tangent to the circle at the point B.
Join OQ and mark the angles as shown in the adjoining figure.

As QA || PO,
∠1 = ∠2 (alt. ∠s)
and ∠4 = ∠3 (corres. ∠s)
.But ∠2 = ∠3 (∵ in ∆OAQ, OA = OQ being radii)
∴ ∠1 = ∠4.
In ∆OPB and ∆OPQ,
OB = OQ (radii of same circle)
∠1 = ∠4 (proved above)
OP = OP (common)
∴ ∆OPB ≅ ∆OPQ (SAS congruence rule)
∴ ∠OBP = ∠OQP (c.p.c.t)
=> ∠OBP = 90° (tangent is ⊥ to radius ,OQ⊥PQ)
=> OB ⊥ PB i.e. radius is perpendicular to PB at point B.
Therefore, PB is tangent to the circle at the point B.

Given h = 15 cm, R = 14 cm, ‘r’ = r cm and volume of bucket = 5390 cm³
∵Volume of bucket = volume of frustum of cone

∴r cannot be negative
∵Radius = r = 7 cm.

Given r = 35 cm and ∠AOB = 90°
Area of minor segment = area of minor sector – area (∆OAB)

Let √5 be a rational number, then
√5 = $$\frac { p }{ q }$$, where p, q are integers, q ≠ 0 and p, q have no common factors (except 1)
=>$$5=\frac { { p }^{ 2 } }{ { q }^{ 2 } }$$ => p² = 5q²
As 5 divides 5q², so 5 divides p², but 5 is prime.
=> 5 divides p
Let p = 5m, where m is an integer.
Substituting this value of p in (i), we get
(5m)² = 5q² => 25m² = 5q² => 5m² = q²
As 5 divides 5m², so 5 divides q², but 5 is prime
=> 5 divides q
Thus p and q have a common factor 5. This contradicts that p and q have no common factors (except 1)
Hence, √5 is not a rational number.
So, we conclude that √5 is an irrational number.
Let 2 + √5 be a rational number, say r
Then, 2 + √5 = r => √5 = r – 2
As r is rational, r – 2 is rational => √5 is rational
But this contradicts the fact that √5 is irrational.
Hence, our assumption is wrong. Therefore, 2 + √5 is an irrational number.

Given vertices are A(3, 0) and B(6, 0) and let third vertex be C(x, y), then

OR
Let the vertices A, B and C of the triangle ABC be (x1 y1), (x2, y2) and (x3 y3) respectively.
Since points D and F are mid-points of the sides AB and
AC respectively, by mid-point theorem, DF || BC and
DF = $$\frac { 1 }{ 2 }BC$$ but E is mid-point of BC, so DF || BE and
DF = BE.
Therefore, DBEF is a parallelogram.
Similarly, DECF and DEFA are parallelograms.
Since the diagonals of a parallelogram bisect each other, mid-points of diagonals BF and DE are same.

Radius of pipe = $$\frac { 14 }{ 2 }$$ cm = 7 cm = $$\frac { 7 }{ 100 }$$ m = 0.07 m
As the water is flowing at the rate of 15 km per hour,
the length of water delivered by the circular pipe in 1 hour
= 15 km = 15000 m
Volume of water delivered by the circular pipe in 1 hour

Hence, the level of water in the pond rise by 21 cm in 2 hours.

Let the original speed of the aeroplane be x km/h.
Time taken to cover the distance of 1500 km = $$\frac { 1500 }{ x }$$ hours
New speed of the aeroplane = (x + 100) km/h.
Time taken to cover the distance of 1500 km at new speed = $$\frac { 1500 }{ x+100 }$$ hours

=> x² + 100x – 300000 = 0
=> x² + 600x – 500x – 300000 = 0 => (x – 500) (x + 600) = 0
=> x = 500 or x = -600
But speed cannot be negative.
Hence, the original speed of the aeroplane = 500 km/h.
Yes, I appreciate the values shown by the pilot. Along with showing concern for the injured passenger he did not fail to perform his duty, by increasing the speed of the plane, he reached the destination on time.

Steps of construction:
1. Draw a circle of radius 3 cm with O as its centre.
3. At O, construct ∠AOC = 120° to meet the circle at B.
4. At A, construct ∠OAR = 90°.
5. At B, construct ∠OBQ = 90° to meet AR at P.

Then PA and PB are tangents to the circle inclined at an angle of 60° to each other.
Justification:
As ∠APB and ∠AOB are supplementary, so ∠APB = 60°.

Let CD = h metres be the height of the building and AB be the tower, then AB = 50 m.
Let BD = d metres be the distance between the foot of the tower and the foot of the building.
Given, ∠CBD = 30° and ∠ADB = 60°.
From right angled ∆CBD, we get

OR
Let the height of the lighthouse AB be h metres and C, D be the positions of two ships. The angles of depressions are shown in the adjoining figure.
Then ∠ACB = 45° and ∠ADB = 30°
Given CD = 200 m, let BC = x metres.
From right angled ∆ABC, we get

When three coins are tossed simultaneously, the outcomes of the random experiment are:
HHH, HHT, HTH, THH, HTT, THT, TTH, TIT
It has 8 equally likely outcomes.
(i) The outcomes favourable to the event ‘atleast one head’ are
HHH, HHT, HTH, THH, HTT, THT, TTH; which are 7 in number.
∴ P(atleast one head) = $$\frac { 7 }{ 8 }$$
(ii) The outcomes favourable to the event ‘atmost two heads’ are
HHT, HTH, THH, THT, HTT, TTH, TTT; which are 7 in number.
∴P(atmost two heads) = $$\frac { 7 }{ 8 }$$
(iii) The outcomes favourable to the event ‘exactly 2 heads’ are
HHT, HTH, THH; which are 3 in number.
∴ P(exactly two heads) = $$\frac { 3 }{ 8 }$$
(iv) The only outcome favourable to the event ‘no head’ is TTT.
∴P(no head) = $$\frac { 1 }{ 8 }$$