CBSE Sample Papers for Class 9 Maths Paper 2 is part of CBSE Sample Papers for Class 9 Maths . Here we have given CBSE Sample Papers for Class 9 Maths Paper 2

## CBSE Sample Papers for Class 9 Maths Paper 2

Board |
CBSE |

Class |
IX |

Subject |
Maths |

Sample Paper Set |
Paper 2 |

Category |
CBSE Sample Papers |

Students who are going to appear for CBSE Class 9 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 9 Maths is given below with free PDF download solutions.

**Time: 3 Hours**

**Maximum Marks: 80**

**General Instructions**

- All questions are compulsory.
- The question paper consists of 30 questions divided into 4 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 of 4 marks each.
- There is no overall choice. However, internal choice has been provided in four questions of 3 marks each and three questions of 4 marks each. You have to attempt only one of the alternatives in all such questions.
- Use of calculator is not permitted.

**Section A**

Question 1.

Find the range of the following data.

36, 34, 48, 56, 72, 82

Question 2.

The perimeter of a triangle is 70 cm and its sides are in the ratio 2:3:5. Find the length of sides of triangle.

Question 3.

In the given figure, ∠COE = 90°. Find the value of x.

Question 4.

Two points with coordinates (4, 3) and (4, – 2) lie on a line, parallel to which axis?

Question 5.

If two triangles have equal area and one side of the one triangle is equal to one side of the other triangle, then prove that their corresponding altitudes are equal.

Question 6.

Express \(0.\overline { 47 } \) in the form \(\frac { p }{ q }\), where p and q are integers and q≠0.

**Section B**

Question 7.

The sum of a two-digit number and the number obtained by reversing the order of its digits is 121. If unit’s and ten’s digits of the number are x and y respectively, then write the linear equation representing the above statement.

Question 8.

Three coins are tossed simultaneously 400 times and following frequencies of the outcomes were recorded

(i) Find the probability of getting no head.

(ii) Find the probability of getting atleast 2 heads.

Question 9.

If the circumference of the base of a cone is 44 cm and its height is 25 cm, then find the volume of the cone.

Question 10.

In the given figure, if ∠POQ = 80°, then find ∠PAQ and ∠PCQ.

Question 11.

If each side of a cube is tripled, then find its volume.

Question 12.

ABC is an equilateral triangle and L, M and N are the mid-points of the sides AB, BC and CA, respectively. Prove that ∆LMN is an equilateral triangle.

**Section C**

Question 13.

In the adjoining figure, ABCD is a parallelogram and a line through A cuts DC at P and BC produced at Q. Prove that

ar (∆BPC) = ar (∆DPQ)

Question 14.

In a circle of radius 5 cm, AB and AC are two equal chords, such that, AB = AC = 6 cm. Find the length of the chord BC.

OR

Prove that the line joining the mid-points of two equal chords of circle subtends equal angles with the chord.

Question 15.

Construct the angle of 15°. Do you think that the angle of \(7\frac { 1 }{ 2 }\)° can be constructed by the bisector of 15°?

Or

Construct a ∆ABC in which AB = 5 cm, ∠A = 45° and ∠B = 60°

Question 16.

A Survey of 200 people was conducted about their preference of visiting various pavilions.

Find the probability that selected person visited

(i) both good living and Delhi pavilion.

(ii) only defence pavilion.

(iii) both toy and defence pavilion.

OR

Probability of getting a blue ball is 2/3 from a bag containing 6 blue and 3 red balls 12 red balls are added in the bag, then find the probability of getting.

(i) a blue ball.

(ii) a red ball.

Question 17.

If two isosceles triangles have a common base, then prove that the line segment joining their vertices bisects the common base at right angles.

Question 18.

PQ is a line segment 12 cm long and R is a point in its interior such that PR = 8 cm. Then, find QR, PQ² – PR² and PR² + QR² + 2PR .QR using Euclid’s axiom.

Question 19.

Plot the points (x, y) given in the following table on the plane choosing suitable units of distances on the axes.

Question 20.

If (ax^{3} + bx^{2} + x – 6) has (x + 2) as a factor and leaves remainder 4, when divided by (x – 2), find the values of a and b.

Or

The polynomial ax^{3} + 3x^{2} – 13 and 2x^{3} – 5x + a leave the same remainder in each case, when divided by (x – 2). Find the value of a.

Question 21.

If a and b are rational numbers and

then find the values of a and b.

Question 22.

In the given figure, prove that x – a + b + c.

**Section D**

Question 23.

The area of the curved surface of a right circular cylinder is 4400 cm² and the circumference of its base is 110 cm. Find the height and volume of the cylinder. [π=\(\frac { 22 }{ 7 }\)]

OR

The volume of a right circular cone is cm² and its diameter is 10 cm. Find its curved and total surface area.[π=\(\frac { 22 }{ 7 }\)]

Question 24.

Two years later, a father will be 8 yr more than three times the age of his son. Taking the present ages of father and son as x and y respectively, write a linear equation for the above and draw its graph. From the graph, find the age of the father, when the son’s age is 10 yr.

Question 25.

How much paper of each shade is needed to make kite given in figure, in which ABCD is a square with diagonal 44 cm?

Question 26.

The ratio of girls and boys in a class is 1 : 3. Set-up an equation between the students of a class and boys and then draw its graph. Also, find the number of boys in a class of 20 students from the graph.

Question 27.

Factorise

Question 28.

Question 29.

Hindustan Metro Company manufactures car batteries of a particular type. The life of 32 batteries (in years) were recorded as follows

(i) Find the chance that the life of a battery randomly selected is less than or equal to 3 yr.

(ii) If the company gives the warranty of a battery which is less than or equal to 4 yr, then what is the chances, that the selected battery is under warranty?

(iii) As per the given data, if the company gives warranty of less than or equal to 2 yr, then Sumit decided to purchase the battery. What his decision shows?

Question 30.

Prove that the sum of the angles in the four exterior segments of a cyclic quadrilateral is equal to 6 right angles.

OR

In the given figure, AB and CD are two chords of a circle, with centre O, intersecting each other at point E, prove that ∠AEC to \(\frac { 1 }{ 2 }\)(angle subtended by arc CXA at the centre + angle subtended by arc DYB at centre)

**Solutions**

Solution 1.

Given data, 36, 34, 48, 56, 72, 82

Here, highest data = 82, Lowest data = 34

∴ Range = Highest data – Lowest data

= 82 – 34 = 48

Solution 2.

Given, perimeter of a triangle = 70 cm and ratio of its sides is 2 : 3 : 5

Let the sides of a triangle be 2x, 3x and 5%.

∴ Perimeter of triangle = 2x+3x+5x

Solution 3.

Here, ∠BOD = ∠AOC

=> 2x = y

Now, ∠COE + ∠EOB + ∠AOC = 180°

90° + x + 2x = 180°

=> 3x = 90°

∴ x = \(\frac { 90 }{ 3 }\) = 30°

Solution 4.

As x-coordinate of both points is 4.

So, both points lie on the line x = 4, which is parallel to Y-axis.

Solution 5.

Let, ABC and DEF are two triangles such that ar (∆ABC) = ar (∆DEF) and BC = EF

Also, let AL and DM are altitudes of ∆ABC and ∆DEF, respectively.

To prove Altitude AL = Altitude DM

Solution 6.

Solution 7.

Given, unit’s digit is x and ten’s digit is y.

Then, number = 10y + x

and new number after reversing the digits

= 10x + y

According to the question,

10y +x +10x + y = 121

=> 11x + 11y = 121

=> x + y = 11 [dividing both sides by 11]

Which is the required linear equation.

Solution 8.

Given, three coins are tossed simultaneously 400 times.

∴ Total number of trials = 400

(i) Let E1 be the event of getting no head.

Thus, number of trials in which the event

E1 happened = 400 -(103 + 124 + 98) = 75

Solution 9.

Let r be the radius of the base of a cone.

Given, height, h = 25cm

and circumference of the base of a cone = 44 cm

∴2πr = 44 [∴circumference of circular base = 2πr]

Solution 10.

Given ∠POQ = 80

From the given figure ,we have

∠POQ = 2 x ∠PAQ

[∵ angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle]

Solution 11.

Let the original side of a cube be a units.

Then, original volume of a cube

= (Side)^{3} = a^{3}cu units

According to the given condition,

New side of a cube = 3a units

∴ New volume of a cube = (Side)^{3}

= (3a)^{3} = 27a^{3}

= 27 x Original volume of cube

Hence, the volume of the cube becomes 27 times

Solution 12.

Given, ∆ABC is an equilateral triangle and L, M, N are the mid-points of the sides AB, BC and CA respectively.

To prove: ∆LMN is an equilateral triangle.

Proof Since, L and N are the mid-points of AB and AC, respectively.

Solution 13.

Given ABCD is a parallelogram, so AB || CD and

AD || BC Lines AQ and DC intersect at P

To prove ar (∆BPC) = ar (∆DPQ)

Construction Join AC.

Proof As ∆ADC and ∆ADQ being on the same base AD and between the same parallel lines AD and BO,

ar (∆MDC) = ar (∆ADQ)

On subtracting ar (∆ADP) from both sides, we get

ar (∆ADC) – ar (∆ADP) = ar (∆ADQ) – ar (∆ADP)

ar(∆APC) = ar(∆DPQ) …(i)

Also, ∆APC and ∆BPC being on the same base PC and between the same parallel lines PC and AB.

So, they are equal in area.

ar(∆APC) = ar(∆BPC) …(ii)

From Eqs. (i) and (ii), we get

ar(∆BPC) = ar(∆DPQ) Hence proved.

Solution 14.

Let O be the centre of a circle and AD be the bisector of ∠BAC.

Then. AD is the perpendcuIar bisector of BC and passes through the centre O. Join CO.

Let AD meets BC at M.

Then,MB = CM

In right angled ∆AMC,

Solution 15.

Steps of construction

(i) First, draw a ray OA.

(ii) Taking 0 as centre and some radius, draw an arc of a circle which intersects OA, say at point B.

(iii) Taking B as centre and same radius, draw an arc which intersect the previously drawn arc, at a point C (say)

(iv) Draw the ray OE passing through C, then ∠EOA = 60°.

(v) Again, taking B and C as centres and radius more than \(\frac { 1 }{ 2 }\) BC, draw two arcs which intersect each other at D.

(vi) Draw the ray OD intersecting an arc drawn in step (ii) at F. This ray OD is the bisector of the ∠EOA.

∠EOD = ∠AOD = \(\frac { 1 }{ 2 }\) ∠EOA = \(\frac { 1 }{ 2 }\) (60°) = 30°

(vii) Now, taking B and F as centres and radius more than \(\frac { 1 }{ 2 }\) BF, draw two arcs which intersect each other at G.

Solution 16.

Total number of people = 200

Solution 17.

Let ∆ABC and ∆DBC are the given triangles with same base BC. in which AB = AC and DB = DC.

Also, let AD meets BC at E.

To prove BE = CE and ∠AEB = ∠AEC = 90°

Proof In ∆ABD and ∆ACD,

we have

AB = AC [given]

DB = DC [given]

and AD = AD [common side]

∆ABD ≅ ∆ACD [by SSS congruence rule]

Solution 18.

Given, PQ = 12 cm and PR = 8 cm

Here, point R lies in the interior of PQ.

So, PR + QR = PQ

=> 8 + QP = 12

On subtracting 8 from both sides, we get

8 + QR – 8 = 12 – 8 [byaxiom3]

=> QR = 4cm

Now, PQ² – PR² = (12)² – (8)²

= 144 – 64 = 80 cm

Also, PP² + QR² + 2PR • QP

= (PR + OR)² = (8 + 4)² = (12)² = 144 cm

Solution 19.

The given points are A(-3,4),B(0,-2 5),C(1,3), D(8,-2), E(6,5), F(3,-4), G(-7,-3.5)and H(-4,0).

Now, we plot the points on the graph paper as shown by dots in the figure below

Solution 20.

Let f(x) = ax^{3} + bx^{2} + x – 6

Since, (x + 2) is a factor of f(x).

f(-2) = 0

=> a(-2)^{3} + b(-2)^{2} + (-2) – 6 = 0

=> -8a + 4b – 2 – 6 = 0

=> -8a + 4b = 8

=> -2a + b = 2 [dividing by 4].. (i)

Also, when divided by (x – 2) of f(x) it leaves a reminder 4.

f(2) = 4

Solution 21.

Solution 22.

In the given figure, join B and D and produce to E

Solution 23.

Given, curved surface area of cylinder = 4400 cm²

and circumference of base of cylinder = 110 cm

Let the radius of cylinder is r and height h.

∴ Circumference of cylinder = 2πr

Solution 24.

Let the present ages of father and son be x yr and y yr, respectively.

Two years later,

Age of father = (x + 2) yr and age of son = (y + 2) yr

According to the question,

Age of father after two years

= 3 (Age of son after two years) + 8

=> x + 2 = 3(y + 2) + 8 => x + 2 = 3y + 6 + 8

=> x – 3y – 12 = 0

which is the required linear equation in two variables.

To draw the graph, we need atleast two solutions of the above equation.

Now, the above equation can be written as

x = 3y + 12 ..(i)

When y = 4,then x = 3 x 4 + 12 = 12 + 12 = 24

When y = 8,then x = 3 x 8 + 12 = 24 + 12 = 36

Thus, we have the following table

Here, we get the points 4(24, 4) and 6(36, 8). Now, plot the points on the graph paper and join them to get a line AB, which represents the required graph.

Now taking point y = 10 on Y-axis draw a line parallel to X-axis, which meets the line at point C and from point C, draw a perpendicular, which intersects X-axis at 42 units distance from Y-axis.

Thus, we get point C(42,10), i.e. when son’s age is 10 yr, then the age of father is 42 yr

Solution 25.

We know that all the sides of a square are equal.

=> AB = BC = CD = DA

In ∆ACD, we have

AC = 44 cm, ∠D = 90°

Using Pythagoras theorem,

(AC)² = (AD)² + (DC)²

=> (44)² = (AD)² + (AD)²

[∵ AD = DC, sides of a square]

=> 2AD² = 44 x 44

Solution 26.

Let the number of boys be x and the number of students be y.

∴ Number of girls = (y – x)

According to the question,

Solution 27.

Solution 28.

Solution 29.

Here, total number of batteries = 32

Total number of trials = 32

(i) Let E be the event of getting the battery whose life is less than or equal to 3 yr.

Number of trials in which event E happened = 7

Now,

Hence, required chance is 0.219.

(ii) Let E be the event of getting the battery whose life is less than or equal to 4 yr.

Number of trials in which event F happened = 27

Now,

Hence, the required chance is 0.844.

(iii) As the given data shows that life of each battery is from 2.7 to 4.6 yr. So, he takes a good decision.

Solution 30.

Let, ABCD be a cyclic quadrilateral and let, angles ∠P,∠Q,∠R and ∠S are in the four external segments.

To prove ∠P + ∠Q + ∠R + ∠S = 6 right angles.

Construction Join SB and SC.

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