CBSE Sample Papers for Class 9 Maths Paper 5

CBSE Sample Papers for Class 9 Maths Paper 5 is part of CBSE Sample Papers for Class 9 Maths . Here we have given CBSE Sample Papers for Class 9 Maths Paper 5

CBSE Sample Papers for Class 9 Maths Paper 5

Board	CBSE
Class	IX
Subject	Maths
Sample Paper Set	Paper 5
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 5 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.
• Questions 1-6 in Section-A are Very Short Answer Type Questions carrying 1 mark each.
• Questions 7-12 in Section-B are Short Answer (SA-I) Type Questions carrying 2 marks each.
• Questions 13-22 in Section-C are Short Answer (SA-II) Type Questions carrying 3 marks each.
• Questions 23 -30 in Section-D are Long Answer Type Questions carrying 4 marks each.

SECTION-A

Question 1.

If √5ⁿ = 125, then find ({ 4 }^{ sqrt [ n ]{ 64 } }) =?

Question 2.

Find the remainder when x³ – ax² + 6x – a is divided by x – a.

Question 3.

What is the distance of point P(4, 3) from origin?

Question 4.

AOB is a straight line. If ∠AOC + ∠BOD = 85°, find the measure of ∠COD.

Question 5.

A bag contains 2 red, 3 green and 1 white ball, what is the probability that the ball picked up is black?

Question 6.

The ratio of heights of two cylinders is 5 : 3, as well as the ratio of their radii is 2 : 3. Find the ratio of the volumes of the cylinder.

SECTION-B

Question 7.

If a + b + c = 3x, then find the value of (x – a)³ + (x – b)³ + (x – c)³ – 3(x – a)(x – b) (x – c).

Question 8.

The difference of two supplementary angles is 34°. Find the angles.

Question 9.

In the given figure, AC > AB, and D is a point on AC such that AB = AD. Prove that BC > CD.

Question 10.

Prove that the area of triangle is half of the product of its base and corresponding height.

Question 11.

An isosceles triangle has perimeter 30 cm and each of equal side is 12 cm. Find the area of the triangle.

Question 12.

Find the mean of the first ten prime numbers.

SECTION-C

Question 13.

Find three rational numbers between (frac { 2 }{ 5 }) and (frac { 3 }{ 5 }).

Question 14.

Factorise: (a² – 2d)² – 23(a² – 2d) + 120

Question 15.

Without plotting the points indicate the quadrant in which they will lie, if

(i) ordinate is 5 and abscissa is – 3.

(ii) abscissa is – 5 and ordinate is – 3.

(iiI) abscissa is – 5 and ordinate is 3.

Question 16.

Draw the graph of two lines, whose equations are 3x – 2y + 6 = 0 and x + 2y – 6 = 0 on the same graph paper. Find the area of the triangle formed by the two lines and x-axis.

Question 17.

In the given figure, ABCD is a square. If ∠ PQR = 90° and PB = QC = DR, prove that ∠ QPR = 45°.

Question 18.

Show that the diagonals of a rhombus are perpendicular to each other.

Question 19.

Given a quadrilateral ABCD in which AB = 6.3 cm, BC = 5.2 cm, CD = 5.6 cm, DA = 7.1 cm and ∠ B = 60°. Construct a triangle equal in area to this quadrilateral.

Question 20.

The diameter of a hemisphere is decreased by 30%. What will be the percentage change in its total surface area?

Question 21.

A right circular cylinder just encloses a sphere of radius r. Find

(i) Surface area of the sphere.

(ii) Curved surface area of the cylinder.

(iii) Ratio of the areas obtained in (i) and (ii)

Question 22.

100 students of a school are selected at random. The marks scored (out of 50 marks) by them in the recently held unit test are tabulated below.





If a student is chosen randomly, then what is the probability that the randomly chosen student scores less than 40% marks?

SECTION-D

Question 23.

Question 24.

If x² – 3x + 2 is a factor of x⁴ – ax² + b, find a and b.

Question 25.

In a residential society rain water is stored in underground water tank. If the water stored at the rate of 30 cubic cm per second and water stored in ‘x’ second and ‘y’ cubic cm.

(i) Write this statement in linear equation in two variables.

(ii) Write this equation in the form of ax + by + c = 0

(iii) What value of the society members shows in rain water storage?

Question 26.

A right circular cone of diameter r cm and height 12 cm rests on the base of a right circular cylinder of radius r cm. Their bases are in the same plane and the cylinder is filled with water upto a height of 12 cm. If the cone is removed, find the height to which water level fall.

Question 27.

In the given figure, ray OS stands on a line POQ, ray OR and ray OT are angle bisectors of ∠POS and ∠SOQ respectively. If ∠ POS = x, find ∠ROT.

Question 28.

∆ABC and ∆DBC are two isosceles triangle on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at P. Show that

(i) ∆ ABD ≅ ∆ ACD

(ii)∆ ABP ≅ ∆ ACP

(iii) AP bisects ∠ A as well as ∠ D.

(iv) AP is the perpendicular bisector of BC.

Question 29.

(i) Prove that angles in the same segment of a circle are equal,

(ii) Using the statement of part (i) prove that ∠x + ∠y = ∠z., where O is the centre of the circle in Fig. (i).

Question 30.

Draw a histogram for the following data

Solutions

Solution 1.

Solution 2.

When x – a = 0 => x = a

P(x) = x³ – ax² + 6x – a

Putting x = a

Remainder = P(a) = (a)³ – a x a² + 6a – a = a³ – a³ + 5a = 5a

P(a) = 5a.

Solution 3.

Coordinates of origin = (0, 0)

Distance = OP



= 5 unit.

Solution 4.

∠AOC + ∠COD + ∠BOD = 180° (Straight line angle)

(∠AOC + ∠BOD) + ∠COD = 180°

85° + ∠COD = 180°

∠COD = 180° – 85° = 95°

∠COD = 95°

Solution 5.

Total number of possible events = Total number of balls = 2 + 3 + 1 = 6.

Number of favourable outcomes = Number of black balls = 0

P(Black ball) = (frac { 0 }{ 6 }) = 0

Zero black balls.

Solution 6.

Solution 7.

∵ (x – a) + (x – b) + (x – c) = 3x – (a + b + c) = 0 [ ∴ a + b + c = 3x]

∴(x – a)³ + (x – b)³ + (x – c)³ – 3(x – a) (x – b) (x – c) = 0

[ ∴ a + b + c = 0 => a³ + b³ + c³ – 3abc = 0]

Solution 8.

Let the first angles be x°.

Second angle = (x + 34)°

∴ x° + (x + 34)° = 180° => 2x + 34° = 180°

2x = 180° – 34° =>2x = 146°

x = 73°

The first angle = 73°

Second angle = 73° + 34° = 107°.

Solution 9.

In ∆ ABD, it is given that

AB = AD ….(i)

In ∆ ABC, AB + BC > AC

=> AB + BC > AD + CD

=> AB + BC > AB + CD [∵AD = AB]

=> BC > CD

Solution 10.

In ∆ABC, AL is the corresponding height.

Base = BC

To prove: ar(∆ABC) = (frac { 1 }{ 2 }) (BC x AL)

Construction: From point C and A draw CD || BA and AD || BC, which meet at D

Solution 11.

Let the third side of isosceles triangle be x.

Perimeter = 30 cm => x + 12 + 12 = 30

=> x + 24 = 30 => x = 6 cm

=> 2s = 30 cm => s = 15 cm

Solution 12.

The first 10 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29

Solution 13.

n = 3, n + 1 = 3 + 1 = 4

Solution 14.

Let a² – 2a = x

(a² – 2d)² – 23 (a² – 2d) + 120 = x² – 23x + 120

= x² – 15x – 8x + 120

= x(x – 15) – 8(x – 15)

= (x – 15)(x – 8)

= (a² – 2a – 15)(a² – 2a – 8) [Put x = a² – 2a]

= (a² – 5a + 3a – 15) (a² – 4a + 2a – 8)

= [a(a – 5) + 3(a – 5)] x [a(a – 4) + 2(a – 4)]

= (a – 5) (a + 3) x (a – 4) (a + 2)

= (a + 2) (a + 3) (a – 4) (a – 5)

Solution 15.

Abscissa means x-axis (points on xx’ or x-axis) and ordinate means y-axis (points on yy’ or y-axis)

(i) If ordinate is 5 and abscissa is – 3 => It represents the point (- 3, 5)

=> Which is in II quadrant.

(ii) If abscissa is -5 and ordinate is – 3 => It represents the point (-5, – 3) = III quadrant.

(iii) If abscissa is – 5 and ordinate is 3 => It represents the point (- 5, 3) = II quadrant.

Solution 16.

By the help of graph.

Solution 17.

PB = QC = DR => AP = PB, DR = RC, BQ = QC

In ∆ PBQ and ∆ QCR

Solution 18.

ABCD is a rhombus.

OA = OC [Diagonals of rhombus bisect each other]

In ∆’s AOB and COB





So, diagonals of rhombus are perpendicular to each other.

Solution 19.

Given: AB = 6.3 cm, BC = 5.2 cm, CD = 5.6 cm, DA = 7.1 cm and ∠ B = 60°.

Steps of constructions:

Step 1: Draw line segment AB = 6.3 cm.

Step 2: Construct ∠ ABC = 60°.

Step 3: With centre B, draw an arc of radius 5.2 cm cutting the ray BX at the point C.

Step 4: With centre C and radius 5.6 cm draw an arc.

Step 5: With centre A and radius 7.1 cm draw an arc cutting the arc drawn in step no. 4 at D.

Step 6: Join C and D, A and D, we get the quadrilateral ABCD.

Step 7: Join B and D.

Step 8: Draw a line CE || BD cutting ray AP at E.

Step 9: Join D and E. We get the required triangle ADE.

Solution 20.

Let r be the radius of the hemisphere and S be its total surface area. Then S = 3πr²

The diameter of the hemisphere is decreased by 30%



Thus if the diameter of a hemisphere is decreased by 30%, then its total surface area is decreased by 51%.

Solution 21.

(i) Surface area of sphere = 4π²

(ii) Height of cylinder = r + r = 2r

Radius of cylinder = r

C.S.A of the cylinder = 2πrh = 2πr(2r) = 4πr²

Solution 22.

Total number of students among whom the survey was conducted =100

40% of 50= (frac { 40 }{ 100 }) x 50 = 20

The students scoring less than 40% marks are the students who scored less than 20 marks. From the given table it is observed that there are5 + 10 = 15 students who scored less than 40% marks.

∴ Required probability = (frac { 15 }{ 100 }) = (frac { 3 }{ 20 }).

Solution 23.

Solution 24.

If x² – 3x + 2 is a factor of x⁴ – ax² + b, then the remainder must be zero.

x² – 3x + 2 = 0 =>x² – 2x – x + 2 = 0

=> (x – 2)(x – 1) = 0 =>x – 2 = 0 or x – 1 = 0

=> x – 1 = 0 => x = 1

or x – 2 = 0 => x = 2

P(x) = x⁴ – ax² + b

Putting x = 1,

Remainder = P(1) = (1)⁴ – a x (1)² + b = 0

=> 1 – a + b = 0 =>a – b = 1 …(1)

Putting x = 2,

Remainder = P(2) = (2)⁴ – a x (2)² + b = 0

=> 16 – 4a + b = 0 =>4a – b = 16 …(2)

Subtracting eq. (1) from eq. (2) we get



Putting the value of a = 5 in equation (1)

a – b = 1

5 – b = 1

– b = – 4

b = 4

a = 5, b = 4

Solution 25.

Storage time of water = ‘x’ seconds

Amount of stored water = y cubic cm

Water stored per second = 30 cubic cm

Water stored in ‘x’ second = 30x

Solution 26.



Radius of base of cone = (frac { r }{ 2 }) = R

Radius of the base of cylinder = r

Height of conical portion = 12 cm

=> Height of water in cylinder before cone taken out = 12 cm.

∴ Volume of water left in the cylinder when cone is taken out



= πr² (12 – 1)

Height of the water left in cylinder = (12 – 1) = 11 cm.

Solution 27.

Ray OS stands on the line POQ.

∠POS = x

Therefore ∠POS + ∠SOQ = 180°

∠POS = x

Solution 28.



(i) In ∆ ABD and ∆ ACD

AB = AC (given)

BD = CD (given)

AD = DA (common)

∆ ABD ≅ ∆ ACD (sss congruency)

(ii) In ∆ ABP and ∆ ACP

AB = AC

∠BAP = ∠CAP

AP = AP

=> ∆ ABP ≅ ∆ ACP (sss congruency)

(iii) From (i)

∆ABD ≅ ∆ACD

=> ∠BAD = ∠CAD (CPCT)

=> ∠BAP = ∠CAP

=> AP bisects ∠A.

In ∆BDP and ∆CDP

BD = CD (given)

BP = CP [∵ ∆ABP ≅ ∆ACP ∵ BP = CP]

DP = DP (Common)

∆BDP ≅ ∆CDP (SSS congruency)

=> ∠BDP = ∠CDP (CPCT)

=> DP is bisector of ∠ D.

=> AP and DP are on same line segment AP.

So AP bisect ∠A as well as ∠D.

(iv) ∵ ∆BDP ≅ ∆CDP

=> BP = CP and ∠BPD = ∠CPD (CPCT)

=> BP = CP and ∠BPD = ∠CPD = 90°

[ ∵ ∠BPD and ∠CPD are linear pair]

=> DP is the perpendicular bisector of BC.

=> AP is the perpendicular bisector of BC.

Solution 29.



Given: An arc PQ of a circle C(0, r) and two angles ∠PRQ and ∠PSQ are in the same segment of the circle.

To prove: ∠PRQ = ∠PSQ

Construction: Join OP and OQ

Proof: In figure 2

∠POQ = 2∠PRQ

[ ∵ The angle subtended by an arc at the centre is twice the angle subtended by it any point in the remaining part of the circle]

∠POQ = 2∠PSQ

=> ∠ PRQ = ∠PSQ

Solution 30.

Here the classes are of unequal width, so we use the adjusted frequencies instead of frequency.

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