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.
- 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 mode of the following marks (out of 10) obtained by 20 students.

4, 6, 5,9, 3,2, 7, 7, 6, 5,4, 9, 10, 10,3, 4, 7, 6, 9, 9

Question 2.

If x^{40} + x^{51} + 5 is divisible by x + 1, then find the remainder.

Question 3.

The diagonals of a parallelogram are perpendicular to each other. Is this statement true? Give reason for your answer.

Question 4.

If x = 2α + 1 and y = α – 1 is a solution of the equation 2x – 3y + 5 = 0, then find the value of α.

Question 5.

The total surface area of a cube is 864 cm². Find its edge and lateral surface area.

Question 6.

Write two solutions of the linear equation x + 2y = 1.

**Section B**

Question 7.

A die is thrown 300 times and the outcomes are noted as given below

If a die is thrown at random, then find the probability of getting prime number

(i) 1

(ii) 2

(iii) 3

(iv) 5

Question 8.

Find the value of a, if (x + 1) is a multiple of 2x^{4} – 4x^{3} + ax^{2} – 3x – 14.

Question 9.

Express \(1.\overline { 27 } \) as a vulgar fraction.

Question 10.

Factorize 1 + 3ab – a² – b² – ab.

Question 11.

Simplify

Question 12.

A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side a. Find the area of the signal board, using Heron’s formula.If its perimeter is 180 cm, what will be the area of the signal board?

**Section C**

Question 13.

In 50 throws of a die, the outcomes were noted as under

A die is thrown at random. What is the probability of getting

(i) even number?

(ii) odd number?

(iii) greater than 4?

Or

The distance (in km) of 40 engineers from their residence to their place of work were found as follows

What is the empirical probability that an engineer lives

(i) less than 7 km from her place of work?

(ii) more than or equal to 7 km from her place of work?

(iii) within \(\frac { 1 }{ 2 }\) km from her place of work?

Question 14.

Two plane mirrors m and n are placed perpendicular to each other, as shown in figure. An incident ray AB to the first mirror is first reflected in the direction of BC and then reflected by the second mirror in the direction of CD. Prove that AB || CD.

OR

In the given figure, two parallel lines l and m are intersected by a transversal p. Show that the quadrilateral formed by the bisectors of interior angles is a parallelogram.

Question 15.

In the given figure, D and E are the points on the base BC of ∆ABC such that BD = CE, AD = AE and ∠ADE = ∠AED. Prove that ∆ABE ≅ ∆ACD.

Question 16.

The cost of levelling the ground in the form of a triangle having the sides 51 m, 37m and 20 m at the rate of Rs 3 per m² is Rs 918. Is this statement true? Give reason for your answer.

Or

If the curved surface area of a cylinder is 94.2 cm² and height is 5 cm, then find radius of its base and volume of the cylinder, [use π = 3.14]

Question 17.

Express y in terms of x for the equation 4x + 7y = 14. Check which of the points (0, 2), [\(\frac { 7 }{ 2 }\),0] and (2, 1) lie on graph of this equation.

Question 18.

In the given figure, QP||ML and other angles are shown. Find the value of x.

Question 19.

A barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used up when writing 310 words on an average. How many words would use up a bottle of ink containing one-fifth of a litre?

Or

Shashi has a piece of canvas whose area is 1102 m². She uses it to make a conical tent with base radius 14 m. Wastage in stitching margin and cutting amounts to 2 m². Find the volume of the tent that can be made with it.

Question 20.

How many solution(s) of the equation 2(x + y) + 3 = 2y are there on the

(i) number line.

(ii) cartesian plane?

Also, show it by the graph.

Question 21.

In the given figure, O is the centre of the circle. Then, find ∠CBD.

Question 22.

In the given figure a transversal l cuts two lines AB and CD at E and F, respectively. EG is the bisector of ∠AEF and FH is the bisector of ∠EFD such that ∠a = ∠b. Show that EG||FH and AB||CD.

**Section D**

Question 23.

Construct a ∆PQR in which QR = 6cm, ∠Q = 60° and PR – PQ = 2 cm.

Question 24.

In VNSD school, there is a ground, where the students are playing. The shape of the ground is circular with diameter of 50 m. One line segment is drawn in the field of length 14 m. [use, π =3.14]

(i) Determine the perimeter of the circle.

(ii) Determine the perpendicular distance from the centre to this line segment.

(iii) Suppose, one student Neeraj is standing at the end point A of a line. He wants to reach the centre of a circle O in shortest time. So, he consider by the route AO. Find the value of route AO.

(iv) Which quality is shown by the Neeraj?

Question 25.

The sides of a triangle are x, x + 1,2x – 1 and its area is x√10. What is the value of x ?

Or

The sides of a triangular field are 51 m, 37 m and 20 m. Find the number of rose beds that can be prepared in the field, if each rose bed occupies a space of 6 sq m.

Question 26.

Calculate the mean and mode for the following data.

Question 27.

Four friends Ashwani, Tarun, Shiva and Deepak are standing at the points A, B, C and D reference to a well situated at the origin (0, 0) with the respective coordinates (1, 2), (-1,2), (-1,-2) and (1,-2).

(i) Plot these points on a single sheet of a graph paper.

(ii) Are they at equal distance from the well?

Question 28.

Prove that

Question 29.

The diagonals of a parallelogram ABCD intersect at O. A line through O meets AB at X and CD at Y. Show that ar (parallelogram AXYD) = \(\frac { 1 }{ 2 }\) ar (parallelogram ABCD)

Question 30.

Draw the graph of the straight line given by the equation \(\frac { x }{ 2 } +\frac { y }{ 3 } =1\) Calculate the area of triangle formed by the line drawn and the coordinate axis.

Or

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 40 students from the graph.

**Solutions**

Solution 1.

On arranging the given data in ascending order, we get

2, 3, 3, 4, 4, 4, 5, 5, 6, 6, 6, 7, 7, 7, 9, 9, 9, 9, 10, 10

Here, 9 occurs most frequently, i.e. four times. So, the required mode is 9.

Solution 2.

Let f(x) = x^{40} + x^{51} + 5

If f(x) is divisible by (x + 1), then remainder is f(- 1)

Solution 3.

No, diagonals of a parallelogram are not perpendicular to each other. They only bisect each other.

Solution 4.

Since, x = 2α + 1 and y = α – 1 is a solution of the equation 2x – 3y + 5 = 0. So, it must satisfy the equation.

On putting x = 2α + 1 and y = α – 1 in the given equation, we get 2(2α + 1) – 3 (α – 1) + 5 = 0

=> 4α + 2 – 3α + 3 + 5 = 0

=> α + 10 = 0

α = -10

Solution 5.

Given, total surface area of the cube = 864cm²

Let each side of the cube be a cm.

Then, 6 a² = 864 [∵ total surface area of cube = 6a²]

⇒ a² = \(\frac { 864 }{ 6 }=144\)

⇒ a = √144 [taking positive square root]

= 12 cm

Lateral surface area of the cube = 4a² = 4 x (12)² = 4 x 144 = 576cm²

Hence, edge and lateral surface area of a cube are 12 cm and 576 cm², respectively.

Solution 6.

The given equation x + 2y = 1 can be written as 2y = 1 – x

\(y=\frac { 1-x }{ 2 }\)

When x = 1, then y = \(\frac { 1-1 }{ 2 }\) = \(\frac { 0 }{ 2 }\) = 0

Thus, x = 1 and y = 0 is a solution of x + 2y = 1. (1/2)

When x = 3, then y = \(\frac { 1-3 }{ 2 }\) = \(\frac { -2 }{ 2 }\) = -1

Thus x = 3 and y = -1 is another solution of x + 2y = 1.

Hence, the two solutions of x + 2y = 1, out of infinitely solutions, are (1,0) and (3, -1).

Solution 7.

Here, total number of trials = 300

(i) We know that, 1 is not a prime number, so we cannot determine the probability of prime number.

(ii) P (getting 2 on a die)

Solution 8.

Let f(x) = 2x^{4} – 4x^{3} + ax^{2} – 3x – 14

As, (x + 1) is a multiple of f(x).

∴ f(-1) = 0 [∵ x + 1 = 0 =>x = -1]

=> 2(-1)^{4} – 4(-1)^{3} + a(-1)^{2} – 3 (-1) – 14 = 0

=> 2 + 4 + a + 3 – 14 = 0

=> 9 + a – 14 = 0

=> a – 5 = 0

=> a = 5

Solution 9.

Let x = \(1.\overline { 27 } \)

=> x = 1.2727…..(i)

On multiplying both sides by 100 in Eq. (i),we get

100x= 100 x 1.2727…

=> 100x = 127.2727…

=> 100x = 126+ 1.2727…

=> 100x = 126 + x [from eq(i)]

=> 99x = 126 =>\(x=\frac { 126 }{ 99 }\)

=> \(x=\frac { 14 }{ 11 }\)

Hence \(1.\overline { 27 } \) = \(x=\frac { 14 }{ 11 }\)

Solution 10.

We have, 1 + 3ab – a² – b² – ab = 1 – (a² + b² – 2ab)

= (1)² – (a – b)²

[∵ (x – y)² = x² + y² – 2xy]

= [1 + (a – b)] [1 – (a – b)]

[∵ x² – y² = (x – y)(x + y)]

= (1 + a – b) (1 – a + b)

Hence, 1 + 2ab – a² – b² = (1 + a – b) (1 – a + b)

Solution 11.

Solution 12.

We know that, an equilateral triangle has equal sides. So, all sides are equal to a.

Perimeter of triangle = 180 cm [given]

=> a + a + a = 180

3a = 180

a = 60 cm

Solution 13.

Given, total number of throws = 50

(i) Number of trials getting an even number = 9 + 7 + 8

Or

∵ Total number of engineers lives = 40 [given]

(i) The number of engineers whose residence is less than 7 km from their place = 9

∴The probability that an engineer lives less than 7 km from their place of work = \(\frac { 9 }{ 40 }\)

(ii) The number of engineers whose residence is more than or equal to 7 km from their place of work = 40 – 9 = 31.

∴The probability that an engineer lives more than or equal to 7 km from their place of work = \(\frac { 31 }{ 40 }\)

(iii) The number of engineers whose residence within \(\frac { 1 }{ 2 }\) km from their place of work = 0.

∴The probability that an engineer lives within \(\frac { 1 }{ 2 }\) km from their place = \(\frac { 0 }{ 40 }\) = 0

Solution 14.

Let BM and CN be the normals to mirrors m and n, respectively and BM and CN intersect at O.

Now, BM ⊥ m,CN ⊥ n and m⊥n

=> BM⊥CN

∴ ∠BOC = 90°

=> ∠2 + ∠3 = 90° …(i)

By laws of reflection, we have

∠1 = ∠2 and ∠3 = ∠4

[∵ angle of incidence = angle of reflection]

=> ∠1 + ∠4 = ∠2 +∠3

=> ∠1 + ∠2 + ∠3 +∠4 = 2(∠2 +∠3)

[∵ adding ∠2 and ∠3 both sides]

=> (∠1 + ∠2) + (∠3 + ∠4) = 2(90°)

= 180° [fromEq. (i)]

=> ∠ABC + ∠BCD – 180°

But ∠ABC and ∠BCD are consecutive interior angles formed when the transversal BC cuts AB and CD.

∴ AB||CD Hence proved.

Or

l||m

∴ ∠XCA = ∠YAC [alternate interior angles]

=> ∠XCA = ∠YAC

Similarly, AB||DC

Therefore, ABCD is a parallelogram.

Solution 15.

Given In ∆ABC, D and E are the points on the base BC and BD = CE, AD = AE and ∠ADE = ∠AED.

To prove ∆ABE = ∆ACD

Proof In ∆ABD and ∆ACE, we have

AD = AE [given]

BD=CE [given]

and ∠ADB = ∠AEC

[∵∠ADE = ∠AED =>180° – ∠ADE

= 180° – ∠AED => ∠ADB = ∠AEC]

∴∆ABD = ∆ACE [by SAS congruence rule]

Hence, AB = AC [by CPCT]

Now, in ∆ABE and ∆ACD, we have

AS = AC [proved above]

BE = CD

[v BD = CE =>BD + DE = CE + ED => BE = CD]

and ∠ABE = ∠ACD

[∵ angles opposite to equal sides are equal]

∴∆ABE = ∆ACD [by SAS congruence rule]

Hence proved.

Solution 16.

True, given sides of a triangle are 51 m, 37 m and 20 m.

Solution 17.

We have, 4x + 7y = 14 => 7y = 14 – 4x

which is not true.

So, the point (2,1) will not lie on the line 4x + 7y = 14.

Solution 18.

Draw AC||QP and BD||LM

∵ QP||LM=>AC||BD

and AC || QP => ∠4 = 15° …(i)

[alternate interior angles]

Also, BD|| ML=>∠1 = 10°

[alternate interior angles]

and AC || BD => ∠2 = ∠3

[alternate interior angles]

Now, as ∠3 + ∠A = 35°

∴ ∠3 + 15° = 35° [from Eq. (i)]

=> ∠3 = 20°

=> ∠2 = 20° [∵∠2 = ∠3]

Now, x = ∠1 + ∠2 = 10° + 20° = 30°

Hence, the value of x is 30°.

Solution 19.

Given, diameter of cylindrical barrel = 5 mm

∴ Radius of cylindrical barrel,

r = \(\frac { 5 }{ 2 }\) = 2.5 mm = 0.25 cm

and its height, h = 7 cm

Now, volume of cylindrical barrel = πr²h

Solution 20.

Given equation is 2(x + y) + 3 = 2y

=> 2x + 2y + 3 = 2y => 2x = – 3 => \(x=\frac { -3 }{ 2 }\)

which is a linear equation in one variable.

(i) On number line, the solution of the given equation is unique (one solution).

Solution 21.

We know that the 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;

Then, ∠AOC = 2∠APC

=> ∠APC = \(\frac { 1 }{ 2 }\) ∠AOC = \(\frac { 1 }{ 2 }\) x 100° = 50°

Now, ∠APC + ∠ABC = 180°

[∵ sum of opposite angles of a cyclic quadrilateral is 180°]

=> 50° + ∠ABC = 180°

=> ∠ABC = 180°- 50° = 130°

Also, ∠ABC + ∠CBD = 180° [linear pair axiom]

=> 130° + ∠CBD =180°

=> ∠CBD = 180° – 130° = 50°

Solution 22.

∵ EG is the bisector of ∠AEF.

∴∠AEG = ∠GEF = a

Similarly, ∠EFH = ∠HFD = b

∴ ∠GEF = ∠EFH

∴ EG || FH

Again ∠AEF = 2a

and ∠EFD = 2b

∴ ∠AEF = ∠EFD

∴ AB||CD

Solution 23.

Given, QR = 6cm, ∠Q = 60° and PR – PQ = 2 cm.

Steps of construction

(i) First, draw the base, QR = 6cm.

(ii) At the point Q, make ∠XQR = 60°.

(iii) Cut line segment as QS = PR – PQ = 2 cm from the line QX extended on opposite side of line segment QX.

(iv) Join SR.

(v) Now, take S and R as the centre respectively and draw the arcs of circle of radius more than \(\frac { 1 }{ 2 }\)SR

These arcs intersect each other at point L and M.

(vi) Join DM. When LM extend, it intersects QX at P.

(vii) Join PR.

Hence, ∆PQR is the required triangle.

Solution 24.

Given, diameter of circle, d = 50 m

∴ Radius of circle, r = \(\frac { 50 }{ 2 }\) = 25m

(i) Now, perimeter of circle = 2πr

= 2 x 3.14 x 25 = 157 m

Solution 25.

Given, sides of a triangle are x,x + 1,2x – 1 and its area = x√10

Let a = x, b = x + 1, c = 2x – 1

Solution 26.

The table for marks and number of students is given below

We know that mode is the value, which contain highest frequency.

Here, we see in the given table that 30 is the highest frequency and its corresponding marks is 25.

Hence, mode of the given data is 25.

Solution 27.

(i) The points A (1,2), B(-1,2), C(—1, -2) and D(1, -2) can be plotted as shown below

(ii) Joining all these points on a graph paper, we get a rectangle.

We know that diagonals of a rectangle are equal and bisect each other.

∴ OA = OB = OC = OD

Hence, all four friends are at equal distance from the well.

Solution 28.

L.H.S

Solution 29.

Since, a diagonal of a parallelogram divides it into two triangles of equal area. Therefore, diagonal AC of parallelogram ABCD divides it into two triangles of equal area.

ar(∆ACD) = \(\frac { 1 }{ 2 }\)ar(parallelogram ABCD) ..(i)

In ∆AOX and ∆COY,

∠AOX = ∠COY [vertically opposite angles]

AO = CO

[∵ diagonals of a parallelogram bisect each other] and ∠OAX = ∠OCY

[since AB|| DC and AC is transversal line, so , ∠CAB = ∠ACD => ∠OAX = ∠OCY]

∆AOX ≅ ∆COY [by ASA congruence criterion]

∴=> ar(∆AOX) = ar(∆COY) [by congruent area axiom]

On adding ar(AOYD) both sides, we get

ar(AOX) + ar (AOYD) = ar (∆COY) + ar (AOYD)

=> ar (parallelogram AXYD) = ar (∆ACD) ..(ii)

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

ar (parallelogram AXYD) = \(\frac { 1 }{ 2 }\)ar (parallelogram ABCD)

Hence proved

Solution 30.

Given equation is \(\frac { x }{ 2 } +\frac { y }{ 3 } =1\)

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