CBSE Sample Papers for Class 9 Maths Paper 3 is part of CBSE Sample Papers for Class 9 Maths . Here we have given CBSE Sample Papers for Class 9 Maths Paper 3

## CBSE Sample Papers for Class 9 Maths Paper 3

Board |
CBSE |

Class |
IX |

Subject |
Maths |

Sample Paper Set |
Paper 3 |

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 3 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.

The statement “Decimal expression of a rational number can be non-terminating non-recurring.” Is it true or false?

Question 2.

Each side of an equilateral triangle is 2x cm. If x√3 = 48, then find its area.

Question 3.

Does Euclid’s fifth postulate imply the existence of parallel lines? Explain.

Question 4.

Plot the points (2,2), (4,-2) and (-2,-2) and check whether they are collinear or not.

Question 5.

For what value of (BC + AC), the construction of a ∆ABC is possible, if AB = 6 cm and ∠A = 45°?

Question 6.

The frequency distribution has been represented graphically as follows

Do you think this representation is correct? Why?

**Section B**

Question 7.

The diameters of two right circular cones are equal. If their slant heights are in the ratio 3:2, then what is the ratio of their curved surface areas?

Question 8.

Construct an equilateral triangle, given its side AB = 6 cm and justify the construction.

Question 9.

In a cricket match, a batsman hits the boundary 5 times out of 40 balls played by him. Find the probability that the boundary is not hit by the batsman. Also, determine the probability of hitting the sixes.

Question 10.

In the given figure, ∠1 = ∠3 and ∠2 = ∠4. Using Euclid’s axiom, show that ∠A = ∠C.

Question 11.

Factorize of 125x^{3} + y^{3}.

Question 12.

A hemispherical bowl is made from a metal sheet having thickness 0.3 cm. The inner radius of the bowl is 24.7 cm. Find the cost of polishing its outer surface at the rate of Rs 4 per 100 cm², [take, π = 3.14]

**Section C**

Question 13.

In a quadrilateral ABCD, there is a point O inside it such that OB – OD. Also, AB = AD and BC = DC. Prove that O lies on AC.

Question 14.

In the given figure, AB is a diameter of a circle with centre O and DO || CB.

If ∠BCD = 120°, then calculate

(i) ∠DAB

(ii) ∠DBA

(iii) ∠CBD

Also, show that ∆AOD is an equilateral triangle.

Or

It two circles intersect in two points, prove that the line through their centres is the perpendicular bisector of the common chord.

Question 15.

Construct a ∆PQR, in which QR = 6.5 cm, Q = 60° and PR – PQ = 1.5 cm and justify it.

Question 16.

The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed. Show that ar (parallelogram ABCD) = ar (parallelogram PBQR)

Or

Diagonals PR and QS of quadrilateral PQRS intersect each other at A. Show that ar (∆PSA) x ar (∆QAR) = ar (∆PAQ) x ar (ASAR).

Question 17.

Find three irrational numbers between \(\frac { 5 }{ 7 }\) and \(\frac { 9 }{ 11 }\)

OR

Evaluate

Question 18.

Write the coordinates of the vertices of a rectangle, whose length and breadth are 6 units and 3 units respectively, one vertex at the origin, the longer side lies on the E-axis and one of the vertices lies in the II quadrant. Also, find the area of the rectangle.

Question 19.

The mean monthly salary of 10 members of a group is Rs 1445. One more member whose monthly salary is Rs 1500 has joined the group. Find the mean monthly salary of 11 members of the group.

Or

Find the mean of the following distribution.

Question 20.

A student wrote the equations of the lines a and b drawn in the following graph as y = 1 and 2x + 3y = 6, respectively. Is he right? If yes, then write the coordinates of point of intersection of lines a and b.

Also, find the area enclosed between these lines and Y-axis

Question 21.

Line l is the bisector of ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A. Show that

(i) ∆APB = ∆AQB

(ii) BP = BQ or B is equidistant from the arms of ∠A

Question 22.

AB and CD are respectively, the smallest and the longest sides of a quadrilateral ABCD. Show that ∠A > ∠C and ∠B > ∠D.

Question 23.

Mukesh asked the teacher whether the two lines which are perpendicular to the same line, are perpendicular to each other or not. His teacher replied that yes, they will be perpendicular to each other giving the reason that two lines parallel to the same line are parallel to each other. His classmate Seema told him that he is wrong and explain him the correct statement. Mukesh thanked Seema for this.

(i) Write the solution given by Seema.

(ii) Identify the value depicted from this action.

Question 24.

The parking charges of a car in a parking lot is Rs 30 for the first two hours and Rs 10 per hour for subsequent hours. Taking total parking time to be xh and total charges as Rs y, write a linear equation in two variables to express the above statement. Draw a graph for the linear equation and read the charges for five hours.

Question 25.

The following observed values of x and y are thought to satisfy a linear equation. Write the linear equation.

Draw the graph, using the values x and y as given in the above table. At what points, the graph of the linear equation

(i) cuts the X-axis?

(ii) cuts the Y-axis?

Question 26.

Following table shows the marks scored by a group of 90 students in a mathematics test of 100 marks

A student is selected at random. Find the probability that student has obtained.

(i) Less than 30.

(ii) 60 or more marks.

(iii) between 40 and 70 marks

(iv) 70 or more

Question 27.

A solid cylinder has a total surface area of 231m². Its curved surface area is 2/3 of the total surface area. Then, find the volume of the cylinder.

Or

The external and internal diameters of a hollow hemispherical vessel are 25 cm and 24 cm, respectively. The cost of painting 1cm² of the surface is Rs 0.05. Find the total cost to paint the vessel all over.

Question 28.

A park in the shape of quadrilateral ABCD has ∠C = 90°, AB = 9 m, BC = 12 m, CD =5m and AD -8m. How much area does it occupy?

Or

A kite in the shape of a square with each diagonal 36 cm and having a tail in the shape of an isosceles triangle of base 10 cm and equal side 6 cm, is made of three different shades as shown in the figure. How much paper of each shade has been used in it? [given, √11 = 3.31]

Question 29.

There are two friends Swati and Sapna who live in a village. Their common friend Monika fell ill. She was admitted in a hospital. Swati and Sapna decided to help Monika. Swati contributed as much money as the fifth root of the cube of amount contributed by Sapna. If the product of amount distributed by two friends is Rs 390625, then find the contribution of each friend.

Question 30.

**Solutions**

Solution 1.

False; because a number whose decimal expression is non-terminating non- recurring, is known as irrational number.

Solution 2.

Given, side of an equilateral triangle = 2x

and x√3 = 48

∴ \(x=\frac { 48 }{ \sqrt { 3 } } \)

Area of equilateral triangle = \(\frac { \sqrt { 3 } }{ 4 } \) (side)²

Solution 3.

Yes, if a straight line l falls on two lines m and n such that sum of the interior angles on one side of l is two right angles, then by Euclid’s fifth postulate, the lines will not meet on this side of l.Also, we know that the sum of the interior angles on the other side of the line l will be two right angles too. Thus, they will not meet on the other side also.

Hence, the lines m and n never meet, i.e. they are parallel.

Solution 4.

Plotting the points on the graph, we get the points P(2,2), 0(4, -2) and R(-2, -2). On joining these three points, we get three lines, i.e. the given points do not lie on the same line. So, given points are not collinear.

Solution 5.

The value of (BC + AC) must be greater than 6 cm, because the sum of any two sides of a triangle is always greater than the third side.

Solution 6.

No, because the width of the rectangles are varying, so we need to make certain modifications in the length of the rectangles, so that the areas are proportional to the frequencies.

Solution 7.

Let the radii and slant heights of two right circular cones be r_{1}, l_{1}, and r_{2}, l_{2}, respectively.

Since, the diameters are equal, therefore

2r_{1} = 2r_{2}

=> r_{1} = r_{2}

Also, ratio of slant heights = l_{1} : l_{2} = 3:2

Now, ratio of curved surface areas of two right circular cone

Hence, the ratio of curved surface areas of cones is 3:2.

Solution 8.

Given, length of equilateral side = 6 cm

Steps of construction

(i) First, draw a line segment BC = 6 cm.

(ii) Taking B and C as centre and radius 6 cm, draw two arcs.

(iii) The arc of step (ii) cut any point A join the lines. AB and AC.

Thus, ∆ABC is an equilateral triangle.

Solution 9.

Given, total number of balls thrown = 40

∴ Total number of trials = 40

Let E be the event of getting boundary

∴Number of trials, in which E happened = 5

We cannot determine the probability of hitting sixes, because their is no information given in the question.

Solution 10.

Given, ∠1 = ∠3 ..(i)

and ∠2 = ∠4 …(ii)

Euclid’s second axiom says that, if equals are added to equals, then the wholes are equal.

On adding Eqs. (i) and (ii), we get

∠1 + ∠2 = ∠3 + ∠4

=> ∠4 = ∠C

Hence proved.

Solution 11.

We have, 125x^{3} + y^{3} = (5x) + (y)^{3}

= (5x + y){(5x)^{2} – (5x)(y) + (y)^{2}}

[∵ a^{3} + b^{3} = (a + b)(a^{2} – ab + b^{2})]

= (5x + y)(25x^{2} – 5xy + y^{2})

Solution 12.

Given, inner radius of the hemispherical bowl = 24.7 cm

Thickness of metal sheet = 0.3 cm

Now, outer radius of the hemispherical bowl

r = 24.7 + 0.3 = 25 cm

∴ Outer surface area of the hemispherical bowl = 2πr²

= 2 x 3.14 x (25)²

= 157 x 25 = 3925 cm²

Now, cost of polishing 100 cm² = Rs4

∴ Cost of polishing 3925 cm² = \(\frac { 4X3925 }{ 100 }\) = Rs 157

Solution 13.

In ∆AOB and ∆AOD,

AB = AD [given]

OB = OD [given]

and OA = OA [common side]

∴ ∆AOB ≅ ∆AOD [by SSS congruence rule]

=> ∠1 = ∠2 [byCPCT] … (i)

Similarly, in ∆CBO and ∆CDO,

BC = DC [given]

OB = OD [given]

and OC = OC [common side]

∴ ∆CSO ≅ ∆CDO [by SSS congruence rule]

=> ∠3 = ∠4 [by CPCT]… (ii)

On adding Eqs. (i) and (ii), we get

∠1 + ∠3 = ∠2 + ∠4

We know that, ∠1 + ∠3 + ∠2 + ∠4 = 360°

[∵ sum of all angles around a point is 360°]

=> ∠1 + ∠3 + ∠1 + ∠3 = 360° [from Eqs. (i) and (ii)]

=> 2(∠1 + ∠3) = 360°

=> ∠1 + ∠3 = 180° => ∠AOC = 180°

Hence, O lies on AC. Hence proved.

Solution 14.

(i) It is clear from the figure that ABCD is a cyclic quadrilateral.

∴∠DCB + ∠DAB = 180°

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

=> 120° + ∠DAB = 180°

[∵∠DCB = ∠BCD = 120°]

=> ∠DAB = 180° – 120° = 60° …(i)

(ii) Given, AB is a diameter.

∴ ∠ADB = 90°

[∵ angle in a semi-circle is right angle] …(ii)

In ∆ABD, ∠ADB + ∠DBA + ∠DAB = 180°

[∵ sum of all angles of a triangle is 180°]

=> 90° + ∠DBA + 60° = 180°

[from Eqs. (i) and (ii)]

=>∠DBA = 180° -150° = 30°

(iii) In, ∆ODB,OB = OD [∵ radii of same circle]

∴ ∠ODB = ∠OBD

[∵ angles opposite to equal sides are equal]

=> ∠ODB = 30° [∵ ∠OBD = ∠DBA = 30°]

Since, DO||CB

∵∠ODB = ∠CBD = 30°

[by alternate interior angles]

Now, in ∆AOD, OD = OA [radii of a circle]

=> ∠OAD = ∠ODA = 60° [from Eq. (i)]

[∵ angles opposite to equal sides are equal]

Now, ∠OAD + ∠O’DA + ∠AOD = 180°

[∵ sum of all angles of a triangles is 180°]

=> 60°+ 60 °+∠AOD = 180°

=> ∠AOD = 60°

∴ ∠OAD = ∠ODA = ∠AOD = 60°

Hence, MOD is an equilateral triangle.

Hence proved.

Or

Let C(O, r) and C(O’, s) be two circles intersecting at points A and B.

Solution 15.

Given, in ∆PQR, QR = 6.5 cm, ∠Q = 60° and PR – PQ = 1.5 cm

Steps of construction

(i) Draw the base, QR = 6.5 cm.

At pointO, draw a ray QX making an ∠XQR = 60°.

Here, PR – PQ = 1.5 cm

PR > PQ

i. e. the side containing the base angle 0 is less than third side, so it is the case II.

(ii) Cut line segment OS equal to PR – PQ, i.e. OS = 1.5 cm from the ray OX extended on opposite side of base QR.

(iii) Join SR and draw its perpendicular bisector ray AB, which intersects SR at M (say).

(iv) Let P be the intersection point of SX and perpendicular bisector AB. Then, join PR.

Thus, ∆PQR is the required triangle.

Justification

Base QR and ∠Q are drawn as given.

Since, AB is the perpendicular bisector of SR and P lies on it.

∴ PS = PR

Now, QS = PS – PQ

=> QS = PR – PQ

Thus, construction justified.

Solution 16.

Given Side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and PBQR is a parallelogram.

To prove ar (parallelogram ABCD) = ar (parallelogram PBQR)

Construction Join AC and PQ.

Proof Since, AC is a diagonal of parallelogram ABCD.

∴ ar (AABC) = \(\frac { 1 }{ 2 }\)ar (parallelogram ABCD) …(i)

Since, PQ is a diagonal of parallelogram PBQR.

∴ ar(ABPQ) = \(\frac { 1 }{ 2 }\)ar (parallelogram PBQR) …(ii)

Since, AQ ||CP and AACQ and AAPQ are on the same base AQ and between the same parallel sides AQ and CP.

∴ ar (AACQ) = ar (AAPQ)

[∵ two triangles on the same base and between the same parallel sides are equal in areas]

=> ar (AACQ) – ar (AABQ)

= ar (AAPQ) – ar (AABQ)

[on subtracting ar (AABQ)from both sides]

=> ar (AABC) = ar (ABPQ)

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

= \(\frac { 1 }{ 2 }\) ar (parallelogram PBQR)

[from Eqs. (i) and (ii)]

=> ar (parallelogram ABCD)

= ar (parallelogram PBQR) Hence proved.

Or

We draw PM⊥QS and RN⊥QS.

Solution 17.

Solution 18.

Let us draw the coordinate axes XOX’and YOY’ and choose a suitable units of distance on the axes.

Let 1cm = 1 unit.

Mark a point A on Y-axis at a 6 units distance in positive direction of Y-axis and mark a point C on X-axis at a 3 units distance in negative direction of X-axis.

Now, draw a perpendicular lines from C and A, which meets at a point B.

Thus, we get the rectangle QABC whose vertices are

O(0, 0),A(0, 6),B(-3, 6) and C(-3, 0)

∵Area of rectangle = Length x Breadth

∴Area of rectangle OABC = 6 x 3 = 18 sq units

Solution 19.

Given, mean monthly salary of 10 members of a group = Rs 1445

Solution 20.

Clearly, line a is parallel to X-axis at a distance of 1 unit in positive direction of Y-axis, therefore its equation is y = 1.

Also, if we draw the graph of line 2x + 3y = 6, then its graph should intersect X-axis at (3, 0) and Y-axis at (0, 2) and so the given equation of line b is correct. Thus, the student is right.

Now, substituting y = 1 in equation of line b, we get

2x + 3(1) = 6 => 2x = 6 – 3 = 3 => \(x=\frac { 3 }{ 2 }\)

Hence, the point of intersection is (3/2,1).

∴ \(AC=\frac { 3 }{ 2 }\)

Clearly, area enclosed between these lines and Y-axis = Area of AABC = \(\frac { 1 }{ 2 }\) x Base x Height

= \(\frac { 1 }{ 2 }\) x BC x AC = \(\frac { 1 }{ 2 }\) x 1 x \(\frac { 3 }{ 2 }\) = \(\frac { 3 }{ 4 }\) sq units

Solution 21.

Given Line l is the bisector of ∠A,

i.e. ∠QAB = ∠PAB

Also, BP and 60 are perpendiculars to the arms of ∠A.

i.e. ∠BQA = ∠BPA = 90°

To prove

(i) ∆APB = ∆AQB

(ii) BP = BQ or 6 is equidistant from the arms of ∠A.

Proof (i) In ∆APB and ∆AQB,

∠BAP = ∠BAQ

∠BPA = ∠BQA [each 90°]

and AB = AB [common side]

∴ ∆APB ≅ ∆AQB

[by MS congruence rule]

(ii) As, ∆APB ≅ ∆AQB

=> BP =BQ [by CPCT]

∴ B is equidistant from the arms of ∠A.

Hence proved.

Solution 22.

First, draw the diagonal AC.

In ∆ABC, AB is the shortest side.

∴ AB < BC => ∠2 < ∠1 …(i)

[∵ angle opposite to the shortest side is smaller]

In ∆ACD, CD is the longest side.

∴ AD < CD => ∠4 < ∠3 …..(ii)

[∵ angle opposite to the longest side is longer]

Solution 23.

(i) Let l and m be two lines perpendicular to the same line n as shown below.

Here, ∠1 = 90° and ∠2 = 90° [given]

∴ ∠1 = ∠2

Also, ∠1 and ∠2 are corresponding angles for lines l and m intersected by the traversal line n.

∴ l||m

Hence, lines l and m are parallel to each other.

(ii) The value has been shown in this question is cooperative learning among students.

Solution 24.

Given, parking charges for the first two hours = Rs 30 and for subsequent hours = Rs 10

Total parking time = x and total charges = Rs y

Then, according to the given condition,

30 + 10(x – 2) = y => 30 + 10x – 20 = y

=> 10x + 10 – y = 0 => 10x – y + 10 = 0

which is the required linear equation in two variables.

It can also be written as

y = 10x + 10 ..(i)

Now, for drawing the graph, we need atleast two solutions of the equation.

When x = 0, then y = 10(0) +10 = 10

When x = -1, then y = -10 + 10 = 0

When x = 1, then y = 10(1) + 10 = 20

So, we have the following table to draw the graph

Here, we have three points A(-1, 0), 6(0,10) and C(1,20). By plotting these points on the graph paper and joining them, we get a straight line AC, which represents the required graph of linear equation.

From the graph, charges for the hours 1,2, 3, 4 and 5 are Rs 20, Rs 30, Rs 40, Rs 50 and Rs 60, respectively.

Solution 25.

Let the linear equation y = mx + c satisfies the points (6,-2)and (-6,6).

– 2 = 6 m + c …(i)

and 6 = – 6 m + c …(ii)

On subtracting Eq. (ii) from Eq. (i), we get

12m = – 8 => m = \(-\frac { 2 }{ 3 }\)

On putting the value of m = \(-\frac { 2 }{ 3 }\) in EQ (i) we get

On putting y=0 in eq(iii) we get

0 + 2x = 6

=> x = 3

(3, 0) is a solution of the equation.

Again, putting x = 0 in Eq. (iii), we get

3y + 0 = 6

=> y = 2

∴ (0,2) is a solution of the equation.

(i) The graph of the linear equation will cut X-axis at (3,0).

(ii) The graph of the linear equation will cut Y-axis at (0,2).

Solution 26.

Solution 27.

Given, total surface area of cylinder = 231 m²

Let the radius and height of cylinder are r and h.

According to the question,

Curved surface area of cylinder

= \(\frac { 2 }{ 3 }\) x Total surface area of cylinder

Solution 28.

Given, a quadrilateral ABCD in which AB = 9 m, BC = 12 m, CD = 5 m, AD = 8 m, and ∠C = 90°.

Join the diagonal BD.

Solution 29.

Let the amount contributed by Sapna be Rs x.

Then, the amount contributed by Swati be (x)^{3/5}.

According to the question,

x(x)^{3/5} = 390625

=> x^{1+3/5} = 390625 [∵ x^{a} – x^{b} = x^{a+b}]

Solution 30.

We hope the CBSE Sample Papers for Class 9 Maths Paper 3 help you. If you have any query regarding CBSE Sample Papers for Class 9 Maths Paper 3, drop a comment below and we will get back to you at the earliest.