CBSE Sample Papers for Class 9 Maths Paper 1

CBSE Sample Papers for Class 9 Maths Paper 1 is part of CBSE Sample Papers for Class 9 Maths . Here we have given CBSE Sample Papers for Class 9 Maths Paper 1

CBSE Sample Papers for Class 9 Maths Paper 1

Board	CBSE
Class	IX
Subject	Maths
Sample Paper Set	Paper 1
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 1 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.
Simplify

Question 2.
Find the area of a triangle whose sides are 3 cm, 4 cm and 5 cm.

Question 3.
On which axes, do the given points lie?
(i) (6, 0)
(ii) (0,-6)

Question 4.
The diameter of a right circular cone is 14 m and its slant height is 10 m. Find its curved surface area.

Question 5.
Find the length of a chord which is at a distance of 4 cm from the centre of a circle of radius 5 cm.

Question 6.
It is given that, the probability of winning a game is 0.8. What is the probability of losing the game?

Section B

Question 7.
In a parallelogram, show that the angle bisectors of two adjacent angles intersect at right angles.

Question 8.
The heights of two cylinder are in the ratio 5 : 3 and their radii are in the ratio 2:3. Then, find the ratio of their volumes.

Question 9.
ABCD is a parallelogram and X is the mid-point of AB. If ar (∆ADC) = 24cm², then find ar (AXCD).

Question 10.
The cost of a ball pen is Rs 5 less than half of the cost of a fountain pen. Write this statement as a linear equation in two variables.

Question 11.
Find the mean of the following distribution.

Question 12.
The length, breadth and height of a room are 5 m, 4 m and 3 m, respectively. Find the cost of white-washing the walls of the room and the ceiling at the rate of Rs. 7.50 per sq m.

Section C

Question 13.
Find the value of 64x³ – 125y³, if 4x – 5y = 16 and xy = 12.
Or
Check whether polynomial P(x) = 2x³ – 9x² + x + 12 is a multiple of (2x – 3) or not.

Question 14.
Represent √7.47 on the number line.
Or
Find the value of .

Question 15.
In the given figure, ABC is an isosceles triangle with AB = AC. BD and CE are two medians of the triangle.
Prove that BD = CE.

Question 16.
In the given figure, side BC of ∆ABC is produced to form ray BD and CE || BA. Show that ∠ACD = ∠A + ∠B. Deduce that ∠A + ∠B + ∠C = 180°.

Question 17.
The three vertices of a rectangle ABCD are A(2, 2), B(-3,2) and C(-3,5). Plot these points on a graph paper and find the area of rectangle ABCD.

Question 18.
If C is the mid-point of the line segment AB, L and M are the mid-points of the line segment AC and BC, respectively, then
prove that AL = LC = CM – MB = \(\frac { 1 }{ 4 }\) AB

Also, state which Euclid’s axiom is applied for proving result.

Question 19.
The mean of 50 items was found to be 40. If at the time of calculation two items were wrongly taken as 32 and 12 instead of 23 and 11, then find the correct mean.
Or
Draw a histogram of the following data

Question 20.
ABCD is a cyclic quadrilateral, whose diagonals intersect at a point E. If ∠DBC = 80° and ∠BAC = 40°, then find ∠BCD. Further, if AB = BC, then find ∠ECD.

Question 21.
In the given figure, two circles intersect at A, B and AC, AD are respectively the diameters of the circles. Prove that the points C, B and D are collinear.

Question 22.
An umbrella is made by stitching 8 triangular pieces of cloth of two different colours. Each piece measuring 20 cm, 40 cm and 40 cm. How much cloth of each colour is required for the umbrella?

OR
The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white washing its curved surface area at the rate of Rs 210 per 100 m²?

Section D

Question 23.
Rationalise

Question 24.
For what value of a is
p(x) = 4x³ + ax² – x – 1 exactly divisible by g(x) = 2x + 1?
Or
Factorize 125x³ – 27y³ + z³ + 45xyz

Question 25.
Under tree plantation programme students of class VIII and IX planted (3x² – 4x – 4) trees in school.
(a) If number of students in those classes are (x – 2) then find out number of trees planted by each student. (Assuming each student planted equal number of trees).
(b) What values of students are exhibited here?

Question 26.
Three coins are tossed simultaneously 180 times and it is found that 3 tails appeared 34 times, 2 tails appeared 55 times, 1 tail appeared 72 times and no tail appeared 19 times. Find the probability of getting
(i) 3 tails
(ii) 2 tails
(iii) 1 tail
(iv) 0 tail.
Or
On a busy road, following data was observed about cars passing through it and number of occupants:

Find the chance that it has
(i) exactly 5 occupants.
(ii) more than 2 occupants.
(iii) less than 5 occupants.
(iv) greater than 5 occupants.

Question 27.
Construct a ∆ABC in which BC = 4.6 cm, ∠B =45° and AB + CA =8.2 cm.

Question 28.
The diameter of a sphere is decreased by 25%. By what per cent does its curved surface area decrease?
Or
In the given figure, ABCD is a rectangle of dimension 4 cm and 6 cm. E and F are the mid-points of AB and BC, respectively. Find the area of the shaded portion.

Question 29.
For the given pairs of linear equations, find solutions of the form x = a, y = 0 and x = 0, y = b respectively.
3x + 2y = 6 and 5x – 2y = 10

Question 30.
A point O inside a rectangle ABCD is joined to the vertices. Prove that the Sum of the areas of a pair of opposite triangles so formed is equal to the sum of the areas of other pair of triangles.

Solutions

Solution 1.

Solution 2.
Let sides of a triangle be a = 3cm, b = 4cm and c = 5cm.

Solution 3.
(i) In point (6, 0), we have the ordinate = 0 and abscissa = 6
∴Point (6, 0) lies on the X-axis.
(ii) In point (0, -6), we have the abscissa = 0 and ordinate = -6
∴Point (0, -6) lies on the Y-axis.

Solution 4.
Given, diameter of cone = 14 m
Radius of cone, r = \(\frac { 14 }{ 2 }\) = 7 m
and slant height, l = 10 m
Curved surface area = πrl = \(\frac { 22 }{ 2 }\) x 7 x 10
= 22 x 10
= 220 m²

Solution 5.
Given, the perpendicular distance, OP = 4 cm
and radius of circle, OA = 5 cm
In right angled ∆APO,
(AO)² = (AP)² + (OP)²
[by Pythagoras theorem]
=> (5)² = (AP)² + (OP)²
=> (AP)² = 25 – 16 = 9
=> 3 cm
[taking positive square root]
We know that, the perpendicular drawn from the centre of a circle to a chord divides the chord into two equal parts.
∴ AP = \(\frac { 1 }{ 2 }\) x AB
=> AB = 2AP
=> AB = 2 x 3 = 6cm

Solution 6.
Given, P (winning a game) = 0.8
We know that,
P (winning a game) + P (not winning a game) = 1
=> 0.8 + P (not winning the game) = 1
=> P (not winning the game) = 1 – 0.8
= 0.2

Solution 7.
Given, ABCD is a parallelogram such that angle bisectors of adjacent angles A and 6 intersect at point P.
We have to show that, ∠APB = 90°
Here, ∠A + ∠B = 180°

Solution 8.
Let the heights of cylinders are h1,h2 and radii r1, r2.

Hence, required ratio of volumes of cylinders is 20:27.

Solution 9.
Given, ABCD is a parallelogram and X is the mid-point of AB. Draw the diagonal AC.

We know that, diagonal of a parallelogram divides it into two triangles of equal areas.
So, ar (parallelogram ABCD) = 2 x ar (∆ADC)
= 2 x 24 = 48cm²
Since, CX is the median of ∆ABC.
We know that a median of a triangle divides it into two triangles of equal areas.
∴ ar (∆BCX) = \(\frac { 1 }{ 2 }\) ar (∆ABC) = \(\frac { 1 }{ 2 }\) x 24 = 12 cm² [∵ ar (∆ABC) = ar (∆ADC)]
Now, ar(AXCD) = ar (parallelogram ABCD) – ar (∆BCX)
= 48 – 12 = 36cm²

Solution 10.
Let cost of a ball pen = Rs x
and cost of a fountain pen = Rs y
According to the question,
cost of a ball pen=half of the cost of a fountain pen – 5
=> x = \(\frac { y }{ 2 }-5\)
=> x = \(\frac { y-10 }{ 2 }\)
=> 2x = y – 10
=> 2x – y + 10 = 0
which is the required linear equation in two variables.

Solution 11.
Table for mean distribution is given below:

= 24.3

Solution 12.
Given, length of a room, l = 5 m
breadth of a room, b = 4 m
and height of a room, h = 3 m
∴ Area of four walls of the room
= 2 h (l + b) = 2 x 3(5 + 4)
= 6 x 9 = 54m²
Area of ceiling = l x b = 5 x 4 = 20 m²
∵ Total area of white-washing the room
= Area of four walls + Area of the ceiling
∴ Total area of the room to be white-washed
= 54 + 20 = 74 m²
∵ Cost of whitewashing 1 m² = Rs 7.50
∴ Cost of whitewashing 74 m² = Rs 7.50 x 74
= Rs 555

Solution 13.
Given, 4x – 5y = 16
On cubing both sides, we get
(4x – 5y)³ = (16)³
=> (4x)³ – (5y)³ – 3 × 4x × 5y(4x – 5y) = (16)³
[∵ (a – b)³ = a³ – b³ – 3 ab(a – b)]
=> 64x³ – 125y³ – 60xy(4x – 5y) = 4096
=> 64x³ – 125y³ – 60 × 12 × 16 = 4096
[∵ xy = 12 and 4x – 5y = 16, given]
=> 64x³ – 125y³ – 11520 = 4096
=> 64x³ – 125y³ = 4096 + 11520
Hence, 64x³ – 125y³ = 15616
Or
The polynomial P(x) will be multiple of (2x – 3), if (2x – 3) divides P(x) completely.
Now, 2x – 3 = 0 => x = \(\frac { 3 }{ 2 }\)

Here we see that remainder is zero. So, (2x – 3)divides P(x) completely.
Therefore, P(x) is a multiple of (2x – 3).

Solution 14.
First, we draw a line AB = 7.47 units.
Now, from point S, mark a distance of 1 unit. Let this point be C. Let O be the mid-point of AC.
Now, draw a semi-circle with centre O and radius OA. Let us draw a line perpendicular to AC passing through point B and intersecting the semi-circle at point D.

Solution 15.
Given ABC is an isosceles triangle with AB = AC. BD and CE are two medians.
To prove BD = CE
Proof In ∆ABC,
AB = AC => ∠B = ∠C ..(i)

Solution 16.
Given, CE || BA, AC and BC are the transversal.
∴ ∠4 = ∠1 …(i)
[alternate interior angles]
and ∠5 = ∠2 …(ii)
[corresponding angles]
On adding Eqs. (i) and (ii), we get
∠4 + ∠5 = ∠1 + ∠2
Hence, ∠ACD = ∠A + ∠B
Now, ∠1 + ∠2 = ∠4 + ∠5
=> ∠1 + ∠2 + ∠3 = ∠3 + ∠4 + ∠5
[adding ∠3 on both sides]
=> ∠A + ∠B + ∠C = 180°
[∵ ∠3 + ∠4 + ∠5 = ∠BCD = a straight angle]
Hence proved.

Solution 17.
Given, three vertices of rectangle ABCD are A(2,2), B(-3,2) and C(-3, 5).
Now, plotting the points on the graph paper, as shown below, we find that the coordinates of D are (2, 5).

Then, AB = 5 units and BC = 3 units
Area of a rectangle = Length x Breadth
= 3 x 5 = 15 sq units

Solution 18.
Given, C is the mid-point of AB.

Here, Euclid’s first and seven axioms are used for proving result. Hence proved.

Solution 19.

Solution 20.
Given, in quadrilateral, ∠BAC = 40° and ∠DBC = 80°

Solution 21.
Given Two circles with AC and AD as diameters respectively intersecting each other at A and B.
To prove Points C, B and D are collinear.
Construction Join CB, BD and AB.

Proof We know that the angle in a semi-circle is 90°.
∴∠ABC = 90° and ∠ABD = 90°
[∵ AC and AD are diameters of circles]
Then, ∠ABC + ∠ABD = 90° + 90° = 180°
∴ CBD is a straight line.
Hence, points C, B and D are collinear.
Hence proved.

Solution 22.
Here, each triangular piece is an isosceles triangle with sides a = 40 cm, b = 40 cm and c = 20 cm.

Solution 23.

Solution 24.
Given polynomials p(x) = 4x³ + ax² – x – 1 and g(x) = 2x + 1
By the factor theorem, p(x) will be exactly divisible by

Solution 25.
(a) Clearly, the number of trees planted by each student is given by

(b) The values exhibited by student are
(i) Scientific attitude
(ii) Dutiful
(iii) Environment awareness
(iv) Social values

Solution 26.
Given, total number of trials = 180
Number of times 3 tails appeared = 34
Number of times 2 tails appeared = 55
Number of times 1 tail appeared = 72
Number of times 0 tail appeared = 19
(i) P (getting 3 tails)

Solution 27.
Given, BC = 4.6 cm, ∠B = 45° and AB + CA = 8.2 cm
In horizontal axis take 1 block is equal to 10 units and in vertical axis, take 1 block is equal to 5 units.
Steps of construction
(i) Draw the base line segment BC = 4.6 cm.
(ii) At the point S, make ∠XBC = 45°
(iii) Now, cut a line segment BD equal to AB + AC = 8.2 cm from the ray BX.
(iv) Join DC.

(v) Draw perpendicular bisector MN of CD which meets BX at A.
(vi) On joining AC, we get the required ∆ASC.

Solution 28.
Let the original diameter of sphere be 2x.
Then, original radius of the sphere, r = x
Original curved surface area of sphere

Solution 29.
Given equation 3x + 2y = 6 can be written

Hence, the required solution of 3x + 2y = 6 is x = 2, y = 0, which is the form of x = a, y = 0. and the solution of 5x – 2y = 10 is x = 0, y = -5, which is the form of x = 0 ,y = b.

Solution 30.
Given A rectangle ABCD and O is a point inside it. OA, OB, OC and OD have been joined.
To prove ar(∆AOD) + ar(∆BOC) = ar(∆AOS) + ar(∆COD)

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