CBSE Sample Papers for Class 9 Maths Paper 3

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.
• 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.
Write a rational number between √2 and √3

Question 2.
Find the remainder when P(x) = x³ – 3x² + 4x + 32 is divided by (x + 2).

Question 3.
What is the equation of y-axis?

Question 4.
If 3∠A = 4∠B = 6∠C, then ∠A : ∠B : ∠C = ?

Question 5.
In which quadrant point (x, y) lies if x < 0, y > 0?

Question 6.
The radii of two cylinders are in the ratio of 2:3 and their heights are in the ratio of 5:3, then find the ratio of their volumes.

SECTION-B

Question 7.
Factorize: x(x – y)³ + 3x²y (x -y).

Question 8.
The base of an Isosceles triangle measure 24 cm and its area is 192 cm². Find its perimeter.

Question 9.
The runs scored by two batsmen in a cricket match is 164. Write a linear equation in two variables x and y. Also write a solution of this equation.

Question 10.
12 packets of salts, each marked 2 kg, actually contained the following weights (in kg) of salt.
1.950, 2.020, 2.060, 1.980, 2.030, 1.970
2.040, 1.990, 1.985, 2.025, 2.000, 1.980
Out of these packets one packet is chosen at random. What is the probability that the chosen packet contains more than 2 kg of salt?

Question 11.
The percentage of marks obtained by a student in six unit test are given below.

The unit test is selected at random. What is the probability that the student gets more than 60% marks in the test?

Question 12.
Evaluate (104)³ using suitable identity.

SECTION-C

Question 13.
If x = 7 + 4√3 , then find the value of \(\left( \sqrt { x } +\frac { 1 }{ \sqrt { x } } \right) \)
OR
Find the value of √4.3 geometrically and represent √4.3 on a given real line.

Question 14.
a³ – b³ + 1 + 3ab

Question 15.
In the figure, AB || CD. Prove that ∠p + ∠q – ∠r = 180°.

Question 16.
In the given figure, ABC is a triangle in which AB = AC. Side BA is produced to D, such that AB = AD. Prove that ∠BCD = 90°.

Question 17.
Mid Point Theorem: Prove that the line segment joining the midpoints of any two sides of a triangle is parallel to the third side and equal to half of it.
OR
Prove that parallelograms on the same base and between same parallels are equal in area.

Question 18.
In the given figure O is the centre of circle, prove that ∠x + ∠y = ∠z

Question 19.
Construct a ∆ABC in which BC = 6 cm, ∠B = 60° and sum of other two sides is 9 cm [i.e., AB + AC = 9 cm].

Question 20.
Plot the points A(3, 3), B(2, 4), C(5, 5), D(0, 2), E(3, -3) and F(-5, -5) on a graph paper. Which of these points are the mirror images in
(i) x-axis,
(ii) y-axis?

Question 21.
A cylindrical metallic pipe is 14 cm long. The difference between the outside and inside surface is 44 cm². If the pipe is made up of 99 cubic cm of metal, find the outer and inner radii of the pipe.

Question 22.
The total surface area of a solid cylinder is 231 cm² and its curved surface area is \(\frac { 2 }{ 3 }\) of the total surface area. Find the volume of the cylinder.

SECTION-D

Question 23.
Express \(0.6+0.\overline { 7 } +0.\overline { 47 } \) in the form \(\frac { p }{ q }\), wherep and q are integers and q ≠ 0.

Question 24.
If both (x – 2) and \(x-\frac { 1 }{ 2 }\) are factors of polynomial Px² + 5x + r show that \(\frac { P }{ r }\) = 1.

Question 25.
A pharmacist needs to strengthen a 15% alcoholic solution to one of 32% alcohol. How much pure alcohol should be added to 800 ml of 15% solution?

Question 26.
In the figure, POQ is a line, Ray OR is perpendicular to line PQ, OS is another ray lying between rays OP and OR. Prove that
∠ROS = \(\frac { 1 }{ 2 }\)(∠QOS – ∠POS)

Question 27.
In the given figure, C is the mid point of AB. If ∠DCA = ∠ECB and ∠DBC = ∠EAC, prove that
(i) ∆DBC ≅ ∆EAC
(ii) DC = EC

Question 28.
In the adjoining figure, D, E, F are mid points of the sides BC, CA and AB of ∆ABC. If BE and DF intersect at X while CF and DE intersect at Y, prove that XY = \(\frac { 1 }{ 4 }\)BC

Question 29.
Draw a histogram to represent the following frequency distribution:

Question 30.
Kartikeya wanted to make a temporary shelter for street dogs by making a box like structure with tarpaulin that covers all the four sides and the top of the house. How much tarpaulin would be referred to make the shelter of height 2.5 m with base dimensions 4m x 3m. Assuming stitching margin is negligible. Which values are depicted in this question?

Solutions

Solution 1.
√2 = 1.414, √3 = 1.732
Rational number between √2 and √3 = \(\frac { \sqrt { 2 } +\sqrt { 3 } }{ 2 } \) = 1.573 ~ 1.6 upto one place of decimal = \(\frac { 8 }{ 5 }\)

Solution 2.
P(x) = x³ – 3x² + 4x + 32,x + 2 = 0 => x = -2
When P(x) is divided by (x + 2), we get
Remainder = P(-2) = [(-2)³ – 3(-2)² + 4(-2) + 32] = -8 – 12 – 8 + 32 = 4
Remainder = 4

Solution 3.
x = 0.

Solution 4.
Let 3∠A = 4∠B = 6∠C = k°

Solution 5.
If x < 0, y > 0 => x = -ve and y = +ve => P(x, y) lies in quadrant II because it is (-, +).

Solution 6.
Let their radius be 2x cm and 3x cm and their height be 5y cm and 3y cm respectively and volume of cylinder = πr²h

Solution 7.
x(x – y)³ + 3x²y (x – y) = x(x – y) [(x – y)² + 3xy] = x(x – y) [(x² + y² – 2xy + 3xy)]
= x (x – y) (x² + y² + xy).

Solution 8.
AL ⊥ BC

Hence perimeter of Isosceles triangle = 20 + 20 + 24 = 64 cm.

Solution 9.
Both Batsmen make runs = x + y = 164
So, linear equation = x + y = 164
If x = 100, then y = 164 – 100 = 64
One solution is

Solution 10.
Total number of packets of salts = 12
Number of packets containing more than 2 kg of salt = 5.

Solution 11.
Number of test in which student gets more than 60% marks = 2.
Total number of test = 6

Solution 12.
(104)³ = (100 + 4)³ = (100)³ + (4)³ + 3(100)(4)(100 + 4)
= 1000000 + 64 + 1200 (100 + 4)
= 1000000 + 64 + 120000 + 4800 = 1124864

Solution 13.

Steps of Construction:
Draw a line segment AB = 4.3 units and extend it to C such that BC = 1 unit.
Find the mid point O of AC.
Draw a semicircle with O as centre and OA as radius.
Draw BD ⊥ AC intersecting the semicircle at D.
Then BD = √4.3 units
with B as centre and BD as radius, draw an arc meeting with AC produced at E. Then BE = BD = √4.3 units.

Solution 14.
a³ – b³ + 1 + 3ab = a³ + (-b)³ + 1 + 3ab = a³ + (-b)³ + (1)³ – 3 x a x (-b) x 1
a³ – b³ + 1 + 3ab = (a – b + 1) (a² + b² + 1 + ab + b – a)
= (a – b + 1) (a² + b² + ab – a + b + 1)
a³ – b³ + 1 + 3 ab = (a – b + 1) (a² + b² + ab – a + b + 1)

Solution 15.

AB || CD.
Through F, draw PFQ || AB || CD
So, q = ∠1 + ∠2
AB || PF and EF is transversal line
∠p° + ∠1 = 180° …(i)
Again
PF || CD and GF is transversal line
∠2 = ∠r [Alternate Interior angles] …(ii)

Adding (i) and (ii)
∠p + ∠1 + ∠2 = 180° + ∠r
∠p + (∠1 + ∠2) = 180° + ∠r
∠p + ∠q = 180° + ∠r
∠p + ∠q – ∠r = 180°

Solution 16.

AB = AC => ∠ABC = ∠ACB
AC = AD => ∠ADC = ∠ACD
∴ ∠ABC + ∠ADC = ∠ACB + ∠ACD = ∠BCD
∠DBC + ∠BDC = ∠BCD
[∵ ∠ABC = ∠DBC, ∠ADC = ∠BDC]
Adding ∠BCD on both sides
∠DBC + ∠BDC + ∠BCD = 2∠BCD
180° = 2 ∠BCD[Angle sum property of A]
=> ∠BCD = 90°
Hence, ∠BCD = 90°.

Solution 17.

Given: A ∆ABC in which D and E are the mid points of AB and AC respectively, DE is joined.
To prove: DE || BC and DE = \(\frac { 1 }{ 2 }\)BC
Construction: Draw CF || BA, meeting DE produced in F.
Proof: In ∆AED and ∆CEF
AE = CF (given)
∠AED = ∠CED (V.O.A.)
∠DAE = ∠FCE (alternate interior angles) BD || CF
∴ ∆AED ≅ ∆CEF (By ASA congruency)
=> AD = CF (CPCT)
But AD = BD
=> BD = CF and BD || CF (by construction)
∴ BCFD is a parallelogram.∴DF || BC and DF = BC
=> DE || BC [∵ DF || BC]
DE = \(\frac { 1 }{ 2 }\) DF = \(\frac { 1 }{ 2 }\) BC [∵ DF = BC]
DE || BC and DE = \(\frac { 1 }{ 2 }\) BC
OR
Given: Two parallelograms ABCD and ABEF on the same base AB and between the same parallel lines AB and FC.

Solution 18.
In ∆ACF, side CF is produced to B
∴ ∠y = ∠1 + ∠3 …(i)
[exterior angle = sum of interior opposite angles]
In ∆AED, side ED is produced to B
∠1 + ∠x = ∠4 …(ii)
Adding eq. (i) and (ii)
∠1 + ∠x + ∠y = ∠1 + ∠3 + ∠4
⇒∠x + ∠y = ∠3 + ∠4 [∵∠4 = ∠3 angles in the same segment ]
⇒∠x + ∠y = 2∠3 [∵∠AOB = 2∠ACB]
∠x + ∠y = ∠z
Hence ∠x + ∠y = ∠z

Solution 19.
Steps of Construction:
(i) Draw BC = 6 cm
(ii) Construct ∠CBX = 60°
(iii) Along BX, set off BP = 9 cm
(iv) Join CP
(v) Draw the perpendicular bisector of CP to intersect BP at A.
(vi) Join AC, then ∆ABC is the required triangle.

Solution 20.
(i) (3, -3) is the mirror image of (3, 3) in x-axis.
(ii) No mirror image in y-axis.

Solution 21.
Let the outer and inner radii be R cm and r cm respectively.
h = 14 cm, outer surface area = 2πRh = 2 x \(\frac { 22 }{ 7 }\) x R x 14 = (88R) cm²

Solution 22.
TSA of cylinder = 2πrh + 2πr² = 231 …(1)
CSA of cylinder = 2πrh = \(\frac { 2 }{ 3 }\) of (TSA of cylinder)

Solution 23.

Solution 24.
Let f(x) = Px² + 5x + r
If (x – 2) is a factor of P(x), then remainder = 0
x – 2 = 0 => x = 2
Remainder = f(2) = P(2)² + 5x² + r = 0
4P + 10 + r = 0
4P + r = -10 ..(i)

Solution 25.
Let x ml of the pure alcohol be added to be 15% solution to get 32% alcoholic solution. New volume of alcoholic solution = (800 + x) ml
Quantity of pure alcohol in (800 + x) ml of 32% solution = Quantity of pure alcohol in 800 ml of 15% solution + x ml of pure alcohol

Solution 26.
∠QOS = ∠ROQ + ∠ROS …(i)
∠POS = ∠POR – ∠ROS …(ii)
Subtracting eqn. (i) by (ii)
∠QOS – ∠POS = ∠ROQ – ∠POR + 2∠ROS
∠QOS – ∠POS = (∠ROQ – ∠ROQ) + 2∠ROS (∠ROQ = ∠POR = 90°)
=> 2∠ROS = ∠QOS – ∠POS
=> ∠ROS = \(\frac { 1 }{ 2 }\) [∠QOS – ∠POS]

Solution 27.

(i) ∠DCA = ∠ECB
=> ∠DCA + ∠DCE = ∠ECB + ∠DCE
=> ∠ACE = ∠BCD
Now in AAEC and ABCD
∠DBC = ∠EAC (given)
AC = BC (C is the mid point)
∠ACE = ∠BCD (Proved)
∴ ADBC ≅ AEAC (ASA congruency)
(ii) ADBC ≅ AEAC
DC = EC (CPCT)

Solution 28.

In ∆ABC, F and E are the mid-points of AB and AC respectively.
∴FE || BC and FE = \(\frac { 1 }{ 2 }\) BC = BD
∴FE || BD and FE = BD [∵FE||BC => FE||BD]
=> BDEF is a parallelogram whose diagonals BE and DF intersect each other at X.
∴X is the midpoint of DF.
Similarly, Y is the mid point of DE.
Thus, in ∆DEF, X and Y are the midpoints of DF and DE respectively.
So, XY || FE

Solution 29.
In the given frequency distribution, class sizes are different. So, to calculate the adjusted frequency for each class, Minimum class size = 5

Solution 30.
Required tarpaulin = Area of shelter = Curved surface area of cuboid + Top area
= 2 (l + b) x h + lb
= 2(4 + 3) x 2.5 + 4 x 3
= 35 + 12
= 47 m²
Values: (i) Care for animal, (ii) Kindness.

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