CBSE Sample Papers for Class 9 Maths Paper 2 is part of CBSE Sample Papers for Class 9 Maths . Here we have given CBSE Sample Papers for Class 9 Maths Paper 2

## CBSE Sample Papers for Class 9 Maths Paper 1

 Board CBSE Class IX Subject Maths Sample Paper Set Paper 2 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 2 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 √2 = 1.414 , then $$\sqrt { \frac { (\sqrt { 2 } -1) }{ (\sqrt { 2 } +1) } } =$$ ?

Question 2.
If $$\frac { x }{ y } +\frac { y }{ x } =-1$$, where x ≠ 0,y ≠ 0, then find the value of x3 – y3.

Question 3.
Who is known as the “Father of Geometry”?

Question 4.
Find the area of the ∆OAB with 0(0, 0). A(4, 0) and B(0, 6).

Question 5.
Two cubes have their volumes in the ratio 1 : 27. Find the ratio of their surface area.

Question 6.
In 50 tosses of a coin, tail appears 32 times. If a coin is tossed at random, what is the probability of getting a head?

SECTION-B

Question 7.
If a + b + c = 0, then $$\left( \frac { { a }^{ 2 } }{ bc } +\frac { { b }^{ 2 } }{ ca } +\frac { c^{ 2 } }{ ab } \right) =?$$

Question 8.
Find the measure of an angle, if six times its complement is 12° less than twice its supplement.

Question 9.
If an angle of a parallelogram is two third of its adjacent angle, find the smallest angle of the parallelogram.

Question 10.
In the adjoining figure, the point D divides the side BC of ∆ABC in the ratio m : n. Prove that $$\frac { ar(\Delta ABD) }{ ar(\Delta ADC) } =\frac { m }{ n }$$ Question 11.
The sides of a triangle are in the ratio 5:12:13 and its perimeter is 150 cm. Find the area of triangle.

Question 12.
Find the mean of the first six prime numbers.

SECTION-C

Question 13.
Rationalise the denominator of $$\frac { \left( \sqrt { 3 } +\sqrt { 2 } \right) }{ \left( \sqrt { 3 } -\sqrt { 2 } \right) }$$
OR
If a and b are rational numbers and $$\frac { \left( \sqrt { 11 } -\sqrt { 7 } \right) }{ \left( \sqrt { 11 } +\sqrt { 7 } \right) } =a-b\sqrt { 77 }$$
Find the value of a and b.

Question 14.
Factorize (p – q)3 + (q – r)3 + (r – p)3

Question 15.
Draw the graph of the equation 2x + 3y = 11. From your graph, find the value of y, when x = -2.

Question 16.
In a quadrilateral ABCD the line segment bisecting ∠C and ∠D meet at E.
Prove that ∠A + ∠B = 2∠CED.

Question 17.
In the adjoining figure, BM ⊥ AC and DN ⊥ AC. If BM = DN, then prove that AC bisect BD. Question 18.
Without plotting the given points on a graph paper, indicate the quadrants in which they lie
(a) ordinate = 6, abscissa = -3
(b) ordinate = -6, abscissa = 4
(c) abscissa = -5, ordinate = -7
(d) ordinate = 3, abscissa = 5
(e) (-6, 5)
(f) (2,-9)

Question 19.
A well of inner diameter 14 m is dug to a depth of 15 m. Earth taken out of it has been evenly spread all around it to a width of 7 m to form an embankment. Find the height of embankment so formed.

Question 20.
Find the number of coins 1.5 cm in diameter and 0.2 cm thick to be melted to form a right circular cylinder of height 5 cm and diameter of 4.5 cm.

Question 21.
The table given below shows the ages of 75 teachers in a school. Here 18-29 means from 18 to 29 including both. A teacher from the school is chosen at random. What is the probability that the teacher chosen is
(i) 40 years or more than 40 years old?
(ii) 49 years or less than 49 years old?
(iii) 60 years or more than 60 years old?

Question 22.
Construct a ∆ABC in which BC = 5.6 cm, ∠B = 30° and the difference between the other two sides is 3 cm.

SECTION D

Question 23.
If a and b are rational numbers and $$\frac { 4+3\sqrt { 5 } }{ 4-3\sqrt { 5 } } =a+b\sqrt { 5 }$$ Find the value of a and b.

Question 24.
If x + y + z = 0, prove that x3 + y3 + z3 = 3xyz. Without actual calculation find the value of (-12)3 + 73 + 53 using above identity.

Question 25.
A taxi charges RS 20 for the first km and @ RS 12 per km for subsequent distance covered. Taking ‘the distance covered as x km and total fare RS y, write a linear equation depicting the relation in ,x and y
(i) Draw the graph between x and y.
(ii) From your graph find the taxi charges for covering 14 km.

Question 26.
In the given figure, AB || CD. Find the value of x. Question 27.
In a ∆ABC, the bisectors of ∠B and ∠C intersect each other at a point O. Prove that ∠BOC = 90° + $$\frac { 1 }{ 2 }$$∠A

Question 28.
Prove that the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Question 29.
The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to the base. If its volume be $$\frac { 1 }{ 27 }$$ of the volume of the given cone, at what height above the base, the section has been made?

Question 30.
A doctor suggests two ways for treatment of a particular disease one is by taking medicine only and other is by doing meditation and yoga. (i) Draw frequency polygon for the given data on the same graph.
(ii) What is the importance of yoga and meditation in our life?

Solutions

Solution 1. Solution 2.
$$\frac { x }{ y } +\frac { y }{ x } =-1$$ ⇒ x² + y² = -xy
⇒ x² + y² + xy = 0
⇒ x3 – y3 = (x – y)(x² + y² + xy) = (x – y) x 0 = 0
⇒ x3 – y3 = 0

Solution 3.
Mathematician Euclid is known as “Father of Geometry’”.

Solution 4.
OA = 4 units, OB = 6 units
ar(∆OAB) = $$\frac { 1 }{ 2 }$$ x OA x OB = $$\frac { 1 }{ 2 }$$ x 4 x 6 = 12 sq. unit Solution 5.
Volume of cube = (side)3, surface area of cube = 6(side)2
Let the volume of first cube = a3; and second cube = b3  Solution 6. Solution 7.
a + b + c = 0 ⇒ a3 + b3 + c3 = 3abc Solution 8.
Let the measure of the required angle = x°
Measure of its complement = (90° – x).
Measure of its supplement = (180° – x)
6(90° – x) = 2(180° – x) – 12
⇒ 540° – 6x = 360° – 2x – 12 ⇒ 4x = 192°
⇒ x = 48°
Hence the measure of required angle = 48°.

Solution 9.
Let one angle of the parallelogram be x°.
We know that sum of two adjacent angles of a parallelogram is equal to 180°. Solution 10. Solution 11.
Let the sides of the triangle be 5x, 12x and 13x as a, b and c. Then
5x + 12x + 13x = 150 => 30x =150 => x = 5
∴ a = 25 cm, b = 60 cm and c = 65 cm
given 2s = 150 cm => s = 75 cm
(s – a) = 50, (s – b)= 15, (s – c) = 10 Solution 12.
The first six prime numbers are 2, 3, 5, 7, 11 and 13. Solution 13.  Solution 14.
Putting p – q = x,q – r = y and r – p = z
x + y + z = p – q + q – r + r – p = O if x + y + z = 0 then x3 + y3 + z3 = 3xyz
∴ (p – q)3 + (q – r)3 + (r – p)3 = 3(p – q)(q – r)(r – p)

Solution 15.
2x + 3y = 11  Line AB is the required graph of 2x + 3y= 11.
Reading the Graph: Given x = -2. Take a point M on the x-axis such that OM = -2.
Draw PM, parallel to they-axis, meeting the line AB at P, it is PM = 5 = y.
So, when x = -2 then y = 5.

Solution 16.
Let CE and DE be the bisector of ∠C and ∠D respectively. Solution 17.
We want to prove that AC bisect BD i.e., OB = OD
Let AC and BD intersect at O and BM = DN
Now ∴ ∆OND ≅ ∆OMB (By ASA)
=> OD = OB (CPCT)
[ ∵ ∠OND = ∠OMB (each 90°), DN = BM (given), ∠DON = ∠BOM (V.O.A.)]
=> AC bisect BD.

Solution 18.
(a) A(-3, 6) lies in quadrant II
(b) B(4, -6) lies in quadrant IV
(c) C(-5, -7) lies in quadrant III
(d) D(5, 3) lies in quadrant I
(e) E(-6, 5) lies in quadrant II
(f) F(2, -9) lies in quadrant IV

Solution 19.
For well: r = 7m, h = 15m = depth
Volume of the earth dug out = Volume of the well
= Volume of cylinder = πr²h  Solution 20.
Each coin is cylindrical in shape. = 45 x 5 = 225
The number of coins required = 225

Solution 21.
Total number of teachers = 75
(i) Number of teachers who are 40 years or more than 40 years old = 35 + 10 = 45 (ii) Number of teachers who are 49 years or less than 49 years old = 5 + 25 + 35 = 65 (iii) Number of teachers who are 60 years or more = 0 Solution 22.
Steps of Construction:
(i) Draw BC = 5.6 cm
(ii) Construct ∠CBX = 30°
(iii) Set off BP = 3 cm
(iv) Join PC
(v) Draw the right bisector of PC, meeting BP produced at A.
(vi) Join AC. Then ∆ABC is the required triangle. Solution 23. Solution 24.
x + y + z = 0 => x + y = – z => (x+y)3 = (-z)3
=> x3 + y3 + 3xy (x + y) = -z3
=> x3 + y3 + 3xy (-z) = -z3
=> x3 + y3 – 3xyz = -z3
=> x3 + y3 + z3 – 3xyz = 0
=> x3 + y3 + z3 = 3xyz
Putting x = -12, y = 7, z = 5
We get x + y + z = -12 + 7 + 5 = -12 + 12 = 0 then x3 + y3 + z3 = 3xyz
=> (-12)3 + (7)3 + (5)3 = 3 x (-12) x 7 x 5
= -3 x 7 x 60
= -21 x 60
= -1260
(-12)3 + 73 + 53 = -1260.

Solution 25.
y = 20 + 12 (x – 1) = 20 + 12x – 12
=> y = 12x + 8 On the graph paper take distance along x-axis and fare (in Rs) along y-axis. We plot the points A(6, 80) and B(11, 140) on the graph paper. Join AB and produce it on both sides to obtain the required graph.
From the graph, when x = 14, we find y = Rs 176. Solution 26. Through E draw a line GEH || AB || CD
Now GE || AB and EA is transversal
∴ ∠GEA = ∠EAB = 50°
[Alternate Interior Angles]
Again EH || CD and EC is a transversal
∠HEC + ∠ECD = 180°
[Cointerior angles of the same side of transversal]
=> ∠HEC + 100° = 180°
=> ∠HEC = 80°
Now GEH is a straight angle
∠GEA + ∠AEC + ∠HEC = 180°
=> 50° + x° + 80° =180° => x = 50°

Solution 27.
In ∆ABC we have
∠A + ∠B + ∠C = 180° Solution 28.
Given: A circle C(O, r) in which arc AB subtends ∠AOB at the centre and ∠ACB at any point C on the remaining part of the circle.
To prove: (i) and (ii) ∠AOB = 2 ∠ACB, when AB is a minor arc or a semicircle.
(iii) Reflex ∠AOB = 2∠ACB, when AB is a major arc.
Construction: Join AB and CO. Produce CO to a point D outside the circle. We know that when one side of a triangle is produced then the exterior angle so formed is equal to the sum of the interior opposite angles.
∴ ∠AOD = ∠OAC + ∠OCA
∠BOD = ∠OBC + ∠OCB
[∵ OC = OA = r ,OC = OB = r]
∠AOD = 2∠OCA and ∠BOD = 2∠OCB
In Figure (i) and Figure (ii) i.e., Case (i) and Case (ii)
∠AOD + ∠BOD = 2∠OCA + 2∠OCB => ∠AOB = 2(∠OCA + ∠OCB)
=> ∠AOB = 2∠ACB
In Figure (iii) i.e., Case (iii)
∠AOD + ∠BOD = 2∠OCA + 2∠OCB
=> Reflex ∠AOB = 2 (∠OCA + ∠OCB) => Reflex ∠AOB = 2∠ACB

Solution 29.
Let the smaller cone have radius r and height h cm and let the radius of the given original cone be R cm.  Solution 30.
(i) We take two imagined classes, one at the beginning, namely (10-20) and the other at the end, namely (70-80) each with frequency zero.
With these two classes, we have the following frequency table: Importance of Yoga and Meditation:
1. Meditation gives you an experience, an inner state, where what is you and what is your is separated.
2. By meditation life seems to be flowing without conflict and just feel more cohesive.
3. Yoga is important for us to achieve:
(i) Good Health
(ii) Success
(iii) Overall well being
(iv) For peace, joy and love
(v) For inner exploration.

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