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 2

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.

Express 0.99999 …. in the form of \(\frac { p }{ q }\), where p and q are integers and q ≠ 0.

Question 2.

Prove that zeroes of 2x^{3} – 3x^{2} + 7x – 6 is 1

Question 3.

In ∆ABC, if ∠A = 45° and ∠B = 70°, then find smallest and longest sides of triangle.

Question 4.

In the given figure, ABCD is a rhombus. If ∠A = 70°, then find ∠CDB.

Question 5.

The dimensions of a cuboid are a, b, c units, its volume is V cubic units and its whole surface area is S sq. units. Then find \(\frac { 1 }{ V }\).

**SECTION-B**

Question 7.

Find the value of 1.5^{3} – 0.9^{3} – 0.6^{3}

Question 8.

If the ratio of two supplementary angles are 2 : 3, find the angles.

Question 9.

Write whether the following statement are true or false? Justify your answer.

(i) Point (0, -2) lies on y-axis.

(ii) (-1, 7) is a point in the II quadrant.

Question 10.

(i) In which quadrant or on which axis does the point (0, -8) lie?

(ii) What is the name of each part of the plane formed by horizontal and vertical lines in Cartesian plane?

Question 11.

A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.

Question 12.

Find the mean of first five prime numbers.

**SECTION-C**

Question 13.

Prove that

Question 14.

Question 15.

Find the value of ‘a‘ and ‘b’ for which (x – 1) and (x + 3) are two factors of polynomial x^{3} – ax^{2} – 13x + b.

Question 16.

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

Question 17.

In figure, O is the centre of circle. Prove that

∠XOZ = 2 (∠XZY + ∠YXZ)

Question 18.

Prove that the sum of angles of a triangle is 180°.

Question 19.

A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.

Question 20.

Construct a ∆ABC in which BC = 7 cm, ∠B = 75° and AB + AC = 13 cm.

Question 21.

1500 families with 2 children were selected randomly, and following data were recorded.

Compute the probability of a family, chosen at random having

(i) 2 girls

(ii) 1 girl

(iii) No girl

Question 22.

If h, C and V are the height, the curved surface area and volume of the cone. Prove that 3πVh^{3} – C²h² + 9V² = 0

**SECTION-D**

Question 23.

The difference between the outside and inside surfaces of a cylindrical metallic 14 cm long pipe is 44 cm². If the pipe is made of 99 cm^{3} of metal, find the outer and inner radii of the pipe.

Question 24.

Simplify

Question 25.

Show √9.3 on number line. Also verify it.

Question 26.

OR

Pallavi participates in Diwali Mela with her friends for the charity to centre of handicapped children. They donate Rs 3600 to the centre from the amount earned in Mela. If each girl donates Rs 150 and each boy donates Rs 200.

(i) Form the linear equation in two variables.

(ii) If number of girls is 8, find number of boys.

(iii) What values of Pallavi and her friends have depicted here?

Question 27.

The medians of a triangle ABC intersect at G. Show that ar (∆BGC) = \(\frac { 1 }{ 3 }\) ar (∆ABC).

Question 28.

The following observations have been arranged in ascending order. If the median of the data is 63. Find the value of x.

29, 32,48, 50, x,x + 2, 72, 78, 84, 95

Question 29.

In figure, BC is a chord of circle. If A is a point on arc BC, then prove that

(i) ∠BAC + ∠OBC = 90° if A is on major arc.

(ii)∠BAC – ∠OBC = 90° if A is on minor arc.

Question 30.

ABCD is a trapezium in which side AB is parallel to DC and E is the mid-point of side AD. If F is the point on BC such that the line segment EF is parallel to DC, then prove that EF = \(\frac { 1 }{ 2 }\) (AB + DC)

**Solutions**

Solution 1.

Let x = 0.9999 …. _(1)

10x = 9.9999 …. _(2)

Subtracting Eqn. (1) by Eqn. (2)

10x – x = (9.9999….) – (0.9999…)

9x = 9 ⇒ \(x=\frac { 9 }{ 9 }=1\) , x = 1

Solution 2.

Let f(x) = 2x^{3} – 3x^{2} + 7x – 6

Putting x = 1, f(1) = 2 x (1)^{3} – 3 x (1)^{2} + 7 x 1 – 6

f(1) = 2 – 3 + 7 – 6 = 9 – 9 = 0

Hence x = 1,is zero of f(x)

Solution 3.

∠A = 45°,∠B = 70°, ∠A + ∠B + ∠C = 180°

∠C= 180° – ∠A – ∠B = 180° – 45° – 70° = 65°

(i) Because the side opposite to the largest angle is longest. So longest side is AC (opposite to ∠B)

(ii) The side opposite to the smallest angle is smallest so BC is the smallest side (opposite to ∠A).

Solution 4.

Let ∠CDB = x°

Then CD = CB

=> ∠CBD = x°

∠BCD = ∠BAD = 70°

(Opposite angles of rhombus)

∴ x + x + 70° = 180° (sum of angles of ∆ equal to 180°)

2x = 110°

x = 55° ⇒ ∠CDB = 55°

Solution 5.

Solution 7.

Let a = 1.5, b = -0.9, c = – 0.6

a + b + c = 1.5 – 0.9 – 0.6 = 1.5 – 1.5 = 0

∴ a^{3} + b^{3} + c^{3} = 3 abc

(1.5)^{3} + (-0.9)^{3} + (-0.6)^{3} = 3 x 1.5 x (-0.9) x (-0.6)

⇒ (1.5)^{3} – 0.9^{3} – 0.6^{3} = 2.430

Solution 8.

Let the measure of two angles be 2x° and 3x°.

So sum of two supplementary angles = 180°

2x + 3x = 180°

5x = 180° => x = 36°

Required angles 2x = 2 x 36° = 72°

3x = 3 x 36° = 108°

Solution 9.

(i) True. The point (0, -2) lies on y-axis, because the coordinates of y-axis is (0, b).

(ii) True. (-1, 7) is a point in the II quadrant, because the coordinates of II quadrant are (-x,y) or (-,+).

Solution 10.

(i) The point (0, -8) lies on the y-axis, because the coordinates of y-axis is (0, b). This point lies on y-axis below the origin at point 8.

(ii) The name of each part of the plane formed by horizontal line (x-axis or abscissa) and vertical line (y-axis or ordinate) are quadrants.

(i) I quadrant => XOY

(ii) II quadrant => X’OY

(iii) III quadrant => X’OY’

(iv) IV quadrant => XOY’

Solurion 11.

When a right ∆ABC revolves about its side AB =12 cm. The solid thus obtained is single cone having the base radius, r = 5 cm and height of the cone h = 12 cm.

Volume of cone = \(\frac { 1 }{ 2 }\) πr²h unit^{3}

V = \(\frac { 1 }{ 2 }\) x π x (5)² x 12

V = 100π cm^{3}

Hence, the volume of solid cone = 100π cm^{3}.

Solution 12.

The first five prime numbers are 2, 3, 5, 7, 11

Hence, the required mean of first 5 prime numbers is 5.6.

Solution 13.

Solution 14.

Solution 15.

Let p(x) = x^{3} – ax^{2} – 13x + b

If (x – 1) and (x + 3) are factors of p (x), then remainders

p (1) = 0 and p (-3) 0

p(1) = 1^{3} – a x 1^{2} – 13 x 1 + b = 0

=> 1 – a – 13 + b = 0

Solution 16.

Given: In quadrilateral ABCD, CD is the longest and AB is the smallest sides.

To prove: (i) ∠A > ∠C (ii) ∠B > ∠D

Construction: Join BD and AC.

Proof: In ∆ABC

BC > AB [∵ AB is the smallest side]

=> ∠BAC > ∠BCA …(i)

In ∆ACD, CD > AD [∵ CD is the longest side]

=> ∠CAD > ∠ACD …(ii)

Adding Eqn. (i) & (ii)

∠BAC + ∠CAD = ∠BCA + ∠ACD

=> ∠BAD > ∠BCD

=> ∠A > ∠C

In ∆ABD AD > AB [∵AB is the smallest side]

=> ∠ABD > ∠ADB …(iii)

In ∆BCD CD > BC

=> ∠DBC > ∠BDC …(iv) [∵CD is the longest side]

Adding Eqn. (iii) & (iv)

∠ABD + ∠DBC > ∠ADB + ∠BDC

=> ∠ABC > ∠ADC

=> ∠B > ∠D

So ∠A > ∠C and ∠B > ∠D.

Solution 17.

In circle C (O, r); join O, Y.

In circle C (O, r), subtends ∠XOY at the centre O and ∠XZY at a point Z on the remaining part of the circle.

By Central Angle theorem

∠XOY = 2 ∠XZY ….(1)

Similarly, ∠YOZ = 2 ∠YXZ …(2)

Adding Eqn. (1) & (2)

∠XOY + ∠YOZ = 2 (∠XZY + ∠YXZ)

=> ∠XOZ = 2 (∠XZY + ∠YXZ)

Hence proved.

Solution 18.

Given: In ∆PQR, ∠1, ∠2 and ∠3 are the angles of ∆PQR.

To prove: ∠1 + ∠2 + ∠3 = 180°

Construction: We draw a line XPY parallel to QR through the opposite vertex P.

Proof: Line XPY || QR and XPY is a line

Therefore ∠4 + ∠1 + ∠5 = 180° …(1)

But XPY || QR and PQ, PR are transversal.

So ∠4 = ∠2 and ∠5 = ∠3

(Pairs of alternate angles)

Putting values of ∠4 and ∠5 in Eqn. (1)

∠2 + ∠1 + ∠3 = 180°

=> ∠1 + ∠2 + ∠3 = 180°

Solution 19.

Draw CE || DA and DC || AE in such a way that AD = 13 m, and DC = 10 m.

=> ADCE is a parallelogram.

=> AD || CE and DC || AE and AD = 13 m, DC = 10 m

∴ AE = DC = 10 m, CE = AD = 13 m

=> BE = AB – AE = 25 – 10 = 15 m

In ∆BCE

BC = 14 m, CE =13 m, BE = 15 m

Area of trapezium ABCD = Area of parallelogram + ar (∆BCE)

= (112 + 84)m² = 196 m²

Area of trapezium ABCD = 196 m²

Solution 20.

Steps of Construction:

1. Draw a line segment BC = 7 cm and at point B, make ∠CBX = 75°.

2. Cut off BD = 13 cm from the ray BX and join D and C.

3. Draw the perpendicular bisector PQ on CD to intersect BD at the point A. Join A,C.

4. ∆ABC is the required triangle.

5. Here AB + AC = 13 cm.

Solution 21.

Total number of families = 475 + 814 + 211 = 500

Solution 22.

Let l be the slant height and r be the radius of the cone.

Solution 23.

Length of a cylindrical pipe, h = 14 cm.

Let inner radius of pipe = 4 cm.

Outer radius of pipe = R cm

Then, outer surface area of pipe = S1 = 2πRh unit²

Inner surface area of pipe = S2 = 2πrh unit²

Difference between outer and inner surfaces

Hence, the outer radius = 2.5 cm

The inner radius = 2 cm

Solution 24.

(a² – b²) + (b² – c²) + (c² – a²) = 0

∴ (a² – b²)^{3} + (b² – c²)^{3} + (c² – a²)^{3} = 3 (a² – b²) (b² – c²) (c² – a²)

=> (a² – b²)^{3} + (b² – c²)^{3} + (c² – a²)^{3} = 3 (a – b) (a + b) (b – c) (b + c) (c – a) (c + a)

Similarly, (a – b) + (b – c) + (c – a) = 0

(a – b)^{3} + (b – c)^{3} + (c – a)^{3} = 3 (a – b) (b – c)(c – a)

Solution 25.

Steps of Construction:

1. Draw a line and mark a point A on it.

2. Draw AB = 9.3 cm.

3. On line segment AB take a point C such that BC = 1 cm.

4. Find the mid-point of AC.

5. Draw a semicircle having radius OA = OC. Draw a line segment perpendicular from point B which cuts semicircle at D. BD = √9.3

6. Draw an arc of radius BD = √9.3 on line which cuts in at E.

7. Point E represents √9.3 on number line.

Verification: Let x = 9.3 = a positive number.

Solution 26.

We know that

(i) Let number of girls = x

Donation by each girl = Rs 150

Number of boys = y

Donation by each boy = Rs 200

Total collection = 3600

A.T.Q.

Number of girls x donation by each girl + number of boys x donation by each boy

= Total collection

x × 150 + y × 200 = 3600

150x + 200y = 3600

(ii) 150x + 200y = 3600

(iii) Values depicted here:

(1) Co-operation

(2) Sincerity

(3) Concerned

(4) Helpfulness.

Solution 27.

Given: A ∆ABC in which the median AD, BE and CF intersected at G.

To prove: ar (∆BGC) = \(\frac { 1 }{ 3 }\) ar(∆ABC)

Proof: In ∆ABC,

=> ar (∆ABD) = ar (∆ACD) …(1)

In ∆GBC,

=> ar (∆GBD) = ar (∆GCD) …(2)

Subtracting Eqn. (2) from Eqn. (1)

ar (∆ABD) – ar (∆GBD) =ar (∆ACD) – ar (∆GCD)

=> ar (∆AGB) = ar (∆AGC) …(3)

Similarly ar (∆AGB) = ar (∆BGC) …(4)

From Eqn. (3) & (4)

ar (∆AGB) = ar (∆BGC) = ar (∆AGC)

ar (∆ABC) = ar (∆AGB) + ar (∆BGC) + ar (∆AGC)

= ar (∆BGC) + ar (∆BGC) + ar (∆BGC)

ar (∆ABC) = 3 x ar (∆BGC)

ar (∆BGC) = \(\frac { 1 }{ 3 }\) ar (∆ABC)

Solution 28.

The given data is arranging in ascending order

Solution 29.

The arc BC for the angle ∠BOC = z on the centre of circle and ∠BAC = x.

z = 2x

(i) In ∆BOC, ∠OBC + ∠OCB + ∠BOC = 180° (Angle sum property of ∆)

⇒ y + y + z = 180°

⇒ 2y + z = 180° ⇒ 2y + 2x = 180° (∵z = 2x)

⇒ x + y = 90°

∠BAC + ∠OBC = 90°

(ii) In ∆BOC, ∠OBC + ∠OCB + ∠BOC = 180° (Angle sum property of ∆)

⇒ y + y + t = 180°

Solution 30.

Trapezium ABCD in which AB || DC and E is the mid-point of AD.

To Prove: EF = \(\frac { 1 }{ 2 }\)(AB+DC)

Proof: In ∆ADC, E is the mid-point of AD and EG || DC

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