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.

If a = 2 and b = 3, find the value of ({ left( frac { 1 }{ a } +frac { 1 }{ b } right) }^{ a })

Question 2.

Find the value of (frac { { left( 2.3 right) }^{ 3 }-0.027 }{ { left( 2.3 right) }^{ 2 }+0.69+0.09 } )

Question 3.

Two complementary angles are in such a way that twice of one angle is thrice the another. Find the smaller angle.

Question 4.

In a parallelogram ABCD, if ∠DAB = 75°, ∠DBC = 60° then find ∠CDB.

Question 5.

If the areas of three adjacent faces of a cuboid are x, y, z respectively, then find the volume of cuboid.

Question 6.

Find the mean of first five natural numbers.

**SECTION-B**

Question 7.

If a² + b² + c² = 250 and ab + bc + ca = 3, then find the value of a + b + c.

Question 8.

In the given figure, if x + y = w + z, then prove that AOB is a straight line.

Question 9.

Name the quadrant in which the following points lie.

(i) (1, 1), (ii) (3,4), (iii) (-2, 4), (iv) (-3, -7)

Question 10.

Plot the points A, B, C, D on the coordinate axes and name the figure formed by joining the points in order.

Question 11.

If right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 5 cm, then find the volume of the solid so formed.

Question 12.

For what value of x, 7 is the mode of the following data:

3, 5, 6, 7, 5,4, 7, 5, 6, x, 8, 7

**SECTION-C**

Question 13.

Question 14.

If a² + b² + c² – ab – bc – ca = 0, then prove that a = b = c.

Question 15.

Prove that

Question 16.

In the given figure, if AB || CD and EF || DQ, then find the value of angles ∠PDQ, ∠AED and ∠DEF.

Question 17.

Prove that angles opposite to equal sides of an isosceles triangle are equal.

Question 18.

If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

OR

Prove that an isosceles trapezium is always cyclic.

Question 19.

Construct a ∆XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm. How will you verify your answer?

Question 20.

The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs 10 per m² is Rs 15,000, find the height of the hall.

Question 21.

Fill in the blanks:

1. Probability of Sure Event =___

2.Probability of Impossible Event =____

3.Probability of an Event exist between___and 1.

Question 22.

Find the area of triangle two sides of which are 8 cm and 11 cm and the perimeter is 32 cm.

**SECTION-D**

Question 23.

If (x=frac { sqrt { 3 } +1 }{ 2 } ) , then find the value of 4x^{3} + 2x^{2} – 8x + 7.

Question 24.

If p = 2 – a, then prove that a^{3} + 6ap + p^{3} – 8 = 0.

Question 25.

In a residential society, rain water is stored in underground water tank. If the water stored at the rate of 30 cubic cm per second and water stored in ‘x’ seconds and ‘y’ cubic cm, then

(i) Write this statement in linear equation in two variables.

(ii) Write this equation in the form of ax + by + c = 0.

(iii) What value of the society members shows in Rain Water Storage?

Question 26.

Prove that the sum of three angles of a triangle is 180°. What is the name of this property of triangle? Using this property if the angle bisectors OB and OC of ∠ABC and ∠ACB of ∆ABC meet at O, then prove that ∠BOC = 90° + (frac { 1 }{ 2 })∠A.

Question 27.

AB and CD are two parallel lines. A transversal line l intersects line AB at X and CD at Y. Prove that the bisectors of interior angles so formed is a parallelogram in which all angles are right angles i.e., it is a rectangle.

OR

Two parallel lines AB and CD are intersected by a transversal l. Show that the quadrilateral formed by the bisectors of interior angles is a rectangle.

Question 28.

A circle with centre O and radius 5 cm have two chords AB and CD which are parallel to each other. If OP ⊥ AB, OQ ⊥ CD, AB = 8 cm and CD = 6 cm, then find the length PQ.

Question 29.

A wooden bookshelf has external dimensions as follows: Height =110 cm, Depth = 25 cm, Breadth = 85 cm. (see figure) The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is 20 paise per cm² and the rate of painting is 10 paise per cm2, find the total expenses required for polishing and painting the surface of the bookshelf.

Question 30.

If observations x_{1}, x_{2}, x_{3} … x_{n} for the variable x has n values is such that (sum _{ i=1 }^{ n }{ left( { x }_{ i }-2 right) =110 } ) and (sum _{ i=1 }^{ n }{ left( { x }_{ i }-5 right) =20 } ) then find the value of n and its mean.

**Solutions**

Solution 1.

Solution 2.

Solution 3.

2x = 3(90° – x) ⇒ 2x = 270° – 3x

⇒ 2x + 3x = 270° ⇒ 5x = 270° ⇒ (x=frac { { 270 }^{ o } }{ 5 } ={ 54 }^{ o })

Smaller angle = 90° – x = 90° – 54° = 36°

Smaller angle = 36°

Solution 4.

In ||gm, ∠DCB = ∠DAB = 75°

(Opposite angles of parallelogram are equal)

In ∆DCB, ∠DBC + ∠DCB + ∠CDB = 180°

60° + 75° + ∠CDB = 180°

∠CDB = 180 ° – 135° = 45°

∠CDB = 45°

Solution 5.

Let for cuboid, length = l, breadth = b, height = h

lb = x, bh = y and hl = z

xyz = (lb) x (bh) x (hl) = l²b²h² = (lbh)²

Ibh = √xyz

Volume = √xyz

Solution 6.

Solution 7.

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

=> (a + b + c)² = 250 + 2 x 3 = 250 + 6 = 256

=> (a + b + c)² = (± 16)² =>a + b + c = ± 16

Solution 8.

The sum of angles all around a point is 360°.

so x + y + z + w = 360° [By angles sum property of ∆]

=> (x + y) + (z + w) = 360° [∵x+y=z+w]

=> (x + y) + (x + y) = 360°

=> 2(x + y) = 360°

=> x + y = 180°

=> AOB is a straight line.

Solution 9.

Solution 10.

Solution 11.

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

V = volume of cone = (frac { 1 }{ 3 }) πr^{2} h units^{3}

V = (frac { 1 }{ 3 }) x π x (12)^{2} x 5 = 240 π cm^{3}.

Hence, the required volume of the solid cone is 240 π cm^{3}.

Solution 12.

Here mode of the given data = M_{o} = 7

The frequency distribution of data (without) x is

Mode = the value corresponding to the height frequency in data

= 5 and 7 ( ∵ 3 is the highest frequency)

Since mode of the given data = 7

=> The frequency of occurrence of 7 should be more than that of 5 in the data.

Hence, x = 1 the required value of x in the data is 7.

Solution 13.

Solution 14.

a² + b²+ c² – ab – bc – ca = 0

Multiplying by 2 on both sides

2a² + 2b² + 2c² – 2ab – 2bc – 2ca = 0

=> (a² – 2ab + b²) + (b² – 2bc + c²) + (c² – 2ac + a²) = 0

=> (a – b)² + (b – c)² + (c – a)² = 0

[The sum of positive quantities will be zero only when, each quantity will be zero]

=> a – b = 0, b – c = 0, c – a = 0

=> a = b,b = c,c = a

=> a = b = c.

Solution 15.

Solution 16.

∵ AB || CD and transversal line intersect E and D respectively.

∴ ∠AED = ∠CDP

(corresponding angles)

=> ∠AED = 34°

Ray EF is inclined on line AB at point E

∴ ∠AEF + ∠BEF = 180°

=> (∠AEP + ∠PEF) + ∠BEF = 180° [ ∵ ∠AEF = ∠AEP + ∠PEF]

=> 34° + ∠PEF + 78° = 180°

=> ∠PEF = 180°- 112° = 68°

=> ∠DEF = 68° …(1)

∵ EF || DQ and transversal line DE intersects at E and D respectively.

∵ ∠FED = ∠PDQ = 68° [ ∵ ∠FED = ∠PEF = 68°]

=> ∠PDQ = 68°

So ∠PDQ = ∠DEF = 68°, ∠AED = 34°

Solution 17.

Given: In isosceles triangle ABC, AB = AC

To prove: ∠B = ∠C

Construction: Draw the bisector AD of ∠A, which meets BC at D.

In ∆BAD and ∆CAD

AB = AC (given)

∠BAD = ∠CAD (by construction)

AD = AD (common)

∆BAD ≅ ∆CAD (by SAS Rule)

=> ∠ABD = ∠ACD (CPCT)

=> ∠B = ∠C

Solution 18.

Given: A trapezium ABCD in which AB || CD and AD = BC

To prove: ∠A + ∠C = 180°

∠B + ∠D = 180°

Construction: Through C, draw CE || AD

Proof: In trapezium ABCD

∵AB || CD and CE || AD

By definition trapezium AECD is a parallelogram.

AD = EC (Opposite sides of a ||gm are equal) …(1)

∠AEC = ∠ADC (Opposite sides of a ||gm are equal) …(2)

AD = BC given …(3)

From equation (1) and (3)

BC = EC

=> ∠CBE = ∠CEB [angles opposite to equal sides of a triangle are equal]

∠AEC + ∠CEB = 180° (by linear pair axiom)

=> ∠ADC + ∠CBE = 180° [∠AEC = ∠ADC and ∠CBE = ∠CEB]

Similarly,

∠BAD + ∠DCB = 180°

By converse of cyclic quadrilateral theorem

□ABCD is cyclic.

Solution 19.

Steps of Construction:

1. Draw a line segment AB = 11 cm (XY + YZ + ZX= 11 cm)

2. Construct an angle of 30° at the point A.

3. Construct an angle of 90° at the point B.

4. Bisect these two angles at A and B respectively and let their bisectors meet at the point X.

5. Draw a perpendicular bisector CD of AX and EF of XB to intersect AB at point Y and Z respectively. Join X, Y and X, Z.

Thus, ∆XYZ is the required triangle.

Verification: Measure the sides XY, YZ and ZX of ∆XYZ. It is found that XY + YZ + ZX= 11 cm.

Solution 20.

Perimeter of the floor of hall = 250 m …(1)

Cost of painting the four walls = Rs 15000 … (2)

Let h be the required height, l and b are length and breadth of the rectangular hall.

∴ Area of four walls = A = 2(1 + b) x h = {2(1 + b) } x h = (Perimeter of the floor) x h

A = 250 h m²

Cost of painting of area 1 m² = Rs 10

Total cost of painting 250 h m² = (250 h x 10) = 2500 h.

2500 h = 15,000 => (h=frac { 15,000 }{ 2500 }) = 6m

Hence the required height of the hall = 6m

Solution 21.

1.1; 2.0; 3.0.

Solution 22.

Let the sides of the given triangle are a, b and c and its perimeter is 2s.

a = 8cm, b = 11cm, 2s = 32cm ⇒ (s=frac { 32 }{ 2 }=16cm)

a + b + c = 2s ⇒ 8 + 11 + c = 32

c = 32 -19 = 13 cm

s – a = 16 – 8 = 8cm

s – b = 16 – 11 = 5 cm

s – c = 16 – 13 = 3 cm

Solution 23.

Solution 24.

p = 2 – a => a + p – 2 = 0

∴ a^{3} + 6ap + p^{3} – 8 = (a)^{3} + (p)^{3} + (-2)^{3} – 3a x p x (-2)

a^{3} + 6ap + p^{3} – 8 = {a + p + (-2)} {a^{2} + p^{2} + (-2)^{2} – ap – p(-2) – a(-2)}

= (a + p – 2) (a^{2} + p^{2} + 4 – ap + 2p + 2a)

= 0 x (a^{2} + p^{2} + 4 – ap + 2p + 2a) = 0

So, a^{3} + 6ap + p^{3} – 8 = 0

Solution 25.

(i) If water is stored in x seconds, total quantity of water = y cm^{3} and the water stored in one second = 30 cm^{3}.

So, the water stored in one second x total seconds = total quantity of water

30 × x = y or y = 30x

(ii) y = 30x can be written in the form of ax + by + c = 0 as

30x – y + 0 = 0

(iii) Value depicted here:

(a) Environmental safety, (b) Cooperation, (c) Money saving

Solution 26.

To prove Angle sum property of triangle.

Given: A triangle ∆ABC.

To prove: ∠A + ∠B + ∠C = 180° or ∠1 + ∠2 + ∠3 = 180°

Construction: From point A, draw a line l || BC.

Proof: l || BC

∴∠2 = ∠4 and ∠3 = ∠5 [Alternate Interior Angles]

∴∠2 + ∠3 = ∠4 + ∠5

=> ∠1 + ∠2 + ∠3 = ∠1 + ∠4 + ∠5

[Adding l on both sides]

∠1 + ∠2 + ∠3 = ∠4 + ∠1 + ∠5 = 180°

∠1 + ∠2 + ∠3 = 180

[The sum of angles on a point of a line is 180° => Straight line angle]

∠A + ∠B + ∠C= 180°

The sum of three angles of triangle is 180°.

This is known as angle sum property of triangle.

(ii) Given: OB and OC are the angle bisectors of ∠ABC and ∠ACB of ∆ABC which meet at point O.

Solution 27.

Given: AB and CD are two parallel lines. A transversal line l cuts them at point X and Y. The bisectors of interior angles intersect at point P and Q.

To prove: XPYQ is a rectangle.

Proof: AB || CD and transversal line l intersects them.

∴ ∠AXY = ∠DYX

[Alternate Interior Angles]

Solution 28.

On joining OA and OC, we get two right triangles ∆OPA and ∆OQC.

∵ Perpendicular drawn from the centre of a circle to the chord bisects the chord.

∴ AP = PB = (frac { 1 }{ 2 }) AB = (frac { 1 }{ 2 }) x 8 = 4 cm

and CQ = DQ = (frac { 1 }{ 2 }) CD = (frac { 1 }{ 2 }) x 6 = 3 cm

In right triangle OPA

OA² = OP² + AP² =>OP² = OA² – AP²

OP² = 5² – 4² => OP² = 25 – 16 = 9 = (3)²

OP = 3 cm

In right triangle OQC,OC² = OQ² + CQ²

OQ² = OC² – CQ²

OQ² = 5² – 3²

OQ² = 25 – 9 = 16 = (4)²

OQ = 4 cm

PQ = OP + OQ = (3 + 4) = 7 cm

Solution 29.

Area of external faces to be polished = Total surface area of the cuboid – 3 x area of one rectangular (open) surface

= [2(85 x 110 + 110 x 25 + 85 x 25) – 3 x 30 x 75] cm^{2}

= [2(9350 + 2750 + 2125) – 6750] cm^{2}

= 21700 cm^{2}

Dimensions of inner boxes are 75 cm x 30 cm x 20 cm

∴ Area of 3 inner boxes to be painted

= 3 x [2(75 + 30) x 20 + 75 x 30] cm^{2}

= 3 [4200 + 2250] cm^{2}

= 19350 cm^{2}

Total expenses for polishing and painting

= Rs (4340 + 1935)

= Rs 6,275.

Solution 30.

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