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.
- 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 \(0.\overline { 001 } \) in the simplest form.

Question 2.

Find the remainder when x^{3} – ax^{2} + 6x – a is divided by x – a.

Question 3.

If an angle is five times its complementary angle, find the measure of angle.

Question 4.

Two parallel sides of a trapezium are 1m and 2m respectively and the perpendicular distance between them is 6m. Then find the area of the trapezium

Question 5.

Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)

Question 6.

The marks obtained by 17 students in Mathematics test (out of 100) are given below: ‘ 91, 82, 100, 100, 96, 65, 82, 76, 79, 90,46, 64, 72, 68, 66, 68,49. Find the range of the data.

OR

What is the probability that two friends have different birthday in the year 2017?

**SECTION-B**

Question 7.

If x + y = 12, and xy = 27, then find the value of x^{3} + y^{3}.

Question 8.

In figure, if ∠POR and ∠QOR form a linear pair and a – b = 80° then find the value of a and b.

Question 9.

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

(i) Point (1,-1) and (-1, 1) lie in the same quadrant.

(ii) The coordinates of the point whose ordinate is \(-\frac { 1 }{ 2 }\) and abscissa is are [\(-\frac { 1 }{ 2 }\),1]

Question 10.

Plot the vertices A (2, -3), B (4, 3) and C (-3, 6) of ∆ABC.

Question 11.

The areas of three adjacent faces of a cuboid are x, y and z. If the volume is v, prove that y² = xyz.

Question 12.

If the mean of 6,4, 7, P and 10 is 8, find the value of P.

**SECTION-C**

Question 13.

Prove that

(a + b)^{3} + (b + c)^{3} + (c + a)^{3} – 3 (a + b) (b + c)(c + a) = 2 (a^{3} + b^{3} + c^{3} – 3 abc)

Question 14.

Using suitable identity find the value of (999)^{3}.

Question 15.

If 5^{x-3} x 3^{2x-8} = 225, then find the value of x.

Question 16.

In figure if AB || CD, EF ⊥ CD and ∠GED = 126°, then find angles ∠AGE, ∠GEF and ∠FGE.

Question 17.

P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar(∆APB) = ar(∆BQC).

Question 18.

Two chords AB and AC of a circle are equal. Prove that the centre of the circle lies on the angle bisector of ∠BAC.

Question 19.

Construct a ∆PQR in which QR = 6 cm, ∠Q = 60° and PR – PQ = 2cm. Also justify it.

Question 20.

If each side of triangle become twice the original sides then find the percentage increase in its area.

OR

What will the percentage increase in area of triangle if each side of triangle become double?

Question 21.

A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.

Question 22.

The record of a weather station shows that out of the past 250 consecutive days, its weather forecast were correct 175 times.

(i) What is the probability that on a given day it was correct?

(ii) What is the probability that it was not correct on a given day?

**SECTION-D**

Question 23.

A and B together can do a piece of work in 10 days, but A alone can do it in 15 days. How many days would B alone take to do the same piece of work?

(i) Write the name of two work which you do in your home to help your family except study.

(ii) Why your parents do hard work for you? Which values depicted here?

Question 24.

Question 25.

On dividing polynomials x^{3} + 2x^{2} – 5ax – 7 and x^{3} + ax^{2} – 12x + 6 by (x + 1) and (x – 2) respectively remainders R1 and R2 are obtained. If 2R1 + R2 = 6, then find the value of ‘a’.

Question 26.

The internal and external diameter of a hollow hemispherical vessel are 24 cm and 25 cm respectively. The cost to paint 1 cm^{2} the surface is Rs 0.05. Find the total cost to paint the vessel all over.

Question 27.

Prove that the algebraic sum of the deviations of individual values/observations from their mean (\(\overline { x } \)) is always zero, i.e.,

Question 28.

The sum of either pair of opposite angles of a cyclic quadrilateral is 180°, prove it. Also prove that if one side of a cyclic quadrilateral is produced, then the exterior angle so formed is equal to the interior opposite angles.

Question 29.

ABCD is a trapezium in which AB || DC. BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F. Show that F is the mid-point of BC.

Question 30.

∆ABC is an isosceles triangle in which AB = AC, side BA is produced to D such that AD = AB. Show that ∠BCD is a right angle.

**Solutions**

Solution 1.

Solution 2.

By the Remainder theorem, required remainder will be p(a) because x – a = 0 => x = a

p (x) = x^{3} – ax^{2} + 6x- a

p(a) = a^{3} – a x a^{2} + 6a – a = 5a

Required remainder = 5a.

Solution 3.

Let x be the angle then from the question

angle = 5 x complementary angle

x = 5 (90 – x) => 6x = 450

x = 75°

The required angle is 75°.

Solution 4.

= 9 m²

Solution 5.

Radius of hemisphere = r = 10 cm

Total surface area of hemisphere = S = 3 πr² = 3 x 3.14 x (10)²

S = 9.42 x 100 = 942 cm²

Required surface area of hemisphere = 942 cm².

Solution 6.

Range = R = Highest value – Lowest value

= 100 – 46

R = 54

OR

For normal year = 365 days

If two friends have birthday in the same day

P(E) = \(\frac { 1 }{ 365 }\)

If two friends have birthday in the different days

then P(E) = 1 – P(E) = \(1-\frac { 1 }{ 365 }\) = \(\frac { 364 }{ 365 }\)

Required probability = \(\frac { 364 }{ 365 }\)

Solution 7.

(x + y)^{3} = x^{3} + y^{3} + 3xy (x + y)

(12)^{3} = x^{3} + y^{3} + 3 x 27 (12)

1728 = x^{3} + y^{3} + 81 x 12

1728 = x^{3} + y^{3} + 972 => x^{3} + y^{3} = 1728 – 972

x^{3} + y^{3} = 756

Solution 8.

∵ ∠POR and ∠QOR form the linear pair.

=> ∠POR + ∠QOR = 180°

=> a + b = 180° …(1)

a – b = 80° …(2)

Adding Eqn. (1) & (2) 2a = 260 => a = 130°

From Eqn. (1) 130° + b = 180° => b = 50°

a = 130°, b = 50°

Solution 9.

(i) The point (1, -1) lie in the IV quadrant because the coordinates of IV quadrant are (x, -y) and point (-1, 1) lie in the second quadrant because its coordinate are (-x, y), so both are not lie in the same quadrant.

IV quadrant=>(1, -1) => (+, -)

II quadrant=>(-1, 1) => (-, +)

(ii) Ordinate represents y-axis and abscissa represents x-axis. So, the coordinate must be [1,\(\frac { -1 }{ 2 }\)] instead of [\(\frac { -1 }{ 2 }\),1] because any point is represented by (x, y) and not by (y, x)

xx’ = x-axis => abscissa

yy’ = y-axis => ordinate

Solution 10.

The vertices are plotted on graph paper.

Solution 11.

Let l, b and h be the length, breadth and height of a cuboid.

Volume of cuboid = v = l x b x h = lbh …(1)

Since x, y and z represent areas of three adjacent faces of the cuboid,

then x = lb, y = bh, z = lh

=> xyz = (lb) x (bh) x (hl)

=> x × y × z = l² x b² x h²

=> xyz = (l x b x h)²

=> xyz = v²

Hence v² = xyz

Solution 12.

Mean of the given data = x = 8

27 + P = 40

P = 40 – 27

P = 13

Solution 13.

Let a + b = x,b + c = y and c + a = z

∴ (a + b)^{3} + (b + c)^{3} + (c + a)^{3} – 3 (a + b) (b + c) (c + a) = x^{3} + y^{3} + z^{3} – 3xyz

But x^{3} + y^{3} + z^{3} – 3xyz = (x + y + z)(x^{2} + y^{2} + z^{2} – xy – yz – zx)

=> {(a + b) + (b + c) + (c + a)} {(a + b)^{2} + (b + c)^{2} + (c + a)^{2} – (a + b) (b + c) – (b + c) x (c + a) – (c + a) (a + b)}

= 2 (a + b – c) {a^{2} + b^{2} + 2ab + b^{2} + c^{2} + 2 bc + c^{2} + a^{2} + 2 ca – (ab + ac + b^{2} + bc) – (bc + ba+ c^{2} + ac) – (ca + cb + a^{2} + b^{2})}

= 2 (a + b + c) (a^{2} + b^{2} + c^{2} – ab – bc – ca)

= 2 (a^{3} + b^{3} + c^{3} – 3 abc)

Solution 14.

∵ 999 = (1000 – 1)

∴ 999^{3} = (1000 – 1)^{3} = (1000)^{3} – 1^{3} – 3 x 1000 x 1 (1000 – 1)

999^{3} = 1000000000 – 1 – 3000 (1000-1)

999^{3} = 1000000000 – 1 – 3000000 + 3000

=>1000003000 – 3000001

999^{3} = 997002999

Solution 15.

5^{x-3} x 3^{2x-8} = 225 = 5^{2} x 3^{2}

5^{x-3} x 3^{2x-8} = 5^{2} x 3^{2}

On comparing the power of each base

x – 3 = 2 and 2x – 8 = 2

x = 2 + 3 and 2x = 2 + 8

x = 5 and x = 5

x = 5

Solution 16.

∠GED = 126°

∠AGE = ∠GED = 126° [Alternate interior angles]

∠AGE = 126°

∠GED = 126°

∠GEF + ∠FED = 126° [∵∠FED = 90°]

∠GEF + 90° = 126°

∠GEF = 126° – 90° = 36°

∠GEF = 36°

∠GEC + ∠GED = 180° (Linear pair)

∠GEC + 126° = 180°

∠GEC = 180°- 126° = 54°

∠FGE = 54°

Solution 17.

Given: A ||gm ABCD in which P and Q are two points Q lying on the side DC and AD.

To prove: ar (∆APB) = ar (∆BQC)

Proof: Now ∆APB and ||gm ∆BCD have the same base AB and lie between the same parallels.

[AB || DC]

ar (∆APB) = \(\frac { 1 }{ 2 }\)ar (||gm ABCD)

Similarly, ∆BQC and ||gm ABCD have the same base BC and lie between the same parallels (BC ||AD)

ar (∆BQC) = \(\frac { 1 }{ 2 }\)ar (||gm ABCD) …(2)

From (1) & (2)

ar (∆APB) = ar (∆BQC)

Solution 18.

Given: A circle C (O, r) in which AB and AC are two equal chords. AD is the bisector of ∠BAC.

To prove: O lies on AD.

Construction: O lies on AD.

Proof: In ∆BAM and ∆CAM,

AB = AC (Given)

∠BAM = ∠CAM (Given)

AM = AM (common)

∴ ∆BAM ≅ ∆CAM (By SAS congruency Rule)

and BM = CM (By CPCT)

∠BMA = ∠CMA (By CPCT)

∠BMA = ∠CMA = 90° [∵∠BMA + ∠CMA = 180°]

=> AM is a perpendicular bisector of chord BC.

=> AD is the perpendicular bisect of the chord BC.

But the perpendicular bisector of a chord always passes through the centre of the circle.

=> AD passes through the centre O of the circle.

=> O lies on AD.

Solution 19.

Steps of Construction: Here PR > PQ

1. Draw a line segment QR = 6 cm and at the point Q make an angle of 60°.

∠RQX = 60°

2. Cut off QS = PR – PQ = 2 cm from the line QX, which is produced on other side of the line segment QX.

3. Join S, R and draw the perpendicular bisector ∠M of SR.

4. Produce ML to intersect QX at the point P. Join P, R.

5. ∆PQR is the required triangle.

Justification: The point of intersection lies on the perpendicular bisector of SR.

=> SP = PR …(1)

=> SQ = SP – PQ

But SP = PR

∴SQ = PR – PQ => Hence the construction is justified.

Solution 20.

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

Solution 21.

Here, a joker’s cap is in the form of a right circular cone whose radius of the base, r = 7 cm

Height = 24 cm

l = √625 = 25 cm

Curved surface area of cone = πrl = \(\frac { 22 }{ 7 }\) x 7 x 25 = 550 cm²

Curved surface area of 10 such caps= (550 x 10) = 5500 cm²

Solution 22.

Total possible outcomes = 250

Number of favourable outcomes = 175

P (not correct recording) = 0.3.

Solution 23.

Let B alone take x days to do the work.

Since A alone can complete the work in 15 days.

Therefore one day work of A = \(\frac { 1 }{ 15 }\).

A and B together complete the work in one day = \(\frac { 1 }{ 10 }\) part of the work.

Now A.T.Q.

A’s one day work + B’s one day work = (A + B)’s one day work

x = 30 days

B can alone do the work in 30 days.

(i) To help your family, you can do the following works.

(a) Cleaning inside and outside the house.

(b) Store clean water.

(c) Save Electricity and Water.

(d) Plantation in your garden or pot.

(e) Marketing with parents to purchase the goods.

(f) Deposit electricity, water, telephone etc. bills and Banking work.

(ii) Your parent do hardwork to secure your future and for your smoothing life.

Values: (1) Care for the family (caring)

(2) Cooperation.

(3) Social Responsibility.

(4) Team Work

(5) Sensitivity

Solution 24.

Solution 25.

Let p(x) = x^{3} + 2x^{2} – 5ax – 7 and q(x) = x^{3} + ax^{2} – 12x + 6 are given polynomials.

If x + 1 = 0, x = -1,p (-1) = R1

R1 = p (-1) = (-1)^{3} + 2 (-1)^{2} – 5a x (-1) – 7

R1 = -1 + 2 + 5a – 7 = 5a – 6 [ ∵ p(x) = x^{3} + 2x^{2} – 5ax – 7]

If x – 2 = 0 =>x = 2,p (2) = R^{2} [∵ q(x) = x^{3} + ax^{2} – 12x + 6]

R2 = p(2) = (2)^{3} + a x (2)^{2} – 12 x 2 + 6 = 8 + 4a – 24 + 6 = 4a – 10

R1 = 5a – 6, R2 = 4a – 10

Putting the values of R1 and R2 in Eqn. 2R1 + R2 = 6

2 (5a – 6) + (4a – 10) = 6

10a – 12 +4a – 10 = 6

14a – 22 = 6

14a = 6 + 22 = 28

a = \(\frac { 28 }{ 2 }\) = 2

a = 2

Solution 26.

Internal diameter of a hollow hemispherical vessel, d = 24 cm.

Internal radius of the hollow hemispherical vessel = r = \(\frac { 24 }{ 2 }\) = 12 cm

External diameter of the hollow hemispherical vessel, D = 25 cm.

External radius of a hemispherical vessel.

Solution 27.

Solution 28.

Given: A cyclic □ PQRS in which ∠P, ∠R and ∠Q, ∠S are two pairs of opposite angles.

To prove: ∠P + ∠R = 180°

∠Q + ∠S = 180°

Construction: Join O, P and O, R.

Proof: By Central angle theorem

∠POR = 2 ∠PSR

[For the same segment angle formed at centre of circle is twice the angle formed at remaining part of the circle]

Solution 29.

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

Let EF intersects BD at P.

Then in ∆DAB

Through E, EP || AB intersects BD at P.

By the converse of mid-point theorem.

=> P is the mid-point of BD.

Now in ∆BCD, P is the mid-point of BD.

Through P, PF || DC intersects BC at P.

By the converse of mid-point theorem.

F is the mid-point of BC.

Hence proved.

Solution 30.

Given: In ∆ABC, AB = AC, side BA is produced to D such that AB = AD.

To prove: ∠BCD = 90°

Construction: Join CD.

Proof: In ∆ABC given that

AB = AC

=> ∠ACB = ∠ABC

Now

AB = AD

AD = AC => ∠ACD = ∠ADC …(2)

[Opposite angles of equal sides]

Adding Eqn. (1) & Eqn. (2)

∠ACB + ∠ACD = ∠ABC + ∠ADC

=> ∠BCD = ∠ABC + ∠BDC [ v ∠ADC = ∠BDC]

=> ∠BCD + ∠BCD = ∠ABC + ∠BCD + ∠BDC

=> 2∠BCD = 180° [Adding ∠BCD on both sides]

[By angle sum property of ∆]

∠BCD = \(\frac { 180 }{ 2 }\) [∠ABC + ∠BCD + ∠BDC = 180°]

∠BCD = 90°

So ∠BCD is a right angle.

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