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.

Question 2.

What is the degree of zero polynomial?

Question 3.

In ∆ABC, BC is produced to D. If ∠ABC = 40° and ∠ACD = 120°, then ∠A = ?

Question 4.

Write the signs of abscissa and ordinate of a point in quadrant (II).

Question 5.

The total surface area of a cube is 216 cm². Find its volume.

Question 6.

Find the slope of the line 2x + 3y – 4 = 0.

**SECTION-B**

Question 7.

If a + b + c = 0, then a³ + b³ + c³ = ?

Question 8.

Find four different solutions of 2x + y = 6

Question 9.

The base of an isosceles triangle is 16 cm and its area is 48 cm². Find the perimeter of a triangle.

Question 10.

If the mean of five observations x, x + 2, x + 4, x + 6 and x + 8 is 13. Find the value of x.

Question 11.

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

Out of these families, one family is selected at random. What is the probability that the selected family has

(i) 2 girls

(ii) no girl.

Question 12.

Two coins are tossed 1000 times and the outcomes are recorded as under.

A coin is thrown at random. What is the probability of getting

(i) at most one head ?

(ii) atleast one head ?

**SECTION-C**

Question 13.

If x = 3 + √8 , find the value of \(\left( { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } \right) \)

Question 14.

If p = 2 – a, prove that a³ + 6ap + p³ – 8 = 0.

Question 15.

In the given figure, prove that x = α + β + γ.

Question 16.

If the bisector of the vertical angle of a triangle bisect the base, prove that the triangle is Isosceles.

Question 17.

The sides BA and DC of quad. ABCD are produced as shown in the given figure. Prove that x° + y° = a° + b°.

Question 18.

In an equilateral triangle, prove that the centroid and the circumcentre coincide.

Question 19.

Construct a Δ ABC whose perimeter is 14 cm and sides are in the ratio 2:3:4.

Question 20.

Plot the points A(-5, 2), B(-4, -3), C(3, -2) and D(6, 0) on a graph paper. Write the name of figure formed by joining them.

Question 21.

The diameter of a roller, 120 cm long is 84 cm. If it takes 500 complete revolutions to level a playground. Find the cost of levelling it at 75 paise per square metre.

Question 22.

The internal and external diameters of a hollow hemispherical vessel are 20 cm and 28 cm respectively. Find the cost of painting the vessel all over at 14 paise per cm².

**SECTION-D**

Question 23.

Show that

Question 24.

Prove that x^{3} + y^{3} + z^{3} – 3xyz =(x + y + z) (x^{2} + y^{2} + z^{2} – xy – yz – zx). Hence find the value of x^{3} + y^{3} + z^{3} if x + y + z = 0.

Question 25.

In a ∆ABC, the side AB and AC are produced to P and Q respectively. The bisector of ∠PBC and ∠QCB intersect at a point O. Prove that ∠BOC = 90° – \(\frac { 1 }{ 2 }\) ∠A.

Question 26.

In each of the following figures, AB || CD. Find the value of x in each case.

Question 27.

Draw the graph of the equation 3x + 2y = 12. At what points does the graph cut the x-axis and the y-axis.

Question 28.

A river 2m deep and 45 m wide is flowing at the rate of 3 km per hour. Find the volume of water that runs into the sea per minute.

Now answer these questions.

1. Why we should preserve and not waste the water and other natural resources? Write one method to restore the water at your home.

2. Why it is necessary to clean all rivers in the country?

Question 29.

The mean of 25 observations is 36. If the mean of first 13 observations is 32 and that of last 13 observations is 39, find the 13th observation.

Question 30.

Plot the points A(2, 5), B(-2, 2) and C(4, 2) on a graph paper. Join AB, BC and AC. Calculate the area of ∆ABC.

**Solutions**

Solution 1.

Solution 2.

Degree of zero polynomial is not defined.

Solution 3.

∠A + ∠B = ∠ACD

⇒ ∠A + 40° = 120°

⇒ ∠ A = 80°

Solution 4.

The points in quadrant II are in the form (-, +).

Solution 5.

The Total Surface Area of cube (T.S.A.) = 6a² = 216

=> a² = 36 => a = √36 = 6 cm

Volume of cube = (a)^{3} = (6)^{3} = 216 cm^{3}

Volume of cube = 216 cm^{3}

Solution 6.

2x + 3y – 4 = 0 ⇒ 3y = – 2x + 4

Solution 7.

a + b + c = 0 ⇒ a + b = – c

⇒ (a + b)^{3} = (- c)^{3} ⇒ a^{3} + b^{3} + 3ab (a + b) = -c^{3}

⇒ a^{3} + b^{3} + 3ab (- c) = -c^{3} ⇒ a^{3} + b^{3} + c^{3} = 3abc

Solution 8.

y = 6 – 2x

(i) If x = 1,y = 6 – 2 = 4

(ii) If x = 2,y = 6 – 4 = 2

(iii) If x = 3,y = 6 – 6 = 0

(iv) If x = 4,y = 6 – 8 = -2

Solution 9.

b =16 cm, area = 48 cm²

Perimeter of Isosceles triangle = 2a + b = 20 + 16 = 36 cm.

Solution 10.

Mean of the given observations

Solution 11.

Total number of families = 1500

Solution 12.

Total number of tosses = 266 + 540 + 194 = 1000

(i) At most one head = Number of times one head or no head appears

= 540+ 194 = 734

P(E1) = P(at most one head) = \(\frac { 734 }{ 1000 }\) =0.734

(ii) Atleast one head = Number of times one head or two head appears

= 540 + 266 = 806

P(E2) = P(atleast one head) = \(\frac { 806 }{ 1000 }\) = 0.806

Solution 13.

x + 3√8

Solution 14.

We have p = 2 – a => a + p + (-2) = 0

=> Putting a = x,p = y and (-2) = z.

we get a + p + (-2) = 0 =>x + y + z = 0

⇒a^{3} + p^{3} + (-2)^{3} = 3 × a × p × (-2) [If x + y + z = 0 then x^{3} + y^{3} + z^{3} = 3xyz]

⇒a^{3} + p^{3} + (-2)^{3} = -6ap

⇒a^{3} + 6ap + p^{3} – 8 = 0

Solution 15.

Join B and D and produce BD to E.

p + q = β and s + t = x

Side BD of A ABD is produced to E

∴ p + α = s ..(1)

Side BD of A CBD is produced to E.

∴ q + γ = t ..(2)

Adding (1) and (2)

(p + α) + (q + γ) = s + 1

(p + q) + (α) + (γ) = x

β + α + γ = x

⇒ x = α + β + γ

Solution 16.

Given: ∆ AABC in which AD is the bisector of ∠ A, which meets BC in D such that BD = DC.

To prove: AB = AC

Construction: Produce AD to E such that AD = DE and join EC.

Proof: In ∆ABD and ∆ECD

BD = DC (given)

AD = DE (by construction)

∠ADB = ∠EDC (V.O.A)

=> ∴ ∆ABD ≅ ∆ECD

∴ ∆ABC ≅ ∆ECD

=> AB = EC and ∠1 = ∠3 …(2) (CPCT)

Also ∠1 = ∠2 …(3) [∵AD bisects ∠A]

∴ ∠2 = ∠3 [using (2)and(3)]

=> EC = AC …(4) [side opposite to equal angles]

=> AB = AC [using (1)and(4)]

=> ∆ABC is Isosceles.

Solution 17.

We have ∠ A + b° = 180° (Linear pair)

=> ∠A = 180° – b° …(1)

Also ∠C + a° = 180° (Linear pair)

∠C = 180° – a° …(2)

Solution 18.

Let ∆ ABC be the given equilateral triangle and let its median AD,

BE and CF intersect at G.

Then G is the centroid of ∆ ABC

In ∆ BCE and ∆ CBF

BC = CB (common)

∠B = ∠C [each 60°]

This shows that G is the circumcentre of ∆ ABC

=> G is the centroid as well as circumcentre of ∆ ABC

Note: Centroid: The point of intersection of the medians of a triangle is called centroid.

The centroid of the triangle is the point located at \(\frac { 2 }{ 3 }\) of the distance from a vertex along the median.

Circumcentre: The centre of the circumcircle of the triangle is called the circumcentre.

Solution 19.

Steps of construction:

(i) Draw a line segment PQ =14 cm. Draw a ray PX making an acute angle with PQ and draw in the downward direction.

(ii) From P, mark set of (2 + 3 + 4) = 9 equal distance along PX.

(iii) Mark points L, M, N on PX such that PL = 2 cm, LM = 3 cm and MN = 4 cm.

(iv) Join NQ. Through L and M draw LB || NQ and MC || NQ. Cutting PQ at B and C respectively.

(v) With B as centre and radius BP draw an arc. With C as centre and radius CQ draw another arc, cutting the previous arc at A.

(vi) Join AB and AC. ∆ ABC is the required triangle.

Solution 20.

Points A(-5, 2), B(-4, -3), C (3, -2) and D(6, 0) are shown on graph paper.

The figure formed by joining them is called quadrilateral.

Solution 21.

Radius of roller = r = 42 cm, Length = h = 120 cm

Area covered by the roller in 1 revolution = C.S.A. of the roller = 2πrh sq. unit.

Solution 22.

Outer radius of the vessel R = 14 cm

Inner radius of the vessel = r = 10 cm

Area of the outer surface = 2πR² cm²

= (2π x 14 x 14) cm² = 392 π cm²

Area of the inner surface = 2πt² cm²

= (2π x 10 x 10) cm² = 200 π cm²

Area of the ring (shaded) at the top = π(R² – r²)

= π[( 14)² – (10)²] = π(14 + 10) (14 – 10)

= 96 π cm²

Total area to be painted = 392 π + 200 π + 96 π

= 688 π cm²

Solution 23.

Solution 24.

x^{3} + y^{3} + z^{3} – 3xyz = (x^{3} + y^{3}) + z^{3} – 3xyz

x^{3} + y^{3} + z^{3} – 3xyz = [(x + y)^{3} – 3xy(x + y)] + z^{3} – 3xyz [let x + y = u]

= u^{3} – 3xyu + z^{3} – 3xyz

= u^{3} + z^{3} – 3xy (u + z) = (u^{3} + z^{3}) – 3xy (u + z)

= (u + z) (u^{2} – uz + z^{2}) – 3xy (u + z) = (u + z) [u^{2} + z^{2} – uz – 3xy]

= (x + y + z) [(x + y)^{2} + z^{2} – (x + y)z – 3xy] [u = x + y]

= (x + y + z) [x^{2} + y^{2} + z^{2} + 2xy – xz -yz – 3xy]

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

if x + y + z = 0

x^{3} + y^{3} + z^{3} – 3xyz = 0 × (x^{2} + y^{2} + z^{2} – xy – yz – zx)

x^{3} + y^{3} + z^{3} – 3xyz = 0

x^{3} + y^{3} + z^{3} = 3xyz

Solution 25.

We have ∠B + ∠CBP = 180° (Linear pair)

Solution 26.

(i) Draw EF || AB || CD

=> ∠1 + ∠2 = x°

Now AB || EF and AE is transversal.

∠1 + ∠BAE = ∠1 + 104° = 180°

(co interior angles)

∠1 = 180°- 104° = 76°

Again EF || CD and EC is the transversal

∠2 + ∠ECD = ∠2 + 116° = 180°

=> ∠2 = 64°

Hence x = ∠1 + ∠2 = 76° + 64° = 140°

=> x = 140°

(ii) Draw EO || AB || CD

x° = ∠1 + ∠2

Now EO || AB and OB is transversal

∠1 + ∠ABO = 180° [Co-interior angles]

∠1 + 40°= 180°

=> ∠1 = 180°- 40° = 140° => ∠1 = 140°

Again EO || CD and DO is transversal

∠2 + ∠CDO = 180°[Co-interior angles]

∠2 + 35° = 180°

=> ∠2 = 180° – 35° = 145°

=> ∠2 = 145°

∠1 + ∠2 = (140° + 145°) = 285°

=> x = ∠1 + ∠2 = 285°

=> x = 285°

Solution 27.

Linear equation 3x + 2y = 12

The point A(4, 0) cuts on the x-axis and point B(0, 6) cuts on the y-axis.

Solution 28.

Volume of the water running into the sea per minute

= Volume of cuboid

= l x bx h

= 50 x 45 x 2

= 4500 m^{3}

(i) We should preserve the water and other natural resources for ourself and for our next generation because they can not be polluted or finished very fast.

(ii) Method of restoring fresh water is by ‘rain water harvesting’ in home.

(iii) Our rivers are very polluted and causing the problems in ecosystem and becoming the shortage of fresh water at large scale. So, it is necessary to clean all rivers of our country.

Solution 29.

Mean of first 13 observations = 32

Sum of the first 13 observations = 32 x 13 = 416

Mean of last 13 observations =39

Sum of last 13 observations = 39 x 13 = 507

Mean of 25 observations = 36

Sum of all the 25 observations = 36 x 25 = 900

∴ the 13th observation = (416 + 507 – 900) = 23

Hence the 13th observation = 23

Solution 30.

From Graph:

BC = CE + BE = (4 – 0) + (0 – (-2))

Base = BC = 4 – (-2) = 4 + 2 = 6 units

Height = AD = OF – OE = 5 – 2 = 3 units

ar (∆ ABC) = \(\frac { 1 }{ 2 }\) x Base x height

= \(\frac { 1 }{ 2 }\) x 6 x 3 = \(\frac { 18 }{ 2 }\) = 9 sq units

ar (∆ ABC) = 9 sq. units.

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