# HBSE Class 9th Maths Solved Question Paper 2021

HBSE Class 9th Maths Solved Question Paper 2021

HBSE Class 9 Maths Previous Year Question Paper with Answer. HBSE Board Solved Question Paper Class 9 Maths 2021. HBSE 9th Question Paper Download 2021. HBSE Class 9 Maths Paper Solution 2021. Haryana Board Class 9th Maths Question Paper 2021 Pdf Download with Answer.

1.(1) Which of the following is not a rational number ?

(A) √2

(B) 0

(C) √4

(D) √-16

Ans. (A) √2

(2) The p/q form of 0.47 (bar) is …………

Ans. 47/(100-1) = 47/99

(3) Find one rational number between 3 and 4.

Ans. 3.5 (infinite rational numbers between 3 & 4)

(4) Which of the following number is not in between 4/5 and 9/5 ?

(A) 3/5

(B) 5/5

(C) 6/5

(D) 8/5

Ans. (A) 3/5 = 0.6  (4/5 = 0.8 and 9/5 = 1.8)

(5) The coefficient of x² in 2 – x² + x³ is ………….

Ans-1

(6) Factors of 4y² – 4y + 1 is :

(A) (2y + 1)²

(B) (4y – 1)²

(C) (2y – 1)²

(D) (2y – 2)²

Ans. (C) (2y – 1)²

4y² – 4y + 1 = (2y)² – 2(2y)(1) + 1² = (2y – 1)²

(7) Factors of 27 – 125a³ – 135a + 225a² is :

(A) (3 + 5a)²

(B) (3a + 5)³

(C) (3a – 5)²

(D) (3 – 5a)³

Ans. (D) (3 – 5a)³

Using formula, a³ – b³ – 3ab(a – b) = (a – b)³

(3 – 5a)³ = 27 – 125a³ – 135a + 225a²

(8) Zero of p(x) = 3x + 1 is …………

Ans.

3x + 1 = 0

3x = -1

x = – 1/3

Zero is -1/3

(9) What is product of (x + 4)(x + 10) ?

Ans. (x + 4)(x + 10) = x² + 10x + 4x + 40 = x² + 14x + 40

(10) What is solution of (2x + 1) = x + 3 ?

Ans.

2x + 1 = x + 3

2x – x = 3 – 1

x = 2

(11) The solution of equation x − 2y = 4 is :

(A) (0, 2)

(B) (4, 0)

(C) (1, 1)

(D) (2, 0)

Ans. (B) (4, 0)

x – 2y = 4

4 – 2(0) = 4

4 – 0 = 4

4 = 4  (L.H.S = R.H.S)

(12) Point (4, 1) satisfies to which equation of line ?

(A) x + 2y = 5

(B) x + 2y = −6

(C) x + 2y = 6

(D) x + 2y = 16

Ans. (C) x + 2y = 6

4 + 2(1) = 6

4 + 2 = 6

6 = 6  (L.H.S = R.H.S)

(13) Possible dimensions of Cuboid whose volume 3x² – 12x is :

(A) 3, x, x + 4

(B) 3, x – 4, x

(C) -3, -x, -x – 4

(D) None of these

Ans. (B) 3, x-4, x

3x² – 12x = 3x(x-4)

(14) Point (5, −7) lies in which Quadrant ?

(15) The point (0, 0) where x-axis and y-axis intersect is called ……………

Ans. Origin

(16) What is abscissa and ordinate of point (-4, -3) ?

(A) x = -4, y = -3

(B) x = -3, y = -4

(C) x = 4, y = 3

(D) None

Ans. (A)  x = -4, y = -3

(17) On joining the points (0, 0), (0, 2), (2, 2) and (2, 0) we obtain a :

(A) Square

(B) Rectangle

(C) Rhombus

(D) Parallelogram

Ans. (A) Square

(18) The sum of any two sides of a triangle is …………… than the third side.

Ans. Greater

(19) The diagonals of a rectangle are …………..

Ans. Equal

(20) If diagonals of quadrilateral bisect each other at right angles, then it is a :

(A) Parallelogram

(B) Rectangle

(C) Rhombus

(D) Trapezium

Ans. (C) Rhombus

(21) Angles in the  same segment of a circle are ……………

Ans. Equal

(22) In cyclic quadrilateral ABCD, AOC is the diameter of circle. If ∠CAD = 50°, then ∠ACD is :

Ans.

∠D = 90° ( because AC = diameter and all angles in semi circle are 90°)

∠ACD = 180° – (90°+50°) = 180° – 140° = 40°

(23) The centre of a circle lies in …………… of the circle :

(A) exterior

(B) circumference

(C) interior

(D) perimeter

Ans. (C) interior

(24) In a pair of set, a triangle is with angles :

(A) 30°, 60°, 90°

(B) 30°, 30°, 45°

(C) 75°, 25°, 80°

(D) 65°, 15°, 100°

Ans. (A) 30°, 60°, 90°

(25) To construct a triangle we must know at least its …………. parts.

Ans. Three

(26) To construct an angle of 22½° which angle we bisect ?

Ans. 22½° × 2 = 45°

(27) Perimeter of an isosceles triangle is 30 cm and its equal sides are of 12 cm, then area of the triangle is :

(A) 8√15 cm²

(B) 7√12 cm²

(C) 9√15 cm²

(D) 15√15 cm²

Ans. (C) 9√15 cm²

Perimeter= 30 cm

Semi Perimeter, s = 30/2 = 15 cm

a = b = 12 cm

c = 30 – (12+12) = 30 – 24 = 6 cm

Using Heron’s Formula,

Area = √s(s-a)(s-b)(s-c) = √15(15-12)(15-12)(15-6) = √15(3)(3)(9) = 9√15 cm²

(28) Base of a triangle is 12 cm and its height is 8 cm, then what will be its area ?

Ans. Area = ½ × base × height = ½ × 12 × 8 = 48 cm²

(29) Area of equilateral triangle with side a is ………….

Ans. (√3/4)a²

(30) The surface area of a cuboid is ………….

Ans. S.A or T.S.A = 2(lb + bh + bl)

(31) Volume of a hemisphere with radius r will be …………..

Ans. Volume of a hemisphere = ⅔πr³

Q32. Find the volume of right circular cone with radius 6 cm and height 7 cm.

Ans. Volume = ⅓πr²h = ⅓ × 22/7 × 6 × 6 × 7 = 264 cm³

(33) What will be the total surface area of a cylinder, when radius r and height h is given ?

(A) 2πrh

(B) 2πr(r+h)

(C) πr²h

(D) 2πr²

Ans. (B) 2πr(r+h) = 2πr² + 2πrh

(34) A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic meters of a liquid ?

(A) 4.50 m

(B) 3.75 m

(C) 4.75 m

(D) 3.50 m

Ans. (C) 4.75 m

Volume = length × breadth × height

V = lbh

h = V ÷ lb = 380 ÷ (10×8) = 380 ÷ 80 = 4.75 m

(35) The mean of all possible factors of 20 will be ………….

Sol.

Factors of 20 are = 1, 2, 4, 5, 10, 20

Mean = (1+2+4+5+10+20)/6 = 42/6 = 7

(36) Class mark of class 150–160 is :

(A) 145

(B) 310

(C) 10

(D) 155

Ans. (D) 155

Class Mark = Mid Value = (150+160)/2 = 310/2 = 155

(37) The heights of 9 students of a class are given (in cm) as follows :

155, 160, 145, 149, 150, 147, 152, 144, 148

The median of this data is :

(A) 150

(B) 147

(C) 149

(D) 148

Ans. (C) 149

Ascending Order  144, 145, 147, 148, 149, 150, 152, 155, 160

Median = (n+1)/2 = (9+1)/2 = 10/2 = 5th Term = 149

(38) In a cricket match a lady batsman hits a boundary 6 times out of 30 balls she plays. The probability that she will not hit a boundary in the next ball is …………..

Ans.

Total = 30

Boundary hits = 6

Not Boundary hits = 30 – 6 = 24

P(not Boundary hits) = 24/30 = 4/5

Q39. The probability of an event lies between :

(A) 0 and 1

(B) −1 and 1

(C) 1 and 2

(D) 2 and 3

Ans. (A) 0 and 1 (0 ≤ P ≤ 1)

(40) What will be the probability of a prime number when a fair die is tossed ?

Ans.

Total numbers = 6 (1,2,3,4,5,6)

Prime numbers = 3 (2,3,5)

P(prime number) = 3/6 = 1/2

Subjective Questions

(2) Express 0.6(bar) in the form p/q , where p and q  are integers and q ≠ 0.

Sol.

x = 0.6 (bar)

x = 0.666……… (i)

10x = 6.666………(ii)

Subtract eqn.(i) from (ii),

10x – x = 6.666….. – 0.666……

9x = 6

x = 6/9 = 2/3

So,  0.6 (bar) = 2/3

(3) Draw the graph of linear equation 3 = 2x + y  in two variables.

Sol.

y = 3 – 2x

when x = 0, then y = 3

when x = 1, then y = 1

using co-ordinates (0,3) and (1,1)

(4) Write the answer of each of following from figure :

(i) The coordinates of B

Ans. (-5, 2)

(ii) The point identified by the coordinates (−3, −5)

Ans. E

(iii) The abscissa of the point D

(iv) The coordinate of point H

Ans. (-5, -3)

(5). Find the value of x and y from given figure, AB || CD.

Sol.

x + 50° = 180° (Linear Pair)

x = 180° – 50° = 130°

y = 130° (Vertical Opposite Angle)

(6) What will be mean of first six odd numbers ?

Sol.

First six od numbers = 1,3,5,7,9,11

Mean = (1+3+5+7+9+11)/6 = 36/6 = 6

(7) Factorise :  x³ – 2x² – x + 2.

Sol.

x³ – 2x² – x + 2 = x²(x – 2) -1(x – 2) = (x – 2)(x² – 1) = (x – 2)(x – 1)(x + 1)

(8) Show that the diagonals of a rhombus are perpendicular to each other.

Sol.

In ΔAOD and ΔCOD

OA = OC [Diagonal of a parallelogram bisect each other]

OD = OD [Common side]

AD = CD [Sides of a rhombus]

∴ΔAOD ≅ ΔCOD [SSS congruency rule]

∠AOD = ∠COD [CPCT]

But, ∠AOD + ∠ COD = 180° [Linear pair, since AOC is a straight line]

So, 2∠AOD = 180°

∠AOD = 180°/2 = 90°

the diagonals AC and BD are perpendicular.

(9) In figure ∠ABC = 69° and ∠ACB = 31°, find ∠BDC.

Sol.

∠A = 180° – (69°+31°) = 180° – 100° = 80°

∠A = ∠BDC = 80° (angles in same segment)

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

Sol.

Steps of construction :

(i) Draw BC = 7 cm

(ii) Construct ∠XBC = 75°

(iii) From ray BX, cut-off line segment BD = AB + BC = 13cm

(iv) Join CD

(v) Draw the perpendicular bisector of CD meeting BX at A

(vi) Join AC to obtain the required triangle ABC

(vii) ABC is the required triangle.

(11) Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.

Sol.

Perimeter= 42 cm

Semi Perimeter, s = 42/2 = 21 cm

a = 18 cm

b = 10 cm

c = 42 – (18+10) = 42 – 28 = 14 cm

Using Heron’s Formula,

Area = √s(s-a)(s-b)(s-c) = √21(21-18)(21-10)(21-14) = √21(3)(11)(7) = 21√11 cm²

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

Sol.

Consider ΔABC an isosceles triangle, AB = AC

In ∆ABD and ∆ACD,

AB = AC (given)

So, ∆ABD ≅ ∆ACD  (RHS)

hence, ∠B = ∠C (CPCT)

OR

ABCD is a quadrilateral in which P, Q, R and S are mid points of the sides AB, BC, CD and DA. AC is a diagonal, show that :

(i) SR || AC and SR = ½AC

(ii) PQ = SR

(iii) PQRS  is a parallelogram

Sol.

(i) In ΔADC, S and R are the mid-points of sides AD and CD respectively. Thus, by using the mid-point theorem

SR || AC and SR = ½AC …..(i)

(ii) In ΔABC, P and Q are mid-points of sides AB and BC. Therefore, by using the mid-point theorem,

PQ || AC and PQ = ½AC ……(ii)

(iii) Using eqn.(i) and (ii), we obtain

PQ || SR and PQ = SR

Clearly, one pair of opposite sides of quadrilateral PQRS is parallel and equal. Hence, PQRS is a parallelogram.

(13) A conical tent is 10 m high and the radius of its base is 24 m. Find :

(i) Slant height of the tent.

(ii) Cost of the canvas required to make the tent, if the cost of 1 m² canvas is ₹70.

Sol.

(i) h = 10 m

r = 24 m

l² = h² + r² = 10² + 24² = 100 + 576 = 676

l = √676 = 26 m

(ii) C.S.A = πrl = 22/7 × 24 × 26 = 13728/7 m²

Cost of 13728/7 m² = 70 × 13728/7 = ₹137280

OR

The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm³ of wood has a mass of 0.6 g.

Sol.
Outer diameter, D = 28 cm

outer radius, R = 28/2 = 14 cm

inner diameter, d = 24 cm

inner radius, r = 24/2 = 12 cm

length or height, h = 35 cm

Volume = πR²h – πr²h = πh(R²-r²) = 22/7 × 35 × (14² – 12²) = 5720 cm²

Mass of 5720 cm² = 5720 × 0.6 = 3432 g = 3.432 kg

(14) An insurance company selected 2000 drivers at random in a particular city to find a relationship between age and accidents. The data obtained are given in the following table :

Find the probabilities of the following events for a driver chosen at random from the city :

(i) being 18-29 years of age and having exactly 3 accidents in one year,

Ans. Probability (P) = 61/2000

(ii) being 30-50 years of age and having one or more accidents in a year,

Ans. Probability (P) = 225/2000

(iii) having no accidents in one year,

Ans. Probability (P) = 1305/2000

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