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 ?  

Ans. 4th 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 

Answer. 6

(iv) The coordinate of point H 

Ans. (-5, -3)

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

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.         

∠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

Construction : AD ⊥ BC

In ∆ABD and ∆ACD, 

AB = AC (given) 

∠ADB = ∠ADC (construction)

AD = AD (common) 

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.