Haryana Board Board Solved Question Paper Class 9 Maths 2023. HBSE 9th Question Paper Download 2023. HBSE Class 9 Maths Paper Solution 2023. Haryana Board Class 9th Maths Question Paper 2023 Pdf Download with Answer. HBSE Class 9 Maths Previous Year Question Paper with Answer.

**HBSE Class 9th Maths Question Paper 2023 Answer**

Q1.(i) Which of the following is an irrational number ?

(A) √4/9

(B) √12/3

(C) √7

(D) √81

Answer : (C) √7

(ii) What will be the zero of the polynomial p(x) = 5x – 1 ?

(A) -1/5

(B) 1/5

(C) 4

(D) -4

Answer : (B) 1/5

Here, p(x) = 0

5x – 1 = 0

5x = 1

x = 1/5

(iii) Write the coefficient of x³ in – 3x⁴ – 7x³ + 4x + 2 :

(A) -3

(B) 3

(C) -7

(D) 4

Answer : (C) -7

(iv) Point (0, 5) lies on ………..

(A) x-axis

(B) y-axis

(C) Origin

(D) None of these

Answer : (B) y-axis

(v) How many straight lines may be drawn from any one point to any other point ?

(A) 0

(B) 1

(C) 2

(D) 3

Answer : (B) 1

(vi) A ………… is breadthless length.

(A) Point

(B) Square

(C) Rectangle

(D) Line

Answer : (D) Line

(vii) Which of the following needs a proof ?

(A) Definition

(B) Axiom

(C) Theorem

(D) Postulate

Answer : (C) Theorem

(viii) Each angle in a semicircle is a ………….

(A) Acute angle

(B) Right angle

(C) Obtuse angle

(D) Straight angle

Answer : (B) Right angle

(ix) Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is :

(A) 1 : 2

(B) 1 : 1

(C) 2 : 1

(D) 3 : 1

Answer : (B) 1 : 1

(x) Volume of the cone is :

(A) πr²h

(B) ⅔πr³

(C) 4/3πr³

(D) ⅓πr²h

Answer : (D) ⅓πr²h

(xi) The value of y is ……….. on x-axis.

(A) 1

(B) 0

(C) 2

(D) 4

Answer : (B) 0

(xii) Probability of an event lies between 0 and 1. (True / False)

Answer : True (0 ≤ P ≤ 1)

(xiii) Every rational number is an integer. (True / False)

Answer : False

(xiv) A terminated line can be produced indefinitely. (True / False )

Answer : True

(xv) A circle has only finite numbers of equal chords. (True / False)

Answer : False

(xvi) A median of a triangle divides it into two triangles of equal areas. (True / False)

Answer : True

Q2. Find the remainder when x³ + 3x² + 3x + 1 is divisible by x + 1.

Solution :

Take x + 1 = 0 so x = -1

p(x) = x³ + 3x² + 3x + 1

p(-1) = (-1)³ + 3(-1)² + 3(-1) + 1 = – 1 + 3 – 3 +1 = 0

Remainder = 0

Q3. Using suitable identity, find the value of (102)².

Solution :

Using Formula, (a+b)² = a² + 2ab + b²

(102)² = (100+2)² = 100² + 2(100)(2) + 2²

= 10000 + 400 + 4 = 10404

Q4. Find the value of ‘k’ for which x – 1 is a factor of the polynomial x² + x + k.

Solution :

Here, x – 1 = 0 so x = 1

p(x) = x² + x + k

p(1) = (1)² + (1) + k = 0

1 + 1 + k = 0

k = -2

Q5. Find the four solutions for the equation 2x + y = 7.

Solution :

There are infinite solutions of this equation y = 7 – 2x.

When x = 0, then y = 7

When x = 1, then y = 5

When x = 2, then y = 3

When x = 3, then y = 1

Four solutions are (0, 7), (1, 5), (2, 3), (3, 1)

Q6. Find the value of P if x = 2, y = 1 is a solution of the equation 2x + 3y = P.

Solution :

2x + 3y = P

2(2) + 3(1) = P

4 + 3 = P

P = 7

Q7. If ∆PRQ ≅ ∆CBA, then find :

(i) QR = ……..

Answer : AB

(ii) PR = ……..

Answer : CB

(iii) PQ = ……..

Answer : CA

(iv) ∠B = ………

Answer : ∠R

Q8.(i) Rationalize the denominator of 1/(√5+√2)

Solution :

1/(√5+√2)

Rationalize the denominator,

= 1/(√5+√2) × (√5-√2)/(√5-√2)

= (√5-√2)/(√5²-√2²)

= (√5-√2)/(5-2)

= (√5-√2)/3

(ii) Simplify (32)^2/5.

Solution : (2⁵)^2/5 = 2² = 4

Q9. Look at the above figure and answer the following :

(i) Coordinates of A, B, C, D, E

Answer : A(3, 0), B(5, 0), C(0, 3), D(-5, 0), E(0, -4)

(ii) Abscissa of point D

Answer : D(-5, 0), Abscissa (x) = -5

(iii) Ordinate of point E

Answer : E(0, -4), Ordinate = -4

(iv) Coordinates of point P, Q

Answer : P(3, 5), Q(5, -4)

Q10. In the given figure, AB || CD || EF and y : z = 3 : 7 are given, then find the value of x.

Solution :

Take, y = 3a, z = 7a

y = ∠DQR (Vertical opposite angle)

∠DQR + z = 180° (Co-interior angle)

y + z = 180°

3a + 7a = 180°

10a = 180°

a = 18°

z = 7a = 7 × 18° = 126°

x = z = 126°

Q11. The angles of a quadrilateral are in the ratio 3 : 6 : 7 : 4. Find all the angles of the quadrilateral.

Solution :

Take, angles are 3x, 6x, 7x, 4x

3x + 6x + 7x + 4x = 360°

20x = 360°

x = 360°/20 = 18°

3x = 3 × 18° = 54°

6x = 6 × 18° = 108°

7x = 7 × 18° = 126°

4x = 4 × 18° = 72°

Therefore, all angles of quadrilateral are 18°, 54°, 108° and 72°.

Q12. In the given figure ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.

Solution :

Reflex ∠POR = 2 × 100° = 200° (angle subtend at centre is double)

∠POR = 360° – 200° = 160°

∠POR + ∠OPR + ∠ORP = 180°

160° + ∠OPR + ∠OPR = 180° (Here ∠OPR = ∠ORP because PO = RO = Radius)

2∠OPR = 180° – 160° = 20°

∠OPR = 20°/2 = 10°

Q13. The floor of a rectangular hall has the perimeter 250 m. If the cost of the painting the four walls at the rate ₹ 12 per m² is ₹ 18000, find the height of the hall.

Solution :

Perimeter of rectangle (P) = 2(L+B) = 250 m

Area of 4 walls = 2(L+B)H = 18000/12

2(L+B)H = 1500

250 × H = 1500

H = 1500/250 = 6 m

so, height of hall = 6 m

Q14. The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x :

29, 32, 48, 50, x, x+2, 72, 78, 84, 95

Solution :

Median = [x+(x+2)] ÷ 2 = 63

2x + 2 = 63 × 2

2x + 2 = 126

2x = 126 – 2

2x = 124

x = 124/2 = 62

Q15. Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes :

Find the probability of the following :

(i) Two heads

Answer : P(Two heads) = 72/200 = 9/25

(ii) Atmost two heads

Answer : P(Atmost two heads) = (72+77+28)/200 = 177/200

(iii) Three tails

Answer : P(Three tails or No head) = 28/200 = 7/50

Q16. Construct a triangle ABC in which ∠B = 60°, ∠C = 45° and perimeter is 10 cm.

Solution :

Steps of Construction :

(i) Draw a line segment XY equal to AB + BC + AC = 10 cm.

(ii) Make angle ∠LXY = 60° which is equal to ∠B = 60° .

(iii) Make angle ∠MYX = 45° which is equal to ∠C = 45°.

(iv) Bisect angles ∠LXY and ∠MYX. These bisectors intersect at a point A.

(v) Make perpendicular bisector of AX which intersect XY at point B.

(vi) Make perpendicular bisector of AY which intersect XY at point C.

(vii) Join AB & AC.

So, ∆ABC is the required triangle.

Q17. Sides of a triangle are in the ratio 3 : 5 : 7 and its perimeter is 300 m. Find the area of the triangle.

Solution :

Take, a = 3x, b = 5x, c = 7x

Perimeter (P) = a + b + c

3x + 5x + 7x = 300

15x = 300

x = 300/15 = 20

a = 3x = 3 × 20 = 60 m

b = 5x = 5 × 20 = 100 m

c = 7x = 7 × 20 = 140 m

Semi Perimeter (s) = 300/2 = 150 m

Using Heron’s Formula,

Area of triangle = √s(s-a)(s-b)(s-c)

A = √150(150-60)(150-100)(150-140)

= √150(90)(50)(10)

= 1500√3 m²

Q18. A solid cube of side 12 cm cut into 8 cubes of equal volumes. Find the surface area of both the cubes.

Solution :

Let side of small cube = a

8 × (Volume of small cube) = Volume of large cube

8 × a³ = 12³

8a³ = 1728

a³ = 1728/8

a³ = 216 = 6³

a = 6 cm

Surface area of large cube = 6(side)² = 6(12)² = 6(144) = 864 cm²

Surface area of small cube = 6(side)² = 6(6)² = 6(36) = 216 cm²

OR

The capacity of a closed cylindrical vessel of height 1 m is 15.4 liters. How many square meters of metal sheet would be needed to make it.

Solution :

Volume or capacity of cylinder = 15.4 litres

V = 15.4 ÷ 1000 = 0.0154 m³ (1m³ = 1000 liters)

Volume of cylinder = πr²h

πr²h = 0.0154

22/7 × r² × 1 = 0.0154

r² = 0.0154 × 7/22 = 0.0049

r = √0.0049 = 0.07

Total surface area (TSA) = 2πr(r+h)

TSA = 2 × 22/7 × 0.07 × (0.07+1) = 2 × 22/7 × 0.07 × 1.07 = 0.4708 m²

Metal Sheet needed = 0.4708 m²

Q19. The distance (in km) of 40 Engineers from their residence to their place of work were found as follows :

5, 3, 10, 20, 25, 11, 13, 7, 12, 31, 19, 10, 12, 17, 18, 11, 32, 17, 16, 2, 7, 9, 7, 8, 3, 5, 12, 15, 18, 3, 12, 14, 2, 9, 6, 15, 15, 7, 6, 12

Construct a grouped frequency distribution table with class-size 5 for the data given above taking the first interval as 0-5 (5 not included).

Solution :

OR

A family with a monthly income of ₹6400 plans his budget for a month as given below :

(i) Represent the above data by a bar-graph.

Solution :

(ii) On which item they spent more ?

Solution : Grocery

**HBSE Class 9th Maths Question Paper 2023 Answer**