HBSE Class 9 Maths Question Paper 2024 Answer Key

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HBSE Class 9 Maths Question Paper 2024 Answer Key

Section – A (1 Mark)

1. Which of the following numbers is an irrational number?
(a) √16
(b) √36
(c) √48
(d) √64
Solution : (c) √48
Here √16 = ±4, √36 = ±6, √64 = ±8 are rational numbers.

2. Which of the following is not a criterion for congruence of triangles?
(a) SSS
(b) SSA
(c) SAS
(d) ASA
Solution : (b) SSA
Five criterion for congruence of triangles are SSS, SAS, ASA, AAS, RHS.

3. Each angle of any equilateral triangle is :
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Solution : (c) 60°
All angles of any equilateral triangle are equal,
x + x + x = 180°
3x = 180°
x = 180°/3 = 60°

4. Two sides of a triangle are of lengths 5 cm and 1.5 cm. The length of the third side of the triangle cannot be :
(a) 3.6 cm
(b) 4.1 cm
(c) 3.8 cm
(d) 3.4 cm
Solution : (d) 3.4 cm
Sum of any two sides of a triangle is always greater than the third side.
But, 1.5 cm + 3.4 cm = 4.9 cm < 5 cm

5. If diagonal of a quadrilateral bisect each other, then quadrilateral is a parallelogram. (True / False)
Solution : True

6. Equal chords of a circle (or of congruent circles) subtend equal angles at the centre. (True / False)
Solution : True

7. If two sides of a triangle are equal, then it is called :
(a) Equilateral triangle
(b) Scalene triangle
(c) Isosceles triangle
(d) Right angled triangle
Solution : (c) Isosceles triangle

8. Volume of the cone is :
(a) πr²h
(b) ⅔ πr³
(c) 4/3 πr³
(d) ⅓πr²h
Solution : (d) ⅓πr²h

9. The class mark of the class 100-120 is :
(a) 100
(b) 110
(c) 120
(d) None of these
Solution : (b) 110
Class mark (mid value) = (lower limit + upper limit) / 2 = (100 + 120) / 2 = 220/2 = 110

10. The smallest prime number is :
(a) Zero
(b) 1
(c) 2
(d) 3
Solution : (c) 2

11. To draw a histogram to represent the following frequency distribution :

The adjusted frequency for class 20-25 is :
(a) 6
(b) 12
(c) 2
(d) None of these
Solution : (d) None of these
Adjusted frequency = (minimum width / class width) × frequency = 5/5 × 8 = 8

12. The coefficient of x² in x³ + 3x² + 2 is :
(a) 1
(b) 3
(c) 2
(d) None of these
Solution : (b) 3

13. Find the value of polynomial x² – x + 1 at x = 1 :
(a) 1
(b) 2
(c) 3
(d) 0
Solution : (a) 1
x² – x + 1 = (1)² – (1) + 1 = 1 – 1 + 1 = 1

14. Volume of sphere whose radius is R :
(a) 4πR²
(b) 4πR³
(c) 4/3 πR³
(d) 2/3 πR³
Solution : (c) 4/3 πR³

15. In ∆ABC, AB = AC and ∠B = 70° then ∠C is :
(a) 40°
(b) 50°
(c) 60°
(d) 70°
Solution : (d) 70°
Given AB = AC
so, ∠B = ∠C = 70° (opposite angles of equal sides are equal)

16. Sum of the interior angles of quadrilateral is :
(a) 180°
(b) 240°
(c) 360°
(d) 300°
Solution : (c) 360°

17. If diagonal of a parallelogram are equal, then it is a ……………..
(a) Rhombus
(b) Rectangle
(c) Square
(d) None of these
Solution : (b) Rectangle

18. Angles in the same segment of a circle are …………….
Solution : Equal

19. Assertion (A) : If √2 = 1.414, √3 = 1.732, then √5 = √2 + √3.
Reason (R) : Square root of positive number always exists.
(a) Both (A) and (R) are true, (R) is the correct explanation of (A).
(b) Both (A) and (R) are true but (R) is not correct explanation of (A).
(c) (A) is true but (R) is false.
(d) (A) is false but (R) is true.
Solution : (d) (A) is false but (R) is true.

20. Assertion (A) : A chord of a circle, which is twice as long as its radius, is a diameter of the circle.
Reason (R) : The longest chord of a circle is a diameter of the circle.
(a) Both (A) and (R) are true, (R) is the correct explanation of (A).
(b) Both (A) and (R) are true but (R) is not correct explanation of (A).
(c) (A) is true but (R) is false.
(d) (A) is false but (R) is true.
Solution : (a) Both (A) and (R) are true, (R) is the correct explanation of (A).

Section – B (2 Marks)

21. Find five rational numbers between 1 and 2.
Solution : Multiply and divide the both numbers by 6.
1 × 6/6, 2 × 6/6
6/6, 12/6
So, the five rational numbers between 1 and 2 are 7/6, 8/6, 9/6, 10/6, 11/6.

22. Simplify : (3 + √3)(3 – √3)
Solution : Using identity, (a + b)(a – b) = a² – b²
(3 + √3)(3 – √3)
= (3)² – (√3)²
= 9 – 3
= 6

OR

Simplify : (32)^⅕
Solution : (32)^⅕ = (2⁵)^⅕ = 2⁵×⅕ = 2

23. Rationalise the denominator of 1/√7-2.
Solution : Divide and multiply by √7+2
(1/√7-2) × (√7+2/√7+2)
= (√7+2)/(√7²-2²)
= (√7+2)/(7-4)
= (√7+2)/3

24. Evaluate 103 × 107.
Solution : 103 × 107 = (100 + 3)(100 + 7)
Using identity (x + a)(x + b) = x² + (a + b)x + ab
= (100)² + (3 + 7)100 + 3×7
= 10000 + 1000 + 21
= 11021

25. Find the value of k, if x – 1 is a factor of P(x) = 2x² + kx + √2.
Solution : Here x -1 = 0 so x = 1
Put x = 1 in P(x) = 2x² + kx + √2
2(1)² + k(1) + √2 = 0
2 + k + √2 = 0
k = -(2+√2)

OR

Use the factor theorem factorize of x² – 5x + 6.
Solution : x² – 5x + 6 = x² – 3x – 2x + 6
= x(x – 3) – 2(x – 3)
= (x – 3)(x – 2)

Section – C (3 Marks)

26. Factorize : 2x² + 7x + 3
Solution : 2x² + 7x + 3 = 2x² + 6x + x + 3
= 2x(x + 3) + 1(x + 3)
= (x + 3)(2x + 1)

27. A hemispherical bowl has a radius of 3 cm. What would be the volume of water it would contain.
Solution : Volume of hemisphere = ⅔πr³
= 2/3 × 22/7 × (3)³
= 396/7
= 56.57 cm³ (approx.)

OR

Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Solution : Diameter (d) = 24 m, Radius (r) = d/2 = 24/2 = 12 m
Total surface area of cone (TSA) = πr² + πrl = πr(r + l)
= 22/7 × 12 (12 + 21)
= 22/7 × 12 × 33
= 8712/7
= 1244.57 m²

28. Factorize : 64x³ – 343y³
Solution : Using identity, a³ – b³ = (a – b)(a² + ab – b²)
64x³ – 343y³ = (4x)³ – (7y)³
= (4x – 7y)[(4x)² + (4x)(7y) + (7y)²]
= (4x – 7y)(16x² + 28xy + 49y²)

29. Find three different solutions of the equation 2x + y = 7.
Solution : y = 7 – 2x
Taking x = 0 then y = 7 – 2(0) = 7
Taking x = 1 then y = 7 – 2(1) = 5
Taking x = 2 then y = 7 – 2(2) = 3
Hence (0,7), (1,5) and (2,3) are three solutions of the given equation.

30. Find the value of k, if x = 2 y = 1 is a solution of the equation 2x + 3y = k.
Solution : Here 2x + 3y = k
2(2) + 3(1) = k
4 + 3 = k
k = 7

31. Factorize : 8x³ + y³ + 27z³ – 18xyz.
Solution : Using identity, a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)
8x³ + y³ + 27z³ – 18xyz
= (2x)³ + (y)³ + (3z)³ – 3(2x)(y)(3z)
= (2x + y + 3z)[(2x)² + (y)² + (3z)² – (2x)(y) – (y)(3z) – (3z)(2x)]
= (2x + y + 3z)(4x² + y² + 9z² – 2xy – 3yz – 6zx)

OR

Using suitable identity, find the value of (999)³.
Solution : (999)³ = (1000 – 1)³
Using identity, (a – b)³ = a³ – b³ – 3ab(a – b)
= (1000)³ – (1)³ – 3(1000)(1)(1000 – 1)
= 1000000000 – 1 – 3000(999)
= 1000000000 – 1 – 299700
= 1000000000 – 299701
= 997002999

Section – D (5 Marks)

32. In fig. PQ and RS lines are intersect at point O. If ∠POR : ∠ROQ = 5 : 7, then find the all angles.

Solution : Here ∠POR = 5x, ∠ROQ = 7x
∠POR + ∠ROQ = 180° (Linear Pair)
5x +7x = 180°
12x = 180°
x = 180°/12 = 15°
∠POR = 5x = 5 × 15 = 75°
∠ROQ = 7x = 7 × 15 = 105°
∠POR = ∠QOS = 75° (V.O.A)
∠ROQ = ∠POS = 105° (V.O.A)

33. If a point C lies between two points A and B such that AC = BC, then prove that AC = ½AB. Explain by drawing the figure.
Solution : Given AC = BC

Adding AC on both sides, we get
AC + AC = AC + BC
2AC = AB
AC = ½AB
Hence Proved.

OR

In fig. lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.

Solution : Here a = 2x, b = 3x
a + b + ∠POY = 180° (Linear Pair)
2x + 3x + 90° = 180°
5x + 90° = 180°
5x = 180° – 90° = 90°
x = 90°/5 = 18°
a = 2x = 2 × 18° = 36°
b = 3x = 3 × 18° = 54°
b + c = 180° (Linear Pair)
54° + c = 180°
c = 180° – 54° = 126°

34. Find the area of triangle whose sides are 40 m, 24 m and 32 m.
Solution : Here a = 40m, b = 24m, c = 32m
Semi perimeter (s) = (a+b+c)/2 = (40+24+32) = 96/2 = 48 m
Using Heron’s formula, area of triangle is
A = √s(s-a)(s-b)(s-c)
= √48(48-40)(48-24)(48-32)
= √48(8)(24)(16)
= √147456
= 384 m²

OR

Find the volume of a right circular cone whose radius 6 cm and height is 7 cm.
Solution : Volume of cone = ⅓πr²h
= 1/3 × 22/7 × (6)² × 7
= 264 cm³

35. Show that diagonal of a rhombus perpendicular to each other.
Solution :

Given : ABCD is a rhombus; AC and BD intersect at E.
To prove : AC ⊥ BD
Proof : In ΔABE and ΔADE
AB = AD (sides of a rhombus)
BE = DE (diagonals bisect each other)
AE = AE (common)
∆ABE ≅ ∆ADE (by SSS)
∠AEB = ∠AED (CPCT)
∠AEB + ∠AED = 180° (Linear pair)
∠AEB + ∠AEB = 180° (∠AEB = ∠AED)
2∠AEB = 180°
∠AEB = 180°/2 = 90°
∠AEB = ∠AED = 90°
Hence, AC and BD are perpendicular to each other.

Section – E (4 Marks)

36. See the fig. and answer the following questions :

(i) The co-ordinate of B.
Solution : (-5, 2)

(ii) The point identified by the co-ordinate (-3, -5).
Solution : E

(iii) (a) Find the abscissa of point D and ordinate of point H.
Solution : Abscissa of point D = 6 and ordinate of point H = -3.

OR

(b) Find the area of rectangle formed by the line segment BD and X-axis in fig.
Solution : Area = 11 × 2 = 22 sq. unit

37. The triangular side walls of the flyover have been used for advertising. The sides of the walls are 122 m, 22 m and 120 m. Earning of Rs. 5000 per m² per year from advertisements is. 

On the basis of the above information and given figure, answer the following questions :
(i) Perimeter of wall is ……………
Solution : P = 122 + 22 + 120 = 264 m

(ii) Write down the Heron’s formula.
Solution : Area = √s(s-a)(s-b)(s-c)

(iii) (a) Area of triangular wall is ……………
Solution : Semi perimeter (s) = 264/2 = 132 m
A = √s(s-a)(s-b)(s-c)
= √132(132-122)(132-22)(132-120)
= √132(10)(110)(12)
= 1320 m²

OR

(b) If company hired one of its walls with area 1680 m² for 3 month, then how much rent did it pay?
Solution : Rent = 1680 × 5000 × 3/12 = ₹ 2100000

38. In a Particular section of IX Class, 40 students were asked about the months of their birth and the following graph was prepared for the data so obtained :

Observe the bar graph given above and answer the following questions :
(i) How many students were born in the month of Novemeber?
Solution : 4 students

(ii) In which month were the maximum number of students born?
Solution : August

(iii) (a) Name the month in which 4 students were born.
Solution : Feb, Oct, Nov, Dec

OR

(b) Find the total number of students born from May to August.
Solution : 5 + 1 + 2 + 6 = 14 students

 

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