**Haryana Board (HBSE)** Class 10 Maths Pre Board Question Paper 2024 Answer Key. Haryana Board Class 10th Pre Board Question Paper PDF Download 2024. Haryana Board Class 10th Pre Board Question Paper Maths 2024. HBSE Class 10th Mathematics Pre Board Question Paper Solution 2024. HBSE Maths Pre Board Question Paper 2024 Class 10.

**HBSE Class 10 Maths Pre-Board Question Paper 2024 Answer Key**

Section – A (1 Mark)

1. LCM of 6, 9 and 15 is ………….

(a) 30

(b) 60

(c) 90

(d) 120

Solution : (c) 90

2. Which of the following is a rational number?

(a) √2

(b) √3

(c) √4

(d) √5

Solution : (c) √4

3. Find the zeros of quadratic polynomial x² – 7x + 12.

(a) 2, 5

(b) -2, -5

(c) 4, 3

(d) -4, -3

Solution : (c) 4, 3

x² – 7x + 12 = 0

x² – 4x – 3x + 12 = 0

x(x – 4) – 3(x – 4) = 0

(x – 4)(x – 3) = 0

x = 4, 3

4. Values of k for which the quadratic equation 2x² + kx + 2 = 0 has equal roots.

(a) ± 2

(b) ± 4

(c) ± 3

(d) ± 5

Solution : (b) ±4

Discriminant (D) = b² – 4ac = 0

k² – 4(2)(2) = 0

k² – 16 = 0

k = √16 = ± 4

5. Find the 15th term of AP: 3, 8, 13, ……………

(a) 53

(b) 63

(c) 73

(d) 83

Solution : (c) 73

First term (a) = 3

Common difference (d) = 8 – 3 = 5

Number of terms (n) = 15

nth term of AP, an = a + (n-1)d

a15 = a + 14d = 3 + (14)5 = 3 + 70 = 73

6. All …………… triangles are similar. (Isosceles / equilateral)

Solution : equilateral

7. The distance between the points (-2, 4) and (5, 7) is ……………

(a) √18

(b) √28

(c) √48

(d) √58

Solution : (d) √58

Taking A(-2, 4) and B(5, 7)

Here x1 = -2, y1 = 4, x2 = 5, y2 = 7

Distance formula = √(x2-x1)²+(y2-y1)²

AB = √(5+2)²+(7-4)² = √7²+3² = √49+9 = √58

8. Find the value of (1-cos²45°)/(1+cos²45°)

(a) 1/2

(b) 1/3

(c) 1/4

(d) 1

Solution : (b) 1/3

(1-cos²45°) / (1+cos²45°) = [1-(1/√2)²] / [1+(√2)²)]

= (1-1/2) / (1+1/2)= (1/2) / (3/2) = 1/3

9. 7cosec²A – 7cot²A equal to :

(a) 1

(b) 3

(c) 5

(d) 7

Solution : (a) 1

7cosec²A – 7cot²A = 7(cosec²A – cot²A) = 7(1) = 7

10. If cosA = 3/4 then what is the value of tanA?

(a) √7

(b) √7/3

(c) √7/4

(d) √3/5

Solution : (b) √7/3

sinA = √1-cos²A = √1-(3/4)² = √1-9/16 = √7/4

tanA = sinA/cosA = (√7/4) / (3/4) = √7/3

11. From a point Q, the length of tangent to a circle is 7 cm and distance of Q from center is 25 cm. The radius of circle is :

(a) 24 cm

(b) 12 cm

(c) 15 cm

(d) 24.5 cm

Solution : (a) 24 cm

Using Pythagoras theorem, H² = P² + B²

PO² + PQ² = QO²

r² + 7² = 25²

r² = 625 – 49 = 576

r = √576 = 24 cm

12. How many tangents can a circle have?

(a) 1

(b) 2

(c) 4

(d) infinitely many

Solution : (d) infinitely many

13. Find the area of sector of a circle with radius 7 cm if angle of sector is 30°.

(a) 77/6 cm²

(b) 77/3 cm²

(c) 77/2 cm²

(d) 154/3 cm²

Solution : (a) 77/6 cm²

Area of sector = θ/360 × πr² = 30/360 × 22/7 × 7² = 77/6 cm²

14. Area of sector of angle P (in degrees) of circle with radius R is …………

(a) P/180 × 2πR

(b) P/180 × πR²

(c) P/360 × 2πR

(d) P/720 × 2πR²

Solution : (d) P/720 × 2πR²

Area of sectror = P/360 × πR²

15. A cylindrical pencil sharpened at one edge is the combination of :

(a) a cone and a cylinder

(b) a hemi sphere and a cylinder

(c) two cylinders

(d) a cone and two cylinder

Solution : (a) a cone and a cylinder

16. What would be the median of first five natural numbers?

(a) 3

(b) 5

(c) 7

(d) 10

Solution : (a) 3

Middle term of 1, 2, 3, 4, 5 in ascending order

17. Find the mode of the following data.

2, 3, 2, 4, 1, 2, 3

(a) 1

(b) 2

(c) 3

(d) 4

Solution : (b) 2

18. A number is selected from numbers 1 to 30. The probability that it is prime, is :

(a) 11/30

(b) 1/29

(c) 1/3

(d) 1/6

Solution : (c) 1/3

Numbers are 1, 2, 3, 4, ………., 30

Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29

P(prime number) = 10/30 = 1/3

Directions for question 19 and 20: In question no. 19 and 20 a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option from (a), (b), (c) and (d).

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion(A).

(b) Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).

(c) Assertion (A) is true but Reason (R) is false.

(d) Assertion (A) is false but Reason (R) is true.

19. Assertion (A) : √5 is an example of Irrational number.

Reason (R) : The square root of all positive integers are irrational numbers.

Solution : (c) Assertion (A) is true but Reason (R) is false.

20. Assertion (A) : The point (-2, 0) lies on x-axis.

Reason (R) : The y-coordinate on the point on x-axis is zero.

Solution : (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion(A).

Section – B (2 Marks)

21. Solve the following pair of linear equations :

3x/2 – 5y/3 = -2 and x/3 + y/2 = 13/6

Solution : Multiply both eqn. by 6 bothside,

9x – 10y = -12 ………(i)

2x + 3y = 13 ……….(ii)

Multiply eq.(i) by 2 and multiply eqn.(ii) by 9,

18x – 20y = -24 ……….(iii)

18x + 27y = 117 ………..(iv)

Subtract eqn.(iii) from eqn.(iv), we get

47y = 141

y = 3

Put y = 3 in eqn.(i),

9x – 10(3) = -12

9x = -12 + 30

9x = 18

x = 2

So, x = 2 and y = 3

22. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA² = CB.CD

Solution :

In ΔABC and ΔDAC,

∠BAC = ∠ADC (Given)

∠ACB = ∠ACD (Common angles)

so, ΔABC ∼ ΔDAC (AA criterion)

If two triangles are similar, then their corresponding sides are proportional

CA/CD = CB/CA

CA² = CB.CD

Hence, proved.

23. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Solution :

To prove: PQ || RS

Given: A circle with centre O and diameter AB and let PQ and RS be the tangent at point A and B respectively.

Proof: AB ⊥ PQ and AB ⊥ RS (tangent is perpendicular to radius)

∠PAB = 90°

∠ABS = 90°

so, PQ || RS (∠PAB and ∠ABS are alternate angles equal)

24. Evaluate the following :

3cot²45° + 2cos²30° – sin²60°

Solution : 3(1)² + 2(√3/2)² – (√3/2)²

= 3(1) + 2(3/4) – (3/4)

= 3 + 6/4 – 3/4 = 15/4

25. Find the area of quadrant of circle whose circumference is 22 cm.

Solution : Circumference = 2πr = 22

2 × 22/7 × r = 22

r = 7/2 cm

Area of quadrant (4th part of circle) = 1/4 × πr² = 1/4 × 22/7 × (7/2)² = 77/8 cm²

OR

The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Solution : Area swept by minute hand in 60 minutes = πr²

Area swept by minute hand in 1 minute = πr²/60

Area swept by minute hand in 5 minute = πr²/60 × 5 = 1/12 × πr² = 1/12 × 22/7 × 14 × 14 = 154/3 cm²

Section- C (3 Marks)

26. Prove that √2 + √3 is irrational number.

Solution : Let us assume, to the contrary, that √2 + √3 is rational.

So, we can find integers a and b (≠ 0) such that √2 + √3 = p/q, where p and q are co-prime integers and q ≠ 0.

√3 = p/q – √2

Squaring both side,

(√3)² = (p/q – √2)²

3 = (p/q)² + (√2)² – 2(√2)p/q

2(√2)p/q = p²/q² + 2 – 3

2(√2)p/q = p²/q² – 1

2(√2)p/q = (p²-q²)/q²

√2 = (p²-q²)/q² × q/2p

√2 = (p²-q²)/2pq

Here (p²-q²)/2pq is a rational number but √2 is irrational number.

This contradiction has arisen because of our incorrect assumption that √2 + √3 is rational.

So, this proves that √2 + √3 is an irrational number.

27. Find the zeros of the quadratic polynomial x² – 9x + 20 and verify the relationship between zeros and coefficients.

Solution : Compare with ax² + bx + c.

Here a = 1, b = -9, c = 20

x² – 9x + 20 = 0

x² – 5x – 4x + 20 = 0

x(x – 5) – 4(x – 5) = 0

(x – 5)(x – 4) = 0

x = 5, 4

So, α = 5 and β = 4

α + β = 5 + 4 = 9 = -b/a = -(-9)/1 = 9

αβ = 5 × 4 = 20 = c/a = 20/1 = 20

Thus, the basic relationships are verified.

28. The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them manages to save 2000 per month, find their monthly income.

Solution : Let the two persons be A and B.

Incomes ratio = 9 : 7

Expenditures = 4 : 3

Income of A = 9x

Expenditure of A = 4y

Saving of A = 9x – 4y = 2000 ……..(i)

Income of B = 7x

Expenditure of B = 3y

Saving of B = 7x – 3y = 2000 ……..(ii)

Multiply eqn.(i) by 3, eqn.(ii) by 4, we get

27x – 12y = 6000 ……….(iii)

28x – 12y = 8000 ……….(iv)

Subtract eqn.(iii) from eqn.(iv), we get

x = 2000

Income of A = 9x = 9 × 2000 = ₹18000

Income of B = 7x = 7 × 2000 = ₹14000

OR

Five years ago, Nuri was thrice as old as Sonu. Two years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

Solution : Let the present age of Nuri = x

And, the present age of Sonu = y

5 years ago, age of Nuri = x – 5

5 years ago, age of Sonu = y – 5

ATQ,

x – 5 = 3(y – 5)

x – 3y = – 10 ……….(i)

10 years later, age of Nuri = x + 10

10 years later, age of Sonu = y + 10

ATQ,

x + 10 = 2(y + 10)

x – 2y = 10 …….(ii)

Subtract eqn.(i) from eqn.(ii), we get

y = 20

Put y = 20 in eqn.(ii),

x – 2(20) = 10

x – 40 = 10

x = 10 + 40 = 50

Hence, age of Nuri (x) = 50 years and age of Sonu (y) = 20 years.

29. Prove that the parallelogram circumscribing a circle is a rhombus.

Solution :

Given: A circle with centre O. A parallelogram ABCD touching the circle at points P, Q, R and S.

To prove: ABCD is a rhombus.

Proof: A rhombus is a parallelogram with all sides equal, So, we have to prove all sides equal.

In parallelogram ABCD, AB = CD & AD = BC (opposite sides of parallelogram are equal)

Hence,

AP = AS (tangents are equal)

BP = BQ (tangents are equal)

CR = CQ (tangents are equal)

DR = DS (tangents are equal)

Adding,

AP + BP + CR + DR = AS + BQ + CQ + DS

(AP+BP) + (CR+DR) = (AS+DS) + (BQ+CQ) AB + CD = AD + BC

AB + AB = AD + AD (given AB = CD & AD = BC)

2AB = 2AD

AB = AD

so, AB = BC = CD = DA

Hence, ABCD is a rhombus.

30. Prove that :

cos/(1+sinA) + (1+sinA)/cosA = 2secA

Solution : LHS. cos/(1+sinA) + (1+sinA)/cosA

= [cos²A+(1+sinA)²] / (1+sinA)cosA

= [cos²A+1+sin²A+2sinA] / (1+sinA)cosA

= (2+2sinA) / (1+sinA)cosA

= 2(1+sinA) / (1+sinA)cosA

= 2/cosA

= 2secA = RHS

OR

(cosecθ-cotθ)² = (1-cosθ)/(1+cosθ)

Solution : RHS. (1-cosθ)/(1+cosθ)

Rationalisation of denominator,

= (1-cosθ)/(1+cosθ) × (1-cosθ)/(1+cosθ)

= (1-cosθ)² / (1²-cos²θ)

= (1+cos²θ-2cosθ) / sin²θ

= 1/sin²θ + cos²θ/sin²θ – 2cosθ/sin²θ

= cosec²θ + cot²θ – 2cotθ.cosecθ

= (cosecθ – cotθ)² = LHS

31. One card is drawn from a well shuffled pack of 52 cards. Find the probability of getting :

(i) a queen of red colour

Solution : P(E) = 2/52 = 1/26

(ii) a face card

Solution : P(E) = 12/52 = 3/13

(iii) 2 of hearts

Solution : P(E) = 1/52

Section – D (5 Marks)

32. Ritu can row downstream 20 km in 2 hours and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

Solution : Let Ritu’s speed of rowing in still water = x km/h

Speed of the stream = y km/h

Ritu’s speed of rowing, upstream = (x – y) km/h

Ritu’s speed of rowing, downstream = (x + y) km/h

Using formula, Speed × Time = Distance

According to question,

Ritu can row downstream 20 km in 2 hours,

2(x + y) = 20

x + y = 10 ………(i)

Ritu can row upstream 4 km in 2 hours,

2(x – y) = 4

x – y = 2 ………(ii)

Adding eqn.(i) and (ii), we get

2x = 12

x = 6

Put x = 6 in eqn.(i),

6 + y = 10

y = 4

Ritu’s speed of rowing in still water (x) = 6 km/h

Speed of the stream (y) = 4 km/h

33. Prove that if a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then other two sides are divided in same ratio.

Solution :

To Prove: AD/DB = AE/EC

Construction: Draw EM ⊥ AB and DN ⊥ AC. Join B to E and C to D.

Proof: In ∆ADE and ∆BDE,

ar(∆ADE) / ar(∆BDE) = (½×AD×EM) / (½×DB×EM) = AD/DB ……….(i)

In ∆ADE and ∆CDE,

ar(∆ADE) / ar(∆CDE) = (½×AE×DN) / (½×EC×DN) = AE/EC ……….(ii)

Since, DE || BC [Given]

∴ ar(∆BDE) = ar(CDE) ……….(iii)

[∆’s having same base and between the same parallel lines then they are equal in area]

From eqn.(i), eqn.(ii) and eqn.(iii),

AD/DB = AE/EC

Hence Proved.

This is also known as Thales theorem or Basic Proportionality Theorem (BPT).

34. A gulab jamun contains sugar syrup upto 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns which shape like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm?

Solution :

Radius of cylindrical part (r) = Radius (r) of hemispherical part = 2.8/2 = 1.4 cm

Length of each hemispherical part = Radius of hemispherical part = 1.4 cm

Length of cylindrical part (h) = 5 – (2 × length of hemispherical part) = 5 – (2 × 1.4) = 5 – 2.8 = 2.2 cm

Volume of one gulab jamun = Volume of cylindrical part + (2 × Volume of hemispherical part)

We know that, volume of a cylinder is πr²h and volume of a hemisphere is ⅔πr³

∴ Volume of one gulab jamun πr²h + (2 × ⅔πr³) = πr²h + 4/3 πr³ = [22/7 × (1.4)² × 2.2] + [4/3 × 22/7 × (1.4)³] = 13.552 + 11.498 = 25.05 cm³

Hence, volume of one gulab jamun = 25.05 cm³

So, volume of 45 gulab jamuns = 45 × 25.05 = 1127.25 cm³

Also, given that, volume of sugar syrup = 30% of volume of gulab jamun

∴ Volume of the sugar syrup of 45 gulab jamuns = 30/100 × 1127.25 = 338.17 cm³ ≈ 338

Hence, the amount of syrup in 45 gulab jamuns is approximately 338 cm³.

OR

From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm².

Solution :

CSA of the cylinder = 2πrh

Area of the base of the cylinder = πr²

CSA of the cone = πrl

Height of the cylinder = Height of the cone = h = 2.4 cm

Diameter of the cylinder = diameter of the cone = d = 1.4 cm

Radius of the cylinder = radius of the cone = r = d/2 = 1.4/2 cm = 0.7 cm

Slant height of the cone, l = √r²+h² = √(0.7)²+(2.4)² = √0.49+5.76 = √6.25 = 2.5 cm

TSA of the remaining solid = CSA of the cylindrical part + CSA of conical part + Area of one cylindrical base = 2πrh + πrl + πr² = πr(2h + l + r)

= 22/7 × 0.7 × (2×2.4 + 2.5 + 0.7) = 17.6 cm²

Hence, the total surface area of the remaining solid to the nearest 18 cm².

35. The median of the following data is 525. Find the values of x and y if total frequency is 100.

Solution :

n = 100 ⇒ n/2 = 50

76 + x + y = 100

x + y = 24 ……..(i)

Median = 525 (median class is 500-600)

l = 500, f = 20, cf = 36+x, h = 100

Median = l + (n/2 – cf)/f × h

525 = 500 + (50-36-x)/20 × 100

525 – 500 = (14-x) × 5

25 = 70 – 5x

5x = 70 – 25 = 45

x = 45/5 = 9

From eqn.(i), we get 9 + y = 24

y = 24 – 9 = 15

So, the values of x = 9 and y = 15.

Section – E (4 Marks : Case Study)

36. In a classroom 4 friends are seated at the points A, B, C and D as shown in figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.

Solution : Here A(3,4), B(6,7), C(9,4), D(6,1)

To check ABCD is a square check AB = BC = CD = DA and AC = BD

AB = √(x2-x1)²+(y2-y1)² = √(6-3)²+(7-4)² = √3²+3² = √9+9 = √18

Similarly BC = √18, CD = √18, DA = √18

AC = √(9-3)²+(4-4)² =√6² = 6

Similarly BD = 6

So, AB = BC = CD = DA = √18 unit, AC = BD = 6 unit

Hence, ABCD is a square.

37. Two poles of equal height are standing opposite each other on either side of road which is 80m wide. From a point between them on the road, the angles of elevation of the top of poles are 60° and 30° respectively. Find the height of the poles and the distance of the point from the poles.

Solution :

In right angled ∆DCP,

tanθ = P/B

tan30° = h/x

1/√3 = h/x

h = x/√3 ………(i)

In right angled ∆ABP,

tanθ = P/B

tan60° = h/(80-x)

√3 = h/(80-x)

h = √3(80-x) ………(ii)

From eqn.(i) and (ii), we get

x/√3 = √3(80-x)

x = 60 m

CP = x = 60 m

PB = 80 – x = 80 – 60 = 20 m

Put x = 60 m in eqn.(ii), we get

h = √3(80-60) = √3(20)

h = 20√3 m

Thus, position of the point P is 60 m from C and P is 20 m from B and height of each pole is 20√3 m.

38. India is competitive manufacturing location due to low cost of manpower and strong technical and engineering capabilities contributing to higher quality production runs. The production of TV sets in a factory increases uniformly by a fixed number every year. It produced 16000 sets in 6th year and 22600 in 9th year. On the basis of above information, answer the following questions.

Solution :

nth term of AP, an = a + (n-1)d

a6 = a + 5d = 16000 …….(i)

a9 = a + 8d = 22600 …….(ii)

Subtract eqn.(i) from (ii), we get

3d = 6600

d = 2200

Put d = 2200 in eqn.(i),

a + 5(2200) = 16000

a + 11000 = 16000

a = 16000 – 11000 = 5000

(i) What is the production during 8th year? (1 Mark)

Solution : a8 = a + 7d = 5000 + 7(2200) = 5000 + 15400 = 20400

(ii) Write the total production in (during) first 3 years. (1 Mark)

Solution : Sn = n/2 [2a + (n-1)d]

S3 = 3/2 [2×5000 + 2×2200] = 3/2 [10000 + 4400] = 3/2 × 14400 = 21600

(iii) In which year, the production is 29,200? (2 Marks)

Solution : an = a + (n-1)d

5000 + (n-1)2200 = 29200

5000 + 2200n – 2200 = 29200

2200n + 2800 = 29200

2200n = 29200 – 2800

2200n = 26400

n = 12

In 12th year, the production is 29200.