# HBSE Class 10 Maths Question Paper 2020 Answer Key

HBSE Class 10 Maths Question Paper 2020 Answer Key

HBSE Class 10 Maths Previous Year Question Paper with Answer. HBSE Board Solved Question Paper Class 10 Maths 2020. HBSE 10th Question Paper Download 2020. HBSE Class 10 Maths Paper Solution 2020. Haryana Board Class 10th Maths Question Paper 2020 Pdf Download with Answer.

Q1. Express 0.375 in the form p/q.

Sol.

0.375 = 375/1000 = 3/8

Q2. The zeroes of 6x² – 7x – 3 are :

(A) -1/3, 3/2

(B) -7/3, -3/6

(C) 7/6, -3/6

(D) None of these

Ans. (A) -1/3, 3/2

6x² – 7x – 3 = 0

6x² – 9x + 2x – 3 = 0

3x(2x – 3) +1(2x – 3) = 0

(2x – 3)(3x + 1) = 0

Either 3x + 1 = 0  or  2x – 3 = 0

x = -1/3, 3/2

Q3. Solve :  x + y = 14,  x – y = 4

Sol.

x + y= 14 ……..(i)

x – y= 4 ………(ii)

Add eqn.(i) and eqn.(ii),

2x = 18

x = 18/2 = 9

Put x = 9 in eqn.(i),

9 + y = 14

y = 14 – 9 = 5

So, x = 9 and y = 5

Q4. Which one is an A. P. series ?

(A) 2, 4, 8, 12, ………

(B) 0.2, 0.22, 0.222, …….

(C) -10, -6, -2, 2, ………

(D) 1, 3, 9, 27, ……….

Ans. (C) -10, -6, -2, 2, ………

Common difference = d

d = -6 – (-10) = -2 – (-6) = 2 – (-2) = 4

Q5. Find the 10th term of A. P. 2, 7, 12, ……..

Sol.

Here, a = 2,  d = 7 – 2 = 5,  n = 10

Tn or an = a + (n-1)d

a10 = 2 + (10-1)5 = 2 + (9)5 = 2 + 45 = 47

Q6. Fill in the blank using correct word given in bracket :

All circles are …………. (congruent, similar)

Ans. similar

Q7. Sides of two similar triangles are in the ratio 4 : 9. Areas of their triangles are in the ratio :

(A) 16 : 81

(B) 8 : 18

(C) 81 : 16

(D) 12 : 27

Ans. (A) 16 : 81

If two triangles are similar to each other, then the ratio of the area of triangles will be equal to the square of the ratio of the corresponding sides of triangle.

So, Areas of their triangles are in the ratio = (4)² : (9)² = 16 : 81

Q8. If the areas of two similar triangles are 36 m² and 121 m² respectively, the ratio of there corresponding sides is :

(A) 11 : 6

(B) 6 : 11

(C) 9 : 11

(D) None of these

Ans. (B) 6 : 11

If two triangles are similar to each other, then the ratio of the corresponding sides of triangles will be equal to the square root of the ratio of the corresponding areas of triangle.

So, Sides of triangles are in the ratio = √36 : √121 = 6 : 11

Q9. From a point Q, the length of tangent to a circle is 24 cm and distance of Q from centre is 25 cm. Find the radius of circle.

Sol.

Using Pythagoras Theorem,

H² = P² + B²

25² = r² + 24²

r² = 25² – 24² = 625 – 576 = 49

r = √49 = 7 cm

Q10. Find the distance between the points (2, 3) and (4, 1).

Sol.

A(2, 3) and B(4, 1)

Here x1 = 2, x2 = 4, y1 = 3, y2 = 1

Using Distance Formula,

AB = √x2-x1)²+(y2-y1

= √(4-2)²+(1-3)²

= √(2)²+(-2)²

= √4+4 = √8 = 2√2 unit

Q11. If (3, 4) is mid point of the line segment whose one end is (7, -2), then find the coordinates of the other end point.

Sol.

O is Mid point of AB,

A(x, y) and O(3,4) and B(7, -2)

Using Mid Point Formula,

O(3, 4) = (x+7)/2 , (y-2)/2

(x+7)/2 = 3 and (y-2)/2 = 4

x + 7 = 6 and y – 2 = 8

x = 6 – 7 = -1 and y = 8 + 2 = 10

Co-ordinates of A(x, y) are (-1, 10).

Q12. Find the value of  tan65°/cot25°.

Sol.

tan65°/cot25° = tan(90°-25°)/cot25° = cot25°/cot25° = 1

Q13. If sinA = 3/4 , then cosA is :

(A) 4/√7

(B) √7/4

(C) 3/√7

(D) None of these

Ans. (B) √7/4

sin²A + cos²A = 1

cos²A = 1 – sin²A

cos²A = 1 – (3/4)² = 1 – 9/16 = 7/16

cosA = √7/4

Q14. Find the area of a sector of a circle with radius 4 cm, if angle of the sector is 30°. (Use π = 3.14)

Sol.

Area = θ/360 × πr² = 30/360 × 3.14 × 4 × 4 = 4.19 cm²

Q15. The diameter of the base of right circular cylinder is 2r and its height is h. The curved surface area is :

(A) πr²h

(B) 2πrh

(C) 2πr(r+h)

(D) None of these

Ans. (B) 2πrh

Q16. A bag contains 3 blue balls, 2 white balls and 4 red balls. If one ball is taken out at random from the bag. What is the probability that it will be White ?

Sol.

Total balls = 3 + 2 + 4 = 9

White balls = 2

P(white) = 2/9

Q17. Prove that √2 is an irrational number.

Sol.

Let us assume that √2 is a rational number.

∴ √2 = p/q,  where p, q are co-prime integers and q ≠ 0

p = √2q

Squaring both side, we get

p² = 2q² ……….. (i)

∴ 2 is a factor of p²

∴ 2 divides p²

∴ 2 divides p

Put, p = 2m, m is integer

Now from (i),

4m² = 2q²

q² = 2m²

∴ 2 is factor of q²

∴ 2 divides q²

∴ 2 divides q

Hence, p,q have a common factor 2. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number

So, √2 is irrational number.

Q18. Divide the polynomial p(x) = 3x³ + x² + 2x + 5 by the polynomial q(x) = x² + 2x + 1. Find the quotient and remainder.

Sol.

Q19. A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds.

Sol.

ED = height of girl = 0.9 m

BD = 1.2 × 4 = 4.8 m

AB = 3.6 m

AF = 3.6 – 0.9 = 2.7 m

EF = BD = 4.8 m

Let CD = x m

In similar ∆ABC and ∆AFE,

BC/EF = AB/AF

(x+4.8)/4.8 = 3.6/2.7

x = 1.6 m

Q20. Prove that :

secA(1-sinA)(secA+tanA) = 1

Sol.

LHS = secA(1-sinA)(secA+tanA)

= 1/cosA × (1-sinA) × (1/cosA + sinA/cosA)

= 1/cosA × (1-sinA) × [(1+sinA)/cosA]

= (1-sin²A)/cos²A

= cos²A/cos²A = 1 = RHS

Q21. Find the circumference of a circle whose area is 6.16 cm².

Sol.

Area of circle = 6.16 cm²

πr² = 6.16

r² = 6.16 × 7/22 = 196/100 = 1.96

r = √1.96 = 1.4 m

Circumference of circle = 2πr = 2 × 22/7 × 1.4 = 8.8 m

Q22. Solve :

5/(x-1) + 1/(y-2) = 2  and  6/(x-1) – 3/(y-2) = 1

Sol.

5/(x-1) + 1/(y-2) = 2 ……..(i)

6/(x-1) – 3/(y-2) = 1 ……..(ii)

Take 1/(x-1) = p and 1/(y-2) = q

5p + q = 2 ………(iii)

6p – 3q = 1 ……..(iv)

Multiply eqn.(iii) by 3 and add in eqn.(iv),

21p = 7

p = 1/3

Putting p = 1/3 in eqn.(iii),

q = 1/3

Now 1/(x-1) = p and 1/(y-2) = q

1/(x-1) = 1/3  and  1/(y-2) = 1/3

x – 1 = 3 and y – 2 = 3

So, x = 4 and y = 5

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