HBSE Class 10 Maths Question Paper 2019 Answer Key

Q1. Express 0.375 in the form p/q.   

Sol. 

0.375 = 375/1000 = 3/8 

 

Q2. Find the sum of zeroes of quadratic polynomial x² + 7x + 10. 

Sol. 

Sum of zeroes = -(coefficient of x) ÷ (coefficient of x²) 

α + β = = -b/a = -7/1 = -7

 

Q3. The values of x and y from the equations x + y = 14 and x – y = 4 are :    

(A) x = 9, y = 4 

(B) x = 9, y = 5  

(C) x = 5, y = 9 

(D) None of these  

Ans. (B) x = 9, y = 5 

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. 30th term of the A.P. 10, 7, 4, ……….. is : 

(A) 77 

(B) 87

(C) –77   

(D) None of these 

Ans. (C) –77 

Here, a = 10,  d = 7 – 10 = -3,  n = 30 

an = a + (n-1)d 

a30 = 10 + (30-1)(-3) = 10 + (29)(-3) = 10 – 87 = -77 

 

Q5. Find the common difference of the A.P. :  3, 1, –1, –3, ……………. 

Sol. Common difference (d) = a2 – a1 = 1 – 3 = -2 

 

Q6. Fill in the blank using the 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. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of circle is :  

(A) 15 cm 

(B) 12 cm

(C) 24.5 cm  

(D) 7 cm 

Ans. (D) 7 cm

Using Pythagoras Theorem, 

H² = P² + B² 

25² = r² + 24² 

r² = 25² – 24² = 625 – 576 = 49 

r = √49 = 7 cm 

 

Q9. Fill in the blank : 

A tangent to a circle intersects it in ……… point(s).  

Ans. One

 

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

Sol.

A(2, 3) and B(4, 1) 

x1 = 2, x2 = 4, y1 = 3, y2 = 1 

Using Distance Formula, 

AB = √x2-x1)²+(y2-y1)²

      = √(4-2)²+(1-3)²

      = √(2)²+(-2)²

      = √4+4 = √8 = 2√2 unit 

 

Q11. Find the mid point of the line joining the points (7, 6) and (–3, –4).  

Sol. 

Take C is Mid point of AB, 

A(7, 6) and B(-3, -4) and C(x, y) 

Here x1 = 7, x2 = -3, y1 = 6, y2 = -4

Using Mid Point Theorem, 

C(x, y) = (x1+x2)/2, (y1+y2)/2 

            = (7-3)/2, (6-4)/2 

            = 4/2, 2/2 

            = 2, 1 

Co-ordinates of C(x, y) is (2, 1). 

 

Q12. Evaluate : sin18°/cos72°

Sol. sin18°/cos72° = sin(90°-72°)/cos72° = cos72°/cos72° = 1

 

Q13. In ∆ABC, right-angled at B, AB = 24 cm, BC = 7 cm. The value of sinA is : 

(A) 7/25 

(B) 7/24

(C) 24/25 

(D) None of these 

Ans. (A) 7/25 

AC² = AB² + BC² = 24² + 7² = 576 + 49 = 625 

AC = √625 = 25 

sinA = BC/AC = 7/25

 

Q14. Find the area of a sector of a circle with radius 6 cm, if angle of the sector is 60°. 

Sol. 

Area = θ/360 × πr² = 60/360 × 22/7 × 6 × 6 = 18.84 cm² 

 

Q15. The volume of the cuboid, whose length, breadth and height are 12 m, 10 m and 8 m respectively is : 

(A) 592 m³

(B) 960 m³

(C) 480 m³ 

(D) None of these 

Ans. (B) 960 m³

Volume of cuboid = length × breadth × height = 12 × 10 × 8 = 960 m³

 

Q16. Two dice are thrown at the same time. Find the probability of getting the sum on the dice is 8. 

Sol. 

(2, 6), (3, 5), (4, 4), (5, 3), (6, 2) 

P(sum 8) = 5/36 

 

Q17. Prove that √2 is irrational.  

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) = x³ – 3x² + 5x – 3 by the polynomial g(x) = x² – 2. Find the quotient and remainder. 

Sol. 

Quotient = x – 3

Remainder = 7x – 9

 

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. In ∆OPQ, right-angled at P, OP = 7 cm and OQ – PQ = 1 cm. Determine the values of sinQ and cosQ. 

Sol. 

OQ – PQ = 1 

In ∆OPQ, 

OQ² = OP² + PQ² 

(1+PQ)² = OP² + PQ² 

1 + PQ² + 2PQ = OP² + PQ² 

1 + 2PQ= 49 

2PQ = 49 – 1 = 48

PQ = 48/2 = 24 cm 

OQ = 1 + PQ = 1 + 24 = 25 cm 

sinQ = 7/25 

cosQ = 24/25 

 

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