# CBSE Class 10 Maths Standard Previous Question Paper 2020 Solution

## CBSE Class 10 Maths Standard Previous Question Paper 2020 Solution SECTION – A

Q. No. 1 to 10 are multiple choice type questions of 1 mark each. Select the correct option.

(1) If one of the zeroes of the quadratic polynomial x2 + 3x + k is 2, then the value of K is

(a) 10

(b) – 10

(c) -7

(d) -2

Solution: –10

(2) The total number of factors of a prime number is

(a) 1

(b) 0

(c) 2

(d) 3

Solution: 2

(3) The quadratic polynomial, the sum of whose zeroes is -5 and their product is 6, is

(a) x2 + 5x + 6

(b) x2 – 5x + 6

(c) x2 – 5x – 6

(d) –x2 + 5x +6

Solution: x2 + 5x + 6

(4) The value of k for which the system of equations x + y -4 = 0 and 2x + ky = 3, has no solution, is

(a) -2

(b) ≠ 2

(c) 3

(d) 2

Solution: 2

(5) The HCF and the LCM of 12, 21, 15 respectively are

(a) 3,140

(b) 12, 420

(c) 3, 420

(d) 420,3

Solution: 3,420

(6) The value of x for which 2x, (x + 10) and (3x + 2) are the three consecutive terms of an AP, is

(a) 6

(b) -6

(c) 18

(d) – 18

Solution: 6

(7) The first term of an AP is ρ and the common difference is q, then its 10th term is

(a) q + 9p

(b) p – 9q

(c) p +9q

(d) 2p + 9q

Solution: p +9q

(8) The distance between the points (a cos θ + b sin θ, 0) and (0, a sin θ – b cos θ), is

(a) a2 + b2

(b) a2 – b2

(c) √ a2 + b2

(d) √ a2 – b2

Solution: √ a2 + b2

(9) If the point P(k, 0) divides the line segment joining the points A(2, -2) and B(-7, 4) in the ratio 1: 2, then the value of k is

(a) 1

(b) 2

(c) -2

(d) -1

Solution: -1

(10) The value of p, for which the points A(3,1), B(5, p) and C(7, -5) are collinear, is

(a) -2

(b) 2

(c) -1

(d) 1

Solution: -2

In Q. Nos. 11 to 15, fill in the blanks, Each question is of 1 mark.

(11) In Fig. 1, △ ABC is circumscribing a circle, the length of BC is ……. cm. Solution: 10

(12) Given △ ABC ~ △ PQR, if AB/PQ = 1/3, then ar(△ ABC) /ar(△ PQR) = ………

Solution: 1/9

(13) ABC is an equilateral triangle of side 2a, then length of one of its altitude is………..

Solution: √  3 a

(14) Cos 800/sin 100 + Cos 590 cosec 310 = ………

Solution: 2

(15) The Value of (Sin2 θ + 1/ 1+ tan2 θ)=……….

Solution: 1

OR

The value of (1 + tan2 θ) (1 – sin θ) (1 + sin θ) = ………..

Solution: 1

Q. Nos. 16 to 20 are short answer type questions of 1 mark each.

(16) The ratio of the length of a vertical rod and the length of its shadow is 1: √3. Find the angle of elevation of the sum at that moment?

Solution: tan θ = 1/ √ 3 ⇒ θ= 300

(17) Two cones have their heights in the ratio 1:3 and radii in the ratio 3:1. What is the ratio of their volumes?

Solution: r1/r2 = 3/1, h1/h2 = 1/3

∴ Ratio of volumes = 1/3 πr21 h1/ 1/3 πr 22 h2

(18) A letter of English alphabet is chosen at random. What is the probability that the chosen letter is a consonant.

Solution: P (consonant) = 21/26

(19) A die is thrown once. What is the probability of getting a number less than 3?

Solution: P (number less than 3) = 2/6 or 1/3

OR

If the probability of winning a game is 0.07, what is the probability of losing it?

Solution: P (losing) = 1 – 0.07

= 0.93

(20) If the mean of the first n natural number is 15, then find n.

Solution: If the mean of first n natural number is 15, then find n.

n (n + 1)/2/n = 15

∴ n = 29

SECTION – B

Q. Nos. 21 to 26 carry 2 marks each.

(21) Show that (a – b)2, (a2 + b2) and (a + b)2 are in AP.

Solution: (a2 + b2) – (a – b)2 = 2ab

(a + b)2 – (a2 + b2) = 2ab

Common difference is same. ∴ given terms are in AP

(22) In Fig. 2, DE ∥ AC and DC ∥ AP. Prove that BE/EC = BC/CP Solution: In ΔABC, DE || AC, ∴ BD/DA = BE/EC …(i)

In ΔABP, DC || AP, ∴ BD/ DA = BC/CP …. (ii)

From (i) & (ii) BE/EC = BC/CP

OR

In Fig. 3, two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that <PTQ = 2 <OPQ. Solution: Let ∠OPQ = θ

∴ ∠TPQ = ∠TQP = 90° – θ

In ΔTPQ, 2(90° – θ) + ∠PTQ = 180°

∴ ∠PTQ = 2θ

= 2∠OPQ

(23) The rod AC of a TV disc antenna is fixed at right angles to the wall AB and a rod CD is supporting the disc as shown in Fig. 4. If AC = 1.5 m long and CD = 3 m, find (i) tan θ (ii) sec θ + cosec θ

Solution: AC/CD = Sin θ ⇒ Sin θ = 1/2 ⇒ θ = 300

(i) tan θ – tan 300 – 1/ √3

(ii) secθ + cosec θ = sec 300 + cos ec 300

= 2/ √3 + 2 or 2(3 + √3)/3

(24) A piece of wire 22 cm long is bent into the form of an arc of circle subtending an angle of 60° at its centre. Find the radius of the circle.

(Use π = 22/7)

Solution: 2 x 22/7 x r x 600/3600 = 22

∴ r = 21 cm

(25) If a number x is chosen at random from the numbers –3, –2, –1, 0, 1, 2, 3. What is the probability that x2 ≤ 4?

Solution: Total number of outcomes = 7

Favourable outcomes are –2, –1, 0, 1, 2, i.e., 5

∴ P(x2 ≤ 4) = 5/7

(26) Find the mean of the following distribution:

 Class: 3-5 5-7 7-9 9-11 11-13 Frequency: 5 10 10 7 8

Solution:

 Classes Xi fi fxxi 3 – 5 5 – 7 7 – 9 9 – 11 11 – 13 4 6 8 10 12 5 10 10 7 8 20 60 80 70 96 Total 40 326

X = Σ fi xi/fi / Σ fi= 326/40 = 8.15

OR

Find the mode of the following data:

 Class: 0-20 20-40 40-60 60-80 80-100 110-120 120-140 Frequency: 6 8 10 12 6 5 3

Solution:

Modal class : 60 – 80

Mode = l + f1 – f0/2f1 – f0 – f2 x h = 60 + 12 – 10/ 24 – 10 – 6 x 20

= 60 + 5 = 65

SECTION – C

Question numbers 27 to 34 carry 3 marks each.

(27) Find the quadratic polynomial whose zeroes are reciprocal of the zeroes of the polynomial f (X) = ax2 + bx + c, a ≠ 0, c ≠ 0

Solution: f (x) = ax2 + bx + c

α + β = b/a, αβ = c/a

New sum of zeroes = 1/ α+1/ β = b/c

New product of zeroes = 1/ α x 1/ β = a/c

∴ Required quardratic polynomial = x2 + b/cx + a/c or ( cx2 + bx +a)

OR

Divide the polynomial f(x) = 3x2 – x3 – 3x + 5 by the polynomial g(x) = x – 1 – x2 and verify the division algorithm.

Solution: Divisor × Quotient + Remainder

= (- x2 + x -1) (x – 2) + 3

= -x3 + 3x2 – 3x + 5 = Divided

(28) Determine graphically the coordinates of the vertices of a triangle, the equations of whose sides are given by 2y – x = 8, 5y – x = 14 and y – 2x = 1.

Solution: 2y – x = 8

 x 0 2 -4 y 4 5 2

5y – x = 14

 x 1 6 -4 y 3 4 2

y – 2x = 1

 x 1 2 0 y 3 5 1

Drawing 3 lines

Coordinates of the vertices of the triangle are A(–4, 2),

B(1, 3) and C(2, 5)

OR

If 4 is the zero of the cubic polynomial x3 – 3x2 – 10x + 24, find its other two zeroes.

Solution: x – 4 is a factor of given polynomial. (29) In a flight of 600 km, an aircraft was slowed due to bad weather. Its average speed for the trip was reduced to 200 km/hr and time of flight increased by 30 minutes. Find the original duration of flight.

Solution: Let the speed of aircraft be x km/hr

∴ 600/ x – 200 – 600/x = 30/60

⇒ x2 – 200x – 240000 = 0

(x – 600) (x + 400) = 0

X = 600, – 400 (Rejected)

Speed of aircraft = 600 km/hr

∴ Duration of flight = 1 hr

(30) Find the area of triangle PQR formed by the points P(–5, 7), Q(– 4, –5) and R(4, 5).

Solution:

ar (PQR) = 1/2[-5 (-5 -50 -4 (5 – 7) + 4 (7+ 5)] sq. units

= 1/2[50 + 8 + 48] sq. units

= 53 sq. units

OR

If the point C(–1, 2) divides internally the line segment joining A(2, 5) and B(x, y) in the ratio 3 : 4, find the coordinates of B.

Solution:

Coordinates of C arc ( 3x + 8/7, 3y + 20/7) = (-1, 2)

⇒ x = –5, y = –2 ∴ Coordinates of B are (–5, –2)

(31) In Fig.5, D = E and AD/ DB = AE/EC. prove that BAC is an isosceles triangle.

Solution: ∠D = ∠E ⇒ AE = AD

∴ AD /DB = AE/EC ⇒ DB = EC

⇒ AD + DB = AE + EC

∴ AB = AC

Hence ΔBAC is an isosceles triangle.

(32) In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then prove that the angle opposite to the first side is a right angle.

Solution:

For correct given, To prove, construction and figure.

For correct proof.

(33) If sinθ + cosθ =√ 3 , then prove that tanθ + cotθ = 1.

Solution:

sin θ + cosθ = √3 ⇒ (sin θ + cosθ) 2  = (√ 3)2

sin2 θ + cos 2 θ + 2sin θ cosθ = 3 ⇒ sin θ cosθ = 1

L.H.S = tanθ + cot θ = sin θ / cosθ + cosθ/ sin θ = 1/ cosθ sin θ = 1 = R. H. S

(34) A cone of base radius 4 cm is divided into two parts by drawing a plane through the mid-point of its height and parallel to its base. Compare the volume of the two parts.

Solution: ΔABC ~ ΔADE, h/2h = BC/4

∴ BC = 2 cm

Ratio of volumes of two parts

= 1/3 π x 22 x h/ 1/3 π x(22 + 42 + 2 x 4) xh

= 4/28 = 1/7 or 1 : 7 (accept 7 : 1 also)

SECTION – D

(35) Show that the square of any positive integer cannot be of form (5q + 2) or (5q +3) for any integer q.

Solution:

Let a be any positive integer. Take b = 5 as the divisor.

∴ a = 5m + r , r = 0,1,2,3,4

Case-1 : a = 5m ⇒ a2 = 25m2 = 5(5m2) = 5q

Case-2 : a = 5m+1 ⇒ a2 = 5(5m2 + 2m) + 1 = 5q + 1

Case-3 : a = 5m+2 ⇒ a2 = 5(5m2 + 4m) + 4 = 5q + 4

Case-4 : a = 5m+3 ⇒ a2 = 5(5m2 + 6m + 1) + 4 = 5q + 4

Case-5 : a = 5m+4 ⇒ a2 = 5(5m2 + 8m + 3) + 1 = 5q + 1

Hence square of any positive integer cannot be of the form (5q + 2) or (5q + 3) for any integer q.

OR

Prove that one of every three consecutive positive integers is divisible by 3.

Solution:

Let n be any positive integer. Divide it by 3.

∴ n = 3q + r, r = 0, 1, 2

Case-1 : n = 3q (divisible by 3)

n + 1 = 3q + 1, n + 2 = 3q + 2

Case-2 : n = 3q + 1 ⇒ n + 1 = 3q + 2, n + 2 = 3q + 3 (divisible by 3)

Case-3 : n = 3q + 2 ⇒ n + 1 = 3q + 3 (divisible by 3), n + 2 = 3q + 4

(36) The sum of four consecutive numbers in AP is 32 and the ratio of product of the first and last terms to the product of two middle terms is 7:15. Find the numbers.

Solution:

Let four consecutive number be a – 3d, a – d, a + d, a + 3d

Sum = 32 ∴ 4a = 32 ⇒ a = 8

(a – 3d) (a  + 3d) (a- d)(a+ d) = 7/15 ⇒ 15 (64 – 9d2) = (64 – d2)

∴ d2 = 4 ⇒ d = ±2

Four numbers are 2, 6, 10, 14.

OR

Solve: 1+ 4 + 7 + 10 +…+ x = 287

Solution:

x = an = 1 +3n – 3 = 3n – 2

Sn = 287 ⇒ n/2 [1 + 3n – 2] = 287

∴ 3n2– n – 574 = 0

(n -14)(3n+ 41) =0 ⇒ n 14

∴ x = 3n – 2 = 40

(37) Draw a line segment AB of length 7 cm. Taking A as centre, draw a circle of radius 3 cm and taking B as centre, draw another circle of radius 2 cm. Construct tangents to each circle from the centre of the other circle.

Solution:

Constructing the circles of radii 3 cm and 2 cm.

Constructing the tangents.

(38) A vertical tower stands on a horizontal plane and is surmounted by a vertical flag-staff of height 6 m. At a point on the plane, the angle of elevation of the bottom and top of the flag-staff are 30° and 45° respectively. Find the height of the tower. (Take √3 = 1.73)

Solution: h/x = tan 300

⇒ x = h √3

6 + h/x = tan 450 ⇒ 6 + h = x

∴ h = 6/√3 – 1 = 3(√3 + 1) = 3 x 2.73 m

= 8.19 m

(39) A bucket in the form of a frustum of a cone of height 30 cm with radii of its lower and upper ends as 10 cm and 20 cm respectively. Find the capacity of the bucket. Also find the total cost of milk that can completely fill the bucket at the rate of  ₹40 per litre. (Use  π = 22/7)

Solution:

Capacity of bucket = 1/3  πh (r21 + r22 + R1 r2)

= 1/3 x 22/7x 30(102 + 202 + 10 x 20) cm3

= 22000 cm3

= 22l

Cost of milk = ₹ 40 x 22 = ₹ 880

(40) The following table gives production yield per hectare (in quintals) of wheat of 100 farms of a village:

 Production yield/hect. 40-45 45-50 50-55 55-60 60-65 65-70 No. of farms 4 6 16 20 30 24

Change the distribution to ‘a more than’ type distribution and draw its ogive.

Solution:

 Production yield/hectare No. of forms More than or equal to 40 100 More than or equal to 45 96 More than or equal to 50 90 More than or equal to 55 74 More than or equal to 60 54 More than or equal to 65 24 Total OR

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

 Class Frequency 0 – 100 2 100 – 200 5 200 – 300 x 300 – 400 12 400 – 500 17 500 – 600 20 600 – 700 y 700 – 800 9 800 – 900 7 900 – 1000 4

Solution:

 Class Frequency Cumulative frequency 0 – 100 2 2 100 – 200 5 7 200 – 300 x 7 +x 300 – 400 12 19 + x 400- 500 17 36 + x 500 – 600 20 56 + x 600 – 700 y 56 + x + y 700 – 800 9 65 + x + y 800 – 900 7 72 + x + y 900 – 1000 4 76 + x + y Total 100

76 x y 100 x y 24 ++= ⇒+= … (i)

500 – 600 is the median class

Median =      Photo

⇒ 525 = 500 + 50 – 36 –x/20 x 100

Solving we get, x = 9

From (i), y = 15