Solutions of Question Paper (2020)

Time : 3 HoursMaximum Marks : 80 MATHEMATICS (Standard)

CBSE Sample Question Paper

Solutions of Question Paper (2020)

Section ‘A’

1. Correct option : (d)

Explanation : According to Euclid’s division lemma, a = bq + r, where 0 ≤ <r b .

If b = 2, then a = 2q or a = 2q + 1

Since a = 2q + 1 is not divisible by 2, then 2q + 1 is an odd integer. 1

2. Correct option : (a)

Explanation : By Euclid's division algorithm, HCF of (65, 117) = 13.

Since, 65m – 117 = 13 Þ m = 2 1

3. Correct option : (a)

Explanation : Let f(x) = x3 + ax2 + bx + c

∵ One of the zeroes of f(x) is –1.

So

, ( )

( ) ( ) ( )

f

a b ca b ca b c

c b a

− =

− + − + − + =− + − + =

− + == + −

1 0

1 1 1 0

1 0

1

1

3 2

Now, αβγ =

-DA

[∵ a = –1, D = c, A=1]

−1βγ = −c1

βγ = c

βγ = b – a + 1 1 4. Correct option : (b) Explanation : x2 + x – 5 = 0 On comparing with ax2 +bx +c = 0 a = 1, b= 1, c = – 5 For real distinct roots b2 – 4ac > 0 (1)2 – 4(1)(–5) > 0 21 > 0 Hence, the equation has two distinct real roots.

1

5. Correct option : (c)

Explanation : According to the question, a rectangle can be represented as :

A C

O B

∴ Distance between the points A(0, 3) and B(5, 0) is

AB = −( ) + −( )= +

=

5 0 0 3

25 9

34

2 2

Hence, the required length of diagonal is 34. 1

6. Correct option : (b)A

CB

D

FE

7.5 cm

8 cm

5 cm

30°

50°

Explanation : ∠B = 180° – ∠A – ∠C

= 180° – 30° – 50°

= 100°

In DABC ~ DDFE,

∴ ∠B = ∠F = 100°

ABDF

= ACDE

57 5.

= 8

DE

DE = 8 7 55

× .

= 12 cm 1

lh-ch-,l-bZ- lkWY;w'ku isij

This solution has been provided by Oswaal Books

OSWAAL CBSE Sample Question Paper, MATHEMATICS (Standard), Class-X2


7. Correct option : (d) Explanation : Since, quadrilateral circumscri-

bing a circle subtends supplementary angles at the centre of the circle.

∴ ∠AOB + ∠COD = 180° 125° + ∠COD = 180° ∠COD = 180° - 125° = 55° 1 8. Correct option : (d) Explanation : Let r1 = 24 cm and r2 = 7 cm Area of first circle = pr1

2 = p(24)2 = 576p cm2

Area of second circle = p p pr2

2 27 49= =( ) cm2

According to question, Area of circle = Area of first circle + Area of second circle

p p pR2 = 576 +49

[where, R be radius of circle] R2 = 625 Þ R = 25 cm Diameter of a circle = 2R = 2 × 25 = 50 cm. 1 9. Correct option : (a) Explanation : When two hemispheres of

equal radii are joined base to base, new solid becomes sphere and curved surface area of sphere is 4pr2. 1

10. Correct option : (a) Explanation : The probability of the event,

which is very unlikely to happen, will be very close to zero. So it’s probability is 0.0001 which is minimum amongst the given values. 1

11. (0, 0) Explanation : The perpendicular bisector of any

line segment always passes through the mid-point of the line segment.

Mid-point of any line segment which passes through the points (x1, y1) and (x2, y2) is given

by x x y y1 2 1 2

2 2+ +

, .

So, the mid-point of the line segment joining the points A(– 2, – 5) and B(2, 5) is

− + − +

=

2 22

5 52

0 0, ( , )

1

12. Same Explanation : If f(x) = ax3 + bx2 + cx + d = 0.

Then, for all negative roots, a, b, c and d must have same sign. 1

13. No Explanation : Since the equation of the form

y = a is a line parallel to x-axis at a distance 'a' from it. y = 0 is the solution of the x-axis and y = – 7 is the equation of the line parallel to x-axis. So the given pairs of parallel lines has no solution. 1

OR

154

Explanation : For parallel lines (or no solution)

aa

bb

cc

k

k

k

1

2

1

2

1

2

3

2

2

5

2

1

4 15

15

4

= ≠

⇒ = ≠

⇒ =

⇒ =

-

1

14. n = 13

Explanation : Here, an = a + (n – 1)d

Þ 63 + (n – 1)2 = 3 + (n – 1)7 ½

Þ 5n = 65

Þ n = 13 ½

15. Inside

Explanation : Distance of the point (5, 8) from the centre

a = ( ) ( )5 3 8 42 2- -+

= | 4 16+ | = | 20 |

= 2 5

Q 2 5 is less than 7.

∴ The point lies inside the circle. 116. Since, tangent touches a circle on a distinct

point. On the diameter of a circle only two parallel tangents can be drawn. 1

17. Given, sin A = cos B or, sin A = sin (90° – B) or, A = 90° – B ∴ A + B = 90° 118. Circumference of the outer circle, 2pr1 = 88 cm

∴ r1 = 88 72 22

××

= 14 cm.

Circumference of the inner circle, 2pr2 = 66 cm

∴ r2 = 66 7

2 22212

××

= cm

= 10.5 cm

∴ Width of the ring = r1 – r2

= (14 – 10.5) cm = 3.5 cm. 1

3Solutions


19. Let radius of hemisphere be r. Given, Volume of hemisphere = Surface area of

hemisphere

or,

23

3πr = 3pr2

or, r= 92

units

\ Diameter = 92

2 9× = units

1

20. MedianOR

Median = 13

Mode +

23

Mean

= 13

(12.4) +

23

(10.5)

=

12 43

213

.+

= 12 4 213

. +

=

33 4

3.

= 11.3 1

B21. Here, y = 5 × 13 = 65 1 and 3 × 65 = 195

and x = 3 × 195 = 585 1 Hence, x = 585 2y = 65

22. Let CD ⊥ AB, then CD = p

Area of DABC =

12

× base × height 1

or, Area of DABC with base AB

=

12

× AB × CD =

12

cp

Also, Area of DABC with base BC

=

12

× BC × AC =

12

ab

So,

12

cp =

12

ab

or, cp = ab. Hence proved. 1OR

In DGEQ and DGFS ∠EGQ = ∠FGS

(vert. opp. angles)

∠EQG = ∠FSG (alt. angles)

∴ DGEQ ~ DGFS (AA similarity) 1

or,

EQFS

= GQGS

or, EQ × GS = GQ × FS.

Hence Proved. 1

23. sin (36 + θ)° = cos (16 + θ)° or, cos [90° – (36 + θ)°] = cos (16 + θ)° 1

or, 90° – 36° – θ = 16° + θ

or, 2θ = 90 °– 36° – 16° = 38°

∴ θ =

38°2

= 19°. 1

OR

LHS =

(sin cos )sin cos

4 4

2 21 2θ θ

θ θ+

-

=

(sin ) (cos )sin cos

2 2 2 2

2 21 2θ θ

θ θ+

-

=

(sin cos ) sin cossin cos

2 2 2 2 2

2 22

1 2θ θ θ θ

θ θ+ -

-

= 1 2

1 2

2 2

2 2--

sin cossin cos

θ θθ θ

= 1 = RHS 2

Commonly Made Error

Common errors are found in simplification.

Answering Tip

Follow step by step simplification to avoid errors.

24. Let AB be the light house and C be the position of the boat.

Since, ∠PAC = 60° ∴ ∠ACB = 60° 1 Let BC be x m.

In DABC,

ABBC

= tan 60°

⇒

40x

= 3

A

B C

D

P

S

E Q

FR

G



⇒ x = 40

3 ×

33

= 40 3

3 m

Hence, the boat is

40 33

m away from the

foot of the light house. 1

25. Perimeter of shaded region = Perimeter of semi-circles PSR + Perimeter of semi-circle RTQ + Perimeter of semi circle PAQ

= p × 5 + p(1.5) + p(3.5) 1 = (5 + 5)p = 10p = 3.14 × 10 = 31.4 cm 1

26.

MarksNumber of

students (fi)

Mid Values

(xi)u

xhi

i=− A

fiui

30 - 35 14 32.5 –3 –42

35 - 40 16 37.5 –2 –32

40 - 45 28 42.5 –1 –28

45 - 50 23 47.5 0 0

50 - 55 18 52.5 1 18

55 - 60 8 57.5 2 16

60 - 65 3 62.5 3 9

N = Sfi = 110

Sfiui = –59

1

Mean = 47.5 + ( )-59110

×5

= 47.5 – 2.682 = 44.82 1

Section ‘C’

27. To get share of each member, we have to divide x3 + 8x2 + 16x + 9 by x + 1.

x + 1 x x x3 2

+8 + 16 + 9 x x2

+ 7 + 9

x3 + x2

– – 7x2 + 16x + 9 7x2 + 7x – –

9x + 9 9x + 9 – – 0 3

Hence, share of each member is x2 + 7x + 9.OR

p(x) = 4x4 + 2x3 – 2x2 + x – 1

g(x) = rate of the each box = x2 + 2x – 3

q(x) = number of boxes

r(x) = amount of money he donated to child.

By using long division method

We need to find q(x) and r(x)

On dividing p(x) by g(x)

x2 + 2x – 3) 4x4 + 2x3 – 2x2 + x – 1 (4x2 – 6x + 22

4x4 + 8x3 – 12x2

– – +

– 6x3 + 10x2 + x – 1

– 6x3 – 12x2 + 18x

+ + –

22x2 – 17x – 1

22x2 + 44x – 66

– – +

– 61x + 65 2

Number of boxes q(x) = 4x2 – 6x + 22

Amount of money he donated to child

= r(x) = – 61x + 65 1

Commonly Made Error

Many candidates make mistakes in dividing the polynomial.

Answering Tip

Adequate practice is needed for division of polynomials.

28. For equation, (2m – 1)x + 3y – 5 = 0 ...(i) a1 = 2m – 1, b1 = 3 and c1 = – 5 ½

and for equation

3x + (n – 1)y – 2 = 0 ...(ii)

a2 = 3, b2 = (n – 1) and c2 = –2 ½

5Solutions


For a pair of linear equations to have infinite number of solutions

aa

1

2

= bb

1

2

= cc

1

2

or 2 13

m - = 31

52n -

= 1

or 2(2m –1) = 15 and 5(n–1) = 6

Hence, m = 174

and n = 115

1

29. Given, equation x2 + px + 16 = 0 have equal roots,

if D = p2 – 4 × 16 = 0 1 p2 = 64 Þ p = ± 8 ½ \ x2 ± 8x + 16 = 0 Þ (x ± 4)2 = 0 x ± 4 = 0 1 \ Roots are x = – 4 and x = 4 ½30. Let a and A be the first terms and d and D be the

common difference of two A.P.'s

SS

n

n' =

na n d

nn

22 1

22 1

+ −( )[ ]

+ −( )[ ]A D

1

or,

2 12 1a n d

n+ −( )+ −( )A D

=

7 14 27

nn

++

or

an

d

n

+ −

+ −

12

12

A D =

7 14 27

nn

++

...(i) ½

To get the ratio of 9th terms,

n − 12 = 8 or n = 17 ½

Hence, tt

9

9 ' =

a d++

88A D

=

7 17 14 17 27

× +× +

=

12095

=

2419

1

31. (i) Let AB be the vertical tree and AC be its shadow. Also, let DE be the vertical tower and DF be its shadow. Join BC and EF. Let DE= x.

20 m

48 m

We have AB = 20 m

AC = 16 m and DF = 48 m 1 In DABC and DDEF, ∠A = ∠D = 90° Since, the Sun casts equal angle at the same

time ∠C = ∠F Therefore by AA criterion of similarity, we

have DABC ~ DDEF

or, AB

DE = AC

DF

or, 20x

= 1648

or, x = 20 48

16´

or, x = 60 m. or,height of tower =60 m. 1 (ii) Similar Triangles. 1

32. Given, AB is a chord of circle with centre O and tangent PB = 24 cm, OP = 26 cm.

Construction : Join O to B and draw OC ̂ AB. By Pythagoras theorem,

A

P

BC

O

1

OB = 26 242 2( ) ( )-

= 676 576 100- =

= 10 cm 1

Now, in DOBC, BC = 12

AB = 162

= 8 cm

(Perpendicular drawn from the centre to a chord bisects it.)

OB = 10 cm

OC2 = OB2 – BC2

= 102 – 82

OC2 = 36

OC = 6 cm 1

∴ Distance of the chord from the centre

= 6 cm.



Commonly Made Error

Some candidates are not versed with the circle properties e.g., could not well identify ÐOBP = 90º (angle between radius and tangent)

Answering Tip

Candidates should be familiar with the proper-ties of circle.

OR

A

B

C

D

E

P

Q

R

S

O

12

34

½

In DAOD and DAOC

OD = OC (Radius of circle)

OA = OA (Common)

ÐADO =ÐOCA (90°, Q angle between radius and tangent]

DAOD = DAOC [RHS] 1

Þ Ð1 = Ð2 (By cpct) ...(i) ½

Similarly, Ð4 = Ð3 ...(ii) ½ On adding (i) and (ii), we get

Ð1 + Ð4 = Ð2 + Ð3 = 12

× 180°

Þ Ð2 + Ð3 = 90° or ÐAOB = 90° ½

33. 1

p =

1

sec tansec tansec tanθ θ

θ θθ θ+

×−−

½

1p

= sec tansec tan

θ θθ θ

--2 2

= sec θ – tan θ

½

Solving sec θ + tan θ = p and sec θ – tan θ = 1p

We get sec θ =

12

1p

p+

=

pp

2 12

+

½

and tan θ = 1

21

pp

- =

pp

2 12

-

½

∴ cos θ =

212

pp +

and sin θ = pp

2

211

-+

1

34. Total number of all possible outcomes

= 62 = 36

(i) The sum less than 7 = (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3),

(2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5, 1)

No. of favourable outcomes = 15

P(have sum less than 7) = 1536

512

=

1

(ii) Product less than 16 = (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (6, 1), (6, 2)


\ P(have a product less than 16)

= 2536

1

(iii) Doublet of odd numbers = (1, 1), (3, 3), (5, 5)


\ P(a doublet of odd number)

= 3

36 =

1

12 1

OR

(a) Total number of cards = 52

Number of non face cards = 52 – 12

= 40

P(non-face cards) =

4052

1013

=

1

(b) Number of black kings = 2

Number of red queens = 2

P(a black King or a red queen) =

452

= 1

13 1

(c) Number of spade cards = 13

P (Spade cards) =

1352

= 14

1

7Solutions


Section ‘D’

35. Let the smaller tap fills the tank in x hours

\The larger tap fills the tank in (x – 2) hours. Time taken by both the taps together

=

158

hours.

Therefore, 1 1

2x x+

− =

815

2

Þ 4x2 – 23x + 15 = 0 ½

Þ (4x – 3)(x – 5) = 0

Þ x =

34

or x = 5

Now, x ¹

34

therefore x = 5 1

Hence, smaller and larger tap can fill the tank separately is 5 hours and 3 hours. respectively. ½

36. AD is the median of DABC from vertex AA(4, –6)

B(3, –2) D(4, 0) C(5, 2)

D (x, y) =

3 52

2 22

+ − +

, = (4, 0) 1

Area of DADB

= 12

×| 4 (–6 + 2) + 4 (– 2 + 0) + 3 (0 + 6)|

= 12

× |– 16 – 8 + 18|

= 12

× 6 = 3 square units ...(i) 1

Area of DADC

= 12

× |4 (0 – 2) + 4 (2 + 6) + 5 (– 6 – 0)|

= 12

× |– 8 + 32 – 30|

= 12

× |(– 6)| = 3 1

Hence, area of DADC = 3 square units. ...(ii)

From (i) and (ii) Area of DADB = Area of DADC

It is verified that median of DABC divides it

into two triangles of equal areas. 1

Commonly Made Error

Many candidates do not know that median divides the opposite side of triangle in equal parts i.e., D is the mid point of BC.

Answering Tip

Concept of median and mid point formula should be understood adequately.

OR

Area of triangle with vertices (a, a2), (b, b2) and (0, 0) is

12

02 2a b b a( ) + −( ) +

2

12

0ab b a− ≠ 2

[Q a ¹ b ¹ 0]

\ Given points are not collinear.

37. Steps of Constructions :

1. Draw a line segment BC = 6 cm.

2. Draw a perpendicular bisector of BC which intersects the line BC at Q.

A'

B

B1

B2

B3

B4

X

QC

C'

A

6 cm

4 cm

2

3. Mark A on the line such that QA = 4 cm.

4. Join A to B and C.

5. Draw a ray BX making an acute angle with BC.

6. Mark four points B1, B2, B3 and B4 on the ray BX. Such that BB1 = B1B2 = B2B3 = B3B4.

7. Join B4C. Draw a line parallel to B4C through B3 intersecting line segment BC at C’.

8. Draw C’A’ || CA from point C’

Hence D A’BC’ is the required triangle. 2



38.

C x A y B 1

In DDCA, DCCA

= tan 60°

Þ 15x

= 3

Þ x =

153

Þ x = 5 3 m 1

In DDCB, DCCB

= ° =+

=tan 4515

1x y

Þ x + y = 15 1

Þ 5 3 + y = 15

Þ y = 15 5 3−

= 5 3 3−( ) m Hence, the distance between the points

= 5 3 3−( ) m.

1

Commonly Made Error

The concept of angle of depression and elevation are not clear to many students. That’s why they are not able to draw the diagram correctly.

Answering Tip

The concept of angle of depression and angle of elevation must be understood deeply and clearly.

OR

A

B Cyx O

60º 45º

60º 45º

300 m

1

In DAOC, tan 45º =

300y

Þ 1 =

300y

or, y = 300 1

In DAOB, tan 60º =

300x

Þ 3 =

300x

or, x =

3003

100 3=

1

\ width of river = 300 100 3+

= 300 + 173.2 = 473.2 m 1

39.

20 cm

12 cm

h

20 cm

12 cm

h

Volume of bucket = 12308.8 cm3

Let r1 = 20 cm, r2 = 12 cm

\ V =

π hr r r r

3 12

22

1 2+ +( )

1

\ 12308.8 =

3 143

400 144 240. × + +( )h

Þ h =

12308 8 33 14 784

15.

.×

×= cm 1

Now, l2 = h2 + (r1 – r2)2 = 225 + 64 = 289 Þ l = 17 cm 1 Surface area of metal sheet used = pr2

2 + pl(r1 +r2) = 3.14 (144 + 17 × 32) = 2160.32 cm2 1

OR

In DABC, ÐA = 90°, AB = 3 cm, and AC = 4 cm

A

B C

9Solutions


\ BC = AB AC22 2 23 4+ = + = 5 cm. ½

Area of semicircle with radius 32

cm + Area

of semi circle with radius 42

cm

=p p2

32 2

42

2 2

+

=π π2

32

+2

22

2

( )

Area of semicircle with radius 52

cm – Area

of DABC =π2

52

12

3 42

− × ×

= 258

6 2π −

cm

...(i) 1

Area of shaded region

= π π

π2

32 2

2258

62

2 2

+ ( ) − −

cm cm2

1

= π2

94

4254

6+ −

+

= π2

94

16 254

6+−

+

= π2

94

94

6−

+

1

= 6 cm2 ½

Commonly Made Error

In such type of problems, mostly students write incorrect formulae of volume of cylinder, frustum and hemisphere, and also they do error in calculations.

Answering Tip

Adequate practice and remembering of formulae is necessary.

40.

Class Frequency (f) Cumulative Frequency (cf )

0 - 10 f1 f110 - 20 5 5 +f120 - 30 9 14 + f130 - 40 12 26 + f140 - 50 f2 26 + f1+f250 - 60 3 29 + f1+f260 - 70 2 31 + f1+f2

401

Given that median is 32.5

Þ median class 30 – 40 ½

Now, 32.5 = 30

1012

20 14 1+ − −( )f 1

[Using Median = lc f

fh+

−

×

N2

. .]

Þ f1 = 3 1

Also, 31 + f1 + f2 = 40

Þ f2 = 6½

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Solutions of Question Paper (2020)