CBSE Class 10 Marking Scheme 2021 for Maths Basic

Class- X

Mathematics-Basic (241)

Marking Scheme SQP-2020-21

Max. Marks: 80 Duration:3hrs

1
1 156 = 22 x 3 x 13

2 Quadratic polynomial is given by x2 - (a +b) x +ab 1
x2 -2x -8

3 HCF X LCM =product of two numbers ½
LCM (96,404) = 96 𝑋 404 = 96 𝑋 404
𝐻𝐶𝐹(96,404) 4 ½
LCM = 9696

OR

Every composite number can be expressed (factorized) as a product 1
of primes, and this factorization is unique, apart from the order in
which the factors occur.

4 x – 2y =0

3x + 4y -20 =0

1 −2
≠ ½
3 4

𝑎1 𝑏1
As, 𝑎2 ≠ 𝑏2 is one condition for consistency.

Therefore, the pair of equations is consistent. ½

5 1 1

6 ɵ = 60°
ɵ
Area of sector =360° Πr2
60° 22 ½
A = 360° X 7 X (6)2 cm2
1 22
A = 6 X 7 X36 cm2
= 18.86cm2 ½

OR

Another method-
Horse can graze in the field which is a circle of radius 28 cm. ½
So, required perimeter = 2Πr= 2.Π(28) cm
22
=2 x X (28)cm
7
½
= 176 cm

7 By converse of Thale’s theorem DE II BC
∟ADE = ∟ABC = 70° ½
Given ∟BAC = 50°
∟ABC + ∟BAC +∟BCA =180° (Angle sum prop of triangles)
700 + 500 + ∟BCA = 180°
∟BCA = 180° - 120° = 60° ½

OR

EC = AC – AE = (7- 3.5) cm = 3.5 cm
𝐴𝐷 2 𝐴𝐸 3.5 1 ½
𝐵𝐷
= 3 and 𝐸𝐶 =3.5 = 1
𝐴𝐷 𝐴𝐸
So, ≠
𝐵𝐷 𝐸𝐶
Hence, By converse of Thale’s Theorem, DE is not Parallel to BC. ½

8 𝑇𝑜𝑡𝑎𝑙 𝑐𝑜𝑠𝑡
Length of the fence = 𝑅𝑎𝑡𝑒
𝑅𝑠.5280
= 𝑅𝑠 24/𝑚𝑒𝑡𝑟𝑒 = 220 m ½
So, length of fence = Circumference of the field
22
∴ 220m= 2 Π r=2 X xr
7
220 𝑥 7
So, r = 2 𝑥 22 m =35 m
½

9

𝐴𝐵
Sol: tan 30 ° =𝐵𝐶 ½
𝐴𝐵
1/√3 = 8
AB = 8 / √3 metres
½
Height from where it is broken is 8/√3 metres

10 Perimeter = Area 1
2Πr = Πr2
r = 2 units

11 3 median = mode + 2 mean 1

12 8 1

13 𝑎1 𝑏1 ½
𝑎2
≠ 𝑏2 is the condition for the given pair of equations to have unique
solution.

4 𝑝
≠
2 2

p ≠4
½
Therefore, for all real values of p except 4, the given pair of equations
will have a unique solution.

OR

𝑎1 2 1
Here, = =
𝑎2 4 2

𝑏1 3 1 𝑐1 5
= = and =
𝑏2 6 2 𝑐2 7

1 1 5
= ≠
2 2 7

𝑎1 𝑏1 𝑐1
½
= ≠ is the condition for which the given system of equations
𝑎2 𝑏2 𝑐2
will represent parallel lines.

So, the given system of linear equations will represent a pair of parallel ½
lines.
14 No. of red balls = 3, No.black balls =5 ½
Total number of balls = 5 + 3 =8
3 ½
Probability of red balls =
8

OR

Total no of possible outcomes = 6
½
There are 3 Prime numbers, 2,3,5.
3 1 ½
So, Probability of getting a prime number is 6 = 2

15

½

ℎ
tan 60° = 15
ℎ
√3 =
15
h = 15√3 m ½

16 1 1

17 i) Ans : b) 1
Cloth material required = 2X S A of hemispherical dome
= 2 x 2Π r2
22
= 2 x 2x 7 x (2.5)2 m2
= 78.57 m2

ii) a) Volume of a cylindrical pillar = Π r2h 1

iii) b) Lateral surface area = 2x 2Πrh 1
22
=4x x 1.4 x 7 m2
7
= 123.2 m2

iv) 2 1
d) Volume of hemisphere =3 Π r3
2 22
= 3 7 (3.5)3 m3
= 89.83 m3

v)
b)
2
Sum of the volumes of two hemispheres of radius 1cm each= 2 x 3 Π13
4 ½
Volume of sphere of radius 2cm = 3 Π 23
2
2 x 3 Π 13
So, required ratio is 4 = 1:8 ½
3
Π 23

18 i) c) (0,0) 1

ii) a) (4,6) 1

iii) a) (6,5) 1

iv) a) (16,0) 1

v) b) (-12,6) 1

19 i) c) 90° 1

ii) b) SAS 1

iii) b) 4 : 9 1

iv) d) Converse of Pythagoras theorem 1

v) a) 48 cm2 1

20 i) d) parabola 1

ii) a) 2 1

iii) b) -1, 3 1

iv) c) 𝑥 2 − 2𝑥 − 3 1

v) d) 0 1

21 Let P(x,y) be the required point. Using section formula

𝑚 1𝑥2+𝑚2𝑥1 𝑚1𝑦2+𝑚2𝑦1 1
{ , } = (x, y)
𝑚1+𝑚2 𝑚1+𝑚2
3(8)+1(4) 3(5)+1(−3)
x = 3+1 , y= 3+1
x=7 y= 3 1
(7,3) is the required point

OR

Let P(x, y) be equidistant from the points A(7,1) and B(3,5) 1
Given AP =BP. So, AP2 = BP2
(x-7)2 + (y-1)2 = (x-3)2 + (y-5)2
X2 -14x+49 +y2-2y +1 = x2 -6x +9+y2 -10y+25
1
x – y =2

22 By BPT, ½
𝐴𝑀 𝐴𝐿
= ............(1)
𝑀𝐵 𝐿𝐶

𝐴𝑁 𝐴𝐿 ½
Also, 𝑁𝐷 =𝐿𝐶 .................(2)

𝐴𝑀 𝐴𝑁
By Equating (1) and (2) 𝑀𝐵 =𝑁𝐷 1

23 To prove: AB + CD = AD + BC.

1
Proof: AS = AP ( Length of tangents from an external point to a circle
are equal)
1
BQ = BP
CQ = CR
DS = DR
AS + BQ + CQ + DS = AP + BP + CR + DR
(AS+ DS) + ( BQ + CQ) = ( AP + BP) + (CR + DR)
AD + BC = AB +CD

24 For the correct construction 2

25 15 cot A =8, find sin A and sec A.
Cot A =8/15 1

𝐴𝑑𝑗
=8/15
𝑂𝑝𝑝𝑜
By Pythagoras Theorem

AC2 =AB2 +BC2
½
AC =√(8𝑥)2 + (15𝑥)2
AC= 17x
½
Sin A = 15/17
Cos A =8/17

OR

By Pythagoras Theorem
QR = √(13)2 − (12)2 cm 1
QR = 5cm

Tan P =5/12
Cot R =5/12 1
Tan P -Cot R =5/12 -5/12
=0

26 9,17,25, .......
Sn = 636
a=9 ½
d = a2 -a1
= 17 – 9 = 8

𝑛
Sn = 2 [ 2a + (n-1) d]
𝑛
Sn = [ 2a + (n-1) d]
2 ½

𝑛
636 = [ 2x 9 + (n-1) 8]
2
1272 = n [ 18 + 8n -8]
1272 = n [10 +8n]
8n2 +10n -1272 =0
4n2 + 5n -636 =0

−𝑏±√𝑏2 −4𝑎𝑐
½
n= 2𝑎
−5±√52 −4𝑥 4𝑥(−636)
n=
2𝑥4
−5±101
n =- 8
96 −106
n= 8 n= 8
−53
n=12 n=- 4
½
n=12 (since n cannot be negative)

27 Let √3 be a rational number.
Then √3 = p/q HCF (p,q) =1 1
Squaring both sides
(√3)2 = (p/q)2
3 = p2/ q2
3q2 = p2
3 divides p2 » 3 divides p
3 is a factor of p
Take p = 3C ½
3q2 = (3c)2
3q2 = 9C2
3 divides q2 » 3 divides q ½
3 is a factor of q
Therefore 3 is a common factor of p and q
It is a contradiction to our assumption that p/q is rational. 1
Hence √3 is an irrational number.

28

Required to prove -: ∟PTQ = 2∟OPQ 1
Sol :- Let ∟PTQ = ɵ
Now by the theorem TP = TQ. So, TPQ is an isosceles triangle
∟TPQ = ∟TQP = ½ (180° -ɵ) 1
= 90° - ½ ɵ
∟OPT = 90° ½
∟OPQ =∟OPT -∟TPQ =90° -(90° - ½ ɵ)
=½ɵ
= ½ ∟PTQ ½

∟PTQ = 2∟OPQ

29 Let Meena has received x no. of 50 re notes and y no. of 100 re 1
notes.So,
50 x + 100 y =2000
x + y =25
multiply by 50
1
50x + 100y =2000
50 x + 50 y = 1250
- - -
50y =750
Y= 15
1
Putting value of y=15 in equation (2)
x+ 15 =25
x = 10

Meena has received 10 pieces 50 re notes and 15 pieces of 100 re
notes

30 (i) 10,11,12…90 are two digit numbers. There are 81
numbers.So,Probability of getting a two-digit number 1
= 81/90 =9/10

(ii) 1, 4, 9,16,25,36,49,64,81 are perfect squares. So, 1
Probability of getting a perfect square number.
= 9/90 =1/10

(iii) 5, 10,15….90 are divisible by 5. There are 18 outcomes.. 1
So,Probability of getting a number divisible by 5.
= 18/90 =1/5

OR

(i) Probability of getting A king of red colour. 1

P (King of red colour) = 2/52 =1/26

(ii) Probability of getting A spade 1
P ( a spade) = 13/52 = 1/4

(iii) Probability of getting The queen of diamonds 1
P ( a the queen of diamonds) = 1/52

31 r1 = 6cm
r2 = 8cm
r3 = 10cm
1
Volume of sphere = 4/3 Π r3
Volume of the resulting sphere = Sum of the volumes of the smaller
spheres.
3Π r = 4/3 Π r 13 + 4/3 Π r 23 +4/3 Π r 3 3
4/ 3
1
3Π r = 4/3 Π (r 13 + r23 + r3 3)
4/ 3

r 3 = 63 + 83 + 103
r3 = 1728
3
r = √1728
r = 12 cm
1
Therefore, the radius of the resulting sphere is 12cm.

32 (sin A-cos A+1)/ (sin A+cosA-1) = 1/(sec A-tan A)

L.H.S. divide numerator and denominator by cos A
1
= (tan A-1+secA)/ (tan A+1-sec A)

= (tan A-1+secA)/(1-sec A + tan A)

We know that 1+tan2 A=sec 2A 1
Or 1=sec2 A-tan2 A = (sec A + tan A)(sec A – tan A)

=( sec A + tan A-1)/[(sec A + tan A)(sec A-tan A)-(sec A-tan A)]

=( sec A + tan A-1)/(sec A-tan A)(sec A + tan A-1)
1

= 1/(sec A-tan A) , proved.

33 Given:-

Speed of boat =18km/hr
Distance =24km

Let x be the speed of stream. ½
Let t1 and t2 be the time for upstream and downstream.
As we know that,

speed= distance / time
⇒time= distance / speed
½
For upstream,
Speed =(18−x) km/hr
Distance =24km
Time =t1
Therefore,

24
t1 =
18−𝑥

For downstream,
Speed =(18+x)km/hr
Distance =24km
Time =t2
Therefore,

24
t2 =
18+𝑥
Now according to the question-

t1=t2+1

24 24
= +1
18−𝑥 18+𝑥
½
24(18+𝑥)− 24 (18−𝑥 )
⇒ =1
(18−𝑥)(18+𝑥)

⇒48x=(18−x)(18+x)

⇒48x=324+18x−18x− x2

⇒ x2 +48x−324=0
⇒ x2+54x−6x−324=0
⇒x(x+54)−6(x+54)=0
⇒(x+54)(x−6)=0

½
⇒x=−54 or x=6

Since speed cannot be negative.

⇒x=−54 will be rejected

∴x=6

Thus, the speed of stream is 6km/hr. 1

OR

Let one of the odd positive integer be x
then the other odd positive integer is x+2 1
their sum of squares = x² +(x+2)²
= x² + x² + 4x +4
= 2x² + 4x + 4
Given that their sum of squares = 290
⇒ 2x² +4x + 4 = 290
⇒ 2x² +4x = 290-4 = 286
⇒ 2x² + 4x -286 = 0 1
⇒ 2(x² + 2x - 143) = 0
⇒ x² + 2x - 143 = 0
⇒ x² + 13x - 11x -143 = 0
⇒ x(x+13) - 11(x+13) = 0
⇒ (x -11)(x+13) = 0
⇒ (x-11) = 0 , (x+13) = 0
Therefore , x = 11 or -13
According to question, x is a positive odd integer.
Hence, We take positive value of x 1
So , x = 11 and (x+2) = 11 + 2 = 13
Therefore , the odd positive integers are 11 and 13 .

34

1

Let AB and CD be the multi-storeyed building and the building
respectively.

Let the height of the multi-storeyed building= h m and

the distance between the two buildings = x m.

AE = CD = 8 m [Given]

BE = AB – AE = (h – 8) m

and

AC = DE = x m [Given]

Also,

∠FBD = ∠BDE = 30° ( Alternate angles)

∠FBC = ∠BCA = 45° (Alternate angles) ½

Now,

In Δ ACB,

1

In Δ BDE,

1

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

h =√3h -8√3

√3h – h =8√3

h (√3 -1) =8√3

8√3
h = 3−1
√

8√3 √3+1 1
h= 3−1 x 3+1
√ √

h-= 4√3 (√3 +1)

h = 12 +4√3 m

Distance between the two building
½

OR

E C

D
A B

From the figure, the angle of elevation for the first position of the
balloon ∟EAD = 60° and for second position ∟BAC = 30°.The 1
vertical distance

ED = CB = 88.2-1.2 =87m.

Let AD = x m and AB = y m.
𝐷𝐸
Then in right Δ ADE, tan60° = 𝐴𝐷

87
√3 = 𝑋 1

87
X =√3 ……….(i)

𝐵𝐶
In right ΔABC, tan 30° = 𝐴𝐵

1 87
=
√3 𝑦

Y = 87√3 ……….(ii)
1
Subtracting(i) and (ii)
87
y-x =87√3 -- -√3
1
87 .2.√3
y-x = √3.√3

y-x = 58√3 m

Hence, the distance travelled by the balloon is equal to BD

y-x =58√3 m.
1

35 Let A be the first term and D the common difference of A.P.

Tp=a=A+(p−1)D=(A−D)+pD (1)
½
Tq=b=A+(q−1)D=(A−D)+qD ..(2) ½

Tr=c=A+(r−1)D=(A−D)+rD ..(3) ½

Here we have got two unknowns A and D which are to be eliminated.

We multiply (1),(2) and (3) by q−r,r−p and p−q respectively and add:

a (q-r) = (A – D )(q-r) + D p(q-r) ½
b(r-p) = (A-D) (r-p) + Dq (r-p) ½
c(p-q) = (A-D) (p-q) + Dr (p-q) ½

a(q−r)+b(r−p)+c(p−q)
1
=(A−D)[q−r+r−p+p−q]+D[p(q−r)+q(r−p)+r(p−q)]
= (A – D ) ( 0 ) + D [ pq-pr + qr – pq + rp – rq ) 1
=0

36 Height (in cm) f C.F.
below 140 4 4
140-145 7 11 1
145-150 18 29
150-155 11 40
155-160 6 46
160-165 5 51
N=51⇒

N/2=51/2=25.5

As 29 is just greater than 25.5, therefore median class is 145-150.
𝑁
( −𝐶)
2
Median= l + 𝑓
Xh

Here, l= lower limit of median class =145
½
C=C.F. of the class preceding the median class =11

h= higher limit - lower limit =150−145=5

f= frequency of median class =18

∴median=
( 25.5−11)
= 145 + X5 ½
18
=149.03
Mean by direct method
1
f xi fxi
Height (in cm)
below 140 4 137.5 550
140-145 7 142.5 997.5
145-150 18 147.5 2655
150-155 11 152.5 1677.5
155-160 6 157.5 945 1
5 162.5 812.5
160-165
∑ 𝑓𝑥
Mean = ________
N

=7637.5/51
= 149.75 1