CBSE Class 10 Marking Scheme 2021 for Maths Standard

MARKING SCHEME SQP
MATHEMATICS (STANDARD)
2020-21
CLASS X

S.NO. ANSWER MARKS
Part-A
1. (LCM)(3) =180 ½
LCM=60 ½

OR

Four decimal places 1

2. α+β=k/3 ½
3=k/3
K=9 ½

3. 3 1 3 ½
= ≠
6 𝑘 8
3 1
=
6 𝑘
½
K=2
4. Let the cost of 1 chair=Rs.x ½
And the cost of 1 table=Rs. y
3x+y=1500 ½
6x+y=2400

5. an=a+(n-1)d
0=27+(n-1)(-3) ½
30=3n
n =10 ½
10th

OR

an=a+(n-1)d
4=a+6x(-4) ½
a=-28 ½

6. 9x2+6kx+4=0
(6k)2-4X9X4=0 ½
36k2=144
K2=4
K=±2 ½

Page 1 of 10

7. x2+7x+10=0
x2+5x+2x+10=0 ½
(x+5)(x+2)=0
X=-5, x= - 2 ½

OR

3ax2-6x+1=0 ½
(-6)2-4(3a) (1)<0

12a>36 =>a>3 ½

8. PQ=PT
PL+LQ=PM+MT
PL+LN=PM+MN
Perimeter(∆PLM)
=PL+LM+PM ½
=PL+LN+MN+PM
=2(PL+LN)
=2(PL+LQ)
=2X28=56cm ½

9.

In ∆PAO ½
Tan30˚=AO/PA
1/√3 =3/PA ½
PA=3√3 cm

OR

In ∆OPQ
˂P+˂Q+˂O=180˚
2˂Q+˂P=180˚ ½
2˂Q+90˚=180˚
2˂Q=90˚
˂Q= 45˚ ½

Page 2 of 10

10. 𝐴𝐷 𝐴𝐸
=
𝐵𝐷 𝐶𝐸
½
3 2
= ½
4.5 𝐶𝐸
CE=3cm

11. 8:5 1

12. Sin30˚+cosB=1
½+cosB=1 ½
CosB=1/2
B=60˚ ½

13. x+y
=2sin2Ɵ +2cos2Ɵ+1 ½
=2(sin2Ɵ +cos2Ɵ)+1
=3 ½

14. length of arc=Ɵ/360˚(2∏r) ½
= 60/360(2X22/7X21)
=22 cm ½

15. ∏R2H=12X4/3∏r3

1X1x16=4/3Xr3 X12 ½
r3=1
r=1
d=2cm ½

16. probability of getting a doublet=1/6 1

OR

probability of getting a black queen=2/52=1/26

17. (a) iii)(15/2,33/2) 1x4=4
(b) i) 4
(c) iii)16
(d) iv)(2.0,8.5)
(e) ii) x-13=0
18. (a) iii)15 cm 1x4=4
(b) iv)They are not the mirror image of one another
(c) ii)Their altitudes have a ratio a:b
(d) iv) 5m
(e) iii)6m
19. (a) ii) (4,-2) 1x4=4
(b) i) Intersects x-axis
(c) iii) parabola

Page 3 of 10

(d) ii) x2 – 36
(e) iii) 0
20. (a) iii)43 1x4=4
(b) iii)60
(c) ii)Median
(d) iii)80
(e) iii)31

Part-B
21. 4=2X2 ½
7=7X1
½
14=2X7
½
LCM=2X2X7=28
½
The three bells will ring together again at 6:28 am
22. Let P(x,0) be a point on X-axis
PA=PB ½
PA2=PB2 ½
(x-2)2+(0+2)2=(x+4)2+(0-2)2
X2+4-4x+4=x2+16+8x+4
-4x+4=8x+16
X=-1 ½
P(-1,0) ½

OR

PR:QR=2:1 ½
1(−2)+2(3) 1(5)+2(2) 1
R( , )
2+1 2+1
R(4/3, 3) ½

23. Sum of zeroes= 5-3√2+5+3√2=10 ½
Product of zeroes= (5-3√2)(5+3√2)= 7 1
P(x)= X2-10x+7 ½

24. Line
seg=1/2

Circles=1
/2

Tangents
=1/2+
½

Page 4 of 10

25. tanA=3/4=3k/4k ½
sinA=3k/5k=3/5,cosA=4k/5k=4/5 ½
1/sinA+1/cosA
=5/3+5/4 ½
=(20+15)/12 ½
=35/12

OR

√3 sinƟ=cosƟ ½
sinƟ/cosƟ=1/√3 ½
tanƟ=1/√3 ½
Ɵ=30˚ ½

26. ˂A = ˂OPA =˂OSA = 90˚ ½
Hence, ˂SOP=90˚
Also, AP=AS
Hence, OSAP is a square
AP=AS=10cm ½
CR=CQ=27cm
BQ=BC-CQ=38-27=11cm ½
BP=BQ=11 cm
X=AB=AP+BP=10+11=21 cm ½

27. Let 2-√3 be a rational number ½
We can find co-prime a and b (b≠0) such that
2-√3=a/b ½
2-a/b=√3 ½
So we get,(2a-b)/b=√3
Since a and b are integers, we get (2a-b)/b is irrational and so
√3 is rational. But √3 is an irrational number ½
Which contradicts our statement ½
Therefore 2-√3 is irrational ½

28. 3x2+px+4=0 ½
3(2/3)2+p(2/3)+4=0
4/3+2p/3+4=0 ½
P=-8 ½
3x2-8x+4=0
3x2-6x-2x+4=0 ½
X=2/3 or x=2 ½
Hence, x=2 ½

Page 5 of 10

OR
α+β=5 ----(1) ½
α-β=1 ----(2) ½
Solving (1) and (2), we get
α=3 and β=2 ½
also αβ=6 ½
or 3(k-1)=6 ½
k-1=2
k=3 ½

29.
Area of 1 segment = area of sector –area of triangle ½
=( 90˚/360˚)πr2 – ½ x7x7
=1/4x22/7x72 – ½ x7x7 ½
= 14cm2 ½
Area of 8 segments=8x14= 112 cm2 ½
Area of the shaded region = 14x14-112 ½
=196-112=84cm2 ½
(each petal is divided into 2 segments)

30. ∆ABC~∆DEF
𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 (∆𝐴𝐵𝐶) 𝐴𝐵+𝐵𝐶+𝐶𝐴 𝐴𝐵 1
= =
𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 (∆𝐷𝐸𝐹) 𝐷𝐸+𝐸𝐹+𝐹𝐷 𝐷𝐸
25 9
½
= ½
15 𝑋
X=5.4cm 1
DE=5.4cm

OR

½

Construction-Draw AM ḻ BC
½
BD ꓕ 1/3 BC , BM=1/2 BC
In ∆ABM,
AB2=AM2+BM2 ½
=AM2+(BD+BM)2
=AM2+DM2+BD2+2BD. DM ½
=AD2+BD2+2BD(BM-BD)
=AD2+(BC/3)2+2. BC/3.(BC/2-BC/3)
=AD2+2BC2/9 ½
=AD2+2AB2/9
Hence,7AB2=9AD2 ½

Page 6 of 10

31. Class Frequency Cumulative 1
frequency
0-5 12 12
5-10 a 12+a
10-15 12 24+a
15-20 15 39+a
20-25 b 39+a+b
25-30 6 45+a+b
30-35 6 51+a+b
35-40 4 55+a+b
Total 70

55+a+b=70 ½
a+b=15
𝑁
−𝑐𝑓
median=l+ 2 𝑓 X h ½
35−24−𝑎
16 =15+ 15 X5
1=(11-a)/3
A=8
½
55+a+b=70 ½
55+8+b=70
B=7

32.
½

½
Let AB=candle
C and D are coins
Tan60˚=AB/BC=h/b
√3=h/b
H=b√3 --------------(1) ½
Tan30˚=AB/BD=h/a
1/√3=h/a
H=a/√3 --------------(2) ½
Multiplying (1) and (2), we get
H2= b√3X a/√3 ½
H2= b a
H=√ab m ½

Page 7 of 10

33.
𝑓1−𝑓0 ½
Mode= l+ xh
2𝑓1−𝑓2−𝑓0 ½
15−𝑥
67 = 60+ x 10
30−12−𝑥 ½
15−𝑥
7 = x 10
18−𝑥
7x(18-x)=10(15-x) ½
126-7x=150-10x
3x=150-126 ½
3x=24
X=8 ½

34. 1

Let BD=river ½
AB=CD=palm trees=h
BO=x ½
OD=80-x
In ∆ABO,
Tan60˚=h/x ½
√3=h/x -----------------------(1)
H=√3x ½
In ∆CDO,
Tan 30˚=h/(80-x)
1/√3= h/(80-x) ---------------------(2) ½
Solving (1) and (2), we get
X=20
H=√3x=34.6 ½
the height of the trees=h=34.6m
BO=x=20m ½
DO=80-x=80-20=60m
½

Page 8 of 10

OR

1

Let AB=Building of height 50m
RT= tower of height= h m ½
BT=AS=x m
AB=ST=50 m ½
RS=TR-TS=(h-50)m
In ∆ARS, tan30˚=RS/AS
1/√3 = (h-50)/x -------------(1)
In ∆RBT, tan60˚=RT/BT ½
√ 3 = h/x --------------(2)
½
Solving (1) and (2), we get ½
h= 75
from (2) ½
x=h/√3
=75/√3 ½
=25√3
Hence, height of the tower=h=75m
Distance between the building and the tower=25√3=43.25m ½

35. For pipe , r = 1cm ½
Length of water flowing in 1 sec, h=0.7m=7cm ½
Cylindrical Tank,R=40 cm , rise in water level=H ½
Volume of water flowing in 1 sec= ∏r2h=∏x1x1x70
=70∏ ½
Volume of water flowing in 60 sec=70∏x60
1
Volume of water flowing in 30 minutes=70∏x60x30 ½

Volume of water in Tank=∏r2H=∏x40x40xH ½
½
Volume of water in Tank= Volume of water flowing in 30 ½
minutes
∏x40x40xH = 70∏x60x30
H=78.75cm

Page 9 of 10

36. Let speed of the boat in still water =x km/hr, and ½
Speed of the current =y km/hr
Downstream speed =(x+y) km/hr ½
Upstream speed =(x−y) km/hr ½
24 16 ½
+ = 6--------(1)
𝑥+𝑦 𝑥−𝑦

36 12
+ = 6----------(2) ½
𝑥+𝑦 𝑥−𝑦

1 1
Let = u and =v
𝑥+𝑦 𝑥−𝑦

Put in the above equation we get,
24u+16v=6 ½
Or, 12u+8v=3 ... (3)
36u+12v=6
Or, 6u+2v=1 ... (4)
Multiplying (4) by 4, we get,
24u+8v=4v … (5)
Subtracting (3) by (5), we get, ½
12u=1
⇒u=1/12
Putting the value of u in (4), we get, v=1/4 ½
1 1 1 1
⇒ 𝑥+𝑦 = 12 and 𝑥−𝑦 = 4
⇒x+y=12 and x−y=4
Thus, speed of the boat in still water = 8 km/hr, ½
Speed of the current = 4 km/hr ½

Page 10 of 10