Download CBSE Class 10 Maths Standard Marking Scheme 2021 Model Paper

Download Central Board of Secondary Education (CBSE) Class 10th (10th Board Exam) Maths Standard Marking Scheme 2021 Model Paper

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

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

Four decimal places
1
2.
+=k/3
?
3=k/3
K=9
?
3.
3=13
?
6 8
3=1
6
K=2
?
4.
Let the cost of 1 chair=Rs.x
?
And the cost of 1 table=Rs. y
3x+y=1500
?
6x+y=2400
5.
an=a+(n-1)d
0=27+(n-1)(-3)
?
30=3n
n =10
?
10th

OR

an=a+(n-1)d
4=a+6x(-4)
?
a=-28
?
6.
9x2+6kx+4=0
(6k)2-4X9X4=0
?
36k2=144
K2=4
K=?2
?
Page 1 of 10

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

3ax2-6x+1=0
?
(-6)2-4(3a) (1)<0
12a>36 =>a>3
?
8.
PQ=PT
PL+LQ=PM+MT
PL+LN=PM+MN
Perimeter(PLM)
=PL+LM+PM
?
=PL+LN+MN+PM
=2(PL+LN)
=2(PL+LQ)
=2X28=56cm
?
9.
In PAO
?
Tan30?=AO/PA
1/3 =3/PA
?
PA=33 cm
OR
In OPQ
P+Q+O=180?
2Q+P=180?
?
2Q+90?=180?
2Q=90?
Q= 45?
?
Page 2 of 10

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10.
=
?
3 = 2
4.5
?
CE=3cm
11.
8:5
1
12.
Sin30?+cosB=1
?+cosB=1
?
CosB=1/2
B=60?
?
13.
x+y
=2sin2 +2cos2+1
?
=2(sin2 +cos2)+1
= 3
?
14.
length of arc=/360?(2r)
?
= 60/360(2X22/7X21)
=22 cm
?
15.
R2H=12X4/3r3
1X1x16=4/3Xr3 X12
?
r3=1
r=1
d=2cm
?
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

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(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
?
7=7X1

?
14=2X7
?
LCM=2X2X7=28
?
The three bells will ring together again at 6:28 am

22.
Let P(x,0) be a point on X-axis

PA=PB
?
PA2=PB2
?
(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
?
P(-1,0)
?

OR



PR:QR=2:1
?
1
R(1(-2)+2(3) , 1(5)+2(2) )
2+1
2+1

R(4/3, 3)
?

23.
Sum of zeroes= 5-32+5+32=10
?
Product of zeroes= (5-32)(5+32)= 7
1
P(x)= X2-10x+7
?


24.

Line
seg=1/2
Circles=1
/2
Tangents
=1/2+
?

Page 4 of 10

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25.
tanA=3/4=3k/4k
?
sinA=3k/5k=3/5,cosA=4k/5k=4/5
?
1/sinA+1/cosA

=5/3+5/4
?
=(20+15)/12
?
=35/12


OR



3 sin=cos
?
sin/cos=1/3
?
tan=1/3
?
=30?
?

26.
A = OPA =OSA = 90?
?
Hence, SOP=90?

Also, AP=AS

Hence, OSAP is a square

AP=AS=10cm
?
CR=CQ=27cm

BQ=BC-CQ=38-27=11cm
?
BP=BQ=11 cm

X=AB=AP+BP=10+11=21 cm
?

27.
Let 2-3 be a rational number
?
We can find co-prime a and b (b0) such that

2-3=a/b
?
2-a/b=3
?
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
?
Which contradicts our statement
?
Therefore 2-3 is irrational
?


28.
3x2+px+4=0
?
3(2/3)2+p(2/3)+4=0

4/3+2p/3+4=0
?
P=-8
?
3x2-8x+4=0

3x2-6x-2x+4=0
?
X=2/3 or x=2
?
Hence, x=2
?














Page 5 of 10

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OR

+=5 ----(1)
?
-=1 ----(2)
?
Solving (1) and (2), we get

=3 and =2
?
also =6
?
or 3(k-1)=6
?
k-1=2

k=3
?

29.

Area of 1 segment = area of sector ?area of triangle
?
=( 90?/360?)r2 ? ? x7x7

=1/4x22/7x72 ? ? x7x7
?
= 14cm2
?
Area of 8 segments=8x14= 112 cm2
?
Area of the shaded region = 14x14-112
?
=196-112=84cm2
?
(each petal is divided into 2 segments)


30.
ABC~DEF

() =++=
1
()
++
?
25=9
?
15
X=5.4cm
1
DE=5.4cm
OR

?

Construction-Draw AM BC
?
BD 1/3 BC , BM=1/2 BC
In ABM,
AB2=AM2+BM2
?
=AM2+(BD+BM)2

=AM2+DM2+BD2+2BD. DM
?
=AD2+BD2+2BD(BM-BD)

=AD2+(BC/3)2+2. BC/3.(BC/2-BC/3)

=AD2+2BC2/9
?
=AD2+2AB2/9

Hence,7AB2=9AD2
?



Page 6 of 10

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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
?
a+b=15




-
median=l+2
X h
?

16 =15+35-24- X 5
15

1=(11-a)/3

A=8

?
55+a+b=70
?
55+8+b=70

B=7


32.

?






?
Let AB=candle

C and D are coins

Tan60?=AB/BC=h/b

3=h/b

H=b3 --------------(1)
?
Tan30?=AB/BD=h/a

1/3=h/a

H=a/3 --------------(2)
?
Multiplying (1) and (2), we get

H2= b3X a/3
?
H2= b a

H=ab m
?


Page 7 of 10

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33.

Mode= l+ 1-0 xh
?
21-2-0
?
67 = 60+ 15- x 10

30-12-
?
7 = 15-x 10

18-
7x(18-x)=10(15-x)
?
126-7x=150-10x

3x=150-126
?
3x=24

X=8
?
34.

1






Let BD=river
?
AB=CD=palm trees=h

BO=x
?
OD=80-x

In ABO,

Tan60?=h/x
?
3=h/x -----------------------(1)

H=3x
?
In CDO,

Tan 30?=h/(80-x)

1/3= h/(80-x) ---------------------(2)
?
Solving (1) and (2), we get

X=20

H=3x=34.6
?
the height of the trees=h=34.6m

BO=x=20m
?
DO=80-x=80-20=60m

?















Page 8 of 10

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OR
1



Let AB=Building of height 50m

RT= tower of height= h m
?
BT=AS=x m

AB=ST=50 m
?
RS=TR-TS=(h-50)m

In ARS, tan30?=RS/AS

1/3 = (h-50)/x -------------(1)

In RBT, tan60?=RT/BT
?
3 = h/x --------------(2)

?
Solving (1) and (2), we get
?
h= 75

from (2)
?
x=h/3

=75/3
?
=253

Hence, height of the tower=h=75m

Distance between the building and the tower=253=43.25m
?

35.
For pipe , r = 1cm
?
Length of water flowing in 1 sec, h=0.7m=7cm
?
Cylindrical Tank,R=40 cm , rise in water level=H
?
Volume of water flowing in 1 sec= r2h=x1x1x70

=70
?
Volume of water flowing in 60 sec=70x60

1
Volume of water flowing in 30 minutes=70x60x30
?

Volume of water in Tank=r2H=x40x40xH
?
?
Volume of water in Tank= Volume of water flowing in 30
?
minutes

x40x40xH = 70x60x30

H=78.75cm
Page 9 of 10

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36.
Let speed of the boat in still water =x km/hr, and
?
Speed of the current =y km/hr

Downstream speed =(x+y) km/hr
?
Upstream speed =(x-y) km/hr
?
24 + 16 =
?
6--------(1)
+
-




36 + 12 =

6----------(2)
+
-
?

Let 1 = u and 1 = v

+
-


Put in the above equation we get,

24u+16v=6
?
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,
?
12u=1

u=1/12

Putting the value of u in (4), we get, v=1/4
?
1 = 1 and 1 = 1

+
12
-
4

x+y=12 and x-y=4

Thus, speed of the boat in still water = 8 km/hr,
?
Speed of the current = 4 km/hr
?


Page 10 of 10

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