Download Central Board of Secondary Education (CBSE) Class 10th (10th Board Exam) Maths Standard Marking Scheme 2021 Model Paper
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
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6 8
3=1
6
K=2
?
4.
Let the cost of 1 chair=Rs.x
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And the cost of 1 table=Rs. y
3x+y=1500
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6x+y=2400
5.
an=a+(n-1)d
0=27+(n-1)(-3)
?
30=3n
n =10
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10th
OR
an=a+(n-1)d
4=a+6x(-4)
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a=-28
?
6.
9x2+6kx+4=0
(6k)2-4X9X4=0
?
36k2=144
K2=4
K=?2
?
Page 1 of 10
7.
x2+7x+10=0
x2+5x+2x+10=0
?
(x+5)(x+2)=0
X=-5, x= - 2
?
OR
3ax2-6x+1=0
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(-6)2-4(3a) (1)<0
12a>36 =>a>3
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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
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
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=2(sin2 +cos2)+1
= 3
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14.
length of arc=/360?(2r)
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= 60/360(2X22/7X21)
=22 cm
?
15.
R2H=12X4/3r3
1X1x16=4/3Xr3 X12
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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
(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
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(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
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P(-1,0)
?
OR
PR:QR=2:1
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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
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
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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
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2-a/b=3
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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
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Which contradicts our statement
?
Therefore 2-3 is irrational
?
28.
3x2+px+4=0
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3(2/3)2+p(2/3)+4=0
4/3+2p/3+4=0
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P=-8
?
3x2-8x+4=0
3x2-6x-2x+4=0
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X=2/3 or x=2
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Hence, x=2
?
Page 5 of 10
OR
+=5 ----(1)
?
-=1 ----(2)
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Solving (1) and (2), we get
=3 and =2
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also =6
?
or 3(k-1)=6
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k-1=2
k=3
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29.
Area of 1 segment = area of sector ?area of triangle
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=( 90?/360?)r2 ? ? x7x7
=1/4x22/7x72 ? ? x7x7
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= 14cm2
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Area of 8 segments=8x14= 112 cm2
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Area of the shaded region = 14x14-112
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=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
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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
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
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55+a+b=70
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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)
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Multiplying (1) and (2), we get
H2= b3X a/3
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H2= b a
H=ab m
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Page 7 of 10
33.
Mode= l+ 1-0 xh
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21-2-0
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67 = 60+ 15- x 10
30-12-
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7 = 15-x 10
18-
7x(18-x)=10(15-x)
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126-7x=150-10x
3x=150-126
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3x=24
X=8
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34.
1
Let BD=river
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AB=CD=palm trees=h
BO=x
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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
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DO=80-x=80-20=60m
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Page 8 of 10
OR
1
Let AB=Building of height 50m
RT= tower of height= h m
?
BT=AS=x m
AB=ST=50 m
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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
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Cylindrical Tank,R=40 cm , rise in water level=H
?
Volume of water flowing in 1 sec= r2h=x1x1x70
=70
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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
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
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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
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Page 10 of 10
This post was last modified on 07 March 2021