Strictly Confidential : (For Internal and Restricted Use Only)
Secondary School Examination
MARKING SCHEME
General Instructions :
1.
The Marking Scheme provides general guidelines to reduce subjectivity and maintain uniformity.
The answers given in the marking scheme are the best suggested answers.
2.
Marking be done as per the instructions provided in the marking scheme. (It should not be done
according to one’s own interpretation or any other consideration). Marking Scheme be strictly
adhered to and religiously followed.
3.
Alternative methods be accepted. Proportional marks be awarded.
4.
If a question is attempted twice and the candidate has not crossed any answer, only first attempt be
evaluated and ‘EXTRA’ written with second attempt.
5.
In case where no answers are given or answers are found wrong in this Marking Scheme, correct
answers may be found and used for valuation purpose.
Marking Scheme
Mathematics
Class – X(SA-1)
560011
Section-A
1.
(D)
1
2.
(C)
1
3.
(B)
1
4.
(B)
1
5.
(B)
1
6.
(A)
1
7.
(B)
1
8.
(D)
1
9.
(B)
1
10. (D)
Section-B
11.
12.
1
Divide 2x2x20 by x3
Quotient  2x5, remainder5
Verification
(x37x6)  (x1)  x2x6}
1
1
Page 1 of 7
 (x2x6)(x3) (x2)
other zeros are 3, 2

13.
14.
a 1 b1 c1
from Ist two
 
a2 b2 c2
3
1

2 p 1 p 1
½+½
1
½
3p32p1
p2
AB60 ; AB30
A45 ; B15
½
½+½
½+½
OR
Writing equations
AB60 ; AB30
Solving for A and B and
Finding A45 and B15
15.
1
1
PX
4
8
PY
8

 and

XQ
4.5
9
YR
9
PX
PY

In PQR,

XQ
YR

1
XYQR
1
[ By converse of BPT ]
16.
1
ABC  DEF

x
50

10
2
[ AA similarity ]
1

x250 m

the height of tower is 250 m
17. Age
1723 2329 2935
No. of
2
5
6
employees
(f)
c.f.
2
7
13
3541
4
4147
2
4753
1
17
19
20
N  20  10 Modal class  2935
2
2
10 7
Median  29
6
6
18.
½
1
½
 29332
Modal class  1530
½
15  8
Mode  15
 15
30 7  8
1
½
 15722
Section-C
Page 2 of 7
19.
Let
1
a
2 3 a
2 3
be a rational no. 

5
b
5
b
where a and b are coprime and b  0

3
5a
2b
As a and b are integers 
1
5a
2b
1
is a rational no.  3 is rational but
3 is irrational which is a contradiction 
is irrational
20.
If a number 4 n , for any natural number n ends with digit 0, then it is divisible by 5.
n
The prime factorization of 4 must contain the prime factor 5.
2 3
5








This is not possible because prime factors of 4n is 2 only and the uniqueness of
Fundamental theorem of arithmetic guarantees that there are no other prime in
factorisation of 4n.
Hence 4n can never end with the digit zero for n N.
OR
Applying Euclid’s division Lemma to 27 and 45 we get
4527118
271819
18920
HCF  9
927181
27(45271)1
27451271
274527
5442272451
x2, y1, d9
21.
2b
a4b 2a
2b
2a
b4a 2b
2a

x
y
 8b  a 
2  b2  a 2 
1
x

2
2
2
4 a  b 
8  b2  a 2 
y
 2
4  a 2  b2 
2
2 b  a
2
2
2
1
2
1
1½
½
1

1
2
4 a  b2

1
½
½
OR
Since BC DE and BE CD with BCCD, BCDE is a rectangle
 opposite sides are equal
i.e BECD  xy5 ______ (1)
Page 3 of 7
½
and DEBC  xy
Since perimeter of ABCDE is 21
ABBCCDDEEA21
½
3  x  y  x  y  x  y  3  21
22.
23.
½
63xy21
3xy15 _______ (2)
Adding (1) and (2) : 4x20
x5
Putting in (1) y0.
p(x)  q (x)  g (x)r (x) (By division algorithm)
 (2x22x1) (4x23x2)  14x10
 8x46x34x28x8x36x24x4x23x214x10
 8x414x32x27x8
½
½
½
1
1
1
½
 1
 1

LHS  
 sinA 
 cosA 
 sinA
 cosA

2
2
 1  sin A  1  cos A 



 sinA  cosA 
½
cos 2 A sin 2 A
 cosAsinA
sinA cosA
1
cosAsinA
RHS 

sinA
cosA
sin 2 A  cos 2 A

cosA
sinA
 cosA sinA
LHS  RHS
1
cosec (AB)  2 and cot (AB) 
3

24.
sin(AB) 
½
½+½
½
1

2
AB30
—
Solving (1) and (2) we get
A  45 and B  15 
25. In right ABC
AC2AB2BC2 (by PT)
In right ABD
AD2AB2BD2 (by PT)
1

BC 
2

AD2AB2 
BC24AD24AB2
Substituting (2) in (1)
AC24AD23AB2
2
(1)
AB60

-
—
(2)
1½
1½
(1)
½
½
1


 BD  BC 
2


-
½
½
(2)
1
Page 4 of 7
26.
In right ABC




AC 2  AB2  BC 2
AC 2  AB2  4BD 2
BC  2 BD 
AC2AB24 [AD2AB2]
AD2AB2BD2]
[
AC 2  AB 2  4AD 2  4AB 2
AC 2  4AD 2  3AB 2
27.




Making correct table
1
1
1
1½
  f i xi 
x a 
 h
  fi 
½
½
Substituting correct values
Finding x 211
½
OR
C.I
010
1020
2030
3040
4050
f
12
16
6
f
9
43f
xi
5
15
25
35
45
fi xi
60
240
150
35 f
405
85535 f
1½
½
 f i xi
 fi
855  35 f
22
43  f
x




28.
94622f85535 f
9113 f
f7
1
f
Classes
120130 2
130140 8
140150 12
150160
f
160170 8
170180 7
Mode  154
 Modal class 150160
½
f 1  f 0
Mode  l
h
2 f 1  f 0  f 2
Page 5 of 7
l150, f1f, f012, f28
h10

154150
4
1
f  12
10
2 f  20
½
 f  12 
5
 f  10 
4 f405 f60
20f
1
Section-D
29.
Since 2 and  2 are zeroes of f (x)
 (x 2 ) and (x 2 ) are factors of f (x)
 (x 2 ) (x 2 ) is a factor of f (x)
(x22) is a factor of f (x)
x 27 x12
1
x 22 x 4  7 x 3  10 x 2  14 x  24
x4
2 x2


7 x 3  12 x 2  14 x
 14 x
7 x3





12 x 2 24
24
12 x2 


X
1
(x27x12) is a factor of f (x)
(x3) (x4) is a factor of f (x)
x3 and x4 are other zeroes of f (x)
all the zeroes of f (x) are 2 ,  2 , 3 and 4
1
½
½
30. Making correct fig.
1
Given, To prove, Construction (if any)
1
Correct Proof
2
OR
For writing given, to prove, figure and construction (if any)
For proving the result correctly.
31.
LHS 

cotA  cosecA  cosec2A  cot 2A

cotA  cosecA  1
 cotA  cosecA    cosecA  cotA  cosecA  cotA 

cotA  cosecA  1
 cotA  cosecA    1  cosecA  cotA 

cotA  cosecA  1


cotA  cosecA
Page 6 of 7
1½
2½
1
1
1
cosA
1

sinA
sinA
1  cosA

 RHS
sinA

1
OR
sin 2 
cos 2 

2
cos 2 
sin 2 
LHS 

1
sin 4  cos 4   2cos 2  sin 2 
cos 2  sin 2 
sin 2  cos 2  



½
2
1
sin 2  cos 2 
1
1
sin 2  cos2 
 cosec2 sec2  RHS
32. LHS (sincosec)2(cossec)2
sin2cosec22sincoseccos2sec22cossec
sin2cos222cosec2sec2




51cot21tan2 
7tan2cot2 







RHS
33.
Representing both equations
Correctly on graph
Correct solution as (1, 4)
Shading the region bounded by the given lines and x-axis
1
2
Area of shaded part  44
8 sq. units
34.
C.I
2030
3040
4050
5060
f
4
12
14
16
c.f.
4
16
30
46
60 70
20
66
7080
8090
90100
16
10
8
100
82
92
100
N
50
2
Median  60
 50 46 
20
Median  60  2  62
½
1


1
1
1
1+1
½
½
1
1½
1½
10
1
Page 7 of 7