JAP22X3
Strictly Confidential: (For Internal and Restricted Use Only)
MARKING SCHEME
â´·¤çÜÌ ÂÚUèÿææ - I, 2013
SUMMATIVE ASSESSMENT – I, 2013
»ç‡æÌ / MATHEMATICS
·¤ÿææ - X / Class – X
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.
not
Marking be done as per the instructions provided in the marking scheme. (It should
be done according to one’s own interpretation or any other consideration).
3.
Alternative methods be accepted. Proportional marks be awarded.
4.
first
If a question is attempted twice and the candidate has not crossed any answer, only
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.
¹‡ÇU-¥ /
SECTION – A
ÂýàÙ â´•Øæ 1 âð 8 Ì·¤ ÂýˆØð·¤ ÂýàÙ 1 ¥´·¤ ·¤æ ãñÐ
Question numbers 1 to 8 carry 1 mark each.
1
(b)
1
2
(d)
1
3
(d)
1
Page 1 of 12
4
(c)
5
(b)
6
(A)
7
(D)
greater than one
(C)
X1
8
1
1
3 2 cm
1
n1 1
2
1
…………….1
1
¹‡ÇU-Õ / SECTION – B
ÂýàÙ â´•Øæ 9 âð 14 Ì·¤ ÂýˆØð·¤ ÂýàÙ 2 ¥´·¤æð´ ·¤æ ãñÐ
Question numbers 9 to 14 carry 2 marks each.
9
85517
119717
LCM5717
595
Page 2 of 12
2
10
p(x)x34x23x18
3
2
2
p(2)2 4(2) 3(2)18
816618
0
 2 is zero of the given polynomial
x 1
11
2 3
  2x 1

3 5 2 x2 1
3x 1 4
3x 1 6
2
The polynomial is 2x25 3 x6
12
DE AC
BE
BD
5
_____ (1)
EC
DA
DF AE
BF
BD
5
_____ (2)
FE
DA
from (1) & (2)
BE
BF
5
EC
FE
EC
FE
5
BE
BF
2
13
C90 (angle in a semi circle)
opp.
2
tanA 5
3
adj.
opp.
3
tanB 5
2
adj.
2 3
 tanA. tanB . 1
3 2
2
14
Page 3 of 12
f
c.f
2
010
3
3
1020
4
7
2030
2
9
3040
5
14
4050
6
20
Median30
Median class 30 – 40
102 9
3 10
5
………… ½
……………………1
30232
…………………… ½
¹‡ÇU-â / SECTION – C
ÂýàÙ â´•Øæ
15 âð 24 Ì·¤
ÂýˆØð·¤ ÂýàÙ 3 ¥´·¤ ·¤æ ãñ´Ð
Question numbers 15 to 24 carry 3 marks each.
15
•
By
Euclid’s
division
lemma,
correct
value
of
‘b’8
and 3
‘r’0, 1, 2, 3, 4, 5, 6, 7
•
for all forms ‘a’ and rejecting which are not even with correct reason and thus given
forms of even integer.
2 x2 7
16
2
3
2
x 2 2x1 1 2x 2 11x 1 16 x2 4
2 x 3 2 4x 2 1 2 x
2 
2  1 
2 7x 21 14x2 4
2 7 x 2 1 14x 2 7
2  2 
1 
3
q(x)2x7
r(x)3 to obtain the quotient and remainder
g(x) q(x)r(x)
 (x22x1) (2x7)3
 2x34x22x7x214x73
 2x311x216x4P(x)
Hence the division algorithm is verified
Page 4 of 12
3
17
3
Graph of each line
Since the lines are co-incident the system of linear equations are consistent with infinite
solutions
18
For infinite number of solutions
a1
b
c
5 1 5 1
a2
b2
c2
k
4
k2 4
5
5
16
k
k
3
2
4kk 4k
 k0, 8
k264
k8 but k 8
19
3
figure
Page 5 of 12
Proof : ABCD is a gmAC
In 's ABE and CFB
ABECFB (alternate angles)
AC
ABE ~ CFB (AA similarity)
AB
AE

5
CF
BC
 ABBC  AECF
(property)
20
3
Given, to prove
Construction : AEBC and DFBC
Proof :
In AOE and DOF
AOEDOF (Vertically Opp. s)
AEODFO 90 (const.)
AOE ~ DOF
(AA)
AO
AE

5
--------- (1)
DO
DF
1
3 BC 3 AE
ar  DABC 
AE
2
5
5
1
ar  DBCD 
DF
3 BC 3 DF
2
AO
from (1)
5
DO
21
3
1
sin (AB)
2
2
sin(AB)sin60
sin (AB)sin 30
A 1 B 5 608 eqn. (1)
sin(AB)
A 2 B 5 308
2A
eqn. (2)
5 908
5 9082 5 458
Applying A45 in ---------- (1)
AB60
A
Page 6 of 12
3
45B60
 B604515
1 2

 tanu1
5
tanu 

22
5 tan 2 u1
5 tan 2 u1
23
1
2
tan u
1
tan 2 u
 2
3
2
1 2 tanu3
1
5 2
tanu
5 22 25 0
C.I
f
x
v5
0 - 10
10 - 20
20 - 30
30 - 40
40 - 50
Total
5
11
19
30
15
80
05
15
25
35
45
2
1
0
1
2
x ah
S f i xi
S fi
xi 2 a
h
fv
3
10
11
0
30
30
39
39
3 10 
80
 254.875
29.875
5 25 1
24
Since 340 is mode

300 – 400 is modal class


f1 2 f0
Mode l 1 
3 h
 2 f1 2 f0 2 f 2 
 20 2 x 
340 5 300 1 
3 100
 40 2 x 2 14 
2
 20 2 x 
405 
 3 100
 26 2 x 
522x 1005x
Page 7 of 12
3
3x 48 
x 16
¹‡ÇU-Î / SECTION – D
ÂýàÙ â´•Øæ
25 âð 34 Ì·¤
ÂýˆØð·¤ ÂýàÙ 4 ¥´·¤æð ·¤æ ãñÐ
Question numbers 25 to 34 carry 4 marks each.
25
Time required by 3 children to complete a card together
LCM of 10, 16, 20
2222580
These children possess following values :
Caring, respect for elders, creative and helpful
26
4
4
Solution
x3
y5
27
ABBCAC
 3x12x3y5
Page 8 of 12
4
x3y4
---------- (1)
3x1x9y6
2x9y5
---------- (2)
Multiply (1) by 3
3 x 2 9 y 5 12
2 x2 9 y 5 5
2
1
2
x
5 7
 y1  AB22 cm
Side of 22 cm
28
4
In ABC
2
2
2
AC AB BC -----------(1)
In BAD
2
2
2
AD AB BD -----------(2)
(1) - (2)
AC2AD2BC2BD2
3
 BC2
4
3
AC2AD2 (AC2AB2)
4
2
 AC 4AD23AB2
29
Construction : Join AC
2
BC ACCD
Page 9 of 12
4
BC AC
5
      (i)
CD BC
In D ABC and D BDC

AC BC
5
BC CD
and C 5 C (common)
 D ABC ~ D BDC
AB BC
5
       (ii)
BD CD
from (i) and (ii), we get

AC AB
5
BC BD
 BD 5 BC
30
LHS 
5
[ Q AB 5 AC]
 sec2 u 2
tan 2 u  1  secu 2 tanu 
1 1 secu 1 tanu
4
 secu 1 tanu  secu 2 tanu  1  secu 2 tanu 
 1 1 secu 1 tanu 
5  secu 2 tanu 
 secu 1 tanu 1 1
 1 1 secu 1 tanu 
sinu 
1 2 sinu
 1
RHS
5 
2
5
cosu 
cosu
 cosu
 1  2 2 
 1  4  1  4 
4   1    2 3 
2 1 
 2  
 2 

 2 
31
1
1
1

5 4 1
2 3  2 1

 16 16 
2

1 3
5 1
2 2
2
Page 10 of 12
4
32
2
2
We have cos 1sin 
4
c2
5 12
c2 1 d2
d
So, cosu5
c2 1 d2
sinu
cosu
Then tanu5
5
c
d
33
Classes
0 – 10
10 – 20
20 – 30
30 – 40
40 – 50
50 – 60
f
Cum. freq.
5
5
p
5p
20
25p
15
40p
q
40pq
5
45pq
n60
 45pq60  pq15
n
30
2
Median class 20 – 30
l20, h10, f20, Cf5p
n2 Cf


Medianl  2
 h
f


25
2
p


28.520 
  1725p
 2 
p8
pq15  q15p7
q7
4
34
Here h10
4
Page 11 of 12
xi 2 75
fi
fi
xi
Class mark
ui 5
10
25
28
45
55
65
3
2
1
30
50
28
12
75 A
0
0
10
15
85
95
1
2
10
30
f100
x 5 A1
fiui
zfiui68
S f i ui
68 3 10
3 h 5 75 2
S fi
100 10
x 5 68.2
*****
Page 12 of 12