Strictly Confidential : (For Internal and Restricted Use Only)
MA1-043
Secondary School Examination
SUMMATIVE ASSESSMENT – I, 2012
MARKING SCHEME
MATHEMATICS
Class – IX
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.
SECTION - A
1.
(B)
1
2.
(C)
1
3.
(B)
1
4.
(C)
1
5.
(A)
1
6.
(D)
1
7.
(C)
1
8.
(A)
1
Page 1 of 11
SECTION - B
9.
10.
4
3  3 5

2

 
2
 4 3  3 5

2


 484524 15


 9324 15


 3 31 8 15

 24 15
½

½
p(x)kx2 x4
1
½
½
p(1)k (1)2(1)40
 k140
k3
11.
12.
1
a 3 b3 c3 3 abc, when abc0
ax2y
b2yz
czx
abc x2 y 2 y  z zx0
1
½
(x2y)3  (2yz)3 (zx)3  3 (x2y) (2yz) (zx)
½
Given Data ACBD
Subtract BC from both sides
½
ACBCBDBC
ABCD
Euclid axiom used is “ Equals are subtracted from equals then difference are equal”.
½
1
13.
1
x35115
x1153580
y18011565
1
OR
1
xyzw360
(xy)(zw)360
xyxy360
[since zwxy]
½
Page 2 of 11
 2(xy)360
xy180
AOB is a straight line
14.
½
a15 cm
b15 cm
c12 cm
a  b  c
15  15  12
42
S


 21
2
2
2
A  s  sa  sb sc
 21  2115  2115  2112 
½
1
 21
6 6 9
 3 7 2  3  2  3  3  3
½
2
21 cm
18
SECTION - C
15.
7
 51 7 2  2  52 7 3 
 52 74    53 75 
7
 7 7 
 7 7 
 52  51    53  52 
4
2
7
5
2
7 
7 
 53    55 
6
8
2
1
5
7 
5 
 521    7 40 
42
2
 7 2  54 
25
1
3
5
7
5
2
½
2
2
7 
5 
 3  8
5 
7 
1
 7  5 
  21

 5  7 40 
2
6
2
42
5
25
5
1
1
2
½
2
½
2
½
2
 75 725
175
OR
5.8 correct construction, neatness and accuracy.
16.
3
2 6  5
 a  b 30
3 5  2 6
2 6  5
3 5  2 6

3 5  2 6
3 5  2 6
30  24152 30
4524
9  4 30

21
9
3
4
 a  
b 
21
7
21

6
1
1
½
½
Page 3 of 11
17.
a3b3  a  b  a2  b  b2 
3
y
y
x
x
      
3
5
3
5
3
2
y
y   x
y
y
x
x
x





    
 3

5
3
5  3
5
3
5
2 y  x2
y2
2 xy
y2
x2


 

 
5  9
25
25
9
25
2
y
y 
x
x






 
3
5
3
5 
y2
2 xy 
x2
 

9
25
15 
2 y  3 x2
y2 
2y  x2
y2 







5  9
25 
5 3
25 

1
1
1
OR
(3x2y)20 xy 
11
9
(3x2y)3 203
27x 38y 318xy (3x2y)  8000
2
27x 38y 3 18 
11
 20  8000
9
27x 38y 38000440
27x 38y 37560
18.
½
½
1
½
½
 a  b    b  c   c a 
2
2 3
2
2 3
2
2 3
 ab 3  bc 3 ca 3
Both Numerator and Denominator are of the form a 3 b3 c3 .
We know that when abc0
then a 3b3 c3 3abc
For Numerator a2 b2 b2  c2  c2  a2 0
For Denominator abbcca0


3  a 2 b2   b2 c 2   c 2 a 2 
3  a  b  b  c  c  a 
 ab  ab  bc  bc  ca   ca 
 ab  bc  ca 
 (ab) (bc) (ca)
½
½
1
½
½
Page 4 of 11
19.
Construct l to AB and CD.
EOMOMD180
EOM180  45  135
and ALOLOE180
LOE180  30  150
Co-interior angles are supplementary
LOE EOM
x135150285
1
1
OR
½
PQRS
Extend RS (to PQ) to M
TMR35 (Corresponding angles)
TRM  18011070 [L.P]
In TMR, TRM70 and TMR35
 y180(7035)
18010575
20.
Proof : ABC isosceles le
ABAC
ABC  ACB
Angles opp. to equal sides are equal.
DBC is an isosceles le.
BDDC and also DBCDCB
Add (1) and (2)
ABCDBCACBDCB
ABDACD
1
½
½
1
½
__________(1)
___________ (2)
1
1
1
Page 5 of 11
21.
Proof : PQRS is a square.
(i)



(ii)
SRT is an equilateral le
 PSR90, TSR60
PSRTSR150, similarly QRT150
PST and QRT, we‟ve PSQR (S)
PST  QRT  150 (A) and STRT (S)
PST  QRT  PTQT
n TQR, QRRT
TQR  QTR  x
xxQRT180
2x150180  2x30  x15
1
1
½
½
1
22.
Proof : r line is the shortest
CFAB
 CF < AC and CF < BC __________ (1)
Similarly BC a line segment. A is the point not on it. ADBC
 AD < AB and AD < AC __________ (2)
Also AC, a line segment B is a point not on it. BEAC
 BE < AB and BE < BC __________ (3)
Add (1), (2) and (3)
2 AD BE CF > 2 AB BC CA
 ABBCCA < ADBECF
Perimeter is greater than sum of three altitudes.
½
½
½
1
½
Page 6 of 11
23.
QRT 9040x180
x18013050
PSR Ext. PSQ
yx30
80
z180y
18080100
1
1
1
24.
Rhombus
Perimeter  52 cm
52
Side 
 13 cm
4
Diagonal  24 cm
OBOD  12 cm
1
1
½
OA 132 12 2  169144  25  5 cm
2
1
Area of rhombus  4 512
2
½
 120 cm2
SECTION - D
25.
7  3 5
7  3 5

3  5
3  5
7  3 5 3  5
7  3 5
3  5


3  5 3  5
3  5
3  5





 21 7




5  9 5  15
95
6  2 5  6  2 5

4
4 5

 5
4




   21 7
1


5  9 5  15
9  5
1
1

1
1
OR
1
Page 7 of 11
1
1
3  2 2


x
3  2 2
3  2 2
1
3  2 2
 
 3  2 2
x
9  8
x3  2 2 ,


1
1
2
x 2 3  2 2  9  8  12 2


2
1
 3  2 2  9  8  12 2
2
x
1
x 2  2  17  12 2  17  12 2
x
 34
1
x 2  2  34
x
26.
4
3
2187

7

4  2187

1
7
 4   37 
5
1
256

1
4

 1331 
2
1
1
3

 5  256 4  2  1331

½
2
3
7
½
3

1
3

 5  4 4  4  2   113 
1
2
1
3
1½

2
427542121
 10820242330
27.
1
2x3 2y3 2z36xyz
2
2
2
LHS =  xyz  xy    yz   zx  


3
3
3
2  x  y  z  3xyz
 2  xyz   x 2 y 2 z 2 xyyzzx  
  xyz   2 x 2 2y 2 2z 2 2xy2yz2zx  
  xyz  x 2 2 xy  y 2 y 2 2yz  z 2 z 22 zx  x 2  
2
2
2
  xyz  xy    yz  zx   



RHS Hence proved.
2 (13)3 2 (14)3 2 (15)3 6131415
2
2
2
  13  14  15  13  14    14  15   15  13  
28.
1½
1
1
1
½
 42  (114)
 42  6  252
½
p(x)4x3 20x 233x18
(x2) is a factor
p(x)(x2) (4x 2 12x9)
1
1
Page 8 of 11
4x2 6x6x9
= 2x (2x3)3 (2x3)
= (2x3) (2x3)
Factors are (2x3) (2x3) (x2)
29.
x 2 4 is a factor of p(x)ax 42x 33x 2bx4
 (x2) (x2) is a factor of p(x)
 p(2)0p(2)
p(2)16 a82122b40

16 a2b0
p(2)16 a16122b40

16 a2b32
a1, b8
p(x)x 4 2x 3 3x 2 8x4
 (x 24) (x 2 2x1)
 (x2) (x2) (x1)
2
1
1
½
1
1
½
½
½
30.
A (3, 4)
B (2, 0)
C (1, 4)
D (1, 0)
IIIrd quad.
x-axis
IInd quad.
x-axis
1
1
1
1
31.
½
½
Page 9 of 11
Triangle ABC.
Sum of all the angles of ABC is 180
Construction Draw a line l parallel to BC.
Given to prove
Since lBC, We have 2y
(1) (Alternate angles are equal)
Similarly lBC
1z
(2) (Alternate angles are equal)
Also, sum of angle at a point A on line l is 180.
 2x1180
i.e. yxz180( from (1) and (2) )
 xyz180
ABC180
Sum of all angles of a  is 180
5x6x7x180
 x=10
Angles are 50, 60 and 70
32.
Proof : In ADE, we have
ADAE,
ADEAED
180ADE180AED
ADBAEC
__________ (1)
Consider ABD and ACE
ADAE
(side)
ADBAEC
(Angle ) from (1)
BDEC
(side)
By SAS congruence ABD  ACE
By cpct ABAC
 ABC is an isosceles triangle
OR
Proof : In ABC, BC > AB (AB smallest side)
BAC > BCA
__________ (1)
ACD CD > AD
CAD >ACD
__________ (2)
Adding (1) and (2),
BAC CAD > BCA ACD
 BAD >BCA
A >C
In ABD, AD > AB
 ABD >ADB
____________ (3)
In BCD, CD > BC
 DBC >BDC
____________ (4)
Add (3) and (4) ABD DBC > ADB BDC
ABC > ADC
B > D
Hence A > C and B > D
1
1
½
½
1
½
½
½
1½
1
½
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1
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1
Page 10 of 11
33.
Proof : Triangle inequality property
In ABC,
ABBC > AC
_________ (1)
In BCD,
BCCD > BD
_________ (2)
In CDA,
CDDA > AC
_________ (3)
In DAB,
DAAB > BD
_________ (4)
Adding (1), (2), (3) and (4), we get
ABBCBCCDCDDADAAB > ACBDACBD
2 AB BC CD DA > 2 AC BD
Perimeter > sum of its diagonals
34.
½
½
½
½
1
1
AOC  DOB
(SAS)
ACBD
(cpct)
OAC  ODB
(cpct)
and OCAOBD (cpct)
AC will be parallel to BD only when OAC OBD, But OAC may not be equal
to OBD, so AC may not be parallel to BD.
1½
1½
1
- oO o -
Page 11 of 11