1. No category

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SUMMER – 2013 EXAMINATION
MODEL ANSWER
Subject: BASIC MATHEMATICS
Subject Code: 17104
Important Instructions to examiners:

The model answer shall be the complete solution for each and every question on the question
paper.

Numerical shall be completely solved in a step by step manner along with step marking.

All alternative solutions shall be offered by the expert along with self-explanatory comments
from the expert.

In case of theoretical answers, the expert has to write the most acceptable answer and offer
comments regarding marking scheme to the assessors.

In should offer the most convincing figures / sketches / circuit diagrams / block diagrams /
flow diagrams and offer comments for step marking to the assessors.

In case of any missing data, the expert shall offer possible assumptions / options and the
ensuing solutions along with comments to the assessors for effective assessment.

In case of questions which are out of the scope of curricular requirement, the expert examiner
shall solve the question and mention the marking scheme in the model answer. However, the
experts are requested to submit their clear cut opinion about the scope of such question in the
paper separately to the coordinator.

Experts shall cross check the DTP of the final draft of the model answer prepared by them.
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
Subject Code: (17104)
Que.
Sub.
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Que.
Summer 2013
Page No: 2/28
Model answers
Marks
Attempt any ten of the following:
2 k 7
1)
(a)
Ans.
Total
Marks
20
Find k, if 3 4 13  0
8 11 33
2
k
7
3
4
13  0
8 11 33
2  4  33  1113  k  3  33  13  8   7  33  32   0
1
22  k  5   7  0
15  5k  0
½
15  5k
02
k 3
(b)
Ans.
½
 1 3  5 7 
Find A if , 2 A  3 


 2 5  6 3 
 1 3  5 7 
2A  3


 2 5  6 3 
3
2A  
6
5
2A  
6
9  5

15 6
7  3

3  6
7
3
9
15
 2 2 
2A  

 0 12 
1 1
A

0 6 
c)
Ans.
d)
Ans.
½
1
½
1 4 
Prove that the matrix 
 is non-singular matrix.
6 9 
1 4
Consider,
6 9
 9  24
 15  0
 Given matrix is non-singular matrix.
2 2
 2 3
 4 5
,B  
If A  
and C  


 then show that AB  AC
2 2
 4 5
 2 3
 2 2   2 3
AB  
 

 2 2   4 5
02
02
2
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Subject Code: (17104)
Que.
Sub.
No.
Que.
Summer 2013
Model answers
12 16 


12 16 
 2 2   4 5
AC  
 

 2 2   2 3
12 16 


12 16 
 AB=AC
1.
(e)
Ans.
Resolve into partial fractions:
2x  3
2x  3

x  2 x  3  x  3 x  1
Page No: 3/28
Marks
Total
Marks
1
1
02
2x  3
x  2x  3
2
2
Let,

2x  3
A
B


 x  3 x  1 x  3 x  1
½
2 x  3   x  1 A   x  3 B
Put x  1
9  4A
9
A
4
Put x  3
1  4 B
1
B
4
2x  3
9 / 4 1/ 4 1  9
1 


 

 x  3 x  1 x  3 x  1 4  x  3 x  1
2x  3
2x  3

x  2 x  3  x  3 x  1
½
½
02
½
OR
2
Let

2x  3
A
B


 x  3 x  1 x  3 x  1
2 x  3   x  1 A   x  3 B
2 x  3   A  B  x  A  3B
By equating equal power coefficients
A  3B  3
A B  2
By solving above equations:
½
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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______________________________________________________________________________________________
Subject Code: (17104)
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Sub.
No.
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1.
Summer 2013
Page No: 4/28
Model answers
Ans.
Total
Marks
½
9
4
1
B 
4
2x  3
9 / 4 1/ 4


 x  3 x  1 x  3 x  1
A
(f)
Marks
½
½
02
Prove that 2sin 2   1  cos 2
1  cos 2  1  cos    
 1   cos  .cos   sin  .sin  
1
 1   cos2   sin 2  
 1  cos2   sin 2 
 sin 2   sin 2 
=2sin 2 
(g)
Ans.
Ans.
02
2
02
Define Allied Angles
Allied Angle : Any two angles whose sum or difference is either zero
or an integral multiple of
(h)
½
½

2
are called as allied angle.
If 2cos 600.cos100  cos A  cos B , then find A and B
cos 600.cos100  cos A  cos B
cos  600  100   cos  600  100   cos A  cos B
1
cos 700  cos 500  cos A  cos B
 A  700 and B  500
1
02
OR
2 cos 60 .cos10  cos A  cos B
0
0
A B
A B
.cos
2
2
A B
 100
2
2 cos 600.cos100  2 cos
1
A B
 600 and
2
A  B  1200
½
A  B  200
 A  700
(i)
and B  500
Evaluate without using calculator
½
tan850  tan400
1+tan850 .tan400
02
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Subject Code: (17104)
Que.
Sub.
No.
Que.
1.
Ans.
Summer 2013
Page No: 5/28
Model answers
Marks
tan850  tan400
1+tan850 .tan400
 tan  850  400 
Total
Marks
1
½
 tan 450
½
1
(j)
Ans.
1
Prove that cos 1 x  sec1  
 x
1
Let cos x  
 cos   x
1
 sec  
x
1
  sec 1  
x
(k)
Ans.
02
½
1
½
02
1
cos 1 x  sec 1  
x
Find the perpendicular distance between the point (3,4) and the line
3x  4 y  5.
Perpendicular distance=
=
ax1  by1  c
a 2  b2
3  3  4  4   5
1
32  42
20
5
=4 units
=
(l)
Ans.
02
Range =Largest value-Smallest value
Attempt any four of the following:
16
Solve the following equations by using Cramers rule:
x  y  2 z  1, 2 x  3 y  4 z  4,3x  2 y  6 z  5
1
Ans.
1
1
3,6,10,1,15,16,21,19,18.
=20
(a)
02
Find the range of the following distribution:
=21-1
2.
1
Let D  2
1 2
3
4
3 2 6
 1 18  8   1 12  12   2  4  9 
 8
1
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Subject Code: (17104)
2.
1
Dx  4
Summer 2013
Page No: 6/28
1 2
3
4
5 2 6
 1 18  8   1 24  20   2  8  15 
 8
½
1 1 2
Dy  2 4
4
3 5 6
 1 24  20   1 12  12   2 10  12 
 16
½
1
Dy  2
1 1
3
4
3 2 5
 115  8   110  12   1 4  9 
8
½
x
½
Dx 8

1
D 8
Dy 16
y

2
D
8
D
8
z z 
 1
D 8
(b)
Ans.
½
04
½
 2 -3  x 1
Find x and y if 
 
1 5   1 0
 2 x  3 2 4  3 y   2 x  3
 x  5 1 2  5 y    x  5

 
 2x  3  2x  3
and x  5  x  5
2   2 x  3 2 4  3 y 

y   x  5 1 2  5 y 
2 4  3 y 
1 2  5 y 
, 4  3y  4  3y
, 2  5y  2  5y
2
2
04
In above all the cases finding the values of x and y are not possible.
 it has no solution.
(c)
 1 3 2
1 0 0 
2 1 2




If A   1 2 0  , B  1 2 0  and C   2 2 1 
 4 0 3 
1 0 3
1 2 2 
then find the matrix D such that 2A-3B-D=C
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
Subject Code: (17104)
Que.
Sub.
No.
Que.
2.
Ans.
Ans.
3 2  1
2 0   3 1
0 3  1
6 4  3 0
4 0   3 6
0 6  3 0
5 2
4 1
2 5
0 0 2
2 0  -  2
0 3 1
0 2 1
0  -  2 2
9  1 2
1 2
2 1 
2 2 
2
1 
2 
Total
Marks
x
2x  3

x  x  2  x  2  x  1
1
2
 3
 2 3 1
If A  
and B   5

1 0 4 
 4
 3
 2 3 1 
Consider AB  
  5
1 0 4  
 4
Resolve into partial fractions:
1
04
7
'
6  , then show that  AB   B ' A '
4
7
6 
4 
1
½
½
1
 17 19 
B ' A'  
23 
 28
 AB  '  B ' A '
Ans.
Marks
Given, 2 A  3B  D  C
D  2 A  3B  C
 6  15  4 14  18  4 
AB  
7  16 
 3  16
 17 28
AB  

 19 23
 17 19 
 AB  '  

 28 23 
 2 1
 3 5 4  
B ' A'  
3 0 


 7 6 4 
 1 4 
(e)
Page No: 7/28
Model answers
1
D  2  1
 4
2
D   2
 8
 3
D   7
 4
(d)
Summer 2013
1
x
x  x2
2
2
Let,

x
 x  2  x  1

A
B

x  2 x 1
1
04
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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______________________________________________________________________________________________
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Sub.
No.
Que.
Summer 2013
Model answers
Page No: 8/28
Marks
Total
Marks
x   x  1 A   x  2  B
2.
Put x  2
2  A  3
2
3
Put x  1
1  3B
A
B
1
3
1
x
 x  2  x  1
(f)
1

Resolve into partial fractions:
9
Ans.
2 / 3 1/ 3

x  3 x 1

1
9
 x  1 x  2 
2
A
B
C


x  1 x  2  x  2 2
 x  1 x  2 
2
9   x  2  A   x  2  x  1 B   x  1 C
2
04
½
½
Put x  1
9  9A
½
 A=1
Putx  2
9=  3C
 C  3
½
Put x  0
9=4A  2 B  C
9=4 1  2 B  3
9  7  2B
2 B  2
 B  1
9
 x  1 x  2 
3.
1
2

1
1
3


x  1 x  2  x  2 2
Attempt any four of the following:
(a)
Using matrix inversion method, solve the following equations:
x  y  z  3, x  2 y  3z  4, x  4 y  9 z  6
1
04
16
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Subject Code: (17104)
Que.
Sub.
No.
Que.
3.
Ans.
Summer 2013
Model answers
Page No: 9/28
Marks
1 1 1 
 x
 3




Let A  1 2 3 , X   y  and B   4 
1 4 9 
 z 
 6 
1 1 1
Consider, A  1 2 3
1 4 9
 118  12   1 9  3  1 4  2 
 662
½
20
 A1exists
2

4
1
Matrix of minors= 
4

1
 2
3
1 3
9
1 9
1
1 1
9
1 9
1
1 1
3
1 3
1 2

1 4
1 1

1 4

1 1
1 2 
6 6 2
 5 8 3 
1 2 1 
 6 6 2 
matrix of cofactors   5 8 3
 1 2 1 
 6 5 1 
Adj.A   6 8 2 
 2 3 1 
1
A1  .adj.A
A
 6 5 1 
1
=  6 8 2
2
 2 3 1 
X  A1 B
 6 5 1   3 
1
=  6 8 2  4
2
 2 3 1   6 
1
½
½
½
Total
Marks
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
Subject Code: (17104)
Que.
Sub.
No.
Que.
Summer 2013
Model answers
x 
4 2
 y  = 1  2  1 
  2   
 z 
 0   0 
3.
Page No: 10/28
Marks
1
OR
Matrix of cofactors can be evaluated by following method:
1
(b)
Ans.
Resolve into partial fractions:
x3  2
x2 1
x
x 1 x  2
2
3
x3  x
 +
x2
3
x 2
x2
 2
 x 2
x 1
x 1
x2
x2
Consider, 2

x  1  x  1 x  1
1
x2
A
B


 x  1 x  1 x  1 x  1
½

x  2   x  1 A   x  1 B
Total
Marks
04
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Que.
Sub.
No.
Que.
Summer 2013
Model answers
Page No: 11/28
Marks
Total
Marks
Put x  1
3.
1  2 A
1
2
Put x  1
A 
1
3=2B
B 
(c)
Ans.
3
2
1

x2
1/ 2 3/ 2


 x  1 x  1 x  1 x  1

x3  2
1/ 2 3/ 2
 x

2
x 1
x  1 x 1
Resolve into partial fractions:
½
04
ex
e2 x  4e x  3
put e x  t
t
t

t  4t  3  t  3 t  1
2

t

A
B

t  3 t 1
 t  3 t  1
 t   t  1 A   t  3 B
½
½
put t  3
3  2 A
3
A
2
put t  1
1
1  2 B
1
 B= 
2


t
 t  3 t  1
1

3 / 2 1/ 2

t  3 t 1
ex
3 / 2 1/ 2
 x

x
x
 e  3 e  1 e  3 e x  1
½
½
04
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Subject Code: (17104)
Que.
Sub.
No.
Que.
3.
(d)
Page No: 12/28
Model answers
Marks
Total
Marks
In any ABC,prove that
sin
Ans.
Summer 2013
A B C
BC  A
C  A B
 sin
 sin
 cos A  cos B  cos C
2
2
2
A  B  C  180
A  B  180  C
A  B  C  180  2C
A B C
 90  C
2
 A B C 
sin 
  sin  90  C 
2


1
 A B C 
sin 
  cos C
2


1
Similarly,
 BC  A
sin 
  cos A
2


 C  A B 
sin 
  cos B
2


½
½
 A B C 
 BC  A
 C  A B 
 sin 
  sin 
  sin 

2
2
2






= cosA+cosB+cosC
(e)
If A and B are both obtuse angles and sin A 
1
5
4
and cos B   ,
13
5
then find sin( A  B).
Ans.
Given sin A 
5
5
13
cos B  
13
4
5
3
5
A
12
 cos A 
12
13
,
B
4
sin B 
3
5
1
04
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Que.
Sub.
No.
Que.
Summer 2013
Model answers
Page No:13/28
Marks
Total
Marks
 A and B are obtuse
3.
12
3
,
sin B 
13
5
 sin( A  B)  sin A.cos B  cos A.sin B
cos A 
5 4 12 3
. 
.
13 5
13 5
56
=
65

1
1
1
04
OR
Given sin A 
5
13
 cos A   1  sin 2 A
=  1
25
169
12
13
A and B are obtuse
=
12
13
4
and cosB=
5
 cos A = 
1
sin B   1  cos 2 B
=  1
=
sin B 
16
25
3
5
3
5
 B is obtuse
1
sin  A  B   sin A.cos B  cos A.sin B
5 4 12 3
. 
.
13 5
13 5
-56
=
65
=
(f)
4
8
84
Prove that sin 1  sin 1  sin 1
5
17
85
1
1
04
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Subject Code: (17104)
Que.
Sub.
No.
Que.
3.
Summer 2013
Model answers
sin A 
Page No: 14/28
Marks
Total
Marks
4
5
 cos A  1  sin 2 A
= 1
=
16
25
3
5
1
and sin B =
8
17
 cos B  1  cos 2 B
= 1
64
289
15
17
LHS=A  B
1
=
RHS= sin 1
84
85
Consider,
sin( A  B)  sin A.cos B  cos A.sin B
4 15 3 8 84
= .  . 
5 17 5 17 85
1
4
8
84
84
A  B  sin 1  sin 1  sin 1
A  B  sin 1
5
17
85
85
OR
4
A
5
4
 sin A 
5
Let sin 1
,
,
8
=B
17
8
sin B 
17
sin -1
1
04
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Sub.
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Que.
Summer 2013
Page No: 15/28
Model answers
Marks
3.
4
5
8
17
3
tan A 
4
3
A  tan 1
 sin 1
15
,
4
3
4
4
 tan 1
5
3
1
,
,
8
15
8
B  tan 1
15
8
8
 sin 1  tan 1
17
15
tan B 
1
4
8
 sin 1
5
17
4
8
 tan 1  tan 1
3
15
 4 8 
 

 tan 1  5 15 
4 8
 1 . 
 5 15 
 60  24 


 tan 1  45 
45  32


 45 
 84 
 tan 1  
 13 
LHS  sin 1
84
85
84
Let sin 1
C
85
84
 sin C 
85
1
½
RHS= sin 1
84
85
13
Total
Marks
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
(Autonomous)
(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
Subject Code: (17104)
Que.
Sub.
No.
Que.
3.
Summer 2013
Model answers
Page No: 16/28
Marks
Total
Marks
84
13
84
C  tan 1
13
84
84
 sin 1
 tan 1
85
13
 tan C 
½
84
 RHS= tan
13
LHS=RHS
04
1
16
Attempt any four of the following:
4.
(a)
Ans.
Prove that tan  A  B  
tan A  tan B
1  tan A.tan B
½
2
1
½
04
(b)
Ans.
1
1
1
1
04
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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______________________________________________________________________________________________
Subject Code: (17104)
Que.
Sub.
No.
Que.
4.
(c)
Summer 2013
Model answers
Page No: 17/28
Marks
Total
Marks
Without using calculator show that
Ans.
1
1
½
½
1
Note: give appropriate marks for another method.
(d)
Prove that
Ans.
1
04
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Sub.
No.
Que.
Summer 2013
Model answers
Page No: 18/28
Marks
Total
Marks
4.
1
1
1
(e)
04
Prove that
Ans.
2
1
½
½
(f)
Ans.
Prove that
04
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Que.
Summer 2013
Page No: 19/28
Model answers
Marks
Total
Marks
4.
2
1
1
04
OR
Note : This example can be solved by another method , see question
no.3 (f) on page.no.15.
Attempt any Four of the following:
5.
(a)
Prove that
1
Ans.
1
1
1
04
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Que.
Sub.
No.
Que.
5.
Summer 2013
Page No: 20/28
Model answers
Marks
Total
Marks
(b)
1
1
1
1
04
(c)
Ans.
1
-(1)
1
1
from(1)
(d)
1
04
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Subject Code: (17104)
Que.
Sub.
No.
Que.
5.
Ans.
Summer 2013
Page No: 21/28
Model answers
Marks
Total
Marks
1
Y-axis
P(
)
X-axis
is
1
1
1
(e)
Ans.
1
1
04
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Sub.
No.
Que.
Summer 2013
Model answers
Page No: 22/28
Marks
Total
Marks
5.
1
1
04
(f)
Ans.
3x
= 9
1
1
1
1
2
1
04
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Subject Code: (17104)
Que.
Sub.
No.
Que.
Summer 2013
Model answers
5.
Page No: 23/28
Marks
1
Total
Marks
04
Attempt any Four of the following:
6.
(a)
Ans.
1
from fig.
1
1
.
1
(b)
Ans.
04
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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No.
Que.
Summer 2013
Page No: 24/28
Model answers
Marks
Total
Marks
6.
1
1
½
1
½
(c)
Find the range and coefficient of range of the following data:
Age
(in years)
Frequency
1019
2029
3039
4049
5059
6069
7079
03
61
223
137
53
19
04
1
1
04
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Subject Code: (17104)
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Sub.
No.
Que.
Summer 2013
Page No: 25/28
Model answers
Marks
Total
Marks
6.
1
1
(d)
04
Find mean deviation from median for the following data:
Class
0 10
10 20
20 30
30 40
40 50
5
8
15
16
6
interval
Freauency
Ans.
Class
0
5
5
25
22
110
10
15
8
120
12
96
20
25
15
375
2
30
30
35
16
560
8
12
40
45
6
270
18
108
50
1350
1
472
1
1
1
04
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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______________________________________________________________________________________________
Subject Code: (17104)
Que.
Sub.
No.
Que.
6.
e)
Summer 2013
Page No: 26/28
Model answers
Marks
Total
Marks
Find variance and coefficient of variance for the following data:
Class
0-
Interval
5
Frequency
3
510
1015
1520
2025
2530
3035
3540
5
9
15
20
16
10
2
Class
Ans.
0-5
2.5
3
7.5
6.25
18.75
5-10
7.5
5
37.5
56.25
281.25
10-15
12.5
9
112.5
156.25
1406.25
15-20
17.5
15
262.5
306.25
4593.75
20-25
22.5
20
450
506.25
10125
25-30
27.5
16
440
756.25
12100
30-35
32.5
10
325
1056.25
10562.5
35-40
37.5
2
75
1406.25
2812.5
80
1710
1
41900
`1
1
04
1
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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Subject Code: (17104)
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Sub.
No.
Que.
6.
Summer 2013
Page No: 27/28
Model answers
Marks
Total
Marks
Class
0-5
2.5
3
-3
-9
9
27
5-10
7.5
5
-2
-10
4
20
10-15
12.5
9
-1
-9
1
9
15-20
17.5 a
15
0
0
0
0
20-25
22.5
20
1
20
1
20
25-30
27.5
16
2
32
4
64
30-35
32.5
10
3
30
9
90
35-40
37.5
2
4
8
16
32
80
4
62
1
262
1
1
1
04
MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION
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(ISO/IEC - 27001 - 2005 Certified)
______________________________________________________________________________________________
Subject Code: (17104)
Que.
Sub.
No.
Que.
Summer 2013
Page No: 28/28
Model answers
Marks
Total
Marks
6.
Note: Students may take any another value for A in the above/below
example. So the above table and corresponding values vary
accordingly. But the final answer will be the same.
(f)
The two sets of observations are given below:
Set-I
Set-II
Which of the two sets is more consistent?
1½
1½
1
Important Note
In the solution of the question paper, wherever possible all the
possible alternative methods of solution are given for the sake of
convenience. Still student may follow a method other than the given
herein. In such case, first see whether the method falls within the scope
of the curriculum, and then only give appropriate marks in accordance
with the scheme of marking.
04