CAT Past Year Question Paper Solution - 2008 | Additional Study Material for CAT PDF Download

 Page 1


Page 1
Total 
Questions
Time Taken 
(Min)
Total 
Attempts
Correct 
Attempts
Incorrect 
Attempts
Net 
Score
TOTAL 90 150
MY PERFORMANCE
Section I 25
Section II 25
Language Comprehension 
and English Usage
Quantitative Ability
Section III 40
Data Interpretation and 
Reasoning
1 3 2 3 3 2 4 2 5 2 6 5 7 3 8 4 9 4 10 1
11 2 12 5 13 5 14 3 15 5 16 1 17 4 18 3 19 4 20 4
21 5 22 2 23 1 24 1 25 1 26 4 27 5 28 2 29 5 30 3
31 4 32 2 33 3 34 3 35 2 36 4 37 5 38 5 39 1 40 1
41 4 42 3 43 5 44 5 45 1 46 1 47 1,4 48 3 49 3 50 2
51 1 52 4 53 3 54 1 55 3 56 2 57 2 58 1 59 3 60 4
61 2 62 5 63 3 64 4 65 2 66 4 67 2 68 4 69 2 70 5
71 3 72 4 73 2 74 2 75 1 76 5 77 2 78 2 79 1 80 4
81 1 82 4 83 3 84 4 85 5 86 3 87 5 88 1 89 4 90 5
Disclaimer:  There are mismatches in our VA Answer key (Question nos. 54, 60, 66, 67 & 69) with the
solutions that IIMs have provided. However, all these questions are quite controversial and Career Launcher
stands by its answer key as we have debated, discussed and 'googled' it time and again.

	






Page 2


Page 1
Total 
Questions
Time Taken 
(Min)
Total 
Attempts
Correct 
Attempts
Incorrect 
Attempts
Net 
Score
TOTAL 90 150
MY PERFORMANCE
Section I 25
Section II 25
Language Comprehension 
and English Usage
Quantitative Ability
Section III 40
Data Interpretation and 
Reasoning
1 3 2 3 3 2 4 2 5 2 6 5 7 3 8 4 9 4 10 1
11 2 12 5 13 5 14 3 15 5 16 1 17 4 18 3 19 4 20 4
21 5 22 2 23 1 24 1 25 1 26 4 27 5 28 2 29 5 30 3
31 4 32 2 33 3 34 3 35 2 36 4 37 5 38 5 39 1 40 1
41 4 42 3 43 5 44 5 45 1 46 1 47 1,4 48 3 49 3 50 2
51 1 52 4 53 3 54 1 55 3 56 2 57 2 58 1 59 3 60 4
61 2 62 5 63 3 64 4 65 2 66 4 67 2 68 4 69 2 70 5
71 3 72 4 73 2 74 2 75 1 76 5 77 2 78 2 79 1 80 4
81 1 82 4 83 3 84 4 85 5 86 3 87 5 88 1 89 4 90 5
Disclaimer:  There are mismatches in our VA Answer key (Question nos. 54, 60, 66, 67 & 69) with the
solutions that IIMs have provided. However, all these questions are quite controversial and Career Launcher
stands by its answer key as we have debated, discussed and 'googled' it time and again.

	






1.  3 Total sum of the numbers written on the blackboard
=
40 41
820
2
×
=
When two numbers ‘a’ and ‘b’ are erased and replaced
by a new number a + b – 1, the total sum of the
numbers written on the blackboard is reduced by 1.
Since, this operation is repeated 39 times, therefore,
the total sum of the numbers will be reduced by
1 × 39 = 39.
Therefore, after 39 operations there will be only 1
number that will be left on the blackboard and that will
be 820 – 39 = 781.
2. 3 The last two digits of any number in the form of 7
4n
will always be equal to 01.
For example: 7
4 
 = 2401 and 7
8 
= 5764801.
3. 2 x
3
 – ax
2
 + bx – c = 0
Let the roots of the above cubic equation be
(a – 1), a, (a + 1)
? a (a – 1) + a (a + 1) + (a + 1) (a –1) = b
? a
2
 –  + a
2
 + a + a
2
 – 1 = b ? 3a
2
 – 1 = b
Thus, the minimum possible value of ‘b’ will be equal
to  – 1 and this value is attained at a = 0.
4. 2 Amount of rice bought by the first customer
x1
kgs
22
??
=+
??
??
Amount of rice remaining 
x1 x 1
xkgs
22 2
- ??
=- + =
??
??
Amount of rice bought by the second customer
1x 1 1 x 1
kgs
22 2 4
-+ ??
=× + =
??
??
Amount of rice remaining
x1 x 1 x 3
kgs
24 4
-+ - ??? ?
=- =
??? ?
??? ?
Amount of rice bought by the third customer
1x 3 1 x 1
kgs
24 2 8
-+ ??
=× + =
??
??
As per the information given in the question
x1 x 3
84
+-
= because there is no rice left after the
third customer has bought the rice.
Therefore, the value of ‘x’ = 7 kgs.
5. 2 Given that f(x) = ax
2
  + bx + c
Also, f(5) = –3f(2) ? f(5) + 3f(2) = 0
? (25a + 5b + c) + 3(4a + 2b + c) = 0
? 37a + 11b + 4c = 0 …(i)
Also, as 3 is a root of f(x) = 0, thus, f(3) = 0.
Therefore, 9a + 3b + c = 0 …(ii)
Using equation (i) and (ii), we get that a = b
Therefore, c = –12a
? f(x) = a(x
2 
+ x –12) = a(x + 4) (x – 3)
Therefore, the other root of f(x) = 0 is –4.
6. 5 f(x) = a(x
2
 + x –12)
Therefore, the value of a + b + c cannot be uniquely
determined.
7. 3 Total number of terms in the sequence 17, 21, 25 …
417 is equal to 
417 17
1 101
4
-
+= .
Total number of terms in the sequence 16, 21, 26 …
466 is equal to 
466 16
191
5
-
+= .
n
th
 term of the first sequence = 4n + 13.
m
th
 term of the second sequence = 5m + 11.
As per the information given in the question 4n + 13
= 5m + 11
? 5m – 4n = 2.
Possible integral values of n that satisfy 5m = 2 + 4n
are (2, 7, 12 … 97)
Therefore, the total number of terms common in both
the sequences is 20.
8. 4 In other words we need to find the total number of
4-digit numbers not more than 4000 using the digits
0, 1, 2, 3 and 4.
The digit at the thousands place can be selected in
3 ways.
The digits at the hundreds place can be selected in
5 ways.
The digits at the tens place can be selected in 5 ways.
The digits at the units place can be selected in 5 ways.
Therefore, the total number of 4-digit numbers less
than 4000 is equal to
3 × 5 × 5 × 5 = 375.
Therefore, the total number of 4-digit numbers not
more than 4000 is equal to 375 + 1 = 376.
 9. 4
A
E
P
F
C
Y
B
D
X
For the shortest route, Neelam follows the following
path:
AE F B ?? ?
No. of ways to reach from A to E: 
() 22!
6
2! 2!
+
=
×
No. of ways to reach from E to F:  1
No. of ways to reach from F to B: 
() 42!
15
4! 2!
+
=
×
Page 3


Page 1
Total 
Questions
Time Taken 
(Min)
Total 
Attempts
Correct 
Attempts
Incorrect 
Attempts
Net 
Score
TOTAL 90 150
MY PERFORMANCE
Section I 25
Section II 25
Language Comprehension 
and English Usage
Quantitative Ability
Section III 40
Data Interpretation and 
Reasoning
1 3 2 3 3 2 4 2 5 2 6 5 7 3 8 4 9 4 10 1
11 2 12 5 13 5 14 3 15 5 16 1 17 4 18 3 19 4 20 4
21 5 22 2 23 1 24 1 25 1 26 4 27 5 28 2 29 5 30 3
31 4 32 2 33 3 34 3 35 2 36 4 37 5 38 5 39 1 40 1
41 4 42 3 43 5 44 5 45 1 46 1 47 1,4 48 3 49 3 50 2
51 1 52 4 53 3 54 1 55 3 56 2 57 2 58 1 59 3 60 4
61 2 62 5 63 3 64 4 65 2 66 4 67 2 68 4 69 2 70 5
71 3 72 4 73 2 74 2 75 1 76 5 77 2 78 2 79 1 80 4
81 1 82 4 83 3 84 4 85 5 86 3 87 5 88 1 89 4 90 5
Disclaimer:  There are mismatches in our VA Answer key (Question nos. 54, 60, 66, 67 & 69) with the
solutions that IIMs have provided. However, all these questions are quite controversial and Career Launcher
stands by its answer key as we have debated, discussed and 'googled' it time and again.

	






1.  3 Total sum of the numbers written on the blackboard
=
40 41
820
2
×
=
When two numbers ‘a’ and ‘b’ are erased and replaced
by a new number a + b – 1, the total sum of the
numbers written on the blackboard is reduced by 1.
Since, this operation is repeated 39 times, therefore,
the total sum of the numbers will be reduced by
1 × 39 = 39.
Therefore, after 39 operations there will be only 1
number that will be left on the blackboard and that will
be 820 – 39 = 781.
2. 3 The last two digits of any number in the form of 7
4n
will always be equal to 01.
For example: 7
4 
 = 2401 and 7
8 
= 5764801.
3. 2 x
3
 – ax
2
 + bx – c = 0
Let the roots of the above cubic equation be
(a – 1), a, (a + 1)
? a (a – 1) + a (a + 1) + (a + 1) (a –1) = b
? a
2
 –  + a
2
 + a + a
2
 – 1 = b ? 3a
2
 – 1 = b
Thus, the minimum possible value of ‘b’ will be equal
to  – 1 and this value is attained at a = 0.
4. 2 Amount of rice bought by the first customer
x1
kgs
22
??
=+
??
??
Amount of rice remaining 
x1 x 1
xkgs
22 2
- ??
=- + =
??
??
Amount of rice bought by the second customer
1x 1 1 x 1
kgs
22 2 4
-+ ??
=× + =
??
??
Amount of rice remaining
x1 x 1 x 3
kgs
24 4
-+ - ??? ?
=- =
??? ?
??? ?
Amount of rice bought by the third customer
1x 3 1 x 1
kgs
24 2 8
-+ ??
=× + =
??
??
As per the information given in the question
x1 x 3
84
+-
= because there is no rice left after the
third customer has bought the rice.
Therefore, the value of ‘x’ = 7 kgs.
5. 2 Given that f(x) = ax
2
  + bx + c
Also, f(5) = –3f(2) ? f(5) + 3f(2) = 0
? (25a + 5b + c) + 3(4a + 2b + c) = 0
? 37a + 11b + 4c = 0 …(i)
Also, as 3 is a root of f(x) = 0, thus, f(3) = 0.
Therefore, 9a + 3b + c = 0 …(ii)
Using equation (i) and (ii), we get that a = b
Therefore, c = –12a
? f(x) = a(x
2 
+ x –12) = a(x + 4) (x – 3)
Therefore, the other root of f(x) = 0 is –4.
6. 5 f(x) = a(x
2
 + x –12)
Therefore, the value of a + b + c cannot be uniquely
determined.
7. 3 Total number of terms in the sequence 17, 21, 25 …
417 is equal to 
417 17
1 101
4
-
+= .
Total number of terms in the sequence 16, 21, 26 …
466 is equal to 
466 16
191
5
-
+= .
n
th
 term of the first sequence = 4n + 13.
m
th
 term of the second sequence = 5m + 11.
As per the information given in the question 4n + 13
= 5m + 11
? 5m – 4n = 2.
Possible integral values of n that satisfy 5m = 2 + 4n
are (2, 7, 12 … 97)
Therefore, the total number of terms common in both
the sequences is 20.
8. 4 In other words we need to find the total number of
4-digit numbers not more than 4000 using the digits
0, 1, 2, 3 and 4.
The digit at the thousands place can be selected in
3 ways.
The digits at the hundreds place can be selected in
5 ways.
The digits at the tens place can be selected in 5 ways.
The digits at the units place can be selected in 5 ways.
Therefore, the total number of 4-digit numbers less
than 4000 is equal to
3 × 5 × 5 × 5 = 375.
Therefore, the total number of 4-digit numbers not
more than 4000 is equal to 375 + 1 = 376.
 9. 4
A
E
P
F
C
Y
B
D
X
For the shortest route, Neelam follows the following
path:
AE F B ?? ?
No. of ways to reach from A to E: 
() 22!
6
2! 2!
+
=
×
No. of ways to reach from E to F:  1
No. of ways to reach from F to B: 
() 42!
15
4! 2!
+
=
×
? Total number of possible shortest paths
= 6 × 1 × 15 = 90
10. 1 Neelam has to reach C via B.
From A to B, the number of paths are 90, as found in
question 9.
From B to C, Neelam follows the route:
Case I: BX C ??
OR Case II: BY C ?? .
Case I: BX C ??
No. of ways to reach from B to X: 
(5 1)!
6
5! 1!
+
=
×
No. of ways to reach from X to C : 2
So, total number of paths are 6 × 2 = 12 ways.
Case II: BY C ?? :
There is just one way.
Therefore, from B to C, there are 6 × 2 + 1 = 13
ways
?Total number of ways of reaching from A to C,
via B = 90 × 13 = 1170.
11. 2 f(x).f(y) = f(xy)
Given, f(2) = 4
We can also write,
f(2) = f(2 × 1) = f(2) ×  f(1)
OR f(1) × 4 = 4
? f(1) = 1
Now we can also write,
11
f(1) f 2 f(2) f
22
?? ??
=× = ×
?? ??
?? ??
?
1f(1) 1
f
2f(2) 4
??
==
??
??
12. 5 seed(n) function will eventually give the digit-sum of
any given number, n.
All the numbers ‘n’ for which seed(n) = 9 will give
the remainder 0 when divided by 9.
For all positive integers n, n < 500, there are 55 multiples
of 9.
13. 5 We can use the formula for the circum radius of a
triangle:
ab c
R
4 (Area of the triangle)
××
=
×
or 
ab c a c
R
1 2AD
4bAD
2
×× ×
==
× ??
×××
??
??
17.5 9
26.25 cm
23
×
==
×
14. 3 The three sides of the obtuse triangle are 8 cm, 15 cm
and x cm. As 15 is greater than 8, hence either x or 15
will be the largest side of this triangle. Consider two
cases:
Case I:
15 cm
C
B 8 cm
x
A
90°
Consider the right ABC ? above,
22
x 15 8 12.68 cm =- =
For all values of x < 12.68, the ABC ? will be obtuse.
But as the sum of two sides of triangle must be greater
than the third side, hence (x + 8) > 15 or x > 7.
Thus, the permissible values of x are 8, 9, 10, 11 and
12.
Case II:
15 cm
C
B 8 cm
x
A
90°
In the right ABC ? above, 
22
x15 8 17 =+ = .
For all values of x  > 17, ABC ? will be obtuse. But, as
the length of third side should be less than the sum of
other two sides, hence x < (15 + 8) or x < 23. The
permissible values of x are: 18, 19, 20, 21 and 22.
From Case I and II, x can take 10 values.
Page 4


Page 1
Total 
Questions
Time Taken 
(Min)
Total 
Attempts
Correct 
Attempts
Incorrect 
Attempts
Net 
Score
TOTAL 90 150
MY PERFORMANCE
Section I 25
Section II 25
Language Comprehension 
and English Usage
Quantitative Ability
Section III 40
Data Interpretation and 
Reasoning
1 3 2 3 3 2 4 2 5 2 6 5 7 3 8 4 9 4 10 1
11 2 12 5 13 5 14 3 15 5 16 1 17 4 18 3 19 4 20 4
21 5 22 2 23 1 24 1 25 1 26 4 27 5 28 2 29 5 30 3
31 4 32 2 33 3 34 3 35 2 36 4 37 5 38 5 39 1 40 1
41 4 42 3 43 5 44 5 45 1 46 1 47 1,4 48 3 49 3 50 2
51 1 52 4 53 3 54 1 55 3 56 2 57 2 58 1 59 3 60 4
61 2 62 5 63 3 64 4 65 2 66 4 67 2 68 4 69 2 70 5
71 3 72 4 73 2 74 2 75 1 76 5 77 2 78 2 79 1 80 4
81 1 82 4 83 3 84 4 85 5 86 3 87 5 88 1 89 4 90 5
Disclaimer:  There are mismatches in our VA Answer key (Question nos. 54, 60, 66, 67 & 69) with the
solutions that IIMs have provided. However, all these questions are quite controversial and Career Launcher
stands by its answer key as we have debated, discussed and 'googled' it time and again.

	






1.  3 Total sum of the numbers written on the blackboard
=
40 41
820
2
×
=
When two numbers ‘a’ and ‘b’ are erased and replaced
by a new number a + b – 1, the total sum of the
numbers written on the blackboard is reduced by 1.
Since, this operation is repeated 39 times, therefore,
the total sum of the numbers will be reduced by
1 × 39 = 39.
Therefore, after 39 operations there will be only 1
number that will be left on the blackboard and that will
be 820 – 39 = 781.
2. 3 The last two digits of any number in the form of 7
4n
will always be equal to 01.
For example: 7
4 
 = 2401 and 7
8 
= 5764801.
3. 2 x
3
 – ax
2
 + bx – c = 0
Let the roots of the above cubic equation be
(a – 1), a, (a + 1)
? a (a – 1) + a (a + 1) + (a + 1) (a –1) = b
? a
2
 –  + a
2
 + a + a
2
 – 1 = b ? 3a
2
 – 1 = b
Thus, the minimum possible value of ‘b’ will be equal
to  – 1 and this value is attained at a = 0.
4. 2 Amount of rice bought by the first customer
x1
kgs
22
??
=+
??
??
Amount of rice remaining 
x1 x 1
xkgs
22 2
- ??
=- + =
??
??
Amount of rice bought by the second customer
1x 1 1 x 1
kgs
22 2 4
-+ ??
=× + =
??
??
Amount of rice remaining
x1 x 1 x 3
kgs
24 4
-+ - ??? ?
=- =
??? ?
??? ?
Amount of rice bought by the third customer
1x 3 1 x 1
kgs
24 2 8
-+ ??
=× + =
??
??
As per the information given in the question
x1 x 3
84
+-
= because there is no rice left after the
third customer has bought the rice.
Therefore, the value of ‘x’ = 7 kgs.
5. 2 Given that f(x) = ax
2
  + bx + c
Also, f(5) = –3f(2) ? f(5) + 3f(2) = 0
? (25a + 5b + c) + 3(4a + 2b + c) = 0
? 37a + 11b + 4c = 0 …(i)
Also, as 3 is a root of f(x) = 0, thus, f(3) = 0.
Therefore, 9a + 3b + c = 0 …(ii)
Using equation (i) and (ii), we get that a = b
Therefore, c = –12a
? f(x) = a(x
2 
+ x –12) = a(x + 4) (x – 3)
Therefore, the other root of f(x) = 0 is –4.
6. 5 f(x) = a(x
2
 + x –12)
Therefore, the value of a + b + c cannot be uniquely
determined.
7. 3 Total number of terms in the sequence 17, 21, 25 …
417 is equal to 
417 17
1 101
4
-
+= .
Total number of terms in the sequence 16, 21, 26 …
466 is equal to 
466 16
191
5
-
+= .
n
th
 term of the first sequence = 4n + 13.
m
th
 term of the second sequence = 5m + 11.
As per the information given in the question 4n + 13
= 5m + 11
? 5m – 4n = 2.
Possible integral values of n that satisfy 5m = 2 + 4n
are (2, 7, 12 … 97)
Therefore, the total number of terms common in both
the sequences is 20.
8. 4 In other words we need to find the total number of
4-digit numbers not more than 4000 using the digits
0, 1, 2, 3 and 4.
The digit at the thousands place can be selected in
3 ways.
The digits at the hundreds place can be selected in
5 ways.
The digits at the tens place can be selected in 5 ways.
The digits at the units place can be selected in 5 ways.
Therefore, the total number of 4-digit numbers less
than 4000 is equal to
3 × 5 × 5 × 5 = 375.
Therefore, the total number of 4-digit numbers not
more than 4000 is equal to 375 + 1 = 376.
 9. 4
A
E
P
F
C
Y
B
D
X
For the shortest route, Neelam follows the following
path:
AE F B ?? ?
No. of ways to reach from A to E: 
() 22!
6
2! 2!
+
=
×
No. of ways to reach from E to F:  1
No. of ways to reach from F to B: 
() 42!
15
4! 2!
+
=
×
? Total number of possible shortest paths
= 6 × 1 × 15 = 90
10. 1 Neelam has to reach C via B.
From A to B, the number of paths are 90, as found in
question 9.
From B to C, Neelam follows the route:
Case I: BX C ??
OR Case II: BY C ?? .
Case I: BX C ??
No. of ways to reach from B to X: 
(5 1)!
6
5! 1!
+
=
×
No. of ways to reach from X to C : 2
So, total number of paths are 6 × 2 = 12 ways.
Case II: BY C ?? :
There is just one way.
Therefore, from B to C, there are 6 × 2 + 1 = 13
ways
?Total number of ways of reaching from A to C,
via B = 90 × 13 = 1170.
11. 2 f(x).f(y) = f(xy)
Given, f(2) = 4
We can also write,
f(2) = f(2 × 1) = f(2) ×  f(1)
OR f(1) × 4 = 4
? f(1) = 1
Now we can also write,
11
f(1) f 2 f(2) f
22
?? ??
=× = ×
?? ??
?? ??
?
1f(1) 1
f
2f(2) 4
??
==
??
??
12. 5 seed(n) function will eventually give the digit-sum of
any given number, n.
All the numbers ‘n’ for which seed(n) = 9 will give
the remainder 0 when divided by 9.
For all positive integers n, n < 500, there are 55 multiples
of 9.
13. 5 We can use the formula for the circum radius of a
triangle:
ab c
R
4 (Area of the triangle)
××
=
×
or 
ab c a c
R
1 2AD
4bAD
2
×× ×
==
× ??
×××
??
??
17.5 9
26.25 cm
23
×
==
×
14. 3 The three sides of the obtuse triangle are 8 cm, 15 cm
and x cm. As 15 is greater than 8, hence either x or 15
will be the largest side of this triangle. Consider two
cases:
Case I:
15 cm
C
B 8 cm
x
A
90°
Consider the right ABC ? above,
22
x 15 8 12.68 cm =- =
For all values of x < 12.68, the ABC ? will be obtuse.
But as the sum of two sides of triangle must be greater
than the third side, hence (x + 8) > 15 or x > 7.
Thus, the permissible values of x are 8, 9, 10, 11 and
12.
Case II:
15 cm
C
B 8 cm
x
A
90°
In the right ABC ? above, 
22
x15 8 17 =+ = .
For all values of x  > 17, ABC ? will be obtuse. But, as
the length of third side should be less than the sum of
other two sides, hence x < (15 + 8) or x < 23. The
permissible values of x are: 18, 19, 20, 21 and 22.
From Case I and II, x can take 10 values.
15. 5
A B
D
E
F
G
P
Q
H L
C
Let , the length of AH = ‘x’ cm
By symmetry of the figure given above, we can
conclude that APD ? and BQC ? will have the same
area.
APD ? 
 is 120° and line ‘L’ divides the square
ABCD in 2 equal halves, therefore
APH HPD 60 ?=? = °
In 
AH x
AHP : tan60 3 HP cm
HP 3
?= °=?=
Area of APD 2 area( AHP) ?=× ? 2
1x x
2x cm
2 33
=× × × =
Area of ABQCDP = area (ABCD) – 2 area ( APD) ?                         
()
2
2
2
2x 2 3 1
2x
4x
33
-
=- =
Required Ratio 
()
2
2
2x 2 3 1
3
23 1
2x
3
-
==-
Alternate method:
Concepts used:
A
B C
a
b
c
abc
sinA sinB sinC
?= =
Also, area of ? ABC =
1
2
ab Sin C
= 
1
2
bc Sin A  = 
1
2
ac Sin B
In the given figure
AB
C D
E
G
H L
F
P Q
a
a
x
x
x x
120 120
For APD, ? Let AP = PD = x cms
ax x
sin120 sin30 sin30
?= =
°°°
sin120 ?° = sin (90 + 30) = cos 30 = 
3
,
2
sin 30 = 
1
2
ax
1
3
2
2
?=
 
a
xcms
3
?=
Thus, area of APD ? is 
1
AP PD sin120
2
×× × °
1a a 3
22 33
=× × ×
 = 
2
2
a
cm
43
by symmetry, Area of APD ? = Area of BQC ? Area of ABQCDP
Thus, ratio of 
[Removing area inside square ABCD
Area of square ABCD 2 (Area of APD)
2 (Areaof APD)
-× ? =
×?
23 1 =-
Page 5


Page 1
Total 
Questions
Time Taken 
(Min)
Total 
Attempts
Correct 
Attempts
Incorrect 
Attempts
Net 
Score
TOTAL 90 150
MY PERFORMANCE
Section I 25
Section II 25
Language Comprehension 
and English Usage
Quantitative Ability
Section III 40
Data Interpretation and 
Reasoning
1 3 2 3 3 2 4 2 5 2 6 5 7 3 8 4 9 4 10 1
11 2 12 5 13 5 14 3 15 5 16 1 17 4 18 3 19 4 20 4
21 5 22 2 23 1 24 1 25 1 26 4 27 5 28 2 29 5 30 3
31 4 32 2 33 3 34 3 35 2 36 4 37 5 38 5 39 1 40 1
41 4 42 3 43 5 44 5 45 1 46 1 47 1,4 48 3 49 3 50 2
51 1 52 4 53 3 54 1 55 3 56 2 57 2 58 1 59 3 60 4
61 2 62 5 63 3 64 4 65 2 66 4 67 2 68 4 69 2 70 5
71 3 72 4 73 2 74 2 75 1 76 5 77 2 78 2 79 1 80 4
81 1 82 4 83 3 84 4 85 5 86 3 87 5 88 1 89 4 90 5
Disclaimer:  There are mismatches in our VA Answer key (Question nos. 54, 60, 66, 67 & 69) with the
solutions that IIMs have provided. However, all these questions are quite controversial and Career Launcher
stands by its answer key as we have debated, discussed and 'googled' it time and again.

	






1.  3 Total sum of the numbers written on the blackboard
=
40 41
820
2
×
=
When two numbers ‘a’ and ‘b’ are erased and replaced
by a new number a + b – 1, the total sum of the
numbers written on the blackboard is reduced by 1.
Since, this operation is repeated 39 times, therefore,
the total sum of the numbers will be reduced by
1 × 39 = 39.
Therefore, after 39 operations there will be only 1
number that will be left on the blackboard and that will
be 820 – 39 = 781.
2. 3 The last two digits of any number in the form of 7
4n
will always be equal to 01.
For example: 7
4 
 = 2401 and 7
8 
= 5764801.
3. 2 x
3
 – ax
2
 + bx – c = 0
Let the roots of the above cubic equation be
(a – 1), a, (a + 1)
? a (a – 1) + a (a + 1) + (a + 1) (a –1) = b
? a
2
 –  + a
2
 + a + a
2
 – 1 = b ? 3a
2
 – 1 = b
Thus, the minimum possible value of ‘b’ will be equal
to  – 1 and this value is attained at a = 0.
4. 2 Amount of rice bought by the first customer
x1
kgs
22
??
=+
??
??
Amount of rice remaining 
x1 x 1
xkgs
22 2
- ??
=- + =
??
??
Amount of rice bought by the second customer
1x 1 1 x 1
kgs
22 2 4
-+ ??
=× + =
??
??
Amount of rice remaining
x1 x 1 x 3
kgs
24 4
-+ - ??? ?
=- =
??? ?
??? ?
Amount of rice bought by the third customer
1x 3 1 x 1
kgs
24 2 8
-+ ??
=× + =
??
??
As per the information given in the question
x1 x 3
84
+-
= because there is no rice left after the
third customer has bought the rice.
Therefore, the value of ‘x’ = 7 kgs.
5. 2 Given that f(x) = ax
2
  + bx + c
Also, f(5) = –3f(2) ? f(5) + 3f(2) = 0
? (25a + 5b + c) + 3(4a + 2b + c) = 0
? 37a + 11b + 4c = 0 …(i)
Also, as 3 is a root of f(x) = 0, thus, f(3) = 0.
Therefore, 9a + 3b + c = 0 …(ii)
Using equation (i) and (ii), we get that a = b
Therefore, c = –12a
? f(x) = a(x
2 
+ x –12) = a(x + 4) (x – 3)
Therefore, the other root of f(x) = 0 is –4.
6. 5 f(x) = a(x
2
 + x –12)
Therefore, the value of a + b + c cannot be uniquely
determined.
7. 3 Total number of terms in the sequence 17, 21, 25 …
417 is equal to 
417 17
1 101
4
-
+= .
Total number of terms in the sequence 16, 21, 26 …
466 is equal to 
466 16
191
5
-
+= .
n
th
 term of the first sequence = 4n + 13.
m
th
 term of the second sequence = 5m + 11.
As per the information given in the question 4n + 13
= 5m + 11
? 5m – 4n = 2.
Possible integral values of n that satisfy 5m = 2 + 4n
are (2, 7, 12 … 97)
Therefore, the total number of terms common in both
the sequences is 20.
8. 4 In other words we need to find the total number of
4-digit numbers not more than 4000 using the digits
0, 1, 2, 3 and 4.
The digit at the thousands place can be selected in
3 ways.
The digits at the hundreds place can be selected in
5 ways.
The digits at the tens place can be selected in 5 ways.
The digits at the units place can be selected in 5 ways.
Therefore, the total number of 4-digit numbers less
than 4000 is equal to
3 × 5 × 5 × 5 = 375.
Therefore, the total number of 4-digit numbers not
more than 4000 is equal to 375 + 1 = 376.
 9. 4
A
E
P
F
C
Y
B
D
X
For the shortest route, Neelam follows the following
path:
AE F B ?? ?
No. of ways to reach from A to E: 
() 22!
6
2! 2!
+
=
×
No. of ways to reach from E to F:  1
No. of ways to reach from F to B: 
() 42!
15
4! 2!
+
=
×
? Total number of possible shortest paths
= 6 × 1 × 15 = 90
10. 1 Neelam has to reach C via B.
From A to B, the number of paths are 90, as found in
question 9.
From B to C, Neelam follows the route:
Case I: BX C ??
OR Case II: BY C ?? .
Case I: BX C ??
No. of ways to reach from B to X: 
(5 1)!
6
5! 1!
+
=
×
No. of ways to reach from X to C : 2
So, total number of paths are 6 × 2 = 12 ways.
Case II: BY C ?? :
There is just one way.
Therefore, from B to C, there are 6 × 2 + 1 = 13
ways
?Total number of ways of reaching from A to C,
via B = 90 × 13 = 1170.
11. 2 f(x).f(y) = f(xy)
Given, f(2) = 4
We can also write,
f(2) = f(2 × 1) = f(2) ×  f(1)
OR f(1) × 4 = 4
? f(1) = 1
Now we can also write,
11
f(1) f 2 f(2) f
22
?? ??
=× = ×
?? ??
?? ??
?
1f(1) 1
f
2f(2) 4
??
==
??
??
12. 5 seed(n) function will eventually give the digit-sum of
any given number, n.
All the numbers ‘n’ for which seed(n) = 9 will give
the remainder 0 when divided by 9.
For all positive integers n, n < 500, there are 55 multiples
of 9.
13. 5 We can use the formula for the circum radius of a
triangle:
ab c
R
4 (Area of the triangle)
××
=
×
or 
ab c a c
R
1 2AD
4bAD
2
×× ×
==
× ??
×××
??
??
17.5 9
26.25 cm
23
×
==
×
14. 3 The three sides of the obtuse triangle are 8 cm, 15 cm
and x cm. As 15 is greater than 8, hence either x or 15
will be the largest side of this triangle. Consider two
cases:
Case I:
15 cm
C
B 8 cm
x
A
90°
Consider the right ABC ? above,
22
x 15 8 12.68 cm =- =
For all values of x < 12.68, the ABC ? will be obtuse.
But as the sum of two sides of triangle must be greater
than the third side, hence (x + 8) > 15 or x > 7.
Thus, the permissible values of x are 8, 9, 10, 11 and
12.
Case II:
15 cm
C
B 8 cm
x
A
90°
In the right ABC ? above, 
22
x15 8 17 =+ = .
For all values of x  > 17, ABC ? will be obtuse. But, as
the length of third side should be less than the sum of
other two sides, hence x < (15 + 8) or x < 23. The
permissible values of x are: 18, 19, 20, 21 and 22.
From Case I and II, x can take 10 values.
15. 5
A B
D
E
F
G
P
Q
H L
C
Let , the length of AH = ‘x’ cm
By symmetry of the figure given above, we can
conclude that APD ? and BQC ? will have the same
area.
APD ? 
 is 120° and line ‘L’ divides the square
ABCD in 2 equal halves, therefore
APH HPD 60 ?=? = °
In 
AH x
AHP : tan60 3 HP cm
HP 3
?= °=?=
Area of APD 2 area( AHP) ?=× ? 2
1x x
2x cm
2 33
=× × × =
Area of ABQCDP = area (ABCD) – 2 area ( APD) ?                         
()
2
2
2
2x 2 3 1
2x
4x
33
-
=- =
Required Ratio 
()
2
2
2x 2 3 1
3
23 1
2x
3
-
==-
Alternate method:
Concepts used:
A
B C
a
b
c
abc
sinA sinB sinC
?= =
Also, area of ? ABC =
1
2
ab Sin C
= 
1
2
bc Sin A  = 
1
2
ac Sin B
In the given figure
AB
C D
E
G
H L
F
P Q
a
a
x
x
x x
120 120
For APD, ? Let AP = PD = x cms
ax x
sin120 sin30 sin30
?= =
°°°
sin120 ?° = sin (90 + 30) = cos 30 = 
3
,
2
sin 30 = 
1
2
ax
1
3
2
2
?=
 
a
xcms
3
?=
Thus, area of APD ? is 
1
AP PD sin120
2
×× × °
1a a 3
22 33
=× × ×
 = 
2
2
a
cm
43
by symmetry, Area of APD ? = Area of BQC ? Area of ABQCDP
Thus, ratio of 
[Removing area inside square ABCD
Area of square ABCD 2 (Area of APD)
2 (Areaof APD)
-× ? =
×?
23 1 =-
16. 1 Number of terms in the given expansion is nothing but
the non-negative integral solutions of the equation
a + b + c = 20.
Total number of non-negative integral solutions
20 3 1 22
31 2
CC231
+-
-
===
Alternative Method:
{}
20 20
(a b c) (a b) c ++ = + +
  
20 20 0 20 19 1 20 0 20
01 20
C (a b) .C C (a b) .C ... C (a b) .C =+ + + + +
Number of  terms = 21 + 20 + 19 + ….. + 1 = 231
For questions 17 to 18:
Raju bets on the horses as follows:
Red – Rs.3000 , White – Rs.2000 and Black – Rs.1000 =
Total of
Rs.6000
He makes no profit no loss in the game. So the possible
ways of recovering his money (Rs.6000) is as follows:
Case (i): 3000 + 3(1000)
Case (ii): 2000 + 4(1000)
Case (iii): 3(2000) + 0
Case (a):  A breakup of 3000 + 3(1000) can be arrived at if
the Black horse finished at 2
nd
 and the Red horse at 3
rd
positions.
Then the White horse is either on the 4
th
 or 5
th
 position.
 1
st
 2
nd
 3
rd
 4
th
 5
th
 
I Grey/Spotted Black Red White Spotted/Grey 
II Grey/Spotted Black Red Spotted/Grey White 
Case (b): A breakup of 2000 + 4(1000) can be arrived at if
the Black horse finished at 1
st
 and the White horse at 3
rd
positions.
Then the Red horse is either on the 4
th
 or 5
th
 position.
 1
st
 2
nd
 3
rd
 4
th
 5
th
 
I Black Grey/Spotted White Red Spotted/Grey 
II Black Grey/Spotted White Spotted/Grey Red 
Case (c): A breakup of 3(2000) + 0 can be arrived at if the
White horse finished at 2
nd
 position.
Then the Red and Black horses must have finished at the 4
th
and 5
th
 positions, not necessary in that order.
 1
st
 2
nd
 3
rd
 4
th
 5
th
 
I Spotted/Grey White Grey/Spotted Red/Black Black/Red 
17. 4 None of the cases has three horses between White
and Red horses.
18. 3 If Grey came fourth, we consider cases (a) and (b).
All the options except (c) can hold true for these
cases. White horse can either be 2
nd
 or 5
th
 in the
race.
19. 4 Statement A: If the number of players at the entry
level is 83, we can get the following table.
Round Number of 
players 
Pair of 
players 
Byes Number of 
matches 
1 83 41 1 41 
2 41 + 1 = 42 21 0 21 
3 21 10 1 10 
4 10 + 1 = 11 5 1 5 
5 5 + 1 = 6 3 0 3 
6 3 1 1 1 
7 1 + 1 = 2 1 0 1 
Since we do not know the number of byes given to
the champion, we cannot ascertain the number of
matches played by him.
Hence, statement A alone is not sufficient.
Statement B: The champion received one bye, but
no information is given regarding the number of entrants
in the tournament.
Hence, statement B alone is not sufficient.
Combining statements A and B, we get that the
total number of matches played by the champion = 7 -
1 = 6
Hence, statements A and B both are required to
answer.
20. 4 Using statement A:
When n = 127, exactly one bye is given in round 1.
When = 96, exactly one bye is given in round 6.
As no unique value of n can be determined, hence,
statement A alone is not sufficient.
Using statement B:
As we do not know exactly how many bye 5 are
given in total, we cannot determine the value of n,
uniquely.
Combining statement A and B:
There is a unique value of n = 124, for which exactly
1 bye is given from the third round to the fourth round.