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# CAT Past Year Question Paper Solution - 2006 CAT Notes | EduRev

## CAT Mock Test Series 2020

Created by: Bakliwal Institute

## CAT : CAT Past Year Question Paper Solution - 2006 CAT Notes | EduRev

``` Page 1

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For questions 1 to 5:
From statement one, team would include exactly one among P,
R, S
? P (or) R (or) S.
From statement two, team would include either M, or Q
? M but not Q
(or) Q but not M
From statement three, if a team includes K, it will include L or
vice versa.
? K, L always accompany each other.
From statement four, if one of S, U, W is included, then the
other two also have to be included.
? S, U, W are always together.
From statement five, L and N cannot be included together
? L, N are never together.
From statement six, L and U cannot be included together.
? L, U are never together.
1. 1 From statements one and two;
one of P, R, S and
one of M, Q are to be selected. We require one more
member.
But from statement three; (K, L) are always together.
Hence 'L'  cannot be included in a team of 3 members.
2. 3 Again, from statement one;
one of P, R, S has to be selected.
To make a team of '5'
'S' will be chosen (which leaves out P and R)
? If 'S' is chosen 'U' and ‘W’ have to be chosen
(statement four)
? If 'U' is chosen 'L' cannot be chosen (statement
five)
? K cannot be chosen (statement three)
And from statement two; one of M (or) Q has to be
chosen.
3. 4 From statements one and two
Two members are to be selected.
Of the remaining seven;
To maximize the size of the team.
We would chose S,
? U and W are included in the team (statement four)
We cannot include K (or) L because we would then
have to leave out N and U (from statements five and
six)
4. 5 If 'K' is included, 'L' has to be included (statement (3))
If 'L' is chosen, neither N nor U can be chosen
(statements (5) and (6))
? S, W are also not included because S, U, W have to
be always together. (Statement (4))
Hence one of P (or) R would be selected (statement
(1)) and one of M (or) Q would be selected (statement
(2))
(K, L) and two of the above five have to be included.
5. 5 If a team includes N, it cannot include 'L',
and therefore, not even 'K'. (from statement five and
three)
According to statement (1), one of P or R or S has to
be included.
According to statement (2), one of M or Q has to be
selected.
So the following cases are possible
P Q N,
R Q N
P M N,
R M N
If 'S' is selected, then S U W M N and S U W Q N  are the
only possible cases.
Hence, in all 4 + 2 = 6 teams can be constituted.
For questions 6 to 10:
6. 3 Let Dipan get x marks in paper II.
Dipan's average in PCB group = 98
Maths group = 95
S.S. group = 95.5
Vernacular group = 95
English group =
+ ??
??
??
96 x
2
Sum of all = 96 × 5
So 95.5 + 96 × 3 +
+
x
48
2
= 96 × 5
?= × - -
x
96 2 95.5 48
2
() =- =× = x 2 96.5 48 2 48.5 97
So (3) is the correct option.
7. 1 The only boy getting 95 in atleast one of the subjects
of the group among all the groups is Dipan.
So (1) is the correct option.
8. 1 A group score of 100 in Social Science would have
increased the scores as follows:
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EU + RC 26 to 50 25
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T otal 75
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For questions 1 to 5:
From statement one, team would include exactly one among P,
R, S
? P (or) R (or) S.
From statement two, team would include either M, or Q
? M but not Q
(or) Q but not M
From statement three, if a team includes K, it will include L or
vice versa.
? K, L always accompany each other.
From statement four, if one of S, U, W is included, then the
other two also have to be included.
? S, U, W are always together.
From statement five, L and N cannot be included together
? L, N are never together.
From statement six, L and U cannot be included together.
? L, U are never together.
1. 1 From statements one and two;
one of P, R, S and
one of M, Q are to be selected. We require one more
member.
But from statement three; (K, L) are always together.
Hence 'L'  cannot be included in a team of 3 members.
2. 3 Again, from statement one;
one of P, R, S has to be selected.
To make a team of '5'
'S' will be chosen (which leaves out P and R)
? If 'S' is chosen 'U' and ‘W’ have to be chosen
(statement four)
? If 'U' is chosen 'L' cannot be chosen (statement
five)
? K cannot be chosen (statement three)
And from statement two; one of M (or) Q has to be
chosen.
3. 4 From statements one and two
Two members are to be selected.
Of the remaining seven;
To maximize the size of the team.
We would chose S,
? U and W are included in the team (statement four)
We cannot include K (or) L because we would then
have to leave out N and U (from statements five and
six)
4. 5 If 'K' is included, 'L' has to be included (statement (3))
If 'L' is chosen, neither N nor U can be chosen
(statements (5) and (6))
? S, W are also not included because S, U, W have to
be always together. (Statement (4))
Hence one of P (or) R would be selected (statement
(1)) and one of M (or) Q would be selected (statement
(2))
(K, L) and two of the above five have to be included.
5. 5 If a team includes N, it cannot include 'L',
and therefore, not even 'K'. (from statement five and
three)
According to statement (1), one of P or R or S has to
be included.
According to statement (2), one of M or Q has to be
selected.
So the following cases are possible
P Q N,
R Q N
P M N,
R M N
If 'S' is selected, then S U W M N and S U W Q N  are the
only possible cases.
Hence, in all 4 + 2 = 6 teams can be constituted.
For questions 6 to 10:
6. 3 Let Dipan get x marks in paper II.
Dipan's average in PCB group = 98
Maths group = 95
S.S. group = 95.5
Vernacular group = 95
English group =
+ ??
??
??
96 x
2
Sum of all = 96 × 5
So 95.5 + 96 × 3 +
+
x
48
2
= 96 × 5
?= × - -
x
96 2 95.5 48
2
() =- =× = x 2 96.5 48 2 48.5 97
So (3) is the correct option.
7. 1 The only boy getting 95 in atleast one of the subjects
of the group among all the groups is Dipan.
So (1) is the correct option.
8. 1 A group score of 100 in Social Science would have
increased the scores as follows:

Score
Increase
Group
Score
Final
Score
Increase
Final
group
Score
Pritam 22 11
=
11
2.2
5

96.1
Joseph 9 4.5
=
4.5
.9
5

95.9
Tirna 21 10.5
=
10.5
2.1
5

95.8
Agni 9 4.5
=
4.5
.9
5

95.2
So the order is Pritam > Joseph > Trina > Agni.
So option (1) is the correct choice.
9. 4 The student having atleast 95 in every group is Dipan,
so the answer is Dipan, option (4).
10. 5 Let us increase the score in one of the subjects of the
following candidates
Least Scores Contributio
n in net
Score
Final Score
Ram 94 in group of 2 3 in 5
groups
96.1 + .6 =
96.7
Agni 82 in group of 2 9 in 5
groups
94.3 + 1.8 =
96.1
Pritam 83 in group of 2 8.5 in 5
groups
93.9 + 1.7 =
95.6
Ayesha 93 in group of 2 3.5 in 5
groups
96.2 + .7 =
96.9
Dipan 95 in group of 1 5 in 5
groups
96 + 1 =
97.0
So, Dipan will end with a highest total.
So the answer is option (5).
For questions 11 to 15:
As only Paul Erdös was having an Erdös number of zero, so
the minimum Erdös  number among A, B, C, D, E, F , G, H should
be 1 or greater than one. At the end of the third day, F co-
authored a paper with A and C. F had the minimum Erdös
number among the 8 people. So if F's Erdös number is y, then
A and C's Erdös number should change to (y + 1) after third
day. As A and C decreased the average by maximum possible
extent, it means C had the second-height Erdös number among
all eight, as A had an Erdös number of infinity. Suppose Erdös
numbers of  A, B, C, D, E, F, G, H are y + 1, b, y + 1, c, d, e, y,
g, h  respectively at the end of third day.
? (y + 1 + b + y + 1 + c + d + e + y + g + h) = 24 = (3 × 8)
? 3y + 2 + b + d + e + g + h = 24
When E co-authored with F , the average Erdös number reduced
again, it means, E's Erdös number was not the same with A &
C initially. As at the end of third day, 5 people had same Erdös
number, they should be A, C and any 3 out of B, D, G, H.
Suppose those 3 people are B, D, G. Then
(3y + 2 + y + 1 + y + 1 + y + 1 + e + h) = 24
? 6y + h + e = 19 …(i)
On the fifth day, E co-authored a paper with F and hence,
Erdös number of E changed to (y + 1). Also the average
decreased by 0.5 which means the total decreased by 4.
Hence, e - (y + 1) = 4
? e – y = 5
Putting the value of e in equation (i), we get
6y + h + (5 + y) = 19
? 7y + h = 14
Only possible value of y = 1 as h cannot be zero.
So after 3rd round Erdös number of A, C, E, F were 2, 2, 6, 1
respectively.
11. 4 Only A, C, E changed their Erdös number, rest 5 did not
change their Erdös number.
12. 2 At the end of conference 6 people including E were
having an Erdös number of 2 and F was having 1 as
Erdös number. So 8th person was having an Erdös
number of [20 – (2 × 6 + 1)] = 7
13. 2 At the end of 3rd round, 5 people were having same
Erdös number. A and C changed their Erdös number
after coauthoring with F. So, the other 3 will have
same Erdös number in the beginning.
14. 2 2
15. 3 After co-authoring with F, E was having Erdös number
of 2, which was 4 less than initial Erdös number of E.
So answer is 2 + 4 = 6.
For questions 16 to 20:
The MCS share price at the beginning of first day is Rs.100
and at the close of day 5 is Rs.110.
The following cases of the closing prices can be derived.

At the end of Day 1Day 2Day 3Day 4Day 5
1 90 80 90 100 110
2 90 100 90 100 110
3 90 100 110 120 110
4 90 100 110 100 110
5 110 100 90 100 110
6 110 100 110 100 110
7 110 120 110 100 110
8 110 120 110 120 110
9 110 120 130 120 110
10 110 120 110 100 110
16. 3 As Chetan sold 10 shares on three consecutive days,
therefore, of the five days, there must be an increase
for three of the five days and a decrease for the
remaining two days. It is given that Michael sold 10
shares only once. Hence, the price is more than 110
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41 3 42 1 43 4 44 2 45 4 46 3 47 5 48 4 49 2 50 1
51 1 52 2 53 1 54 2 55 5 56 2 57 2 58 4 59 2 60 4
61 2 62 4 63 1 64 4 65 4 66 3 67 2 68 5 69 1 70 5
71 5 72 5 73 2 74 5 75 3



LRDI  1 to 25 25
EU + RC 26 to 50 25
QA 51 to 75 25
T otal 75
T otal
questions
T otal
attempted
T otal
correct
T otal
wrong
Net
Score
Time
T aken
Question
number
	
For questions 1 to 5:
From statement one, team would include exactly one among P,
R, S
? P (or) R (or) S.
From statement two, team would include either M, or Q
? M but not Q
(or) Q but not M
From statement three, if a team includes K, it will include L or
vice versa.
? K, L always accompany each other.
From statement four, if one of S, U, W is included, then the
other two also have to be included.
? S, U, W are always together.
From statement five, L and N cannot be included together
? L, N are never together.
From statement six, L and U cannot be included together.
? L, U are never together.
1. 1 From statements one and two;
one of P, R, S and
one of M, Q are to be selected. We require one more
member.
But from statement three; (K, L) are always together.
Hence 'L'  cannot be included in a team of 3 members.
2. 3 Again, from statement one;
one of P, R, S has to be selected.
To make a team of '5'
'S' will be chosen (which leaves out P and R)
? If 'S' is chosen 'U' and ‘W’ have to be chosen
(statement four)
? If 'U' is chosen 'L' cannot be chosen (statement
five)
? K cannot be chosen (statement three)
And from statement two; one of M (or) Q has to be
chosen.
3. 4 From statements one and two
Two members are to be selected.
Of the remaining seven;
To maximize the size of the team.
We would chose S,
? U and W are included in the team (statement four)
We cannot include K (or) L because we would then
have to leave out N and U (from statements five and
six)
4. 5 If 'K' is included, 'L' has to be included (statement (3))
If 'L' is chosen, neither N nor U can be chosen
(statements (5) and (6))
? S, W are also not included because S, U, W have to
be always together. (Statement (4))
Hence one of P (or) R would be selected (statement
(1)) and one of M (or) Q would be selected (statement
(2))
(K, L) and two of the above five have to be included.
5. 5 If a team includes N, it cannot include 'L',
and therefore, not even 'K'. (from statement five and
three)
According to statement (1), one of P or R or S has to
be included.
According to statement (2), one of M or Q has to be
selected.
So the following cases are possible
P Q N,
R Q N
P M N,
R M N
If 'S' is selected, then S U W M N and S U W Q N  are the
only possible cases.
Hence, in all 4 + 2 = 6 teams can be constituted.
For questions 6 to 10:
6. 3 Let Dipan get x marks in paper II.
Dipan's average in PCB group = 98
Maths group = 95
S.S. group = 95.5
Vernacular group = 95
English group =
+ ??
??
??
96 x
2
Sum of all = 96 × 5
So 95.5 + 96 × 3 +
+
x
48
2
= 96 × 5
?= × - -
x
96 2 95.5 48
2
() =- =× = x 2 96.5 48 2 48.5 97
So (3) is the correct option.
7. 1 The only boy getting 95 in atleast one of the subjects
of the group among all the groups is Dipan.
So (1) is the correct option.
8. 1 A group score of 100 in Social Science would have
increased the scores as follows:

Score
Increase
Group
Score
Final
Score
Increase
Final
group
Score
Pritam 22 11
=
11
2.2
5

96.1
Joseph 9 4.5
=
4.5
.9
5

95.9
Tirna 21 10.5
=
10.5
2.1
5

95.8
Agni 9 4.5
=
4.5
.9
5

95.2
So the order is Pritam > Joseph > Trina > Agni.
So option (1) is the correct choice.
9. 4 The student having atleast 95 in every group is Dipan,
so the answer is Dipan, option (4).
10. 5 Let us increase the score in one of the subjects of the
following candidates
Least Scores Contributio
n in net
Score
Final Score
Ram 94 in group of 2 3 in 5
groups
96.1 + .6 =
96.7
Agni 82 in group of 2 9 in 5
groups
94.3 + 1.8 =
96.1
Pritam 83 in group of 2 8.5 in 5
groups
93.9 + 1.7 =
95.6
Ayesha 93 in group of 2 3.5 in 5
groups
96.2 + .7 =
96.9
Dipan 95 in group of 1 5 in 5
groups
96 + 1 =
97.0
So, Dipan will end with a highest total.
So the answer is option (5).
For questions 11 to 15:
As only Paul Erdös was having an Erdös number of zero, so
the minimum Erdös  number among A, B, C, D, E, F , G, H should
be 1 or greater than one. At the end of the third day, F co-
authored a paper with A and C. F had the minimum Erdös
number among the 8 people. So if F's Erdös number is y, then
A and C's Erdös number should change to (y + 1) after third
day. As A and C decreased the average by maximum possible
extent, it means C had the second-height Erdös number among
all eight, as A had an Erdös number of infinity. Suppose Erdös
numbers of  A, B, C, D, E, F, G, H are y + 1, b, y + 1, c, d, e, y,
g, h  respectively at the end of third day.
? (y + 1 + b + y + 1 + c + d + e + y + g + h) = 24 = (3 × 8)
? 3y + 2 + b + d + e + g + h = 24
When E co-authored with F , the average Erdös number reduced
again, it means, E's Erdös number was not the same with A &
C initially. As at the end of third day, 5 people had same Erdös
number, they should be A, C and any 3 out of B, D, G, H.
Suppose those 3 people are B, D, G. Then
(3y + 2 + y + 1 + y + 1 + y + 1 + e + h) = 24
? 6y + h + e = 19 …(i)
On the fifth day, E co-authored a paper with F and hence,
Erdös number of E changed to (y + 1). Also the average
decreased by 0.5 which means the total decreased by 4.
Hence, e - (y + 1) = 4
? e – y = 5
Putting the value of e in equation (i), we get
6y + h + (5 + y) = 19
? 7y + h = 14
Only possible value of y = 1 as h cannot be zero.
So after 3rd round Erdös number of A, C, E, F were 2, 2, 6, 1
respectively.
11. 4 Only A, C, E changed their Erdös number, rest 5 did not
change their Erdös number.
12. 2 At the end of conference 6 people including E were
having an Erdös number of 2 and F was having 1 as
Erdös number. So 8th person was having an Erdös
number of [20 – (2 × 6 + 1)] = 7
13. 2 At the end of 3rd round, 5 people were having same
Erdös number. A and C changed their Erdös number
after coauthoring with F. So, the other 3 will have
same Erdös number in the beginning.
14. 2 2
15. 3 After co-authoring with F, E was having Erdös number
of 2, which was 4 less than initial Erdös number of E.
So answer is 2 + 4 = 6.
For questions 16 to 20:
The MCS share price at the beginning of first day is Rs.100
and at the close of day 5 is Rs.110.
The following cases of the closing prices can be derived.

At the end of Day 1Day 2Day 3Day 4Day 5
1 90 80 90 100 110
2 90 100 90 100 110
3 90 100 110 120 110
4 90 100 110 100 110
5 110 100 90 100 110
6 110 100 110 100 110
7 110 120 110 100 110
8 110 120 110 120 110
9 110 120 130 120 110
10 110 120 110 100 110
16. 3 As Chetan sold 10 shares on three consecutive days,
therefore, of the five days, there must be an increase
for three of the five days and a decrease for the
remaining two days. It is given that Michael sold 10
shares only once. Hence, the price is more than 110
for only one day and on all the remaining days, it
cannot exceed 110. The only satisfying case is (3).
Hence, the price at the end of Day 3 is Rs.110.
17. 2 The satisfying cases are (1), (2), (4), (5), (6). Hence,
the price at the end of Day 4 is Rs.100.
18. 1 Let Chetan and Michael start with x number of shares
initially.
From case (1), we get that the number of shares with
Michael = x + 10
and number of shares with Chetan = x + 10 + 10 – 10
– 10 – 10 = x – 10.
So Michael has 20 more shares than Chetan. This is
the only satisfying case.
Hence, the share price at the end of Day 3 is Rs.90.
19. 5 Consider cases (3) and (7). Only these two satisfies
the condition that Michael had Rs.100 less than Chetan
at the end of day 5.
For case (3),
Number of shares with Chetan = x + 10 – 10 – 10 – 10
+ 10 = x – 10
And with Michael = x – 10
For case (7),
Number of shares with Chetan = x – 10 – 10 + 10 + 10
– 10 = x – 10
And with Michael = x – 10
In either case, number of shares with Michael and
Chetan are the same.
20. 4 To maximise the amount gathered by both of them, we
need to look into those cases wherein we have
maximum number of 110 excess figures. It is only then
that Michael and Chetan both will make money. So we
check for case (9).
For case (9),
Extra cash with Chetan by the end of day 5 = 1100 +
1200 + 1300 – 1200 – 1100 = Rs.1300
And that with Michael = 1200 + 1300 + 1200 = Rs.3700
Total extra cash with both of them = 1300 + 3700 =
Rs.5000
For questions 21 to 25:
In this set, the fuel cost for each of the path is given. In addition,
there are four toll collection junctions.
21. 5 No traffic flows on the street from D to T.
Now, we have fuel cost on different paths as
SAT   : 9 + 5 = Rs. 14 + toll at junction A
SBAT : 2 + 2 +  5 = Rs. 9 + toll at junction B and A
SBCT : 2 3 + 2 = Rs. 7 + toll at junction B and C
SDCT : 7 + 1 + 2 = Rs. 10 + toll at junction D and C
Now, checking the options we find that toll at junction
A is 0 or 1.
When toll is 0, fuel cost on SAT = 14 + 0  = Rs. 14
When toll is 1, fuel cost on SAT = 14 + 1 = Rs. 15
The fuel cost on all the paths should be equal.
Options (1), (2), (3) can be ruled out as in all these
options toll at C and D add up to more than Rs. 5. As
fuel cost on SDCT is Rs. 10 without toll, so with toll it
cannot exceed Rs. 15 (i.e. toll of path SAT).
Option (4) is ruled out as in this option SAT comes out
to be Rs. 14 and SDCT sums up to Rs.15.
So correct answer is option (5).
22. 2 & 3
Note: Both the options b and c are correct.
Available routes are:
SAT ?  Rs. 14
SBAT ? Rs. 9
SDCT ? Rs. 10
SDT ? Rs. 13
Now, fuel cost of SAT - fuel of SDT = 14 – 13 = Rs. 1.
Hence toll at junction D should be 1 more than the toll at
A. So option (a), (d) and (e) are ruled out.
Now, fuel cost of SAT - fuel cost of SBAT = 14 – 9 =
Rs. 5. So toll at junction B should be Rs. 5. So answer
could be either (2) or option (3).
23. 1 Available paths considering no toll are
SAT ?  Rs. 14
SBCT ?  Rs. 7
SBAT ?  Rs. 9
SDCT ?  Rs. 10
SDT ?  Rs. 13
It is very likely that option (4) is selected. But, if all the
five routes have the same cost, then there will be an
equal flow on all the five routes i.e., 20% on each
route. But, then the percentage of traffic. on
S – A ? 20%
S – B ? 40% (As there are two routes involving S – B.)
S – D ? 40%  (As there are two routes involving S –
D.)
But, it is given that traffic on S – A = traffic on S – B =
traffic on S – D.
24. 4 Available routes are
SAT ?  Rs. 14
SBAT ?  Rs. 9
SBCT ?  Rs. 7
SDCT ?  Rs. 10
SDT ?  Rs. 13
Fuel cost on path SAT - fuel cost on path SDT = 14 - 13
= Rs. 1.
So the toll at junction D should be 1 more than toll at
junction A. So option a and c are ruled out.
Fuel cost on path SAT - fuel cost on path SBCT = 14 -
7 = Rs. 7.
So sum of toll at junction B and C should be 7 more
than the toll at A. Hence, only option (d) matches.
25. 3 We have to find a path on which minimum cost is
incurred and such that total traffic through B does not
exceed 70%.
So option (5) is ruled out because we can send all the
traffic through SDCT or SDT and meet all conditions.
Option (1) is also ruled out as in that case all traffic will
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EU + RC 26 to 50 25
QA 51 to 75 25
T otal 75
T otal
questions
T otal
attempted
T otal
correct
T otal
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Net
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Time
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For questions 1 to 5:
From statement one, team would include exactly one among P,
R, S
? P (or) R (or) S.
From statement two, team would include either M, or Q
? M but not Q
(or) Q but not M
From statement three, if a team includes K, it will include L or
vice versa.
? K, L always accompany each other.
From statement four, if one of S, U, W is included, then the
other two also have to be included.
? S, U, W are always together.
From statement five, L and N cannot be included together
? L, N are never together.
From statement six, L and U cannot be included together.
? L, U are never together.
1. 1 From statements one and two;
one of P, R, S and
one of M, Q are to be selected. We require one more
member.
But from statement three; (K, L) are always together.
Hence 'L'  cannot be included in a team of 3 members.
2. 3 Again, from statement one;
one of P, R, S has to be selected.
To make a team of '5'
'S' will be chosen (which leaves out P and R)
? If 'S' is chosen 'U' and ‘W’ have to be chosen
(statement four)
? If 'U' is chosen 'L' cannot be chosen (statement
five)
? K cannot be chosen (statement three)
And from statement two; one of M (or) Q has to be
chosen.
3. 4 From statements one and two
Two members are to be selected.
Of the remaining seven;
To maximize the size of the team.
We would chose S,
? U and W are included in the team (statement four)
We cannot include K (or) L because we would then
have to leave out N and U (from statements five and
six)
4. 5 If 'K' is included, 'L' has to be included (statement (3))
If 'L' is chosen, neither N nor U can be chosen
(statements (5) and (6))
? S, W are also not included because S, U, W have to
be always together. (Statement (4))
Hence one of P (or) R would be selected (statement
(1)) and one of M (or) Q would be selected (statement
(2))
(K, L) and two of the above five have to be included.
5. 5 If a team includes N, it cannot include 'L',
and therefore, not even 'K'. (from statement five and
three)
According to statement (1), one of P or R or S has to
be included.
According to statement (2), one of M or Q has to be
selected.
So the following cases are possible
P Q N,
R Q N
P M N,
R M N
If 'S' is selected, then S U W M N and S U W Q N  are the
only possible cases.
Hence, in all 4 + 2 = 6 teams can be constituted.
For questions 6 to 10:
6. 3 Let Dipan get x marks in paper II.
Dipan's average in PCB group = 98
Maths group = 95
S.S. group = 95.5
Vernacular group = 95
English group =
+ ??
??
??
96 x
2
Sum of all = 96 × 5
So 95.5 + 96 × 3 +
+
x
48
2
= 96 × 5
?= × - -
x
96 2 95.5 48
2
() =- =× = x 2 96.5 48 2 48.5 97
So (3) is the correct option.
7. 1 The only boy getting 95 in atleast one of the subjects
of the group among all the groups is Dipan.
So (1) is the correct option.
8. 1 A group score of 100 in Social Science would have
increased the scores as follows:

Score
Increase
Group
Score
Final
Score
Increase
Final
group
Score
Pritam 22 11
=
11
2.2
5

96.1
Joseph 9 4.5
=
4.5
.9
5

95.9
Tirna 21 10.5
=
10.5
2.1
5

95.8
Agni 9 4.5
=
4.5
.9
5

95.2
So the order is Pritam > Joseph > Trina > Agni.
So option (1) is the correct choice.
9. 4 The student having atleast 95 in every group is Dipan,
so the answer is Dipan, option (4).
10. 5 Let us increase the score in one of the subjects of the
following candidates
Least Scores Contributio
n in net
Score
Final Score
Ram 94 in group of 2 3 in 5
groups
96.1 + .6 =
96.7
Agni 82 in group of 2 9 in 5
groups
94.3 + 1.8 =
96.1
Pritam 83 in group of 2 8.5 in 5
groups
93.9 + 1.7 =
95.6
Ayesha 93 in group of 2 3.5 in 5
groups
96.2 + .7 =
96.9
Dipan 95 in group of 1 5 in 5
groups
96 + 1 =
97.0
So, Dipan will end with a highest total.
So the answer is option (5).
For questions 11 to 15:
As only Paul Erdös was having an Erdös number of zero, so
the minimum Erdös  number among A, B, C, D, E, F , G, H should
be 1 or greater than one. At the end of the third day, F co-
authored a paper with A and C. F had the minimum Erdös
number among the 8 people. So if F's Erdös number is y, then
A and C's Erdös number should change to (y + 1) after third
day. As A and C decreased the average by maximum possible
extent, it means C had the second-height Erdös number among
all eight, as A had an Erdös number of infinity. Suppose Erdös
numbers of  A, B, C, D, E, F, G, H are y + 1, b, y + 1, c, d, e, y,
g, h  respectively at the end of third day.
? (y + 1 + b + y + 1 + c + d + e + y + g + h) = 24 = (3 × 8)
? 3y + 2 + b + d + e + g + h = 24
When E co-authored with F , the average Erdös number reduced
again, it means, E's Erdös number was not the same with A &
C initially. As at the end of third day, 5 people had same Erdös
number, they should be A, C and any 3 out of B, D, G, H.
Suppose those 3 people are B, D, G. Then
(3y + 2 + y + 1 + y + 1 + y + 1 + e + h) = 24
? 6y + h + e = 19 …(i)
On the fifth day, E co-authored a paper with F and hence,
Erdös number of E changed to (y + 1). Also the average
decreased by 0.5 which means the total decreased by 4.
Hence, e - (y + 1) = 4
? e – y = 5
Putting the value of e in equation (i), we get
6y + h + (5 + y) = 19
? 7y + h = 14
Only possible value of y = 1 as h cannot be zero.
So after 3rd round Erdös number of A, C, E, F were 2, 2, 6, 1
respectively.
11. 4 Only A, C, E changed their Erdös number, rest 5 did not
change their Erdös number.
12. 2 At the end of conference 6 people including E were
having an Erdös number of 2 and F was having 1 as
Erdös number. So 8th person was having an Erdös
number of [20 – (2 × 6 + 1)] = 7
13. 2 At the end of 3rd round, 5 people were having same
Erdös number. A and C changed their Erdös number
after coauthoring with F. So, the other 3 will have
same Erdös number in the beginning.
14. 2 2
15. 3 After co-authoring with F, E was having Erdös number
of 2, which was 4 less than initial Erdös number of E.
So answer is 2 + 4 = 6.
For questions 16 to 20:
The MCS share price at the beginning of first day is Rs.100
and at the close of day 5 is Rs.110.
The following cases of the closing prices can be derived.

At the end of Day 1Day 2Day 3Day 4Day 5
1 90 80 90 100 110
2 90 100 90 100 110
3 90 100 110 120 110
4 90 100 110 100 110
5 110 100 90 100 110
6 110 100 110 100 110
7 110 120 110 100 110
8 110 120 110 120 110
9 110 120 130 120 110
10 110 120 110 100 110
16. 3 As Chetan sold 10 shares on three consecutive days,
therefore, of the five days, there must be an increase
for three of the five days and a decrease for the
remaining two days. It is given that Michael sold 10
shares only once. Hence, the price is more than 110
for only one day and on all the remaining days, it
cannot exceed 110. The only satisfying case is (3).
Hence, the price at the end of Day 3 is Rs.110.
17. 2 The satisfying cases are (1), (2), (4), (5), (6). Hence,
the price at the end of Day 4 is Rs.100.
18. 1 Let Chetan and Michael start with x number of shares
initially.
From case (1), we get that the number of shares with
Michael = x + 10
and number of shares with Chetan = x + 10 + 10 – 10
– 10 – 10 = x – 10.
So Michael has 20 more shares than Chetan. This is
the only satisfying case.
Hence, the share price at the end of Day 3 is Rs.90.
19. 5 Consider cases (3) and (7). Only these two satisfies
the condition that Michael had Rs.100 less than Chetan
at the end of day 5.
For case (3),
Number of shares with Chetan = x + 10 – 10 – 10 – 10
+ 10 = x – 10
And with Michael = x – 10
For case (7),
Number of shares with Chetan = x – 10 – 10 + 10 + 10
– 10 = x – 10
And with Michael = x – 10
In either case, number of shares with Michael and
Chetan are the same.
20. 4 To maximise the amount gathered by both of them, we
need to look into those cases wherein we have
maximum number of 110 excess figures. It is only then
that Michael and Chetan both will make money. So we
check for case (9).
For case (9),
Extra cash with Chetan by the end of day 5 = 1100 +
1200 + 1300 – 1200 – 1100 = Rs.1300
And that with Michael = 1200 + 1300 + 1200 = Rs.3700
Total extra cash with both of them = 1300 + 3700 =
Rs.5000
For questions 21 to 25:
In this set, the fuel cost for each of the path is given. In addition,
there are four toll collection junctions.
21. 5 No traffic flows on the street from D to T.
Now, we have fuel cost on different paths as
SAT   : 9 + 5 = Rs. 14 + toll at junction A
SBAT : 2 + 2 +  5 = Rs. 9 + toll at junction B and A
SBCT : 2 3 + 2 = Rs. 7 + toll at junction B and C
SDCT : 7 + 1 + 2 = Rs. 10 + toll at junction D and C
Now, checking the options we find that toll at junction
A is 0 or 1.
When toll is 0, fuel cost on SAT = 14 + 0  = Rs. 14
When toll is 1, fuel cost on SAT = 14 + 1 = Rs. 15
The fuel cost on all the paths should be equal.
Options (1), (2), (3) can be ruled out as in all these
options toll at C and D add up to more than Rs. 5. As
fuel cost on SDCT is Rs. 10 without toll, so with toll it
cannot exceed Rs. 15 (i.e. toll of path SAT).
Option (4) is ruled out as in this option SAT comes out
to be Rs. 14 and SDCT sums up to Rs.15.
So correct answer is option (5).
22. 2 & 3
Note: Both the options b and c are correct.
Available routes are:
SAT ?  Rs. 14
SBAT ? Rs. 9
SDCT ? Rs. 10
SDT ? Rs. 13
Now, fuel cost of SAT - fuel of SDT = 14 – 13 = Rs. 1.
Hence toll at junction D should be 1 more than the toll at
A. So option (a), (d) and (e) are ruled out.
Now, fuel cost of SAT - fuel cost of SBAT = 14 – 9 =
Rs. 5. So toll at junction B should be Rs. 5. So answer
could be either (2) or option (3).
23. 1 Available paths considering no toll are
SAT ?  Rs. 14
SBCT ?  Rs. 7
SBAT ?  Rs. 9
SDCT ?  Rs. 10
SDT ?  Rs. 13
It is very likely that option (4) is selected. But, if all the
five routes have the same cost, then there will be an
equal flow on all the five routes i.e., 20% on each
route. But, then the percentage of traffic. on
S – A ? 20%
S – B ? 40% (As there are two routes involving S – B.)
S – D ? 40%  (As there are two routes involving S –
D.)
But, it is given that traffic on S – A = traffic on S – B =
traffic on S – D.
24. 4 Available routes are
SAT ?  Rs. 14
SBAT ?  Rs. 9
SBCT ?  Rs. 7
SDCT ?  Rs. 10
SDT ?  Rs. 13
Fuel cost on path SAT - fuel cost on path SDT = 14 - 13
= Rs. 1.
So the toll at junction D should be 1 more than toll at
junction A. So option a and c are ruled out.
Fuel cost on path SAT - fuel cost on path SBCT = 14 -
7 = Rs. 7.
So sum of toll at junction B and C should be 7 more
than the toll at A. Hence, only option (d) matches.
25. 3 We have to find a path on which minimum cost is
incurred and such that total traffic through B does not
exceed 70%.
So option (5) is ruled out because we can send all the
traffic through SDCT or SDT and meet all conditions.
Option (1) is also ruled out as in that case all traffic will
be passed through SBCT [not possible as traffic at B
can't be more than 70%]
Option (2) is also ruled out as it is possible only when
toll at junction C is 2. In that case also all traffic will
pass through B.
Option (3) can be the answer, when toll at junction B
is 4 and toll at junction C is 0. Then SDCT will have toll
equal to Rs. 10.
As Rs. 10 is less than Rs. 13, so option (4) is also
ruled out.
Hence, option (3) is the correct choice.
26. 5 The paragraph stresses on the relationships between
the factories, dealers and the consumers. Every entity
has certain short-term expectations from each other.
This makes these relationships strenuous. This strain
leads to feelings of mistrust and lack of commitment.
So the longer this continues, the more the chances of
everyone succumbing to this vicious trap and they
would soon realize that they have sacrificed long-
term stability and gain for short-term benefits. Hence
Option (5). Option (4) is too specific to industry (at the
cost of the other players – dealers and customers),
option (2) suffers from the same short-comings
together with throwing the technical (unexplained)
jargon ‘supply chain’ to us. Option (1) takes into account
only 2 players and repeats what is stated in the
deals” in the term ‘Deal making’; option (3) seems close
but can be eliminated as the word ‘adversary’ is too
strong. The passage implies that everyone tries to
maximize his benefits, not that they ‘oppose’ one
another.
27. 1 The passage heads towards describing the functions
that bad / good maps (and therefore theories) serve.
Just as a ‘Bad theory’ does not help us understand a
problem, a ‘good theory’ is invaluable to us, though it
may be simplified. ‘Simplified’ here implies that less
valuable information is left out.  According to this logic,
option (2), (3), (4), get eliminated. Option (5) is close
but  more negative in tone than required. The word
‘limitation’ here indicates a short coming whereas the
passage implies that it is a simplification as it would
not be of practical use otherwise.
28. 2 Going with the direction of the passage, the last line is
stating ‘now all players “profess” to seek only peace’.
Profess means to mask or to pretend. Thus option (2)
which talks about the veil being lifted is the most logical
statement that completes the passage. More so this
also follows from the source of the text.
29. 3 The answer is very direct. With every statement of
his, the author seeks to show how foolish those people
are who call his advice ‘rules’. After his first statement
he has posed the rhetorical question “Call that a rule?”
The same should follow after his second “scarcely a
rule!”
30. 4 In the first part of the passage, the author seeks to
explain why one who is young would exploit an
entrepreneurial opportunity. Thus, in the second part
of the passage once the “however” is established,
evidence will seek to show how older people will be
reluctant to exploit entrepreneurial opportunity. Option
(2) seems correct but it only gives a general statement
that with age, people become reluctant to new ideas.
Between option (2) and (4), option (4) goes in
continuation with the text as it states that at a mature
age, people are unwilling to utilize entrepreneurial
opportunities. So option (4) is correct.
31. 2 According to the passage, "A critical attitude needs
for its raw material, as it were, theories or beliefs
which are held more or less dogmatically". Therefore,
our critical attitude is the tool by which we shape our
dogmatic beliefs. Thus, the relationship of dogmatic
beliefs and critical attitude is equivalent to that of a
chisel and that of a marble stone.
32. 1 Option (3), (4) and (5) are ruled out because they are
not supported by the passage. (negative, neutral,
inferior) - Option (1) and (2) are close but (1) is better
because the question is about the role of dogmatic
behaviour with respect to the development of science.
In the third paragraph, 8
th
line, it is mentioned that
dogmatic attitude is pseudo/pre-scientific attitude.
Science needs dogmatic beliefs for their critical
revision. Beginning of fourth paragraph states that
science begins with myths and criticism of myths.
Thus, dogmatic behavior is required to develop science
because the former serves as the base on which
33.  4 Refer to the last sentence of the second paragraph. It
is clear from the context (especially from the words -
'experience', 'maturity') that time has a direct effect on
the evolution of thinking. Option (4) is the only option
which takes into account the element of time (the word
- 'stages').
34. 5 Option (5) is correct because this statement suggests
that critical attitude is a process of questioning which
leads to tentative hypothesis. A critical attitude by itself
is not opposed to conviction, but it tries to modify the
conviction according to reason.
35. 3 Refer to the third last paragraph of the passage;
dogmatic attitude is pseudo-scientific because its aim
is only to verify its laws and schemata even if it has to
neglect the refutations. Whereas critical attitude is
flexible enough to change, refute or falsify its tenets
and therefore has a questioning attitude.
36. 3 We refer to the tenth line of the third paragraph. Here
brutalities but it also had positive consequences like
rapid industrialization. Hence option (3) is the best
```
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