CAT Past Year Question Paper Solution - 2002 CAT Notes | EduRev

CAT Mock Test Series 2020

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CAT : CAT Past Year Question Paper Solution - 2002 CAT Notes | EduRev

 Page 1


Page 1
CAT 2002 Actual Paper
1 3 2 4 3 4 4 3 5 3 6 1 7 1 8 3 9 3 10 3
11 2 12 4 13 1 14 4 15 1 16 1 17 1 18 4 19 2 20 3
21 3 22 3 23 4 24 1 25 3 26 3 27 2 28 2 29 2 30 2
31 2 32 4 33 4 34 2 35 2 36 2 37 3 38 2 39 1 40 4
41 2 42 2 43 2 44 4 45 3 46 1 47 3 48 2 49 4 50 3
51 3 52 2 53 1 54 2 55 1 56 4 57 4 58 2 59 4 60 4
61 3 62 2 63 4 64 3 65 3 66 2 67 2 68 4 69 3 70 1
71 3 72 2 73 3 74 4 75 4 76 1 77 1 78 4 79 2 80 4
81 4 82 3 83 4 84 3 85 1 86 3 87 *2 88 3 89 2 90 2
91 4 92 4 93 4 94 3 95 4 96 2 97 2 98 4 99 3 100 3
101 3 102 2 103 4 104 2 105 4 106 3 107 1 108 3 109 4 110 2
111 3 112 1 113 4 114 3 115 1 116 4 117 3 118 2 119 3 120 2
121 1 122 4 123 1 124 3 125 1 126 3 127 2 128 3 129 4 130 1
131 1 132 4 133 4 134 4 135 2 136 4 137 2 138 2 139 4 140 4
141 4 142 1 143 1 144 2 145 1 146 3 147 4 148 3 149 1 150 3
Scoring table
    Section
DI 1 to 50 50
QA 51 to 100 50
EU + RC 101 to 150 50
T otal 150
T otal
questions
T otal
attempted
T otal
correct
T otal
incorrect
Net
score
 Time
taken
Question
number
	





Page 2


Page 1
CAT 2002 Actual Paper
1 3 2 4 3 4 4 3 5 3 6 1 7 1 8 3 9 3 10 3
11 2 12 4 13 1 14 4 15 1 16 1 17 1 18 4 19 2 20 3
21 3 22 3 23 4 24 1 25 3 26 3 27 2 28 2 29 2 30 2
31 2 32 4 33 4 34 2 35 2 36 2 37 3 38 2 39 1 40 4
41 2 42 2 43 2 44 4 45 3 46 1 47 3 48 2 49 4 50 3
51 3 52 2 53 1 54 2 55 1 56 4 57 4 58 2 59 4 60 4
61 3 62 2 63 4 64 3 65 3 66 2 67 2 68 4 69 3 70 1
71 3 72 2 73 3 74 4 75 4 76 1 77 1 78 4 79 2 80 4
81 4 82 3 83 4 84 3 85 1 86 3 87 *2 88 3 89 2 90 2
91 4 92 4 93 4 94 3 95 4 96 2 97 2 98 4 99 3 100 3
101 3 102 2 103 4 104 2 105 4 106 3 107 1 108 3 109 4 110 2
111 3 112 1 113 4 114 3 115 1 116 4 117 3 118 2 119 3 120 2
121 1 122 4 123 1 124 3 125 1 126 3 127 2 128 3 129 4 130 1
131 1 132 4 133 4 134 4 135 2 136 4 137 2 138 2 139 4 140 4
141 4 142 1 143 1 144 2 145 1 146 3 147 4 148 3 149 1 150 3
Scoring table
    Section
DI 1 to 50 50
QA 51 to 100 50
EU + RC 101 to 150 50
T otal 150
T otal
questions
T otal
attempted
T otal
correct
T otal
incorrect
Net
score
 Time
taken
Question
number
	





Page 2
CAT 2002 Actual Paper
1. 3 Statement I tells us that
(1) Ashish is not an engineer, (2) Ashish got more
offers than the engineers.
Hence, Ashish did not have 0 offers.
After this the following table can be achieved.
Profession N ames Offers
3 2 1 0 X P rofession 
CA Ash ish × ×  × X E ngineer
M D D hanraj  × × × X E ngineer
E conom ist Sam eer ×  ××
E ngineer × × × 
From statement IV, Dhanraj is not at 0 and 1.
2. 4 Option (3) is ruled out by statement VII.
Option (1) is ruled out by statements VII and VIII.
From statement IV, Sandeep had Rs. 30 to start and
Daljeet Rs. 20.
From statement II, option (2) is not possible as Sandeep
was left with Re 1, he spent Rs. 29. But according to
(2) he spent Rs. 1.50 more than Daljeet. But Daljeet
had only Rs. 20. Hence option (4) is correct.
3. 4 Data insufficient, please check the question.
4. 3 Statements V and VI rule out options (1) and (2). Since
contestants from Bangalore and Pune did not come
first, school from Hyderabad can come first. Convent
is not in Hyderabad which rules out option (4).
5. 3 The only two possible combinations are:
Younger Older
24
39
Cubes of natural numbers are 1, 8, 27, 64, ... . Here,
64 and above are not possible as the age will go
above 10 years.
If younger boy is 2 years old, then older boy is 4
years
old. Then, Father’s age is 24 years and Mother’s age
is 
42
21 years.
2
=
Also, 24 – 21 = 3
?Age of younger boy = 2 years
6. 1 Total seats in the hall     200
Seats vacant 20
Total waiting 180
Ladies 72
Seating capacity of flight
2
180 120
3
×=
Number of people in flight A = 100
For flight B = 180 – 100 = 80
Thus, airhostess for A = 
80
4
20
=
Empty seats in flight B = 120 – 80 = 40
40 : 4 = 10 : 1
7. 1
N
W E
S
S
F
M oves @ 20 km ph 
t = ½ hr = 30 m inutes
 s = 20 ×        = 10 km ?
30
60
10 km
10 km
20 km
40 km
10 km
@ 100 kmph
t = 24 minutes
s = 40 km ?
@ 40 km ph
t = 30 minutes
 s = 20 km ?
@ 40 km ph
t = 15 minutes
 s = 10 km ?
@ 40 km ph
t = 15 minutes
 s = 10 km ?
START
IInd
Signal
IIIrd
Signal
IVTH
Signal
Vth Signal
FINISH
I  Signal
Note: s = Distance covered; v = Velocity (km/hr)
           t = Time taken;  s = v × t
The total distance travelled by the motorist from the starting
point till last signal = 10 + 10 + 20 + 40 + 10 = 90 km.
8. 3
N
W E
S
10 km
10 km
20 km
40 km
10 km
S
II
III
IV
F
40 km
30 km
T
I
By Pythagoras’ Theorem,
SF =
22
ST TF +
 = 
22
40 30 2500 += = 50 km
Page 3


Page 1
CAT 2002 Actual Paper
1 3 2 4 3 4 4 3 5 3 6 1 7 1 8 3 9 3 10 3
11 2 12 4 13 1 14 4 15 1 16 1 17 1 18 4 19 2 20 3
21 3 22 3 23 4 24 1 25 3 26 3 27 2 28 2 29 2 30 2
31 2 32 4 33 4 34 2 35 2 36 2 37 3 38 2 39 1 40 4
41 2 42 2 43 2 44 4 45 3 46 1 47 3 48 2 49 4 50 3
51 3 52 2 53 1 54 2 55 1 56 4 57 4 58 2 59 4 60 4
61 3 62 2 63 4 64 3 65 3 66 2 67 2 68 4 69 3 70 1
71 3 72 2 73 3 74 4 75 4 76 1 77 1 78 4 79 2 80 4
81 4 82 3 83 4 84 3 85 1 86 3 87 *2 88 3 89 2 90 2
91 4 92 4 93 4 94 3 95 4 96 2 97 2 98 4 99 3 100 3
101 3 102 2 103 4 104 2 105 4 106 3 107 1 108 3 109 4 110 2
111 3 112 1 113 4 114 3 115 1 116 4 117 3 118 2 119 3 120 2
121 1 122 4 123 1 124 3 125 1 126 3 127 2 128 3 129 4 130 1
131 1 132 4 133 4 134 4 135 2 136 4 137 2 138 2 139 4 140 4
141 4 142 1 143 1 144 2 145 1 146 3 147 4 148 3 149 1 150 3
Scoring table
    Section
DI 1 to 50 50
QA 51 to 100 50
EU + RC 101 to 150 50
T otal 150
T otal
questions
T otal
attempted
T otal
correct
T otal
incorrect
Net
score
 Time
taken
Question
number
	





Page 2
CAT 2002 Actual Paper
1. 3 Statement I tells us that
(1) Ashish is not an engineer, (2) Ashish got more
offers than the engineers.
Hence, Ashish did not have 0 offers.
After this the following table can be achieved.
Profession N ames Offers
3 2 1 0 X P rofession 
CA Ash ish × ×  × X E ngineer
M D D hanraj  × × × X E ngineer
E conom ist Sam eer ×  ××
E ngineer × × × 
From statement IV, Dhanraj is not at 0 and 1.
2. 4 Option (3) is ruled out by statement VII.
Option (1) is ruled out by statements VII and VIII.
From statement IV, Sandeep had Rs. 30 to start and
Daljeet Rs. 20.
From statement II, option (2) is not possible as Sandeep
was left with Re 1, he spent Rs. 29. But according to
(2) he spent Rs. 1.50 more than Daljeet. But Daljeet
had only Rs. 20. Hence option (4) is correct.
3. 4 Data insufficient, please check the question.
4. 3 Statements V and VI rule out options (1) and (2). Since
contestants from Bangalore and Pune did not come
first, school from Hyderabad can come first. Convent
is not in Hyderabad which rules out option (4).
5. 3 The only two possible combinations are:
Younger Older
24
39
Cubes of natural numbers are 1, 8, 27, 64, ... . Here,
64 and above are not possible as the age will go
above 10 years.
If younger boy is 2 years old, then older boy is 4
years
old. Then, Father’s age is 24 years and Mother’s age
is 
42
21 years.
2
=
Also, 24 – 21 = 3
?Age of younger boy = 2 years
6. 1 Total seats in the hall     200
Seats vacant 20
Total waiting 180
Ladies 72
Seating capacity of flight
2
180 120
3
×=
Number of people in flight A = 100
For flight B = 180 – 100 = 80
Thus, airhostess for A = 
80
4
20
=
Empty seats in flight B = 120 – 80 = 40
40 : 4 = 10 : 1
7. 1
N
W E
S
S
F
M oves @ 20 km ph 
t = ½ hr = 30 m inutes
 s = 20 ×        = 10 km ?
30
60
10 km
10 km
20 km
40 km
10 km
@ 100 kmph
t = 24 minutes
s = 40 km ?
@ 40 km ph
t = 30 minutes
 s = 20 km ?
@ 40 km ph
t = 15 minutes
 s = 10 km ?
@ 40 km ph
t = 15 minutes
 s = 10 km ?
START
IInd
Signal
IIIrd
Signal
IVTH
Signal
Vth Signal
FINISH
I  Signal
Note: s = Distance covered; v = Velocity (km/hr)
           t = Time taken;  s = v × t
The total distance travelled by the motorist from the starting
point till last signal = 10 + 10 + 20 + 40 + 10 = 90 km.
8. 3
N
W E
S
10 km
10 km
20 km
40 km
10 km
S
II
III
IV
F
40 km
30 km
T
I
By Pythagoras’ Theorem,
SF =
22
ST TF +
 = 
22
40 30 2500 += = 50 km
Page 3
CAT 2002 Actual Paper
9. 3 For the case when 1st signal were 1 red and 2 green
lights, the surface diagram will be as given below.
N
W E
S
10 km
10 km
20 km
40 km
10 km
S
II
III
IV
F
50 km
T
I
40 km
TF = 50 km;  ST = 40 km
Considering the above figure, option (3) is correct,
50 km to the east and 40 km to the north.
10. 3 If the car was heading towards South from the start
point, then the surface diagram will be as given below.
N
W E
S
10 km
20 km
40 km
10 km
II
III
IV
S
I
40 km
10 km
F
START
FINISH
30 km
Hence, we can see that option (3) is correct.
11. 2 Total five lie between 10 E and 40 E.
Austria, Bulgaria, Libya, Poland, Zambia
N                N           N           N        S
1
20%
5
=
12. 4 Number of cities starting with consonant and in the
northern hemisphere = 10.
Number of countries starting with consonant and in
the east of the meridien = 13.
Hence, option (4) is the correct choice. The difference
is 3.
13. 1 Three countries starting with vowels and in southern
hemisphere — Argentina. Australia and Ecuador and
two countries with capitals beginning with vowels —
Canada and Ghana.
14. 4 Let us consider two cases:
(a) If 5 min remaining the score was 0 – 2. Then final
score could have been 3 – 3. [Assuming no other
Indian scored]
(b) But if the score before 5 min was 1 – 3, then final
score could have been 4 – 3.
14. 4 From statement A, we know only the number of goals
made by India is the last 5 minutes. But, as we don’t
know what the opponent team did in the last 5 minutes,
we can’t conclude anything. So statement A alone is
not sufficient.
Similarly, statement B does not talk about the total
number of goals scored by India. So statement B is not
sufficient.
Using both the statements, we have two possibilities:
(I) If Korea had scored 3 goals 5 minutes before the
end of the match India would have scored 1 goal. In
the last 5 minutes as India made 3 goals and Korea on
the whole made 3 goals, we can conclude that India
had won the game.
(II) If Korea had scored 3 goals 5 minutes before the
end of the match, India would have scored zero goals.
In the last 5 minutes, as India made 3 goals and Korea
on the whole made 3 goals, we can say the match
was drawn.
Hence, we cannot answer the question even boy
using both the statements together.
15. 1 From A, if by adding 12 students, the total number of
students is divisible by 8. By adding 4 students, it will
be divisible by 8.
16. 1 From (A), (x + y)
11
xy
??
+
??
??
 = 4 or (x + y)
yx
xy
?? +
??
??
 = 4
? (x + y)
2
 = 4xy
? (x – y)
2
 = 0
? x = y ... (i)
From (B), (x – 50)
2
 = (y – 50)
2
On solving
x(x – 100) = y(y – 100) ... (ii)
This suggests that the values of x and y can either be
0 or 100.
Page 4


Page 1
CAT 2002 Actual Paper
1 3 2 4 3 4 4 3 5 3 6 1 7 1 8 3 9 3 10 3
11 2 12 4 13 1 14 4 15 1 16 1 17 1 18 4 19 2 20 3
21 3 22 3 23 4 24 1 25 3 26 3 27 2 28 2 29 2 30 2
31 2 32 4 33 4 34 2 35 2 36 2 37 3 38 2 39 1 40 4
41 2 42 2 43 2 44 4 45 3 46 1 47 3 48 2 49 4 50 3
51 3 52 2 53 1 54 2 55 1 56 4 57 4 58 2 59 4 60 4
61 3 62 2 63 4 64 3 65 3 66 2 67 2 68 4 69 3 70 1
71 3 72 2 73 3 74 4 75 4 76 1 77 1 78 4 79 2 80 4
81 4 82 3 83 4 84 3 85 1 86 3 87 *2 88 3 89 2 90 2
91 4 92 4 93 4 94 3 95 4 96 2 97 2 98 4 99 3 100 3
101 3 102 2 103 4 104 2 105 4 106 3 107 1 108 3 109 4 110 2
111 3 112 1 113 4 114 3 115 1 116 4 117 3 118 2 119 3 120 2
121 1 122 4 123 1 124 3 125 1 126 3 127 2 128 3 129 4 130 1
131 1 132 4 133 4 134 4 135 2 136 4 137 2 138 2 139 4 140 4
141 4 142 1 143 1 144 2 145 1 146 3 147 4 148 3 149 1 150 3
Scoring table
    Section
DI 1 to 50 50
QA 51 to 100 50
EU + RC 101 to 150 50
T otal 150
T otal
questions
T otal
attempted
T otal
correct
T otal
incorrect
Net
score
 Time
taken
Question
number
	





Page 2
CAT 2002 Actual Paper
1. 3 Statement I tells us that
(1) Ashish is not an engineer, (2) Ashish got more
offers than the engineers.
Hence, Ashish did not have 0 offers.
After this the following table can be achieved.
Profession N ames Offers
3 2 1 0 X P rofession 
CA Ash ish × ×  × X E ngineer
M D D hanraj  × × × X E ngineer
E conom ist Sam eer ×  ××
E ngineer × × × 
From statement IV, Dhanraj is not at 0 and 1.
2. 4 Option (3) is ruled out by statement VII.
Option (1) is ruled out by statements VII and VIII.
From statement IV, Sandeep had Rs. 30 to start and
Daljeet Rs. 20.
From statement II, option (2) is not possible as Sandeep
was left with Re 1, he spent Rs. 29. But according to
(2) he spent Rs. 1.50 more than Daljeet. But Daljeet
had only Rs. 20. Hence option (4) is correct.
3. 4 Data insufficient, please check the question.
4. 3 Statements V and VI rule out options (1) and (2). Since
contestants from Bangalore and Pune did not come
first, school from Hyderabad can come first. Convent
is not in Hyderabad which rules out option (4).
5. 3 The only two possible combinations are:
Younger Older
24
39
Cubes of natural numbers are 1, 8, 27, 64, ... . Here,
64 and above are not possible as the age will go
above 10 years.
If younger boy is 2 years old, then older boy is 4
years
old. Then, Father’s age is 24 years and Mother’s age
is 
42
21 years.
2
=
Also, 24 – 21 = 3
?Age of younger boy = 2 years
6. 1 Total seats in the hall     200
Seats vacant 20
Total waiting 180
Ladies 72
Seating capacity of flight
2
180 120
3
×=
Number of people in flight A = 100
For flight B = 180 – 100 = 80
Thus, airhostess for A = 
80
4
20
=
Empty seats in flight B = 120 – 80 = 40
40 : 4 = 10 : 1
7. 1
N
W E
S
S
F
M oves @ 20 km ph 
t = ½ hr = 30 m inutes
 s = 20 ×        = 10 km ?
30
60
10 km
10 km
20 km
40 km
10 km
@ 100 kmph
t = 24 minutes
s = 40 km ?
@ 40 km ph
t = 30 minutes
 s = 20 km ?
@ 40 km ph
t = 15 minutes
 s = 10 km ?
@ 40 km ph
t = 15 minutes
 s = 10 km ?
START
IInd
Signal
IIIrd
Signal
IVTH
Signal
Vth Signal
FINISH
I  Signal
Note: s = Distance covered; v = Velocity (km/hr)
           t = Time taken;  s = v × t
The total distance travelled by the motorist from the starting
point till last signal = 10 + 10 + 20 + 40 + 10 = 90 km.
8. 3
N
W E
S
10 km
10 km
20 km
40 km
10 km
S
II
III
IV
F
40 km
30 km
T
I
By Pythagoras’ Theorem,
SF =
22
ST TF +
 = 
22
40 30 2500 += = 50 km
Page 3
CAT 2002 Actual Paper
9. 3 For the case when 1st signal were 1 red and 2 green
lights, the surface diagram will be as given below.
N
W E
S
10 km
10 km
20 km
40 km
10 km
S
II
III
IV
F
50 km
T
I
40 km
TF = 50 km;  ST = 40 km
Considering the above figure, option (3) is correct,
50 km to the east and 40 km to the north.
10. 3 If the car was heading towards South from the start
point, then the surface diagram will be as given below.
N
W E
S
10 km
20 km
40 km
10 km
II
III
IV
S
I
40 km
10 km
F
START
FINISH
30 km
Hence, we can see that option (3) is correct.
11. 2 Total five lie between 10 E and 40 E.
Austria, Bulgaria, Libya, Poland, Zambia
N                N           N           N        S
1
20%
5
=
12. 4 Number of cities starting with consonant and in the
northern hemisphere = 10.
Number of countries starting with consonant and in
the east of the meridien = 13.
Hence, option (4) is the correct choice. The difference
is 3.
13. 1 Three countries starting with vowels and in southern
hemisphere — Argentina. Australia and Ecuador and
two countries with capitals beginning with vowels —
Canada and Ghana.
14. 4 Let us consider two cases:
(a) If 5 min remaining the score was 0 – 2. Then final
score could have been 3 – 3. [Assuming no other
Indian scored]
(b) But if the score before 5 min was 1 – 3, then final
score could have been 4 – 3.
14. 4 From statement A, we know only the number of goals
made by India is the last 5 minutes. But, as we don’t
know what the opponent team did in the last 5 minutes,
we can’t conclude anything. So statement A alone is
not sufficient.
Similarly, statement B does not talk about the total
number of goals scored by India. So statement B is not
sufficient.
Using both the statements, we have two possibilities:
(I) If Korea had scored 3 goals 5 minutes before the
end of the match India would have scored 1 goal. In
the last 5 minutes as India made 3 goals and Korea on
the whole made 3 goals, we can conclude that India
had won the game.
(II) If Korea had scored 3 goals 5 minutes before the
end of the match, India would have scored zero goals.
In the last 5 minutes, as India made 3 goals and Korea
on the whole made 3 goals, we can say the match
was drawn.
Hence, we cannot answer the question even boy
using both the statements together.
15. 1 From A, if by adding 12 students, the total number of
students is divisible by 8. By adding 4 students, it will
be divisible by 8.
16. 1 From (A), (x + y)
11
xy
??
+
??
??
 = 4 or (x + y)
yx
xy
?? +
??
??
 = 4
? (x + y)
2
 = 4xy
? (x – y)
2
 = 0
? x = y ... (i)
From (B), (x – 50)
2
 = (y – 50)
2
On solving
x(x – 100) = y(y – 100) ... (ii)
This suggests that the values of x and y can either be
0 or 100.
Page 4
CAT 2002 Actual Paper
17. 1 Statement:
A. Let the wholesale price is x.
Thus, listed prices = 1.2x
After a discount of 10%, new price = 0.9 × 1.2x
             = 1.08x
? 1.08 – x = 10$.
Thus, we know x can be found.
B. We do not know at what percentage profit, or at
what amount of profit the dress was actually
sold.
18. 4 A gives 500 as median and B gives 600 as range.
A and B together do not give average. Therefore, it
cannot be answered from the given statements.
19. 2 From statement A, we know that for all –1 < x < 1,
we can determine |x – 2| < 1 is not true. Therefore,
statement A alone is sufficient.
From statement B, –1 < x < 3, we cannot determine
whether |x – 2| < 1 or not. Therefore, statement B
alone is sufficient.
20. 3 From statement A, we cannot find anything.
From B alone we cannot find.
From A and B,
300
F
R
X
58
196
x + 196 + 58 = 300. Thus, x can be found.
21. 3 Jagdish (J), Punit (P), Girish (G)
(A) J = 
2
9
 [P + G]
P + G + J = 38500
Thus, only J can be found.
(B) Similarly, from this only P can be found.
Combining we know J, P and G can be found.
22. 3 Emp. numbers 51, 58, 64, 72, 73 earn more than 50
per day in complex operations.
Total = 5
23. 4 80% attendance = 80% of 25 = 20 days
Emp. numbers 47, 51, 72, 73, 74, 79, 80.
Thus, total = 7
24. 1
Emp. No. Earnings No. of daysE/D
ED
(medium) (medium)
2001151 159.64 13.33 11.97
2001158 109.72 9.61 11.41
2001164 735.22 12.07 60.91
2001171 6.10 4.25 -
2001172 117.46 8.50 13.81
2001179 776.19 19.00 40.85
2001180 1262.79 19.00 66.46
Hence, Emp. number 2001180 earns the maximum
earnings per day.
25. 3 Emp. numbers 51, 58, 64, 71, 72 satisfy the
condition.
[For emp. 64, you see 12 is not the double of 5. And
735 is not even double of 402.
Hence, 
402 735
.
512
>
Note: Emp. numbers 48, 49, 50 are not eligible for
earnings. Hence, they are not counted.
26. 3 Total revenue of 1999 = 3374
5% of 3374 = 3374 × 
5
100
 = 168.7
For 1999, revenue for Spain is 55, Rest of Latin America
is 115, North Sea is 140, Rest of the world is 91.
So total four operations of the company accounted
for less than 5% of the total revenue earned in the
year 1999.
27. 2 The language in the question is ambiguous.
Taking the question to be more than 200% growth in
revenue, the revenue in 2000 will be more than 3
times that in 1999. Hence, (2) is the answer.
Taking the revenue in 2000 to be more than 200% of
that in 1999, the revenue in 2000 should be more than
twice of that in 1999. Then there will be 4 operations.
28. 2 Four operations, as given below:
(1) North Africa and Middle-East
(2) Argentina
(3) Rest of Latin America
(4) Far East
have registered yearly increase in income before taxes
and charges from 1998 to 2000.
29. 2 Percentage increase in net income before tax and
charges for total world (1998-99)
= 
1375 248
100
248
-
× = 454.4%
Spain is making loss.
Page 5


Page 1
CAT 2002 Actual Paper
1 3 2 4 3 4 4 3 5 3 6 1 7 1 8 3 9 3 10 3
11 2 12 4 13 1 14 4 15 1 16 1 17 1 18 4 19 2 20 3
21 3 22 3 23 4 24 1 25 3 26 3 27 2 28 2 29 2 30 2
31 2 32 4 33 4 34 2 35 2 36 2 37 3 38 2 39 1 40 4
41 2 42 2 43 2 44 4 45 3 46 1 47 3 48 2 49 4 50 3
51 3 52 2 53 1 54 2 55 1 56 4 57 4 58 2 59 4 60 4
61 3 62 2 63 4 64 3 65 3 66 2 67 2 68 4 69 3 70 1
71 3 72 2 73 3 74 4 75 4 76 1 77 1 78 4 79 2 80 4
81 4 82 3 83 4 84 3 85 1 86 3 87 *2 88 3 89 2 90 2
91 4 92 4 93 4 94 3 95 4 96 2 97 2 98 4 99 3 100 3
101 3 102 2 103 4 104 2 105 4 106 3 107 1 108 3 109 4 110 2
111 3 112 1 113 4 114 3 115 1 116 4 117 3 118 2 119 3 120 2
121 1 122 4 123 1 124 3 125 1 126 3 127 2 128 3 129 4 130 1
131 1 132 4 133 4 134 4 135 2 136 4 137 2 138 2 139 4 140 4
141 4 142 1 143 1 144 2 145 1 146 3 147 4 148 3 149 1 150 3
Scoring table
    Section
DI 1 to 50 50
QA 51 to 100 50
EU + RC 101 to 150 50
T otal 150
T otal
questions
T otal
attempted
T otal
correct
T otal
incorrect
Net
score
 Time
taken
Question
number
	





Page 2
CAT 2002 Actual Paper
1. 3 Statement I tells us that
(1) Ashish is not an engineer, (2) Ashish got more
offers than the engineers.
Hence, Ashish did not have 0 offers.
After this the following table can be achieved.
Profession N ames Offers
3 2 1 0 X P rofession 
CA Ash ish × ×  × X E ngineer
M D D hanraj  × × × X E ngineer
E conom ist Sam eer ×  ××
E ngineer × × × 
From statement IV, Dhanraj is not at 0 and 1.
2. 4 Option (3) is ruled out by statement VII.
Option (1) is ruled out by statements VII and VIII.
From statement IV, Sandeep had Rs. 30 to start and
Daljeet Rs. 20.
From statement II, option (2) is not possible as Sandeep
was left with Re 1, he spent Rs. 29. But according to
(2) he spent Rs. 1.50 more than Daljeet. But Daljeet
had only Rs. 20. Hence option (4) is correct.
3. 4 Data insufficient, please check the question.
4. 3 Statements V and VI rule out options (1) and (2). Since
contestants from Bangalore and Pune did not come
first, school from Hyderabad can come first. Convent
is not in Hyderabad which rules out option (4).
5. 3 The only two possible combinations are:
Younger Older
24
39
Cubes of natural numbers are 1, 8, 27, 64, ... . Here,
64 and above are not possible as the age will go
above 10 years.
If younger boy is 2 years old, then older boy is 4
years
old. Then, Father’s age is 24 years and Mother’s age
is 
42
21 years.
2
=
Also, 24 – 21 = 3
?Age of younger boy = 2 years
6. 1 Total seats in the hall     200
Seats vacant 20
Total waiting 180
Ladies 72
Seating capacity of flight
2
180 120
3
×=
Number of people in flight A = 100
For flight B = 180 – 100 = 80
Thus, airhostess for A = 
80
4
20
=
Empty seats in flight B = 120 – 80 = 40
40 : 4 = 10 : 1
7. 1
N
W E
S
S
F
M oves @ 20 km ph 
t = ½ hr = 30 m inutes
 s = 20 ×        = 10 km ?
30
60
10 km
10 km
20 km
40 km
10 km
@ 100 kmph
t = 24 minutes
s = 40 km ?
@ 40 km ph
t = 30 minutes
 s = 20 km ?
@ 40 km ph
t = 15 minutes
 s = 10 km ?
@ 40 km ph
t = 15 minutes
 s = 10 km ?
START
IInd
Signal
IIIrd
Signal
IVTH
Signal
Vth Signal
FINISH
I  Signal
Note: s = Distance covered; v = Velocity (km/hr)
           t = Time taken;  s = v × t
The total distance travelled by the motorist from the starting
point till last signal = 10 + 10 + 20 + 40 + 10 = 90 km.
8. 3
N
W E
S
10 km
10 km
20 km
40 km
10 km
S
II
III
IV
F
40 km
30 km
T
I
By Pythagoras’ Theorem,
SF =
22
ST TF +
 = 
22
40 30 2500 += = 50 km
Page 3
CAT 2002 Actual Paper
9. 3 For the case when 1st signal were 1 red and 2 green
lights, the surface diagram will be as given below.
N
W E
S
10 km
10 km
20 km
40 km
10 km
S
II
III
IV
F
50 km
T
I
40 km
TF = 50 km;  ST = 40 km
Considering the above figure, option (3) is correct,
50 km to the east and 40 km to the north.
10. 3 If the car was heading towards South from the start
point, then the surface diagram will be as given below.
N
W E
S
10 km
20 km
40 km
10 km
II
III
IV
S
I
40 km
10 km
F
START
FINISH
30 km
Hence, we can see that option (3) is correct.
11. 2 Total five lie between 10 E and 40 E.
Austria, Bulgaria, Libya, Poland, Zambia
N                N           N           N        S
1
20%
5
=
12. 4 Number of cities starting with consonant and in the
northern hemisphere = 10.
Number of countries starting with consonant and in
the east of the meridien = 13.
Hence, option (4) is the correct choice. The difference
is 3.
13. 1 Three countries starting with vowels and in southern
hemisphere — Argentina. Australia and Ecuador and
two countries with capitals beginning with vowels —
Canada and Ghana.
14. 4 Let us consider two cases:
(a) If 5 min remaining the score was 0 – 2. Then final
score could have been 3 – 3. [Assuming no other
Indian scored]
(b) But if the score before 5 min was 1 – 3, then final
score could have been 4 – 3.
14. 4 From statement A, we know only the number of goals
made by India is the last 5 minutes. But, as we don’t
know what the opponent team did in the last 5 minutes,
we can’t conclude anything. So statement A alone is
not sufficient.
Similarly, statement B does not talk about the total
number of goals scored by India. So statement B is not
sufficient.
Using both the statements, we have two possibilities:
(I) If Korea had scored 3 goals 5 minutes before the
end of the match India would have scored 1 goal. In
the last 5 minutes as India made 3 goals and Korea on
the whole made 3 goals, we can conclude that India
had won the game.
(II) If Korea had scored 3 goals 5 minutes before the
end of the match, India would have scored zero goals.
In the last 5 minutes, as India made 3 goals and Korea
on the whole made 3 goals, we can say the match
was drawn.
Hence, we cannot answer the question even boy
using both the statements together.
15. 1 From A, if by adding 12 students, the total number of
students is divisible by 8. By adding 4 students, it will
be divisible by 8.
16. 1 From (A), (x + y)
11
xy
??
+
??
??
 = 4 or (x + y)
yx
xy
?? +
??
??
 = 4
? (x + y)
2
 = 4xy
? (x – y)
2
 = 0
? x = y ... (i)
From (B), (x – 50)
2
 = (y – 50)
2
On solving
x(x – 100) = y(y – 100) ... (ii)
This suggests that the values of x and y can either be
0 or 100.
Page 4
CAT 2002 Actual Paper
17. 1 Statement:
A. Let the wholesale price is x.
Thus, listed prices = 1.2x
After a discount of 10%, new price = 0.9 × 1.2x
             = 1.08x
? 1.08 – x = 10$.
Thus, we know x can be found.
B. We do not know at what percentage profit, or at
what amount of profit the dress was actually
sold.
18. 4 A gives 500 as median and B gives 600 as range.
A and B together do not give average. Therefore, it
cannot be answered from the given statements.
19. 2 From statement A, we know that for all –1 < x < 1,
we can determine |x – 2| < 1 is not true. Therefore,
statement A alone is sufficient.
From statement B, –1 < x < 3, we cannot determine
whether |x – 2| < 1 or not. Therefore, statement B
alone is sufficient.
20. 3 From statement A, we cannot find anything.
From B alone we cannot find.
From A and B,
300
F
R
X
58
196
x + 196 + 58 = 300. Thus, x can be found.
21. 3 Jagdish (J), Punit (P), Girish (G)
(A) J = 
2
9
 [P + G]
P + G + J = 38500
Thus, only J can be found.
(B) Similarly, from this only P can be found.
Combining we know J, P and G can be found.
22. 3 Emp. numbers 51, 58, 64, 72, 73 earn more than 50
per day in complex operations.
Total = 5
23. 4 80% attendance = 80% of 25 = 20 days
Emp. numbers 47, 51, 72, 73, 74, 79, 80.
Thus, total = 7
24. 1
Emp. No. Earnings No. of daysE/D
ED
(medium) (medium)
2001151 159.64 13.33 11.97
2001158 109.72 9.61 11.41
2001164 735.22 12.07 60.91
2001171 6.10 4.25 -
2001172 117.46 8.50 13.81
2001179 776.19 19.00 40.85
2001180 1262.79 19.00 66.46
Hence, Emp. number 2001180 earns the maximum
earnings per day.
25. 3 Emp. numbers 51, 58, 64, 71, 72 satisfy the
condition.
[For emp. 64, you see 12 is not the double of 5. And
735 is not even double of 402.
Hence, 
402 735
.
512
>
Note: Emp. numbers 48, 49, 50 are not eligible for
earnings. Hence, they are not counted.
26. 3 Total revenue of 1999 = 3374
5% of 3374 = 3374 × 
5
100
 = 168.7
For 1999, revenue for Spain is 55, Rest of Latin America
is 115, North Sea is 140, Rest of the world is 91.
So total four operations of the company accounted
for less than 5% of the total revenue earned in the
year 1999.
27. 2 The language in the question is ambiguous.
Taking the question to be more than 200% growth in
revenue, the revenue in 2000 will be more than 3
times that in 1999. Hence, (2) is the answer.
Taking the revenue in 2000 to be more than 200% of
that in 1999, the revenue in 2000 should be more than
twice of that in 1999. Then there will be 4 operations.
28. 2 Four operations, as given below:
(1) North Africa and Middle-East
(2) Argentina
(3) Rest of Latin America
(4) Far East
have registered yearly increase in income before taxes
and charges from 1998 to 2000.
29. 2 Percentage increase in net income before tax and
charges for total world (1998-99)
= 
1375 248
100
248
-
× = 454.4%
Spain is making loss.
Page 5
CAT 2002 Actual Paper
Percentage increase for North Africa and Middle-East
341 111
111
-
× 100 = 207.2%
Percentage increase for Argentina
838 94
100
94
-
=×
     = 791.5%
From the table one can directly say that there is no
operation other than Argentina, whose percentage
increase in net income before taxes and charges is
higher than the average (world).
30. 2 Statement 1 is obviously wrong.
(2) 
54 20
65 52
>
. Hence, (2) is correct.
(3) 
500 61
1168 187
>
. Hence (3) is wrong.
31. 2 Profitability of North Africa and Middle-East in 2000
= 
356
530
= 0.67
Profitability of Spain in 2000 = 
225
43
= 5.23
Profitability of Rest of Latin America in 2000 = 
169
252
,
i.e. < 1.
Profitability of Far East in 2000 = 
189
311
= < 1
32. 4 Except Rest of Latin America and Rest of the World all
the operations are greater than 2.
33. 4 Options (1), (2) and (3), are ruled out. So the correct
option is (4).
34. 2 It can be easily observed from the two charts that
Switzerland’s ratio of chart 1 to chart 2 is 
20
11
 has the
highest price per unit kilogram for its supply. Finding
the ratio of the value and quantity is enough to reach
the solution.
35. 2 Total value of distribution to Turkey is 16% of 5760
million Euro.
Total quantity of distribution to Turkey is 15% of 1.055
million tonnes.
So the average price in Euro per kilogram for Turkey is
16
5760
100
5.6
15
1055
100
??
×
??
??
??
×
??
??

36. 2 BC ? AC ? AAC = 0
37. 3
095.2
BD AE AAB ?????? ?
?Least cost of sending one unit from any refinery
to AAB
= 0 + 95.2 = 95.2.
38. 2 BB ? AB ? AAG = 311.1
Same as above.
39. 1 First we will have to check the minimum cost for
receiving at AAA. This is 0 for AE. But, BB to AE is
very high. Next is AC [314.5]. BB to AC is 451.1. After
AC, the others are high. Hence, 314.5 + 451.1 = 765.6
is the least cost.
40. 4 Number of refineries = 6
Number of depots = 7
Number of districts = 9
Therefore, number of possible ways to send petrol
from any refinery to any district is 6 × 7 × 9 = 378.
41. 2 The highest cost is for the route
BE ? AE ? AAH = 2193.0
For questions 42 to 47:
  
Position 
of 
States 
(Rank)
96-97 97-98 98-99 99-00 00-01
1 MAMA MAMA MA
2TN TN TN TN TN
3GU AP AP AP AP
4AP GU GUGU UP changed
5 KAUP UP UPGU tw ice
6 UPKA KAKA KA
7WB WB WBWB WB
}
Year
42. 2 From above table, we can conclude that option (2) is
correct.
43. 2 On referring to the table, we can see that UP is the
state which changed its relative ranking most number
of times.
44. 4 We can say directly on observing the graph that the
sales tax revenue collections for AP has more than
doubled from 1997 to 2001.
45. 3 Growth rate of tax revenue can be calculated as:
(Sales tax revenue of correct year – Sales tax revenue
of previous year)
For year 1997-98 
7826 7290
7826
-
 = 0.068
For year 1998-99  
8067 7826
7826
-
 = 0.030
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