CAT Past Year Question Paper Solution - 2001 | Additional Study Material for CAT PDF Download

 Page 1


Page 1
CAT 2001 Actual Paper
1 c 2 b 3 a 4 d 5 a 6 c 7 b 8 a 9 d 10 a
11 a 12 d 13 a 14 c 15 d 16 d 17 d 18 d 19 a 20 d
21 c 22 b 23 d 24 c 25 b 26 a 27 c 28 c 29 a 30 c
31 c 32 a 33 a 34 c 35 a 36 b 37 c 38 d 39 b 40 b
41 c 42 d 43 d 44 b 45 a 46 b 47 b 48 c 49 b 50 b
51 a 52 d 53 c 54 b 55 d 56 c 57 a 58 c 59 d 60 a
61 d 62 c 63 b 64 c 65 d 66 c 67 d 68 a 69 d 70 a
71 a 72 c 73 a 74 b 75 b 76 a 77 d 78 d 79 b 80 b
81 a 82 c 83 d 84 b 85 c 86 a 87 d 88 c 89 d 90 c
91 c 92 a 93 a 94 a 95 d 96 c 97 b 98 b 99 a 100 b
101 d 102 b 103 d 104 b 105 b 106 a 107 c 108 b 109 a 110 c
111 c 112 a 113 b 114 a 115 d 116 d 117 d 118 b 119 a 120 d
121 c 122 d 123 a 124 a 125 c 126 b 127 d 128 b 129 a 130 c
131 c 132 c 133 c 134 b 135 b 136 c 137 b 138 c 139 d 140 b
141 c 142 b 143 b 144 b 145 b 146 d 147 a 148 c 149 a 150 c
QA  1 to 50 50
EU + RC 51 to 100 50
DI + DS + AR 101 to 150 50
T otal 150
T otal
questions
T otal
attempted
T otal
correct
T otal
wrong
Net
Score
 Time
T aken
Question
number

	






Page 2


Page 1
CAT 2001 Actual Paper
1 c 2 b 3 a 4 d 5 a 6 c 7 b 8 a 9 d 10 a
11 a 12 d 13 a 14 c 15 d 16 d 17 d 18 d 19 a 20 d
21 c 22 b 23 d 24 c 25 b 26 a 27 c 28 c 29 a 30 c
31 c 32 a 33 a 34 c 35 a 36 b 37 c 38 d 39 b 40 b
41 c 42 d 43 d 44 b 45 a 46 b 47 b 48 c 49 b 50 b
51 a 52 d 53 c 54 b 55 d 56 c 57 a 58 c 59 d 60 a
61 d 62 c 63 b 64 c 65 d 66 c 67 d 68 a 69 d 70 a
71 a 72 c 73 a 74 b 75 b 76 a 77 d 78 d 79 b 80 b
81 a 82 c 83 d 84 b 85 c 86 a 87 d 88 c 89 d 90 c
91 c 92 a 93 a 94 a 95 d 96 c 97 b 98 b 99 a 100 b
101 d 102 b 103 d 104 b 105 b 106 a 107 c 108 b 109 a 110 c
111 c 112 a 113 b 114 a 115 d 116 d 117 d 118 b 119 a 120 d
121 c 122 d 123 a 124 a 125 c 126 b 127 d 128 b 129 a 130 c
131 c 132 c 133 c 134 b 135 b 136 c 137 b 138 c 139 d 140 b
141 c 142 b 143 b 144 b 145 b 146 d 147 a 148 c 149 a 150 c
QA  1 to 50 50
EU + RC 51 to 100 50
DI + DS + AR 101 to 150 50
T otal 150
T otal
questions
T otal
attempted
T otal
correct
T otal
wrong
Net
Score
 Time
T aken
Question
number

	






Page 2 CAT 2001 Actual Paper
1. c Let the marks scored in five subjects be 6x, 7x, 8x, 9x
and 10x (on a scale of 1).
Average score = 60%
100
60
5
x 10 x 9 x 8 x 7 x 6
=
+ + + +
? ? 8x = 0.6
? x = 0.075
So the marks are 0.45, 0.525, 0.6, 0.675 and 0.75.
Number of times the marks exceed 50% is 4.
2. b
(2 – 2x) x
x
x
x
2 m
x 2
Let the length of the edge cut at each corner be
x m. Since the resulting figure is a regular octagon,
? x 2 – 2 x x
2 2
= + x 2 – 2 2 x = ?
2x (1 2) 2 ?+ =
2
x
21
?=
+
2
22x .
21
?- =
+
3. a Check the answer choices basis the fact that:
Odd × Odd = Odd
Odd × Even = Even
Even × Even = Even
4. d x > 5, y < –1
Use answer choices.
Take x = 6, y = –6. We see none of the statements
(1, 2 and 3) is true. Hence the correct option is (d).
5. a First light blinks after 20 s.
Second light blinks after 24 s.
They blink together after LCM (20 and 24) = 120 s = 2
min.  Hence, the number of times they blink together in
an hour = 30.
6. c We can put a minimum of 120 oranges and a maximum
of 144 oranges, i.e., 25 oranges need to be filled in
128 boxes.
There are 25 different possibilities if there are 26
boxes. In such a case, at least 2 boxes contain the
same number of oranges. (i.e., even if each of the 25
boxes contain a different number of oranges, the 26th
must contain one of these numbers).
Similarly, if there are 51 boxes, at least 3 boxes contain
the same number of oranges.
Hence, at least 6 boxes have the same number of
oranges in case of 128 boxes.
7. b
N
W
S
E
C
P
A
3 km
9 km
O
r
? APS and ? AOC are similar triangles.
Where OC = r
?
2
r9
r3
81 (2r 3)
=
+
++
Now use the options. Hence, the diameter is 9 km.
8. a Let BC = y and AB = x.
Then area of ? CEF = Area(? CEB) – Area(? CFB)
                                = y .
3
x
.
2
1
– y .
3
x 2
.
2
1
 = 
6
xy
Area of ABCD = xy
? Ratio of area of ? CEF and area of ABCD is
6
1
xy :
6
xy
=
Alternate method:
Join AC, therefore Area of 
1
ABC Area of ABCD
2
?=
Also,
Area of CAE Area of CEF Area of CFB ?= ? = ? 1
Area of CEF Area of ABCD
6
?? =
9. d Work done in one day by A, B, C and D are
32
1
and
16
1
,
8
1
,
4
1
respectively.
Using answer choices, we note that the pair of B and
C does
16
3
of work in one day; the pair of A and D
does 
32
9
32
1
4
1
= +
 of the work in one day.
Hence, A and D take 
9
32
days.
B and C take 
6
32
3
16
=
days.
Hence, the first pair must comprise of A and D.
10. a Let the four-digit number be abcd.
a + b = c + d       ... (i)
b + d = 2(a + c)   ... (ii)
a + d = c              ... (iii)
From (i) and (iii), b = 2d
Page 3


Page 1
CAT 2001 Actual Paper
1 c 2 b 3 a 4 d 5 a 6 c 7 b 8 a 9 d 10 a
11 a 12 d 13 a 14 c 15 d 16 d 17 d 18 d 19 a 20 d
21 c 22 b 23 d 24 c 25 b 26 a 27 c 28 c 29 a 30 c
31 c 32 a 33 a 34 c 35 a 36 b 37 c 38 d 39 b 40 b
41 c 42 d 43 d 44 b 45 a 46 b 47 b 48 c 49 b 50 b
51 a 52 d 53 c 54 b 55 d 56 c 57 a 58 c 59 d 60 a
61 d 62 c 63 b 64 c 65 d 66 c 67 d 68 a 69 d 70 a
71 a 72 c 73 a 74 b 75 b 76 a 77 d 78 d 79 b 80 b
81 a 82 c 83 d 84 b 85 c 86 a 87 d 88 c 89 d 90 c
91 c 92 a 93 a 94 a 95 d 96 c 97 b 98 b 99 a 100 b
101 d 102 b 103 d 104 b 105 b 106 a 107 c 108 b 109 a 110 c
111 c 112 a 113 b 114 a 115 d 116 d 117 d 118 b 119 a 120 d
121 c 122 d 123 a 124 a 125 c 126 b 127 d 128 b 129 a 130 c
131 c 132 c 133 c 134 b 135 b 136 c 137 b 138 c 139 d 140 b
141 c 142 b 143 b 144 b 145 b 146 d 147 a 148 c 149 a 150 c
QA  1 to 50 50
EU + RC 51 to 100 50
DI + DS + AR 101 to 150 50
T otal 150
T otal
questions
T otal
attempted
T otal
correct
T otal
wrong
Net
Score
 Time
T aken
Question
number

	






Page 2 CAT 2001 Actual Paper
1. c Let the marks scored in five subjects be 6x, 7x, 8x, 9x
and 10x (on a scale of 1).
Average score = 60%
100
60
5
x 10 x 9 x 8 x 7 x 6
=
+ + + +
? ? 8x = 0.6
? x = 0.075
So the marks are 0.45, 0.525, 0.6, 0.675 and 0.75.
Number of times the marks exceed 50% is 4.
2. b
(2 – 2x) x
x
x
x
2 m
x 2
Let the length of the edge cut at each corner be
x m. Since the resulting figure is a regular octagon,
? x 2 – 2 x x
2 2
= + x 2 – 2 2 x = ?
2x (1 2) 2 ?+ =
2
x
21
?=
+
2
22x .
21
?- =
+
3. a Check the answer choices basis the fact that:
Odd × Odd = Odd
Odd × Even = Even
Even × Even = Even
4. d x > 5, y < –1
Use answer choices.
Take x = 6, y = –6. We see none of the statements
(1, 2 and 3) is true. Hence the correct option is (d).
5. a First light blinks after 20 s.
Second light blinks after 24 s.
They blink together after LCM (20 and 24) = 120 s = 2
min.  Hence, the number of times they blink together in
an hour = 30.
6. c We can put a minimum of 120 oranges and a maximum
of 144 oranges, i.e., 25 oranges need to be filled in
128 boxes.
There are 25 different possibilities if there are 26
boxes. In such a case, at least 2 boxes contain the
same number of oranges. (i.e., even if each of the 25
boxes contain a different number of oranges, the 26th
must contain one of these numbers).
Similarly, if there are 51 boxes, at least 3 boxes contain
the same number of oranges.
Hence, at least 6 boxes have the same number of
oranges in case of 128 boxes.
7. b
N
W
S
E
C
P
A
3 km
9 km
O
r
? APS and ? AOC are similar triangles.
Where OC = r
?
2
r9
r3
81 (2r 3)
=
+
++
Now use the options. Hence, the diameter is 9 km.
8. a Let BC = y and AB = x.
Then area of ? CEF = Area(? CEB) – Area(? CFB)
                                = y .
3
x
.
2
1
– y .
3
x 2
.
2
1
 = 
6
xy
Area of ABCD = xy
? Ratio of area of ? CEF and area of ABCD is
6
1
xy :
6
xy
=
Alternate method:
Join AC, therefore Area of 
1
ABC Area of ABCD
2
?=
Also,
Area of CAE Area of CEF Area of CFB ?= ? = ? 1
Area of CEF Area of ABCD
6
?? =
9. d Work done in one day by A, B, C and D are
32
1
and
16
1
,
8
1
,
4
1
respectively.
Using answer choices, we note that the pair of B and
C does
16
3
of work in one day; the pair of A and D
does 
32
9
32
1
4
1
= +
 of the work in one day.
Hence, A and D take 
9
32
days.
B and C take 
6
32
3
16
=
days.
Hence, the first pair must comprise of A and D.
10. a Let the four-digit number be abcd.
a + b = c + d       ... (i)
b + d = 2(a + c)   ... (ii)
a + d = c              ... (iii)
From (i) and (iii), b = 2d
Page 3
CAT 2001 Actual Paper
From (i) and (ii), 3b = 4c + d
? 3(2d) = 4c + d
? 5d = 4c
? d
4
5
c =
Now d can be 4 or 8.
But if d = 8, then c = 10 not possible.
So d = 4 which gives c = 5.
11. a Amount of money given to X
= 12 × 300 + 12 × 330 + ... + 12 × 570
= 12[300 + 330 + ... + 540 + 570]
] 30 9 600 [
2
10
12 × + × =
= 52200
Amount of money given to Y is
6 × 200 + 6 × 215 + 6 × 230 + 6 × 245 + ... to 20 terms
= 6[200 + 215 + 280 ... 485]
] 15 19 400 [
2
20
6 × + × =
= 6 × 10[400 + 285]
= 60 × 685 = 41100
? Total amount paid = 52200 + 41100 = Rs. 93,300.
12. d Let the number be x.
Increase in product = 53x – 35x = 18x
                               ? 18x = 540 ? x = 30
Hence new product = 53 × 30 = 1590.
13. a Let x be the total number of people the college will ask
for donations.
? People already solicited = 0.6x
Amount raised from the people solicited
= 600 × 0.6x = 360x
Now 360x constitutes 75% of the amount.
Hence, remaining 25% = 120x
?Average donation from remaining people
= 
x 4 . 0
x 120
= Rs. 300.
14. c The value of y would be negative and the value of x
would be positive from the inequalities given in the
question.
Therefore, from (a), y becomes positive. The value of
xy
2
 would be positive and will not be the minimum.
From (b) and (c), x
2
y and 5xy would give negative
values but we do not know which would be the
minimum.
On comparing (a) and (c), we find that
x
2
 < 5x in 2 < x < 3.
2
xy 5xy ?> [Since y is negative.]
? 5xy would give the minimum value.
15. d Let  y = n
3
 – 7n
2
 + 11n – 5
At n = 1, y = 0
? (n – 1) (n
2
 – 6n + 5)
 = (n – 1)
2
 (n – 5)
Now (n – 1)
2
  is always positive.
For n < 5, the expression gives a negative quantity.
Therefore, the least value of n will be 6.
Hence, m = 6.
16. d
Wall
8 m
2 m
Ground
y
x
Ladder
Let the length of the ladder be x feet. We have
8
2
 + y
2
 = x
2
 and (y + 2) = x
Hence, 64 + (x – 2)
2
 = x
2
? 64 + x
2
 – 4x + 4 = x
2
? 68 = 4x
 
? x = 17
17. d Let there be x mints originally in the bowl.
Sita took 
3
1
, but returned 4. So now the bowl has
4 x
3
2
+ mints.
Fatima took 
4
1
of the remainder, but returned 3.
So the bowl now has 3 4 x
3
2
4
3
+
?
?
?
?
?
?
?
?
+ mints.
Eshwari took half of remainder that is
?
?
?
?
?
?
+
?
?
?
?
?
?
?
?
+ 3 4 x
3
2
4
3
2
1
She returns 2, so the bowl now has
17 2 3 4 x
3
2
4
3
2
1
= +
?
?
?
?
?
?
+
?
?
?
?
?
?
?
?
+
 ? x = 48
Short cut:
Since Sita was the first person to pick and she picks
up 
1
3
 of the mint, but if you see the options, none of
the option is a multiple of 3.
18. d In 30 years from 1971 to 2001, number of odd days
= 30 + (8 from leap years) = 38 and 38 = 3 mod 7
So December 9, 1971 is Sunday – 3 days
= Thursday
19. a The product of 44 and 11 is 484.
If base is x,  then,
3414 = 484 x 4 x 1 x 4 x 3
0 1 2 3
= × + + +
480 x x 4 x 3
2 3
= + + ?
This equation is satisfied only when x = 5.
So base is 5.
In decimal system, the number 3111 can be written as
406.
Page 4


Page 1
CAT 2001 Actual Paper
1 c 2 b 3 a 4 d 5 a 6 c 7 b 8 a 9 d 10 a
11 a 12 d 13 a 14 c 15 d 16 d 17 d 18 d 19 a 20 d
21 c 22 b 23 d 24 c 25 b 26 a 27 c 28 c 29 a 30 c
31 c 32 a 33 a 34 c 35 a 36 b 37 c 38 d 39 b 40 b
41 c 42 d 43 d 44 b 45 a 46 b 47 b 48 c 49 b 50 b
51 a 52 d 53 c 54 b 55 d 56 c 57 a 58 c 59 d 60 a
61 d 62 c 63 b 64 c 65 d 66 c 67 d 68 a 69 d 70 a
71 a 72 c 73 a 74 b 75 b 76 a 77 d 78 d 79 b 80 b
81 a 82 c 83 d 84 b 85 c 86 a 87 d 88 c 89 d 90 c
91 c 92 a 93 a 94 a 95 d 96 c 97 b 98 b 99 a 100 b
101 d 102 b 103 d 104 b 105 b 106 a 107 c 108 b 109 a 110 c
111 c 112 a 113 b 114 a 115 d 116 d 117 d 118 b 119 a 120 d
121 c 122 d 123 a 124 a 125 c 126 b 127 d 128 b 129 a 130 c
131 c 132 c 133 c 134 b 135 b 136 c 137 b 138 c 139 d 140 b
141 c 142 b 143 b 144 b 145 b 146 d 147 a 148 c 149 a 150 c
QA  1 to 50 50
EU + RC 51 to 100 50
DI + DS + AR 101 to 150 50
T otal 150
T otal
questions
T otal
attempted
T otal
correct
T otal
wrong
Net
Score
 Time
T aken
Question
number

	






Page 2 CAT 2001 Actual Paper
1. c Let the marks scored in five subjects be 6x, 7x, 8x, 9x
and 10x (on a scale of 1).
Average score = 60%
100
60
5
x 10 x 9 x 8 x 7 x 6
=
+ + + +
? ? 8x = 0.6
? x = 0.075
So the marks are 0.45, 0.525, 0.6, 0.675 and 0.75.
Number of times the marks exceed 50% is 4.
2. b
(2 – 2x) x
x
x
x
2 m
x 2
Let the length of the edge cut at each corner be
x m. Since the resulting figure is a regular octagon,
? x 2 – 2 x x
2 2
= + x 2 – 2 2 x = ?
2x (1 2) 2 ?+ =
2
x
21
?=
+
2
22x .
21
?- =
+
3. a Check the answer choices basis the fact that:
Odd × Odd = Odd
Odd × Even = Even
Even × Even = Even
4. d x > 5, y < –1
Use answer choices.
Take x = 6, y = –6. We see none of the statements
(1, 2 and 3) is true. Hence the correct option is (d).
5. a First light blinks after 20 s.
Second light blinks after 24 s.
They blink together after LCM (20 and 24) = 120 s = 2
min.  Hence, the number of times they blink together in
an hour = 30.
6. c We can put a minimum of 120 oranges and a maximum
of 144 oranges, i.e., 25 oranges need to be filled in
128 boxes.
There are 25 different possibilities if there are 26
boxes. In such a case, at least 2 boxes contain the
same number of oranges. (i.e., even if each of the 25
boxes contain a different number of oranges, the 26th
must contain one of these numbers).
Similarly, if there are 51 boxes, at least 3 boxes contain
the same number of oranges.
Hence, at least 6 boxes have the same number of
oranges in case of 128 boxes.
7. b
N
W
S
E
C
P
A
3 km
9 km
O
r
? APS and ? AOC are similar triangles.
Where OC = r
?
2
r9
r3
81 (2r 3)
=
+
++
Now use the options. Hence, the diameter is 9 km.
8. a Let BC = y and AB = x.
Then area of ? CEF = Area(? CEB) – Area(? CFB)
                                = y .
3
x
.
2
1
– y .
3
x 2
.
2
1
 = 
6
xy
Area of ABCD = xy
? Ratio of area of ? CEF and area of ABCD is
6
1
xy :
6
xy
=
Alternate method:
Join AC, therefore Area of 
1
ABC Area of ABCD
2
?=
Also,
Area of CAE Area of CEF Area of CFB ?= ? = ? 1
Area of CEF Area of ABCD
6
?? =
9. d Work done in one day by A, B, C and D are
32
1
and
16
1
,
8
1
,
4
1
respectively.
Using answer choices, we note that the pair of B and
C does
16
3
of work in one day; the pair of A and D
does 
32
9
32
1
4
1
= +
 of the work in one day.
Hence, A and D take 
9
32
days.
B and C take 
6
32
3
16
=
days.
Hence, the first pair must comprise of A and D.
10. a Let the four-digit number be abcd.
a + b = c + d       ... (i)
b + d = 2(a + c)   ... (ii)
a + d = c              ... (iii)
From (i) and (iii), b = 2d
Page 3
CAT 2001 Actual Paper
From (i) and (ii), 3b = 4c + d
? 3(2d) = 4c + d
? 5d = 4c
? d
4
5
c =
Now d can be 4 or 8.
But if d = 8, then c = 10 not possible.
So d = 4 which gives c = 5.
11. a Amount of money given to X
= 12 × 300 + 12 × 330 + ... + 12 × 570
= 12[300 + 330 + ... + 540 + 570]
] 30 9 600 [
2
10
12 × + × =
= 52200
Amount of money given to Y is
6 × 200 + 6 × 215 + 6 × 230 + 6 × 245 + ... to 20 terms
= 6[200 + 215 + 280 ... 485]
] 15 19 400 [
2
20
6 × + × =
= 6 × 10[400 + 285]
= 60 × 685 = 41100
? Total amount paid = 52200 + 41100 = Rs. 93,300.
12. d Let the number be x.
Increase in product = 53x – 35x = 18x
                               ? 18x = 540 ? x = 30
Hence new product = 53 × 30 = 1590.
13. a Let x be the total number of people the college will ask
for donations.
? People already solicited = 0.6x
Amount raised from the people solicited
= 600 × 0.6x = 360x
Now 360x constitutes 75% of the amount.
Hence, remaining 25% = 120x
?Average donation from remaining people
= 
x 4 . 0
x 120
= Rs. 300.
14. c The value of y would be negative and the value of x
would be positive from the inequalities given in the
question.
Therefore, from (a), y becomes positive. The value of
xy
2
 would be positive and will not be the minimum.
From (b) and (c), x
2
y and 5xy would give negative
values but we do not know which would be the
minimum.
On comparing (a) and (c), we find that
x
2
 < 5x in 2 < x < 3.
2
xy 5xy ?> [Since y is negative.]
? 5xy would give the minimum value.
15. d Let  y = n
3
 – 7n
2
 + 11n – 5
At n = 1, y = 0
? (n – 1) (n
2
 – 6n + 5)
 = (n – 1)
2
 (n – 5)
Now (n – 1)
2
  is always positive.
For n < 5, the expression gives a negative quantity.
Therefore, the least value of n will be 6.
Hence, m = 6.
16. d
Wall
8 m
2 m
Ground
y
x
Ladder
Let the length of the ladder be x feet. We have
8
2
 + y
2
 = x
2
 and (y + 2) = x
Hence, 64 + (x – 2)
2
 = x
2
? 64 + x
2
 – 4x + 4 = x
2
? 68 = 4x
 
? x = 17
17. d Let there be x mints originally in the bowl.
Sita took 
3
1
, but returned 4. So now the bowl has
4 x
3
2
+ mints.
Fatima took 
4
1
of the remainder, but returned 3.
So the bowl now has 3 4 x
3
2
4
3
+
?
?
?
?
?
?
?
?
+ mints.
Eshwari took half of remainder that is
?
?
?
?
?
?
+
?
?
?
?
?
?
?
?
+ 3 4 x
3
2
4
3
2
1
She returns 2, so the bowl now has
17 2 3 4 x
3
2
4
3
2
1
= +
?
?
?
?
?
?
+
?
?
?
?
?
?
?
?
+
 ? x = 48
Short cut:
Since Sita was the first person to pick and she picks
up 
1
3
 of the mint, but if you see the options, none of
the option is a multiple of 3.
18. d In 30 years from 1971 to 2001, number of odd days
= 30 + (8 from leap years) = 38 and 38 = 3 mod 7
So December 9, 1971 is Sunday – 3 days
= Thursday
19. a The product of 44 and 11 is 484.
If base is x,  then,
3414 = 484 x 4 x 1 x 4 x 3
0 1 2 3
= × + + +
480 x x 4 x 3
2 3
= + + ?
This equation is satisfied only when x = 5.
So base is 5.
In decimal system, the number 3111 can be written as
406.
Page 4 CAT 2001 Actual Paper
20. d Let x be the speed of Rahul, and y be the speed of
current in mph.
4
1
y – x
y
6
y x
12
–
y – x
12
2 2
= ? =
+
4
y – x
y
2 2
= ?        ... (i)
When Rahul doubles his rowing speed, then we have
1
y x 2
12
–
y – x 2
12
=
+
?
12
1
y – x 4
y 2
2 2
=
24
y – x 4
y
2 2
= ?      ... (ii)
? From (i) and (ii), we have 2x
2
 = 5y
2
Putting x
2
 = 
2
y
2
5
in (i), we get 
4
y
2
3
y
2
=
3
8
y = ?
.
21. c Let ‘x’ be the number of males in Mota Hazri.
Chota Hazri Mota Hazri
Males x – 4522 x
Females 2(x – 4522) x + 4020
x + 4020 – 2(x – 4522) = 2910 ? x = 10154
? Number of males in Chota Hazri = 10154 – 4522
                    = 5632
22. b Let the number of students in classes X, Y and Z be
a, b and c respectively. Then,
total of X = 83a
total of Y = 76b
total of Z = 85c
Also 79
b a
b 76 a 83
=
+
+
? 4a = 3b;
and 81
c b
c 85 b 76
=
+
+
? 4c = 5b
Hence, b = a
3
4
, c = 
55 4 5
baa
44 3 3
=× =
Average of X, Y and Z = 
c b a
c 85 b 76 a 83
+ +
+ +
= 
45
83a 76 a 85 a
33
45
aa a
33
+× +×
++
  = 
5 . 81
12
978
=
23. d
25
25
24
32
A
E
B
C D
20 
40
20 
CE = 
22
25 – 20 = 15
(Since DBC is isosceles triangle.)
ABCD is a quadrilateral
where AB = 32 m, AD = 24 m, DC = 25 m, CB = 25 m
and DAB ? is right angle.
By Pythagoras Theorem: DB = 40 m
So area of ? 1
ADB 32 24 384 sq. m
2
=× × =
Now in isosceles ? BCD, perpendicular CE from C to
BD bisects BD.
BE = DE = 
40
20m.
2
=
Now by Pythagoras Theorem:
22
CE 25 20 15m. =- =
So area of ? 1
BCD 2 15 20 300 sq. m
2
=× × × =
Hence, area of ABCD = 384 + 300 = 684 sq. m
24. c Let the total number of pages in the book be n.
Let page number x be repeated.
Then 
?
=
= +
n
1 i
1000 x i
1000 x
2
) 1 n ( n
= +
+
Thus, 1000
2
) 1 n ( n
=
+
gives n = 44
Since 990
2
) 1 n ( n
=
+
(for n = 44), hence, x = 10.
25. b If Shyam takes 1 min for every 3 steps, then he takes
3
1
min for every step.
For 25 steps, he takes 
3
25
min, i.e. 8.33 min.
So Vyom takes 
2
1
min for every step.
For 20 steps, he takes 
2
20
min, i.e. 10 min.
Difference between their time = 1.66 min.
Escalator takes 5 steps in 1.66 min and difference in
number of steps covered = 5
Speed of escalator is 1 step for 0.33 min,
i.e. 3 steps per minute.
If escalator is moving, then Shyam takes 25 steps and
escalator also takes 25 steps.
Hence, total number of steps = 50.
26. a Let the cost of 1 burger, 1 shake and 1 fries be x, y
and z.
Then
3x + 7y + z = 120 ... (i)
4x + 10y + z = 164.5 ... (ii)
x + 3y = 44.5 ... (iii) (ii – i)
Page 5


Page 1
CAT 2001 Actual Paper
1 c 2 b 3 a 4 d 5 a 6 c 7 b 8 a 9 d 10 a
11 a 12 d 13 a 14 c 15 d 16 d 17 d 18 d 19 a 20 d
21 c 22 b 23 d 24 c 25 b 26 a 27 c 28 c 29 a 30 c
31 c 32 a 33 a 34 c 35 a 36 b 37 c 38 d 39 b 40 b
41 c 42 d 43 d 44 b 45 a 46 b 47 b 48 c 49 b 50 b
51 a 52 d 53 c 54 b 55 d 56 c 57 a 58 c 59 d 60 a
61 d 62 c 63 b 64 c 65 d 66 c 67 d 68 a 69 d 70 a
71 a 72 c 73 a 74 b 75 b 76 a 77 d 78 d 79 b 80 b
81 a 82 c 83 d 84 b 85 c 86 a 87 d 88 c 89 d 90 c
91 c 92 a 93 a 94 a 95 d 96 c 97 b 98 b 99 a 100 b
101 d 102 b 103 d 104 b 105 b 106 a 107 c 108 b 109 a 110 c
111 c 112 a 113 b 114 a 115 d 116 d 117 d 118 b 119 a 120 d
121 c 122 d 123 a 124 a 125 c 126 b 127 d 128 b 129 a 130 c
131 c 132 c 133 c 134 b 135 b 136 c 137 b 138 c 139 d 140 b
141 c 142 b 143 b 144 b 145 b 146 d 147 a 148 c 149 a 150 c
QA  1 to 50 50
EU + RC 51 to 100 50
DI + DS + AR 101 to 150 50
T otal 150
T otal
questions
T otal
attempted
T otal
correct
T otal
wrong
Net
Score
 Time
T aken
Question
number

	






Page 2 CAT 2001 Actual Paper
1. c Let the marks scored in five subjects be 6x, 7x, 8x, 9x
and 10x (on a scale of 1).
Average score = 60%
100
60
5
x 10 x 9 x 8 x 7 x 6
=
+ + + +
? ? 8x = 0.6
? x = 0.075
So the marks are 0.45, 0.525, 0.6, 0.675 and 0.75.
Number of times the marks exceed 50% is 4.
2. b
(2 – 2x) x
x
x
x
2 m
x 2
Let the length of the edge cut at each corner be
x m. Since the resulting figure is a regular octagon,
? x 2 – 2 x x
2 2
= + x 2 – 2 2 x = ?
2x (1 2) 2 ?+ =
2
x
21
?=
+
2
22x .
21
?- =
+
3. a Check the answer choices basis the fact that:
Odd × Odd = Odd
Odd × Even = Even
Even × Even = Even
4. d x > 5, y < –1
Use answer choices.
Take x = 6, y = –6. We see none of the statements
(1, 2 and 3) is true. Hence the correct option is (d).
5. a First light blinks after 20 s.
Second light blinks after 24 s.
They blink together after LCM (20 and 24) = 120 s = 2
min.  Hence, the number of times they blink together in
an hour = 30.
6. c We can put a minimum of 120 oranges and a maximum
of 144 oranges, i.e., 25 oranges need to be filled in
128 boxes.
There are 25 different possibilities if there are 26
boxes. In such a case, at least 2 boxes contain the
same number of oranges. (i.e., even if each of the 25
boxes contain a different number of oranges, the 26th
must contain one of these numbers).
Similarly, if there are 51 boxes, at least 3 boxes contain
the same number of oranges.
Hence, at least 6 boxes have the same number of
oranges in case of 128 boxes.
7. b
N
W
S
E
C
P
A
3 km
9 km
O
r
? APS and ? AOC are similar triangles.
Where OC = r
?
2
r9
r3
81 (2r 3)
=
+
++
Now use the options. Hence, the diameter is 9 km.
8. a Let BC = y and AB = x.
Then area of ? CEF = Area(? CEB) – Area(? CFB)
                                = y .
3
x
.
2
1
– y .
3
x 2
.
2
1
 = 
6
xy
Area of ABCD = xy
? Ratio of area of ? CEF and area of ABCD is
6
1
xy :
6
xy
=
Alternate method:
Join AC, therefore Area of 
1
ABC Area of ABCD
2
?=
Also,
Area of CAE Area of CEF Area of CFB ?= ? = ? 1
Area of CEF Area of ABCD
6
?? =
9. d Work done in one day by A, B, C and D are
32
1
and
16
1
,
8
1
,
4
1
respectively.
Using answer choices, we note that the pair of B and
C does
16
3
of work in one day; the pair of A and D
does 
32
9
32
1
4
1
= +
 of the work in one day.
Hence, A and D take 
9
32
days.
B and C take 
6
32
3
16
=
days.
Hence, the first pair must comprise of A and D.
10. a Let the four-digit number be abcd.
a + b = c + d       ... (i)
b + d = 2(a + c)   ... (ii)
a + d = c              ... (iii)
From (i) and (iii), b = 2d
Page 3
CAT 2001 Actual Paper
From (i) and (ii), 3b = 4c + d
? 3(2d) = 4c + d
? 5d = 4c
? d
4
5
c =
Now d can be 4 or 8.
But if d = 8, then c = 10 not possible.
So d = 4 which gives c = 5.
11. a Amount of money given to X
= 12 × 300 + 12 × 330 + ... + 12 × 570
= 12[300 + 330 + ... + 540 + 570]
] 30 9 600 [
2
10
12 × + × =
= 52200
Amount of money given to Y is
6 × 200 + 6 × 215 + 6 × 230 + 6 × 245 + ... to 20 terms
= 6[200 + 215 + 280 ... 485]
] 15 19 400 [
2
20
6 × + × =
= 6 × 10[400 + 285]
= 60 × 685 = 41100
? Total amount paid = 52200 + 41100 = Rs. 93,300.
12. d Let the number be x.
Increase in product = 53x – 35x = 18x
                               ? 18x = 540 ? x = 30
Hence new product = 53 × 30 = 1590.
13. a Let x be the total number of people the college will ask
for donations.
? People already solicited = 0.6x
Amount raised from the people solicited
= 600 × 0.6x = 360x
Now 360x constitutes 75% of the amount.
Hence, remaining 25% = 120x
?Average donation from remaining people
= 
x 4 . 0
x 120
= Rs. 300.
14. c The value of y would be negative and the value of x
would be positive from the inequalities given in the
question.
Therefore, from (a), y becomes positive. The value of
xy
2
 would be positive and will not be the minimum.
From (b) and (c), x
2
y and 5xy would give negative
values but we do not know which would be the
minimum.
On comparing (a) and (c), we find that
x
2
 < 5x in 2 < x < 3.
2
xy 5xy ?> [Since y is negative.]
? 5xy would give the minimum value.
15. d Let  y = n
3
 – 7n
2
 + 11n – 5
At n = 1, y = 0
? (n – 1) (n
2
 – 6n + 5)
 = (n – 1)
2
 (n – 5)
Now (n – 1)
2
  is always positive.
For n < 5, the expression gives a negative quantity.
Therefore, the least value of n will be 6.
Hence, m = 6.
16. d
Wall
8 m
2 m
Ground
y
x
Ladder
Let the length of the ladder be x feet. We have
8
2
 + y
2
 = x
2
 and (y + 2) = x
Hence, 64 + (x – 2)
2
 = x
2
? 64 + x
2
 – 4x + 4 = x
2
? 68 = 4x
 
? x = 17
17. d Let there be x mints originally in the bowl.
Sita took 
3
1
, but returned 4. So now the bowl has
4 x
3
2
+ mints.
Fatima took 
4
1
of the remainder, but returned 3.
So the bowl now has 3 4 x
3
2
4
3
+
?
?
?
?
?
?
?
?
+ mints.
Eshwari took half of remainder that is
?
?
?
?
?
?
+
?
?
?
?
?
?
?
?
+ 3 4 x
3
2
4
3
2
1
She returns 2, so the bowl now has
17 2 3 4 x
3
2
4
3
2
1
= +
?
?
?
?
?
?
+
?
?
?
?
?
?
?
?
+
 ? x = 48
Short cut:
Since Sita was the first person to pick and she picks
up 
1
3
 of the mint, but if you see the options, none of
the option is a multiple of 3.
18. d In 30 years from 1971 to 2001, number of odd days
= 30 + (8 from leap years) = 38 and 38 = 3 mod 7
So December 9, 1971 is Sunday – 3 days
= Thursday
19. a The product of 44 and 11 is 484.
If base is x,  then,
3414 = 484 x 4 x 1 x 4 x 3
0 1 2 3
= × + + +
480 x x 4 x 3
2 3
= + + ?
This equation is satisfied only when x = 5.
So base is 5.
In decimal system, the number 3111 can be written as
406.
Page 4 CAT 2001 Actual Paper
20. d Let x be the speed of Rahul, and y be the speed of
current in mph.
4
1
y – x
y
6
y x
12
–
y – x
12
2 2
= ? =
+
4
y – x
y
2 2
= ?        ... (i)
When Rahul doubles his rowing speed, then we have
1
y x 2
12
–
y – x 2
12
=
+
?
12
1
y – x 4
y 2
2 2
=
24
y – x 4
y
2 2
= ?      ... (ii)
? From (i) and (ii), we have 2x
2
 = 5y
2
Putting x
2
 = 
2
y
2
5
in (i), we get 
4
y
2
3
y
2
=
3
8
y = ?
.
21. c Let ‘x’ be the number of males in Mota Hazri.
Chota Hazri Mota Hazri
Males x – 4522 x
Females 2(x – 4522) x + 4020
x + 4020 – 2(x – 4522) = 2910 ? x = 10154
? Number of males in Chota Hazri = 10154 – 4522
                    = 5632
22. b Let the number of students in classes X, Y and Z be
a, b and c respectively. Then,
total of X = 83a
total of Y = 76b
total of Z = 85c
Also 79
b a
b 76 a 83
=
+
+
? 4a = 3b;
and 81
c b
c 85 b 76
=
+
+
? 4c = 5b
Hence, b = a
3
4
, c = 
55 4 5
baa
44 3 3
=× =
Average of X, Y and Z = 
c b a
c 85 b 76 a 83
+ +
+ +
= 
45
83a 76 a 85 a
33
45
aa a
33
+× +×
++
  = 
5 . 81
12
978
=
23. d
25
25
24
32
A
E
B
C D
20 
40
20 
CE = 
22
25 – 20 = 15
(Since DBC is isosceles triangle.)
ABCD is a quadrilateral
where AB = 32 m, AD = 24 m, DC = 25 m, CB = 25 m
and DAB ? is right angle.
By Pythagoras Theorem: DB = 40 m
So area of ? 1
ADB 32 24 384 sq. m
2
=× × =
Now in isosceles ? BCD, perpendicular CE from C to
BD bisects BD.
BE = DE = 
40
20m.
2
=
Now by Pythagoras Theorem:
22
CE 25 20 15m. =- =
So area of ? 1
BCD 2 15 20 300 sq. m
2
=× × × =
Hence, area of ABCD = 384 + 300 = 684 sq. m
24. c Let the total number of pages in the book be n.
Let page number x be repeated.
Then 
?
=
= +
n
1 i
1000 x i
1000 x
2
) 1 n ( n
= +
+
Thus, 1000
2
) 1 n ( n
=
+
gives n = 44
Since 990
2
) 1 n ( n
=
+
(for n = 44), hence, x = 10.
25. b If Shyam takes 1 min for every 3 steps, then he takes
3
1
min for every step.
For 25 steps, he takes 
3
25
min, i.e. 8.33 min.
So Vyom takes 
2
1
min for every step.
For 20 steps, he takes 
2
20
min, i.e. 10 min.
Difference between their time = 1.66 min.
Escalator takes 5 steps in 1.66 min and difference in
number of steps covered = 5
Speed of escalator is 1 step for 0.33 min,
i.e. 3 steps per minute.
If escalator is moving, then Shyam takes 25 steps and
escalator also takes 25 steps.
Hence, total number of steps = 50.
26. a Let the cost of 1 burger, 1 shake and 1 fries be x, y
and z.
Then
3x + 7y + z = 120 ... (i)
4x + 10y + z = 164.5 ... (ii)
x + 3y = 44.5 ... (iii) (ii – i)
Page 5
CAT 2001 Actual Paper
Multiplying (iii) by 4 and subtracting (ii) from it, we find
2y – z = 13.5  ...(iv)
Subtracting (iv) from (iii), we get x + y + z = 31.
27. c Taking a = b = c = d = 1, we get the minimum value as
(1 + 1) (1 + 1) (1 + 1) (1 + 1) = 2 × 2 × 2 × 2 = 16.
28. c Let ‘t’ be the time taken for all three together, then
11 11
t6 t 1 2t t
++ =
++
Solving the above equation, we get
3t
2
 + 7t – 6 = 0 or t = 
2
3
 hr
= 40 min
29. a
C
A B
10
D
20
Let’s assume AB be the longest side of 20 units and
another side AC is 10 units. Here CD ? AB.
Since area of ? ABC = 80 = 
1
AB CD
2
×
So 
80 2
CD 8
20
×
== .
In ? ACD; AD = 
22
10 –86 =
Hence DB = 20 – 6 = 14.
So CB = 
22
14 8 196 64 260 unit += + =
30. c Let the 6th and the 7th terms be x and y.
Then 8th term = x + y
Also y
2
 – x
2
 = 517
? (y + x)(y – x) = 517 = 47 × 11
So y + x = 47
y – x = 11
Taking y = 29 and x = 18, we have 8th term = 47,
9th term = 47 + 29 = 76 and 10th term = 76 + 47 = 123.
31. c Fresh grapes contain 10% pulp.
?20 kg fresh grapes contain 2 kg pulp.
Dry grapes contain 80% pulp.
?2 kg pulp will contain 
220
2.5
0.8 8
== kg dry
grapes
32. a Total time taken by B to cover 60 km
60 6
hr hr
50 5
==
It stops at station C for 
4
1
hr.
Now, in ?
?
?
?
?
?
?
?
+
4
1
5
6
hr train X travels
29
70 101.5 km
20
×=
This implies that they do not cross each other by the
time train Y finishes its stoppage at station C.
Let they meet after t hr.
Then 70t + 50(t – 
4
1
) = 180  
120
5 . 192
t = ?
hr
Distance from A will be (70 × 
120
5 . 192
) km = 112 km
approximately
33. a Let the highest number be n and x be the number
erased.
Then 
n(n 1)
– x
7 602
2
35
(n – 1) 17 17
+
== .
Here, the denominator (n – 1) must be a multiple of 17.
For n – 1 = 68 ? n = 69, we have 
69(70)
x
602
2
68 17
-
=
?
x = 602 × 68 – 69 × 35 = 2415 – 2408 = 7.
Hence, n = 69 and x = 7 satisfy the above conditions.
34. c
E
A
C
F
B
D
y
y
40°
x
x
Here ?ACE=180 – 2x , ?BCF = 180 – 2y
and x + y + 40° = 180° (In ? DEF)
   So x + y = 140°
   So ?ACB = 180° –  ?ACE –  ?BCF
= 180° – (180° – 2x) – (180° – 2y)
=  2(x + y) – 180°
= 2 × 140 – 180 = 100°
35. a In first updown cycle, the reduction price is Rs. 441.
According to this, (b) and (d) are removed. Now we
have to analyse (c), if the original price is Rs. 2,500,
then after first operation, the price will be  2500 –
441= Rs. 2,059.
In second operation, it will come down to around  Rs.
1688
441
here decrease% 100 18% .
2500
??
=× ˜
??
??
So the value is not equivalent to Rs. 1,944.81.
Hence, option (a) is the answer.