CAT Past Year Question Paper Solution - 2007 CAT Notes | EduRev

CAT Mock Test Series 2020

Created by: Bakliwal Institute

CAT : CAT Past Year Question Paper Solution - 2007 CAT Notes | EduRev

 Page 1


	


QA  1 to 25 25
LRDI 26 to 50 25
EU + RC 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
1 2 2 5 3 1 4 3 5 4 6 5 7 3 8 2 9 1 10 1
11 4 12 2 13 3 14 2 15 4 16 4 17 1 18 5 19 2 20 4
21 3 22 3 23 2 24 1 25 4 26 1 27 4 28 5 29 5 30 1
31 4 32 3 33 5 34 2 35 3 36 5 37 1 38 2 39 5 40 1
41 * 42 3 43 1 44 4 45 2 46 4 47 2 48 3 49 2 50 4
51 5 52 2 53 4 54 1 55 5 56 3 57 2 58 1 59 4 60 5
61 1 62 2 63 4 64 3 65 5 66 4 67 5 68 2 69 3 70 2
71 3 72 4 73 1 74 5 75 3
	
Page 2


	


QA  1 to 25 25
LRDI 26 to 50 25
EU + RC 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
1 2 2 5 3 1 4 3 5 4 6 5 7 3 8 2 9 1 10 1
11 4 12 2 13 3 14 2 15 4 16 4 17 1 18 5 19 2 20 4
21 3 22 3 23 2 24 1 25 4 26 1 27 4 28 5 29 5 30 1
31 4 32 3 33 5 34 2 35 3 36 5 37 1 38 2 39 5 40 1
41 * 42 3 43 1 44 4 45 2 46 4 47 2 48 3 49 2 50 4
51 5 52 2 53 4 54 1 55 5 56 3 57 2 58 1 59 4 60 5
61 1 62 2 63 4 64 3 65 5 66 4 67 5 68 2 69 3 70 2
71 3 72 4 73 1 74 5 75 3
	
1. 2 Sum of the odd integers in the set S
= () ()
n
23 n–1 2
2
×+ ×
() ( )
n
2n 4 n n 2
2
=+=×+
Therefore, the average of the odd integers in set S
= n + 2
Sum of the even integers in the set S
= () ()
n
22 n –12
2
×+ ×
() ( )
n
2n 2 n n 1
2
=+=×+
Therefore, the average of the even integers in the set
S = n + 1
Hence, X – Y = (n + 2) – (n + 1) = 1
2. 5 The total age of all the eight people in the family = 231
As per the information given in the question, the total
age of all the people in the family = 231 + 3 × 8 – 60 +
0 = 195
Similarly, the total age of the people in the family four
years ago = 195 + 3 × 8 – 60 + 0 = 159.
Therefore, the current average age of all the people in
the family = 
159 32
24 years.
8
+
=
3. 1 f(1) + f(2) + f(3) + …. + f(n) = 
2
nf(n)
, f(1) = 3600.
For n = 2,
? f(1) + f(2) = 
2
2f(2)
 ? f(2) = 
2
f(1)
(2 – 1)
For n = 3,
( )
2
2
1
3600 1 f(3) 3 f(3)
2 – 1
??
??
++=
??
??
??
2
22
21
f(3) 3600
2 –13 – 1
??
??
?? ?= × ×
??
??
??
??
Similarly,
( )( )( ) ( )
22 2 2
22 2 2
23 4...8
f(9) 3600
2 –13 –14 –1... 9 – 1
×× ×
=×
Therefore, f(9) = 80
4. 3 Let the number of currency 1 Miso, 10 Misos and 50
Misos be x, y and z respectively.
? x + 10y + 50z = 107
Now the possible values of z could be 0, 1 and 2.
For z = 0: x + 10y = 107
Number of integral pairs of values of x and y that
satisfy the equation x + 10y = 107 will be 11. These
values of x and y in that order are
(7, 10); (17, 9); (27, 8)…
(107, 0).
For z = 1: x + 10y = 57
Number of integral pairs of values of x and y that
satisfy the equation x  + 10y = 57 will be 6. These
values of x and y in that order are (7, 5); (17, 4); (27,
3);
(37, 2); (47, 1) and (57, 0).
For z = 2: x  + 10y = 7
There is only one integer value of x and y that satisfies
the equation x  + 10y = 7 in that order is (7, 0).
Therefore, total number of ways in which you can
pay a bill of 107 Misos = 11 + 6 + 1 = 18
5. 4 Suppose the cheque for Shailaja is of Rs. X and Y
paise
As per the question,
3 × (100X + Y) = (100Y + X) – 50
? 299X = 97Y – 50
299X 50
Y
97
+
?=
Now the value of Y should be a integer.
Checking by options only for X = 18, Y is a integer and
the value of Y = 56
6. 5
14 1
,n 60
mn 12
+= <
11 4 n – 48
–
m12 n 12n
?= =
12n
m
n – 48
?=
Positive integral values of m for odd integral values of
n are for n = 49, 51 and 57.
Therefore, there are 3 integral pairs of values of m
and n that satisfy the given equation.
7. 3 Using A: 
II I
W 45.5 and W 44.5 ==
Using B: Weight of Deepak = 70kg (Only after using
statement A)
This is sufficient to find weight of Poonam using the
data given in the question statement. Hence option (3)
is correct choice.
8. 2 Using A: Inner radius of the tank is atleast 4 m. So
volume 
3
4
r where 4 r 10
3
=p < <
This volume can be greater as well as smaller than
400 for different r.
Using B: The given data gives the volume of the
material of tank, which can be expressed as
33
4
(10 r ),
3
p-
which will give the value of r which is
unique and sufficient to judge if the capacity is
adequate. Hence option (2) is correct choice.
Page 3


	


QA  1 to 25 25
LRDI 26 to 50 25
EU + RC 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
1 2 2 5 3 1 4 3 5 4 6 5 7 3 8 2 9 1 10 1
11 4 12 2 13 3 14 2 15 4 16 4 17 1 18 5 19 2 20 4
21 3 22 3 23 2 24 1 25 4 26 1 27 4 28 5 29 5 30 1
31 4 32 3 33 5 34 2 35 3 36 5 37 1 38 2 39 5 40 1
41 * 42 3 43 1 44 4 45 2 46 4 47 2 48 3 49 2 50 4
51 5 52 2 53 4 54 1 55 5 56 3 57 2 58 1 59 4 60 5
61 1 62 2 63 4 64 3 65 5 66 4 67 5 68 2 69 3 70 2
71 3 72 4 73 1 74 5 75 3
	
1. 2 Sum of the odd integers in the set S
= () ()
n
23 n–1 2
2
×+ ×
() ( )
n
2n 4 n n 2
2
=+=×+
Therefore, the average of the odd integers in set S
= n + 2
Sum of the even integers in the set S
= () ()
n
22 n –12
2
×+ ×
() ( )
n
2n 2 n n 1
2
=+=×+
Therefore, the average of the even integers in the set
S = n + 1
Hence, X – Y = (n + 2) – (n + 1) = 1
2. 5 The total age of all the eight people in the family = 231
As per the information given in the question, the total
age of all the people in the family = 231 + 3 × 8 – 60 +
0 = 195
Similarly, the total age of the people in the family four
years ago = 195 + 3 × 8 – 60 + 0 = 159.
Therefore, the current average age of all the people in
the family = 
159 32
24 years.
8
+
=
3. 1 f(1) + f(2) + f(3) + …. + f(n) = 
2
nf(n)
, f(1) = 3600.
For n = 2,
? f(1) + f(2) = 
2
2f(2)
 ? f(2) = 
2
f(1)
(2 – 1)
For n = 3,
( )
2
2
1
3600 1 f(3) 3 f(3)
2 – 1
??
??
++=
??
??
??
2
22
21
f(3) 3600
2 –13 – 1
??
??
?? ?= × ×
??
??
??
??
Similarly,
( )( )( ) ( )
22 2 2
22 2 2
23 4...8
f(9) 3600
2 –13 –14 –1... 9 – 1
×× ×
=×
Therefore, f(9) = 80
4. 3 Let the number of currency 1 Miso, 10 Misos and 50
Misos be x, y and z respectively.
? x + 10y + 50z = 107
Now the possible values of z could be 0, 1 and 2.
For z = 0: x + 10y = 107
Number of integral pairs of values of x and y that
satisfy the equation x + 10y = 107 will be 11. These
values of x and y in that order are
(7, 10); (17, 9); (27, 8)…
(107, 0).
For z = 1: x + 10y = 57
Number of integral pairs of values of x and y that
satisfy the equation x  + 10y = 57 will be 6. These
values of x and y in that order are (7, 5); (17, 4); (27,
3);
(37, 2); (47, 1) and (57, 0).
For z = 2: x  + 10y = 7
There is only one integer value of x and y that satisfies
the equation x  + 10y = 7 in that order is (7, 0).
Therefore, total number of ways in which you can
pay a bill of 107 Misos = 11 + 6 + 1 = 18
5. 4 Suppose the cheque for Shailaja is of Rs. X and Y
paise
As per the question,
3 × (100X + Y) = (100Y + X) – 50
? 299X = 97Y – 50
299X 50
Y
97
+
?=
Now the value of Y should be a integer.
Checking by options only for X = 18, Y is a integer and
the value of Y = 56
6. 5
14 1
,n 60
mn 12
+= <
11 4 n – 48
–
m12 n 12n
?= =
12n
m
n – 48
?=
Positive integral values of m for odd integral values of
n are for n = 49, 51 and 57.
Therefore, there are 3 integral pairs of values of m
and n that satisfy the given equation.
7. 3 Using A: 
II I
W 45.5 and W 44.5 ==
Using B: Weight of Deepak = 70kg (Only after using
statement A)
This is sufficient to find weight of Poonam using the
data given in the question statement. Hence option (3)
is correct choice.
8. 2 Using A: Inner radius of the tank is atleast 4 m. So
volume 
3
4
r where 4 r 10
3
=p < <
This volume can be greater as well as smaller than
400 for different r.
Using B: The given data gives the volume of the
material of tank, which can be expressed as
33
4
(10 r ),
3
p-
which will give the value of r which is
unique and sufficient to judge if the capacity is
adequate. Hence option (2) is correct choice.
9. 1 Using A: x = 30, y = 30 and z = 29 will give the
minimum value.
Using B: Nothing specific can be said about the relation
between x, y and z.
Hence option (1) is correct choice.
10. 1 Using A: 
OM 2
OL 1
=
But if O lies on JK, maximum possible value of 
OM
OL
 is
2
1
 (when O lies on K)
So, Rahim is unable to draw such a square
Using B: Nothing specific can be said about the
dimensions of the figure.
Hence option (1) is correct choice.
For questions 11 and 12:
Let the cruising speed of the plane and the time difference
between A and B be y km/hr and x hours respectively.
Distance between A and B = 3000 kilometers. For, the plane
moving from city A to City B: 3000 =  (7 – x) × (y – 50). This is
satisfied for x = 1 and y = 550. These are the only values given
in the options that satisfy the above equation.
11. 4 12. 2
For questions 13 and 14:
To maximise Shabnam’s return we need to evaluate all the
given options in the question number 7. Assume Shabnam had
one rupee to invest. Let the return be denoted by ‘r’.
Consider the option (30% in option A, 32% in option B
and 38% in  option  C): If the stock market rises, then
r = 0.1 × 0.3 + 5 × 0.32 – 2.5 × 0.38 = 0.653
If the stock market falls, then
r = 0.1 × 0.3 – 3 × 0.32 + 2 × 0.38 = – 0.197
Consider option (100% in option A): This will give a return
of 0.1%.
Consider option (36% in option B and 64% in option C):
If the stock market rises, then
r = 5 × 0.36 – 2.5 × 0.64 = 0.2
If the stock market falls, then
r = – 3 × 0.36 + 2 × 0.64 = 0.2
Consider option (64% in option B and 36% in option C):
If the stock market rises, then
r = 5 × 0.64 – 2.5 × 0.36 = 2.1
If the stock market falls, then
r = – 3 × 0.64 + 2 × 0.36 = –1.2
Consider option (1/3 in each of the 3 options): If the
stock market rises, then
r = 0.1 × 0.33 + 5 × 0.33 – 2.5 × 0.33 = 0.858
If the stock market falls, then
r = 0.1 × 0.33 – 3 × 0.33 + 2 × 0.33 = –0.297
We can see that only in option (36% in option B and 64% in
option C), Shabnam gets an assured return of 0.2%
irrespective of the behaviour of the stock market. So right
option for questions number 13 is (0.20%) and question number
14 is (36% in option B and 64% in option C).
13. 3 14. 2
For questions 15 and 16:
15. 4 The number of members in the set S = 
n
C
2
, where n is
greater than = 4
Each member of S has two distinct numbers.
Let us say (1, 2) is one of the members of S.
To find the number of enemies each member of S will
have be equal to
2
n–2
2
n – 5n + 6
C = 
2
16. 4 Considering any two members of S, that are friends
there will be 1 number of the pairs that will be common.
The common element of these pairs will have n – 3
pairs, with the remaining  n – 3 elements. There will be
one more member made up of the remaining two
constituent elements which are not same. In total there
are n – 3 + 1 = n – 2 other members of S that are
common friends of the chosen two pairs or numbers.
Alternative Method for questions 15 and 16:
For n = 6, the number of elements in the set S = {(1, 2), (1, 3),
(1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3,
6), (4, 5), (4, 6) and (5, 6)
Lets consider the member (1, 2).
15. 4 Number of enemies for this member is 6, i.e. (3, 4), (3,
5), (3, 6), (4, 5) (4, 6) and (5, 6).
Checking by options, this is only satisfied by
2
n – 5n + 6
2
Hence 
2
n – 5n + 6
2
is the correct choice.
16. 4 For n = 6 lets consider the members (1, 2) and (1, 3)
Friends of the member (1, 2) in the set S are (1, 4), (1,
5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6).
Friends of the member (1, 3) in the set S (1, 4), (1, 5),
(1, 6), (2, 3), (3, 4), (3, 5), (3, 6).
The number of members of S that are common friends
to the above member are 4, i.e. (1, 4), (1, 5), (1, 6),
(2, 3).
So the answer is n – 2.
17. 1 In each team, T
j
 there are two players, one it shares
with T
j – 1
 and other with
T
j – 1
. Other (k – 2) players team T
j
 shares with no
other team. So, total players which play for only one
team.
= (k – 2)n
One player is common in T
1
 and T
2
, one in T
2
 and T
3
and so on.
Number of such players = number of pairs = n
So, total players = (k – 2)n + n = n(k – 1)
Page 4


	


QA  1 to 25 25
LRDI 26 to 50 25
EU + RC 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
1 2 2 5 3 1 4 3 5 4 6 5 7 3 8 2 9 1 10 1
11 4 12 2 13 3 14 2 15 4 16 4 17 1 18 5 19 2 20 4
21 3 22 3 23 2 24 1 25 4 26 1 27 4 28 5 29 5 30 1
31 4 32 3 33 5 34 2 35 3 36 5 37 1 38 2 39 5 40 1
41 * 42 3 43 1 44 4 45 2 46 4 47 2 48 3 49 2 50 4
51 5 52 2 53 4 54 1 55 5 56 3 57 2 58 1 59 4 60 5
61 1 62 2 63 4 64 3 65 5 66 4 67 5 68 2 69 3 70 2
71 3 72 4 73 1 74 5 75 3
	
1. 2 Sum of the odd integers in the set S
= () ()
n
23 n–1 2
2
×+ ×
() ( )
n
2n 4 n n 2
2
=+=×+
Therefore, the average of the odd integers in set S
= n + 2
Sum of the even integers in the set S
= () ()
n
22 n –12
2
×+ ×
() ( )
n
2n 2 n n 1
2
=+=×+
Therefore, the average of the even integers in the set
S = n + 1
Hence, X – Y = (n + 2) – (n + 1) = 1
2. 5 The total age of all the eight people in the family = 231
As per the information given in the question, the total
age of all the people in the family = 231 + 3 × 8 – 60 +
0 = 195
Similarly, the total age of the people in the family four
years ago = 195 + 3 × 8 – 60 + 0 = 159.
Therefore, the current average age of all the people in
the family = 
159 32
24 years.
8
+
=
3. 1 f(1) + f(2) + f(3) + …. + f(n) = 
2
nf(n)
, f(1) = 3600.
For n = 2,
? f(1) + f(2) = 
2
2f(2)
 ? f(2) = 
2
f(1)
(2 – 1)
For n = 3,
( )
2
2
1
3600 1 f(3) 3 f(3)
2 – 1
??
??
++=
??
??
??
2
22
21
f(3) 3600
2 –13 – 1
??
??
?? ?= × ×
??
??
??
??
Similarly,
( )( )( ) ( )
22 2 2
22 2 2
23 4...8
f(9) 3600
2 –13 –14 –1... 9 – 1
×× ×
=×
Therefore, f(9) = 80
4. 3 Let the number of currency 1 Miso, 10 Misos and 50
Misos be x, y and z respectively.
? x + 10y + 50z = 107
Now the possible values of z could be 0, 1 and 2.
For z = 0: x + 10y = 107
Number of integral pairs of values of x and y that
satisfy the equation x + 10y = 107 will be 11. These
values of x and y in that order are
(7, 10); (17, 9); (27, 8)…
(107, 0).
For z = 1: x + 10y = 57
Number of integral pairs of values of x and y that
satisfy the equation x  + 10y = 57 will be 6. These
values of x and y in that order are (7, 5); (17, 4); (27,
3);
(37, 2); (47, 1) and (57, 0).
For z = 2: x  + 10y = 7
There is only one integer value of x and y that satisfies
the equation x  + 10y = 7 in that order is (7, 0).
Therefore, total number of ways in which you can
pay a bill of 107 Misos = 11 + 6 + 1 = 18
5. 4 Suppose the cheque for Shailaja is of Rs. X and Y
paise
As per the question,
3 × (100X + Y) = (100Y + X) – 50
? 299X = 97Y – 50
299X 50
Y
97
+
?=
Now the value of Y should be a integer.
Checking by options only for X = 18, Y is a integer and
the value of Y = 56
6. 5
14 1
,n 60
mn 12
+= <
11 4 n – 48
–
m12 n 12n
?= =
12n
m
n – 48
?=
Positive integral values of m for odd integral values of
n are for n = 49, 51 and 57.
Therefore, there are 3 integral pairs of values of m
and n that satisfy the given equation.
7. 3 Using A: 
II I
W 45.5 and W 44.5 ==
Using B: Weight of Deepak = 70kg (Only after using
statement A)
This is sufficient to find weight of Poonam using the
data given in the question statement. Hence option (3)
is correct choice.
8. 2 Using A: Inner radius of the tank is atleast 4 m. So
volume 
3
4
r where 4 r 10
3
=p < <
This volume can be greater as well as smaller than
400 for different r.
Using B: The given data gives the volume of the
material of tank, which can be expressed as
33
4
(10 r ),
3
p-
which will give the value of r which is
unique and sufficient to judge if the capacity is
adequate. Hence option (2) is correct choice.
9. 1 Using A: x = 30, y = 30 and z = 29 will give the
minimum value.
Using B: Nothing specific can be said about the relation
between x, y and z.
Hence option (1) is correct choice.
10. 1 Using A: 
OM 2
OL 1
=
But if O lies on JK, maximum possible value of 
OM
OL
 is
2
1
 (when O lies on K)
So, Rahim is unable to draw such a square
Using B: Nothing specific can be said about the
dimensions of the figure.
Hence option (1) is correct choice.
For questions 11 and 12:
Let the cruising speed of the plane and the time difference
between A and B be y km/hr and x hours respectively.
Distance between A and B = 3000 kilometers. For, the plane
moving from city A to City B: 3000 =  (7 – x) × (y – 50). This is
satisfied for x = 1 and y = 550. These are the only values given
in the options that satisfy the above equation.
11. 4 12. 2
For questions 13 and 14:
To maximise Shabnam’s return we need to evaluate all the
given options in the question number 7. Assume Shabnam had
one rupee to invest. Let the return be denoted by ‘r’.
Consider the option (30% in option A, 32% in option B
and 38% in  option  C): If the stock market rises, then
r = 0.1 × 0.3 + 5 × 0.32 – 2.5 × 0.38 = 0.653
If the stock market falls, then
r = 0.1 × 0.3 – 3 × 0.32 + 2 × 0.38 = – 0.197
Consider option (100% in option A): This will give a return
of 0.1%.
Consider option (36% in option B and 64% in option C):
If the stock market rises, then
r = 5 × 0.36 – 2.5 × 0.64 = 0.2
If the stock market falls, then
r = – 3 × 0.36 + 2 × 0.64 = 0.2
Consider option (64% in option B and 36% in option C):
If the stock market rises, then
r = 5 × 0.64 – 2.5 × 0.36 = 2.1
If the stock market falls, then
r = – 3 × 0.64 + 2 × 0.36 = –1.2
Consider option (1/3 in each of the 3 options): If the
stock market rises, then
r = 0.1 × 0.33 + 5 × 0.33 – 2.5 × 0.33 = 0.858
If the stock market falls, then
r = 0.1 × 0.33 – 3 × 0.33 + 2 × 0.33 = –0.297
We can see that only in option (36% in option B and 64% in
option C), Shabnam gets an assured return of 0.2%
irrespective of the behaviour of the stock market. So right
option for questions number 13 is (0.20%) and question number
14 is (36% in option B and 64% in option C).
13. 3 14. 2
For questions 15 and 16:
15. 4 The number of members in the set S = 
n
C
2
, where n is
greater than = 4
Each member of S has two distinct numbers.
Let us say (1, 2) is one of the members of S.
To find the number of enemies each member of S will
have be equal to
2
n–2
2
n – 5n + 6
C = 
2
16. 4 Considering any two members of S, that are friends
there will be 1 number of the pairs that will be common.
The common element of these pairs will have n – 3
pairs, with the remaining  n – 3 elements. There will be
one more member made up of the remaining two
constituent elements which are not same. In total there
are n – 3 + 1 = n – 2 other members of S that are
common friends of the chosen two pairs or numbers.
Alternative Method for questions 15 and 16:
For n = 6, the number of elements in the set S = {(1, 2), (1, 3),
(1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3,
6), (4, 5), (4, 6) and (5, 6)
Lets consider the member (1, 2).
15. 4 Number of enemies for this member is 6, i.e. (3, 4), (3,
5), (3, 6), (4, 5) (4, 6) and (5, 6).
Checking by options, this is only satisfied by
2
n – 5n + 6
2
Hence 
2
n – 5n + 6
2
is the correct choice.
16. 4 For n = 6 lets consider the members (1, 2) and (1, 3)
Friends of the member (1, 2) in the set S are (1, 4), (1,
5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6).
Friends of the member (1, 3) in the set S (1, 4), (1, 5),
(1, 6), (2, 3), (3, 4), (3, 5), (3, 6).
The number of members of S that are common friends
to the above member are 4, i.e. (1, 4), (1, 5), (1, 6),
(2, 3).
So the answer is n – 2.
17. 1 In each team, T
j
 there are two players, one it shares
with T
j – 1
 and other with
T
j – 1
. Other (k – 2) players team T
j
 shares with no
other team. So, total players which play for only one
team.
= (k – 2)n
One player is common in T
1
 and T
2
, one in T
2
 and T
3
and so on.
Number of such players = number of pairs = n
So, total players = (k – 2)n + n = n(k – 1)
18. 5 Let the four-digit number be denoted by aabb = 11 ×
(100a + b).
Now since aabb is a perfect square, 100a + b should
be a multiple of 11.
The only pairs of values of a and b that satisfy the
above mentioned condition is a = 7 and  b = 4. Clearly
7744 is a perfect square.
19. 2 Using the given data –
( ) ( )
22
2
240 40b 40 c – 240 20b 20 c
2
3
240 20b 20 c
++ + +
=
++
and
( ) ( )
22
2
240 60b 60 c – 240 40b 40 c
1
2
240 40b 40 c
++ + +
=
++
Solving the above equations, 
1
candb10
10
==
So cost for producing x units = 
2
x
240 10x
10
++
Profit earned from x units
= 
22
xx
30x – 240 10x 20x–– 240
10 10
??
?? ++ =
??
??
()
2 1
760 x 100
10
=- -
For maximum profit, x – 100 = 0
? x = 100.
20. 4 Maximum profit = Rs. 760
21. 3 Price of Darjeeling tea (in rupees per kilo gram) is 100
+ 0.10n
Price of Ooty tea (in rupees per kilo gram) is 89 +
0.15n
Price of the Darjeeling tea on the 100
th
 day = 100 + 0.1
× 100 = 110 89 0.15n 110 n 140 ?+ = ?=
Number of days in the months of January, February,
March and April in the year 2007 = 31 + 28 + 31 + 30 =
120.
Therefore, the price of both the tea will be equal on
20
th
 May.
22. 3
P Q
B
A
If P and Q lie on the intersections of the circles as
shown in the figure given below.
P Q
B
A
In this case triangle APQ is equilateral. So the maximum
possible measure of the angle AQP is 60°. The answer
is between 0 and 60.
23. 2 Let f(x) = 
2
ax bx c ++
At x  = 1, f(1) = a + b + c = 3
At x = 0, f(0) = c = 1
The maximum of the function f(x) is attained at
x = 
b
–
2a
 = 1 = 
a – 2
2a
? a = –2 and b = 4
? f(x) = 
2
–2x + 4x + 1
Therefore, f(10) = –159
For questions 24 and 25:
Using the given expressions —
11
2
22
22
33
22 3
44
32 2 3
55
33 2 4
66
ap b q
apq b q
apq b pq
apq bpq
apq b pq
apq bpq
==
==
==
==
==
==
and so on
24. 1 ()
nn n n
–11
22 2 2
nn
ab niseven pq p q
+
+=+
() ()
n–1
2
qpq p q =+
Page 5


	


QA  1 to 25 25
LRDI 26 to 50 25
EU + RC 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
1 2 2 5 3 1 4 3 5 4 6 5 7 3 8 2 9 1 10 1
11 4 12 2 13 3 14 2 15 4 16 4 17 1 18 5 19 2 20 4
21 3 22 3 23 2 24 1 25 4 26 1 27 4 28 5 29 5 30 1
31 4 32 3 33 5 34 2 35 3 36 5 37 1 38 2 39 5 40 1
41 * 42 3 43 1 44 4 45 2 46 4 47 2 48 3 49 2 50 4
51 5 52 2 53 4 54 1 55 5 56 3 57 2 58 1 59 4 60 5
61 1 62 2 63 4 64 3 65 5 66 4 67 5 68 2 69 3 70 2
71 3 72 4 73 1 74 5 75 3
	
1. 2 Sum of the odd integers in the set S
= () ()
n
23 n–1 2
2
×+ ×
() ( )
n
2n 4 n n 2
2
=+=×+
Therefore, the average of the odd integers in set S
= n + 2
Sum of the even integers in the set S
= () ()
n
22 n –12
2
×+ ×
() ( )
n
2n 2 n n 1
2
=+=×+
Therefore, the average of the even integers in the set
S = n + 1
Hence, X – Y = (n + 2) – (n + 1) = 1
2. 5 The total age of all the eight people in the family = 231
As per the information given in the question, the total
age of all the people in the family = 231 + 3 × 8 – 60 +
0 = 195
Similarly, the total age of the people in the family four
years ago = 195 + 3 × 8 – 60 + 0 = 159.
Therefore, the current average age of all the people in
the family = 
159 32
24 years.
8
+
=
3. 1 f(1) + f(2) + f(3) + …. + f(n) = 
2
nf(n)
, f(1) = 3600.
For n = 2,
? f(1) + f(2) = 
2
2f(2)
 ? f(2) = 
2
f(1)
(2 – 1)
For n = 3,
( )
2
2
1
3600 1 f(3) 3 f(3)
2 – 1
??
??
++=
??
??
??
2
22
21
f(3) 3600
2 –13 – 1
??
??
?? ?= × ×
??
??
??
??
Similarly,
( )( )( ) ( )
22 2 2
22 2 2
23 4...8
f(9) 3600
2 –13 –14 –1... 9 – 1
×× ×
=×
Therefore, f(9) = 80
4. 3 Let the number of currency 1 Miso, 10 Misos and 50
Misos be x, y and z respectively.
? x + 10y + 50z = 107
Now the possible values of z could be 0, 1 and 2.
For z = 0: x + 10y = 107
Number of integral pairs of values of x and y that
satisfy the equation x + 10y = 107 will be 11. These
values of x and y in that order are
(7, 10); (17, 9); (27, 8)…
(107, 0).
For z = 1: x + 10y = 57
Number of integral pairs of values of x and y that
satisfy the equation x  + 10y = 57 will be 6. These
values of x and y in that order are (7, 5); (17, 4); (27,
3);
(37, 2); (47, 1) and (57, 0).
For z = 2: x  + 10y = 7
There is only one integer value of x and y that satisfies
the equation x  + 10y = 7 in that order is (7, 0).
Therefore, total number of ways in which you can
pay a bill of 107 Misos = 11 + 6 + 1 = 18
5. 4 Suppose the cheque for Shailaja is of Rs. X and Y
paise
As per the question,
3 × (100X + Y) = (100Y + X) – 50
? 299X = 97Y – 50
299X 50
Y
97
+
?=
Now the value of Y should be a integer.
Checking by options only for X = 18, Y is a integer and
the value of Y = 56
6. 5
14 1
,n 60
mn 12
+= <
11 4 n – 48
–
m12 n 12n
?= =
12n
m
n – 48
?=
Positive integral values of m for odd integral values of
n are for n = 49, 51 and 57.
Therefore, there are 3 integral pairs of values of m
and n that satisfy the given equation.
7. 3 Using A: 
II I
W 45.5 and W 44.5 ==
Using B: Weight of Deepak = 70kg (Only after using
statement A)
This is sufficient to find weight of Poonam using the
data given in the question statement. Hence option (3)
is correct choice.
8. 2 Using A: Inner radius of the tank is atleast 4 m. So
volume 
3
4
r where 4 r 10
3
=p < <
This volume can be greater as well as smaller than
400 for different r.
Using B: The given data gives the volume of the
material of tank, which can be expressed as
33
4
(10 r ),
3
p-
which will give the value of r which is
unique and sufficient to judge if the capacity is
adequate. Hence option (2) is correct choice.
9. 1 Using A: x = 30, y = 30 and z = 29 will give the
minimum value.
Using B: Nothing specific can be said about the relation
between x, y and z.
Hence option (1) is correct choice.
10. 1 Using A: 
OM 2
OL 1
=
But if O lies on JK, maximum possible value of 
OM
OL
 is
2
1
 (when O lies on K)
So, Rahim is unable to draw such a square
Using B: Nothing specific can be said about the
dimensions of the figure.
Hence option (1) is correct choice.
For questions 11 and 12:
Let the cruising speed of the plane and the time difference
between A and B be y km/hr and x hours respectively.
Distance between A and B = 3000 kilometers. For, the plane
moving from city A to City B: 3000 =  (7 – x) × (y – 50). This is
satisfied for x = 1 and y = 550. These are the only values given
in the options that satisfy the above equation.
11. 4 12. 2
For questions 13 and 14:
To maximise Shabnam’s return we need to evaluate all the
given options in the question number 7. Assume Shabnam had
one rupee to invest. Let the return be denoted by ‘r’.
Consider the option (30% in option A, 32% in option B
and 38% in  option  C): If the stock market rises, then
r = 0.1 × 0.3 + 5 × 0.32 – 2.5 × 0.38 = 0.653
If the stock market falls, then
r = 0.1 × 0.3 – 3 × 0.32 + 2 × 0.38 = – 0.197
Consider option (100% in option A): This will give a return
of 0.1%.
Consider option (36% in option B and 64% in option C):
If the stock market rises, then
r = 5 × 0.36 – 2.5 × 0.64 = 0.2
If the stock market falls, then
r = – 3 × 0.36 + 2 × 0.64 = 0.2
Consider option (64% in option B and 36% in option C):
If the stock market rises, then
r = 5 × 0.64 – 2.5 × 0.36 = 2.1
If the stock market falls, then
r = – 3 × 0.64 + 2 × 0.36 = –1.2
Consider option (1/3 in each of the 3 options): If the
stock market rises, then
r = 0.1 × 0.33 + 5 × 0.33 – 2.5 × 0.33 = 0.858
If the stock market falls, then
r = 0.1 × 0.33 – 3 × 0.33 + 2 × 0.33 = –0.297
We can see that only in option (36% in option B and 64% in
option C), Shabnam gets an assured return of 0.2%
irrespective of the behaviour of the stock market. So right
option for questions number 13 is (0.20%) and question number
14 is (36% in option B and 64% in option C).
13. 3 14. 2
For questions 15 and 16:
15. 4 The number of members in the set S = 
n
C
2
, where n is
greater than = 4
Each member of S has two distinct numbers.
Let us say (1, 2) is one of the members of S.
To find the number of enemies each member of S will
have be equal to
2
n–2
2
n – 5n + 6
C = 
2
16. 4 Considering any two members of S, that are friends
there will be 1 number of the pairs that will be common.
The common element of these pairs will have n – 3
pairs, with the remaining  n – 3 elements. There will be
one more member made up of the remaining two
constituent elements which are not same. In total there
are n – 3 + 1 = n – 2 other members of S that are
common friends of the chosen two pairs or numbers.
Alternative Method for questions 15 and 16:
For n = 6, the number of elements in the set S = {(1, 2), (1, 3),
(1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3,
6), (4, 5), (4, 6) and (5, 6)
Lets consider the member (1, 2).
15. 4 Number of enemies for this member is 6, i.e. (3, 4), (3,
5), (3, 6), (4, 5) (4, 6) and (5, 6).
Checking by options, this is only satisfied by
2
n – 5n + 6
2
Hence 
2
n – 5n + 6
2
is the correct choice.
16. 4 For n = 6 lets consider the members (1, 2) and (1, 3)
Friends of the member (1, 2) in the set S are (1, 4), (1,
5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6).
Friends of the member (1, 3) in the set S (1, 4), (1, 5),
(1, 6), (2, 3), (3, 4), (3, 5), (3, 6).
The number of members of S that are common friends
to the above member are 4, i.e. (1, 4), (1, 5), (1, 6),
(2, 3).
So the answer is n – 2.
17. 1 In each team, T
j
 there are two players, one it shares
with T
j – 1
 and other with
T
j – 1
. Other (k – 2) players team T
j
 shares with no
other team. So, total players which play for only one
team.
= (k – 2)n
One player is common in T
1
 and T
2
, one in T
2
 and T
3
and so on.
Number of such players = number of pairs = n
So, total players = (k – 2)n + n = n(k – 1)
18. 5 Let the four-digit number be denoted by aabb = 11 ×
(100a + b).
Now since aabb is a perfect square, 100a + b should
be a multiple of 11.
The only pairs of values of a and b that satisfy the
above mentioned condition is a = 7 and  b = 4. Clearly
7744 is a perfect square.
19. 2 Using the given data –
( ) ( )
22
2
240 40b 40 c – 240 20b 20 c
2
3
240 20b 20 c
++ + +
=
++
and
( ) ( )
22
2
240 60b 60 c – 240 40b 40 c
1
2
240 40b 40 c
++ + +
=
++
Solving the above equations, 
1
candb10
10
==
So cost for producing x units = 
2
x
240 10x
10
++
Profit earned from x units
= 
22
xx
30x – 240 10x 20x–– 240
10 10
??
?? ++ =
??
??
()
2 1
760 x 100
10
=- -
For maximum profit, x – 100 = 0
? x = 100.
20. 4 Maximum profit = Rs. 760
21. 3 Price of Darjeeling tea (in rupees per kilo gram) is 100
+ 0.10n
Price of Ooty tea (in rupees per kilo gram) is 89 +
0.15n
Price of the Darjeeling tea on the 100
th
 day = 100 + 0.1
× 100 = 110 89 0.15n 110 n 140 ?+ = ?=
Number of days in the months of January, February,
March and April in the year 2007 = 31 + 28 + 31 + 30 =
120.
Therefore, the price of both the tea will be equal on
20
th
 May.
22. 3
P Q
B
A
If P and Q lie on the intersections of the circles as
shown in the figure given below.
P Q
B
A
In this case triangle APQ is equilateral. So the maximum
possible measure of the angle AQP is 60°. The answer
is between 0 and 60.
23. 2 Let f(x) = 
2
ax bx c ++
At x  = 1, f(1) = a + b + c = 3
At x = 0, f(0) = c = 1
The maximum of the function f(x) is attained at
x = 
b
–
2a
 = 1 = 
a – 2
2a
? a = –2 and b = 4
? f(x) = 
2
–2x + 4x + 1
Therefore, f(10) = –159
For questions 24 and 25:
Using the given expressions —
11
2
22
22
33
22 3
44
32 2 3
55
33 2 4
66
ap b q
apq b q
apq b pq
apq bpq
apq b pq
apq bpq
==
==
==
==
==
==
and so on
24. 1 ()
nn n n
–11
22 2 2
nn
ab niseven pq p q
+
+=+
() ()
n–1
2
qpq p q =+
25. 4 ()
n 1 n1 n1 n 1
22 22
nn
a b n is odd p q p q
+- - +
+= +
()()
n–1
2
pq pq =+
Substituting 
12
p and q
33
==
n1
2
nn
2
ab
9
-
??
+=
??
??
Substituting n = 7, 
nn
ab 0.01 +>
Substituting n = 9, 
nn
a b 0.01 +<
Hence smallest value of n is 9
26. 1 The diet should contain 10% minerals and only two
ingredient contain 10% minerals namely O and Q.
Hence, only by mixing O and Q, a diet with 10% minerals
can be formed.
Hence, there is only one way.
27. 4 None of the choices among (1), (2) and (3) can be
used to form the diet with 10% fat and atleast 30%
protein. For Q and S to form the diet with 10% fat and
at least 30% protein, let us suppose that they are
mixed in x : y ration. Then,
() () x50 y0
10
xy
+
=
+
? x : y = 1 : 4
Cost per unit 
() ( ) 1 200 4 100
Rs.120
5
+
==
Similarly, for R and S, cost per unit = Rs. 200
?Cost per unit is lowest for Q and S.
28. 5 To make a diet with P, Q and S having atleast 60%
carbohydrates, the proportion of P should be the
maximum and the other two should be minimum to
get the lowest per unit cost. Options (2) and (5)
satisfies this but the lowest cost per unit can be
achieved when P, Q and S are mixed in the
proportion 4 : 1 : 1.
29. 5 As the ingredients are mixed in equal amounts, so
we can take the average of the constituent
percentage of the elements used.
Only option (5) satisfies all the conditions.
30. 1 From statement A, it is clear that 40% of top academic
performers are athletes and that is equal to 10. So
total number of academic performers can be calculated.
Statement B does not provide any relevant information.
So the answer is (1).
31. 4 Statement A and B alone are not sufficient but if both
are combined, then we can form the following
sequence:
1 2 3 4 5 
D E B C A 
So the answer is (4).
32. 3 Statement A alone is sufficient because 10% of the
female employees have engineering background, 70%
of the employees are females, so 7% of the employees
are female and having engineering background. Hence,
18% of theemployees are male and having
engineering background. From statement B, we know
the number of male employees having engineering
background. So, the percentage of male employees
having engineering background can be calculated. So,
the answer is (3)
33. 5 Statement A alone is not sufficient because it is not
giving any information about the opponent. Statement
B alone is also not sufficient because it is not giving
any information regarding the performance of Mahindra
& Mahindra in the second half. Even if both the
statements are used together, we will get two cases:
M & M 0 1 
Opponent 3 4 
So in one case, match is drawn and in the other case,
it is won by Mahindra & Mahindra.
Hence, the answer is (5)
For questions 34 to 37:
Looking at the values in the table one can easily conclude that
the costs which are directly proportional to the change in
volume of proportion are ‘Material’, ‘Labour’ and ‘Operating
cost of machines’. Rest of the costs are all fixed costs. If ‘x’ is
the number of units produced in 2007, then the total cost of
production would be
C = 9600 (Fixed cost) + 100x (Variable cost),
Variable cost = 100x because as the number of units for 2006
is 1200 and variable cost for that is 120000 i.e. 100 times the
number of units.
34. 2 Total cost = 9600 + 100 × 1400 = 149600
Cost per unit = ()
149600
107 approx.
1400
=
35. 3 To avoid any loss the total selling price should be
equal to the total cost price. If ‘x’ units are produced
and selling price of each unit is 125 Rs.
Therefore, 125x = 9600 + 100x
25x = 9600
x 384 ?=
Hence, 384 units should be produced.
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