CAT Past Year Question Paper Solution - 2005 | Additional Study Material for CAT PDF Download

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QA  1 to 30 30
EU + RC 31 to 60 30
DI + DS + AR 61 to 90 30
T otal 90
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1 1 2 3 3 2 4 4 5 1 6 2 7 2 8 4 9 4 10 3
11 4 12 1 13 2 14 3 15 2 16 1 17 2 18 4 19 4 20 3
21 3 22 2 23 1 24 1 25 1 26 3 27 4 28 4 29 3 30 2
31 3 32 2 33 2 34 3 35 2 36 4 37 2 38 2 39 1 40 3
41 2 42 3 43 1 44 4 45 4 46 3 47 1 48 1 49 1 50 2
51 3 52 1 53 2 54 3 55 2 56 1 57 3 58 3 59 4 60 4
61 3 62 4 63 3 64 3 65 1 66 2 67 4 68 1 69 2 70 2
71 4 72 3 73 1 74 3 75 1 76 2 77 3 78 2 79 4 80 4
81 4 82 1 83 2 84 3 85 3 86 3 87 3 88 1 89 2 90 4

	


	
1. 1
3333
x 16171819 =+ + +
 is even number
Therefore, 2 divides x.
33 2 2
a b (a b)(a – ab b ) += + +
? a + b always divides a
3
 + b
3
.
Therefore, 16
3
 + 19
3
 is divisible by 35.
18
3
 + 17
3
 is divisible by 35.
Thus, x is divisible by 70.
Hence, option (1) is the correct choice.
2. 3
AB C D
–20 20
90 –90
–10 10
–50 50
–100 100
110 –110
Total +60 30 –40 –50
D gets emptied first,  it gets emptied in 20 minutes.
Hence, option (3) is the correct answer.
3. 2 A
C
D B
Shaded area = 2 × (area of sector ADC – area of
? ADC)
2
1
21 – 11
42
p??
=× × × ×
??
??
– 1
2
p
=
Hence option (2)
4. 4 Let r be the radius of the two circular tracks.
? The rectangle has dimensions 4r × 2r.
B
rr
r r
r
r
A
A covers a distance of 2r + 2r + 4r + 4r = 12 r
B covers a distance of 2r 2r 4 r p+ p = p
Time taken by both of them is same.
?
BA
BA
4r 12r
SS
SS 3
pp
=? =
?Required percentage 
BA
A
S – S
100
S
=×
– 3
100 4.72%.
3
p
=× =
5. 1 Let there be m boys and n  girls
n
2
n(n – 1)
C45 n(n – 1) 90 n 10
2
== ? = ? =
     
m
2
m(m – 1)
C 190 190 m(m – 1) 380 m 20
2
=? =? = ? =
Number of games between one boy and one girl
= 
10 20
11
C C 10 20 200 ×=× =
Hence option (1)
Questions 6 and 7:
A B
C
2.5 km
5 km
Ram: A (9:00 AM)
C (9:30)
A
(11:00 A M )
C
(10:30)
B (10:00 AM)
B (10:00 AM)
                   @ 5km/h
Shyam: A (9:45 AM)
C (10:00) AM
A
(10:45 AM )
C
(10:30)
B (10:15 AM)
B (10:15 AM)
@ 10 km/h
6. 2 It is clear that Ram and Shyam shall meet each other
between C & B, sometime after 10:00 AM. At 10:00 AM
they are moving as shown below:
Shyam @ 10 km /h
C B
2.5 km
Ram @ 5 km/h
Page 3


QA  1 to 30 30
EU + RC 31 to 60 30
DI + DS + AR 61 to 90 30
T otal 90
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1 1 2 3 3 2 4 4 5 1 6 2 7 2 8 4 9 4 10 3
11 4 12 1 13 2 14 3 15 2 16 1 17 2 18 4 19 4 20 3
21 3 22 2 23 1 24 1 25 1 26 3 27 4 28 4 29 3 30 2
31 3 32 2 33 2 34 3 35 2 36 4 37 2 38 2 39 1 40 3
41 2 42 3 43 1 44 4 45 4 46 3 47 1 48 1 49 1 50 2
51 3 52 1 53 2 54 3 55 2 56 1 57 3 58 3 59 4 60 4
61 3 62 4 63 3 64 3 65 1 66 2 67 4 68 1 69 2 70 2
71 4 72 3 73 1 74 3 75 1 76 2 77 3 78 2 79 4 80 4
81 4 82 1 83 2 84 3 85 3 86 3 87 3 88 1 89 2 90 4

	


	
1. 1
3333
x 16171819 =+ + +
 is even number
Therefore, 2 divides x.
33 2 2
a b (a b)(a – ab b ) += + +
? a + b always divides a
3
 + b
3
.
Therefore, 16
3
 + 19
3
 is divisible by 35.
18
3
 + 17
3
 is divisible by 35.
Thus, x is divisible by 70.
Hence, option (1) is the correct choice.
2. 3
AB C D
–20 20
90 –90
–10 10
–50 50
–100 100
110 –110
Total +60 30 –40 –50
D gets emptied first,  it gets emptied in 20 minutes.
Hence, option (3) is the correct answer.
3. 2 A
C
D B
Shaded area = 2 × (area of sector ADC – area of
? ADC)
2
1
21 – 11
42
p??
=× × × ×
??
??
– 1
2
p
=
Hence option (2)
4. 4 Let r be the radius of the two circular tracks.
? The rectangle has dimensions 4r × 2r.
B
rr
r r
r
r
A
A covers a distance of 2r + 2r + 4r + 4r = 12 r
B covers a distance of 2r 2r 4 r p+ p = p
Time taken by both of them is same.
?
BA
BA
4r 12r
SS
SS 3
pp
=? =
?Required percentage 
BA
A
S – S
100
S
=×
– 3
100 4.72%.
3
p
=× =
5. 1 Let there be m boys and n  girls
n
2
n(n – 1)
C45 n(n – 1) 90 n 10
2
== ? = ? =
     
m
2
m(m – 1)
C 190 190 m(m – 1) 380 m 20
2
=? =? = ? =
Number of games between one boy and one girl
= 
10 20
11
C C 10 20 200 ×=× =
Hence option (1)
Questions 6 and 7:
A B
C
2.5 km
5 km
Ram: A (9:00 AM)
C (9:30)
A
(11:00 A M )
C
(10:30)
B (10:00 AM)
B (10:00 AM)
                   @ 5km/h
Shyam: A (9:45 AM)
C (10:00) AM
A
(10:45 AM )
C
(10:30)
B (10:15 AM)
B (10:15 AM)
@ 10 km/h
6. 2 It is clear that Ram and Shyam shall meet each other
between C & B, sometime after 10:00 AM. At 10:00 AM
they are moving as shown below:
Shyam @ 10 km /h
C B
2.5 km
Ram @ 5 km/h
Fig. at 10:00 AM
From now, time taken to meet = 
()
2.5
10 5 +
× 60 min
= 10 minutes
So, they meet each other at 10:10 AM.
7. 2 It is clear from the diagram that at 10:30; Shyam
overtakes Ram.
Alternate: At 10:15 the situation is as show:
A B
C
D
1.25 km
Shyam at B
m oving @ 10 km /h
Ram at D 
moving @ 5 km/h
Time taken for Shyam to overtake Ram = 
()
1.25
10 5 -
× 60
= 15 min.
? Shyam overtakes Ram at 10:30 AM.
8. 4 R = 
()
()
65
65 65
65 65
64 64 64
64 64
1
30 30 1
30 30 1
30
30 30 1 1
30 30 1
30
??
--
??
--
??
=
+- ??
+-
??
??
? R = 
65
65
64 64
1
11
36 30
30 1
11
30
??
??
?? --
??
??
??
??
?? ??
+-
?? ??
?? ??
? R = 30
()
()
65
64
10.96
10.96
??
-
??
??
+??
??
In 
()
()
65
64
10.96
,
10.96
-
+
numerator is only slightly less then 1.
and  denominator is only slightly more than 1.
Hence, R is certainly greater than 1.
9. 4 Case I:Chords on same side of the centre.
A
C
12 D
4
B
O
12
20
20 16
OB
2
 = OA
2
 – AB
2
 = 20
2
 – 16
2
 = 144
OB = 12
OD
2
 = 20
2
 – 12
2
 = 400 – 144 = 256
OD = 16
BD = 4 cm
Case II: Chords on opposite side of the centre.
A B
C
D
P
Q
O
AB = 32 cm
CD = 24 cm
OP = 
22
AO AP -
= 
() ( )
22
20 16 -
OP = 12 cm
& OQ = 
() ()
22
OC CQ - = 
() ( )
22
20 12 -
OQ = 16 cm
Distance = PQ = 12 + 16 = 28 cm.
10. 3 y
2
 = x
2
2x
2
 – 2kx + k
2
 – 1 = 0
D = 0
? 4k
2
 = 8k
2
 – 8
? 4k
2
 = 8
?k
2
 = 2 ? k = ±
2
.
k = +
2
 gives
the equation = 2x
2
 – 22x +1 = 0;
Its root is 
b1
,
2a 2
-
=+
 k = –
2
 gives
the equation 2x
2
 + 22x + 1 = 0. Its root is
 –
1
2
this root is –ve, will  reject  k = –
2
.
Only answer is k = + 2 .
Alternate: Graph based.
x
2
 – y
2
 = 0  & (x – k)
2
 + y
2
 = 1 are plotted below.
We are solving for a unique positive x.
x
2
 – y
2
 = 0
is a pair of straight lines
y = x & y = –x
(x – k)
2
 + y
2
 = 1 is a circle
with center (k, 0) & radius 1.
Page 4


QA  1 to 30 30
EU + RC 31 to 60 30
DI + DS + AR 61 to 90 30
T otal 90
T otal
questions
T otal
attempted
T otal
correct
T otal
wrong
Net
Score
 Time
T aken
Question
number
1 1 2 3 3 2 4 4 5 1 6 2 7 2 8 4 9 4 10 3
11 4 12 1 13 2 14 3 15 2 16 1 17 2 18 4 19 4 20 3
21 3 22 2 23 1 24 1 25 1 26 3 27 4 28 4 29 3 30 2
31 3 32 2 33 2 34 3 35 2 36 4 37 2 38 2 39 1 40 3
41 2 42 3 43 1 44 4 45 4 46 3 47 1 48 1 49 1 50 2
51 3 52 1 53 2 54 3 55 2 56 1 57 3 58 3 59 4 60 4
61 3 62 4 63 3 64 3 65 1 66 2 67 4 68 1 69 2 70 2
71 4 72 3 73 1 74 3 75 1 76 2 77 3 78 2 79 4 80 4
81 4 82 1 83 2 84 3 85 3 86 3 87 3 88 1 89 2 90 4

	


	
1. 1
3333
x 16171819 =+ + +
 is even number
Therefore, 2 divides x.
33 2 2
a b (a b)(a – ab b ) += + +
? a + b always divides a
3
 + b
3
.
Therefore, 16
3
 + 19
3
 is divisible by 35.
18
3
 + 17
3
 is divisible by 35.
Thus, x is divisible by 70.
Hence, option (1) is the correct choice.
2. 3
AB C D
–20 20
90 –90
–10 10
–50 50
–100 100
110 –110
Total +60 30 –40 –50
D gets emptied first,  it gets emptied in 20 minutes.
Hence, option (3) is the correct answer.
3. 2 A
C
D B
Shaded area = 2 × (area of sector ADC – area of
? ADC)
2
1
21 – 11
42
p??
=× × × ×
??
??
– 1
2
p
=
Hence option (2)
4. 4 Let r be the radius of the two circular tracks.
? The rectangle has dimensions 4r × 2r.
B
rr
r r
r
r
A
A covers a distance of 2r + 2r + 4r + 4r = 12 r
B covers a distance of 2r 2r 4 r p+ p = p
Time taken by both of them is same.
?
BA
BA
4r 12r
SS
SS 3
pp
=? =
?Required percentage 
BA
A
S – S
100
S
=×
– 3
100 4.72%.
3
p
=× =
5. 1 Let there be m boys and n  girls
n
2
n(n – 1)
C45 n(n – 1) 90 n 10
2
== ? = ? =
     
m
2
m(m – 1)
C 190 190 m(m – 1) 380 m 20
2
=? =? = ? =
Number of games between one boy and one girl
= 
10 20
11
C C 10 20 200 ×=× =
Hence option (1)
Questions 6 and 7:
A B
C
2.5 km
5 km
Ram: A (9:00 AM)
C (9:30)
A
(11:00 A M )
C
(10:30)
B (10:00 AM)
B (10:00 AM)
                   @ 5km/h
Shyam: A (9:45 AM)
C (10:00) AM
A
(10:45 AM )
C
(10:30)
B (10:15 AM)
B (10:15 AM)
@ 10 km/h
6. 2 It is clear that Ram and Shyam shall meet each other
between C & B, sometime after 10:00 AM. At 10:00 AM
they are moving as shown below:
Shyam @ 10 km /h
C B
2.5 km
Ram @ 5 km/h
Fig. at 10:00 AM
From now, time taken to meet = 
()
2.5
10 5 +
× 60 min
= 10 minutes
So, they meet each other at 10:10 AM.
7. 2 It is clear from the diagram that at 10:30; Shyam
overtakes Ram.
Alternate: At 10:15 the situation is as show:
A B
C
D
1.25 km
Shyam at B
m oving @ 10 km /h
Ram at D 
moving @ 5 km/h
Time taken for Shyam to overtake Ram = 
()
1.25
10 5 -
× 60
= 15 min.
? Shyam overtakes Ram at 10:30 AM.
8. 4 R = 
()
()
65
65 65
65 65
64 64 64
64 64
1
30 30 1
30 30 1
30
30 30 1 1
30 30 1
30
??
--
??
--
??
=
+- ??
+-
??
??
? R = 
65
65
64 64
1
11
36 30
30 1
11
30
??
??
?? --
??
??
??
??
?? ??
+-
?? ??
?? ??
? R = 30
()
()
65
64
10.96
10.96
??
-
??
??
+??
??
In 
()
()
65
64
10.96
,
10.96
-
+
numerator is only slightly less then 1.
and  denominator is only slightly more than 1.
Hence, R is certainly greater than 1.
9. 4 Case I:Chords on same side of the centre.
A
C
12 D
4
B
O
12
20
20 16
OB
2
 = OA
2
 – AB
2
 = 20
2
 – 16
2
 = 144
OB = 12
OD
2
 = 20
2
 – 12
2
 = 400 – 144 = 256
OD = 16
BD = 4 cm
Case II: Chords on opposite side of the centre.
A B
C
D
P
Q
O
AB = 32 cm
CD = 24 cm
OP = 
22
AO AP -
= 
() ( )
22
20 16 -
OP = 12 cm
& OQ = 
() ()
22
OC CQ - = 
() ( )
22
20 12 -
OQ = 16 cm
Distance = PQ = 12 + 16 = 28 cm.
10. 3 y
2
 = x
2
2x
2
 – 2kx + k
2
 – 1 = 0
D = 0
? 4k
2
 = 8k
2
 – 8
? 4k
2
 = 8
?k
2
 = 2 ? k = ±
2
.
k = +
2
 gives
the equation = 2x
2
 – 22x +1 = 0;
Its root is 
b1
,
2a 2
-
=+
 k = –
2
 gives
the equation 2x
2
 + 22x + 1 = 0. Its root is
 –
1
2
this root is –ve, will  reject  k = –
2
.
Only answer is k = + 2 .
Alternate: Graph based.
x
2
 – y
2
 = 0  & (x – k)
2
 + y
2
 = 1 are plotted below.
We are solving for a unique positive x.
x
2
 – y
2
 = 0
is a pair of straight lines
y = x & y = –x
(x – k)
2
 + y
2
 = 1 is a circle
with center (k, 0) & radius 1.
y
x
y = –x
y = +x
(2, 0)
(1) k = 2;
clearly, no solution
(2) k = 0
y
x
y = –x
y = +x
x = +a
x = –a
x = a, –a are its two solutions.
– rejected.
(3) k = 
2 +
unique value of x & a positive one as shown.
y
x
y = –x
y = +x
1
(  2,  0)
x =ß
1
(4) k = 
– 2
, also gives the unique value of x but it is
negative one.
y
x
1
(–  2,  0)
x = –ß
2
11. 4 If p = 1! = 1, then
p + 2 = 3 when divided by 2! will give a remainder of 1.
If p = 1! + 2 × 2!  = 5, then
p + 2 = 7 when divided by 3! will give a remainder of 1.
Hence, p = 1! + (2 × 2!) + (3 × 3!) + … + (10 × 10!) when
divided by 11! leaves a remainder 1.
Alternative method:
P = 1 + 2.2! + 3.3!+ ….10.10!
= (2 –1)1! + ( 3 – 1)2! + (4 – 1)3! + ….(11 – 1)10!
=2! – 1! + 3! – 2! + ….. 11! –10! = 1 + 11!
Hence, the remainder is 1.
12. 1
(0, 41)
(41, 0)
A
B
x
y
(0, 0)
equation of line = x + y = 41. If the (x, y) co-ordinates
of the points are integer, their sum shall also be integers
so that x + y = k (k, a variable) as we have to exclude
points lying on the boundary of triangle; k can take all
values from 1 to 40 only. k = 0 is also rejected as at k
= 0 will give the point A; which can’t be taken.
Now, x + y = k, (k = 1, 2, 3, ... 40)
with k = 40; x + y = 40; taking integral solutions.
We get points (1, 39), (2, 38); (3, 37) ...(39, 1)
i.e. 39 points
x + y = 40 will be satisfied by 39 points.
Similarly, x + y = 39 is satisfied by 38 points.
x + y = 38 by 37 points.
x + y = 3 by 2 points.
x + y = 2 is satisfied by 1 point.
x + y = 1 is not satisfied by any point.
So, the total no. of all such points is:
39 + 38 + 37 + 36 + ... 3 + 2 + 1 = 
39 40
2
×
= 780 points.
13. 2 Let A = abc. Then, B = cba.
Given, B > A ? c > a
As B – A = (100c + 10b + a) – (100a + 10b + 1)
? B – A = 100 (c – a) + (a – c)
? B – A = 99 (c – a).  Also, (B – A) is divisible by 7.
But, 99 is not divisible by 7 (no factor like 7 or 7
2
).
Therefore, (c – a) must be divisible by 7 {i.e., (c – a)
must be 7, 7
2
, etc.}. Since c and a are single digits,
value of (c – a) must be 7. The possible values of (c,
a) {with c > a} are (9, 2) and (8, 1). Thus, we can
write A as:
A : abc = 1b8 or 2b9
As b can take values from 0 to 9, the smallest & largest
possible value of A are:
    A
min
 = 108
&  A
max
 = 299
Only option (b) satisfies this. Hence, (2) is the correct
option.
14. 3 a
1
 = 1,        a
n+1
 – 3a
n
 + 2 = 4n
a
n+1
 = 3a
n
 + 4n – 2
when n = 2 then a
2
 = 3 + 4 – 2 = 5
when n = 3 then a
3
 = 3 × 5 + 4 × 2 – 2 = 21
from the options, we get an idea that a
n
 can be
expressed in a combination of some power of
3 & some multiple of 100.
(1) 3
99
 – 200; tells us that a
n
 could be: 3
n–1
 – 2 × n;
     but it does not fit a
1
 or a
2
 or a
3
Page 5


QA  1 to 30 30
EU + RC 31 to 60 30
DI + DS + AR 61 to 90 30
T otal 90
T otal
questions
T otal
attempted
T otal
correct
T otal
wrong
Net
Score
 Time
T aken
Question
number
1 1 2 3 3 2 4 4 5 1 6 2 7 2 8 4 9 4 10 3
11 4 12 1 13 2 14 3 15 2 16 1 17 2 18 4 19 4 20 3
21 3 22 2 23 1 24 1 25 1 26 3 27 4 28 4 29 3 30 2
31 3 32 2 33 2 34 3 35 2 36 4 37 2 38 2 39 1 40 3
41 2 42 3 43 1 44 4 45 4 46 3 47 1 48 1 49 1 50 2
51 3 52 1 53 2 54 3 55 2 56 1 57 3 58 3 59 4 60 4
61 3 62 4 63 3 64 3 65 1 66 2 67 4 68 1 69 2 70 2
71 4 72 3 73 1 74 3 75 1 76 2 77 3 78 2 79 4 80 4
81 4 82 1 83 2 84 3 85 3 86 3 87 3 88 1 89 2 90 4

	


	
1. 1
3333
x 16171819 =+ + +
 is even number
Therefore, 2 divides x.
33 2 2
a b (a b)(a – ab b ) += + +
? a + b always divides a
3
 + b
3
.
Therefore, 16
3
 + 19
3
 is divisible by 35.
18
3
 + 17
3
 is divisible by 35.
Thus, x is divisible by 70.
Hence, option (1) is the correct choice.
2. 3
AB C D
–20 20
90 –90
–10 10
–50 50
–100 100
110 –110
Total +60 30 –40 –50
D gets emptied first,  it gets emptied in 20 minutes.
Hence, option (3) is the correct answer.
3. 2 A
C
D B
Shaded area = 2 × (area of sector ADC – area of
? ADC)
2
1
21 – 11
42
p??
=× × × ×
??
??
– 1
2
p
=
Hence option (2)
4. 4 Let r be the radius of the two circular tracks.
? The rectangle has dimensions 4r × 2r.
B
rr
r r
r
r
A
A covers a distance of 2r + 2r + 4r + 4r = 12 r
B covers a distance of 2r 2r 4 r p+ p = p
Time taken by both of them is same.
?
BA
BA
4r 12r
SS
SS 3
pp
=? =
?Required percentage 
BA
A
S – S
100
S
=×
– 3
100 4.72%.
3
p
=× =
5. 1 Let there be m boys and n  girls
n
2
n(n – 1)
C45 n(n – 1) 90 n 10
2
== ? = ? =
     
m
2
m(m – 1)
C 190 190 m(m – 1) 380 m 20
2
=? =? = ? =
Number of games between one boy and one girl
= 
10 20
11
C C 10 20 200 ×=× =
Hence option (1)
Questions 6 and 7:
A B
C
2.5 km
5 km
Ram: A (9:00 AM)
C (9:30)
A
(11:00 A M )
C
(10:30)
B (10:00 AM)
B (10:00 AM)
                   @ 5km/h
Shyam: A (9:45 AM)
C (10:00) AM
A
(10:45 AM )
C
(10:30)
B (10:15 AM)
B (10:15 AM)
@ 10 km/h
6. 2 It is clear that Ram and Shyam shall meet each other
between C & B, sometime after 10:00 AM. At 10:00 AM
they are moving as shown below:
Shyam @ 10 km /h
C B
2.5 km
Ram @ 5 km/h
Fig. at 10:00 AM
From now, time taken to meet = 
()
2.5
10 5 +
× 60 min
= 10 minutes
So, they meet each other at 10:10 AM.
7. 2 It is clear from the diagram that at 10:30; Shyam
overtakes Ram.
Alternate: At 10:15 the situation is as show:
A B
C
D
1.25 km
Shyam at B
m oving @ 10 km /h
Ram at D 
moving @ 5 km/h
Time taken for Shyam to overtake Ram = 
()
1.25
10 5 -
× 60
= 15 min.
? Shyam overtakes Ram at 10:30 AM.
8. 4 R = 
()
()
65
65 65
65 65
64 64 64
64 64
1
30 30 1
30 30 1
30
30 30 1 1
30 30 1
30
??
--
??
--
??
=
+- ??
+-
??
??
? R = 
65
65
64 64
1
11
36 30
30 1
11
30
??
??
?? --
??
??
??
??
?? ??
+-
?? ??
?? ??
? R = 30
()
()
65
64
10.96
10.96
??
-
??
??
+??
??
In 
()
()
65
64
10.96
,
10.96
-
+
numerator is only slightly less then 1.
and  denominator is only slightly more than 1.
Hence, R is certainly greater than 1.
9. 4 Case I:Chords on same side of the centre.
A
C
12 D
4
B
O
12
20
20 16
OB
2
 = OA
2
 – AB
2
 = 20
2
 – 16
2
 = 144
OB = 12
OD
2
 = 20
2
 – 12
2
 = 400 – 144 = 256
OD = 16
BD = 4 cm
Case II: Chords on opposite side of the centre.
A B
C
D
P
Q
O
AB = 32 cm
CD = 24 cm
OP = 
22
AO AP -
= 
() ( )
22
20 16 -
OP = 12 cm
& OQ = 
() ()
22
OC CQ - = 
() ( )
22
20 12 -
OQ = 16 cm
Distance = PQ = 12 + 16 = 28 cm.
10. 3 y
2
 = x
2
2x
2
 – 2kx + k
2
 – 1 = 0
D = 0
? 4k
2
 = 8k
2
 – 8
? 4k
2
 = 8
?k
2
 = 2 ? k = ±
2
.
k = +
2
 gives
the equation = 2x
2
 – 22x +1 = 0;
Its root is 
b1
,
2a 2
-
=+
 k = –
2
 gives
the equation 2x
2
 + 22x + 1 = 0. Its root is
 –
1
2
this root is –ve, will  reject  k = –
2
.
Only answer is k = + 2 .
Alternate: Graph based.
x
2
 – y
2
 = 0  & (x – k)
2
 + y
2
 = 1 are plotted below.
We are solving for a unique positive x.
x
2
 – y
2
 = 0
is a pair of straight lines
y = x & y = –x
(x – k)
2
 + y
2
 = 1 is a circle
with center (k, 0) & radius 1.
y
x
y = –x
y = +x
(2, 0)
(1) k = 2;
clearly, no solution
(2) k = 0
y
x
y = –x
y = +x
x = +a
x = –a
x = a, –a are its two solutions.
– rejected.
(3) k = 
2 +
unique value of x & a positive one as shown.
y
x
y = –x
y = +x
1
(  2,  0)
x =ß
1
(4) k = 
– 2
, also gives the unique value of x but it is
negative one.
y
x
1
(–  2,  0)
x = –ß
2
11. 4 If p = 1! = 1, then
p + 2 = 3 when divided by 2! will give a remainder of 1.
If p = 1! + 2 × 2!  = 5, then
p + 2 = 7 when divided by 3! will give a remainder of 1.
Hence, p = 1! + (2 × 2!) + (3 × 3!) + … + (10 × 10!) when
divided by 11! leaves a remainder 1.
Alternative method:
P = 1 + 2.2! + 3.3!+ ….10.10!
= (2 –1)1! + ( 3 – 1)2! + (4 – 1)3! + ….(11 – 1)10!
=2! – 1! + 3! – 2! + ….. 11! –10! = 1 + 11!
Hence, the remainder is 1.
12. 1
(0, 41)
(41, 0)
A
B
x
y
(0, 0)
equation of line = x + y = 41. If the (x, y) co-ordinates
of the points are integer, their sum shall also be integers
so that x + y = k (k, a variable) as we have to exclude
points lying on the boundary of triangle; k can take all
values from 1 to 40 only. k = 0 is also rejected as at k
= 0 will give the point A; which can’t be taken.
Now, x + y = k, (k = 1, 2, 3, ... 40)
with k = 40; x + y = 40; taking integral solutions.
We get points (1, 39), (2, 38); (3, 37) ...(39, 1)
i.e. 39 points
x + y = 40 will be satisfied by 39 points.
Similarly, x + y = 39 is satisfied by 38 points.
x + y = 38 by 37 points.
x + y = 3 by 2 points.
x + y = 2 is satisfied by 1 point.
x + y = 1 is not satisfied by any point.
So, the total no. of all such points is:
39 + 38 + 37 + 36 + ... 3 + 2 + 1 = 
39 40
2
×
= 780 points.
13. 2 Let A = abc. Then, B = cba.
Given, B > A ? c > a
As B – A = (100c + 10b + a) – (100a + 10b + 1)
? B – A = 100 (c – a) + (a – c)
? B – A = 99 (c – a).  Also, (B – A) is divisible by 7.
But, 99 is not divisible by 7 (no factor like 7 or 7
2
).
Therefore, (c – a) must be divisible by 7 {i.e., (c – a)
must be 7, 7
2
, etc.}. Since c and a are single digits,
value of (c – a) must be 7. The possible values of (c,
a) {with c > a} are (9, 2) and (8, 1). Thus, we can
write A as:
A : abc = 1b8 or 2b9
As b can take values from 0 to 9, the smallest & largest
possible value of A are:
    A
min
 = 108
&  A
max
 = 299
Only option (b) satisfies this. Hence, (2) is the correct
option.
14. 3 a
1
 = 1,        a
n+1
 – 3a
n
 + 2 = 4n
a
n+1
 = 3a
n
 + 4n – 2
when n = 2 then a
2
 = 3 + 4 – 2 = 5
when n = 3 then a
3
 = 3 × 5 + 4 × 2 – 2 = 21
from the options, we get an idea that a
n
 can be
expressed in a combination of some power of
3 & some multiple of 100.
(1) 3
99
 – 200; tells us that a
n
 could be: 3
n–1
 – 2 × n;
     but it does not fit a
1
 or a
2
 or a
3
(2) 3
99
 + 200; tells us that a
n
 could be: 3
n–1
 + 2 × n;
     again, not valid for a
1
, a
2
 etc.
(3) 3
100
 – 200; tells 3
n
 – 2n: valid for all a
1
, a
2
, a
3
.
(4) 3
100
 + 200; tells 3
n
 + 2n: again not valid.
    so, (3) is the correct answer.
15. 2
right m ost digit (RM D) left m ost digit (LM D)
odd positions can be counted in 2 ways.
(i) Counting from the LMD-end:
odd
positions
We have 1, 2, 3, 4 & 5 to be filled in these blocks. Odd
nos. (1, 3, 5) to be filled in at odd positions. Other
places are to be filled by even nos. (2 or 4) Let’s
count, how many such nos. are there with 2 at the
unit’s digit
odd
2
Odd nos. can be filled in 
3
P
2
 = 6 way.
The remaining two places are to be filled by 2 nos.
(one odd no. left out of 1, 3, 5 & one even i.e. 4) in = 2
ways.
So, there are 6 × 2 = 12 number with 2 at the rightmost
place. Similarly; there are 12 such nos. with 4 at the
rightmost digits.
The sum of rightmost digits in all such number
= 12(2 + 4) = 72
(ii) Now counting from the RMD-end.
Let’s place 1 at the units place and check, how many
nos. are possible with (1, 3) at the odd positions:
5 & (2 or 4)
3 1
(4 or 2)
No. of such cases = 2 × 2 = 4 ways.
5 & (2 or 4)
1
(4 or 2)
3
Here again no. of ways = 2 × 2 = 4 ways
So, there are 4 + 4 = 8 nos, in which (1, 3) are at odd
positions. Similarly there are 8 nos. in which (1, 5) are
at odd positions. So, in all there are 16 nos. where 1 is
at unit’s place. Similarly there are 16 nos. with 3 at
unit’s place and 16 more with 5 at unit’s place.
Summing up all the odd unit’s digits = 16(1 + 3 + 5)
= 144
From (i) and (ii) we can now, sum up all (even or odd)
nos. at units place = 72 +144 = 216
Hence answer is (2)
16. 1 ((30)
4
)
680
 = (8100)
680
.
Hence, the right most non-zero digit is 1.
17. 2
1cm
60° 60°
60°
90°
A
B C
D
P Q
90°
Drawn figure since it have not to be within distance of
1 m so it will go along APQD, which is the path of
minimum distance.
AP = 
90
21
360 2
p
×p× =
Also AP = QD = 
2
p
So the minimum distance = AP + PQ + QD
= 
11
22
pp
++ = +p
18. 4 P = 
xy
xy
log log
yx
?? ??
+
?? ??
?? ??
= 
xx yy
log x –logy logy – log x +
= 
xy
2 – log y – log x
Let 
x
tlog y =
2
11
P2–– t – t
t t
??
== -
??
??
which can never be positive. Out of given options, it
can’t assume a value of +1.
19. 4 It is given that 10 < n < 1000.
Let n be a two digit number. Then,
n = 10a + b ? p
n
 = ab, s
n
 = a + b
Then, ab + a + b = 10a + b
? ab = 9a ? b = 9
? There are 9 such numbers 19, 29, 33, …, 99.
Now, let n be a three digit number.
? n = 100a + 10b + c ? p
n
 = abc, s
n
= a + b + c