Download 51. B is South east of x so From fig. clear that C is farthest east and H

North
51.
B is South east of x so
East
West
South
From fig. clear that C is farthest east and H is farthest south roads.
52.
Ans. (c) Since A and B have played seven games in a row, B cannot partner with A as no pair can together
play more than seven games in a row. If B partners with D, then C will have to play with A which I does
not want to, hence this is also incorrect. Hence, B can only partner with C.
53.
From Given
R
O
6
8
and
C
H
7
3
P
R
9
6
So, Code for
S
E
2
1
S
2
E
1
A
4
I
5
R
6
E
1
A
4
C
7
H
3
A
4
R
6
C
7
H
3
Question 54-56:
Since team in A has beaten R, P and S, these teams cannot be in A and only Q, T can be in A. Since team Q has
beaten teams in A, E, C. Hence A cannot have team Q. Thus, only team T can be in A. R. is given to be in B and S
cannot be in C. Therefore, the only teams that can be in C are P and Q. Since Q has beaten the team in C, hence Q
cannot be in C. Thus P is in C. The only two teams left are Q and S and towns left are D and E. Q has beaten team in
E, hence Q will be D and the team in E in S. From this, the answer will be :
A
B
C
D
E





T
R
P
Q
S
54.
Ans. (d) It is clear that team Q is in town D.
55.
Ans. (c) Team P is in town C.
56.
Ans. (d) Town A have team T.
57.
Ans. (a) 120, 99, 80, 63, 48
120 – 99 = 21
99 – 88 = 19
80 – 63 = 17
63 – 48 = 15
So next is
48 – 13 = 35
58.
Ans. (b) Total sections before new admission = 16 – 3 = 13
Total students before new admission = 13  24 = 312
Total Students after new admission = 16  21 = 336
So no of newly admitted students = 336 – 312 = 24
59.
Ans. (b) 4 is assigned to P and difference between P and T is 5 so 9 is assigned to T. difference between N
and T is 3 so T = 9 and So Hence 6 is assigned to N (because 9 – 3 = 6)
Question 60-63:
It is given A is 2nd to the left of B and C is immediate right of B. D is immediate left of A. Now E will be in place
left between A and B as shown
D
A
E
B
C
It is Given a is facing P there are as many girls P and Q as between R and S and S and R are not facing B or D. So Q
is facing S and remaining T is facing D. as shown
T
P
S/R
Q
S/R
D
A
E
B
C
60.
Ans. (a) C and D are pair of boys standing at ends of row.
61.
Ans. (d) From Given information none option is true.
62.
Ans. (a) E is standing immediate right of A.
63.
Ans. (c) Q is facing B.
64.
Ans. (b) Let mother’s age is x and daughter’s age is 63 – x four year back mother’s age was 4 times that of
daughter’s age at the that time
 (x – 4) = 4(63 – x – 4)
 x – 4 = 4  59 – 4x
 5x = 240  x = 48
So mother’s present age is 48 years
65.
Ans. (a) Clock gains 10 seconds in 5 min
so in 1 hour it gain 120 second = 2 min
from 9:00 AM to 7:00 AM total 10 hours
it gain 20 min in 10 hours = 2 min  10 = 20 min
so correct time 7:20 – 00:20
66.
Ans. (a) Let Ronit’s present age be x years. Then, father’s present age = (a + 3x) years = 4x years.
 (4x + 8) = 5/2(x + 8)  8x + 16 = 5x + 40
 3x = 24
x=8
 4 x  16 48
Hence, required ratio 

2
 x  16 24
67.
Ans. (c) 1, 2, 3, 6, 11, 20, 37, 68, ……….
Every number is the sum of previous three numbers.
1+2+3=6
2 + 3 + 6 = 11
3 + 6 + 11 = 20
6 + 11 + 20 = 37
11 + 20 + 37 = 68
So 20 + 37 + 68 = 125
Hence, c is correct answer.
Questions 68 – 69:
From given information
(i)
F is opposite to A
(ii)
F is immediate right of B
(iii)
D is opposite to C
So B is opposite to E
68.
Ans. (a) from figure it is clear that D and E are sitting next to A
69.
Ans. (c) From figure E is opposite to B
70.
Ans. (d) Numbers are 210 and 220
Arithmetic mean of 210 and 220 
71.
210  220
 29  219
2
Ans. (b) Let the 5 boxes be a, b, c, d and e
Let (a, b) be the 2 boxes with minimum weight and (d, e) be the once with max weight.
Given, a + b = 110 and d + e = 121
‘c’ is the box which has the middle weight.
We know 4(a + b + c + d + e) = 1156  a + b + c + d + e = 289
Therefore, c = 289 – (a + b) – (d + e) = 58
Since we have 10 different pairs, we can say no 2 weights are the same.
a + b = 110
Now look at the next minimum weight which is 112, 2 kgs more than a + b, therefore b must be 2 kgs less
than c,
b = 56, therefore a = 54
Now (a, b, c) = (54, 56, 58)
Also given d + e = 121
Possibilities are (60, 61) or (59, 62) Since both weight should be > 58 (c)
Now if we consider (60, 61) as a possibility glancing through the options we can find 61 + 58(c) = 119 is
not there in the mentioned set, so eliminate the possibility.
Hence (d, e) can only be (59, 62) and the maximum weight is 62 kg.
Question 72-76:
Clearly from statement (1) and (2), figure 1 follows; statement (3), figure 2 follows; from statement (4),
figure 3 follows; from statement (5), figure 4 follows; and from statements (6) and (7), figure 5 follows.
72.
Ans. (d) It follows from figure 5 that distance of M from L = LK + KM = 1/2 + 1 + 1.5 km and M is to the
north of L.
73.
Ans. (b) If E is between B and C, we will have the following figure:
Thus, the statement that D is 2 km west of B is false.
74.
Ans. (d) From the figure, the distance between A and D
1

= AB + ED – BE  1  1    2  .25 i.e. between 1.5 to 2 km
4

75.
Ans. (c)
77.
Ans. (c) Let student in class 1 = 2k
Class 2 = 3k
Class 3 = 5k
So total student = 10 k
Now from given condition
2k  20 4

  k = 10
3k  20 5
Hence total student = 10k = 10  10 = 100
78.
Ans. (c) Let age of Babita is x
So age of Ajith is 3x
Let age of Chetu is y So das age is 2y
From given condition x > y
For find age of Ajith
(i) Statement is not alone sufficient
(ii) Statement is not alone sufficient
But taking together we find actual age of Ajith.
76.
Ans. (d)
Question 79-80:
First we need to find out, which one person can be between R and T. Since Q and T cannot be in the same row,
hence Q cannot be there. Since U is facing R and U is left to S, Hence U and S also cannot be there. Thus only P can
be there between R and T. It is given that P is in the middle of the row, hence there will be 3 members in one row
and three members in the other row. Thus the two rows formed will be RPT and USQ from left to right. Thus the
answers will be -
79.
Ans. (b) From figure it is clear that R, P, T is in same row.
80.
Ans. (b) It is clear that U is left of S.
81.
Ans. (c) From fig S is facing P.
82.
Ans. (d) From figure clear that TQ is face each other.
Question 54-58:
From 1st information F is psychologist and doctor is he Grandfather
Person
Sex
Profession
A
Male
Doctor
B
Female
Advocate
C
Male
Jeweller
D
Female
Manager
E
Male / Female
Engineer
F
Male / Female
Psychologist
Marital Status
Grand of F and
Mother of E and F
Father of E and F
Grand Mother of F and E
54.
Ans. (d) From above information A is doctor
55.
Ans. (b) Professor of E is Engineer
56.
Ans. (c) A is Grand Father of E.
57.
Ans. (d) The number of male member is not determined because Sex of E and F is not given.
58.
Ans. (a) From Given information AD and CB are two married couple.
Question 51-53:
From 1st information Red car first so CEO
From 2nd blue is between Red and Green so blue is of President and Green is of Vice President and name Enid
Executive
Executive Name
Colour of Car
CEO
Cheryl
Red
President
Bert
Blue
Vice President
Enid
Green
Secretary
Alice
Yellow
Treasurer
David
Purple
51.
Ans. (d) From Given information it is clear that Alice is secretary.
52.
Ans. (c) CEO is Cheryl.
53.
Ans. (a) From given information the colour of vice president car is Green.
GENERAL ENGLISH
91.
Ans. (b) Being soaked to the skin means that the kitten doesn’t have a dry part on it, it’s completely
drentched. An abyss (A), i.e. a chasm or deep hole, would not soak anything. Craven (B) is not a noun, but
an adjective meaning cowardly: persons / things, not characteristics / qualities, get soaked. “Hide” [(D),
(E)] is a synonym for skin usually describing tanned leather, not a live kitten’s skin. Hence, only storm can
soak the skin.
92.
Ans. (c) A burglar breaks into houses or other buildings and steals things. Thieves is also used in the same
context. Gangsters are related to crime. Hence, pirates is the correct answer.
93.
Ans. (a) Since the sentence is interrogative, hence c and d options cannot be correct. B option changes the
meaning by excluding the word ‘dare’. Hence, A is the correct answer.
94.
Ans. (a) Option A and B may seem correct but there is a difference between the two. Archeology is the
excavation and study of artefacts in an effort to interpret and reconstruct past human behavior while
Anthropology is the study of ancient societies. Hence, A is the correct answer.
95.
Ans. (c) Line 7 of the passage gives the answer.
96.
Ans. (d) Line 3 says protein rich foods ward off the ills of malnutrition, hence malnutrition relates to lack
of proteins.
97.
Ans. (a) B and D cannot be correct as the subject precedes the object while in passive voice the object
should come before the subject. Between A and C, C is incorrect because the verb write is in simple present
and cannot be changed to wrote. It should be written by. Hence A is correct.
98.
Ans. (d) The only possible sequence is D. Rest of the options do not form meaningful sentences.
99.
Ans. (a) Only option A forms a meaningful sentence. Rest of the options hold no meaning.
100.
Ans. (b) Since the sentence is affirmative, the tag will be negative. Hence, B
101.
Ans. (c) Prudent is used in relation to being wise and careful.
102.
Ans. (c) An cannot be used as ‘flower’ does not start with a vowel. The shows definitions, which does not
hold. Hence, C
103.
Ans. (a)
104.
Ans. (d) A and C show opposite relationship. In B, both the words mean the same. Only D is correct.
105.
Ans. (b)
106.
Ans. (c) Since the verb used is ‘propose’, hence the mode of sentence is suggestion and option C can only
be used.
107.
Ans. (c)
108.
Ans. (c)
109.
Ans. (b)
110.
Ans. (b)
COMPUTER AWARENESS
113.
Decimal number of hexadecimal number A10
= 162  10 + 16  1 + 160  0 = 0 + 16 + 2560
(A10)16 = (2560)10
Decimal number of hexadecimal number B21
= 162  11 + 161  2 + 162  1 = 1 + 32 + 256  11 = 2816 + 33
(B12)16 = (2849)10
Then
(A10)16 + (B12)16 = (2576)10 + (2849)10 = (5425)10
114.
Ans. (c) 2’s complement of 0011010110011100
1’s complement = 1100101001100011
+
1
2’s complement = 1100101001100100
113.
Ans. (b) Multiplication of (111)2 by (101)2
111  101
111
0000
11100
100011
114.
Ans. (a) 8 bit 2’s complement representation of –ve integer – 93
93 can be represent in binary
93
1
46
0
23
1
11
1
50
1
2
0
1
0
2
2
2
2
2
2
(93)10 = (01011101)2
(-93)10 = (101011101)2
1’s complement of 101011101 = 010100010
+
1
2’s complement
010100011
Consider only 8 bit then
115.
Ans. (b) Since A and B are floating point integers, and floating point numbers have precision upto 6 places.
Hence these floating point operations will give the following results X = 0.0
Since C = 1.0 and A has 1030 digits, hence, C becomes negligible and since precision is only upto 6
decimal places, hence it will remain A.
Y=A
X = 0.0 + 1.0 = 1.0
Y = Y + B = A + B = 0.0
116.
Ans. (d) X(X + Y)
X.X+XY
X (1 + Y)
X+X.Y
 X (1 + Y)
X.1=X
X.1
(A) Option
(B) Option
(C) Option
(D) All of the above.
117.
Ans. (b) In a nibble, there are 4 bits = (4)/8 bytes = 1/2 bytes.
120.
Ans. (d) Case 1st : first start with zero
A = 0 2C1 2C1 2C1 2C1 2C1 2C1 2C1 = 27 = 128
Case 2nd : End with 11
B = 2C1 2C1 2C1 2C1 2C1 2C1 2C1 11 = 26 = 64
Case 3rd : Start with zero end with 11
C = 0 2C1 2C1 2C1 2C1 2C1 2C1 11 = 25 =32
Total case = A + B – C = 128 + 64 – 32 = 160
115.
Ans. (c) Multiplication of (10101)2 and (11101)2 in hexadecimal
1010  11101
000000
1010100
10101000
101010000
1000110001
In hexadecimal from 0010 0011 0001
2
6
1
116.
2
2
2
2
2
2
2
2
2
Ans. (d)
531
1
265
1
132
0
66
0
.53125  2
33
1
.6250  2
16
0
.1250  2
8
0
.250  2
4
0
.500  2
2
0
1.000
1
1
100001001
Hence = (1000010011.10001)2
1
0
0
0
0
1