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332 Aptitude Questions and Answers:
1 :: A,B and C are 8 bit nos.they are as follows:
A=11000101
B=00110011
C=00111010
find ((A-C)UB)
THE OPTIONS ARE
1)55 2)64 3)32 4)53
ANSWER=55
1)55 2)64 3)32 4)53
ANSWER=55
2 :: A and B are shooters and having their exam. A and B fall short of 10 and 2 shots respectively to the qualifying mark. If each of them fired atleast one shot and even by adding their total score together, they fall short of the qualifying mark, what is the qualifying mark?
ANS-- 11
as A is short by 10- he has shot 1
as B is short by 2 - he has shot 9
9+1=10 10<11
as A is short by 10- he has shot 1
as B is short by 2 - he has shot 9
9+1=10 10<11
3 :: A, B and C are 8 bit nos. They are as follows:
A -> 1 1 0 1 1 0 1 1
B -> 0 1 1 1 1 0 1 0
C -> 0 1 1 0 1 1 0 1 (n=intersection, u=union)
Find ((A n B) u C) =? . A-B is {A} - {A n B}
this is basically binary operation
n-intersection i.e.LOGICAL AND
U -union i.e. LOGICAL OR
AnB= 01011010
((AnB)uC)=01111111
n-intersection i.e.LOGICAL AND
U -union i.e. LOGICAL OR
AnB= 01011010
((AnB)uC)=01111111
4 :: A non stop bus to Amritsar overtakes an auto also moving towards Amritsar at 10 am. The bus reaches Amritsar at 12.30 pm and starts on the return journey after 1 hr. On the way back it meets the auto at 2 pm. At what time the auto will reach Amritsar?
a) 2.30pm
b) 3.00pm
c) 3.15pm
d) 3.30pm
Answer:b:3 P.M.
Assumptions:
Distance between two points where Bus meets auto=x
Distance between the point where bus meets auto for the second time and Amritsar=y
velocity of bus=Vb
velocity of auto=va
vb=x/2
va=x/4
y=0.5*vb=05*x/2
time taken by auto for covering the distance y=y/va=1 Hour
hence time at which auto reaches Amritsar=2 P.M.+1 hour=3 P.M.
Assumptions:
Distance between two points where Bus meets auto=x
Distance between the point where bus meets auto for the second time and Amritsar=y
velocity of bus=Vb
velocity of auto=va
vb=x/2
va=x/4
y=0.5*vb=05*x/2
time taken by auto for covering the distance y=y/va=1 Hour
hence time at which auto reaches Amritsar=2 P.M.+1 hour=3 P.M.
5 :: A, B and C are 8 bit nos. They are as follows:
A -> 1 1 0 0 0 1 0 1
B -> 0 0 1 1 0 0 1 1
C -> 0 0 1 1 1 0 1 0 ( - =minus, u=union)
Find ((A - C) u B) =?
A - C = 10001011
& (A - C) U B =
1 0 0 0 1 0 1 1
0 0 1 1 0 0 1 1
==========
1 0 1 1 1 0 1 1
HENCE,
ANS = 10111011
& (A - C) U B =
1 0 0 0 1 0 1 1
0 0 1 1 0 0 1 1
==========
1 0 1 1 1 0 1 1
HENCE,
ANS = 10111011
6 :: A building with height D ft shadow up to G. A neighbor building with what height shadow C ft is?
G shadow for height D ftC shadow for height = D*C/G
8 :: My flight takes of at 2am from a place at 18N 10E and landed 10 Hrs later at a place with coordinates 36N70W. What is the local time when my plane landed?
a) 6:00 am
b) 6:40am
c) 7:40
d) 7:00
e) 8:00
The Answer is 6:40 AM
The destination place is 80 degree west to the starting place. Hence the time difference between these two places is 5 hour 20 min. (=24hr*80/360).
When the flight landed, the time at the starting place is 12 noon (2 AM + 10 hours).
Hence, the time at the destination place is 12 noon - 5:20 hours = 6: 40 AM
The destination place is 80 degree west to the starting place. Hence the time difference between these two places is 5 hour 20 min. (=24hr*80/360).
When the flight landed, the time at the starting place is 12 noon (2 AM + 10 hours).
Hence, the time at the destination place is 12 noon - 5:20 hours = 6: 40 AM
9 :: How many positive integer solutions does the equation 2x+3y = 100 have?
a) 50
b) 33
c) 16
d) 35
Its very simple... See by observing that in 2x+3y=100 2x is always even and 100 is also even so u need to have 3y also positive. so within 100 3 can be multiplied by 100/33 times to get an even number. So (b) 33 is the answer.