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55 Quantitative Aptitude Questions and Answers:
Table of Contents:
1 :: A man walks 3km east then 5km north then 8km east. Now how far is he from the starting point
if u draw a rectangle of length of total(3+8) km and breadth 5km and draw a diagonal from the starting to ending point, by Pythagaros Theorem, square(hyp)=square(the both sides)
Therefore, 25+121=sqrt(answer)
it comes to be 12.083km
Therefore, 25+121=sqrt(answer)
it comes to be 12.083km
2 :: One ship start his journey from a port and another ship start his journey from H at the same time, at which place they will meet?
The right question is:A ship started from port and moving with I mph and another ship started from L and moving with H mph. At which place these two ships meet? !_____!_____!_____!_____!_____!_____! port G H I J K L
3 :: Find the fourth row, having the bit pattern as an integer in an 8-bit computer, and express the answer in its decimal value.
A 0 0 0 0 1 1 1 1
B 0 0 1 1 0 0 1 1
C 0 1 0 1 0 1 0 1
(AU(B-C)) ?
223 is answer
B-C is bitwise addition of B and 2s compliment of C
then bitwise union of A with abover result gives the ans as 1101 1111 which
has decimal value of 223.
B-C is bitwise addition of B and 2s compliment of C
then bitwise union of A with abover result gives the ans as 1101 1111 which
has decimal value of 223.
4 :: Which of the following set of numbers has the highest Standard deviation?
1,0,1,0,1,0
-1,-1,-1,-1, -1,-1
1,1,1,1,1,1
1,1,0,-1,0,- 1
-1,-1,-1,-1, -1,-1
5 :: Wind flows 160 miles in 330 min, for 80 miles how much time required?
answer as 165
as
160 miles are covered in 330 min
1 mile will be covered in 330/160 min
Hence 80 miles will be covered in 80*330/160 =165 min.
as
160 miles are covered in 330 min
1 mile will be covered in 330/160 min
Hence 80 miles will be covered in 80*330/160 =165 min.
6 :: A salesperson multiplied a number and got the answer 3, instead of that number divided by 3.What is the answer he actually has to get?
3*1=3
instead of 1 if he divided by 3
then he will get 1only
3*(1/3)=1
instead of 1 if he divided by 3
then he will get 1only
3*(1/3)=1
7 :: For temperature a function is given according to time :((t**2)/6)+4t+12 what is the temperature rise or fall between 5PM to 8PM?
Apply t=5, ans=33.7 for t=8, ans=46.7
The differenc between the two, 46.7-33.7=13 is the answr upto my knowledge.
The differenc between the two, 46.7-33.7=13 is the answr upto my knowledge.
8 :: A building with height D shadow up to G. A neighbor building with what height shadows C feet. |----|----|----|----|----|----|----| A B C D E F G H
this can be solved by using the rule of cross multiplication i guess....
to get a shadow of G ft the ht of the building should be D
hence, for the shadow to be C ft, the height of the building should be CxD/G
to get a shadow of G ft the ht of the building should be D
hence, for the shadow to be C ft, the height of the building should be CxD/G
9 :: 27(deg) 6 43.15\
45(deg) 31 4.3\
find the degrees when the time is 3o clock?
The angel between the minutes and hour hand would be 45 degree at 3'0 clock
10 :: What is the largest prime no. stored in 8bit/6bit/9bit/7bit memory?
2. A,B,C are the mechanisms used separately to reduce the wastage of fuel by 25%,20%,10%. What will be fuel economy if they were used combine?
for 1st machine 100-25 ie 75
for 2nd machine 100-20 ie 80
for 3rd machine its 100-10 ie 95
so efficiency in rev = (90*80*75)/10000
=54
so fuel economy =100-54
=46
for 2nd machine 100-20 ie 80
for 3rd machine its 100-10 ie 95
so efficiency in rev = (90*80*75)/10000
=54
so fuel economy =100-54
=46
11 :: A bus started from bus stand at 8.00am, and after 30 minutes staying at destination, it returned back to the bus stand. The destination is 27 miles from the bus stand. The speed of the bus is 18mph. In return journey bus travels with 50% fast speed. At what time it returns to the bus stand?
As the Bus is travelling 18mph it takes 1 & 1/2 hr to reach destination.
Bus stays there for 1/2 hr.
As in the return journy bus travelled with the 50% fast spped meance with the speed of 27mph. meance it took 1 hr to return
For total journy it took 1&1/2+1/2+1hr=3 hr.
therefore 8+3 =11
Bus returns as 11 a.m.
Bus stays there for 1/2 hr.
As in the return journy bus travelled with the 50% fast spped meance with the speed of 27mph. meance it took 1 hr to return
For total journy it took 1&1/2+1/2+1hr=3 hr.
therefore 8+3 =11
Bus returns as 11 a.m.
12 :: A plane moves from 9ËšN40ËšE to 9ËšN40ËšW. If the plane starts at 10 am and takes 8 hours to reach the destination, find the local arrival time?
80 deg --------> 80*4 = 320 minutes = 5 hours 20 minutes
10 a.m + 8 hours = 6 p.m
6 p.m + 5 hours + 20 minutes = 11.20 p.m = answer
10 a.m + 8 hours = 6 p.m
6 p.m + 5 hours + 20 minutes = 11.20 p.m = answer
14 :: The number 362 in decimal system is given by (1362)x in the X system of numbers find the value of X
a} 5
b) 6
c) 7
d) 8
e) 9
Ans. is 6.
(1362)to base x = (362) to base 10
so
1*x^3 + 3 * x^2 + 6 * x^1 + 2*x^0 = 362
substitute x = 6;
216 + 108 + 36 + 2 = 362.
we'll get 362.
Hence awnser is 6.
(1362)to base x = (362) to base 10
so
1*x^3 + 3 * x^2 + 6 * x^1 + 2*x^0 = 362
substitute x = 6;
216 + 108 + 36 + 2 = 362.
we'll get 362.
Hence awnser is 6.
15 :: A person was fined for exceeding the speed limit by 10mph. Another person was also fined for exceeding the same speed limit by twice the same. If the second person was traveling at a speed of 35 mph, find the speed limit?
Threshold speed = x mph
let us assume that the 1st person was travelling at (x+10)mph
so it's obvious that the 2nd person travels at 2*(x+10)mph = 35mph; which means x = 7.5 which is the threshold or the speed limit.
So according to me the ans is 7.5mph.
let us assume that the 1st person was travelling at (x+10)mph
so it's obvious that the 2nd person travels at 2*(x+10)mph = 35mph; which means x = 7.5 which is the threshold or the speed limit.
So according to me the ans is 7.5mph.
16 :: TCS Quantitative Aptitude Questions:
1. 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
2.in 80 coins one coin is counterfiet what is minimum number of weighings to find out counterfiet coin ?
3.Three men goes to a hotel to stay ,the clerk says $30 per room/day so all the three plans to stay in one room so each pays $10.After some time the clerk realizes that he made a mistake of collecting $30 but the room cost only $25,there fore he decides to return $5 to them so he calls the roomboy and gives him $5 asking him to return.The roomboy keeps $2 with him and he returns only $3($1 for each).Now Totally all have paid $9 each($27)+rommboy $2 which is equal to $27.where did $1 go,who has made the mistake?
a) 6:00 am b) 6:40am c)7:40 d)7:00 e)8:00
2.in 80 coins one coin is counterfiet what is minimum number of weighings to find out counterfiet coin ?
3.Three men goes to a hotel to stay ,the clerk says $30 per room/day so all the three plans to stay in one room so each pays $10.After some time the clerk realizes that he made a mistake of collecting $30 but the room cost only $25,there fore he decides to return $5 to them so he calls the roomboy and gives him $5 asking him to return.The roomboy keeps $2 with him and he returns only $3($1 for each).Now Totally all have paid $9 each($27)+rommboy $2 which is equal to $27.where did $1 go,who has made the mistake?
17 :: A 0 0 0 0 1 1 1 1
B 0 0 1 1 0 0 1 1
C 0 1 0 1 0 1 0 1
Find ( A u B ) n C
To find A-C, We will find 2's compliment of C and them add it with A,
That will give us (A-C)
2's compliment of C=1's compliment of C+1
=10101010+1=10101011
A-C=10101011+00001111
=10111010
Now (A-C) U B is .OR. logic operation on (A-C) and B
10111010.OR . 00110011
The answer is = 10111011,
Whose decimal equivalent is 187.
That will give us (A-C)
2's compliment of C=1's compliment of C+1
=10101010+1=10101011
A-C=10101011+00001111
=10111010
Now (A-C) U B is .OR. logic operation on (A-C) and B
10111010.OR . 00110011
The answer is = 10111011,
Whose decimal equivalent is 187.
18 :: 194 base 10= ___ base 5
The base 5 representation of the decimal number 2048 is______?
2048 to base 10:-
2 ["add" the next (first) digit]
2 * 10 = 20 [multiply by the base]
20 + 0 = 20 [add the next digit]
20 * 10 = 200 [multiply by the base]
200 + 4 = 204 [add the next digit]
204 * 10 = 2040 [multiply by the base]
2040+ 8 = 2048 [add the next digit]
??? to base 5:-
3
3* 5 = 15 [multiply by the base]
15+1 = 16 [add the next digit]
16* 5 = 80 [multiply by the base]
80+ 1 = 81 [add the next digit]
81* 5 = 405 [multiply by the base]
405+ 4 = 409 [add the next digit]
409*5 = 2045
2045+3 = 2048
since both the answers tally i.e., 2048==2048
hence the no. 31143 is the ans..
2 ["add" the next (first) digit]
2 * 10 = 20 [multiply by the base]
20 + 0 = 20 [add the next digit]
20 * 10 = 200 [multiply by the base]
200 + 4 = 204 [add the next digit]
204 * 10 = 2040 [multiply by the base]
2040+ 8 = 2048 [add the next digit]
??? to base 5:-
3
3* 5 = 15 [multiply by the base]
15+1 = 16 [add the next digit]
16* 5 = 80 [multiply by the base]
80+ 1 = 81 [add the next digit]
81* 5 = 405 [multiply by the base]
405+ 4 = 409 [add the next digit]
409*5 = 2045
2045+3 = 2048
since both the answers tally i.e., 2048==2048
hence the no. 31143 is the ans..
19 :: A work is done by two people in 24 min. one of them can do this work a lonely in 40 min. how much time required to do the same work for the second person?
let A does the work in 40 mins (given)
let B does the work in x mins
amt of work done by A in 1 min = 1/40
amt of work done by B in 1 min = 1/x
amt of work done jointly by both A & B = 1/24
which implies (1/40)+(1/x) = (1/24)
on solvin the above eqn. for x we get
x = 60
let B does the work in x mins
amt of work done by A in 1 min = 1/40
amt of work done by B in 1 min = 1/x
amt of work done jointly by both A & B = 1/24
which implies (1/40)+(1/x) = (1/24)
on solvin the above eqn. for x we get
x = 60
20 :: Two trees are there. One grows at 3/5 of the other in 4 years, total growth of trees is 8 ft. what growth will smaller tree will have in 2 years?
Answer should be 1.5 !!!
Let the trees be x,y
from question : -
eq1 is x = 3/5 (y)
eq2 after 4 yrs,
x+y = 8
from the 2 equations we get y = 5 and x = 3
according to our equation x= 3/5(y) -> x is the smaller tree.
In 4 yrs x grows 3ft
In 2 yrs x grows ?ft
Ans = 1.5ft.
Let the trees be x,y
from question : -
eq1 is x = 3/5 (y)
eq2 after 4 yrs,
x+y = 8
from the 2 equations we get y = 5 and x = 3
according to our equation x= 3/5(y) -> x is the smaller tree.
In 4 yrs x grows 3ft
In 2 yrs x grows ?ft
Ans = 1.5ft.
21 :: Suppose Jalia is twice older than qurban. If Jalia was 4 years younger, qurban was 3 years older, the diff. between their ages is 12 years?
Jalia is 38 years old and qurban is 19 years old.
38-4=34 jalia younger
19+3=22 qurban older the difference btw them is 34-22=12.
38-4=34 jalia younger
19+3=22 qurban older the difference btw them is 34-22=12.
22 :: If A,B,C are the mechanism used seperately to reduce the wastage of the fuel by 30%, 40%, 10% What will be the fuel economy if they were used Combined?
A. 68.4
B. 62.2
C. 58
D. 27
Let each mechanism solution is 100 litre.
In "A" mechanism, wastage is (100-30)/100 = 70 %
In "B" mechanism, wastage is (100-40)/100 = 60 %
In "C" mechanism, wastage is (100-10)/100 = 90 %
If all are combined,total solution = 300 litres
Total wastage = 220 litres
Total used fuel = 80 litres
So, wastage in 100 litres = 80/3 litres
% wastage = 26.6666.... %
= 27 % (approx)
In "A" mechanism, wastage is (100-30)/100 = 70 %
In "B" mechanism, wastage is (100-40)/100 = 60 %
In "C" mechanism, wastage is (100-10)/100 = 90 %
If all are combined,total solution = 300 litres
Total wastage = 220 litres
Total used fuel = 80 litres
So, wastage in 100 litres = 80/3 litres
% wastage = 26.6666.... %
= 27 % (approx)
23 :: A super fast bus of KSRTC starting from Trivandrum and reaches Attingal in 45 minutes with an average speed of 40 km/hr. If the speed is increased by 10 km/hr how much time it will take to cover the same distance?
(a) 34 minutes
(b) 36 minutes
(c) 38 minutes
(d) 40 minutes
b) 36 min
d=s*t
s=40 km/hr=40/60km/min
t=45min
d=(4/6)*45=30km
now increased speed= 50 km/hr=50/60 km/min
d=30 km
t=d/s
= 36 min
d=s*t
s=40 km/hr=40/60km/min
t=45min
d=(4/6)*45=30km
now increased speed= 50 km/hr=50/60 km/min
d=30 km
t=d/s
= 36 min
24 :: A ship started from port and moving with I miles per hour and another ship started from L and moving with H miles per hour. At which place these two ships meet? |----|----|----|----|----|----| port G H I J K L
Well ship A moves at 2blks per hr and ship B moves at 4 blks per hr..assuming they move towards each we should have the answer between h and i. Lets take one hr time, ship A will be at I and B at H, it means they have alredy met so just moving them both back we will have their meeting between H&I.