Aptitude Test 8
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 Total number of questions : 15
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 Question 1 of 15
1. Question
1 points.
Find the length of the longest pole that can be placed in a room 12 m long 8m broad and 9m high.
CorrectAnswer : A
Solution : Length of longest pole = Length of the diagonal of the room
= Ö(12^{2}+8^{2}+9^{2}=.Ö(289)= 17 m
IncorrectAnswer : A
Solution : Length of longest pole = Length of the diagonal of the room
= Ö(12^{2}+8^{2}+9^{2}=.Ö(289)= 17 m
 Question 2 of 15
2. Question
1 points.
Nine persons went to a hotel for taking their meals 8 of them spent Rs.12 each on their meals and the ninth spent Rs.8 more than the average expenditure of all the nine.What was the total money spent by them?
CorrectAnswer : B
Solution : Let the average expenditure of all nine be Rs.x
Then 12*8+(x+8)=9x or 8x=104 or x=13.
Total money spent = 9x=Rs.(9*13)=Rs.117.
IncorrectAnswer : B
Solution : Let the average expenditure of all nine be Rs.x
Then 12*8+(x+8)=9x or 8x=104 or x=13.
Total money spent = 9x=Rs.(9*13)=Rs.117.
 Question 3 of 15
3. Question
1 points.
When an amount was distributed among 14 boys, each of them got rs 80 more than the amount received by each boy when the same amount is distributed equally among 18 boys. What was the amount?
CorrectAnswer : C
Solution : Let the total amount be Rs. X the,
x – x = 80 ó 2 x = 80 ó x =63 x 80 = 5040.
14 18 126 63
Hence the total amount is 5040
IncorrectAnswer : C
Solution : Let the total amount be Rs. X the,
x – x = 80 ó 2 x = 80 ó x =63 x 80 = 5040.
14 18 126 63
Hence the total amount is 5040
 Question 4 of 15
4. Question
1 points.
A train starts full of passengers at the first station it drops 1/3 of the passengers and takes 280 more at the second station it drops one half the new total and takes twelve more .on arriving at the third station it is found to have 248 passengers. Find the no of passengers in the beginning?
CorrectAnswer : D
Solution : Let no of passengers in the beginning be x
After first station no passengers=(xx/3)+280=2x/3 +280
After second station no passengers =1/2(2x/3+280)+12
½(2x/3+280)+12=248
2x/3+280=2*236
2x/3=192
x=288
IncorrectAnswer : D
Solution : Let no of passengers in the beginning be x
After first station no passengers=(xx/3)+280=2x/3 +280
After second station no passengers =1/2(2x/3+280)+12
½(2x/3+280)+12=248
2x/3+280=2*236
2x/3=192
x=288
 Question 5 of 15
5. Question
1 pointsCorrectAnswer : A
Solution : Given series is a G.P. with a = 2, r = 2 and n = 8.
\sum = a(r^{n}1) = 2 x (2^{8 }–1) = (2 x 255) =510
(r1) (21)
IncorrectAnswer : A
Solution : Given series is a G.P. with a = 2, r = 2 and n = 8.
\sum = a(r^{n}1) = 2 x (2^{8 }–1) = (2 x 255) =510
(r1) (21)
 Question 6 of 15
6. Question
1 pointsCorrectAnswer : A
Solution : All 2 digit numbers divisible by 3 are :
12, 51, 18, 21, …, 99.
This is an A.P. with a = 12 and d = 3.
Let it contain n terms. Then,
12 + (n – 1) x 3 = 99 or n = 30.
\Required sum = 30/2 x (12+99) = 1665.
IncorrectAnswer : A
Solution : All 2 digit numbers divisible by 3 are :
12, 51, 18, 21, …, 99.
This is an A.P. with a = 12 and d = 3.
Let it contain n terms. Then,
12 + (n – 1) x 3 = 99 or n = 30.
\Required sum = 30/2 x (12+99) = 1665.
 Question 7 of 15
7. Question
1 points.
A man spends 2/5 of his salary on house rent,3/10 of his salary on food and 1/8 of his salary on conveyence.if he has Rs.1400 left with him,find his expenditure on food and conveyence.
CorrectAnswer : B
Solution : Part of salary left=1(2/5+3/10+1/8)
Let the monthly salary be Rs.x
Then, 7/40 of x=1400
X=(1400*40/7)
=8600
Expenditure on food=Rs.(3/10*800)=Rs.2400
Expenditure on conveyence=Rs.(1/8*8000)=Rs.1000
IncorrectAnswer : B
Solution : Part of salary left=1(2/5+3/10+1/8)
Let the monthly salary be Rs.x
Then, 7/40 of x=1400
X=(1400*40/7)
=8600
Expenditure on food=Rs.(3/10*800)=Rs.2400
Expenditure on conveyence=Rs.(1/8*8000)=Rs.1000
 Question 8 of 15
8. Question
1 points.
A can do a piece of work in 80 days. He works at it for 10 days B alone finishes the remaining work in 42 days. In how much time will A and B working together, finish the work?
CorrectAnswer : C
Solution : Work done by A in 10 days =(1/80*10)=1/8
Remaining work = (1 1/8) =7/ 8
Now,7/ 8 work is done by B in 42 days.
Whole work will be done by B in (42 x 8/7) = 48 days.
A’s 1 day’s work = 1/80 and B’s 1 day’s work = 1/48
IncorrectAnswer : C
Solution : Work done by A in 10 days =(1/80*10)=1/8
Remaining work = (1 1/8) =7/ 8
Now,7/ 8 work is done by B in 42 days.
Whole work will be done by B in (42 x 8/7) = 48 days.
A’s 1 day’s work = 1/80 and B’s 1 day’s work = 1/48
 Question 9 of 15
9. Question
1 points.
A train 150 m long is running with a speed of 68 kmph. In what time will it pass a man who is running at 8 kmph in the same direction in which the train is going?
CorrectAnswer : D
Solution : Speed of the train relative to man = (68 – 8) kmph
= (60* 5/18) m/sec = (50/3)m/sec
Time taken by the train to cross the man I
= Time taken by It to cover 150 m at 50/3 m / sec = 150 *3/ 50 sec = 9sec
IncorrectAnswer : D
Solution : Speed of the train relative to man = (68 – 8) kmph
= (60* 5/18) m/sec = (50/3)m/sec
Time taken by the train to cross the man I
= Time taken by It to cover 150 m at 50/3 m / sec = 150 *3/ 50 sec = 9sec
 Question 10 of 15
10. Question
1 points.
The length of a rectangle is twice its breadth. If its length is decreased by 5 cm and breadth is increased by 5 cm, the area of the rectangle is increased by 75 sq. cm. Find the length of the rectangle.
CorrectAnswer : A
Solution : Let breadth = x. Then, length = 2x. Then,
(2x – 5) (x + 5) – 2x * x = 75 <=> 5x – 25 = 75 <=> x = 20.
:. Length of the rectangle = 20 cm.
IncorrectAnswer : A
Solution : Let breadth = x. Then, length = 2x. Then,
(2x – 5) (x + 5) – 2x * x = 75 <=> 5x – 25 = 75 <=> x = 20.
:. Length of the rectangle = 20 cm.
 Question 11 of 15
11. Question
1 points.
In a simultaneous throw of pair of dice .find the probability of getting the total more than 7.
CorrectAnswer : D
Solution : Here n(s)=(6*6)=36
let e=event of getting a total more than 7
={(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)}
p(e)=n(e)/n(s)=15/36=5/12.
IncorrectAnswer : D
Solution : Here n(s)=(6*6)=36
let e=event of getting a total more than 7
={(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)}
p(e)=n(e)/n(s)=15/36=5/12.
 Question 12 of 15
12. Question
1 pointsCorrectAnswer : C
Solution : Let original radius = R. Then, new radius = (150/100)R=(3R/2)
Original volume = (4/3)ÕR^{3}, New volume = (4/3)Õ(3R/2)^{3 }=(9ÕR^{3}/2)
Increase % in volume=((19/6)ÕR^{3})*(3/4ÕR^{3})*100))% = 237.5%
IncorrectAnswer : C
Solution : Let original radius = R. Then, new radius = (150/100)R=(3R/2)
Original volume = (4/3)ÕR^{3}, New volume = (4/3)Õ(3R/2)^{3 }=(9ÕR^{3}/2)
Increase % in volume=((19/6)ÕR^{3})*(3/4ÕR^{3})*100))% = 237.5%
 Question 13 of 15
13. Question
1 points.
I walk a certain distance and ride back taking a total time of 37 minutes. I could walk both ways in 55 minutes. How long would it take me to ride both ways?
CorrectAnswer : B
Solution : Let the distance be x km. Then,
( Time taken to walk x km) + (time taken to ride x km) =37 min.
( Time taken to walk 2x km ) + ( time taken to ride 2x km )= 74 min.
But, the time taken to walk 2x km = 55 min.
Time taken to ride 2x km = (7455)min =19 min.
IncorrectAnswer : B
Solution : Let the distance be x km. Then,
( Time taken to walk x km) + (time taken to ride x km) =37 min.
( Time taken to walk 2x km ) + ( time taken to ride 2x km )= 74 min.
But, the time taken to walk 2x km = 55 min.
Time taken to ride 2x km = (7455)min =19 min.
 Question 14 of 15
14. Question
1 points.
After getting 2 successive discounts, a shirt with a list price of Rs 150 is available at Rs 105. If the second discount is 12.55,find the first discount.
CorrectAnswer : C
Solution : Let the first discount be x%
Then,87.5% of (100x)% of 150= 105
87.5/100*(100x)/100*450=150=>105=>100x=(105*100*100)/(150*87.5)=80
x=(10080)=20
first discount = 20%
IncorrectAnswer : C
Solution : Let the first discount be x%
Then,87.5% of (100x)% of 150= 105
87.5/100*(100x)/100*450=150=>105=>100x=(105*100*100)/(150*87.5)=80
x=(10080)=20
first discount = 20%
 Question 15 of 15
15. Question
1 points.
The average weight of 8 person’s increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What might be the weight of the new person?
CorrectAnswer : C
Solution : Total weight increased = (8 x 2.5) kg = 20 kg.
Weight of new person = (65 + 20) kg = 85 kg.
IncorrectAnswer : C
Solution : Total weight increased = (8 x 2.5) kg = 20 kg.
Weight of new person = (65 + 20) kg = 85 kg.