Aptitude - Pipes and Cistern - Discussion
Discussion Forum : Pipes and Cistern - General Questions (Q.No. 4)
4.
Two pipes A and B can fill a cistern in 37
minutes and 45 minutes respectively. Both pipes are opened. The cistern will be filled in just half an hour, if the B is turned off after:

Answer: Option
Explanation:
Let B be turned off after x minutes. Then,
Part filled by (A + B) in x min. + Part filled by A in (30 -x) min. = 1.
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![]() |
2 | + | 1 | ![]() |
+ (30 - x). | 2 | = 1 |
75 | 45 | 75 |
![]() |
11x | + | (60 -2x) | = 1 |
225 | 75 |
11x + 180 - 6x = 225.
x = 9.
Discussion:
90 comments Page 8 of 9.
Vini said:
6 years ago
How will A become 2/75, can anyone explain?
Nilesh12 said:
6 years ago
Formula(1 - total time/2nd pipe time)*1st pipe time.
then,
(1-30/75/2)*45.
Means,
(1-60/75)*45.
(75-60/75)*45.
(1/5)*45.
Answer 9.
then,
(1-30/75/2)*45.
Means,
(1-60/75)*45.
(75-60/75)*45.
(1/5)*45.
Answer 9.
(1)
Karansingh said:
6 years ago
There is a shorcut trick here.
Find lcm of 75 and 45.
Lcm is 225.
Divide by 75 and 45 to lcm.
Hereafter dividing lcm you will get 3 and 5.
Now given is that cistern will be filled in half hour means 30 min if b turned off.
They do 30/5 = 6.
Now add 3+6 = 9 (why to add this because we asked that if it is closed then we have to add that 3+6).
Find lcm of 75 and 45.
Lcm is 225.
Divide by 75 and 45 to lcm.
Hereafter dividing lcm you will get 3 and 5.
Now given is that cistern will be filled in half hour means 30 min if b turned off.
They do 30/5 = 6.
Now add 3+6 = 9 (why to add this because we asked that if it is closed then we have to add that 3+6).
(1)
Gunjan said:
5 years ago
Y(1-t/x).
X = 75/2.
Y =45.
T = 30.
This is the formula for this question.
X = 75/2.
Y =45.
T = 30.
This is the formula for this question.
OmkarR said:
5 years ago
@All.
According to me, the solution is;
The cistern will be filled in just half an hour, if the B is turned off after:
i.e A work 30 min so,
A total work 75/2 minutes. 30 /75/2=4/5.
Remaining work is 1-4/5=1/5,
So,
B can fill in 45 minute. Then 45/5 =9.
According to me, the solution is;
The cistern will be filled in just half an hour, if the B is turned off after:
i.e A work 30 min so,
A total work 75/2 minutes. 30 /75/2=4/5.
Remaining work is 1-4/5=1/5,
So,
B can fill in 45 minute. Then 45/5 =9.
(1)
Ruby said:
5 years ago
Another solution in term of common sense just for MCQ purpose:
The time that A takes to fill - 371/2= 37.5 min
The time that B takes more than A to fill - 45/37.5=1.2 times
After 30 min extra time would have been needed by A to fill - 7.5 minutes.
The time that B needs to compensate for - 7.5*1.2= 9 minutes.
The time that A takes to fill - 371/2= 37.5 min
The time that B takes more than A to fill - 45/37.5=1.2 times
After 30 min extra time would have been needed by A to fill - 7.5 minutes.
The time that B needs to compensate for - 7.5*1.2= 9 minutes.
Arun said:
5 years ago
Let 'B' be turned off after 'x' minutes.
Then:
Part filled by A and B in 1 minute:
[(2/75) + (1/x) ] = [1/30] minutes.
[(2x + 75) ] 30 = 75x.
x = 225/9 = 9 minutes(Ans).
Then:
Part filled by A and B in 1 minute:
[(2/75) + (1/x) ] = [1/30] minutes.
[(2x + 75) ] 30 = 75x.
x = 225/9 = 9 minutes(Ans).
Shruti said:
5 years ago
We get total amount to be filled by taking the LCM of (45,75/2) = 225litres.
Efficiency of A=225/(75/2)= 6litres/min
Efficiency of B=225/(45) = 5litres/min
So A in 30mins =30* 6 = 180litres
Remaining amount = 225 - 180 = 45L
Now B fills 5L in 1min,
45L it requires 9mins.
Efficiency of A=225/(75/2)= 6litres/min
Efficiency of B=225/(45) = 5litres/min
So A in 30mins =30* 6 = 180litres
Remaining amount = 225 - 180 = 45L
Now B fills 5L in 1min,
45L it requires 9mins.
(10)
Clinton said:
5 years ago
((A - Time of closing)/A) * B = Answer.
Here,
A=75/2.
Time of Closing the pipe=30.
B = 45.
So, Apply in the formula.
(((75/2)-30)/(75/2)) * 45 = 9 (Answer).
Here,
A=75/2.
Time of Closing the pipe=30.
B = 45.
So, Apply in the formula.
(((75/2)-30)/(75/2)) * 45 = 9 (Answer).
(2)
Rohini said:
5 years ago
@Kasi.
Nice explanation.
Nice explanation.
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