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:
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.
 x |
 |
2 | + | 1 |  |
+ (30 - x). | 2 | = 1 |
| 75 | 45 | 75 |
 |
11x | + | (60 -2x) | = 1 |
| 225 | 75 |
11x + 180 - 6x = 225.
x = 9.
Discussion:
91 comments Page 3 of 10.
Vini said:
6 years ago
How will A become 2/75, can anyone explain?
Abey said:
6 years ago
Thanks for explaining the answer @Kasi.
Devi said:
6 years ago
For this type of question, we can approach like;
Here pipe A fill-in 75/2min(let it be 'x'), pipe B in 45 min(let it be 'y').
At what time B pipe should be close to fill a total cistern in 30 min (let it be 't'min)is,
y(1-(t/x))= 45(1-((30*2)/75)) = 9min.
Here pipe A fill-in 75/2min(let it be 'x'), pipe B in 45 min(let it be 'y').
At what time B pipe should be close to fill a total cistern in 30 min (let it be 't'min)is,
y(1-(t/x))= 45(1-((30*2)/75)) = 9min.
Hithma said:
7 years ago
I Can't understand this. Please, someone explain to me. clearly.
Swapnil said:
7 years ago
Pipe A fills the tank in 75/2 Min.
As pipe A runs 30 mins it fills,
75/2 min = 1 tank.
30 min = 1*2/75*30 = 4/5,
Pipe A fills 4/5 in 30 mins, remaining 1/5.
Pipe B fills 1 full tank in 45 mins,so.
1 tank = 45 min.
1/5 tank = 45/5 = 9 min.
As pipe A runs 30 mins it fills,
75/2 min = 1 tank.
30 min = 1*2/75*30 = 4/5,
Pipe A fills 4/5 in 30 mins, remaining 1/5.
Pipe B fills 1 full tank in 45 mins,so.
1 tank = 45 min.
1/5 tank = 45/5 = 9 min.
(1)
Swapna said:
7 years ago
Thank you @Pintu Francis.
Narula said:
7 years ago
The much simple way is the LCM method.
LCM of (75/2 and 45) = 225, assume this is a total unit of work.
Thus, efficiency of A = [(225)/(75/2)] = 6 units/min and B = 225/45 = 5 units/min.
Since A works for 30 mins, he will finish = 6 x 30 = 180 units.
Remaining = 225 - 180 = 45 units to be completed by B in = (45/5) = 9 mins.
LCM of (75/2 and 45) = 225, assume this is a total unit of work.
Thus, efficiency of A = [(225)/(75/2)] = 6 units/min and B = 225/45 = 5 units/min.
Since A works for 30 mins, he will finish = 6 x 30 = 180 units.
Remaining = 225 - 180 = 45 units to be completed by B in = (45/5) = 9 mins.
(1)
Hari said:
7 years ago
@Sarika.
Just take LCM of 37.5 and 45=450(total units).
You need 30mins to fully fill so(450/30)=15mins u have in total if Both worked non stop
now A works for a full period so 450/75 (convert 37.5 into the whole num as used for LCM)=6 mins.
So, B has worked 15-6=9mins.
Just take LCM of 37.5 and 45=450(total units).
You need 30mins to fully fill so(450/30)=15mins u have in total if Both worked non stop
now A works for a full period so 450/75 (convert 37.5 into the whole num as used for LCM)=6 mins.
So, B has worked 15-6=9mins.
Unnimaya said:
7 years ago
Well said @Kasi Srinivas.
Sarika said:
8 years ago
I didn't get this sum please give me some tricks to understand the problem.
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