Aptitude - Pipes and Cistern - Discussion

Discussion Forum : Pipes and Cistern - General Questions (Q.No. 10)
10.
Two pipes A and B can fill a tank in 15 minutes and 20 minutes respectively. Both the pipes are opened together but after 4 minutes, pipe A is turned off. What is the total time required to fill the tank?
10 min. 20 sec.
11 min. 45 sec.
12 min. 30 sec.
14 min. 40 sec.
Answer: Option
Explanation:

Part filled in 4 minutes = 4 1 + 1 = 7 .
15 20 15

Remaining part = 1 - 7 = 8 .
15 15

Part filled by B in 1 minute = 1
20

1 : 8 :: 1 : x
20 15

x = 8 x 1 x 20 = 10 2 min = 10 min. 40 sec.
15 3

The tank will be full in (4 min. + 10 min. + 40 sec.) = 14 min. 40 sec.

Discussion:
37 comments Page 2 of 4.

D Priya said:   8 years ago
@Leela.

To fulfil the condition they equate to 1. A part of work + remaining part of work = 1. To be precise, you and your friend are doing a work by sharing. You completed half the work and your friend completed half the remaining work. Ie 1/2 + 1/2 = 1.

This logic is being used in this kind of problems.

Leela said:   8 years ago
Can anyone tell me why do we equate to one?

Manisha said:   8 years ago
It is clear now Thank you, @Ashim.

Varsha said:   8 years ago
Why equating with 1?

Venky said:   8 years ago
Can anyone explain it in easy method?

Ashim said:   9 years ago
LCM of 15 & 20 is 60.
(4 + 3) * 4 = 28.
60 - 28 = 32.
4 + 32/3 =14min = 40sec.

Sandy said:   9 years ago
Super solution @Kalpna.

Neha said:   9 years ago
How to change the fraction value into minutes?

Ramees said:   10 years ago
Easy to convert 44/3 into 14(2/3).

Divide 44 into 3 we get remainder 2(i.e 14*3 = 42).

Then we can write 14 (remainder/3) that is 14(2/3).

Vikash patel said:   1 decade ago
Why x-4?


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