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 4 of 4.

Anjali said:   1 decade ago
Hi Kalpna,

How you write 44/3 = 14 min 40 sec ?

Talha said:   1 decade ago
How is 44/3 = 14 min 40 sec ?

Satyajit said:   1 decade ago
@thnx kalpna

Naveen said:   1 decade ago
Hi santhu.

Its 44/3 only.

Santhu said:   1 decade ago
Hi kalpana how u solve this x value

4/15+x/20=1

from above equation x val is = 46/3 its not 44/3....

Priya P. said:   1 decade ago
@ kalpna.

Thanks. Really less time consuming.
(1)

Kalpna said:   1 decade ago
Another method could be : 4/15+x/20=1

x=44/3 i.e 14 min 40 sec.

4/15 is the part filled by A in 4 mins

and x is the total time taken to fill the tank by B.
(1)


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