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Munagala Janardhana Reddy

· started a discussion

· 1 Months ago

alternate solution (Question: 271926.0 )

given pipe A fill in20 min
leakage empty in (rate is 1/3) ==> take time 3 times to empty
==>3*20= 60 min
to fill half tank let it takes x minutes
x((1/20) - (1/60)) = 1/2 ==>x = 15min
to empty leakage alone that half tank be y
y(1/60) = 1/2 ==> y =30 min
total time for this process is = x + y = 15 +30 = 45 min

Question:

Tap A fills a tank in 20 min while B empties it at  13   rd the rate at which A fills it. At 5 pm, A and B are simultaneously started and the tank is 50% full, tap A is turned off. At what time will the tank be empty?

Options:
A)

12 : 00 AM

B)

5 : 30 PM

C)

5 : 45 PM

D)

6 : 00 PM

Solution:

Ans: (c)

  

Abhishek

· commented

· 1 Months ago

this is very lengthy method!
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