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kajal jadhav

· started a discussion

· 1 Months ago

kjl (Question: 338600.0 )

i dont understand plz any other idia

Question:
A cistern has three pipes P,Q and R. The pipes P & Q can fill it in 6 hours and 7 hours respectively and R can empty it in 3 hours. If the pipes are opened in the same order at 1 pm,2 pm and 3 pm, when will the cistern be empty?
Options:
A) 9 am
B) 1 am
C) 10 am
D) 11 am
E) None of these
Solution:

By 3 pm work done by pipe ‘P’ for 2 hours and pipe ‘Q’ for 1 hours 

= 2 ×1/6 + 1 ×1/7

= 10/21

At 3pm the work done by P,Q and R in 1 hour

= 1/6 + 1/7 - 1/3

= \(\cfrac{7+6-14}{42}\)

= - 1/42

i.e., in every hour 1/42 of the cistern will be leaked out. Thus, 10/21 of the cistern will be leaked out after 20 hours.

After 3 pm add 20 hours the time will be 11 am.

Knowledge Expert

· commented

· 1 Months ago

Dear student,
Pipe P and Q work for 2 and 1 hour respectively that is 2*(1/6) + 1*(1/7) = 10/21 so P and Q filled 10/21 parts till 3PM, but now pipe R will open at 3pm.. so in one our these three pipes leak 1/42 parts..., ( but we have 10/21 parts to leak/empty the tank from 3pm ) Here we are trying to empty 10/21 parts... ie, 1/42 for 1 hour, 10/21 in 10 hours.. finally 3pm + 10 hours = 11 am.
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