Two pipes, P and Q can fill a cistern in 12 and 15 minutes respectively. Both are opened together, but at the end of 3 minutes, P is turned off. In how many more minutes will Q fill the cistern?
A. 7 minutes
B. $$7\frac{1}{2}$$ minutes
C. 8 minutes
D. $$8\frac{1}{4}$$ minutes
Answer: Option D
Solution(By Examveda Team)
P can fill cistern in 12 minutes P fills cistern in 1 minute = $$\frac{1}{{12}}$$ Q can fill cistern in 15 minutes Q fills cistern in 1 minute = $$\frac{1}{{15}}$$ part P and Q together can fill cistern in 1 minute, $$ = \frac{1}{{12}} + \frac{1}{{15}} \Rightarrow \frac{9}{{60}}$$ part So, they can together fill cistern in 3 minute, $$ = 3 \times \frac{9}{{60}} \Rightarrow \frac{9}{{20}}$$ part Rest Cistern = $$1 - \frac{9}{{20}}$$ = $$\frac{{11}}{{20}}$$ part $$\frac{{11}}{{20}}$$ part cistern must be filled by Q in $$\frac{{\frac{{11}}{{20}}}}{{\frac{1}{{15}}}} = 8\frac{1}{4}$$ minutes Alternatively, Pipe P can fill cistern in one minute = $$\frac{{100}}{{12}}$$ = 8.33% Pipe Q can fill cistern in one minute = $$\frac{{100}}{{15}}$$ = 6.66% (P + Q) together can fill the cistern in one minute = 15%; In 3 minutes cistern is filled = 45% P become off, then rest of cistern will be filled by Pipe Q in = $$\frac{{55}}{{6.66}}$$ = $$8\frac{1}{4}$$ minutes.Join The Discussion
Comments ( 3 )
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D. 40 days
Options answer and explanation are wrong
P 12
Q 15
12,15 lcm =60 (total work)
P 1m work =60÷12
=5
Q 1m work =60÷15
=4
last 3 minute p didnt work in which p + q = 5 + 4 = 9 x 3min = 27 units left
which has to be done by q alone
27 / 4 = 6.75 minuts
P 12
Q 15
12,15 lcm =60 (total work)
P 1m work =60÷12
=5
Q 1m work =60÷15
=4
Two pipes 3m work =3(5+4)
=27
Remaining work (60-27)=43
Then
Q will done the remaining work in= 43/4
=8 1/4 m
Option D is answer
This ans is wrong