Two pipes A and B can fill a tank in 15 min and 20 min, respectively. Both the pipes are opened together but after 4 min, pipe A is turned off. What is the total time required to fill the tank? [FCI (Assistant) Grade III 2015] |
A) 12 min 40 s
B) 11 min 35 s
C) 14 min 40 s
D) 13 min 35 s
Correct Answer: C
Solution :
Part filled by both in 1 min |
\[=\frac{1}{15}+\frac{1}{20}=\frac{4+3}{60}=\frac{7}{60}\] |
Now, part filled in 4 min \[=4\times \frac{7}{60}=\frac{7}{15}\] |
\[\therefore \] Remaining part \[=1-\frac{7}{15}=\frac{8}{15}\] |
Now, let the remaining part is filled by pipe B in x min. |
Then, \[x\times \frac{1}{20}=\frac{8}{15}\] |
\[\Rightarrow \] \[x=\frac{8\times 20}{15}=\frac{8\times 4}{3}\] |
\[\Rightarrow \] \[x=\frac{32}{3}=10\frac{2}{3}=\left( 10+\frac{2}{3} \right)\min \] |
\[\Rightarrow \] \[x=10+\frac{2}{3}\times 60\] |
\[x=10\min \,+40\,\,=\,10\min 40\,s\] |
\[\therefore \] Total time taken to fill tank |
\[=4\min +10\min 40\,s\] |
\[=14\min 40\,s\] |
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