A) 7 : 12 pm
B) 7 : 15 pm
C) 7 : 10 pm
D) 7 : 18 pm
Correct Answer: A
Solution :
[a] Part filled by pipe A in \[2\,h=\frac{2}{3}\] Part filled by pipe B in \[1\,h=\frac{1}{4}\] Total filled part\[=\frac{2}{3}+\frac{1}{4}=\frac{8+3}{12}=\frac{11}{12}\] Now, part emptied by all the three pipes in 1 h \[=1-\frac{1}{3}-\frac{1}{4}=\frac{12-4-3}{12}=\frac{5}{12}\] Since, \[\frac{5}{12}\]part is emptied in 1 h. Therefore, \[\frac{11}{12}\]part is emptied in \[\frac{11}{12}\times \frac{12}{5}\,h=\frac{11}{5}\,h=2\,h+12\min \] Total required time \[=5\,h+2\,h+12\min =7\,h\,\,12\min \] \[=7:12\,pm.\] |
You need to login to perform this action.
You will be redirected in
3 sec