Miscellaneous Question 79
- Two inlet pipes $P$ and $Q$ can fill an empty tank in 18 hours and 20 hours respectively. Both the pipes were opened simultaneously but as soon as the tank was half filled a leak was detected in the bottom of the tank. After closing that leak, the two inlet pipes took 1 hour less to fill the remaining half of the tank than they had taken to fill the first half of the tank. Find how much time the leak would take to empty the filled tank? (in hours)
(1) $54 \frac{126}{361}$
(2) $51 \frac{126}{361}$
(3) $44 \frac{126}{361}$
(4) $58 \frac{126}{361}$
(5) Other than those given as options
IBPS Bank PO/MT CWE (Prelim Exam) 12.10.2019
Show Answer
Correct Answer: 79. (1)
Solution: 79. (1) Let the leak empty the full tank in $x$ hours.
Case I,
Part of tank filled in 1 hour
$=\frac{1}{18}+\frac{1}{20}-\frac{1}{x}$
$=\frac{10 x+9 x-180}{180 x}$
$=\frac{19 x-180}{180 x}$
$\therefore$ Time taken in filling half tank
$=\frac{90 x}{19 x-180}$
Case II,
Time taken in filling remaining half tank
$=\frac{1}{2\left(\frac{1}{18}+\frac{1}{20}\right)}$
$=\frac{1}{2\left(\frac{10+9}{180}\right)}=\frac{90}{19}$
$\therefore \frac{90 x}{19 x-180}-\frac{90}{19}=1$
$\Rightarrow 90\left(\frac{19 x-19 x+180}{19(19 x-180)}\right)=1$ $\Rightarrow 19(19 x-180)=90 \times 180$
$=16200$
$\Rightarrow 19 x-180=\frac{16200}{19}$
$\Rightarrow 19 x=\frac{16200}{19}+180$
$=\frac{16200+3420}{19}=\frac{19620}{19}$
$\Rightarrow x=\frac{19620}{361}$
$=54 \frac{126}{361}$ hours
BETA
