Question:
A tank is emptied everyday at a fixed time point. Immediately thereafter, either pump A or pump B or both start working until the tank is full. On Monday, A alone completed ?lling the tank at 8 pm. On Tuesday, B alone completed filling the tank at 6 pm. On Wednesday, A alone worked till 5 pm, and then B worked alone from 5 pm to 7 pm, to fill the tank. At what time was the tank ?lled on Thursday if both pumps were used simultaneously all along?
- 4:36 pm
- 4:12 pm
- 4:24 pm
- 4:48 pm
Correct Answer: 3
Let x be the time, on a 24 hours clock, at which the tank is empty.
Time taken by pipe A alone to fill the tank is (20 – x) hrs.
Time taken by pipe B alone to fill the tank is (18 – x) hrs.
On the other day, A fill the tank for (15 – x) hrs and B for 2 hrs.
Let A and B be the rate of works of pipe A and B respectively.
$\Rightarrow(20-x) A=(18-x) B=(17-x) A+2 B$
$\Rightarrow \frac{A}{B}=\frac{2}{3}$
$\Rightarrow(20-x) 2=(18-x) 3$
$\Rightarrow(20-x) A=1$
Let $(20-x) A=1$
$A=\frac{1}{6}$
$B=\frac{1}{4}$
When both work simultaneously, time taken
$=\frac{1}{\frac{1}{6}+\frac{1}{4}}=2.4 \mathrm{hrs}=2 \mathrm{hrs} 24 \mathrm{min}$
The tank will be filled by $16 : 24 \mathrm{i.e.} .4 : 24 \mathrm{pm}$
Let x be the time, on a 24 hours clock, at which the tank is empty.
Time taken by pipe A alone to fill the tank is (20 – x) hrs.
Time taken by pipe B alone to fill the tank is (18 – x) hrs.
On the other day, A fill the tank for (15 – x) hrs and B for 2 hrs.
Let A and B be the rate of works of pipe A and B respectively.
$\Rightarrow(20-x) A=(18-x) B=(17-x) A+2 B$
$\Rightarrow \frac{A}{B}=\frac{2}{3}$
$\Rightarrow(20-x) 2=(18-x) 3$
$\Rightarrow(20-x) A=1$
Let $(20-x) A=1$
$A=\frac{1}{6}$
$B=\frac{1}{4}$
When both work simultaneously, time taken
$=\frac{1}{\frac{1}{6}+\frac{1}{4}}=2.4 \mathrm{hrs}=2 \mathrm{hrs} 24 \mathrm{min}$
The tank will be filled by $16 : 24 \mathrm{i.e.} .4 : 24 \mathrm{pm}$
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