A tap can fill a tank in 6 hours. After 4 hours it was found that only 60% of the tank is full due to a leakage at the bottom which was then immediately repaired. In how much time will the tank be completely filled?

Question

Prepp Team · Accepted Answer

Understanding the Tank Filling Problem

This problem involves calculating the total time to fill a tank considering a normal filling rate and an unexpected leakage. We are given the time it takes for a tap to fill the tank completely, the time elapsed before a leakage was noticed and repaired, and the percentage of the tank filled during that time.

Calculating Rates

First, let's determine the rate at which the tap fills the tank and the rate at which the leakage empties it.

Tap's Filling Rate: The tap fills the entire tank in 6 hours. So, the filling rate of the tap is:
Rate_tap = $\frac{1 \text{ tank}}{6 \text{ hours}} = \frac{1}{6}$ tank per hour.
Work Done by Tap in 4 Hours (without leakage): If there were no leakage, in 4 hours, the tap would fill:
Amount_filled_ideal = Rate_tap $\times$ Time = $\frac{1}{6} \times 4 = \frac{4}{6} = \frac{2}{3}$ of the tank.
Actual Work Done in 4 Hours: We are told that after 4 hours, only 60% of the tank was full. Let's convert 60% to a fraction:
60% = $\frac{60}{100} = \frac{3}{5}$ tank.
Amount Leaked in 4 Hours: The difference between the ideal amount filled and the actual amount filled is the amount lost due to leakage.
Amount_leaked = Amount_filled_ideal - Actual_filled = $\frac{2}{3} - \frac{3}{5}$

To subtract these fractions, we find a common denominator, which is 15:

Amount_leaked = $\frac{2 \times 5}{3 \times 5} - \frac{3 \times 3}{5 \times 3} = \frac{10}{15} - \frac{9}{15} = \frac{1}{15}$ tank.
Leakage Rate: This leakage of $\frac{1}{15}$ tank occurred over 4 hours. So, the leakage rate is:
Rate_leak = $\frac{\text{Amount leaked}}{\text{Time}} = \frac{1/15 \text{ tank}}{4 \text{ hours}} = \frac{1}{15 \times 4} = \frac{1}{60}$ tank per hour.

Calculating Time to Fill Remaining Tank

The leakage was repaired immediately after 4 hours. At this point, 60% ($\frac{3}{5}$) of the tank is already full. We need to find the time required to fill the remaining portion.

Remaining Portion of the Tank:
Remaining_portion = $1 - \text{Actual_filled} = 1 - \frac{3}{5} = \frac{5}{5} - \frac{3}{5} = \frac{2}{5}$ tank.
Filling Rate After Repair: Since the leakage is repaired, the tank is now filled only by the tap at its normal rate.
Filling Rate = Rate_tap = $\frac{1}{6}$ tank per hour.
Time to Fill the Remaining Portion:
Time_remaining = $\frac{\text{Remaining_portion}}{\text{Filling Rate}} = \frac{2/5 \text{ tank}}{1/6 \text{ tank per hour}}$

Time_remaining = $\frac{2}{5} \times \frac{6}{1} = \frac{12}{5}$ hours.
Converting Time to Hours and Minutes:
$\frac{12}{5}$ hours = $2 \frac{2}{5}$ hours.

To convert the fraction of an hour to minutes, multiply by 60:

Minutes = $\frac{2}{5} \times 60 = \frac{120}{5} = 24$ minutes.

So, the time needed to fill the remaining portion is 2 hours and 24 minutes.