Answer:
40 hours
Step-by-step explanation:
Let's say that the slower pipe can fill x amount of the tank in one hour. It adds x amount to the tank every hour. Therefore, we can say, for the slower pipe,
1 hour = x amount
Then, the faster pipe empties the tank 5 times faster than the slower pipe fills it, so it removes 5 times the amount that the smaller pipe puts in, so for the faster pipe,
1 hour = -5x amount (negative to symbolize removing).
For the problem at hand, the tank starts at full, or 100%=1. It ends empty, or at 0% = 0. After 10 hours, if we only account for the slower pipe, we add x amount to the tank every hour, so we add 10 times that total, resulting in
1 + 10 * x as the ending result if we don't include the faster pipe. Then, the faster pipe removes 5x every hour, so in 10 hours, it removes 50x, so we have
1+10 * x - 50 * x as the final amount of stuff in the tank, which is equal to 0. Therefore, we have
1 + 10 * x - 50 * x = 0
1 - 40 * x = 0
add 40*x to both sides to isolate the x and its coefficient
1 = 40 * x
divide both sides by 40 to isolate x
x= 0.025
Therefore, the slower pipe adds 0.025, or 1/40 = 2.5% to the tank every hour. We want to figure out how long it would take for the slower pipe to fill up an empty tank, or turn it from 0% to 100% full.
Because the slower pipe adds 2.5% of the tank every hour, we can say that over y hours, it fills up
2.5% * y amount of the tank. We want to figure out how many hours it would take to make it 100% (we need to add 100% of the tank in the problem), so we can say
2.5% * y = 100%
divide both sides by 2.5% to isolate y
100%/2.5% = y = 40
