One pipe can empty a tank 5 times faster than another pipe can fill the tank. Starting with a full tank, if both pipes are turned on, it takes 10 hours to empty the tank. How long does it take the slower pipe working alone to fill an empty tank

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

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step-by-step explanation: