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Stolb23 [73]
3 years ago
11

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 turne

d on, it takes 10 hours to empty the tank. How long does it take the slower pipe working alone to fill an empty tank
Mathematics
1 answer:
xxTIMURxx [149]3 years ago
5 0

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

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