Two water tanks hold 28.2 gallons and 31.2 gallons of water. The larger tank is leaking at a rate of 0.12 gallons per hour. The smaller tank is leaking at a rate of 0.8 gallons per hour. After how many hours will there be less water in the larger tank than the smaller tank?

Question

Wade Dillon

Mathematics

23 March, 16:16

Two water tanks hold 28.2 gallons and 31.2 gallons of water. The larger tank is leaking at a rate of 0.12 gallons per hour. The smaller tank is leaking at a rate of 0.8 gallons per hour. After how many hours will there be less water in the larger tank than the smaller tank?

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Answers (1)

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Nyla Mason · Answer 1

75 hours

Step-by-step explanation:

To solve this question, you need to find when will the volume of water inside the smaller tank equal to the larger tank. Seems there is a typo in the question since the smaller tank have bigger leak than the larger tank and its impossible to solve with this number. So, I will show you how to solve if the leak is 0.08 gallons per hour instead of 0.8 gallons per hour.

The smaller tank holds 28.2 gallons and leaking at 0.08 gallons per hour. If y=volume and x=time, then the equation for water volume inside the smaller tank will be:

y1 = 28.2 - 0.08x

The larger tank holds 31.2 gallons and leaking at 1.2 gallons per hour. The equation for water volume inside the larger tank will be:

y2 = 31.2 - 0.12x

We want to find the time when the volume (y) of both tanks if equal, so

y1=y2

28.2 - 0.08x = 31.2 - 0.12x

-0.08x + 0.12x = 31.2 - 28.2

0.04x = 3

x = 75

The volume of both tanks will be equal after 75 hours, after that the larger tank will have less water compared to the smaller tank because it leaks faster.