Question

A tank holds 28 gallons of water and a larger tank holds 31 gallons of water. The larger tank is leaking at a rate of 0.2 gallons per hour. The smaller tank is leaking at a rate of 0.15 gallons per hour. After how many hours, y, will there be less water in the larger tank than in the smaller tank?

Answer 1

60 hours

Explanation:

To find out when there will be less water in the larger tank than in the smaller tank, we need to calculate how long it takes for the water in the larger tank to reach the same level as the water in the smaller tank while accounting for the leak rates.

Let's denote:

L1 as the initial water level in the smaller tank (28 gallons)

L2 as the initial water level in the larger tank (31 gallons)

R1 as the leak rate of the smaller tank (0.15 gallons per hour)

R2 as the leak rate of the larger tank (0.2 gallons per hour)

t as the number of hours it takes for the two tanks to have the same amount of water.

Now, we can set up an equation to solve for t:

L1 - R1 * t = L2 - R2 * t

Plug in the values:

28 - 0.15t = 31 - 0.2t

Now, let's solve for t:

0.05t = 31 - 28

0.05t = 3

Now, divide both sides by 0.05 to solve for t:

t = 3 / 0.05

t = 60

So, it will take 60 hours for there to be less water in the larger tank than in the smaller tank.