Question

An aquifer holds 146,000 cubic feet of water. Water exits through a well at a steady rate of 500 cubic feet per day, and the aquifer is replenished at a constant rate of x cubic feet per day. If the water in the aquifer is used up in a year (365 days), what is the value of x? Ignore evaporation and other losses.

Answer 1

First, find the amount of water in cubic feet that the well removes in one year:

500 ft3 per day × 365 days per year = 182,500 ft3 per year.

More water exits the aquifer in a year than it can hold at any given time. So, we know that some water enters the aquifer to make up the difference.

182,500 ft3 lost to well – 146,000 ft3 stored in aquifer = 36,500 ft3 added per year.

One year is the same as 365 days, so

Therefore, x = 100 cubic feet of water per day.

Answer 2

The value of x = 100 ft^3/ day

Explanation:

We need to look at what amount of water comes in and what amount of water leaves

What comes in: x ft^3/ day

What leaves : 500x ft^3/ day

We start with 146 000 ft^3

Water in tank = what we start with + what comes in - what leaves

water in tank = 146000 + x*d - 500*d where d is the number of days

After 365 days , the tank has 0 ft^3 of water

0 = 146000 + x*365 - 500*365

Multiply

0 = 146000 + 365x - 182500

Combine like terms

0 = -36500+365x

Add 36500 to each side

36500 = 36500-36500+365x

36500 = 365x

Divide by 365

36500/365 = 365x/365

100=x

The value of x = 100