A 10-gallon water cooler in an office provides water for the whole department Each hour, 30

Question

ounces of water are removed from the cooler and drunk by office workers. Write an equation to
show how long the water in the cooler will last. (10 gallons is 1,280 ounces.)
$1,280-30h=0$
$10-30h=0$
$30h=10$
$1,280+30h=0$

Answer 1

To find out how long the water in the cooler will last, we can start with the total amount of water and the rate at which it is being consumed.

Given:

• The cooler holds 10 gallons of water, which is equal to 1,280 ounces.
• Each hour, 30 ounces of water are drunk by the workers.

We need to find \( h \), the number of hours the water will last. The equation should balance the initial water amount with the water used over time. So we can write:

\[ 1280 - 30h = 0 \]

This equation indicates that the initial amount of water (1,280 ounces) minus the amount used over \( h \) hours (which is \( 30h \) ounces) equals zero when the water runs out.

So, the correct equation is:

\[ 1280 - 30h = 0 \]