Algebra word problem

Question

A truck that hauls water is capable of carrying a maximum of 3000 lb. Water weighs 8.3454 lb gal, and the plastic tank on the truck that holds water weighs 91 lb. What is the maximum amount of water the truck can carry? Create and solve an equation or inequality that represents the situation.

Answer 1

Start with the fixed quantity; 91lb is the weight of the tank. On the other side, the maximum capacity is 3000lbs. So whatever remaining capacity for water is 3000-91. And because we are looking for water quantity in gallons, it will be whatever capacity in lbs divided by the weight of one gallon (8.3454).

So:

91 (tank weight) + 8.3454g (where g is number of gallons of water) <= 3000 (total capacity)

8.3454g <= 3000 - 91

g <= (3000-91)/8.3454

This solution to the inequality represents the maximum capacity of water that the truck can carry, in gallons.

Answer 2

Hi Margret,

Let's start with our fixed values. That is, the weights that won't change regardless of how much water we put in the truck. Our max is 3000 lbs. The tank weighs 91 lbs., so we have, from the start:

3000=91 + some value

Now, let's talk about that value. We are given that water weighs 8.3454 lbs./gallon. So set the unknown number of gallons--the value we are looking for--equal to x. This gives:

8.3454x

Now, our equation is:

3000= 91 + 8.3454x

To solve, subtract 91 on both sides:

2909=8.3454x

Divide both sides by 8.3454

x=348.57 gallons

I hope this helps.

Answer 3

Because you see the word maximum, this should tell you that the weight being carried by the truck has to be less than or equal to 3000 pounds. Without adding any water, the truck has a plastic tank that weighs 91 pounds, so we make sure to include that in the equation. We can then add to it each 8.3454 pound gallon until we hit less than or equal 3000. We don't know how many gallons this will be, so this is what we multiply by x (our unknown number of gallons times the weight of each gallon).

This looks like

91 + 8.3454x ≤ 3000

To solve this, simply subtract 91 from both sides, then divide both sides by 8.3454 to get x by itself. This should give us x ≤ (3000-91)/8.3454, or

x ≤ 348.56 gallons of water

If the truck doesn't accept partially filled gallons, then round down to x ≤ 348 gallons. This means the truck can carry up to 348 gallons of water, but no more than that!