The engine in Paul's weed whacker uses an oil-gas mixture. The owner's manual recommends mixing the oil and gasoline together, using the size of the oil container to guide how much of each to use. It states to use 1/8 part oil with 4 parts gasoline. Paul buys a 32-ounce container of oil. How large of a container will he need to hold the mixture if the amount of oil he uses is 1/8 of the oil container?

Question

Scarlet Owens

Mathematics

5 March, 10:31

The engine in Paul's weed whacker uses an oil-gas mixture. The owner's manual recommends mixing the oil and gasoline together, using the size of the oil container to guide how much of each to use. It states to use 1/8 part oil with 4 parts gasoline. Paul buys a 32-ounce container of oil. How large of a container will he need to hold the mixture if the amount of oil he uses is 1/8 of the oil container?

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Answers (1)

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Aryana Yates · Answer 1

This is how to solve the problem.

First is to get the ounces of oil, and that is

Ounces of oil = (1/8) * 32

Ounces of oil = 4

Then,

= Ounces of oil / ounces of gas

= (1/8) / 4

= 1/32

Then,

= ounces of oil / ounces of gas

= 1/32

Let g = ounces of gas neede

4 / g = 1 / 32

g = 4 * 32

g = 128

Thus, you have 128 ounces of gas and 4 ounces of oil. The total answer is 128 + 4 = 132. So, it holds 132 ounces.