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Question from Aniish, a student:

Hello...I have a question to clarify.
The radiator of a jeep has a capacity of 4 gallons. It is filled with an anti-freeze solution of water and glycol which analyzes 10% glycol.what volume of the mixture must be drawn off and replaced with glycol to obtain a 25% glycol solution?

all percentages are by volume

Aniish,

The current mixture is $1\%$ glycol so at the moment you have $0.1 \times 4 = 0.4$ gallons of glycol in the mixture. You want a $25\%$ solution so that will contain $0.25 \times 4 = 1$ gallon of glycol.

Suppose you remove $x$ gallons of the mixture and replace it with $x$ gallons of glycol. In the $x$ gallons of mixture you remove there are $0.1 \times x$ gallons of glycol so now, in the mixture you have $0.4 - 0.1 \times x + x \mbox{ gallons }$ of glycol. From the previous paragraph you want this to be $1$ gallon. Solve for x. Make sure you check your answer.

Penny

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