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A radiator contains 8 quarts of a mixture of water and antifreeze. If 40% of the mixture is antifreeze, how much of the mixture should be drained and replace by pure antifreeze so that the resultant mixture will contain 60% antifreeze

Verified Answer :

I believe 2 2/3 quarts of the mixture. Cheers!

As per linear equation, 1.6 quarts of mixture should be drained and replace by pure antifreeze so that the resultant mixture will contain 60% antifreeze.

What is a linear equation?

"A linear equation is an equation in which the highest power of the variable is always 1."

Given, the radiator contains 8 quarts of a mixture of water and antifreeze.

40% of the mixture is antifreeze.

Therefore, the amount of antifreeze in radiator is

\(= \frac{(8)(40)}{100}\) quarts

\(= 3.2\) quarts

If some amount of mixture is drained and replace by pure antifreeze, then the total amount mixture will be

\(= (8-x+x)\) quarts

\(= 8\) quarts

The new amount antifreeze in 8 quarts of mixture will be

\(= \frac{(8)(60)}{100}\) quarts

\(= 4.8\) quarts

Therefore, the amount of mixture is drained and replace by pure antifreeze is

\(= (4.8 - 3.2)\) quarts

\(= 1.6\) quarts

Learn more about linear equation here: https://brainly.com/question/11897796

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