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
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.
"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
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