Chapter A.8, Problem 40AYU

To determine

To find: How much mixture should be drained and replaced with pure antifreeze to make desired solution.

Answer to Problem 40AYU

  $5$ litre of mixture was drained and replaced by pure antifreeze to make $60\%$ antifreeze solution.

Explanation of Solution

Given information:

The cooling system capacity is $15$ litre.

Initially system is filled with $40\%$ antifreeze and some mixture is drained and replaced by pure antifreeze to make a $60\%$ antifreeze solution.

Calculation:

As per the given information −

The cooling system capacity is $15$ litre.

Initially system is filled with $40\%$ antifreeze and some mixture is drained and replaced by pure antifreeze to make a $60\%$ antifreeze solution.

Let us suppose $x$ gallon of mixture is drained.

So, total amount of antifreeze drained will be $x\times 40\%=0.4x$ litre.

Amount of antifreeze left in the system $=15\times 40\%-0.4x=6-0.4x$ litre.

Drained amount of mixture is replaced by same amount of pure antifreeze.

Now amount of antifreeze in the system $=6-0.4x+x=6+0.6x$ .

Solution is $60\%$ antifreeze.

  $\begin{array}{l}\text{i}\text{.e}\text{. }6+0.6x=15\times 60\%\\ \Rightarrow 6+0.6x=9\\ \Rightarrow x=\frac{3}{0.6}\\ \Rightarrow x=5\end{array}$

Hence, $5$ litre of mixture was drained and replaced by pure antifreeze to make $60\%$ antifreeze solution.