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From Stewart, Precalculus, $5$th ed, p.$71$, q.$55$

The radiator in a car is filled with a solution of $60\%$ antifreeze and $40\%$ water. The manufacturer of the antifreeze suggests that, for summer driving, optimal cooling of the engine is obtained with only $50\%$ antifreeze. If the capacity of the radiator is $ 3.6$L, how much coolant should be drained and replaced with water to reduce the antifreeze concentration to the recommended level?

I don't think there is a need to set up a model, since the currently concentration of antifreeze is at $60\%$ and desired contraction is $50\%$, the difference is $10\%$ of the total capacity of the radiator.

$10\%\times3.6L = 0.36L$

$0.36$L to be drained from the original $60\%$ concentration of $\dfrac{2.16}{3.6}L$ and replaced with water would give $50\%$ concentration at $\dfrac{1.8}{3.6}L$

The answer at the back of the book says $0.6$L to be drained. I don't understand why this is so, if $0.6$L is drained the new concentration would be $43.333\%$.

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Hint: If you are draining $.36L$ of fluid, what does that fluid include? When draining, both water and antifreeze are present and you need to take that into account. –  Amzoti May 20 '13 at 1:29
    
www.calcchat.com –  yiyi May 20 '13 at 1:32

2 Answers 2

up vote 1 down vote accepted

Let $x$ be the amount in litres of coolant replaced by water.

Amount of water originally in the coolant: $0.4 \times 3.6$L

Amount of water removed when $x$ amount of coolant is removed: $(0.4\times x)$ L

Note that the removal of $x$ amount of coolant would result in the removal of fluids according to their proportion.

Amount of water when $x$ amount of water is added: $\left(0.4 \times 3.6-(0.4\times x)+x\right)$L $=(1.44+0.6x)$L

Amount of antifreeze originally in the coolant: $0.6 \times 3.6$L

Amount of antifreeze removed when $x$ amount of coolant is removed: $(0.6\times x)$ L

Amount of antifreeze when $x$ amount of water is added: $\left(0.6 \times 3.6-(0.6\times x)\right)$L $=(2.16-0.6x)$L

For equal fluid proportions, the amount of fluids should be equal.

$\therefore 1.44+0.6x=2.16-0.6x$

or $x=0.6$L

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Another way of addressing the same problem:

Whatever you do, you need to wind up with 1.8 litres of antifreeze in the car ($50\%$ of $3.6$ litres).

You already have $2.16$ litres of antifreeze in the car ($60\%$ of $3.6$ litres).

So you need to remove $0.36 (2.16 - 1.8)$ litres of antifreeze from the radiator.

The solution in the radiator is a $60\%$ solution: $0.6$ litres of antifreeze in $1$ litre of solution, So:

$0.6$ litres of antifreeze in $1.0$ litre of solution

$1.0$ litres of antifreeze in $\frac{1}{0.6}$ litres of solution

$0.36$ litres of antifreeze in $\frac{0.36}{0.6}$ litres of solution, or $0.6$ litres of solution

So, removing $0.6$ litres of solution and adding $0.6$ litres of water will give the desired result.

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