mixture problem 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 $43.333\%$. 
 A: 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
A: 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.
A: let $x=$liters of coolant to be drained and replaced
$$0.4 \times 3.6-0.4x+x=0.5\times 3.6 $$ 
Hence $x=0.6$ liters.
