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EDIT: I've misread the question, evaluating for 10gal instead of the half and that's why my answer is two times the correct one. Sorry for that! Thanks you Dave for making me realize my mistake.EndOfEDIT

I'm currently going through Serge Lang - Basic Mathematics, so...

The problem:

The radiator of a car can contain 10 gal of liquid. If it is half full with a mixture having 60% antifreeze and 40% water, how much more water must be added so that the resulting mixture has only
a) 40% antifreeze?
The correct answer according to the book: $\frac{5}{2}$gal
The answer I get: 5 gal

What I do to get the answer:

W-watter, A-Antifreeze
The current mixture is 60%A + 40%W
We add more water and get mixture 40%A + 60%W

Let $x$ be the water we add.
Evaluating how much water($x$) should be added to get 60%W of 10+$x$(the mixture after we added the water):

$$\frac{40}{100}(10)+x=\frac{60}{100}(10+x)$$ This gives me result $x=5$.

That's how I check the result:

40%W + 60%A = 10gal (AWmix) = 4galW+ 6galA

The liquid after the water is added(10+$x$) becomes 15gal(AWmix).
We add 5galW and get:
4W+ 6A + 5W = 9W + 6A= 15gal(AWmix)
60% of 15 is 9, 40% is 6. That's the amount of water and antifreeze percentage we want.

That became quite long for simple problem... So, is there flaw in my reasoning or just a typo in the book?
Thanks in advance for any response!

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    So to start you have $5$gal, where the composition is $3$gal antifreeze and $2$gal water. What value $x $ satisfies $3/x=0.4$ (this will be the total amount in the tank to obtain $40%$ antifreeze)? Then simply subtract $5 $ from $x$ to get the amount of water you need to add. – 2017-02-10
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    Thank you for the answer! I was going to ask from where these **5gal** come but then I've realised that I misunderstood what the problem ask for. I was reading it like "we have 10gal and that's the half of the radiator capacity, how much we should add..." It was: the radiator **can** contain, not contain. So we start with 5gal not with 10gal as I started. Sorry guys for bothering you! Obviously I made a mistake. – 2017-02-10
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    No problem. I think the question says "half full" from $10$gal so that you check your answer at the end, to see if the amount of water you add is possible. That is, if you have to add more than $5$gal of water then it can't be done. – 2017-02-10

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