While reading one aptitude book, I came across this question:

$35\%$ alcohol mixed with $60\%$ alcohol to get a $50\%$ alcohol. In what ratio were they mixed?

I spent 20 minutes on this question, but I couldn't even figure out what it means! Could anyone explain this to me?

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    $\begingroup$ this seems more of a homework question. Anyway, 35% alcohol means that in a liter of mixture there is 35 cl of alcohol and 65 cl of other stuff (probably water). $\endgroup$ – mau Aug 2 '10 at 8:47
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    $\begingroup$ This question is asking what this question means, not for a solution. Therefore, I think that it is a perfectly valid question. I can't undo my close vote, but I tidied it up a bit $\endgroup$ – Casebash Aug 2 '10 at 9:25
  • $\begingroup$ so the question is not precalculus algebra, but notation... (and in my previous comment I answered that question :-) $\endgroup$ – mau Aug 2 '10 at 10:10
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    $\begingroup$ Dear Nitesh, The question is asking the following: a person was given one bottle which was 35% alcohol (the rest water, presumably) and another bottle that was 65% alcohol. They then mixed them in some proportion in a new bottle, producing a drink that was 50% alcohol. You have to figure out what the proportion was. (Your answer will be of the form "two litres of bottle one for every litre of bottle two", or, more abstractly "they were mixed in a ratio of 2:1". (I just made up the numbers 2 and 1 here, they are not the correct answer. It is just to show you what the answer will look like.) $\endgroup$ – Matt E Aug 2 '10 at 15:07
  • $\begingroup$ @Matt: That would be a perfect answer, if the question was open $\endgroup$ – Casebash Aug 3 '10 at 7:32

Call the ratio you want to find r=(volume of 35% alcohol):(volume of 60% alcohol). Suppose that there are r liters of the 35% alcohol (so there are 0.35 L of alcohol and 0.65 L of water), 1 liter of the 60% alcohol (so the ratio is r), and 1+r liters of the 50% alcohol. From there, you can construct an equation about the alcohol (or about the water) and solve for r.


Mixture1 : 60%

Mixture2 : 35%

Let us say you take mixture using a pot of 100 units.

also you take Mixture1 'a' times and Mixture2 'b' times using the pot.

$\rightarrow$ $\frac{(a 60 + b 35)}{(a 100 + b 100)} = \frac{50}{100}$

solving this you get

$\frac{a}{b} = \frac{3}{2}$

-> 300 units of Mixture1 and 200 units of Mixture2 will give you 500 units of mixture with alcohol concentration of 50% because 3 x 60 + 2 x 35 = 250.


As Mention mixtures Should be $50$% Assume we take mixture of A is $x$ ($35$% alcohol)and the second mixture in $y$ ($60$% alcohol) volume. So as given

$$\frac{\text{alcohol volume}}{\text{total volume}}=\frac{50}{100}$$

we get $$\frac{35x+60y}{100x+100y}=\frac{50}{100}$$ By simplifying it $$35x+60y=50(x+y)$$ $$15x=10y$$ we get $$x/y=2/3$$


Given a liters of 35% alcohol solution, and b liters of 60% alcohol solution, what would be the alcohol ratio after we'll mix this two solutions?

The a liters are actually a*35/100 alcohol, and a*(1-(35/100)) water. Likewise the b liters contains b*60/100 alcohol and b*(1-(60/100)) water.

Mixing them together would give us a*35/100+b*60/100 alcohol and a*(1-(35/100))+b*(1-(60/100)) water. The question is for which a,b we have the ratio

((a*35+b*60)/100)/((a*65+b*40)/100) = 50% = 1/2


(a*35+b*60)/(a*65+b*40) = 1/2

I believe you can take it from here...

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    $\begingroup$ Where are you getting the = 60 from in your last two equations? $\endgroup$ – Isaac Aug 2 '10 at 9:11
  • $\begingroup$ Oops. The target percentage is 50% and not 60%. A typo. $\endgroup$ – Elazar Leibovich Aug 3 '10 at 6:52
  • $\begingroup$ Even given that, I think what you're computing on the left side of those equations is the ratio of alcohol to water, so I suspect you want = 1. $\endgroup$ – Isaac Aug 3 '10 at 7:24

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