The problem is as follows:
Assume that you have three mugs. Each mug has a capacity of $1\textrm{liter}$. The first mug is full and it has $60\%$ of orange juice and $40\%$ of water. The second mug which is also full is $80\%$ of orange juice and $20\%$ of lemon juice. Finally the third mug is empty. Now, you want to fill the third mug using the contents of the other two mugs mentioned. The condition is that such mug must have the double amount of orange juice that of water. Using this information. What percentage of lemon juice will remain in the third mug?
The alternatives given in my book are as follows:
$\begin{array}{ll} 1.&\textrm{2 percent}\\ 2.&\textrm{4 percent}\\ 3.&\textrm{5 percent}\\ 4.&\textrm{6 percent}\\ \end{array}$
I'm confused exactly how should I approach this question. Is there more than meets the eye here?.
What I tried to do is this:
Assuming $x$ is the volume to be taken from the first mug. Then my equation becomes into:
$$\left(\frac{6}{10}\right)\cdot x + \left(\frac{8}{10}\right)(1-x)=\left(\frac{2}{3}\right)\cdot 1$$
The two thirds comes that the requested concentration of the orange juice is two times that of water. (Or at least that's how I'm understanding it).
Hence
$$2:1\equiv\frac{2}{2+1}$$
Therefore solving all of this yields $x=\frac{2}{3}$, which I believe should the volume of orange juice to be taken from the first mug.
Now, because $1-x=1-\frac{2}{3}=\frac{1}{3}$
Is taken out from the second mug. Then I conclude that one third of liter has in its content one third of the $\frac{2}{10}$ of lemon juice.
Thus:
$$\frac{1}{3}\times\frac{2}{10}=\frac{2}{30}=\frac{1}{15}$$
which in terms of percentage would be 6.67 % but this answer does not appear in any of the choices. What went wrong?. Can someone help me here to spot exactly which part did I derailed from the intended interpretation?.
It would help a lot that an answer could explain this to me so I can be adequately guided.
If $x$ if the percentage of volume you remove from the first mug, the equation you should be solving is (because you want your total volume of orange juice in the third mug [LHS] to be equal to twice the volume of water in the third mug [RHS]) $$\frac{6}{10}x + \frac{8}{10}(1-x) = 2 \ \frac{4}{10}x$$ which yields $x= \frac{4}{5}$ and consequently the percentage of lemon juice is given by $$\frac{1}{5} \ \frac{2}{10} = \frac{1}{25} = 4\%.$$