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Whilst revising, a problem caught my eye and I cannot seem to find an answer. I am usually bad at these types of questions.

On a certain Russian-American committee, $\frac23$ of members are men, and $\frac38$ of the men are Americans. If $\frac35$ of the committee members are Russians, what fraction of the members are American women?

A. $\frac{3}{20}$ B. $\frac{11}{60}$ C. $\frac{1}{4}$ D. $\frac{2}{5}$ E. $\frac{5}{12}$

Could you please explain how to approach and analyse the problem, maybe give some hints or the complete procedure of solving?

I get a bit confused with all those fractions. What I tried was to convert them to percentages but that seemed a bad idea.

Sorry if this question is annoying.

Thank you.

Update: I solved the problem both intuitively and mathematically. Thanks.

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    $\begingroup$ You've got two categories, each with two options. You can find the probabilities of being in the intersection of each of these options (e.g. Russian Women) using the proportions and conditional proportions (like how its not that 3/8 are american, but 3/8 of men are american), and your answer will be there. $\endgroup$ Commented May 24, 2015 at 13:38

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Let me try. We will pick a nice number for the total number of members so that every category of members come out whole numbers. We can pick the least common multiple of all the denominators of the fractions. That gives us $120.$ Of course as long as we are only interested in the ratio, it don't matter what that number is. It makes computation easier to do.

Suppose there were $120$ members. There are $80$ men and $40$ women. Of the $80$ men $30$ are American and $50$ russian. There are $72$ Russians on the committee that leave $22$ Russian women and $18$ American women on the committee. The fraction of American women on the committee is $$\frac{18}{120} = \frac3{20}.$$

Hopefully I did not make any silly arithmetic errors.

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  • $\begingroup$ I'd only suggest you edit to make clear why you chose 120. It may not be obvious to OP. $\endgroup$ Commented May 24, 2015 at 14:36
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Here's a more algebraic approach. We know that $\frac{2}{5}$ of the committee is American (the other $\frac{3}{5}$ is Russian), and that $\frac{2}{3}\cdot\frac{3}{8} = \frac{1}{4}$ of the committee is American men. Thus, if the proportion of American women is $x$, then, $$\frac{1}{4} + x = \frac{2}{5}$$ that is, if you add up the proportion of American men and the proportion of American women (assuming men and women are the only two categories), then you get the proportion of Americans. From here, we can solve for $x$ to get $$x = \frac{2}{5} - \frac{1}{4} = \frac{3}{20},$$ which is in agreement with abel's answer.

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An alternative approach, very similar to @Strants's, with similar logic, but a slightly different way of looking at things:

We know that $\frac{2}{3}$ of the committee members are men and that $\frac{3}{8}$ of them are Americans.

This means that the proportion of all committee members who are American men is:

$\frac{2}{3} \times \frac{3}{8} = \frac{6}{24} = \frac{1}{4}$

We know also that $\frac{3}{5}$ of the members are Russian.

We know that there are only four types of people at the committee:

American men, Russian men, Russian women and American women.

We have found that $\frac{1}{4}$ of the members are American men and $\frac{3}{5}$ of them are Russian - i.e. Russian men and Russian women account for $\frac{3}{5}$ of the members.

Therefore, the proportion of members who are American men, Russian men, or Russian women is:

$\frac{3}{5} + \frac{1}{4} = \frac{12}{20} + \frac{5}{20} = \frac{17}{20}$

Therefore, the proportion of members who are American women is:

$1 - \frac{17}{20} = \frac{3}{20} = \text{A}$.

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I can't improve on @abel 's excellent answer but do want to point out that it's an example of an important strategy. When you're "confused with all those fractions" and converting to percentages doesn't help (because they are just fractions) try natural frequencies - pick a number (in this case 120) that lets you count the cases.

See

http://opinionator.blogs.nytimes.com/2010/04/25/chances-are/

http://www.medicine.ox.ac.uk/bandolier/booth/glossary/freq.html

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