How many audiences at a musical

[Johnx1]

5/8 of the audience at a musical are females. 3/10 of the females are girls and 1/3 of the males are boys. There are 21 more girls than boys. How many people are there in the audience?

My answer. I'm not sure how to do this.

I know that there are 3/8 males.-------------------------------------------------------------------------------------------------------
In the book they did:

For girls, 3/10 * 5/8 = 3/16 girls (why they multiplied both fractions?)
For boys, 1/3 * 3/8 = 1/8 boys (same, why they multiplied both fractions?)

Then they subtracted 3/16 - 1/8 to find the audience. So there is 1/16 audiences. (why subtract both fractions?)

Lastly, they did 1/16 = 21. So they got an another of 336. (why did they equal it to 21?)

I'm not sure how the book did it.
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Is there an easy way to do this creating an algebraic expression?

 

[MarkFL]

We could use variables, although we'll essentially wind up doing what your book did. Let \(A\) be the number of people in the audience, with \(M\) being the number of males, \(B\) the number of boys, \(W\) the number of women and \(G\) the number of girls. From the given information, we may write:

$$G=\frac{3}{10}F=\frac{3}{10}\cdot\frac{5}{8}A=\frac{3}{16}A$$

$$B=\frac{1}{3}M=\frac{1}{3}\cdot\frac{3}{8}A=\frac{1}{8}A$$

We know the number of girls is 21 more than the number of boys:

$$G=B+21$$

Hence:

$$\frac{3}{16}A=\frac{1}{8}A+21$$

Multiply through by 16:

$$3A=2A+336$$

Subtract through by \(2A\):

$$A=336$$

 

[Johnx1]

$$G=\frac{3}{10}F=\frac{3}{10}\cdot\frac{5}{8}A=\frac{3}{16}A$$

$$B=\frac{1}{3}M=\frac{1}{3}\cdot\frac{3}{8}A=\frac{1}{8}A$$

We know the number of girls is 21 more than the number of boys:

$$G=B+21$$

Before I posted the question, I did get somewhat to this part, but I didn't put (Girls and Women) and (Boys and Men) into as an Audience.

So I got stuck here $$G=B+21$$

You made it much more sense. Thank you.