MHB How many audiences at a musical

  • Thread starter Thread starter Johnx1
  • Start date Start date
AI Thread Summary
The discussion revolves around calculating the total audience at a musical based on gender ratios and the difference in numbers between girls and boys. It is established that 5/8 of the audience are females, with 3/10 of those being girls, while 3/8 are males, with 1/3 of them being boys. The equation G = B + 21 is derived from the information that there are 21 more girls than boys. By setting up algebraic expressions for the number of girls and boys in relation to the total audience, the total audience is calculated to be 336. The conversation highlights the importance of understanding how to manipulate fractions and set up equations to solve such problems.
Johnx1
Messages
48
Reaction score
0
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.
-----------------------------------------------------------------------------------------------------

Is there an easy way to do this creating an algebraic expression?
 
Mathematics news on Phys.org
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 \(2A\):

$$A=336$$
 
MarkFL said:
$$G=\frac{3}{10}F=\frac{3}{10}\cdot\frac{5}{8}A=\frac{3}{16}A$$

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

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.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

Similar threads

Back
Top