### So many percent more

**Updated with an answer... see below**

I'm continuing to catch up after vacation, and spotted a good discussion about problems with percent, at MathNotations. (via Let's Play Math blog).

Here's a problem to solve, first of all:

There are 20% more girls than boys in the senior class.

What percent of the seniors are girls?

The answer is NOT that 40% are boys and 60% are girls...

You see, let's say there were 40 boys and 60 girls, 100 students total. If there are 40 boys, then 20% more than that would be 40 + 4 + 4 = 48 girls and not 60!

Try solve it. I'll let you think a little before answering it myself. Don't just rush over to the Mathnotations blog either! Use your thinking caps! I've already given you a big hint!

**Update:**

You can easily solve this problem by taking

*any*example number for the number of boys. Like I did above, if you have 40 boys, you'd need 48 girls and there'd be 88 students total. What percent of the seniors are girls then? It'd be 48/88 = 0.545454... ≈ 54.55%. And 45.45% are boys.

But the same works even if you choose that there'd be 10 boys, which then means that there are 12 girls, and then the percent of girls in the class is 12/22 * 100% = 54.55%.

**NOTE**that this problem includes two DIFFERENT "wholes". First of all, it says "There are 20% more girls than boys in the senior class." This is a comparison, and the total number of boys is the "one whole" or the "100%". The number of girls is 20% more, or 120% of the boys.

In algebra terms, if there are p boys, then there will be 1.2p girls.

Then the final question involves a totally different "one whole" or 100%: it asks how many percent of the

*seniors*are girls.

So the group of seniors becomes the 100% or the "whole", and all percent calculations are based on that. Therefore one will then compare the number of girls to the total number of seniors.

In algebra terms, the final answer as a decimal is 1.2p/(p + 1.2p) = 1.2/2.2

See also Denise's post about searching for the 100% in percent problems.