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I'm working on an 8th grade math book and came across this problem:

In a class, the average marks of the boys and girls are 520 and 420 respectively. And the average marks per student is 500. What is the percentage of boys in the class?

This seemed to be very simple for me but I tried:

Let the number of boys and girls be b and g. $- (520b + 420g)/(b + g) = 500$, then I could not frame the next equation.

I also tried:

Let the total number of students be x and the number of boys be b. $- (520b + 420(x - b))/(x) = 500$;

but could not come to a conclusion.

Shevliaskovic
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Ramana
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1 Answers1

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Suppose there are $N$ students in the class. Let $b$ denote the percentage of boys in the class. In other words, if there are $B$ boys in the class, then $b=\frac{B}{N}$. Similarly, let $g$ denote the percentage of girls. Now, from the information, we have $$500 = \frac{\{\text{sum of students' marks}\}}{N}$$ and $$520=\frac{\{\text{sum of boys' marks}\}}{B},420=\frac{\{\text{sum of girls' marks}\}}{G}\,\,.$$ It follows that $$\frac{520B + 420G}{N}=520b+420g=520b+420(1-b)=500\,\,,$$ as you said, where the second equality is from the fact that there are only two genders in the class, presumably. So, $b=0.8$ .$$$$[Aside: the numerator in the first expression of the equalities above comes from the definition of average in the case of boys and girls, respectively: explicitly, for the case of the boys, $520B=\{\text{sum of boys' marks}\}$ and we also have $\{\text{sum of boys' marks}\} + \{\text{sum of girls' marks}\}=\{\text{sum of students' marks}\}$]

afedder
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  • @ affeder. Thank you very much for you immediate response! :) – Ramana May 16 '14 at 06:09
  • @ affeder. Sorry if I gave you any trouble. I actually got 4g = b, but I was silly to not be able to derive that b was 80%. Thanks! – Ramana May 16 '14 at 06:14