Boy and girl average scores

Question

So I was doing an aptitude test when I stumbled on this question.

I can't make sense of it at all, does this kind of problem has a name I can research? How would you go about solving it?

I went as far as doing -> (girl average score + boy average score)/2 which gives me $\frac{70+64}2 = 67$ and the problem says the whole class average is $66$, not $67$. I'm lost, if anyone has the kindness to explain to me how to work it out I'd be super grateful.

Thank you

Answer 1

You can maybe think of it as the solution to the following equation: $$ 64(x)+70(1-x)=66 \Leftrightarrow x = \frac{2}{3} $$ Where $x$ is the percentage of boys in the class and thus $1-x$ being the percentage of girls.

And since you know the total number of students is 60 the number of boys in the class must be $\frac{2}{3}(60)=40$

Answer 2

Let $b$ be the number of boys and $g$ be the number of girls.

Since the average of the boys is $64$, the total score of all the boys is $64b$, similarly, the total score of all the girls is $70g$. Now, we are given that there are $60$ students and the average is $66$. That means the total score of all the students is $60\cdot66=3960$. This gives us an equation: $$64b+70g=3960.$$ Also, $b+g=60$. Solving this system of equations yields $g=20$ and $b=40$, hence the answer is B.

The average in this problem is weighted to the ratio of boys and girls. Since the number of boys is greater than the number of girls and they have a lower average, the class average of $66$ is lower than the average if the number of boys and girls were the same, which you found is $67$.

Answer 3

The problem can be solved without (much) Math, by using intuition. What effect does the overall average have on each of the gender averages?

$(66)$ moves the girl's average of $(70)$ down by $(4)$ points, and moves the boy's average of $(64)$ up by $(2)$ points. Therefore, the weight of the boy's average must be twice that of the weight of the girl's average.

So, you need to find two fractions, $a$ and $b$ such that $a + b = 1$ and $2a = b$. This is solved by $~\displaystyle a = \frac{1}{3}, ~b = \frac{2}{3}.~$ Therefore, $~\displaystyle \frac{2}{3}~$ of the people are boys, and $~\displaystyle \frac{1}{3}~$ of the people are girls.