Finding Value Given a Ratio - Elementary

Question

I am leading an elementary tutoring session and it appears my approach to the following question does not sit well with my students.

The ratio of boys to girls in a school is 2:3. If there are 600 students in a school, how many boys are in the school?

The solution I presented is to set $x =$ # boys, $y =$ # girls and then solve $$ \frac{x}{y} = 2/3 \hspace{2mm} \text{ and } \, x + y = 600.$$

1. I am sympathetic to students not liking a particular solution for psychological reasons, even if it is logically equivalent to one that sits better with them. We see this even in more advanced mathematics. For example, see exercise 18 in chapter 1 of Introduction to Commutative Algebra.

2. Throughout undergrad and grad school I often found that even some of my peers found my solutions overly formal or systematic. I think there is a bit of a spectrum for mathematicians between formal/systematic to conceptual/intuitional. I admittedly fall to the former, which may not always lend well to educating beginner students.

3. I am interested in seeing what solutions other experienced mathematicians think would be intuitive or conceptual, without completely disembarking from rigor. In order to be within the guidelines of this community, please simply answer the highlighted question above, but do so with a beginner audience in mind. Obviously everyone here can solve that problem with ease, so the selected answer will (subjectively) be based on which presentation of the solution is best.

4. The group of students is aged 18 - 25 and studying for a placement test to re-enter school. They have mixed backgrounds but have all not studied math in several years.

-- Edit: I see there is a math education community now. But scince we also have an education tag, I'll let this fly until someone thinks it is more appropriate elsewhere.

Answer 1

Would a solution that emphasises multiplicative reasoning, i.e., that makes more essential use of ratios and fractions, be more accessible? Recasting the exercise as a probability/gambling question:

  The student population is $600.$
  Everyone is equally likely to be selected to recite a poem.
  The odds of selecting a boy is $2:3.$

Set A

1. How many boys are there? $\quad\leftarrow$interpreting a ratio
2. Hence, what is the probability of selecting a boy? $\quad\leftarrow$a fraction as a part of a whole

Set B

1. What is the probability of selecting a boy? $\quad\leftarrow$interpreting a ratio
2. Hence, how many boys are there? $\quad\leftarrow$a fraction as a part of a whole

Answer 2

Using one variable only. Let $x$ be the number of boys, then $600 - x$ is the number of girls. Thus $\dfrac{x}{600 - x} = \dfrac{2}{3} \implies 3x = 2(600 - x)= 1200 - 2x \implies 5x = 1200\implies x = 240$. So there are $240$ boys and $360$ girls !

Answer 3

For every $2$ boy-somethings there are $3$ girl-somethings so there are $2+3=5$ "somethings" that add up to $600$. Now $\dfrac{600}{5}=120$ meaning that each something is $120$.

$\therefore\quad 2\text{ boy-somethings } =240\qquad 3\text{ girl-somethings } =360$