Middle school problem - Percentage of students passing an exam

Question

This is a middle school problem that my nephew got, the teacher asked to solve it without using proportions:

The $\frac23$ of boys and the $\frac34$ of girls have passed an exam. Knowing that the number of boys enrolled in the exam is three times the number of girls, do we have enough information to calculate the percentage of the group that have passed the exam? [$68.75\%$]

That's how I solved it:

$x =$ number of boys

$y =$ number of girls

We know that $x = 3y$, so the total of students that enrolled in the exam is $4y$.

To calculate the number of students that have passed the exam:$$\left(\frac23\right)\times3y + \left(\frac34\right)\times y = \left(\frac{11}4\right)\times y$$ This is the number of students that have passed the exam related to the number of girls. To calculate the percentage related to the number of boys $+$ girls: $$\frac{\left(\frac{11}4\right)\times y}{4y} = 0.6875 = 68.75\%$$ But to calculate it I have used proportions. Is it even possible to get $68.75\%$ without using proportions?

Answer 1

The answer to the literal question asked is "yes, there is enough information". That answer doesn't use proportions, but it's probably not what the teacher had in mind. Pointing it out might brand you as a troublemaker.

Since as Kanwaljit Singh notes in his answer, the data are given as proportions (fractions, percentages and proportions are different ways to talk about the same idea) you can't really solve the problem without proportions. You can disguise them, as @user26766 does in his answer.

Answer 2

I think its not possible. If we have most of the information in question is in proportion then without using proportion we can't solve it.

Answer 3

Let x = number of boys in the class,

Let y =number of girls in the class.

$x = 3y$

$$\frac 23 x + \frac 34 y = \frac 23 (3y) + \frac 34 y = 2y+ \frac 34 y = \frac 84 y + \frac 34 y = \frac{11}{4} y$$ equals the number of students from a class of $4y$ who passed the test.

The percentage of the class that passed the text is $$\frac{\frac {11y}{4}}{4y} \times 100\% = \frac {11}{16}\times 100\%$$

Answer 4

Obviously you can't solve the problem without using facts about proportions, since all the input data as well as the answer are themselves proportions.

But here's an attempt to do it without performing any division operation, which is the usual way of getting a proportion. Instead, we use a probabilistic argument.

Let us select a student at random from all students who took the test, with uniform probability of choosing each student. The percentage probability that the randomly selected student has passed the exam is the same as the percentage of students who passed.

Let $A$ be the probability that the student passed, $B$ the probability that the student was a boy, $G$ the probability that the student was a girl. Then by the law of total probability, $$ P(A) = P(A\mid B)P(B) + P(A\mid G)P(G). $$ From the problem statement we can "guess" that the proportion of girls in the population is $0.25$ (and confirm that this is correct, because $3\times 0.25 + 0.25 = 1$), and we assume $P(B) = 1 - P(G),$ so we can deduce the probability values \begin{align} P(B) &= 0.75, \\ P(G) &= 0.25, \\ P(A\mid B) &= \frac23, \\ P(A\mid G) &= \frac34. \\ \end{align}

Therefore $$ P(A) = \frac23 \cdot 0.75 + \frac34 \cdot 0.25 = 0.5 + 0.75 \cdot 0.25 = 0.6875. $$ To get a percentage, we multiply by $100\%.$

But this has used a lot of knowledge about proportions (for example, how we know that the solution to $3P(G) + P(G) = 1$ is unique), and it has sneakily performed some division to conclude that $\frac34=0.75$ (and also to "guess" the proportion of girls, we presumably mentally divide $1$ by $4$). So I think the teacher's secret almost surely involves some sneaky way of hiding the operations that produce proportions.

On the other hand, to deduce the direct answer to the question (which is, "Yes, there is enough information"), we only need to know how proportions work; we do not actually need to evaluate any specific proportions.