Help with Mathematics problem! (1 Viewer)

Cryptical · Mar 13, 2017

I need help with this question I was given:

A class went on two trips. Every student in the class attended at least one of the two trips.

In each trip, no more than 2/5 of the class were boys. Prove that no more than 4/7 of the class are boys.

Thanks.

InteGrand · Mar 13, 2017

Cryptical said:
I need help with this question I was given:

A class went on two trips. Every student in the class attended at least one of the two trips.

In each trip, no more than 2/5 of the class were boys. Prove that 4/7 of the class are boys.

Thanks.

A class with for example no boys and only girls would satisfy the hypotheses of the question but not the conclusion.

Cryptical · Mar 13, 2017

I think can assume that the number of boys is non zero

InteGrand · Mar 13, 2017

Cryptical said:
I think can assume that the number of boys is non zero

Even if there's just 1 boy and 9 girls say, the hypotheses can be satisfied but the conclusion not. In general if there's an upper bound on the proportion of boys in each trip, we could just have arbitrarily small proportion of boys in each trip and get an overall arbitrarily small proportion of boys in the class. Is there a typo in the question? Was it meant to be an upper bound on the proportion of girls per trip instead (equivalently, a lower bound on the proportion of boys)?

Cryptical · Mar 13, 2017

I made a mistake in the question; I have edited it so that it is "no more than 4/7"

InteGrand · Mar 13, 2017

Cryptical said:
I need help with this question I was given:

A class went on two trips. Every student in the class attended at least one of the two trips.

In each trip, no more than 2/5 of the class were boys. Prove that no more than 4/7 of the class are boys.

Thanks.

Cryptical said:
I made a mistake in the question; I have edited it so that it is "no more than 4/7"

$\noindent Note that it is equivalent to show that $\frac{B}{G}$ cannot exceed $\frac{4}{3}$, where $B$ is the total number of boys in the class and $G$ is that for girls. (It is clear from the hypotheses that $G > 0$). Suppose $g$ girls go on trip $1$ and $h$ go on trip $2$, and wlog $g \leq h$. (By the hypotheses, $g$ and $h$ are both assumed positive.) The way to upper-bound the ratio $\frac{B}{G}$ is to upper-bound $B$ and lower-bound $G$ simultaneously if we can, in terms of the data $g$ and $h$.$

$\noindent A lower bound $G$ is $h$ (this occurs iff the set of girls who went on the first trip is a subset of the set of girls who went on the second trip). An upper-bound on the number of boys who went on trip $1$ is the $X$ such that $X = \frac{2}{5}\left(X + g\right)\iff X = \frac{2}{3}g$. Similarly, an upper bound on the number of boys in trip $2$ is $\frac{2}{3}h$. An upper bound on $B$ is then $\frac{2}{3}g + \frac{2}{3}h$ (when the set of boys who went on trip $1$ is disjoint from that in trip $2$.) Thus $\frac{B}{G}$ cannot exceed$

$\begin{align*}\frac{\frac{2}{3}g + \frac{2}{3}h}{h} &\leq \frac{\frac{2}{3}h + \frac{2}{3}h}{h} \quad (\text{as }g \leq h)\\ &= \frac{4}{3},\end{align*}$

$\noindent as required. (Of course, this upper bound can be attained, e.g. if the class has $4$ boys and $3$ girls and two of the boys go on trip $1$ and the other two go on trip $2$, with all three girls going on each trip.)$

Cryptical · Mar 13, 2017

Thank you InteGrand. A very nice solution!

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