The King informed the Doctor that exactly two-thirds of the men had received a black eye, three-fourths had sprained a wrist, and four-fifths had stubbed a toe. “It follows, then,” the Doctor surmised, “that at least twenty-six of the men have a black eye, a sprained wrist, and a stubbed toe.” How many injured men were in the wood?
Here are some thoughts:
Suppose we have a total of $x$ injured men. Let $E$ be the number of people with blacked-eye, $W$ be sprained wrist, and $T$ be stabbed-toe. So we have
- $E= \frac{2}{3}x$
- $W= \frac{3}{4}x$
- $T= \frac{4}{5}x$
From here we can make a Venn diagram.
Here are some information we can figure out:
- $E^c=\frac{1}{3}x$, where $E^c$ is the complement of $e$.
- $W^c=\frac{1}{4}x$
- $T^c=\frac{1}{5}x$
- $E-26=E -(E\cap W \cap T ) $
- $W-26=W -(E\cap W \cap T ) $
- $T-26=T -(E\cap W \cap T ) $
From here I played with the Venn diagram, and tried to eliminate some parts. But I failed. Could some please help me to continue?
Thanks!
Let $(E)$ be the number of people with only blacked-eye, $(EW)$ with only blacked-eye and sprained wrist, $(EWT)$ with all three, so on.
We have:
$$\begin{align} (E)+(EW)+(ET)+(EWT)=\frac{2}{3}x \qquad(1)\\ (W)+(EW)+(WT)+(EWT)=\frac{3}{4}x\qquad(2)\\ (T)+(ET)+(WT)+(EWT)=\frac{4}{5}x\qquad(3)\\ (E)+(W)+(T)+(EW)+(ET)+(WT)+(EWT)=x\;\;\,\qquad(4)\\ \text{Minimum of }(EWT)=26\;\qquad(5) \end{align}$$
$$(1)+(2)+(3)-(4): \qquad (EW)+(ET)+(WT)+2(EWT)=\frac{73}{60}x$$
Here, the RHS is fixed, so in order for $(EWT)$ to have its minimum value, $(EW)+(ET)+(WT)$ has to have its maximum value. In $(4)$, $(E),(W),(T) \ge 0$ so the maximum is $x-(EWT)$.
$$\frac{73}{60}x=(EW)+(ET)+(WT)+2(EWT) \le x+(EWT)$$ $$\implies (EWT) \ge\frac{13}{60}x$$ $$\implies x=120$$