2452.33 – Batting Average no. 2


Mike enters the Vermilion game with a batting average of .274. After going 3-for-4 in the game, his batting average jumps up to .289. How many hits did he have for the season before the Vermilion game began?


Solution

We recall that a batting average of .274 means that h=hitsb=bats=0.274\dfrac{h = \text{hits}}{b = \text{bats}}=0.274, which is almost certainly rounded from some long decimal.

So h=0.274bh=0.274b, and h+3b+4=0.289\dfrac{h+3}{b+4} = 0.289 (again, rounded). So we wheel up the algebra machine:

h+3=0.289(b+4)0.274b=0.289b+1.1560.015b=1.844b=1.8440.015=122.933\begin{align*}h+3 &= 0.289(b+4) \\[0.5em]0.274b &= 0.289b + 1.156 \\[0.5em]0.015b &= 1.844 \\[0.5em]b &= \dfrac{1.844}{0.015} \\[1em]&= 122.933\ldots\end{align*}

Let's assume 123123. Then h=0.274×123=33.702h = 0.274 \times 123 = 33.702. Let's assume 3434.

Checking, 34123=0.27642\dfrac{34}{123}= 0.27642\ldots So we've got rounding issues. 0.2760.276 is a bit too big.

Should we try b=124b = 124? 34124=0.27419\dfrac{34}{124}= 0.27419\ldots That's pretty good. And 34+3124+4=0.28906\dfrac{34+3}{124+4} = 0.28906\ldots Also pretty good.

If you start over using these 5-place decimals, you'll get 3434 and 124124 almost exactly.