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 .274.274 means that hb=0.274\frac{h}{b}=0.274, which is almost certainly rounded from some long decimal.

So hb=0.274\frac{h}{b}=0.274, and h+3b+4=0.289\frac{h+3}{b+4} = 0.289 (again, rounded).

So we wheel up the algebra machine:

h=0.274bh = 0.274b and h+3=0.289(b+4)=0.289b+1.156h+3 = 0.289(b+4) = 0.289b + 1.156.

0.274b+3=0.289b+1.1560.015b=1.8440.274b + 3 = 0.289b + 1.156 \rightarrow 0.015b = 1.844 (=31.156= 3 - 1.156)

b=1.8440.015=122.933...\rightarrow b = \frac{1.844}{0.015} = 122.933... Let's assume 123123.

Then h=0.274x123=33.702h = 0.274 x 123 = 33.702. Let's assume 3434.

Checking, 34123=0.27642...\frac{34}{123}= 0.27642... So we've got rounding issues.

0.2760.276 is a bit too big. Should we try b=124b = 124 34124=0.27419...\frac{34}{124}= 0.27419... That's pretty good.

And 34+3124+4=0.28906...\frac{34+3}{124+4} = 0.28906... Also pretty good.

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