I am reading the following simple problem on rounding.
10 people ordered calculators.
The least expensive was \$19.95 and the most expensive \$39.95 Half ordered a \$29.95 calculator.
Select the best estimate of the amount spent:
a) \$240
b) \$310
c) \$345
d) \$355
My approach:
The rest of the 5 could have ordered either \$20 or \$40 or \$10
So we have (on top of the $5$ that the problem mentions they bought the \$30 calculator)
$150 + 5 * 20 = 250$
or
$150 + 5 * 40 = 350$
or
$150 + 5 * 30 = 300$
So we get an average of 250 + 350 + 300 = \$900 which divided by 3 gives \$300
That is not part of the options though and the solution states that the correct answer is b
What am I messing up here?
To round things out, let $$A=20\quad B=30\quad C=40$$
Assuming some kind of bell-curve in purchasing, we have
$$2A+5B+2C+??=40+150+ 80+??=270+??$$
Given that this is already above the $240$-estimate where $??=-30$, we must assume that $??=40$ and the sum $$270+??=310$$