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2008 AMC 10B Problem 4

2008 AMC 10B Problem 4AMC 10 Step by Step

Name: Date: September 13, 2026

Linear Equations & Word Problems · difficulty 1 of 5 · about 1 min · position P1-10

A semipro baseball league has teams with $21$ players each. League rules state that a player must be paid at least $\$15,000$ and that the total of all players' salaries for each team cannot exceed $\$700,000.$ What is the maximum possible salary, in dollars, for a single player?

A)
$270,000$
B)
$385,000$
C)
$400,000$
D)
$430,000$
E)
$700,000$

Answer

Key insight

To maximize one salary, pay the other 20 players the minimum 15,000 each and give the rest of the 700,000 cap to the star.

Solution

The team's payroll is at most $700{,}000$ . To leave as much as possible for one player, every one of the other $20$ players should earn the minimum, $15{,}000$ :
$20 \cdot 15{,}000 = 300{,}000.$
The remaining money goes to the highest-paid player:
$700{,}000 - 300{,}000 = 400{,}000.$
This is attainable (twenty players at $15{,}000$ and one at $400{,}000$ totals exactly $700{,}000$ ), so it is the maximum.

The answer is $\boxed{\textbf{(C)}\ 400,000}$ .

Why this works

When a sum is capped and each part has a floor, one part is largest exactly when every other part sits at its floor. Recognizing "maximize one under a total constraint" as "minimize everyone else" turns the problem into one subtraction.

The trap

Subtracting 21 minimum salaries instead of 20 (giving 385,000) or ignoring the minimum entirely (giving 700,000).

Common mistakes

Subtracting 21 minimum salaries instead of 20 (giving 385,000) or ignoring the minimum entirely (giving 700,000).
Dividing $700{,}000$ evenly or otherwise assuming salaries must be equal.

Techniques

Consider the largest/smallest element or boundary case