Printable

2003 AMC 10B Problem 2

2003 AMC 10B Problem 2AMC 10 Step by Step

Name: Date: September 13, 2026

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

Al gets the disease algebritis and must take one green pill and one pink pill each day for two weeks. A green pill costs $1 $more than a pink pill, and Al's pills cost a total of$ \ $546$ for the two weeks. How much does one green pill cost?

A)
$7
B)
$14
C)
$19
D)
$20
E)
$39

Answer

Key insight

Fourteen days cost 546 dollars, so one green plus one pink pill costs 39, and two prices differing by 1 that sum to 39 are 20 and 19.

Solution

Two weeks is $14$ days, and each day Al buys exactly one pill of each color. The daily cost of the pair is therefore
$\frac{546}{14} = 39 \text{ dollars}.$

Let the pink pill cost $p$ dollars; the green pill costs $p + 1$ . Then $p + (p+1) = 39$ , so $2p = 38$ and $p = 19$ . The green pill costs $19 + 1 = 20$ dollars.

Check: $14 \cdot (20 + 19) = 14 \cdot 39 = 546$ .

The answer is $\boxed{\textbf{(D)}\ \$20}$ .

Why this works

The total is a per-day cost repeated $14$ times, so dividing first reduces the problem to "two numbers with sum $39$ and difference $1$ ." A sum-and-difference pair is solved instantly: the larger is $\frac{\text{sum} + \text{difference}}{2}$ . Reduce a repeated-purchase problem to a single unit before introducing variables.

The trap

Answering 39, the cost of a daily pair, or 19, the pink pill, instead of the green pill.

Common mistakes

Answering $39$ , the cost of a daily pair, or $19$ , the pink pill, instead of the green pill.
Using $7$ days for "two weeks" or dividing by $28$ pills instead of $14$ pairs.

Techniques

Set up the equation/formula and compute; no special trick needed