2003 AMC 10A Problem 13

Call the numbers $a, b, c$. We have $a + b + c = 20$, $a = 4(b + c)$ and $b = 7c$.

Let $s = b + c$. Then $a = 4s$ and $a + s = 20$ give $5s = 20$, so $s = 4$ and $a = 16$.

Now $b + c = 4$ with $b = 7c$ gives $8c = 4$, so $c = \tfrac{1}{2}$ and $b = \tfrac{7}{2}$.

The product is $16 \cdot \tfrac{7}{2} \cdot \tfrac{1}{2} = 28$.

The answer is $\boxed{\textbf{(A)}\ 28}$.

"One quantity is $k$ times the rest" means it is $\frac{k}{k+1}$ of the total, which collapses a three-variable system to two quick ratio splits. Grouping $b + c$ as a single unknown is the substitution that avoids writing out a full system.

• Misreading 'four times the sum of the other two' as four times each of them, or assuming the numbers must be integers.
• Setting the third number as $7$ times the second (reversing the ratio), which gives $b = \tfrac12$, $c = \tfrac72$; the product is unchanged here, but the habit is dangerous.