2007 AMC 10A Problem 5

Work in cents. Let a pencil cost $p$ cents and a notebook $n$ cents:
$$7p + 8n = 415, \qquad 5p + 3n = 177.$$
Eliminate $p$: multiply the first equation by $5$ and the second by $7$:
$$35p + 40n = 2075, \qquad 35p + 21n = 1239.$$
Subtracting, $19n = 836$, so $n = 44$. Then $5p = 177 - 132 = 45$, so $p = 9$.

The requested order costs $16 \cdot 9 + 10 \cdot 44 = 144 + 440 = 584$ cents, that is $\$5.84$.

The answer is $\boxed{\textbf{(B)}\ \$5.84}$.

Two independent purchases of the same two items pin down both unit prices, and any other order is then a direct evaluation. Converting to cents turns a decimal system into an integer one, which is faster and less error-prone under time pressure.

Three copies of the second purchase (15 pencils, 9 notebooks) cost $\$5.31$; the target order is exactly one pencil and one notebook more. One pencil plus one notebook must cost far less than a fifth of the first purchase, so the total is a little above $\$5.31$; only $\$5.84$ fits among the choices.

• Arithmetic slips with the decimals during elimination; working in cents ($415$ and $177$) keeps everything integer.
• Trying to express $16p + 10n$ as a simple combination of the two given bundles; the multipliers are not nice, so solving for $p$ and $n$ is the practical route.