2007 AMC 10A Problem 5

Two different purchases of the same two items are enough to find both prices. Work in cents so there are no decimals.

1. Name the prices. Let a pencil cost $p$ cents and a notebook cost $n$ cents. Then
$$7p + 8n = 415, \qquad 5p + 3n = 177.$$
2. Make the pencil counts match. Multiply the first equation by $5$ and the second by $7$, so both have $35p$:
$$35p + 40n = 2075, \qquad 35p + 21n = 1239.$$
3. Subtract to get rid of the pencils: $19n = 836$, so $n = 44$. A notebook costs $44$ cents.
4. Find the pencil price. Put $n = 44$ into $5p + 3n = 177$: $5p + 132 = 177$, so $5p = 45$ and $p = 9$. A pencil costs $9$ cents.
5. Check with the first purchase: $7 \times 9 + 8 \times 44 = 63 + 352 = 415$. Correct.
6. New order. $16$ pencils and $10$ notebooks cost
$$16 \times 9 + 10 \times 44 = 144 + 440 = 584 \text{ cents} = \$5.84.$$

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

Two unknown prices need two facts, and the problem gives exactly two purchases. Once both prices are known, any other order is simple multiplication. Using cents keeps every number whole, which makes mistakes much less likely.

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 $\boxed{\textbf{(B)}\ \$5.84}$.