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2009 AMC 10B Problem 1

2009 AMC 10B Problem 1AMC 10 Step by Step

Name: Date: September 13, 2026

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

Each morning of her five-day workweek, Jane bought either a $50$ -cent muffin or a $75$ -cent bagel. Her total cost for the week was a whole number of dollars. How many bagels did she buy?

A)
$1$
B)
$2$
C)
$3$
D)
$4$
E)
$5$

Answer

Key insight

Five muffins cost $2.50, and each swap to a bagel adds 25 cents, so the total is a whole number of dollars only after two swaps.

Solution

Suppose Jane bought $b$ bagels and $5 - b$ muffins. In cents her total is
$75b + 50(5 - b) = 250 + 25b.$
This must be a multiple of $100$ . Since $250$ already leaves a remainder of $50$ , the term $25b$ must contribute another $50$ (or $150$ , $250$ , and so on). With $0 \le b \le 5$ the only option is $25b = 50$ , so $b = 2$ .

Check: two bagels and three muffins cost $1.50 + 1.50 = \$3.00$ .

The answer is $\boxed{\textbf{(B)}\ 2}$ .

Why this works

Writing the total as "baseline plus a fixed step per swap" turns the question into a simple divisibility check. Whenever a word problem has two item types with a fixed count, express everything in terms of the number of one type; the constraint then lives in a single linear expression.

Alternative approach

Try the choices in cents: $1$ bagel gives $275$ , $2$ gives $300$ , $3$ gives $325$ , $4$ gives $350$ , $5$ gives $375$ . Only $300$ is a whole number of dollars.

The trap

Forgetting that the muffin-only baseline already ends in 50 cents and concluding four bagels are needed to add a full dollar.

Common mistakes

Forgetting that the muffin-only baseline already ends in 50 cents and concluding four bagels are needed to add a full dollar.
Working in dollars with decimals and making a rounding slip; cents keep everything as integers.

Techniques

Use the answer choices (mod checks, size, form) to eliminate or select · Test small/specific values or special cases to find or verify the answer