AMC 10 · 2009 · #1
Grade 4 arithmeticEach 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?
Pick an answer.
AMC 10 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: On each of $5$ workdays Jane buys exactly one item: a $50$-cent muffin or a $75$-cent bagel. Her total spending for the five days comes out to an exact whole number of dollars. Find how many of the five items were bagels.
Givens: There are $5$ purchases, one per day; A muffin costs $50$ cents; a bagel costs $75$ cents; The five-day total is a whole number of dollars (no leftover cents); Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$
Unknowns: The number of bagels Jane bought during the week
Understand
Restated: On each of $5$ workdays Jane buys exactly one item: a $50$-cent muffin or a $75$-cent bagel. Her total spending for the five days comes out to an exact whole number of dollars. Find how many of the five items were bagels.
Givens: There are $5$ purchases, one per day; A muffin costs $50$ cents; a bagel costs $75$ cents; The five-day total is a whole number of dollars (no leftover cents); Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) $5$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #8 Analyze the Units, #3 Eliminate Possibilities
Naming the number of bagels with a single letter (Tool #4) lets the muffin count and the whole total be written from that one unknown, so the whole-dollar rule becomes one clean condition to check. Working in cents (Tool #8) is the key move: 'a whole number of dollars' just means the cent total ends in $00$, i.e. is a multiple of $100$. With only six possible bagel counts, Tool #3 (Eliminate Possibilities) then finishes by testing which count actually lands on a multiple of $100$.
Execute — Answer: B
4.MD.A.2 Step 1 Work in cents
- Switch every price to cents so the money is whole numbers: a muffin is $50$ and a bagel is $75$.
- 'A whole number of dollars' means the total has no leftover cents — its cent value ends in $00$, so the total must be a multiple of $100$.
💡 Counting in cents turns 'whole number of dollars' into the simple test 'ends in $00$.'
4.OA.A.3 Step 2 Name the bagels, build the total
- Let $b$ be the number of bagels.
- Since there are $5$ purchases, the number of muffins is $5-b$.
- The total cost in cents is the bagels plus the muffins: $75b+50(5-b)$.
- Multiply out $50(5-b)=250-50b$ and combine: $75b+250-50b=250+25b$.
💡 One unknown, $b$, captures the whole week because the muffins are just whatever is left of the five days.
4.OA.B.4 Step 3 Set the whole-dollar condition
- For a whole number of dollars, the total $250+25b$ cents must be a multiple of $100$.
- The base $250$ already ends in $50$, and each extra bagel adds $25$ cents.
- So the question is: how many $25$-cent steps past $250$ land exactly on a multiple of $100$?
💡 A whole-dollar total is just a multiple of $100$, so we hunt for the multiple of $100$ inside the reachable range.
4.OA.B.4 Step 4 Test the six counts
- Plug in $b=0,1,2,3,4,5$: the totals are $250,275,300,325,350,375$ cents.
- The only one that is a multiple of $100$ is $300$ cents $=\$3.00$, which happens when $b=2$. Every other count leaves stray $25$, $50$, or $75$ cents. So Jane bought $2$ bagels, which is choice (B).
💡 With just six cases, checking each is faster than any clever trick and leaves no doubt.
4.MD.A.2 Switch every price to cents so the money is whole numbers: a muffin is $50$ and 4.OA.A.3 Let $b$ be the number of bagels. Since there are $5$ purchases, the number of mu 4.OA.B.4 For a whole number of dollars, the total $250+25b$ cents must be a multiple of $ 4.OA.B.4 Plug in $b=0,1,2,3,4,5$: the totals are $250,275,300,325,350,375$ cents. The onl Review
Reasonableness: The weekly total must sit between all muffins, $5\times50=250$ cents ($\$2.50$), and all bagels, $5\times75=375$ cents ($\$3.75$). The only whole-dollar amount in that range is $\$3.00$. Starting from $\$2.50$, each muffin swapped for a bagel adds $25$ cents, and reaching $\$3.00$ needs $50$ more cents — exactly $2$ swaps, i.e. $2$ bagels. This matches the found answer of $2$.
Alternative: Track only the last two digits (the cents) as bagels are added. All muffins give a total ending in $50$; each added bagel bumps the ending by $25$: $50\to75\to00$. The ending first hits $00$ after $2$ bagels, so the answer is $2$ without computing any full total.
CCSS standards used (min grade 4)
4.MD.A.2Solve word problems involving distances, time, liquid volumes, and money (Converting the $50$-cent and $75$-cent prices to cents and reading 'whole number of dollars' as a multiple of $100$ cents.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Expressing the muffins as $5-b$ and building the weekly total $75b+50(5-b)=250+25b$.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Requiring the total to be a multiple of $100$ and testing which bagel count $0$–$5$ satisfies it.)
⭐ Count the money in cents: a whole number of dollars just means the total ends in $00$, so find which choice lands there.
⭐ Count the money in cents: a whole number of dollars just means the total ends in $00$, so find which choice lands there.
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