AMC 10 · 2006 · #22
Grade 6 arithmeticElmo makes N sandwiches for a fundraiser. For each sandwich he uses B globs of peanut butter at 4\cent per glob and J blobs of jam at 5\cent per blob. The cost of the peanut butter and jam to make all the sandwiches is $$ 2.53.AssumethatB,J,andNarepositiveintegerswithN>1$. What is the cost of the jam Elmo uses to make the sandwiches?
Pick an answer.
AMC 10 2006 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: Elmo makes $N$ sandwiches. Each one uses $B$ globs of peanut butter at $4$ cents per glob and $J$ blobs of jam at $5$ cents per blob. Making all the sandwiches costs $\$2.53$ in peanut butter and jam. With $B$, $J$, $N$ all positive integers and $N>1$, find how much the jam alone costs.
Givens: Peanut butter costs $4$ cents per glob, jam costs $5$ cents per blob; Each sandwich uses $B$ globs of peanut butter and $J$ blobs of jam; There are $N$ sandwiches in all, and the total peanut-butter-plus-jam cost is $\$2.53$; $B$, $J$, and $N$ are positive integers with $N>1$; Answer choices: (A) $\$1.05$, (B) $\$1.25$, (C) $\$1.45$, (D) $\$1.65$, (E) $\$1.85$
Unknowns: The total cost of just the jam Elmo uses
Understand
Restated: Elmo makes $N$ sandwiches. Each one uses $B$ globs of peanut butter at $4$ cents per glob and $J$ blobs of jam at $5$ cents per blob. Making all the sandwiches costs $\$2.53$ in peanut butter and jam. With $B$, $J$, $N$ all positive integers and $N>1$, find how much the jam alone costs.
Givens: Peanut butter costs $4$ cents per glob, jam costs $5$ cents per blob; Each sandwich uses $B$ globs of peanut butter and $J$ blobs of jam; There are $N$ sandwiches in all, and the total peanut-butter-plus-jam cost is $\$2.53$; $B$, $J$, and $N$ are positive integers with $N>1$; Answer choices: (A) $\$1.05$, (B) $\$1.25$, (C) $\$1.45$, (D) $\$1.65$, (E) $\$1.85$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #8 Analyze the Units, #2 Make a Systematic List, #3 Eliminate Possibilities
The costs are spelled out per glob and per blob, so Tool #4 (Introduce a Variable) turns the story into the single equation $N(4B+5J)=253$. Tool #8 (Analyze the Units) is what makes that equation clean: switching from dollars to cents turns $\$2.53$ into the whole number $253$, so everything is an integer. The winning idea is that $253$ barely factors — $253=11\times 23$ — so $N$ has almost no choices. Tool #2 (Make a Systematic List) writes down those few divisors, and Tool #3 (Eliminate Possibilities) throws out the ones that leave no whole-number $B$ and $J$, until only one case survives.
Execute — Answer: D
6.EE.B.7 Step 1 Turn the story into one equation
- Work in cents so every number is a whole number.
- One sandwich uses $B$ globs at $4$ cents and $J$ blobs at $5$ cents, costing $4B+5J$ cents.
- All $N$ sandwiches cost $N$ times that.
- The total is $\$2.53=253$ cents, so $N(4B+5J)=253$.
💡 Counting in cents keeps every quantity a whole number, so the total is a clean integer product.
4.OA.B.4 Step 2 Factor 253 to pin down N
- In the equation $N(4B+5J)=253$, both $N$ and $4B+5J$ are whole numbers, so $N$ must divide $253$.
- Factoring, $253=11\times 23$, and both $11$ and $23$ are prime, so the only divisors of $253$ are $1,\,11,\,23,\,253$.
- Since $N>1$, that leaves just $N=11$, $N=23$, or $N=253$.
💡 A product of two whole numbers can only split the ways its factors allow, and $253$ splits almost no ways.
6.EE.B.5 Step 3 Test each N and drop the dead ends
- For each candidate, $4B+5J$ must equal $253\div N$ with $B,J\ge 1$.
- If $N=253$, then $4B+5J=1$, impossible since $4B+5J\ge 9$.
- If $N=23$, then $4B+5J=11$; trying $J=1$ needs $4B=6$ and no other $J$ works either, so no whole-number answer.
- If $N=11$, then $4B+5J=23$, which does have a solution.
- Only $N=11$ survives.
💡 The smallest possible sandwich cost is $4+5=9$ cents, so any target below that is instantly out.
4.OA.A.3 Step 4 Solve for B and J, then price the jam
- Solve $4B+5J=23$ with whole numbers.
- The jam term $5J$ must leave a multiple of $4$ behind: $J=1$ gives $4B=18$ (no), $J=2$ gives $4B=13$ (no), $J=3$ gives $4B=8$ so $B=2$ (yes).
- So each sandwich uses $J=3$ blobs of jam, across $N=11$ sandwiches.
- The jam costs $5$ cents a blob, so the total jam cost is $11\times 3\times 5 = 165$ cents $=\$1.65$, which is choice (D).
💡 Once $N$ is forced, the leftover equation has a single whole-number fit, and the jam cost falls right out.
6.EE.B.7 Work in cents so every number is a whole number. One sandwich uses $B$ globs at 4.OA.B.4 In the equation $N(4B+5J)=253$, both $N$ and $4B+5J$ are whole numbers, so $N$ m 6.EE.B.5 For each candidate, $4B+5J$ must equal $253\div N$ with $B,J\ge 1$. If $N=253$, 4.OA.A.3 Solve $4B+5J=23$ with whole numbers. The jam term $5J$ must leave a multiple of Review
Reasonableness: Check the numbers against the story: $N=11$ sandwiches, each with $B=2$ globs and $J=3$ blobs. Per sandwich that is $4\cdot 2+5\cdot 3=8+15=23$ cents, and $11\times 23=253$ cents $=\$2.53$, matching exactly. The jam part is $11\times 3\times 5=165$ cents $=\$1.65$, and the peanut butter part is $11\times 2\times 4=88$ cents, and $165+88=253$ — the two pieces add back to the total, so $\$1.65$, choice (D), holds up.
Alternative: Instead of dividing $253$ by each factor, notice $253=11\times 23$ and read it directly as $N=11$ sandwiches costing $23$ cents each, since $N>1$ and the per-sandwich cost $4B+5J$ must be at least $9$ (ruling out $N=253$) while $N=23$ would force a $11$-cent sandwich with no valid $B,J$. Then just split $23=4B+5J$ as $8+15$, giving $B=2$, $J=3$ and jam cost $11\times 15=\$1.65$.
CCSS standards used (min grade 6)
6.EE.B.7Solve real-world problems by writing equations of the form px = q (Translating the sandwich story into the single equation $N(4B+5J)=253$ in cents.)4.OA.B.4Find all factor pairs for a whole number and recognize prime versus composite (Factoring $253=11\times 23$ to list every divisor and narrow $N$ to $\{11,23,253\}$.)6.EE.B.5Understand solving an equation as finding which values make it true (Testing each candidate $N$ to see whether $4B+5J=253/N$ has positive integer solutions.)4.OA.A.3Solve multi-step word problems using the four operations with whole numbers (Solving $4B+5J=23$ for $B=2,J=3$ and computing the jam cost $11\times 3\times 5=165$ cents.)
⭐ When a total is a whole number of cents, factor it: $253=11\times 23$ leaves almost no choices, so $N$ must be $11$ sandwiches — and the jam works out to $\$1.65$.
⭐ When a total is a whole number of cents, factor it: $253=11\times 23$ leaves almost no choices, so $N$ must be $11$ sandwiches — and the jam works out to $\$1.65$.
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