AMC 10 · 2008 · #8
Grade 6 countingA class collects 50 dollars to buy flowers for a classmate who is in the hospital. Roses cost 3 dollars each, and carnations cost 2 dollars each. No other flowers are to be used. How many different bouquets could be purchased for exactly 50 dollars?
Pick an answer.
AMC 10 2008 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: You have exactly $50$ to spend on a bouquet. Roses cost $3$ each and carnations cost $2$ each, and you must spend the whole $50$ with no other flowers. Count how many different combinations of roses and carnations spend exactly $50$.
Givens: Total to spend is exactly $50$ dollars; Each rose costs $3$ dollars, each carnation costs $2$ dollars; Only roses and carnations are allowed; Answer choices: (A) $1$, (B) $7$, (C) $9$, (D) $16$, (E) $17$
Unknowns: The number of different (roses, carnations) combinations that cost exactly $50$ dollars
Understand
Restated: You have exactly $50$ to spend on a bouquet. Roses cost $3$ each and carnations cost $2$ each, and you must spend the whole $50$ with no other flowers. Count how many different combinations of roses and carnations spend exactly $50$.
Givens: Total to spend is exactly $50$ dollars; Each rose costs $3$ dollars, each carnation costs $2$ dollars; Only roses and carnations are allowed; Answer choices: (A) $1$, (B) $7$, (C) $9$, (D) $16$, (E) $17$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #3 Eliminate Possibilities, #14 Extreme Principle, #2 Make a Systematic List
Tool #4 (Introduce a Variable) turns the money story into one clean equation, $3r+2c=50$, which pins down exactly what must be true. From there the choices are still infinite-looking, so Tool #3 (Eliminate Possibilities) uses a parity fact to throw away every odd number of roses at once. Tool #14 (Extreme Principle) finds the biggest number of roses that still leaves room, capping the range. Then Tool #2 (Make a Systematic List) walks the survivors in order so none are missed and none are double-counted.
Execute — Answer: C
6.EE.B.6 Step 1 Turn the money into an equation
- Let $r$ be the number of roses and $c$ be the number of carnations.
- Roses cost $3r$ dollars and carnations cost $2c$ dollars, and together they must be exactly $50$.
- So every valid bouquet is a whole-number solution of $3r+2c=50$ with $r\ge 0$ and $c\ge 0$.
💡 Naming the two counts turns a vague shopping question into one exact equation you can test.
2.OA.C.3 Step 2 Only an even number of roses works
- The carnation part $2c$ is always even, and the total $50$ is even.
- For $3r+2c$ to add up to an even $50$, the rose part $3r$ must also be even.
- Since $3$ is odd, $3r$ is even only when $r$ itself is even.
- That erases every odd number of roses in one stroke, leaving only even $r$ to check.
💡 Even total minus even carnation cost leaves an even amount for roses, and odd-priced roses hit an even amount only in even-sized groups.
6.EE.B.5 Step 3 Find the largest number of roses that fits
- The roses alone cannot cost more than $50$, or there is no money left for a non-negative number of carnations.
- So $3r\le 50$.
- The biggest even $r$ that fits is $16$, since $3\times16=48\le 50$ but the next even value $18$ gives $3\times18=54>50$.
- So $r$ runs through the even whole numbers from $0$ up to $16$.
💡 Push the number of roses to its limit to find where the money runs out.
4.OA.C.5 Step 4 List the survivors and count them
- Step down the even values of $r$ and compute $c=\dfrac{50-3r}{2}$ each time: $r=0\to c=25$, $r=2\to c=22$, $r=4\to c=19$, $r=6\to c=16$, $r=8\to c=13$, $r=10\to c=10$, $r=12\to c=7$, $r=14\to c=4$, $r=16\to c=1$.
- Every one gives a whole, non-negative $c$, so all nine are real bouquets.
- Counting them gives $9$, so the answer is (C).
💡 Walk the allowed rose counts in order so you catch every combination exactly once.
6.EE.B.6 Let $r$ be the number of roses and $c$ be the number of carnations. Roses cost $ 2.OA.C.3 The carnation part $2c$ is always even, and the total $50$ is even. For $3r+2c$ 6.EE.B.5 The roses alone cannot cost more than $50$, or there is no money left for a non- 4.OA.C.5 Step down the even values of $r$ and compute $c=\dfrac{50-3r}{2}$ each time: $r= Review
Reasonableness: The even rose counts from $0$ to $16$ are $0,2,4,6,8,10,12,14,16$; that is $\frac{16-0}{2}+1=9$ values, matching the direct list. Each carnation total $25,22,19,16,13,10,7,4,1$ steps down by $3$ every time $r$ climbs by $2$ (because two more roses cost $6$, which is three fewer carnations), and all stay whole and non-negative, so no combination was wrongly kept or dropped. The answer $9$ is choice (C). It sensibly sits between $1$ (far too few) and $17$ (which would wrongly count odd rose amounts too).
Alternative: Instead count by carnations. Solving for them, $c=\dfrac{50-3r}{2}$, so $50-3r$ must be even, again forcing $r$ even. Or list the multiples of $3$ that are even and at most $50$ ($0,6,12,\dots,48$); each leaves an even remainder to fill with carnations, and there are $9$ such multiples, giving the same count.
CCSS standards used (min grade 6)
6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Letting $r$ and $c$ stand for the counts of roses and carnations and writing $3r+2c=50$.)2.OA.C.3Determine whether a group of objects has an odd or even number (Using even/odd reasoning to show $3r$ must be even, so $r$ must be even.)6.EE.B.5Understand solving an equation or inequality as a process of finding values (Using $3r\le 50$ to bound the number of roses so carnations stay non-negative.)4.OA.C.5Generate a number or shape pattern following a given rule (Listing the even rose counts $0,2,\dots,16$ and counting the nine valid bouquets.)
⭐ Turn the spending into one equation, use even-and-odd to knock out half the cases, then list the rest in order and count.
⭐ Turn the spending into one equation, use even-and-odd to knock out half the cases, then list the rest in order and count.
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