AMC 8 · 2024 · #9
Grade 4 arithmeticProblem
All the marbles in Maria's collection are red, green, or blue. Maria has half as many red marbles as green marbles, and twice as many blue marbles as green marbles. Which of the following could be the total number of marbles in Maria's collection?
Pick an answer.
AMC 8 2024 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: Maria's marbles are all red, green, or blue. The number of red marbles is half the number of green marbles, and the number of blue marbles is twice the number of green marbles. Which of the choices $24, 25, 26, 27, 28$ could be the total number of marbles?
Givens: Three marble colors: red (R), green (G), blue (B); $R = \tfrac{1}{2} G$ (red is half of green); $B = 2G$ (blue is twice green); Answer choices: (A) 24, (B) 25, (C) 26, (D) 27, (E) 28
Unknowns: Which choice can equal the total $T = R + G + B$
Understand
Restated: Maria's marbles are all red, green, or blue. The number of red marbles is half the number of green marbles, and the number of blue marbles is twice the number of green marbles. Which of the choices $24, 25, 26, 27, 28$ could be the total number of marbles?
Givens: Three marble colors: red (R), green (G), blue (B); $R = \tfrac{1}{2} G$ (red is half of green); $B = 2G$ (blue is twice green); Answer choices: (A) 24, (B) 25, (C) 26, (D) 27, (E) 28
Plan
Primary tool: #6 Guess and Check
Secondary: #5 Look for a Pattern, #3 Eliminate Possibilities
Once you notice $G$ must be even, the most natural move is Tool #6 (Guess and Check): try the smallest even green counts $G = 2, 4, 6, 8, \dots$ and just build the collection. After a few tries, Tool #5 (Look for a Pattern) reveals the totals are $7, 14, 21, 28, \dots$ — multiples of 7. Since the problem is multiple choice, Tool #3 (Eliminate Possibilities) finishes the job by keeping only the choice that is a multiple of 7. This avoids algebra entirely; we ‘build’ the answer with concrete numbers.
Execute — Answer: E
4.OA.A.1 Step 1 - Build the smallest possible case.
- ‘Half as many red as green’ means red is one of the two equal pieces of green, and ‘twice as many blue as green’ means blue is two greens stacked together.
- For red to be a whole number, green must split evenly in two — so green must be **even**.
- Try the smallest even number $G = 2$.
💡 ‘Half as many’ and ‘twice as many’ are multiplicative comparison phrases, which are introduced in the Grade 4 multiplicative-comparison standard.
1.OA.A.2 Step 2 - Add the three color counts for $G = 2$ to get the total.
- $R + G + B = 1 + 2 + 4 = 7$.
- So the smallest possible collection has 7 marbles.
💡 Adding three small whole numbers to get a single total is the Grade 1 three-addend addition skill.
4.OA.C.5 Step 3 - Now try the next few even greens $G = 4, 6, 8$ and line up the totals to spot a pattern.
- $G=4 \Rightarrow (R,G,B) = (2,4,8)$, total $14$.
- $G=6 \Rightarrow (3,6,12)$, total $21$.
- $G=8 \Rightarrow (4,8,16)$, total $28$.
- The totals $7, 14, 21, 28, \dots$ go up by 7 every time, so every possible total is a **multiple of 7**.
💡 Generating a number pattern from a rule (each new even $G$ adds 7 to the total) is the Grade 4 ‘generate a pattern from a rule’ standard.
4.OA.B.4 Step 4 - Only choices that are multiples of 7 can be the total.
- Recall the 7s multiplication facts: $7, 14, 21, 28, 35, \dots$.
- Check each choice: (A) 24, (B) 25, (C) 26, (D) 27 are all **not** multiples of 7 and get eliminated.
- (E) 28 $= 7 \times 4$, so it is a multiple of 7.
- Confirm $T = 28$ works: $(R,G,B) = (4,8,16)$ are all whole numbers — perfect.
💡 Recognizing which whole numbers are multiples of 7 is exactly the Grade 4 ‘factors and multiples’ standard.
4.OA.A.1 Build the smallest possible case. ‘Half as many red as green’ means red is one o 1.OA.A.2 Add the three color counts for $G = 2$ to get the total. $R + G + B = 1 + 2 + 4 4.OA.C.5 Now try the next few even greens $G = 4, 6, 8$ and line up the totals to spot a 4.OA.B.4 Only choices that are multiples of 7 can be the total. Recall the 7s multiplicat Review
Reasonableness: Plug the answer $28$ back into the original conditions. Green $8$, red $4$ ($= 8 \div 2$, half of green ✓), blue $16$ ($= 8 \times 2$, twice green ✓), total $4 + 8 + 16 = 28$ ✓. All three counts are whole numbers, so every condition is satisfied. Also, one ‘bundle’ of (1 red, 2 green, 4 blue) is 7 marbles, and the total is just $k$ bundles, so only multiples of 7 are possible — confirming why 24, 25, 26, 27 cannot occur.
Alternative: Tool #13 (Convert to Algebra) is the standard alternative. Let $G = 2x$, then $R = x$ and $B = 4x$, so $T = x + 2x + 4x = 7x$. Hence $T$ must be a multiple of 7, and the only choice that fits is 28. Same answer, but for an elementary student the guess-and-check + pattern combo is more concrete and avoids introducing variables.
CCSS standards used (min grade 4)
1.OA.A.2Solve word problems involving three whole numbers whose sum is within 20 (Adding the three color counts in the smallest case, $1 + 2 + 4 = 7$, to find the total.)4.OA.A.1Interpret a multiplication equation as a comparison (Translating ‘half as many red as green’ and ‘twice as many blue as green’ into multiplicative relationships between the color counts.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Identifying which of the answer choices 24, 25, 26, 27, 28 is a multiple of 7 to lock in the answer.)4.OA.C.5Generate a number or shape pattern following a given rule (Generating the totals $7, 14, 21, 28$ for $G = 2, 4, 6, 8$ to discover the rule ‘the total is always a multiple of 7’.)
⭐ This AMC 8 problem only needs Grade 4 multiplicative comparison and multiples of 7 you already know!
⭐ This AMC 8 problem only needs Grade 4 multiplicative comparison and multiples of 7 you already know!
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