AMC 8 · 2024 · #21
Easy mode Grade 4Problem
Imagine a group of frogs living in a tree. A frog turns green whenever it sits in the shade, and turns yellow whenever it sits in the sun.
At the start, for every 1 yellow frog there are 3 green frogs.
Then two things happen:
- 3 green frogs hop over to the sunny side, so they turn yellow.
- 5 yellow frogs hop over to the shady side, so they turn green.
After these moves, the new ratio of green frogs to yellow frogs is 4 to 1.
How many more green frogs than yellow frogs are there now?
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: An army of frogs lives in a tree. A frog is green in the shade and yellow in the sun. Initially the ratio of green to yellow frogs is $3 : 1$. Then $3$ green frogs move to the sun (turning yellow) and $5$ yellow frogs move to the shade (turning green). After these moves, the ratio is $4 : 1$. We want the difference between the **new** number of green frogs and the **new** number of yellow frogs.
Givens: Initial green : yellow $= 3 : 1$ (3 green for every 1 yellow); $3$ green frogs move to the sun $\Rightarrow$ green drops by $3$, yellow grows by $3$; $5$ yellow frogs move to the shade $\Rightarrow$ yellow drops by $5$, green grows by $5$; After the moves, green : yellow $= 4 : 1$; Answer choices: (A) 10, (B) 12, (C) 16, (D) 20, (E) 24
Unknowns: The difference between the new number of green frogs and the new number of yellow frogs
Understand
Restated: An army of frogs lives in a tree. A frog is green in the shade and yellow in the sun. Initially the ratio of green to yellow frogs is $3 : 1$. Then $3$ green frogs move to the sun (turning yellow) and $5$ yellow frogs move to the shade (turning green). After these moves, the ratio is $4 : 1$. We want the difference between the **new** number of green frogs and the **new** number of yellow frogs.
Givens: Initial green : yellow $= 3 : 1$ (3 green for every 1 yellow); $3$ green frogs move to the sun $\Rightarrow$ green drops by $3$, yellow grows by $3$; $5$ yellow frogs move to the shade $\Rightarrow$ yellow drops by $5$, green grows by $5$; After the moves, green : yellow $= 4 : 1$; Answer choices: (A) 10, (B) 12, (C) 16, (D) 20, (E) 24
Plan
Primary tool: #6 Guess and Check
Secondary: #2 Make a Systematic List, #3 Eliminate Possibilities
The reference solution uses Tool #13 (Algebra) — introduce variables $G, Y$ and solve a system. But "$3:1$" really just means "for every $1$ yellow there are $3$ green," so once we pick the initial yellow count, the initial green count and both new counts are forced. That makes Tool #6 (Guess & Check) a natural fit: try initial yellow $Y = 3, 4, 5, \dots$ in order, compute the post-move counts, and check whether they hit the $4:1$ target. We use Tool #2 (Systematic List) to enumerate candidates without skipping and Tool #3 (Eliminate) to rule out the ones whose new ratio is wrong. This whole path uses only multiplication and subtraction — no variables, no equations — so an elementary student can follow it.
Execute — Answer: E
4.OA.A.2 Step 1 - First, rewrite "$3:1$" in friendlier language — it just means "for every $1$ yellow frog there are $3$ green frogs," which is a **multiplicative comparison**.
- So once we pick the initial yellow count $Y$, the initial green count is automatically $3 \times Y$.
- The possible (yellow, green) pairs at the start are therefore $(1, 3), (2, 6), (3, 9), (4, 12), \dots$ — green is always $3$ times yellow.
💡 Turning "$3:1$" into "green is $3$ times as many as yellow" is exactly Grade 4 multiplicative comparison ("$\sim$ times as many").
2.NBT.B.5 Step 2 - Next, set up what happens after the moves.
- The green side loses $3$ and gains $5$, for a net change of $-3 + 5 = +2$.
- The yellow side loses $5$ and gains $3$, for a net change of $+3 - 5 = -2$.
- So the initial pair $(Y, G)$ becomes $(Y - 2,\; G + 2)$ after the moves.
💡 Combining $+3$ with $-5$ and $-3$ with $+5$ to get the net change $\pm 2$ is just Grade 2 add/subtract within 100.
4.OA.A.2 Step 3 - Now **guess and check** the initial yellow $Y$ from small to large.
- For the new ratio to be $4 : 1$, the new green count must be exactly $4$ times the new yellow count: $G + 2 = 4 \times (Y - 2)$.
- (And we need $Y - 2 \geq 1$, so $Y \geq 3$.) We will make a small table and check each $Y$.
💡 Asking "is new green exactly $4$ times new yellow?" is a Grade 4 multiplicative comparison, and trying candidates in order is the heart of guess-and-check.
3.OA.C.7 Step 4 - Try $Y = 3, 4, 5, 6, 7, 8, 9, 10$ in turn.
- Each line shows $(Y, G) \to (Y-2, G+2)$ and the resulting ratio.
- $Y=3: (3,9)\to(1,11)$, $11/1 = 11$ — way too big.
- $Y=4: (4,12)\to(2,14)$, $14/2 = 7$ — too big.
- $Y=5: (5,15)\to(3,17)$, $17/3$ is not a whole number.
- $Y=6: (6,18)\to(4,20)$, $20/4 = 5$ — still too big.
- $Y=7: (7,21)\to(5,23)$, $23/5$ not whole.
- $Y=8: (8,24)\to(6,26)$, $26/6$ not whole.
- $Y=9: (9,27)\to(7,29)$, $29/7$ not whole.
- $Y=10: (10,30)\to(8,32)$, $32/8 = 4$ — **exactly $4:1$!** The ratios are decreasing $11, 7, 5, 4, \dots$, so $Y = 10$ is the first hit and that is our answer.
💡 Computing $3 \times 10 = 30$, $4 \times 8 = 32$, and $32 \div 8 = 4$ uses Grade 3 fluent multiplication and division within 100.
2.NBT.B.5 Step 5 - Finally, answer the actual question: "difference between the number of green and yellow frogs **now**." After the moves, green $= 32$ and yellow $= 8$, so the difference is $32 - 8 = 24$.
- Among the choices $10, 12, 16, 20, 24$, this matches exactly (E) $24$.
- The other choices are all smaller than $24$ and disagree with the value we computed, so they are eliminated.
💡 Subtracting a one-digit number from a two-digit number ($32 - 8$) is core Grade 2 subtraction within 100.
4.OA.A.2 First, rewrite "$3:1$" in friendlier language — it just means "for every $1$ yel 2.NBT.B.5 Next, set up what happens after the moves. The green side loses $3$ and gains $5 4.OA.A.2 Now **guess and check** the initial yellow $Y$ from small to large. For the new 3.OA.C.7 Try $Y = 3, 4, 5, 6, 7, 8, 9, 10$ in turn. Each line shows $(Y, G) \to (Y-2, G+2 2.NBT.B.5 Finally, answer the actual question: "difference between the number of green and Review
Reasonableness: The new ratio $4 : 1$ means $4$ green for every $1$ yellow, so with $8$ yellow there must be $4 \times 8 = 32$ green, giving a difference of $32 - 8 = 24$. A second check: in the new ratio $4:1$, treating yellow as $1$ group and green as $4$ groups, the difference is $3$ groups. Each group has $8$ frogs, so the difference is $3 \times 8 = 24$. Both views give $24$, and it matches the largest answer choice, which is reasonable since the new ratio is quite lopsided.
Alternative: An alternative is Tool #13 (Algebra): set $G = 3Y$ and $G + 2 = 4(Y - 2)$, then solve the system to get $Y = 10$ directly (the reference solution's path). It gives the same answer, but the guess-and-check path above needs only multiplication and never introduces a variable or equation, so it is accessible to an elementary student.
CCSS standards used (min grade 4)
2.NBT.B.5Fluently add and subtract within 100 (Computing the net change $\pm 2$ from the moves and the final difference $32 - 8 = 24$.)3.OA.C.7Fluently multiply and divide within 100 (Computing $3 \times Y$ for initial green counts and dividing $32 \div 8 = 4$ to verify the $4:1$ ratio.)4.OA.A.2Multiply or divide to solve word problems involving multiplicative comparison (Rewriting "$3:1$" as "green is $3$ times yellow" and "$4:1$" as "green is $4$ times yellow" so each guess can be checked by multiplication.)
⭐ This AMC 8 problem only needs Grade 4 multiplicative comparison ("$\sim$ times as many") you already know!
⭐ This AMC 8 problem only needs Grade 4 multiplicative comparison ("$\sim$ times as many") you already know!
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