AMC 8 · 2024 · #21

Grade 4 rate-ratio

Archetype Multiplicative Ratio with Integrality Constraints └ Sub-archetype Ratio Shift — Two State

ratio-proportionlinear-equations-one-var systematic-enumeration ↑ Prerequisites: ratio-proportionmulti-digit-arithmetic

📏 Medium solution 💡 3 insights

Problem

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.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Guess and Check

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, … 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.

1STEP 1

"3 : 1" just means green is 3 times yellow, so the initial pairs are (1, 3), (2, 6), (3, 9), … — green always triple the yellow.

initial (Y, G) ∈ {(1,3),(2,6),(3,9),(4,12),(5,15),(6,18),…} (green = 3 × yellow)

2STEP 2

Green nets -3 + 5 = +2 and yellow nets +3 - 5 = -2, so after the moves (Y, G) becomes (Y - 2, G + 2).

after moves: (Y_new, G_new) = (Y - 2, G + 2)

3STEP 3

Now guess and check: the new ratio is 4 : 1 exactly when G + 2 = 4 × (Y - 2), and we need Y ≥ 3.

Check condition: G + 2 ?= 4 × (Y - 2)

4STEP 4

Testing Y = 3, 4, 5, … the ratios fall 11, 7, 5, …; only Y = 10 gives (10, 30) → (8, 32) with $\frac{32}{8}$ = 4.

Y=10: (10, 30)→(8, 32),

\frac{32}{8}

= 4 =

\frac{4}{1}

✓

5STEP 5

New green 32 minus new yellow 8 gives the difference 32 - 8 = 24, matching choice (E).

32 - 8 = 24 → (E)

Answer

The new ratio 4 : 1 means 4 green for every 1 yellow, so with 8 yellow there must be 4 × 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 × 8 = 24. Both views give 24, and it matches the largest answer choice, which is reasonable since the new ratio is quite lopsided.

💡Key takeaway

This AMC 8 problem only needs Grade 4 multiplicative comparison (" ∼ times as many") you already know!

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