Competition · AMC preparation · step 4 of 4
AMC 8 · 2024 · #21
Grade 4 rate-ratioPick an answer.
AMC 8 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Read the ratio as a multiple
"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.
Turning "3:1" into "green is 3 times as many as yellow" is exactly Grade 4 multiplicative comparison (" ∼ times as many").
4.OA.A.2Make A Systematic ListWrite the counts after the move
Green nets -3 + 5 = +2 and yellow nets +3 - 5 = -2, so after the moves (Y, G) becomes (Y - 2, G + 2).
Combining +3 with -5 and -3 with +5 to get the net change ± 2 is just Grade 2 add/subtract within 100.
2.NBT.B.5Make A Systematic ListSet up the new-ratio test
Now guess and check: the new ratio is 4 : 1 exactly when G + 2 = 4 × (Y - 2), and we need Y ≥ 3.
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.
4.OA.A.2Guess And CheckTry the yellow counts in turn
Testing Y = 3, 4, 5, … the ratios fall 11, 7, 5, …; only Y = 10 gives (10, 30) → (8, 32) with = 4.
Computing 3 × 10 = 30, 4 × 8 = 32, and 32 ÷ 8 = 4 uses Grade 3 fluent multiplication and division within 100.
Trying the possible starting yellow counts in order, the arrangement that ends with the 4:1 finish is the one whose new counts are 8 yellow and 32 green.
▸ Why?
Every possible starting picture is pinned down by a single number, the starting yellow count, because the starting green count is always three times it.
▸ Why?
A 3:1 start means the green pile is three equal copies of the yellow pile, so you get the green count by taking the yellow count three times.
▸ Why?
Once a starting picture is chosen, the moves force the finish: green ends two higher and yellow ends two lower than they began.
▸ Why?
Three green leave and five green arrive, so three of the arrivals only replace the ones that left and just two are a real gain; the yellow side changes by two the same way in reverse.
▸ Why?
The arrangement that fits is the one where the new green count is exactly four equal copies of the new yellow count, which is what the 4:1 finish demands.
▸ Why?
A 4:1 finish means the green is four equal copies of the yellow, so the test for a match is whether the new green equals the new yellow taken four times.
Take the difference asked for
New green 32 minus new yellow 8 gives the difference 32 - 8 = 24, matching choice (E).
Subtracting a one-digit number from a two-digit number (32 - 8) is core Grade 2 subtraction within 100.
2.NBT.B.5Eliminate PossibilitiesThis AMC 8 problem only needs Grade 4 multiplicative comparison (" ∼ times as many") you already know!
- Read the ratio as a multiple
- Write the counts after the move
- Set up the new-ratio test
- Try the yellow counts in turn
- Take the difference asked for
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