Competition · AMC preparation · step 4 of 4
AMC 8 · 2002 · #25
Grade 6 rate-ratioalgebraPick an answer.
AMC 8 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The problem never gives a dollar amount, only fractions and the rule "each gift is equal." That equal-gift dollar value is the hidden link, so Tool #4 (Introduce a Variable) — call it x — turns every starting balance into a multiple of x. Once Moe has 5x, Loki has 4x, and Nick has 3x, the answer is a ratio of dollar counts. Tool #6 (Guess and Check) gives a quick concrete sanity check: pick x = $5 and confirm the same fraction appears.
Name the equal gift
Let x be the equal gift each friend hands Ott, so Ott collects it three times for 3x.
When a problem says "the same amount each time," giving that amount a single letter usually unlocks the rest.
6.EE.A.2Introduce A VariableWork back each starting amount
Reverse each fraction: Moe started with 5x, Loki with 4x, and Nick with 3x.
"A fraction of my money equals x" reverses into "my money equals x divided by that fraction" — multiply by the reciprocal.
Each friend's starting money is fixed by the shared gift: since the gift x is one-fifth of Moe's money, one-fourth of Loki's, and one-third of Nick's, Moe must have started with 5x, Loki with 4x, and Nick with 3x.
▸ Why?
Take Moe. Saying the gift x is one-fifth of his money means his money was split into five equal parts and x is one of them, so his whole start is those five equal parts of x — that is 5x; reading Loki's one-fourth and Nick's one-third the same way gives 4x and 3x.
▸ Why?
Moe's money was cut into five equal parts with nothing left over and no part counted twice, so adding the five parts back returns his whole starting amount.
▸ Why?
Those five parts are each the same size, the gift x, and five equal parts of x is five groups of x, which is 5x.
Add the group total
Nothing is created or lost, so add the starting balances: the group holds 12x.
Money only moves between friends, so the group total is fixed — adding starting balances is the same as adding final balances.
6.EE.A.3Introduce A VariableForm Ott's fraction
Ott's 3x over the group's 12x reduces to — choice (B).
The x cancels — exactly what we hoped, since the answer cannot depend on the actual dollar amount.
6.RP.A.1Introduce A VariableWhen three different fractions all give the same amount, name that amount x and let each starting balance fall out — the unknown x cancels at the end, so the answer is a ratio of plain counts.
- Name the equal gift
- Work back each starting amount
- Add the group total
- Form Ott's fraction
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