Competition · AMC preparation · step 4 of 4
AMC 8 · 2008 · #20
Grade 6 rate-ratioPick an answer.
AMC 8 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The two fractions describe two counts that the problem says are equal — that is exactly an equation waiting to be written, so Tool #3 (Set Up an Equation) drives the work. Tool #4 (Introduce a Variable) gives us letters for the number of boys and the number of girls so we can write the equation. The word "minimum" is the cue for Tool #14 (Consider Extreme Cases): once the equation forces a ratio between boys and girls, the smallest whole-number pair fitting that ratio gives the answer.
Name the unknowns
Let b be the number of boys and g the number of girls; then boys and girls passed.
Giving the two unknown counts their own letters is the standard Grade 6 move before any equation.
6.EE.A.2Introduce A VariableWrite the passing equation
"Equal numbers passed" means the two passing counts are equal: .
"Equal" in English becomes "=" in math — a one-step Grade 6 translation.
6.EE.B.7Eliminate PossibilitiesClear the fractions
Multiply both sides by the LCM 12 to clear denominators, giving 8b = 9g.
Multiplying by the LCM is the Grade 6 way to turn a fraction equation into a clean whole-number equation.
6.NS.B.4Eliminate PossibilitiesRead off the ratio
Since 8 and 9 are coprime, whole-number solutions force the ratio b : g = 9 : 8.
An equation like 8b = 9g with coprime coefficients 8 and 9 pins the ratio b:g at 9:8.
6.RP.A.1Eliminate PossibilitiesPick the smallest counts
The smallest pair is b = 9, g = 8; check: (9) = 6 boys and (8) = 6 girls passed — equal.
"Smallest counts in a fixed ratio" is the Grade 6 GCF idea — take the numerator and denominator of the already-reduced fraction.
The smallest whole-number class that meets every condition has 9 boys and 8 girls.
▸ Why?
Both sides of 8b = 9g name one common amount M, so M is at the same time a whole-number multiple of 8 and of 9; the smallest number that is a multiple of both is 72, and that forces b = 72 ÷ 8 = 9 and g = 72 ÷ 9 = 8.
▸ Why?
M = 8b counts M as 8 equal groups of b, and M = 9g counts the same M as 9 equal groups of g, so M divides evenly by 8 and by 9 — it is a common multiple of both.
▸ Why?
The smallest amount that is a multiple of both 8 and 9 is their least common multiple; because 8 and 9 share no factor larger than 1, that least common multiple is the whole product 8 × 9 = 72, so no common multiple smaller than 72 exists and the least possible M is 72.
▸ Why?
With M fixed at 72, the boy and girl counts come straight back by undoing each multiplication: b = 72 ÷ 8 = 9 and g = 72 ÷ 9 = 8.
Add boys and girls
Add: 9 boys + 8 girls = 17 students, the smallest possible class.
Once the smallest ratio pair is found, the total is just their sum.
6.RP.A.3Extreme PrincipleWhen two fractions of two groups give the same count, write one equation, clear the fractions, and the smallest whole numbers in the resulting ratio give the answer — here 9 boys and 8 girls, total 17.
- Name the unknowns
- Write the passing equation
- Clear the fractions
- Read off the ratio
- Pick the smallest counts
- Add boys and girls
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