AMC 8 · 2014 · #7
Grade 6 rate-ratioPick an answer.
AMC 8 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The class has 4 extra girls on top of an otherwise equal split. Tool #7 (Identify Subproblems) lets us peel off those 4 extra girls first, so the remaining 28 - 4 = 24 students split evenly into 12 boys and 12 girls. Adding the 4 back gives the counts in one clean step. We keep Tool #6 (Guess and Check) on hand because it's the natural way to double-check the answer against the multiple-choice ratios: only 16:12 satisfies both "sums to 28" and "differs by 4". Algebra (Tool #13) would also work, but for two unknowns connected by a sum and a difference, splitting off the extra is faster and more intuitive.
Set aside the 4 extra girls; that leaves 28 - 4 = 24 students who split evenly.
Subtracting the "extra" first is the classic subproblem move: turn an unequal split into a fair one.
2.OA.A.1Identify SubproblemsHalf of 24 is 12, so there are 12 boys (and 12 girls before the extras go back).
Dividing a total into two equal groups is a Grade 3 "partitive division" idea.
3.OA.A.2Identify SubproblemsPut the 4 extra girls back: 12 + 4 = 16 girls.
Once the equal split is found, the extras only land in the girls' column.
2.OA.A.1Identify SubproblemsWrite girls : boys = 16 : 12, then divide both by gcd(16, 12) = 4 for lowest terms.
A ratio in lowest terms divides both numbers by their greatest common factor, just like reducing a fraction.
6.RP.A.1Identify SubproblemsOnce you set the 4 extra girls aside, the rest of the class splits in half — and the ratio idea you need is right at the Grade 6 level.