AMC 8 · 2013 · #16
Grade 6 rate-rationumber-theoryPick an answer.
AMC 8 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
We are given two ratios that share one term (8^th-graders) but use different numbers for it (5 vs. 8). Tool #7 (Identify Subproblems) splits the work into two clean pieces: (1) rescale each ratio so the 8^th-grader count agrees, then (2) merge into a single three-part ratio and add. The bridge between the two subproblems is finding a common value for 8^th-graders — namely lcm(5, 8). Tool #6 (Guess and Check) is held in reserve for the review phase, since the smaller answer choices (16, 40, 55, 79) can be ruled out by quick divisibility checks.
Subproblem 1: make the 8^th-grade term equal in both ratios — the smallest shared value is lcm(5, 8) = 40.
Finding the least common multiple of two small numbers is a Grade 6 number-system skill.
6.NS.B.4Identify SubproblemsRescale 8^th : 6^th = 5 : 3 by ×8 so the 8^th term becomes 40 — giving 40 : 24.
Multiplying both parts of a ratio by the same number gives an equivalent ratio — the Grade 6 ratio-reasoning move.
6.RP.A.3Identify SubproblemsRescale 8^th : 7^th = 8 : 5 by ×5 so its 8^th term is also 40 — giving 40 : 25.
Same Grade 6 move on the other ratio so the two ratios now speak the same language about 8^th-graders.
6.RP.A.3Identify SubproblemsSubproblem 2: both share 40, so they snap into 40 : 25 : 24; gcd = 1 means it is already in lowest whole-number terms.
Extending a two-term ratio into a three-term ratio by aligning the shared term is the Grade 6 ratio-language skill.
6.RP.A.1Identify SubproblemsAdd the lowest-terms counts: 40 + 25 + 24 = 89 → (E).
Once the ratio is in lowest whole-number form, adding the parts gives the smallest possible total.
6.RP.A.3Identify SubproblemsThis AMC 8 problem only needs Grade 6 ratio reasoning — match the shared term using lcm, then add the parts!