Competition · AMC preparation · step 4 of 4
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.
Match the eighth-grade counts
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 SubproblemsScale the first ratio
Rescale 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 SubproblemsScale the second ratio
Rescale 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 SubproblemsCombine into one ratio
Subproblem 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.
The smallest whole-number counts of 8^th-, 7^th-, and 6^th-graders that satisfy both given ratios at the same time are 40 : 25 : 24.
▸ Why?
Both ratios can be rewritten so the 8^th-graders read 40, and since that is the very same group of students in each, the two separate comparisons lock together into the single list 40 : 25 : 24.
▸ Why?
Multiplying both parts of a ratio by the same whole number gives an equivalent ratio, so 5 : 3 becomes 40 : 24 (times 8) and 8 : 5 becomes 40 : 25 (times 5) while still describing the same groups.
▸ Why?
The 8^th-graders are one single group, so the count read off the first ratio and the count read off the second must be the same number; once both read 40, the 6^th- and 7^th-grader counts hang from that shared anchor on one common scale.
▸ Why?
The 8^th-grader count cannot be smaller than 40, because it must be a number reachable both by fives (from 5 : 3) and by eights (from 8 : 5), and 40 is the first count where those two arrive together.
▸ Why?
The ratio 5 : 3 hands out 8^th-graders in equal groups of 5 and the ratio 8 : 5 hands them out in equal groups of 8, so the count is a whole-number multiple of 5 and also of 8.
▸ Why?
The multiples of 5 (5, 10, …, 40) and the multiples of 8 (8, 16, …, 40) first meet at 40, so 40 is the least common multiple of 5 and 8 — the smallest count both equal-group sizes can reach together.
Add the three counts
Add 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!
- Match the eighth-grade counts
- Scale the first ratio
- Scale the second ratio
- Combine into one ratio
- Add the three counts
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