2013 AMC 8 Problem 16

Attempt Problem 16 of the 2013 AMC 8 below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

16.

A number of students from Fibonacci Middle School are taking part in a community service project. The ratio of 88th-graders to 66th-graders is 5:3,5:3, and the ratio of 88th-graders to 77th-graders is 8:5.8:5. What is the smallest number of students that could be participating in the project?
1616
4040
5555
7979
8989
Answer: E
Concepts:ratio and proportionleast common multiple
Small hint:
The number of eighth-graders must fit both ratios.
Big hint:
Use the least common multiple of 55 and 88.
Video solution:
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Written solution:
The number of 88th-graders must be a multiple of 55 and 8,8, which means that it is at least 40.40. Using this number, we get that the number of 66th-graders is 40⋅35=24.40 \cdot \dfrac{3}{5} = 24. Similarly, the number of 77th-graders is 40⋅58=25.40 \cdot \dfrac{5}{8} = 25.
The total number of students is therefore 40+24+25=89.40 + 24 + 25 = 89.
Thus, E is the correct answer.
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