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.

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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
Difficulty rating: 1370
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 4035=24.40 \cdot \dfrac{3}{5} = 24. Similarly, the number of 77th-graders is 4058=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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