2013 AMC 8 Problems/Problem 16: Difference between revisions

Revision as of 19:35, 27 October 2020

Problem

A number of students from Fibonacci Middle School are taking part in a community service project. The ratio of $8^\text{th}$ -graders to $6^\text{th}$ -graders is $5:3$ , and the the ratio of $8^\text{th}$ -graders to $7^\text{th}$ -graders is $8:5$ . What is the smallest number of students that could be participating in the project?

$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 55 \qquad \textbf{(D)}\ 79 \qquad \textbf{(E)}\ 89$

Video Solution

https://youtu.be/rQUwNC0gqdg?t=949

Solution

Solution 1: Algebra

We multiply the first ratio by 8 on both sides, and the second ratio by 5 to get the same number for 8th graders, in order that we can put the two ratios together:

$5:3 = 5(8):3(8) = 40:24$

$8:5 = 8(5):5(5) = 40:25$

Therefore, the ratio of 8th graders to 7th graders to 6th graders is $40:25:24$ . Since the ratio is in lowest terms, the smallest number of students participating in the project is $40+25+24 = \boxed{\textbf{(E)}\ 89}$ .

Solution 2: Fakesolving

The number of 8th graders has to be a multiple of 8 and 5, so assume it is 40 (the smallest possibility). Then there are $40*\frac{3}{5}=24$ 6th graders and $40*\frac{5}{8}=25$ 7th graders. The numbers of students is $40+24+25=\boxed{\textbf{(E)}\ 89}$

2013 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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 <math>\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 55 \qquad \textbf{(D)}\ 79 \qquad \textbf{(E)}\ 89</math>
+==Video Solution==
+https://youtu.be/rQUwNC0gqdg?t=949
 ==Solution==

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