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Difference between revisions of "2013 AMC 8 Problems/Problem 16"

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==Solution==
 
==Solution==
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The number of 8th graders has to be a multiple of 8 and 5, so assume it is 40. Then there are <math>40*\frac{3}{5}=24</math> 6th graders and <math>40*\frac{5}{8}=25</math> 7th graders. The numbers of students is <math>40+24+25=\boxed{\textbf{(E)}\ 89}</math>
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===Solution 1: Algebra===
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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:
 +
 
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<math>5:3 = 5(8):3(8) = 40:24</math>
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 +
<math>8:5 = 8(5):5(5) = 40:25</math>
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Therefore, the ratio of 8th graders to 7th graders to 6th graders is <math>40:25:24</math>. Since the ratio is in lowest terms, the smallest number of students participating in the project is <math>40+25+24 = \boxed{\textbf{(E)}\ 89}</math>.
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===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 <math>40*\frac{3}{5}=24</math> 6th graders and <math>40*\frac{5}{8}=25</math> 7th graders. The numbers of students is <math>40+24+25=\boxed{\textbf{(E)}\ 89}</math>
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2013|num-b=15|num-a=17}}
 
{{AMC8 box|year=2013|num-b=15|num-a=17}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:45, 29 November 2013

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?

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

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 $40+24+25=\boxed{\textbf{(E)}\ 89}$

See Also

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

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