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Difference between revisions of "2008 AMC 8 Problems/Problem 7"

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==Solution==
 
==Solution==
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Separate into two equations 3 over 5 = m over 45 and 3 over 5 = 60 over N and solve for the unknowns. M=27 and N=100 so M+N=E 127 :D
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Separate into two equations <math>\frac35 = \frac{M}{45}</math> and <math>\frac35 = \frac{60}{N}</math> and solve for the unknowns. <math>M=27</math> and <math>N=100</math>, therefore <math>M+N=\boxed{\textbf{(E)}\ 127}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2008|num-b=6|num-a=8}}
 
{{AMC8 box|year=2008|num-b=6|num-a=8}}
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[[Category:Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 10:21, 27 April 2024

Problem

If $\frac{3}{5}=\frac{M}{45}=\frac{60}{N}$, what is $M+N$?

$\textbf{(A)}\ 27\qquad \textbf{(B)}\ 29 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 105\qquad \textbf{(E)}\ 127$

Solution

Separate into two equations $\frac35 = \frac{M}{45}$ and $\frac35 = \frac{60}{N}$ and solve for the unknowns. $M=27$ and $N=100$, therefore $M+N=\boxed{\textbf{(E)}\ 127}$.

See Also

2008 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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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