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

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\textbf{(D)}\ 105\qquad
 
\textbf{(D)}\ 105\qquad
 
\textbf{(E)}\ 127</math>
 
\textbf{(E)}\ 127</math>
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 +
==Solution==
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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}}

Revision as of 22:03, 24 December 2012

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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All AJHSME/AMC 8 Problems and Solutions