2008 AMC 8 Problems/Problem 7

Revision as of 10:21, 27 April 2024 by Naturalselection (talk | contribs) (See Also)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png