1950 AHSME Problems/Problem 2

Revision as of 10:57, 5 July 2013 by Nathan wailes (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

Let $R=gS-4$. When $S=8$, $R=16$. When $S=10$, $R$ is equal to:

$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 21\qquad\textbf{(E)}\ \text{None of these}$

Solution

Our first procedure is to find the value of $g$. With the given variables' values, we can see that $8g-4=16$ so $g=\frac{20}{8}=\frac{5}{2}$.

With that, we can replace $g$ with $\frac{5}{2}$. When $S=10$, we can see that $10\times\frac{5}{2}-4=\frac{50}{2}-4=25-4=\boxed{\text{(D) 21}}$.

See Also

1950 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions

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