Difference between revisions of "1950 AHSME Problems/Problem 2"

(Created page with "==Problem== Let <math> R=gS-4 </math>. When <math>S=8</math>, <math>R=16</math>. When <math>S=10</math>, <math>R</math> is equal to: <math> \textbf{(A)}\ 11\qquad\textbf{(B)}\ ...")
 
 
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==Solution==
 
==Solution==
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Our first procedure is to find the value of <math>g</math>. With the given variables' values, we can see that <math>8g-4=16</math> so <math>g=\frac{20}{8}=\frac{5}{2}</math>.
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With that, we can replace <math>g</math> with <math>\frac{5}{2}</math>. When <math>S=10</math>, we can see that <math>10\times\frac{5}{2}-4=\frac{50}{2}-4=25-4=\boxed{\text{(D) 21}}</math>.
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==See Also==
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{{AHSME 50p box|year=1950|num-b=1|num-a=3}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 11:57, 5 July 2013

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
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All AHSME Problems and Solutions

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