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

Latest revision as of 10:57, 5 July 2013

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

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

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 $10\times\frac{5}{2}-4=\frac{50}{2}-4=25-4=\boxed{\text{(D) 21}}$ .

1950 AHSC (Problems • Answer Key • Resources)
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

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 ==Solution==
+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>.
+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>.
+==See Also==
+{{AHSME 50p box|year=1950|num-b=1|num-a=3}}
+[[Category:Introductory Algebra Problems]]
+{{MAA Notice}}