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Difference between revisions of "1959 AHSME Problems/Problem 4"

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==Solution==
 
==Solution==
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Let the part proportional to <math>1</math> equal <math>x</math>. Then, the parts are <math>x, \frac{1}{3}x</math>, and <math>\frac{1}{6}x</math>. The sum of these parts should be <math>78</math>, so <math>\frac{9}{6}x=1\frac{1}{2}x=78</math>. Solving for <math>x</math>, <math>x=52</math>. The middle part is <math>\frac{1}{3}x</math>, so the answer is <math>17\frac{1}{3}</math>, or <math>\boxed{\textbf{E}}</math>
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Let the part proportional to <math>1</math> equal <math>x</math>. Then, the parts are <math>x, \frac{1}{3}x</math>, and <math>\frac{1}{6}x</math>. The sum of these parts should be <math>78</math>, so <math>\frac{9}{6}x=1\frac{1}{2}x=78</math>. Solving for <math>x</math>, <math>x=52</math>. The middle part is <math>\frac{1}{3}x</math>, so the answer is <math>17\frac{1}{3}</math>, or <math>\boxed{\textbf{C}}</math>
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==See also==
 
==See also==
 
{{AHSME 50p box|year=1959|num-b=3|num-a=5}}
 
{{AHSME 50p box|year=1959|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:31, 19 August 2021

Problem 4

If $78$ is divided into three parts which are proportional to $1, \frac13, \frac16,$ the middle part is: $\textbf{(A)}\ 9\frac13 \qquad\textbf{(B)}\ 13\qquad\textbf{(C)}\ 17\frac13 \qquad\textbf{(D)}\ 18\frac13\qquad\textbf{(E)}\ 26$


Solution

Let the part proportional to $1$ equal $x$. Then, the parts are $x, \frac{1}{3}x$, and $\frac{1}{6}x$. The sum of these parts should be $78$, so $\frac{9}{6}x=1\frac{1}{2}x=78$. Solving for $x$, $x=52$. The middle part is $\frac{1}{3}x$, so the answer is $17\frac{1}{3}$, or $\boxed{\textbf{C}}$

See also

1959 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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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