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Difference between revisions of "1950 AHSME Problems/Problem 1"

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==Solution==
 
==Solution==
  
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If the three number are in proportion to <math>2:4:6</math>, then they should also be in proportion to <math>1:2:3</math>. This implies that the three numbers can be expressed as <math>x</math>, <math>2x</math>, and <math>3x</math>. Add these values together to get:  
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If the three numbers are in proportion to <math>2:4:6</math>, then they should also be in proportion to <math>1:2:3</math>. This implies that the three numbers can be expressed as <math>x</math>, <math>2x</math>, and <math>3x</math>. Add these values together to get:  
 
<cmath>x+2x+3x=6x=64</cmath>
 
<cmath>x+2x+3x=6x=64</cmath>
 
Divide each side by 6 and get that  
 
Divide each side by 6 and get that  
 
<cmath>x=\frac{64}{6}=\frac{32}{3}=10 \frac{2}{3}</cmath>
 
<cmath>x=\frac{64}{6}=\frac{32}{3}=10 \frac{2}{3}</cmath>
−
which is answer choice <math>\boxed{C}</math>.
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which is <math>\boxed{\textbf{(C)}}</math>.
  
 
==See Also==
 
==See Also==
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{{AHSME 50p box|year=1950|before=First Question|num-a=3}}
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{{AHSME 50p box|year=1950|before=First Question|num-a=2}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 20:20, 10 January 2023

Problem

If $64$ is divided into three parts proportional to $2$, $4$, and $6$, the smallest part is:

$\textbf{(A)}\ 5\frac{1}{3}\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 10\frac{2}{3}\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ \text{None of these answers}$

Solution

If the three numbers are in proportion to $2:4:6$, then they should also be in proportion to $1:2:3$. This implies that the three numbers can be expressed as $x$, $2x$, and $3x$. Add these values together to get: \[x+2x+3x=6x=64\] Divide each side by 6 and get that \[x=\frac{64}{6}=\frac{32}{3}=10 \frac{2}{3}\] which is $\boxed{\textbf{(C)}}$.

See Also

1950 AHSC (Problems • Answer Key • Resources)
Preceded by
First Question 		Followed by
Problem 2
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All AHSME Problems and Solutions

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