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

(Created page with "==Problem== If <math>64</math> is divided into three parts proportional to <math>2</math>, <math>4</math>, and <math>6</math>, the smallest part is: <math> \textbf{(A)}\ 5\frac...")
 
m (→‎Solution)
 
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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:  
+
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>
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which is answer choice <math>\boxed{C}</math>.
+
which is <math>\boxed{\textbf{(C)}}</math>.
 +
 
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==See Also==
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{{AHSME 50p box|year=1950|before=First Question|num-a=2}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 21:20, 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
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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