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

(→‎Solution 1)
(→‎Solution 1)
 
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So, the sum of the ratios is 12  
 
So, the sum of the ratios is 12  
  
−
So, the first number is (64/12)x2  = 32/3 (This is the smallest)
+
So, the first number is <cmath>=\frac{64*2}{12}</cmath>=<cmath>\frac{32}{3}</cmath>
 +
So, the second number is <cmath>=\frac{64*4}{12}</cmath>=<cmath>\frac{64}{3}</cmath>
  
−
So, the second number is (64/12)x4 = 64/3
+
So, the third number is<cmath>=\frac{64*6}{12}</cmath>=32
  
−
So, the third number is (64/12)x6  = 32
+
We can see that the smallest is<cmath>=\frac{32}{3}</cmath>
  
−
<cmath>x=\frac{64}{6}=\frac{32}{3}=10 \frac{2}{3} </cmath>
+
<cmath>x=\frac{32}{3}=10 \frac{2}{3} </cmath>
 
which is <math>\boxed{\textbf{(C)}}</math>.
 
which is <math>\boxed{\textbf{(C)}}</math>.
  

Latest revision as of 11:07, 21 September 2025

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 1

Given, The ratios are 2:4:6 The number is 64

So, the sum of the ratios is 12

So, the first number is \[=\frac{64*2}{12}\]=\[\frac{32}{3}\] So, the second number is \[=\frac{64*4}{12}\]=\[\frac{64}{3}\]

So, the third number is\[=\frac{64*6}{12}\]=32 

We can see that the smallest is\[=\frac{32}{3}\]

\[x=\frac{32}{3}=10 \frac{2}{3}\] which is $\boxed{\textbf{(C)}}$.


~Jayeed Mahmud (Bangladesh) 9/21/25

Solution 2

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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