Difference between revisions of "1950 AHSME Problems/Problem 1"
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<math> \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} </math> | <math> \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} </math> | ||
| − | ==Solution== | + | ==Solution 1== |
| + | Given, | ||
| + | The ratios are 2:4:6 | ||
| + | The number is 64 | ||
| − | 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: | + | So, the sum of the ratios is 12 |
| + | |||
| + | 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 third number is<cmath>=\frac{64*6}{12}</cmath>=32 | ||
| + | |||
| + | We can see that the smallest is<cmath>=\frac{32}{3}</cmath> | ||
| + | |||
| + | <cmath>x=\frac{32}{3}=10 \frac{2}{3} </cmath> | ||
| + | which is <math>\boxed{\textbf{(C)}}</math>. | ||
| + | |||
| + | |||
| + | ~Jayeed Mahmud (Bangladesh) 9/21/25 | ||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | 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 | + | which is <math>\boxed{\textbf{(C)}}</math>. |
==See Also== | ==See Also== | ||
Latest revision as of 11:07, 21 September 2025
Contents
Problem
If 
is divided into three parts proportional to 
, 
, and 
, the smallest part is:
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 ![\[=\frac{64*2}{12}\]](http://latex.artofproblemsolving.com/5/3/8/538bb94fb47723583fc91fe1906ebb021da38371.png)
=![\[\frac{32}{3}\]](http://latex.artofproblemsolving.com/a/4/8/a481b33fbe61bc5e66197e62402a0eca04f16b86.png)
So, the second number is ![\[=\frac{64*4}{12}\]](http://latex.artofproblemsolving.com/1/b/b/1bb891f77dca68db86c7cb8cf6298719b2f91052.png)
=![\[\frac{64}{3}\]](http://latex.artofproblemsolving.com/7/7/5/775096796590a1929220999515dbbef617445098.png)
So, the third number is![\[=\frac{64*6}{12}\]](http://latex.artofproblemsolving.com/6/e/7/6e73258f2cc0acdfc781ce9b6cbac5d940c72b77.png)
=32
We can see that the smallest is![\[=\frac{32}{3}\]](//latex.artofproblemsolving.com/f/b/d/fbde17bcec0b15495ac8fa94c03977ce3379120d.png)
![\[=\frac{32}{3}\]](http://latex.artofproblemsolving.com/f/b/d/fbde17bcec0b15495ac8fa94c03977ce3379120d.png)
![\[x=\frac{32}{3}=10 \frac{2}{3}\]](http://latex.artofproblemsolving.com/1/0/4/1043c6c601ac68380dd19040c4423e0c6dfc3d8e.png)
which is 
.
~Jayeed Mahmud (Bangladesh) 9/21/25
Solution 2
If the three numbers are in proportion to 
, then they should also be in proportion to 
. This implies that the three numbers can be expressed as 
, 
, and 
. Add these values together to get:
![\[x+2x+3x=6x=64\]](http://latex.artofproblemsolving.com/e/c/7/ec7b60c73cb5fd6028a9e6002fbd703e0de99e5d.png)
Divide each side by 6 and get that
![\[x=\frac{64}{6}=\frac{32}{3}=10 \frac{2}{3}\]](http://latex.artofproblemsolving.com/d/9/a/d9a7ed04e834a6040f551968e811172b1bf633f0.png)
which is .
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
