Difference between revisions of "1950 AHSME Problems/Problem 1"
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So, the third number is (64/12)x6 = 32 | So, the third number is (64/12)x6 = 32 | ||
| − | 32/3 = 10 \frac{2}{3} | + | 32/3 = 10 \frac{2}{3} |
which is <math>\boxed{\textbf{(C)}}</math>. | which is <math>\boxed{\textbf{(C)}}</math>. | ||
~Jayeed Mahmud (Bangladesh) 9/21/25 | ~Jayeed Mahmud (Bangladesh) 9/21/25 | ||
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| − | |||
==Solution 2== | ==Solution 2== | ||
Revision as of 10:59, 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 (64/12)x2 = 32/3 (This is the smallest)
So, the second number is (64/12)x4 = 64/3
So, the third number is (64/12)x6 = 32
32/3 = 10 \frac{2}{3}
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. 
