1956 AHSME Problems/Problem 16
Problem 16
The sum of three numbers is 
. The ratio of the first to the second is 
,
and the ratio of the second to the third is 
. The second number is:
Solution
Let the 
numbers be 
and 
. We see that
![\[a+b+c = 98\]](http://latex.artofproblemsolving.com/a/1/9/a19e174a2a977a30668dae420893fae94ac3ca77.png)
and
![\[\frac{a}{b} = \frac{2}{3} \Rrightarrow 3a = 2b\]](http://latex.artofproblemsolving.com/8/6/f/86ff42e081fc9cb125c59ae0f7e7b97a611a7372.png)
![\[\frac{b}{c} = \frac{5}{8} \Rrightarrow 8b = 5c\]](http://latex.artofproblemsolving.com/3/1/3/31371e78ea1eabe0bcae23ba5361493f9eb76102.png)
Writing 
and in terms of 
we have
and 
.
Substituting in the sum, we have
![\[\frac{2}{3} b + b + \frac{8}{5} b = 98\]](http://latex.artofproblemsolving.com/2/e/a/2ea3357eda2600e331064430f70094432acdf7a0.png)
![\[\frac{49}{15} b = 98\]](http://latex.artofproblemsolving.com/f/4/1/f4108342884665c2b91b4ba224902ed450bec928.png)
![\[b = 98 \cdot \frac{15}{49} \Rrightarrow b = 30\]](http://latex.artofproblemsolving.com/d/6/c/d6c05997150a8b5c6aad1e28afa15274b743c759.png)
~JustinLee2017
See Also
| 1956 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
