1992 AHSME Problems/Problem 16

Revision as of 00:31, 28 September 2014 by Timneh (talk | contribs) (Created page with "== Problem == If <cmath>\frac{y}{x-z}=\frac{x+y}{z}=\frac{x}{y}</cmath> for three positive numbers <math>x,y</math> and <math>z</math>, all different, then <math>\frac{x}{y}=</m...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

If \[\frac{y}{x-z}=\frac{x+y}{z}=\frac{x}{y}\] for three positive numbers $x,y$ and $z$, all different, then $\frac{x}{y}=$

$\text{(A) } \frac{1}{2}\quad \text{(B) } \frac{3}{5}\quad \text{(C) } \frac{2}{3}\quad \text{(D) } \frac{5}{3}\quad \text{(E) } 2$

Solution

$\fbox{E}$

See also

1992 AHSME (ProblemsAnswer KeyResources)
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. AMC logo.png