Difference between revisions of "1993 AHSME Problems/Problem 9"
(Created page with "== Problem == Country <math>A</math> has <math>c\%</math> of the world's population and <math>d\%</math> of the worlds wealth. Country <math>B</math> has <math>e\%</math> of the...") |
Logsobolev (talk | contribs) (→Solution) |
||
| (One intermediate revision by one other user not shown) | |||
| Line 10: | Line 10: | ||
== Solution == | == Solution == | ||
| − | <math>\fbox{D}</math> | + | |
| + | Let <math>W</math> be the wealth of the world and <math>P</math> be the population of the world. Hence the wealth of each citizen of <math>A</math> is <math>w_A = \frac{0.01d W}{0.01cP}=\frac{dW}{cP}</math>. Similarly the wealth of each citizen of <math>B</math> is <math>w_B =\frac{eW}{fP}</math>. We divide <math>\frac{w_A}{w_B} = \frac{de}{cf}</math> and see the answer is <math>\fbox{D}</math> | ||
== See also == | == See also == | ||
Latest revision as of 21:59, 27 May 2021
Problem
Country 
has 
of the world's population and 
of the worlds wealth. Country 
has 
of the world's population and 
of its wealth. Assume that the citizens of share the wealth of equally,and assume that those of share the wealth of equally. Find the ratio of the wealth of a citizen of to the wealth of a citizen of .
Solution
Let 
be the wealth of the world and 
be the population of the world. Hence the wealth of each citizen of is 
. Similarly the wealth of each citizen of is 
. We divide 
and see the answer is 
See also
| 1993 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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. 
