Difference between revisions of "1993 AHSME Problems/Problem 9"

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== Solution ==
 
== Solution ==
<math>\fbox{D}</math>
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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 $A$ has $c\%$ of the world's population and $d\%$ of the worlds wealth. Country $B$ has $e\%$ of the world's population and $f\%$ of its wealth. Assume that the citizens of $A$ share the wealth of $A$ equally,and assume that those of $B$ share the wealth of $B$ equally. Find the ratio of the wealth of a citizen of $A$ to the wealth of a citizen of $B$.

$\text{(A) } \frac{cd}{ef}\quad \text{(B) } \frac{ce}{ef}\quad \text{(C) } \frac{cf}{de}\quad \text{(D) } \frac{de}{cf}\quad \text{(E) } \frac{df}{ce}$

Solution

Let $W$ be the wealth of the world and $P$ be the population of the world. Hence the wealth of each citizen of $A$ is $w_A = \frac{0.01d W}{0.01cP}=\frac{dW}{cP}$. Similarly the wealth of each citizen of $B$ is $w_B =\frac{eW}{fP}$. We divide $\frac{w_A}{w_B} = \frac{de}{cf}$ and see the answer is $\fbox{D}$

See also

1993 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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