Difference between revisions of "1973 AHSME Problems/Problem 12"

(Solution to Problem 12)
 
(See Also)
 
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==See Also==
 
==See Also==
{{AHSME 35p box|year=1973|num-b=11|num-a=13}}
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{{AHSME 30p box|year=1973|num-b=11|num-a=13}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Latest revision as of 12:58, 20 February 2020

Problem

The average (arithmetic mean) age of a group consisting of doctors and lawyers in 40. If the doctors average 35 and the lawyers 50 years old, then the ratio of the numbers of doctors to the number of lawyers is

$\textbf{(A)}\ 3: 2\qquad\textbf{(B)}\ 3: 1\qquad\textbf{(C)}\ 2: 3\qquad\textbf{(D)}\ 2: 1\qquad\textbf{(E)}\ 1: 2$

Solution

Let $d$ be the number of doctors and $l$ be the number of lawyers. The average age of the doctors is $35$, so the sum of the doctors' ages is $35d$. The average age of the lawyers is $50$, so the sum of the lawyers' ages is $50l$. With the two information, we can write an equation.

\[\frac{35d+50l}{d+l} = 40\] \[35d+50l=40d+40l\] \[10l=5d\] \[\frac{d}{l} = 2\] The ratio of the number of doctors to the number of lawyers is $\boxed{\textbf{(D)}\ 2: 1}$.

See Also

1973 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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All AHSME Problems and Solutions