Difference between revisions of "1966 AHSME Problems/Problem 37"
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<cmath>\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{B-1}</cmath> | <cmath>\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{B-1}</cmath> | ||
<cmath>\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{2}{C}\Longarrowright \frac{1}{A}+\frac{1}{B}=\frac{1}{C}</cmath> | <cmath>\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{2}{C}\Longarrowright \frac{1}{A}+\frac{1}{B}=\frac{1}{C}</cmath> | ||
| − | Substituting the third equation into the first equation | + | Equating the first <math>2</math> equations gets |
| − | <cmath>\frac{2}{A}+\frac{2}{ | + | <cmath>\frac{1}{A-6}=\frac{1}{B-1}\Longarrowright A=B-5</cmath> |
| + | Substituting the new relation along with the third equation into the first equation gets | ||
| + | <cmath>\frac{2}{A}+\frac{2}{A-5}=\frac{1}{A-6}</cmath> | ||
| + | Solving the quadratic gets <math>A=3,\frac{20}{3}</math><math>. Since </math>B=A-5>0<math>, </math>A=\frac{20}{3}<math> is the only legit solution. | ||
| + | |||
| + | Thus </math>B=\frac{5}{2}<math> and </math>h=\frac{1}{\frac{1}{A}+\frac{1}{B}}=\frac{4}{3}$. | ||
== See also == | == See also == | ||
Revision as of 21:40, 23 December 2019
Contents
Problem
Three men, Alpha, Beta, and Gamma, working together, do a job in 6 hours less time than Alpha alone, in 1 hour less time than Beta alone, and in one-half the time needed by Gamma when working alone. Let 
be the number of hours needed by Alpha and Beta, working together, to do the job. Then equals:
Solution
Solution 2
Let 
,
,
denote the number of hours needed by Alpha, Beta, Gamma, respectively. We also have their respective efficiency 
, 
, and 
. Thus we get the equations
![\[\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{A-6}\]](http://latex.artofproblemsolving.com/e/2/6/e26c40b89eeabe776619cbb25ba5c46c570e35b7.png)
![\[\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{B-1}\]](http://latex.artofproblemsolving.com/f/3/c/f3c35188ae70a4151610aa17bef358284e15aa27.png)
\[\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{2}{C}\Longarrowright \frac{1}{A}+\frac{1}{B}=\frac{1}{C}\] (Error compiling LaTeX. Unknown error_msg)
Equating the first 
equations gets
\[\frac{1}{A-6}=\frac{1}{B-1}\Longarrowright A=B-5\] (Error compiling LaTeX. Unknown error_msg)
Substituting the new relation along with the third equation into the first equation gets
![\[\frac{2}{A}+\frac{2}{A-5}=\frac{1}{A-6}\]](http://latex.artofproblemsolving.com/5/5/d/55d8542d0e43ec1c29da6109de68f6bd013ed73f.png)
Solving the quadratic gets 
B=A-5>0
A=\frac{20}{3}$is the only legit solution.
Thus$ (Error compiling LaTeX. Unknown error_msg)B=\frac{5}{2}
h=\frac{1}{\frac{1}{A}+\frac{1}{B}}=\frac{4}{3}$.
See also
| 1966 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 36 |
Followed by Problem 38 | |
| 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. 
