Difference between revisions of "1966 AHSME Problems/Problem 37"
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== Solution == | == Solution == | ||
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Let <math>A</math>,<math>B</math>,<math>C</math> denote the number of hours needed by Alpha, Beta, Gamma, respectively. We also have their respective efficiency <math>\frac{1}{A}</math>, <math>\frac{1}{B}</math>, and <math>\frac{1}{C}</math>. Thus we get the equations | Let <math>A</math>,<math>B</math>,<math>C</math> denote the number of hours needed by Alpha, Beta, Gamma, respectively. We also have their respective efficiency <math>\frac{1}{A}</math>, <math>\frac{1}{B}</math>, and <math>\frac{1}{C}</math>. Thus we get the equations | ||
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Solving the quadratic gets <math>A=3</math> or <math>\frac{20}{3}</math>. Since <math>B=A-5>0</math>, <math>A=\frac{20}{3}</math> is the only legit solution. | Solving the quadratic gets <math>A=3</math> or <math>\frac{20}{3}</math>. Since <math>B=A-5>0</math>, <math>A=\frac{20}{3}</math> is the only legit solution. | ||
| − | Thus <math>B=\frac{5}{3}</math> and <math>h=\frac{1}{\frac{1}{A}+\frac{1}{B}}=\frac{4}{3}</math>, <math>\ | + | Thus <math>B=\frac{5}{3}</math> and <math>h=\frac{1}{\frac{1}{A}+\frac{1}{B}}=\frac{4}{3}</math>, <math>\fbox{\textbf{(C)}}</math>. |
~ Nafer | ~ Nafer | ||
== See also == | == See also == | ||
| − | {{AHSME box|year=1966|num-b=36|num-a=38}} | + | {{AHSME 40p box|year=1966|num-b=36|num-a=38}} |
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 10:39, 29 July 2024
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
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)
![\[\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{2}{C}\Longrightarrow \frac{1}{A}+\frac{1}{B}=\frac{1}{C}\]](http://latex.artofproblemsolving.com/5/f/5/5f5070052534001d5d3275be6187807bde3f2227.png)
Equating the first 
equations gets
![\[\frac{1}{A-6}=\frac{1}{B-1}\Longrightarrow A=B+5\]](http://latex.artofproblemsolving.com/8/b/a/8ba769ed48f103ed1747abefcb0fb51a0ac26929.png)
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 
or 
. Since 
, 
is the only legit solution.
Thus 
and 
, 
.
~ Nafer
See also
| 1966 AHSC (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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
