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Difference between revisions of "1966 AHSME Problems/Problem 37"

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== Solution ==
 
== Solution ==
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<math>\fbox{C}</math>
 
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== Solution 2 ==
 
  
 
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.
  
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Thus <math>B=\frac{5}{3}</math> and <math>h=\frac{1}{\frac{1}{A}+\frac{1}{B}}=\frac{4}{3}</math>, <math>\boxed{C}</math>.
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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

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 $h$ be the number of hours needed by Alpha and Beta, working together, to do the job. Then $h$ equals:

$\text{(A) } \frac{5}{2} \quad \text{(B) } \frac{3}{2} \quad \text{(C) } \frac{4}{3} \quad \text{(D) } \frac{5}{4} \quad \text{(E) } \frac{3}{4}$

Solution

Let $A$,$B$,$C$ denote the number of hours needed by Alpha, Beta, Gamma, respectively. We also have their respective efficiency $\frac{1}{A}$, $\frac{1}{B}$, and $\frac{1}{C}$. Thus we get the equations \[\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{A-6}\] \[\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{B-1}\] \[\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{2}{C}\Longrightarrow \frac{1}{A}+\frac{1}{B}=\frac{1}{C}\] Equating the first $2$ equations gets \[\frac{1}{A-6}=\frac{1}{B-1}\Longrightarrow A=B+5\] Substituting the new relation along with the third equation into the first equation gets \[\frac{2}{A}+\frac{2}{A-5}=\frac{1}{A-6}\] Solving the quadratic gets $A=3$ or $\frac{20}{3}$. Since $B=A-5>0$, $A=\frac{20}{3}$ is the only legit solution.

Thus $B=\frac{5}{3}$ and $h=\frac{1}{\frac{1}{A}+\frac{1}{B}}=\frac{4}{3}$, $\fbox{\textbf{(C)}}$.

~ 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

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