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

(→‎Solution 2)
(→‎Solution 2)
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<cmath>\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{A-6}</cmath>
 
<cmath>\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{A-6}</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{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}\Longrightarrow \frac{1}{A}+\frac{1}{B}=\frac{1}{C}</cmath>
 
Equating the first <math>2</math> equations gets  
 
Equating the first <math>2</math> equations gets  
−
<cmath>\frac{1}{A-6}=\frac{1}{B-1}\Longarrowright A=B-5</cmath>
+
<cmath>\frac{1}{A-6}=\frac{1}{B-1}\Longrightarrow A=B-5</cmath>
 
Substituting the new relation along with the third equation into the first equation gets
 
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>
 
<cmath>\frac{2}{A}+\frac{2}{A-5}=\frac{1}{A-6}</cmath>

Revision as of 21:42, 23 December 2019

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

$\fbox{C}$

Solution 2

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,\frac{20}{3}$$. Since$B=A-5>0$,$A=\frac{20}{3}$is the only legit solution.

Thus$ (Error compiling LaTeX. Unknown error_msg)B=\frac{5}{2}$and$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

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