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

(→‎Solution 2)
(→‎Solution 2)
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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>
−
Solving the quadratic gets <math>A=3,\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}{2}</math> and <math>h=\frac{1}{\frac{1}{A}+\frac{1}{B}}=\frac{4}{3}</math>.
 
Thus <math>B=\frac{5}{2}</math> and <math>h=\frac{1}{\frac{1}{A}+\frac{1}{B}}=\frac{4}{3}</math>.

Revision as of 21:44, 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 $A=3$ or $\frac{20}{3}$. Since $B=A-5>0$, $A=\frac{20}{3}$ is the only legit solution.

Thus $B=\frac{5}{2}$ 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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