Difference between revisions of "1966 AHSME Problems/Problem 37"
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== Solution 2 == | == Solution 2 == | ||
− | Let A,B, | + | 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 |
+ | <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{2}{C}</cmath> | ||
== See also == | == See also == |
Revision as of 21:33, 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
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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