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:
Answer Choices:
A. 25
B. 23
C. 34
D. 45
E. 43
Solution:
Let a,b,c be the number of hours needed, respectively, by Alpha, Beta, and Gamma to do the job when working alone. Then a1+b1+c1=a−61=b−11=c/21∴b=a−5 and c=2a−12 ∴a1+a−51+2(a−6)1=a−61∴a=320; the value a=3 is rejected.
∴b=35 and c=34∴a1+ b1=20/31+5/31=43∴ h=34