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​