Question

A can do 40% of a job in 12 hours. B is 16.67% more efficient than A. C is 28.57% less efficient than B. How long would the three of them take to do the job together?

• A. 10 hours
• B. 15 hours
• C. 9 hours
• D. 8 hours
• E. 12 hours

Answer 1

The correct option is A 10 hours

If A can do 40% of a job in 12 hours, then A can do 100% of the job in 12 × (100/40) = 30 hours.

Let the efficiency of A be 6.

Efficiency of B = 6 + 16.67% of 6 = 6 + 1 = 7

Efficiency of C = 7 – 28.57% of 7 = 7 – 2 = 5

Total work = Efficiency of A × Time taken by A = 6 × 30 = 180

Hence, the required time = Total work/Sum of efficiencies of A, B, and C = 180/(6 + 7 + 5) = 10 hours