A can complete a piece of work in 15 days. B is 25% more efficient than A. In how many days can both of
them complete the work together, if they work on alternate days starting with A?
Answers
A can do a piece of work in 18 days, B is 50% more efficient than A. The number of days B will take to do the same piece of work is?
If A is 50% more efficient than B, then B can do 1.5 times the amount of work that A can do in the same amount of time as A. Since the amount of time is 18 days, we have the following rate: 1.5 pieces of work / 18 days. But, we want to know how many days it will take B to do 1 piece of work. So, we take the reciprocal of 1.5 pieces of work/ 18 days and simplify to get 12 days/1 piece of work. In other words, B takes 12 days to do one piece of work. If you try different percentages and days, you may infer the following:
If A can do a piece of work in N days, and B is P% more efficient than A, then B can do the same piece of work in N*100/(100+P) days [where P > 0].
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