Q. Working alone, A can complete a task in ‘a’ days and B in ‘b’ days. They take turns in doing the task with each working 2 days at a time. If A starts they finish the task in exactly 10 days. If B starts, they take half a day more. How long does it take to complete the task if they both work together?
Sol: The rate of work and all the fractions might look a little confusing, so I tried to break it down in simple terms.
Scenario 1 - A starts and proceeds to work for two days, followed by B who also works for 2 days. In this case, the work is completed in 10 days. Which means A has worked for 6 days and B has worked for 4 days (A,A,B,B,A,A,B,B,A,A).
Scenario 2 - B starts and proceeds to work for two days, followed by A who also works for 2 days. In this case, the work is completed in 10.5 days. Which means A has worked for 4.5 days and B has worked for 6 days.
Now, if we equate both scenarios, we get - 6A + 4B = 4.5A + 6B. Solve the equation and you get - A:B = 2:1.5 (or 4:3).
Since we have the ratio of their efficiencies, we can use the same get an estimate of the work done. For eq, since the ratio of their efficiencies is 4:3, we can go ahead with the assumption that A has done 4 unites of work per day, while B has done 3 units of work per day. Substitute these values in one of the scenarios, and you get the total work done.
6A + 4B (scenario one) can be written as 6*4 + 4*3 = 36. Now, that we have the total work, simply divide this by the work that can by A & B in a day if they work together, which is 4+3 = 7.
Now, to find the number of days required to complete the task if they both work together, we just have to divide the total work by the work that can by A & B in a day if they work together => 36/7
I tried to make it as elaborate as possible to make it easier to understand each step, so that it'll be easier for you to remember and replicate. Once you get a hang of it, this should not take more than a minute to solve.