Efficiencies are equal within a group
E.g. 5) 2 men and 7 boys can do a piece of work in 14 days. 3 men and 8 boys can do the same piece of work in 11 days. Then 8 men and 6 boys can do 3 Times the work in how many days?
Two steps are to be followed:
i) Find out the relation between the work done by a man and that by a boy. For this equate the work done in 1 day by both groups
ii) Once you have the relation from step (i) convert the group into groups of only men or only boys. This way you will be in a position to compare two groups easily.
2 men and 7 boys are needed to do a piece of work in 14 days. This means to complete the work we need 2 men to work for 14 days and 7 boys to work for 14 days. So if we want to complete the work in one day we need: 2(14) = 28 men and 7(14) = 98 boys => 28M + 98B.
Also, 3 men and 8 boys take 11 days, so if they need to complete the work in 1 day, we need 3(11) =33 men and 8 (11) =88 boys => 33M + 88B
We can now equate the two cases as 28M + 98B = 33M + 88B
Solving we get 5M=10B or 1M=2B i.e. one man is equivalent to two boys.
Converting both the groups as group of only men:
3 men & 8boys = 3men + 4 men = 7 men can finish one work in 11 days
The question is for a group of 8 men and 6 boys which equate to 8 + 6/2 Men = 11 Men.
If 7 men take 11 days, then 11 men will take 7 days to finish one of the work.
To finish 3 Times the work, they will take 7×3=21 days.