We will use the unitary method;
(plz.. understand that their task is considered as 1 work)
Let the time taken by by one women to complete the work be ‘x’
and man be ‘y’
One women in x days=1 work (x/x=1)
One women in 1 day= 1/x work ( only a part of work done)
Now, 2 women in one day= 1/x * 2 = 2/x - 1
Similarly,one man in y days= 1 work
One man in one day=1/y work
5 man in one day=5/y work -2
Acc to case 1: 2 women and 5 men complete the work in 4 days.
[4 day=1 work ; 1 day= ¼ work]
So, 2/x + 5/y = ¼ work (from 1 and 2)
4(2/x + 5/y) = 1
8/x + 20/y = 1 -
3
Acc to case 2: 3 women and 6 men finishes it in 3 days.
3/x + 6/y = 1/3 work
3(3/x + 6/y) = 1
9/x + 18/y = 1 -
4
NOW, LET 1/x BE a AND 1/y BE b
Rewriting;
8a + 20b = 1
(from3)
9a + 18b = 1 (from 4)
Now solve this using any method……(I prefer elimination)
a = 1/18
b= 1/36
putting them in 1/x and 1/y
you ‘ll get ;
