Aptitude - Time and Work - Discussion

Solution by LCM Method:

A does 80% of the work in 20 days => 4/5th of the work in 20 days
=> (4/5)x = 20 => x=25 days (No of days A takes to complete the work)

Remaining work = 1/5 which is completed by A+B in 3 days.
(1/5)x = 3 => x=15 days (No of days A+B takes to complete the work).

Therefore,
A - 25 days - 3 units.
A+B - 15 days - 5 units.
LCM - 75.
A+B = 5 => B=2 units.
Therefore, total work done by B = 75/2.
= 37 1/2.

80% work done by A -------->20 days (given)
1 work ie. 100% work done by A ------->20*100/80 = 25 days
therefore, total work is done by A = 25 days

Now,
20% work done by (A+B) implies,

20/100 work done --------> 3 days.
1 work ie. 100% work done in ------>3*100/20.= 15 days.

Now we have,
A's 1 days work = 1/25 ----------------------> 1
and, (A+B)'s 1 days work = 1/15 -------------------------> 2

Putting A's work in equation 2.
B's one day work = 2/75.

Therefrom total work done by B = 75/2.
= 37 1/2.

Nice explanation, Thanks @Vikram Sihag.

80% = 4/5.

A does 4/5 work in 20 days.

A 1 day work =(4/5)/20 = 1/25,
remaining work = 1/5.

A + B does 1/5 work in 3 days.
A + B 1 day work=(1/5)/3 = 1/15.

Therefore;
B's 1 day work = (a+b)-a = 1/15-1/25 = 2/75.

Thanks @Harish.

1day work of A : (1/20)*80/100 = 1/25,
A+B = 3/25 + 3/x=1/5(for 3days).
Find x = 37.5.

80% is completed so remaining is 20% work.
This 20% work is completed by A&B together in 3 days,
If 20% is 3 days then 100% is 15 days,
A completed 80% in 20days then A can complete 100% in 25 days,
So B's work =1/15 -1/25 =2/75,
2 work in 75 days,
1 work in 75/2 days,
= 37 1/2 days.

@ Rajendra.

Thanks for the clear explanation.

Thank you @Mayuri.

80% work --> 20 Days, Then
100% work --> ?? Days.

i.e 20 * 100/80 = 25.
Similarly (remaining work) = 20% work done in 3 days that means 100% work done in 15 Days.

Now
1/25+B = 1/15,
1/15-1/25 = B,
2/75,
37.5.