Aptitude - Time and Work - Discussion

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.

Nice explanation, Thanks @Vikram Sihag.

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.

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.

Let's take 5 parts total.
A finishes 80% of it in 20 days, A can finish whole work in 25 days.
A one day work is 1/25.
A= 4/5 work in 20 days.
Remaining, 1-(4/5)= 1/5 part.
(A+B)= 1/5 work in 3 days.
A's 3 days work = 3/25.
Therefore,
For 3 days, A+B= 1/5.
(3/25)+B= 1/5.
B= 2/25 in 3 days.
B's one day work, 2/75 work.
Whole work by B= 75/2.
B = 37 1/2 days.

Thanks @Ganesh.

Thank you so much @Baskar.

A do 80%work for 20days.

(Simplify it remaining 20%do in 5days).

Now work for 20%is 1/5 and work are done by both a+b is 1/3. Do lcm we get 15 and subtract 5-3/15=2/15.
Now it is 20% of work done by B is 2/15.
Now 20%*5=100% i.e,2/15*5=2/75 ans is 37 1/2 days.

Thank you @Rajendra.

Thank you @Baskar.