Aptitude - Time and Work - Discussion

A can do 80% of work in 20 days.
A + B 20% of work in 3 days.
A alone can do 20% of work in 5 days and A+B can do 80% in 12 days.

Now, A 100% in 25 days and A + B in 15 days.
L.C.M of 15 and 25 = 75
The efficiency of A is 3/day and of A + B 5/day.
If A efficiency is 3, then B is 2.
B days = work/B efficiency.
=> 75/2 = 37.5.

A's work rate = 80% in 20.days =80%/20=4
Remaining work of A and B = 100%-80%= 20% in 3 days = 20%/3.
B can do work = 20/3 - 4 = 20-12/3 = 8/3%.
B 100% work = 100/8/3 = 37.5 days.

Given
A completes 80% work in 20 days (20% work in 5 days).

A alone can complete 100% work in 25days(20/80*100=25)
A+B together complete 20% of the work in 3 days
A+B together complete 100% of the work in 15 days.
Here we need to consider only 100% work done by A+B&A.
(A+B)-A = [1/15 - 1/25].
LCM of 15, 25 is 75.
(A+B)-A = [5-3]/75,
(A+B)-A = 2/75,
(A+B)-A = 37.5.

A in 80% work done 20 days.
A in 100% work done 100*20/80= 25 days.

A+B in 20% work done 3 days.
A+B in 100% work done 100*3/20= 15 days.
A = 25 days.
A+B = 15 days.

LCM OF 75.
1 DAY OF WORK 75/25 = 3
1 DAY A+B WORK 75/15 = 5.
A + B = 5
3 + B = 5
B = 2
B total working days =>75/2 = 37.5 days.

A -> 80% work in 20 days.
A + B ---> 20% work in 3 days.

A --> 100% work in 25 days.
A+B ->100%work in 15 days
LCM of 25 and 15 is 75.
Capacity of A is -->3 units
Capacity of A+B ---> 5 units.

So, the total work was 75(LCM).
B capacity = A+B (Capacity)-A(capacity).
B(Capacity) = 5-3.
B(Capacity) = 2.
B(work) = totalWork/B(Capacity).
B(Work) = 75/2.
B(Work) = 37.5.

A completes 80% of the work in 20 days.
This means A does 4% of the work each day (80% ÷ 20 days).
In 3 days, A would complete 12% of the work (4% × 3 days).
The remaining work, 8%, is done by B in 3 days.

To find out how long B would take to complete the entire work (100%), we calculate it as follows:

If B does 8% of the work in 3 days, then B's work rate is 8% ÷ 3 days = 2.67% per day.
Therefore, to complete 100% of the work, B would require:
100% ÷ 2.67% per day = 37.5 days.
So, B would take 37.5 days to complete 100% of the work.

(no of days required A to complete 100% work + no of days required B to complete 100% work) * no of days
= no of days required complete whole work by A+B -----> 1
A per-day work = 20/80 = 1/ 4 meaning 4 units/day.

Hence, no of days required A to complete 100% work = 100/4 = 25 days.
No of days are required to complete the whole work by A+B = 100/20 = 5.
The sub in 1.
(1/25 +1/B) *3 = 1/5.
3/B = 2/25.
B = 25*3/2
B = 37.5

Good, Thanks @Keerthan G.

80% of the work in 20 days means 4/5.
100% he will complete in 25 days.
Now, the remaining 20% of the work in 3days means;
1-4/5 = 1/5 this is for one day
For 3days 3*1/5 = 1/15,
Now [1/15 - 1/25] = 2/75.
Rev= 75/2=37.5

This can also be solved in this way.
A - 80% work in 20 days .
=>1/A=1/20 (1 day work of A) ---> 1

Also Given remaining 20% work done in 3 days by help of B.
Let us calculate days if they work for other 80%.

20% = 3 days.
80% = x.
x = 12 days.
=> 1/A + 1/B = 1/12.

Substitute value of 1/A from eq 1
then B = 30

=> 80% = 30 days,
100% = x.
By calculating x = 37.5.
So, B alone can finish work in 37.5 days.