Aptitude - Time and Work - Discussion

Together 1/30*20=2/3.
2parts work completed.
Remaining 1 part 'A'completed in 20days.
1part work 20days.
3 prts --------?.
3 * 20 = 60 days.

Hello,
A+b = 60/30 =>2
B=60/20 =>3

60 is total work.
A+b=2,
A+3=2,
A=1.

So, A divided total work.
A=60/1.
= 60ans.

(A + B) together they can complete work in 30 days.
So on each day they will do (1/30) of the total work.
So it is given that A and B together did for the first 20 days,
For 20 days they can do 20 times of (1/30) i.e. 20/30,

Remaining work is done by A which is 10/30 or 1/3 of total work for 20 days.
(1/3) * (total work) =20 days.
Multiplying 3 on both sides.

You will get,
Total work is done in 20 * 3 = 60 days.

Why the reaming work is -1? Explain it please.

Simplest and straightforward solution:

Let us assume x is the efficiency of A ( units done per day )
y is the efficiency of B ( units done per day )

According to the question;
The total work to be done is the same in any case, whether they do it together or separately.
Work = efficiency * time.
ATQ;
(x + y) * 30 = (x+y)*20 + 20x.

Solve and get x = y.

Which means the efficiency of both A and B are same.
( also could have made a wild guess to assume 1 unit per day to be the efficiency of both it satisfies the constraints in the given question)

Now, since the efficiency of both is 1 unit per day total work to be done = 60 units.
The efficiency of A is 1 unit per day, so it will take 60 days for A to complete the work.

Thank you for the best and easy explanation @Kaaly and @Senchu Thomas.

Nice explanation @Harish.

A+B =30 DAYS.
B LEFT 20 DAYS.
A FINSH WORK IN 20 DAYS.
1-1/20 =19/20.
1/30-19/20 =1/60.
60 DAYS.

What is the total time taken by B? Please answer it.

To make the last step easy we can understand it as follows:

1/3work can be done in 20 days.
So, 1 wor will be don win 20 * 3 60 days (try using direct proportion)