Aptitude - Time and Work - Discussion

Good explanation. Thank you all.

@Ankush

Here is answer for your question.

Take LCM of 36, 42 & 48 i.e. 1008
Now Rehman and Sandeep star work and worked for 6 days.
Now per day work of Rehman = 1008/36 = 28
And per day work of Sandeep = 1008/42=24
Together they have completed per day work = 28+24=52.
Now in 6 days, they completed 52*6 = 312 work out of 1008.
Now last 10 days again Rehman and Sandeep worked so the work completed in 10 days = 52*10=520.
So total work completed by Rehman and Sandeep = 312+520=832.
Remaining work = 1008-832= 176 which is completed by Abbas alone and Abbas per day work is 1008/48= 21.
So Abbas worked for 176/21 Days. Or 8 8/21 days.

[A's 1day work = 1/18.

In how many days(x) =1/3 of total work.]
Cross multiplication:
1*1/3=1/18*x
X=6 days.

I have a question that while B is working for 10 days is A remains idle? because nothing is mentioned about that in the question.

A=> 18days.
B=> 15days.
B left after 10 days of work then the remaining work of B have to be done by A,
L.C.M. of 18 & 15 =>90.
The efficiency of A is 5 and of B is 6,
work done by B is 6*10=60,
then 90-60=30 (here 90 is L.C.M.)
then work done by A is 30/5 = 6days.

1/18 WORK DONE IN 1 DAY.
1/3 WORK DONE IN X DAY.
X*1/18 = 1/3,
X = 18/3 = 6.

A can do it in 18 days.
B can do it in 15 days.

Take LCM of those 18 and 15 I.e., 90.

Consider total work as 90,
ThenA's efficiency = total work/time taken by A.
=>90/18=5.
Similarly for B=>90/15=6.
So B's 10days work is 10*6=60.
Remaining work is 90-60=30.

Time taken by A to complete a work is =remaining work/efficiency of A.
=>30/5 = 6days.

How to calculate the remaining work that is 1/3?

Please, anyone, clear it.

TAKE LCM of 18 & 90 and divide by the days
5-A-18.
6-B-15.

First B do work for 10 days, means 10*6 = 60 already done left work(30) is done by A alone so divide 30/5=6.

And 6 is the answer

Can't get it. Please explain me clearly.