Aptitude - Time and Work - Discussion

First, we took lcm of all 3 that is A, B and C that is 60. On the basis of this A's efficiency is 3, B's efficiency is 2 and C's 1. So first 2 days A will do the work that is 3x2=6. Third-day work will be 6 as all three are doing the work. So 3 day's work will be 12. On the basis of the unitary method in every 3 day, they are doing 12 work so 3 day's efficiency will be 4unit. So, total work that is 60 divide by 4= 15.

That is 1/15 work is done in 1 Day.
Now 4/5 part of work that is (1-1/5) will be done in :15*4/5 =12 days.

One simple way : 1/20 + (1/90 + 1/180) = 1/15.

In 3 days ,the total work = 3/20 + 1/30 + 1/60 = 1/5.
1/5 % done in 3 days.
1% done in 3/(1/5) = 3 * 5 = 15 days.

a b c
20 30 60 (LCM=120)
6 4 2 (a20/120=6, b30/720=4, c60/120=2).

6 + 4 + 2 = 12,
12/120 = 5
5 * 3 = 15.

A=20 A Effi=3.
B=30 LCM=60 B Effi=2.

So,Work=60 => 3+3+6=12(3 days work)=> 12*5=60=>15days.

C=60 C Effi=1.

A complete in 20days.
B complete in 30days.
C complete in 60days.

Lcm(20,30,60)= 60.
So total work = 60.

A's one day work = 60/20 = 3.
B's one day work = 60/30 = 2.
C's one day work = 60/60 = 1.

A+B+C 's one day work = 3+2+1=6.

First day A's complete = 3unit of work.
Sec day A complete = 3unit of work.
3rd day A+B+C = 6unit of work.
.
.
So on

First 3day total work completed= 3+3+6=12 unit of work.

So (total work / first schedule work) x schedule days.
= (60/12) x 3.
= 5 x 3 = 15days.

To complete 1/5th of work - 3 days.

Hence to complete 5/5th of work - x days.

By cross multiply, you will get x value as 15 days.

How got the 3rd step (1/10+1/10)?

3days = 1/5.
6days = 2/5.
9days = 3/5.
12days = 4/5.
15days = 5/5 = 1 total work done is complete.