Aptitude - Time and Work - Discussion

We have A work 2 days alone third-day B&C are added that means A+A+ (A+B+C) THIS is 3days work.

So A can do 2days work=1/10, and A+B+C 1day work=1/10, so add 1/10+1/10=1/5then total day work by A is 3*5=15.

Nice explanation, Thanks @Vijay.

Simple, 3rd day B and C will hep A.
So first 2 days A will be working his work for two days is 2*(1/20) = 1/10.
Now 3rd day he is helped by B and C .
So A's work + B's work + C's work for 1 day is( 1/20 +1/30 +1/60) take LCM and solve we get 1/10.

Now A's 2days of work + (A+B+C's )work for 1 day.
1/10 +1/10 = 1/5.

So 1/5 work is completed which includes A's 2 days of work and Band C assistance for A (previous step same thing)
now
1/5 work is done in --- 3days
1( indicates complete work ) in ---- ? days

Ans is 15 days.

According to me, the answer is 12.

How 6/60 come?

Nice explanation @Nikita.

Why we add 1/10 and 1/10 here? Please explain.

Usually, we are taking 20 days means 1/20. Then why there finding 2 days work is multiplied by 2?

My thoughts are 1/20*1/2. Please explain me.

It can be done in an easier way than the given solution:
Let the number of days required = x.

So basically, B will contribute (1/3)rd of its work in 30 days and C will contribute (1/3)rd of its work in 60 days.
Work by A in one day: 1/20.
Work by B in one day: 1/30.
Work by C in one day: 1/60.

So total work can be easily calculated as:
(x/20) + (1/3)*(x/30) + (1/3)*(x/60) = 1.
So, 12x = 180,

And thus x = 15 days.

Can you explain me how it is 1/5 and 3*15? I'm not getting this concept.