Aptitude - Time and Work - Discussion

Take whole work is 1.B and C assists A only 3rd day.
So A"s two day work is 1/10.
On 3rd day total work is done is 1/5.

That means 1/5 work done taken in 3rd day. So whole work(1) is done in 15 days.

For whole work take lcm of 20,30 and 60 ie 60 so whole work is 60.
A alone work = 60/20 = 3w/d.
B=2w/d.
C=1w/d.
A+b+c= 6w/d.
A work 3 /day so 60/3=10 days.
It means A alone work 10 days and ie,
10*3(A per day work)= 30.
Remaining work 60-30=30.
So remaining 30 work done by;
A+B+C= 30/6(pr day work)= 5 days,
So total days 10(alone A)+ 5(A+B+C)=15 days.

Hi, I'm not getting the below logic in quotes

" Complete 3 days work is 1/10 + 1/10 = 2/10 =1/5"

Shouldn't that be [ (A's ist day)+ (A's 2nd day) + (A+B+C combined on 3rd day)]?

ie = 1/20 + 1/10 + 1/10?

Please correct me if I am wrong.

Works that done till 3rd day:

[ (A's ist day)+ (A's 2nd day) + (A+B+C combined on 3rd day)].

ie = 1/20+1/10+1/10,
= 1/4.

ie remaining pending work = 3/4.

So, the no of days Al required to complete the work= 20*(3/4),
= 15 days.

Here 3'day's work is 1/5 so next 3 day work is also 1/5.

1/5+ 1/5+ 1/5+ 1/5+ 1/5 i.e its happen 5 times so 3x5= 15 days.

Explain the Step 1.

A's 2 day's work = 1/20*2= 1/10. Explain it clearly.

Please explain the last step ( 1/10 + 1/10) = 1/5.

solve :: x * (1/20)+x/3*(1/30+1/60) = 1 for x, where x is number of days A worked.

Please, anyone solve this.

A - 20.
B - 30.
C - 60.
Total work is 60.
Because LCM of ( 20,30,60 ) is 60.
Then A one day work is 60/20 = 3,
B one day work is. 60/30 = 2,
C one day work is 60 /60 = 1,

1st day work is only A = 3.
2nd day work is only A =3.
3rd day work is A+B+C (3+2+1) = 6.

So.,
3 day work 12.
6 day work 24.
9 day work 36.
12 day work 48.
15 day work 60.
Total 60 work is completed in 15 day.
It's simply

It is solved by chocholate method
20 30 60 lcm of those 60.
3 2 1 are their 1 day works.
a a a+b+c a a a+b+c a a a+b+c a a a+b+c a a
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
3 +3 + 7 +3 +3 +7 +3 +3 +7 +3 +3 +7 +3 +3 +7=60 days.