Aptitude - Time and Work - Discussion

From the explaination, we got to know that in 3 day's 1/5 of the work completed.

If A's one day's work is 1/20 then he can complete in 20 days. Same way we should find out 1 day work for the above problem.

i.e., If 3 day's work=1/5.
then one day work=(1/5)/3 =>1/15,
=>so 1 day work=1/15 then he can complete in 15 days.

Why are we multiplying by 5 for whole work? Explain.

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.

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

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.

Since B and C assisting A on every 3rd day, B & C takes 30x3=90 days and 60x3=180 days respectively to do the work.

Therefore

1/20+1/90+1/180 = 12/180 = 1/15 or 15 days.

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

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.

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.