Aptitude - Time and Work - Discussion

Whole work will be done in (3 x 5) = 15 days.

I m unable to understand a last step. Explain me?

How they multiply by 5?

LCM of 20,30,60 is 600
A's work is 600/20
if b and c on third day then 2 days(first and second) for A and 1 day for B and c(third day)
so A's work is 600/(20*2)=600/40=15

We have already calculate 1/5 for 3 days. Then why again multiplying by 3? please help.

If we add the work of b and c for 1 day we get 1/30 + 1/60=1/20.

First day = 1/20.

Second day =2/20.

Third day =4/20 // work of b & c is added 3/20 + 1/20.

Every 3 days 4/20.

For 6 days we get 8/20.

For 9 days we get 12/20.

For 12 days we get 16/20.

For 15 days we get 20/20 = 1. Job over.

In the last step the multiplication of 3 and 5 is for the no.of persons (3) and amount of work done (5).

I didn't understood how did they calculated the work done in 3 days. Can somebody explain me clearly?

If x = A worked day

{x*1/20 + x/3*(1/30+1/20)} = 1

x/20+x/60 = 1

x = 15.

There is confusion because the number is same for A working for 2 days (i.e.1/10 Days) and all three, A+B+C=1/10 days.

Hence its.

=2 working days of A 1/10+ All 3 together 1/10 days.

i.e. (1/10+1/10).

1/5 works is being done in 3 days.

i.e 1 of the 5 parts of work is done in 3 days.

So total there are 5 parts of work which will take 5*3=15 days.

A's one day work is 1/20 ,

Therefore A's 2 day work will be (1/20)*2=(1/10),

Now (A+B+C)'s one day work will be = (1/20)+(1/30)+(1/60)=(1/10).

Since A is assisted on every third day of his work then total work done in three days will include A's 2 day's of work plus the work done on third day, that is (A+B+C)'s one day work,

Therefore work done in 3 days will be,A's two days work (1/10 )+ A,B,C's one day work (1/10)= (1/5).

Now in three days 1 out of 5 parts of work is done , and yet 4 parts are to be done there fore 4*3= 12 days and since we have calculated the earlier 1 parts requirement of days i,e 3 days total days will to finish 5 parts will be (12+3)= 15 days.