Aptitude - Time and Work - Discussion

"A" can complete work in 20 days.

A's 1 day work is 1/20.

A's 2 days work is 2/20.

According to the question A work for 2 day's and on the third day.

B & C assists (**helps, not replaced).

So it goes like this.

A's first day + A's second day + (A + B + C) on third day, B & C helped.

i.e 2 1/20+1/20/+(1/20+1/30+1/60).

=>2/20+(1/20+1/30+1/60).

=>2/20+1/10.

= 1/5.

So for first 2 days and combining B and C on third day they will work 1/5th of a work.

All together (A+A+ (A+B+C) ) works for 3 days to get 1/5th of a work so.

If 1/n work is done in 1 days total work is done in 1*n = n days.

The same here, if 1/5 work is done in 3 days total work is done in 5*3 = 15 days.

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.

LCM method:

The LCM of A,B,C is 60 so consider total work to be 60 unit.
Since A take 20 days, he does 3 unit per day.
Similarly, B does 2 unit per day.
Similarly, C does 1 unit per day.

So total unit of work done by 3 combined is 6 unit.

So by Addition of each day of Unit of work done until reaching total work. So the addition of A alone and A+B+C can be summarized as.
A doing work for first 2 days will be 2*3, total Work 6 done.
On 3rd day work done is 6, total work done 12.
Again A working alone for 4th and 5th is 2*3 =6,
Total work 18
On 6th day work done (A+B+C) 6, Total work 24
On 7th and 8th A is 6, total work 30.
On 9th day again (A+B+C) work done 6, TOTAL 36.
On 10 and 11th A alone work was done 6, Total work 42.
On 12th (A+B+C) work done 6, Total work 48.
On 13 and 24th A alone is 6, Total work 54.
On 15th A+B+C is 6, Total work was done 60 unit, work completed.

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

Here is a solution:

Efficiency A = 3 , B = 2, C = 1 and A+B+C = 6.

Now,
Work done in format ie A can do a work in 2day and next 1day with B&c .
So.
Eff * time = work.

A = 3 * 2 = 6.
A + B + C = 6 * 1 = 6.
Means work done in 3 days is 6 + 6 = 12 work.
Days -------- work
3 -------- 12
? ------- 60 (total work)

Days =(60*3) / 12.
The answer is = 15 days.

Step 1: Work is done by A+B+C in one day can be calculated as 1/20 + 1/30 + 1/60 = 1/10.
Step 2: For the first two days only A is doing work alone so the work is done by A in two days will be 2*1/20= 1/10.
Step 3: Now total work is done in 3 days by A B & C will be = 1/10+1/10= 1/5.
Step 4: But we need a complete work to be done so we have to convert 1/5 to 1 therefore if we do 1/5 work 5 times then work will be completed (1/5 x 5=1).

Therefore 5 x 3 = 15.

Hope you will understand this solution.

How did we get 6/60? Anyone explain about it.

A work in one day = 1/20,
B work in one day = 1/30,
C work in one day = 1/60.

As A work for 2 days and then on third day B and C do work.

Total work done in three days = A 3 days work+B 1-day work+ C one day work,
= 3/20 + 1/30 + 1/60,
= 9/60 + 2/60 + 1/60
= 12/60.

Total work is done in three days =1/5.
3 = 1/5part.

Let x days required to complete work = 1 part.

3/x = 1/5/1.
x = 3 * 5 = 15 days.

How we got only A done 3 days of work? Please explain it.

@Vijay

We know that the whole work done is 1. here 3 days work is 1/5.
So,
if, 1/5 work=3days.
then, 1 work= ?
Cross multiply we get ?= 3 x 5= 15days.