Aptitude - Time and Work - Discussion

A. B. C. Take => a=20, b=3, c=60 days respectively.

The Lcm of 20, 30, 60 is 60.
So the efficiency of a, b, c is :.

Eff days.

3<---a---20---->.
2<---b---30---->
1<---c---60---->.

And 60 (total work).

Now work done by A in 2 days will be => 3*3 = 6.

On the 3rd day a, b, c will work so Work done will be = 3+2+1 = 6.
Work done in 3 days = 6+6 =12.
So, (12 work) is done in 3 days,
(1 work) will be done in = 3/12 days.

(total work is 60) , will be done in = (3/12) *60 = 60/4 = 15 days.

Solution:

A do a piece of work in = 1/20 days.
'A' work alone for 2 days, so the work done by A on the second day is = 2/20 = 1/10
On the third day, A is assisted by B&C, then the work done by three of them is = 1/10 + 1/20 + 1/30 = 3/20.

Total work done by A in 3 days is = 1/20 + 3/20 = 1/5.

As we have to find number of days, so,

1/5 part of work done by A is in = 3 days.
1 part of the work done by A is in = 3x5 = 15 days.

In 15 days, A can do the work if he is assisted by B& C on the third day.

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.

A Little Addition to @Vikas's explanation

A is working alone for two days, 3rd day he is assisted by B and C.

A's 1 day work=1/20.
A working on 2 days=2*1/20=1/10.

A+B+C working on 3rd day, so 1 day of working together = 1/20+1/30+1/60 = 6/60 = 1/10.

So total work done till 3rd day=1/10+1/10=2/10=1/5.
So if in 3 days = 1/5 of work is completed....
Total work is always 1;

So in order to make the value of RHS 1 multiply both side by 5;
Then, 3*5 days = 1/5*5 work will be completed.

= 15 days.

A complete in 20days.
B complete in 30days.
C complete in 60days.

Lcm(20,30,60)= 60.
So total work = 60.

A's one day work = 60/20 = 3.
B's one day work = 60/30 = 2.
C's one day work = 60/60 = 1.

A+B+C 's one day work = 3+2+1=6.

First day A's complete = 3unit of work.
Sec day A complete = 3unit of work.
3rd day A+B+C = 6unit of work.
.
.
So on

First 3day total work completed= 3+3+6=12 unit of work.

So (total work / first schedule work) x schedule days.
= (60/12) x 3.
= 5 x 3 = 15days.

A=30
B=20
C=60

Now take the lcm that is 60 mean A will make 3 chairs for example in one day B will make 2 chairs in one day and will make c will make 1 chair ( chairs example is for understanding).

Now A will make 3 chairs in one day and after 3 days it will make 9chair ( 3 *3) and at the third day ( B +c) will also make 3 chairs mean at the third day A +B +C WILL MAKE 12 chair in 3 days to make 60 chairs which our lcm it will require 15 days ( 12*5 = 60) that is 3*5 is 15 days.

A can do 1/20 of the work per day.
B can do 1/30 of the work per day.
C can do 1/60 of the work per day.

Together, they can do 1/20 + 1/30 + 1/60 of the work per day. But since B and C only help every third day, they can do, on average, 1/20 + 1/3 (1/30 + 1/60) of the work per day.

= 1/20 + 1/3 (1/30 + 1/60).
= 1/20 + 1/3 (3/60).
= 1/20 + 1/3 (1/20).
= 1/20 + 1/60.
= 4/60.
= 1/15.

So if they can do 1/15 of the work per day, they can finish the job in 15 days.

Answer: 15 days.

Worker - TotalDay - OneDay
A - 20 - 1/20
B - 30 - 1/30
C - 60 - 1/60

1st Day - A.
2nd Day- B.
3rd Day - A, B, C.

They said A lonely can work for 2 day.
2A = 2(1/20) =1/10.

Then in third day A, B, C together;
(A+B+C) = 1/20+1/30+1/60,
= 1/10.

3 Days(not 3rd Day) work is;
=2A + (A+B+C).
= 1/10+1/10.
3Days =1/5 work,
6Days =2/5 work,
9Days =3/5 work,
12Days=4/5 work,
15Days= 5/5 =1 work completed.

As A does work in 20 days.
B does in 30 days.
C does in 60 days.

Take the lcm and you will get total work = 60 parts.
No of parts A does work = 60/20 = 3 part / day.
Similarly B = 2 parts /day and c= 1 parts/day.
If A does work for 2 days he covers 3+3 parts = 6 parts.
And every third day B & C joins and then A + B +C = 6 parts on 3rd day.
Total work from first day to third day = 6 + 6 parts= 12 parts.
Total work = 60 parts.

Therefore no of days require = 5 * 3 = 15 days.

A = 20, B = 30 &C = 60.

Total work (LCM of 20, 30, 60) = 60.

A's 1day work= 60/20 = 3units.
B's 1day work= 60/30 = 2.
C's 1day work= 60/60 = 1.

1st day A will do the work =3 unit.
2nd day again A will do the work = 3.
In 3rd-day, A + B + C will do the work = 3 + 2 + 1 = 6.

After 3 days total work = 12 (3 + 3 + 6).

After 6 days total work = 24.

Similarly;

After 12 days total work = 48.

15 days total work = 60.

So total no days to complete the work (60) is 15 days.