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.

Easy way to answer is:

LCM of 20 30 60 = 60.
A = 20 * 3 = 60,
B = 30 * 2 = 60,
C = 60 * 1 = 60,.

A + B + C = 3 + 2 + 1 = 6 ( together)
A's 2 days work = 2 * 3 = 6.

Total work done by them = 6 + 6 = 12 unit
Total work is 60 unit.
So, 60 ÷ 15 = 5 units ( In one day by them )
Then in 3 days,
3*5 = 15 days

@Monoj.

Best explanation.

Mem ------> time ---- efficiency ----work
a------> 20 ---- 3 ---- 60
b------> 30 ---- 2 ---- 60
c------> 60 ---- 1 ---- 60

Now we make a series like :-
Day1 total unit of work will be -> 3 units.
Day2 total unit of work will be-> 3 units.
Day3 total unit of work will be -> 3+1+2 = 6(because b and c helped him).

Day1 Day2 Day3
3 3 6.

Work will be-> 12 units = 3 days.
The total units of work is 60 units then divide 60 with 12(60/12=5).
Now the total units are 5 times the 3 days of work than days will also multiply by 5 which is 3*5="15 days" (Ans).

Why have to take 3 days of work?

A = 20/60 -> 3 units
B = 30/60 -> 2 units
C = 60/60 -> 1 unit
= 6 units.

L.C.M=60
B and C help on every third day, which means A can do the work alone for 2 days.
2 days A work = 3x3 = 9 units + 3rd day(A+B+C) 6 = 15.

For the peopel who are confused with last step.

We knew that 3 days work is 1/5.. the remaining work is 1-1/5=4/5.. ie already worked 3 days here.

Then we are calculating that how many 3 days are required to complete the remaining work 4/5.

Ie 4/5/1/5 = 4/5 * 5/1 = 4.. 4 (three days required).
ie 4*3 = 12 we already worked 3 days then total days will be 12 + 3 = 15.

LCM of A, B, C's works = 60,
And the work done by A in 1 day = 3 (60/20),
so in 2 days work done by A is = 3*2= 6,
And work done by A+B+C in 1 day = 6 [(60/20) + (60/30) + (60/60)],
So total work done for 3 days = 12,
So the total number of days required to complete 60 units of work = 60 * 3/12 = 15.

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.

Answer Explained:

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....
Than, 3*5 days = 1/5*5 work will be completed.
=15days