Aptitude - Time and Work - Discussion

As LCM(20, 30, 60) = 60.

A's rate = 3, B's rate = 2 and C' rate = 1.
As work = 60 units

A will work for 2 days = 2 x 3 = 6.
B and C will join A on third day = (2+3+1) x 1 = 6
Work done in 3 days = 12 units.
Work done each day = 4.

The time required for 60 units of work at a rate of 4 is 15 days.

1/5 work is done in 3 days.

The total work 1 done in x days.
Simply cross multiply;
x*1/5 = 3 * 1.
x/5 = 3,
x = 15.

A = 1/20.
B = 1/30.
C = 1/60.

Now, 1/20+1/30+1/60=5/60 = 1/12.
Here, 1 days efficiency of A is 3, B is 2 and C is 1.
Now,
A can do the work in first day= 3,
" in the second day=3+3=6,
" in third day = 6+6=12,
So, in three days, A completed the work 12.
Now,
12 work = 3 days.
60 work = 3/12*60= 15 days.

Work done by A = 1/20.
Work done by B= 1/30.
Work done by C= 1/60.
Now, work done by A in 2 days is= (1/20) * 2 = 1/10.
And work done by A,B,C in one day= 1/20 + 1/30 + 1/60 = (3+2+1)/60 = 6/60 = 1/10.

And the work done in first 3 days = 1/10 + 1/10 = 1/5.

So we can say that 1/5 part of work is done= in 3 days.
And 1 complete work is done in=. 3/(1/5) = 3*5 = 15 days.

@Monoj.

Best explanation.

LCM of 20,30,60 = 60.
Efficiency of One day work of;

A = 60/20 = 3,
B = 60/30 = 2,
C = 60/60 = 1.

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

Total work done by them = 6 + 6 = 12 unit (in 3 days),

Then,
(Total work)/(Total work done in 3 days),
That is, 60/12 = 5,
So,
1 = 3, then 5 = 5 * 3 = 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

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.

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).

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.