Aptitude - Time and Work - Discussion

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.

@Dhanam.

Because on third day they are doing combined work it means that for 1st 2 days they have worked alone here in question. About A has asked so he is doing only A.

@Kiran.

You didn't account for A's third day of work.

A does 6 units of work in two days and another 3 units on the third day. So in total, A has done 9 units of work in 3 days.

Hi, can anyone tell me what's wrong here?

Given,

A=20days.
B=30days.
C=60days.

I took as 60units of work by taking LCM of A, B, C.

A's one day work=3 units/day.
B's one day work=2 units/day.

A's two days work = 6 units.
B's one day work= 2 units.
C's one day work= 1 unit.

A+B+C's 3 days work = 9 units.

So, I can take A+B+C's one day work = 3 units =>9/3=3 units/day.
So, to complete 60 units of work they will take 20days =>20*3=60 units.
So, I am saying that answer is "20 days".

Correct me if I'm wrong.

Every 3 days 1/5 work is done. Then in how many days 1 whole work is done.

Mathematically = 1/5:3::1:x.

3*1/ (1/5) = 3÷1/5= 3*5/1= 3*5= 15.

3 days ---> work 1/5
x days ---> to get total work =1

Cross multiplication, then we get x = 15.