Aptitude - Time and Work - Discussion

Consider A and B work alone.

1/16 + 1/12.
3/48 and 4/48.
That means A does work 3 units per day and B does 4 units per day.
So, now total work is done by A and B in one-day => 4 + 3 = 7 units per day
The total work is done by A and B in 4 days = 28 units.
Remaining units = 48-28 = 20 units.

That means 20 units work is done by C in 4 days in order to complete the work;
So, WORK DONE BY C IN ONE DAY = 20/4= 5 UNIT PER DAY,
C alone can do work in = 48/5 i.e 9(3/5).

Good explanation, Thanks @Billal Hossain.

Solution:

Let, C did the job in= x days.

ATQ,

1/16+ 1/12+ 1/x = 1/4.
=> 1/x= 1/4- 1/16-1/12,
=> 1/x= (12-3-4)/48,
=> 1/x= 5/48,
=> x= 48/5,
=> x = 9 (3/5).
C alone can do the job in 9(3/5) days.

Thanks all for explaining.

@Neha.

It is 48/5=> 45+3/5 = 9 * 3/5.

I can't understand 9 3/5 last answer.

16,12,4 of lcm=48.
16/48=3.
12/28=4.
4/48=12.


So, total works efficiency =12-7=5 per days;
48/5 is the answer.

@Bhimesh.

Yes, you are correct.

Why 5/48 is 48/5 because A can do work in 16 days so his efficiency for 1 day is 1/16.

(which means (work/time)). So same for C's 1-day work is 5/48 and he can complete the work in 48/5.

@Sai Kumar.

It is 1÷16 + 1÷12 so 1st do LCM of 16,12 then you'll get the answer as 48 later on divide (48÷16, 48÷12) then you'll get answers as 3 and 4 (as multiplied by the numerator 1 then 3*1 will be 3 and 4*1 will be 4) add 3+4 you'll get 7 then the final answer will be 7/48