Aptitude - Time and Work - Discussion

A can do job in 16 days.
B can do in 12 days.

Make the LCM for both A and B.
After LCM we get 48.
This is the total work for both A and B=48.
For one day work of A is 16*3=48.

That means A can do 3 units of work per day,
For one day work of B is 12*4=48.

That means B can do 4 units of work per day.
(A+C)=3+4=7.
(B+C)=4+4=8.

TOTAL work of (A+B)and(B+C) = 7 + 8 = 15.

We find out c alone can do the job in how many days
So we can subtract 48-15=33.

Convert 33 in fraction form that you get the answer.

1/4- (1/16+1/12) = (1/4-7/48) please anyone explain this.

@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

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.

@Bhimesh.

Yes, you are correct.

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.

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

@Neha.

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

Thanks all for explaining.

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.