Aptitude - Time and Work - Discussion

Suppose C can complete work in 18 days then A and B can complete the work in 54 and 27 days respectively.
Now A's 1 day work is 1/54 and B's 1 day work is 1/27 and C's 1/18.
Together 1/54+1/27+1/18=1/9.
Work finishes by together in 9 days.
Now when work complete in 9 days B's days are 27 so when work complete in 2 days B's days=?

27*2/9=6 days.

Hai guys, i think mini is correct. from the question we get if B works for 1 hr, A takes 2 hrs to complete the same work i.e.,2A=B

Suppose A, B and C take x, x/2, x/3 days respectively to finish the work.
work done per day by

A is 1/x
B is 1/(x/2)=2/x
C ix 1/(x/3)=3/x

And the work done by A, B and C together per day is 1/2 (since work is completed in 2 days)

So 1/x + 2/x + 3/x = 1/2
6/x = 1/2
x = 12.

Answer : 6

My thoughts:

Let the time for A to do the job be X.
Let the time for B to do the job be 2X (since A takes twice as much time as B).
Let the time for B to do the job be 3X (since A takes thrice as much time as C).

So it takes all three X+2X+3X = 1/2 (days to finish the job).
This gives 6x=1/2. Therefore x=3.

Therefore the time taken for B to do the work is 2X.
Which is 2*3=6 which is equal to 6days.

It takes 3X that is 3*3 for C to do the work.
Which is equal 9 days.

With the above, you would find out that indeed A takes twice as much time as B and thrice as much time as C.

I hope this would suffice.

Hai friends,

Taking the values of A,B,C as x, x/2, x/3.

1/x + 2/x + 3/x = 1/2.
On calculating we get,

6/x = 1/2.
Therefore, x = 12;

On substituting X = 12 in main equation.

1/12 + 2/12+ 3/12 = 1/2.

Hence, the value of B = 1/6.

Can anyone please explain me that how these x, x/2, x/3 become 1/x + 2/x + 3/x.

Here I don't get is we take 1 day's work of A as 1/x and after that it's taken 2/x + 3/x as B and C's 1 day work but actually after A's 1 day work B's & C's work should be taken as 1/x/2 and 1/x/3.

Which becomes x/2 and x/3 So finally it should look like 1/x + x/2 + x/3.

Why 12 is divided by 2 in last?

This can be solved using physics. Force=Power/time. A, B and C does the same amount of work, hence, force is a constant. The power of three differs. Hence 1/t=f/p where f is a constant and t=2. 1/2=1/x+2/x+3/x.

Lets assume that A completes the work in 'x' days.

B completes the same amount of work in 'x/2' days because A takes twice the time than B.

C completes the same amount of work in 'x/3' days because A takes thrice the time than C.

Now the work completed by A in one day is '1/x'.

Work completed by B in one day is '2/x'
Work completed by C in one day is '3/x'

All three together can complete the work in '2' days as given in the problem. So the work done by all three together in one day is '1/2'.

Hence the equation: '(1/x + 2/x + 3/x) = 1/2'.

Evaluating the above equation gives the value 'x=12' which is the number of days taken by A to complete the work.

As we know that number of days taken by B complete the work is 'x/2', we substitute the 'x' value.

Hence the evaluation '12/2' where 12 is the value of 'x'.

HENCE THE TIME REQUIRED BY B TO COMPLETE THE WORK IS 6 DAYS.

A can complete 1 unit/day.
B can complete 2 unit/day.
C can complete 3 unit/day.

Total works complete together by all in 1 days = 6 unit/day.

So, they complete in two days = 6X2 = 12 units.
So B takes 12/2 = 6 days to complete.