Aptitude - Time and Work - Discussion

Let us understand this in a much simpler way.

Let the time taken by B to finish the work be X days.
Then according to the question Time is taken by the A is (X-60).
Work done by B in one day is (1/X),
Work done by A in one day is (1/X-60),

And we know that A is three times faster than B, So work done by A in one day is three times the work done by B in one day.

So 3*(1/X)=(1/X-60), 3X-180=X, X=90, Therefore B takes 90 days to complete the work hence A take 30 days to complete the work, therefore together they take 1/(1/30+1/90) days, which is equal to 45/2.

Thank you all.

a b
x 3x

2x=60 so x=30

total work=3*30=90
Ans=90/4=22 1/2.

Ratio of times taken by A and B = 1:3.

Means B will take 3 times which A will do in 1 time.
If difference of time is 2 days, B takes 3 days,
If the difference of time is 10 days, B takes (3/2) * 10 =15 days.

Here given k=3, n=60.

So formula is -> kn/ (k^2-1) =3*60/ (3^2-1) =180/8=22 1/2 days.

Please provide the solution of if A is 5 times better than B together they complete the work in 50 min then B alone can do the work in how many minutes.

Can anyone explain me why the step 3/2*60 used. I couldn't get the logic of finding difference as 2.

Let see @Shaman.

A finish work in x day and B can 5x day. Then we can do A+B= 1/x +1/5x.

We have A+B = 50,

50 = 5x/6 after sol x= 60 it means A can do in 60 min and B 300 min.

How 4/90 comes?

From where does this 4/90 comes? How 4?

I have better solution.

A:B-3:1(rate).
A:B-1:3(days).

Diff of 2 between ratio of days which is equivalent to 60 and 2*30=60.
Hence 30 will be multiplied to 1 and 3.
And we get 30 and 90.
Together days taken= 30 * 90/30 + 90 = 22 1/2.