A works 3.5 times as fast as B, and A takes 45 days less than B to complete a job when each works alone. Calculate the number of days taken to complete the same job if A and B work together.

Given:

A works 3.5 times as fast as B.

A takes 45 days less than B to complete the job.

If one person works faster than another, their work rates are inversely proportional to the time they take to complete the same job. 

Efficiency of A = 3.5 times efficiency of B 

​$$\frac{\text{Efficiency of A}}{\text{Efficiency of B}} = \frac {7}{2}$$ 

Now As we know ,

Efficiency $$\propto \frac {1}{Time}$$ 

Then time taken by A and B 

let A's time be 2x and B's time be 7x

According to the question, 

7x - 2x = 45

x = 9 

time taken by A alone to complete the work = 18 days 

For B = 63days

Now , work rate  by A and B together = $$\frac{1}{18}+\frac{1}{63} = \frac { 7 + 2}{126} = \frac{9}{126} = \frac{1}{14}$$ unit

time taken to complete the work together = 14 days

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