Question

A alone can complete a work in 5 days more than A+B together and B alone can complete a work in 45 days more than A+B together. Then in how many days A and B together can complete the work?​

Answer 1

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Answer 2

Therefore in 15 days  A and B together can complete the work.

Step-by-step explanation:

Let A and B can compete the work in x days.

Given that A takes 5 days more  than( A+ B ) together to complete the work.

Then in (x+5) days A can complete the work .

Again given,  B takes 45 days more than ( A+ B ) together to complete the work.

Then in (x+45) days B can complete the work.

A's one day work $\frac{1}{x+5}$

B 's one day work $\frac{1}{x+45}$

Therefore (A+B)'s one day work $(\frac{1}{x+5} +\frac{1}{x+45})$

(A+B)'s x days work $=x(\frac{1}{x+5} +\frac{1}{x+45})$

Since A and B completed the work in x.

The total work is 1

According to the problem,

$x(\frac{1}{x+5} +\frac{1}{x+45})=1$

$\Rightarrow x[\frac{x+45+x+5}{(x+5)(x+45)} ]=1$

$\Rightarrow \frac{x(2x+50)}{x^2+50x+225} =1$

$\Rightarrow 2x^2+50x= x^2+50x+225$

$\Rightarrow x^2+50x-x^2-50x= 225$

$\Rightarrow x^2=225$

$\Rightarrow x=\sqrt{225}$

$\Rightarrow x=15$

Therefore  A and B together can complete the work in 15 days.