blueblac92 wrote:

ganand wrote:

Hi,

The general framework of the above problem:

Total time taken to complete the work (working together) = \(x\) hrs

Time taken by A to complete the work (working alone) = \(a\) hrs more than combined time, i.e. \((x+a)\) hrs

Time taken by B to complete the work (working alone) = \(b\) hrs more than combined time, i.e. \((x+b)\) hrs

Then \(x = \sqrt{a \times b}\)

In the above problem \(a = 8\) hrs, and \(b = 4.5\) hrs. Hence \(x = \sqrt{a \times b} = \sqrt{8 \times 4.5} = \sqrt{36} = 6\) hrs.

Thanks.

I am not quite sure how you came up with this solution.

Can you elaborate please?

Thank you!

Hi @blueblac9,

Ok. I'll derive the relation.

Let A works at the rate of \(p\) units/hr. Let B works at the rate of \(q\) units/hr.

Total time taken to complete the work (working together) = \(x\) hrs.

Time taken by A to complete the work (working alone) = \(a\) hrs more than combined time, i.e. \((x+a)\) hrs.

\(\Rightarrow p(x+a) = (p+q)x \Rightarrow px + pa = px + qx \Rightarrow pa = qx \Rightarrow p = \frac{qx}{a}\) --- (1)

Time taken by B to complete the work (working alone) = \(b\) hrs more than combined time, i.e. \((x+b)\) hrs.

\(\Rightarrow q(x+b) = (p+q)x \Rightarrow qx + qb = px + qx \Rightarrow qb = px \Rightarrow p = \frac{qb}{x}\) --- (2)

From (1) and (2), we have following:

\(\frac{qx}{a} = \frac{qb}{x} \Rightarrow x^2 = ab \Rightarrow x = \sqrt{ab}\).

Hope this helps.

Thanks.