Solution:-

Let us assume the number be B.

As per the condition given in the question, B + 1/B =\[2\frac{9}{40}\]

So, B + 1/B = 89/40

\[({{B}^{2}}~+\text{ }1)/B\text{ }=\text{ }89/40\]

Cross multiplication we get,

\[{{B}^{2}}~+\text{ }1\text{ }=\text{ }\left( 89/40 \right)B\]

By transposing we get,

\[{{B}^{2}}_{~}+\text{ }1-\left( 89/40 \right)B\text{ }=\text{ }0\]

Multiply by \[40\] for both side of each term we get,

\[40{{B}^{2}}~+\text{ }40-89B\text{ }=\text{ }0\]

\[40{{B}^{2}}~-89B\text{ }+\text{ }40\text{ }=\text{ }0\]

\[40{{B}^{2}}-64B-25B\text{ }+\text{ }40\text{ }=\text{ }0\]

Take out common in each terms,

$8B(5B – 8) – 5(5B – 8) = 0$

$(5B – 8) (8B – 5) = 0$

Equate both to zero,

$5B – 8 = 0$

$8B – 5 = 0$

$5B = 8$

$8B = 5$

$B = 8/5$

$B = 5/8$

Therefore, the numbers are $5/8$ and $8/5$.