Let the rate of upstream be x kmph and rate of downstream be y kmph

⇒ Then, $(\frac{{20}}{x}\; + \;\frac{{32}}{y} = 8)$      .....(1)

⇒ Also, $(\frac{{32}}{x}\; + \;\frac{{20}}{y} = \frac{{17}}{2})$     .....(2)

Adding eq (1) and eq(2), we get

⇒ 52(1/x + 1/y) = 33/2

⇒ $(\frac{1}{x}\; + \;\frac{1}{y} = \frac{{33}}{{104}})$      ......(3)

Subtracting eq (1) from eq(2), we get

⇒ 12(1/x – 1/y) = 1/2

⇒ $(\frac{1}{x}\; - \;\frac{1}{y} = \frac{1}{{24}})$     ......(4)

Adding eq (3) and eq (4), we get

⇒ 2/x = 14/39

⇒ x = 5.57 kmph

⇒ Substitute x in eq (3), we get

⇒ 1/(5.57) – 1/y = 1/24

⇒ y = 7.25 kmph

Upstreamrate is 5.57 kmph and rate downstream is 7.25 kmph

∴ Rate of current = $(\frac{{7.25\; - \;5.57}}{2})$ = 0.84 kmph