(b) 24
Let the tens' digit be x. Then, ones' digit = \(\frac8x\)
Original number = 10x + \(\frac8x\) = \(\frac{10x^2+8}{x}\)
Number after reversing
= \(10\times\frac8x+x=\frac{80}x+x=\frac{80+x^2}{x}\)
Given, \(\frac{10x^2+8}{x}\) + 18 = \(\frac{80 +x^2}{x}\)
⇒ \(\frac{10x^2 + 8 + 18x}{x}\) = \(\frac{80 +x^2}{x}\)
⇒ 9x2 + 18x - 72 = 0
⇒ 9x2 + 36x -18x - 72 = 0
⇒ 9x(x + 4) - 18 (x + 4) = 0
⇒ (9x - 18)(x + 4) = 0
⇒ 9x -18 = 0 or x + 4 = 0
⇒ x = 2, - 4
Rejecting the negative value, x = 2
∴ Original number = 10 x 2 + \(\frac82\) = 20 + 4 = 24.