A two-digit number is such that the product of its digits is 8. When 1

1.	A two-digit number is such that the product of its digits is 8. When 18 is added to the number, the digits are reversed. The number is : (a) 18 (b) 24 (c) 42 (d) 81
Answer» (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.

A two-digit number is such that the product of its digits is 8. When 18 is added to the number, the digits are reversed. The number is : (a) 18 (b) 24 (c) 42 (d) 81

Answer»

(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.