Question

In a two digit number, the digit in one's place is 2 more than the digit in ten's place. Find the number if the product of digits is 48.

• A. 46
• B. 86
• C. 68
• D. 64

Answer 1

The correct option is C 68

Let $x$ be the digit in ten's place . Then the digit in one's place will be $(x+2)$.
Given that, $x(x+2)=48$
$\Rightarrow x^2+2x=48$
$\Rightarrow x^2+2x-48=0$
$\Rightarrow x^2+8x-6x-48=0$
$\Rightarrow (x+8)(x-6)=0$
$\therefore x=-8,6$

Value of $x$ cannot be a negative value. So $x=6.$

$\therefore$ The digit in ten's place is 6.
Hence, the digit in one's place is $(x+2)=8$.

So, the required number is 68.