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A $2$-digit number is such that the product of the digits is $14$. When $45$ is added to the number, the digits are reversed. Find the number.
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Let the two digit number be $10x + y$
$xy = 14$ ..... (i) (1 Mark)
$10x + y + 45 = 10y + x$
$y-x = 5$ .....(ii) (1 Mark)
From (i) and (ii)
$x(x+5) = 14 \Rightarrow x^2 + 5x - 14 = 0$ (1/2 Mark)
$(x+7)(x-2) = 0$ (1/2 Mark)
$x = 2$ (as $x \neq -7$) (1 Mark)
Number = $27$ (1 Mark)
$xy = 14$ ..... (i) (1 Mark)
$10x + y + 45 = 10y + x$
$y-x = 5$ .....(ii) (1 Mark)
From (i) and (ii)
$x(x+5) = 14 \Rightarrow x^2 + 5x - 14 = 0$ (1/2 Mark)
$(x+7)(x-2) = 0$ (1/2 Mark)
$x = 2$ (as $x \neq -7$) (1 Mark)
Number = $27$ (1 Mark)