Problems based on Numbers Test 6

$\left(\frac{5}{12}+\frac{17}{16}+\mathit{x}\right)$
$\frac{20+51}{48}=\frac{71}{48}$
Since the addition of the both numbers should be a whole number, $\frac{71}{48}$ is greater than 1 but less than 2.
So, the least value of x will be such that the sum of both will be $=\mathit{x}=2-\frac{71}{48}$ $=\frac{96-71}{48}=\frac{25}{48}$