Three numbers are in the ratio $\dfrac{1}{2} : \dfrac{1}{3} : \dfrac{1}{4}$. If the sum of their squares is 244, find the numbers.

The ratio is $\dfrac{1}{2} : \dfrac{1}{3} : \dfrac{1}{4}$.

L.C.M. of 2, 3, 4 = 12.

$\text{Ratio } = \dfrac{1}{2} \times 12 : \dfrac{1}{3} \times 12 : \dfrac{1}{4} \times 12 \\[0.5em] = 6 : 4 : 3.$

Since the ratio is 6 : 4 : 3, let the numbers be 6x, 4x and 3x.

Given, sum of squares of numbers = 244.

∴ (6x)² + (4x)² + (3x)² = 244
⇒ 36x² + 16x² + 9x² = 244
⇒ 61x² = 244
⇒ x² = $\dfrac{244}{61}$
⇒ x = $\sqrt{4}$ = 2

∴ x = 2, 6x = 12, 4x = 8, 3x = 6.

Hence, the numbers are 12, 8 and 6.