The ratio between three positive numbers is $\dfrac{1}{4} : \dfrac{1}{3} : \dfrac{1}{2}$. When the square of the middle number is subtracted from the sum of the squares of the other, the result is 725. Find the numbers.

It is given that the ratio between three positive numbers = $\dfrac{1}{4} : \dfrac{1}{3} : \dfrac{1}{2}$

Let the three number be $\dfrac{1}{4}x, \dfrac{1}{3}x \text{ and } \dfrac{1}{2}x$.

If the square of the middle number is subtracted from the sum of the squares of the other, the result is 725.

$\Rightarrow \Big[\Big(\dfrac{1}{4}x\Big)^2 + \Big(\dfrac{1}{2}x\Big)^2\Big] - \Big(\dfrac{1}{3}x\Big)^2 = 725\\[1em] \Rightarrow \Big[\dfrac{1}{16}x^2 + \dfrac{1}{4}x^2\Big] - \dfrac{1}{9}x^2 = 725\\[1em] \Rightarrow \Big[\dfrac{1}{16}x^2 + \dfrac{1 \times 4}{4 \times 4}x^2\Big] - \dfrac{1}{9}x^2 = 725\\[1em] \Rightarrow \Big[\dfrac{1}{16}x^2 + \dfrac{4}{16}x^2\Big] - \dfrac{1}{9}x^2 = 725\\[1em] \Rightarrow \dfrac{5}{16}x^2 - \dfrac{1}{9}x^2 = 725\\[1em] \Rightarrow \dfrac{5 \times 9}{16 \times 9}x^2 - \dfrac{1 \times 16}{9 \times 16}x^2 = 725\\[1em] \Rightarrow \dfrac{45}{144}x^2 - \dfrac{16}{144}x^2 = 725\\[1em] \Rightarrow \dfrac{29}{144}x^2 = 725\\[1em] \Rightarrow x^2 = \dfrac{725 \times 144}{29}\\[1em] \Rightarrow x^2 = \dfrac{104,400}{29}\\[1em] \Rightarrow x^2 = 3,600\\[1em] \Rightarrow x = \sqrt{3,600}\\[1em] \Rightarrow x = 60$

So, the numbers = $\dfrac{1}{4}x = \dfrac{1}{4} \times 60 = 15, \dfrac{1}{3}x = \dfrac{1}{3} \times 60 = 20 \text{ and } \dfrac{1}{2}x = \dfrac{1}{2} \times 60 = 30$

Hence, the number = 15, 20 and 30.