Let the smaller diameter pipe take $x$ hours to fill the tank alone.
Then the larger diameter pipe takes $(x - 2)$ hours.
In 1 hour, smaller pipe fills $\dfrac{1}{x}$ and larger pipe fills $\dfrac{1}{x-2}$ of the tank.
Together they fill in $\dfrac{15}{8}$ hours, so:
$$\frac{1}{x} + \frac{1}{x-2} = \frac{8}{15}$$
$$\frac{(x-2)+x}{x(x-2)} = \frac{8}{15}$$
$$15(2x - 2) = 8x(x - 2)$$
$$30x - 30 = 8x^2 - 16x$$
$$8x^2 - 46x + 30 = 0$$
$$4x^2 - 23x + 15 = 0$$
Factorising: $4x^2 - 20x - 3x + 15 = 0$
$$4x(x - 5) - 3(x - 5) = 0$$
$$(4x - 3)(x - 5) = 0$$
So $x = 5$ or $x = \dfrac{3}{4}$.
Since $x = \dfrac{3}{4}$ gives $(x - 2) < 0$, it is rejected.
∴ Smaller pipe takes 5 hours and larger pipe takes 3 hours to fill the tank separately.
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