The sum of two natural numbers is 15 and the sum of their reciprocals is

Let the required natural numbers be x and (15 − x).

According to the given condition,

$\frac{1}{x}+\frac{1}{15-x}=\frac{3}{10}$

$\Rightarrow \frac{15-x+x}{x(15-x)}=\frac{3}{10}$

$\Rightarrow \frac{15}{15 x-x^{2}}=\frac{3}{10}$

$\Rightarrow 15 x-x^{2}=50$

$\Rightarrow x^{2}-15 x+50=0$

$\Rightarrow x^{2}-10 x-5 x+50=0$

$\Rightarrow x(x-10)-5(x-10)=0$

$\Rightarrow(x-5)(x-10)=0$

$\Rightarrow x-5=0$ or $x-10=0$

$\Rightarrow x=5$ or $x=10$

When x = 5,
15 − x = 15 − 5 = 10

When x = 10,
15 − x = 15 − 10 = 5

Hence, the required natural numbers are 5 and 10.