$\frac{7}{15}$ of $\frac{5}{3}$ of a number is greater than $\frac{7}{9}$ of $\frac{3}{5}$ of the same number by $9$. Find the quadruple of the number.

Let the number be $x$.According to the question:
$$\left(\frac{7}{15}\times \frac{5}{3}\times x\right)-\left(\frac{7}{9}\times \frac{3}{5}\times x\right)=9$$
Simplifying the terms:
$$\left(\frac{7}{9}\times x\right)-\left(\frac{7}{15}\times x\right)=9$$
Taking the LCM of $9$ and $15$, which is $45$:
$$\frac{35x-21x}{45}=9$$
$$\frac{14x}{45}=9$$
$$14x=9\times 45$$
$$14x=405$$
$$x=\frac{405}{14}$$
We need to find the quadruple of the number, which is $4x$:
$$4x=4\times \frac{405}{14}=\frac{810}{7}\approx \mathrm{115.71428...}$$
Rounding to two decimal places gives $115.72$.