Question

After being treated with chemotherapy, the radius of the tumor decreased by 23%. What is the corresponding percentage decrease in the volume of the tumor

Answer 1

Answer:

The volume of the tumor experimented a decrease of 54.34 percent.

Step-by-step explanation:

Let suppose that tumor has an spherical geometry, whose volume ($V$) is calculated by:

$V = \frac{4\pi}{3}\cdot R^{3}$

Where $R$ is the radius of the tumor.

The percentage decrease in the volume of the tumor ($\%V$) is expressed by:

$\%V = \frac{\Delta V}{V_{o}} \times 100\,\%$

Where:

$\Delta V$ - Absolute decrease in the volume of the tumor.

$V_{o}$ - Initial volume of the tumor.

The absolute decrease in the volume of the tumor is:

$\Delta V = V_{o}-V_{f}$

$\Delta V = \frac{4\pi}{3}\cdot (R_{f}^{3}-R_{o}^{3})$

The percentage decrease is finally simplified:

$\%V = \left[1-\left(\frac{R_{f}}{R_{o}}\right)^{3} \right]\times 100\,\%$

Given that $R_{o} = R$ and $R_{f} = 0.77\cdot R$, the percentage decrease in the volume of tumor is:

$\%V = \left[1-\left(\frac{0.77\cdot R}{R}\right)^{3} \right]\times 100\,\%$

$\%V = 54.34\,\%$

The volume of the tumor experimented a decrease of 54.34 percent.