Answer
The pressure reduces by a factor of 4.3.
Work Step by Step
$P=F/A \qquad $15-2
In both cases, F is the portion of weight with which the crutch acts on the floor. We assume its equal in both cases.
Use the index "0" for pressure exerted and area without the tip, and index "1" with the rubber tip attached.
We want $k=\displaystyle \frac{P_{0}}{P_{1}}\quad $(how much larger is $\mathrm{P}_{0}$ than $\mathrm{P}_{1}$)
$k=\displaystyle \frac{\frac{F}{A_{0}}}{\frac{F}{A_{1}}}=\frac{A_{1}}{A_{0}}=\frac{r_{1}^{2}\pi}{r_{0}^{2}\pi}=\frac{(2.5\mathrm{c}\mathrm{m})^{2}}{(1.2\mathrm{c}\mathrm{m})^{2}}=4.3$
The pressure reduces by a factor of 4.3.
