8. A wire of length $95 \mathrm{~m}$ is cut into three pieces such that the length of each piece is $50 \%$ greater than the previous piece. What is the length of the largest piece?
A $36 \mathrm{~m}$
B $42 \mathrm{~m}$
C $45 \mathrm{~m}$
D $46 \mathrm{~m}$
E $48 \mathrm{~m}$
SolUtion
C

Let $v$ be the length of the smallest piece. Then the middle piece has length $\frac{3}{2} v$ and the largest has length $\frac{3}{2} \times \frac{3}{2} v=\frac{9}{4} v$. So the total length is $\frac{9}{4} v+\frac{3}{2} v+v=\frac{19}{4} v$ and hence $v=20 \mathrm{~m}$ and the longest piece has length $\frac{9}{4} \nu$, that is $45 \mathrm{~m}$.