The distance from Coe Hill to Calabogie is 150 kilometres. Pat leaves Coe Hill at 1:00 p.m. and drives at a speed of $80 \mathrm{~km} / \mathrm{h}$ for the first $60 \mathrm{~km}$. How fast must he travel for the remainder of the trip to reach Calabogie at 3:00 p.m.?
(A) $65 \mathrm{~km} / \mathrm{h}$
(B) $70 \mathrm{~km} / \mathrm{h}$
(C) $72 \mathrm{~km} / \mathrm{h}$
(D) $75 \mathrm{~km} / \mathrm{h}$
(E) $90 \mathrm{~km} / \mathrm{h}$
Since Pat drives $60 \mathrm{~km}$ at $80 \mathrm{~km} / \mathrm{h}$, this takes him $\frac{60 \mathrm{~km}}{80 \mathrm{~km} / \mathrm{h}}=\frac{3}{4} \mathrm{~h}$.

Since Pat has 2 hours in total to complete the trip, then he has $2-\frac{3}{4}=\frac{5}{4}$ hours left to complete the remaining $150-60=90 \mathrm{~km}$.

Therefore, he must travel at $\frac{90 \mathrm{~km}}{\frac{5}{4} \mathrm{~h}}=\frac{360}{5} \mathrm{~km} / \mathrm{h}=72 \mathrm{~km} / \mathrm{h}$.

ANSWER: (C)