MathDB
Aytown to Beetown

Source:

February 11, 2009

Problem Statement

A train traveling from Aytown to Beetown meets with an accident after 1 1 hr. It is stopped for 12 \frac{1}{2} hr., after which it proceeds at four-fifths of its usual rate, arriving at Beetown 2 2 hr. late. If the train had covered 80 80 miles more before the accident, it would have been just 1 1 hr. late. The usual rate of the train is: <spanclass=latexbold>(A)</span> 20 mph<spanclass=latexbold>(B)</span> 30 mph<spanclass=latexbold>(C)</span> 40 mph<spanclass=latexbold>(D)</span> 50 mph<spanclass=latexbold>(E)</span> 60 mph <span class='latex-bold'>(A)</span>\ \text{20 mph} \qquad <span class='latex-bold'>(B)</span>\ \text{30 mph} \qquad <span class='latex-bold'>(C)</span>\ \text{40 mph} \qquad <span class='latex-bold'>(D)</span>\ \text{50 mph} \qquad <span class='latex-bold'>(E)</span>\ \text{60 mph}