1955 AMC 12 Problem 41

Question

A train traveling from Aytown to Beetown meets with an accident after

1

hr. It is stopped for

\dfrac12

hr., after which it proceeds at four-fifths of its usual rate, arriving at Beetown

2

hr. late. If the train had covered

80

miles more before the accident, it would have been just

1

hr. late. The usual rate of the train is:

20

mph

30

mph

40

mph

50

mph

60

mph

Traveling at

\frac{4}{5}

speed adds one-fourth of the normal time for the affected distance

Compare the two delays; moving the accident point

80

miles changes the delay by

1

hour

Let the usual rate be

R

mph and the total distance be

D.

After the first hour, the normal time for the remaining distance is

\frac{D}{R}-1.

Traveling it at

\frac{4R}{5}

adds one-fourth of that time, so

\frac12+\frac14\left(\frac DR-1\right)=2.

If the accident occurs

80

miles later,

\frac12+\frac14\left(\frac DR-1-\frac{80}{R}\right)=1.

Subtracting the second equation from the first gives

\frac{20}{R}=1,

hence

R=20

mph.

Thus, the correct answer is A.

Problem 41 in other years