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

Answer: A

Concepts:rate-time-distance delay simultaneous conditions

Difficulty rating: 2310

Small Hint:

Traveling at $\frac{4}{5}$ speed adds one-fourth of the normal time for the affected distance

Big Hint:

Compare the two delays; moving the accident point $80$ miles changes the delay by $1$ hour

Solution:

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.

1955 AMC 12 Problem 41

Answer: A

Difficulty rating: 2310

Small Hint:

Big Hint:

Solution:

Problem 41 in Other Years