1954 AMC 12 Problem 48

Question

A train, an hour after starting, meets with an accident which detains it a half hour, after which it proceeds at

\dfrac34

of its former rate and arrives

3\dfrac12

hours late. Had the accident happened

90

miles farther along the line, it would have arrived only

3

hours late. The length of the trip in miles was:

400

465

600

640

550

Let the normal speed be

v

and the trip length be

L

; slowing to

\frac{3v}{4}

adds one-third of the normal time on the affected distance

Use the two late-arrival equations to find first

\frac{L}{v}

, then

\frac{90}{v}

Let the normal speed be

v

and the trip length be

L.

Traveling a distance at

\frac{3v}{4}

rather than

v

adds one-third of its normal travel time. In the first case,

\frac12+\frac13\left(\frac Lv-1\right)=\frac72,

so

\frac{L}{v}=10.

If the accident happens

90

miles farther along,

\frac12+\frac13\left(\frac Lv-1-\frac{90}{v}\right)=3.

Substituting

\frac{L}{v}=10

gives

\frac{90}{v}=\frac{3}{2},

hence

v=60

and

L=10v=600.

Thus, the correct answer is C.

Problem 48 in other years