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:

$400$

$465$

$600$

$640$

$550$

Answer: C

Concepts:distance rate and time relative speed system of equations

Difficulty rating: 2310

Small Hint:

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

Big Hint:

Use the two late-arrival equations to find first $\frac{L}{v}$ , then $\frac{90}{v}$

Solution:

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.

1954 AMC 12 Problem 48

Answer: C

Difficulty rating: 2310

Small Hint:

Big Hint:

Solution:

Problem 48 in Other Years