Question

In order to pass $B$ going $40$ mph on a two-lane highway $A,$ going $50$ mph, must gain $30$ feet. Meantime, $C,$ $210$ feet from $A,$ is headed toward him at $50$ mph. If $B$ and $C$ maintain their speeds, then, in order to pass safely, $A$ must increase his speed by:

$30$ mph

$10$ mph

$5$ mph

$15$ mph

$3$ mph

Answer: C

Concepts:relative speed rate-time-distance passing

Difficulty rating: 1870

Small Hint:

Let $v$ be $A$ ’s new speed and equate the passing time to the time before $A$ and $C$ meet

Big Hint:

Use relative distances and speeds: $\frac{30}{v-40}=\frac{210}{v+50}$

Solution:

Let $v$ be $A$ ’s new speed. Relative to $B,$ $A$ gains at $v-40$ mph, so the passing time is proportional to $\frac{30}{v-40}.$ $A$ and $C$ close their $210$ -foot separation at $v+50$ mph, so their meeting time is proportional to $\frac{210}{v+50}.$ At the limiting safe time, $\frac{30}{v-40}=\frac{210}{v+50}.$ Thus $30v+1500=210v-8400,$ giving $v=55$ mph. $A$ must increase his original speed by $5$ mph.

Therefore, the correct answer is C.

1955 AMC 12 Problem 50

Answer: C

Difficulty rating: 1870

Small Hint:

Big Hint:

Solution:

Problem 50 in Other Years