Question

Cars $A$ and $B$ travel the same distance. Car $A$ travels half that distance at $u$ miles per hour and half at $v$ miles per hour. Car $B$ travels half the time at $u$ miles per hour and half at $v$ miles per hour. The average speed of Car $A$ is $x$ miles per hour and that of Car $B$ is $y$ miles per hour. Then we always have

$x\le y$

$x\ge y$

$x=y$

$x\lt y$

$x\gt y$

Answer: A

Concepts:harmonic mean inequality algebraic manipulation

Difficulty rating: 1800

Small Hint:

Car $A$ ’s average is the harmonic mean of $u$ and $v,$ while Car $B$ ’s is their arithmetic mean

Big Hint:

Subtract the two means and factor the numerator as a square

Solution:

Car $A$ ’s equal-distance average and Car $B$ ’s equal-time average are $x=\frac{2uv}{u+v}, \qquad y=\frac{u+v}{2}.$ Their difference is $\begin{aligned} y-x &=\frac{(u+v)^2-4uv}{2(u+v)}\\ &=\frac{(u-v)^2}{2(u+v)}\\ &\ge0. \end{aligned}$ Thus $x\le y,$ with equality possible when $u=v.$

Therefore, the correct answer is A.

1973 AMC 12 Problem 27

Answer: A

Difficulty rating: 1800

Small Hint:

Big Hint:

Solution:

Problem 27 in Other Years