1973 AMC 12 Problem 27

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

Car

A

’s average is the harmonic mean of

u

and

v,

while Car

B

’s is their arithmetic mean

Subtract the two means and factor the numerator as a square

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.

Problem 27 in other years