1973 AHSME Problems/Problem 27


Cars A and B travel the same distance. Care 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

$\textbf{(A)}\ x \leq y\qquad \textbf{(B)}\ x \geq y \qquad \textbf{(C)}\ x=y \qquad \textbf{(D)}\ x<y\qquad \textbf{(E)}\ x>y$


Let $t$ be the total number of time in hours that Car B took to drive the distance. This means that for have the time Car B traveled $\tfrac{ut}{2}$ miles and for half the time Car B traveled $\tfrac{vt}{2}$ miles. That means the total distance traveled is $\tfrac{ut+vt}{2}$ miles, so \[y = \frac{\frac{ut+vt}{2}}{t}\] \[y = \frac{u+v}{2}\] Because both cars traveled the same distance, for half the distance Car A took $\tfrac{ut+vt}{4u}$ hours and for half the distance Car A took $\tfrac{ut+vt}{4v}$ hours. That means \[x = \frac{\frac{ut+vt}{2}}{\frac{ut+vt}{4u} + \tfrac{ut+vt}{4v}}\] \[x = \frac{2}{\frac{1}{u} + \frac{1}{v}}\] In order to compare $x$ and $y$, we need to compare the values that $x$ and $y$ are equal to. \[\frac{2}{\frac{1}{u} + \frac{1}{v}} \bigcirc \frac{u+v}{2}\] Since the denominators of both numbers are positive, cross-multiplying won’t change the comparison sign. \[4 \bigcirc 2 + \frac{u}{v} + \frac{v}{u}\] \[2 \bigcirc  \frac{u}{v} + \frac{v}{u}\] Because $uv$ is positive, the comparison sign does not need to be changed either. \[2uv \bigcirc u^2 + v^2\] \[0 \bigcirc (u-v)^2\] By the Trivial Inequality, $0 \le (u-v)^2$. All of the steps are reversible, so $\boxed{\textbf{(A)}\ x \leq y}$. This can be confirmed by testing values of $u$ and $v$.

See Also

1973 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 26
Followed by
Problem 28
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