2002 AMC 10A Problems/Problem 12

Problem

Mr. Bird gets up every day at 8:00 AM to go to work. If he drives at an average speed of 40 miles per hour, he will be late by 3 minutes. If he drives at an average speed of 60 miles per hour, he will be early by 3 minutes. How many miles per hour does Mr. Bird need to drive to get to work exactly on time?

$\text{(A)}\ 45 \qquad \text{(B)}\ 48 \qquad \text{(C)}\ 50 \qquad \text{(D)}\ 55 \qquad \text{(E)} 58$

Solution 1

Let the time he needs to get there in be t and the distance he travels be d. From the given equations, we know that $d=\left(t+\frac{1}{20}\right)40$ and $d=\left(t-\frac{1}{20}\right)60$ . Setting the two equal, we have $40t+2=60t-2$ and we find $t=\frac{1}{4}$ of an hour. Substituting t back in, we find $d=12$ . From $d=rt$ , we find that r, and our answer, is $\boxed{\text{(B)}\ 48 }$ .

Solution 2

Since either time he arrives at is 3 minutes from the desired time, the answer is merely the harmonic mean of 40 and 60. The harmonic mean of a and b is $\frac{2}{\frac{1}{a}+\frac{1}{b}}=\frac{2ab}{a+b}$ . In this case, a and b are 40 and 60, so our answer is $\frac{4800}{100}=48$ , so $\boxed{\text{(B)}\ 48}$ .

2002 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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2002 AMC 10A Problems/Problem 12

Contents

Problem

Solution 1

Solution 2

See Also