Difference between revisions of "2002 AMC 12A Problems/Problem 11"

Revision as of 19:44, 25 August 2022

The following problem is from both the 2002 AMC 12A #11 and 2002 AMC 10A #12, so both problems redirect to this page.

Problem

Mr. Earl E. 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?

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

Solution

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 $40t+2=60t-3$ and we find $t=\frac{1}{4}$ of an hour. Substituting t back in, we find $d=12$ . From $d=rt$ , we find that $r$ , our answer, is $\boxed{\textbf{(B) }48 }$ .

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 $\boxed{\textbf{(B)}\ 48}$ .

Solution 3

A more general form of the argument in Solution 2, with proof:

Let $d$ be the distance to work, and let $s$ be the correct average speed. Then the time needed to get to work is $t=\frac ds$ .

We know that $t+\frac 3{60} = \frac d{40}$ and $t-\frac 3{60} = \frac d{60}$ . Summing these two equations, we get: $2t = \frac d{40} + \frac d{60}$ .

Substituting $t=\frac ds$ and dividing both sides by $d$ , we get $\frac 2s = \frac 1{40} + \frac 1{60}$ , hence $s=\boxed{\textbf{(B) }48}$ .

(Note that this approach would work even if the time by which he is late was different from the time by which he is early in the other case - we would simply take a weighted sum in step two, and hence obtain a weighted harmonic mean in step three.)

Video Solution

https://youtu.be/ZpYOnfqm5Bc

2002 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

2002 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 25: / Line 25: @@
 (Note that this approach would work even if the time by which he is late was different from the time by which he is early in the other case - we would simply take a weighted sum in step two, and hence obtain a weighted harmonic mean in step three.)
+== Video Solution ==
+https://youtu.be/ZpYOnfqm5Bc
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