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

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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?
 
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?
  
<math>\text{(A)}\ 45 \qquad \text{(B)}\ 48 \qquad \text{(C)}\ 50 \qquad \text{(D)}\ 55 \qquad \text{(E)} 58</math>
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<math>\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 55 \qquad \textbf{(E)}\ 58</math>
  
 
==Solution==
 
==Solution==
 
===Solution 1===
 
===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 <math>d=\left(t+\frac{1}{20}\right)40</math> and <math>d=\left(t-\frac{1}{20}\right)60</math>. Setting the two equal, we have <math>40t+2=60t-3</math> and we find <math>t=\frac{1}{4}</math> of an hour. Substituting t back in, we find <math>d=12</math>. From <math>d=rt</math>, we find that r, and our answer, is <math>\boxed{\text{(B)}\ 48 }</math>.
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Let the time he needs to get there in be <math>t</math> and the distance he travels be <math>d</math>. From the given equations, we know that <math>d=\left(t+\frac{1}{20}\right)40</math> and <math>d=\left(t-\frac{1}{20}\right)60</math>. Setting the two equal, we have <math>40t+2=60t-3</math> and we find <math>t=\frac{1}{4}</math> of an hour. Substituting t back in, we find <math>d=12</math>. From <math>d=rt</math>, we find that <math>r</math>, our answer, is <math>\boxed{\textbf{(B) }48 }</math>.
  
 
===Solution 2===
 
===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 <math>\frac{2}{\frac{1}{a}+\frac{1}{b}}=\frac{2ab}{a+b}</math>. In this case, a and b are 40 and 60, so our answer is <math>\frac{4800}{100}=48</math>, so <math>\boxed{\text{(B)}\ 48}</math>.
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Since either time he arrives at is <math>3</math> minutes from the desired time, the answer is merely the [[harmonic mean]] of 40 and 60. The harmonic mean of a and b is <math>\frac{2}{\frac{1}{a}+\frac{1}{b}}=\frac{2ab}{a+b}</math>. In this case, a and b are 40 and 60, so our answer is <math>\frac{4800}{100}=48</math>, so <math>\boxed{\textbf{(B)}\ 48}</math>.
  
 
===Solution 3===
 
===Solution 3===
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We know that <math>t+\frac 3{60} = \frac d{40}</math> and <math>t-\frac 3{60} = \frac d{60}</math>. Summing these two equations, we get: <math>2t = \frac d{40} + \frac d{60}</math>.  
 
We know that <math>t+\frac 3{60} = \frac d{40}</math> and <math>t-\frac 3{60} = \frac d{60}</math>. Summing these two equations, we get: <math>2t = \frac d{40} + \frac d{60}</math>.  
  
Substituting <math>t=\frac ds</math> and dividing both sides by <math>d</math>, we get <math>\frac 2s = \frac 1{40} + \frac 1{60}</math>, hence <math>s=\boxed{48}</math>.
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Substituting <math>t=\frac ds</math> and dividing both sides by <math>d</math>, we get <math>\frac 2s = \frac 1{40} + \frac 1{60}</math>, hence <math>s=\boxed{\textbf{(B) }48}</math>.
  
 
(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.)
 
(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.)

Latest revision as of 12:18, 8 November 2021

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.)

See Also

2002 AMC 12A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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