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

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===Solution 2===
 
===Solution 2===
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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>.
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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.  
 +
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>.
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 +
(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.)
  
 
===Solution 3===
 
===Solution 3===
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Let x be equal to the total amount of distance he needs to cover. Let y be equal to the amount of time he would travel correctly.
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Setting up a system of equations, <math>\frac x{40} -3 = y</math> and <math>\frac x{60} +3 = y</math>
  
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A more general form of the argument in Solution 2, with proof:
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Solving, we get x = 720 and y = 15.
  
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Let <math>d</math> be the distance to work, and let <math>s</math> be the correct average speed. Then the time needed to get to work is <math>t=\frac ds</math>.
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We divide x by y to get the average speed, <math>\frac {720}{15} = 48</math>. Therefore, the answer is <math>\boxed{\textbf{(B) }48}</math>.
  
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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>.
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~MathKatana
  
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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>.
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== Video Solution ==
 +
https://youtu.be/ZpYOnfqm5Bc
  
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(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 by Daily Dose of Math==
 +
 
 +
https://youtu.be/S17zidbQYWo
 +
 
 +
~Thesmartgreekmathdude
  
 
==See Also==
 
==See Also==

Latest revision as of 11:34, 10 August 2024

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. 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.)

Solution 3

Let x be equal to the total amount of distance he needs to cover. Let y be equal to the amount of time he would travel correctly. Setting up a system of equations, $\frac x{40} -3 = y$ and $\frac x{60} +3 = y$

Solving, we get x = 720 and y = 15.

We divide x by y to get the average speed, $\frac {720}{15} = 48$. Therefore, the answer is $\boxed{\textbf{(B) }48}$.

~MathKatana

Video Solution

https://youtu.be/ZpYOnfqm5Bc

Video Solution by Daily Dose of Math

https://youtu.be/S17zidbQYWo

~Thesmartgreekmathdude

See Also

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

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