Difference between revisions of "2014 AMC 8 Problems/Problem 17"
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<math>\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }8\qquad\textbf{(D) }10\qquad \textbf{(E) }12</math> | <math>\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }8\qquad\textbf{(D) }10\qquad \textbf{(E) }12</math> | ||
− | ==Solution== | + | ==Solution 1== |
− | Note that on a normal day, it takes him <math>1/3</math> hour to get to school. However, today it took <math>\frac{1/2 \text{ mile}}{2 \text{ mph}}=1/4</math> hour to walk the first <math>1/2</math> mile. That means that he has <math>1/3 -1/4 = 1/12</math> hours left to get to school, and <math>1/2</math> mile left to go. Therefore, his speed must be <math>\frac{1/2 \text{ mile}}{1/12 \text { hour}}=\boxed{6 \text{ mph}}</math>, so <math>\text{(B)}</math> is the answer. | + | Note that on a normal day, it takes him <math>1/3</math> hour to get to school. However, today it took <math>\frac{1/2 \text{ mile}}{2 \text{ mph}}=1/4</math> hour to walk the first <math>1/2</math> mile. That means that he has <math>1/3 -1/4 = 1/12</math> hours left to get to school, and <math>1/2</math> mile left to go. Therefore, his speed must be <math>\frac{1/2 \text{ mile}}{1/12 \text { hour}}=\boxed{6 \text{ mph}}</math>, so <math>\boxed{\text{(B) }6}</math> is the answer. |
− | ==Solution | + | ==Solution 2== |
Using the harmonic mean formula, and making the speed he needs to take to get to school for the last half <math>x</math>, we can make the expression: | Using the harmonic mean formula, and making the speed he needs to take to get to school for the last half <math>x</math>, we can make the expression: | ||
− | <math>\frac{ | + | <math>\frac{2 * 2x}{2 + x}</math> |
Which simplifies to: | Which simplifies to: | ||
− | <math>\frac{ | + | <math>\frac{4x}{2 + x}</math> |
− | Then, because we know that since he goes at <math>3</math> mph on a normal day, we can say that it is the harmonic mean of the two rates he goes at today, so we add that to | + | Then, because we know that since he goes at <math>3</math> mph on a normal day, we can say that it is the harmonic mean of the two rates he goes at today, so we add that to our expression and turn it into something familiar: |
− | <math>\frac{ | + | <math>\frac{4x}{2 + x} = 3</math> |
Solving that equation gives us: | Solving that equation gives us: | ||
− | <math>\boxed{\text { (B)} 6}</math> | + | <math>\boxed{\text {(B) } 6}</math> |
==See Also== | ==See Also== |
Latest revision as of 19:58, 4 November 2019
Contents
Problem
George walks mile to school. He leaves home at the same time each day, walks at a steady speed of miles per hour, and arrives just as school begins. Today he was distracted by the pleasant weather and walked the first mile at a speed of only miles per hour. At how many miles per hour must George run the last mile in order to arrive just as school begins today?
Solution 1
Note that on a normal day, it takes him hour to get to school. However, today it took hour to walk the first mile. That means that he has hours left to get to school, and mile left to go. Therefore, his speed must be , so is the answer.
Solution 2
Using the harmonic mean formula, and making the speed he needs to take to get to school for the last half , we can make the expression:
Which simplifies to:
Then, because we know that since he goes at mph on a normal day, we can say that it is the harmonic mean of the two rates he goes at today, so we add that to our expression and turn it into something familiar:
Solving that equation gives us:
See Also
2014 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 AJHSME/AMC 8 Problems and Solutions |
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