Difference between revisions of "2014 AMC 10A Problems/Problem 15"
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Note that he drives at <math>50</math> miles per hour after the first hour and continues doing so until he arrives. | Note that he drives at <math>50</math> miles per hour after the first hour and continues doing so until he arrives. | ||
− | Let <math>d</math> be the distance still needed to travel after | + | Let <math>d</math> be the distance still needed to travel after <math>1</math> hour. We have that <math>\dfrac{d}{50}+1.5=\dfrac{d}{35}</math>, where the <math>1.5</math> comes from <math>1</math> hour late decreased to <math>0.5</math> hours early. |
Simplifying gives <math>7d+525=10d</math>, or <math>d=175</math>. | Simplifying gives <math>7d+525=10d</math>, or <math>d=175</math>. | ||
− | Now, we must add an extra <math>35</math> miles traveled in the first hour, giving a total of <math>\boxed{\textbf{(C)} | + | Now, we must add an extra <math>35</math> miles traveled in the first hour, giving a total of <math>\boxed{\textbf{(C) } 210}</math> miles. |
==Solution 2 (Answer Choices)== | ==Solution 2 (Answer Choices)== | ||
Instead of spending time thinking about how one can set up an equation to solve the problem, one can simply start checking the answer choices. | Instead of spending time thinking about how one can set up an equation to solve the problem, one can simply start checking the answer choices. | ||
Quickly checking, we know that neither choice <math>\textbf{(A)}</math> or choice <math>\textbf{(B)}</math> work, but <math>\textbf{(C)}</math> does. We can verify as follows. After <math>1</math> hour at <math>35 \text{ mph}</math>, David has <math>175</math> miles left. This then takes him <math>3.5</math> hours at <math>50 \text{ mph}</math>. But <math>210/35 = 6 \text{ hours}</math>. Since <math>1+3.5 = 4.5 \text{ hours}</math> is <math>1.5</math> hours less than <math>6</math>, our answer is <math>\boxed{\textbf{(C)} \: 210}</math>. | Quickly checking, we know that neither choice <math>\textbf{(A)}</math> or choice <math>\textbf{(B)}</math> work, but <math>\textbf{(C)}</math> does. We can verify as follows. After <math>1</math> hour at <math>35 \text{ mph}</math>, David has <math>175</math> miles left. This then takes him <math>3.5</math> hours at <math>50 \text{ mph}</math>. But <math>210/35 = 6 \text{ hours}</math>. Since <math>1+3.5 = 4.5 \text{ hours}</math> is <math>1.5</math> hours less than <math>6</math>, our answer is <math>\boxed{\textbf{(C)} \: 210}</math>. | ||
+ | |||
+ | ==Solution 3 (similar to Solution 1 but using a different equation)== | ||
+ | |||
+ | Let the total distance be <math>d</math>. Then <math>d=35(t+1)</math>. Since 1 hour has passed, and he increased his speed by <math>15</math> miles per hour, and he had already traveled <math>35</math> miles, the new equation is <math>d-35=50(t-1-\frac{1}{2})=50(t-\frac{3}{2})</math>. Solving, <math>d=35t+35=50t-40</math>, <math>15t=75</math>, <math>t=5</math>, and <math>d=35(5+1)=35\cdot 6=210 \Longrightarrow \boxed{\textbf{(C) } 210}</math>. | ||
+ | |||
+ | ~JH. L | ||
==See Also== | ==See Also== |
Latest revision as of 04:18, 20 June 2022
- The following problem is from both the 2014 AMC 12A #11 and 2014 AMC 10A #15, so both problems redirect to this page.
Contents
Problem
David drives from his home to the airport to catch a flight. He drives miles in the first hour, but realizes that he will be
hour late if he continues at this speed. He increases his speed by
miles per hour for the rest of the way to the airport and arrives
minutes early. How many miles is the airport from his home?
Solution 1
Note that he drives at miles per hour after the first hour and continues doing so until he arrives.
Let be the distance still needed to travel after
hour. We have that
, where the
comes from
hour late decreased to
hours early.
Simplifying gives , or
.
Now, we must add an extra miles traveled in the first hour, giving a total of
miles.
Solution 2 (Answer Choices)
Instead of spending time thinking about how one can set up an equation to solve the problem, one can simply start checking the answer choices.
Quickly checking, we know that neither choice or choice
work, but
does. We can verify as follows. After
hour at
, David has
miles left. This then takes him
hours at
. But
. Since
is
hours less than
, our answer is
.
Solution 3 (similar to Solution 1 but using a different equation)
Let the total distance be . Then
. Since 1 hour has passed, and he increased his speed by
miles per hour, and he had already traveled
miles, the new equation is
. Solving,
,
,
, and
.
~JH. L
See Also
2014 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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 |
2014 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.