Difference between revisions of "2014 AMC 10A Problems/Problem 15"
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| − | ==Solution 1 | + | ==Solution 1== |
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>\textbf{(C) }210</math> miles. | + | 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 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>. |
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| + | ==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== | ||
{{AMC10 box|year=2014|ab=A|num-b=14|num-a=16}} | {{AMC10 box|year=2014|ab=A|num-b=14|num-a=16}} | ||
| + | {{AMC12 box|year=2014|ab=A|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
| + | |||
| + | [[Category: Introductory Algebra Problems]] | ||
Latest revision as of 03:18, 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. 
