Difference between revisions of "2014 AMC 10A Problems/Problem 15"

m (Solution 1)
(Solution 2 (Answer Choices))
Line 23: Line 23:
 
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==
 +
 +
Let us call the total number of miles after David changes speed <math>x</math>. We realize that if David is driving 50 miles an hour, and he arrives 30 minutes early, <math>x\equiv 25\pmod{50}</math>. The only solution that fits the description is <math>\boxed{210}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 20:16, 6 April 2018

The following problem is from both the 2014 AMC 12A #11 and 2014 AMC 10A #15, so both problems redirect to this page.

Problem

David drives from his home to the airport to catch a flight. He drives $35$ miles in the first hour, but realizes that he will be $1$ hour late if he continues at this speed. He increases his speed by $15$ miles per hour for the rest of the way to the airport and arrives $30$ minutes early. How many miles is the airport from his home?

$\textbf{(A) }140\qquad \textbf{(B) }175\qquad \textbf{(C) }210\qquad \textbf{(D) }245\qquad \textbf{(E) }280\qquad$


Solution 1

Note that he drives at $50$ miles per hour after the first hour and continues doing so until he arrives.

Let $d$ be the distance still needed to travel after the first $1$ hour. We have that $\dfrac{d}{50}+1.5=\dfrac{d}{35}$, where the $1.5$ comes from $1$ hour late decreased to $0.5$ hours early.

Simplifying gives $7d+525=10d$, or $d=175$.

Now, we must add an extra $35$ miles traveled in the first hour, giving a total of $\boxed{\textbf{(C)} \: 210}$ 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 $\textbf{(A)}$ or choice $\textbf{(B)}$ work, but $\textbf{(C)}$ does. We can verify as follows. After $1$ hour at $35 \text{ mph}$, David has $175$ miles left. This then takes him $3.5$ hours at $50 \text{ mph}$. But $210/35 = 6 \text{ hours}$. Since $1+3.5 = 4.5 \text{ hours}$ is $1.5$ hours less than $6$, our answer is $\boxed{\textbf{(C)} \: 210}$.



Solution 3

Let us call the total number of miles after David changes speed $x$. We realize that if David is driving 50 miles an hour, and he arrives 30 minutes early, $x\equiv 25\pmod{50}$. The only solution that fits the description is $\boxed{210}$.

See Also

2014 AMC 10A (ProblemsAnswer KeyResources)
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 (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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS