Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2011 AMC 10A Problems/Problem 15"

(→‎Solution 2)
(→‎Solution 2)
Line 21: Line 21:
 
Multiplying both sides by <math>0.02 (d - 40)</math>, we get
 
Multiplying both sides by <math>0.02 (d - 40)</math>, we get
 
<cmath>d = 55 \cdot 0.02 \cdot (d - 40) = 1.1d - 44.</cmath>
 
<cmath>d = 55 \cdot 0.02 \cdot (d - 40) = 1.1d - 44.</cmath>
−
Then <math>0.1d = 44</math>, so <math>d = \boxed{440}</math>. The answer is <math>\mathbf{(C)}}</math>
+
Then <math>0.1d = 44</math>, so <math>d = \boxed{440}</math>. The answer is <math>\mathrm{(C)}\\qquad</math>
  
 
==Video Solution==
 
==Video Solution==

Revision as of 13:53, 12 July 2021

Problem 15

Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first $40$ miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of $0.02$ gallons per mile. On the whole trip he averaged $55$ miles per gallon. How long was the trip in miles?

$\mathrm{(A)}\ 140 \qquad \mathrm{(B)}\ 240 \qquad \mathrm{(C)}\ 440 \qquad \mathrm{(D)}\ 640 \qquad \mathrm{(E)}\ 840$

Solution 1

We know that $\frac{\text{total miles}}{\text{total gas}}=55$. Let $x$ be the distance the car traveled during the time it ran on gasoline, then the amount of gas used is $0.02x$. The total distance traveled is $40+x$, so we get $\frac{40+x}{0.02x}=55$. Solving this equation, we get $x=400$, so the total distance is $400 + 40 = \boxed{440 \ \mathbf{(C)}}$.

Solution 2

Let $d$ be the length of the trip in miles. Roy used no gasoline for the 40 first miles, then used 0.02 gallons of gasoline per mile on the remaining $d - 40$ miles, for a total of $0.02 (d - 40)$ gallons. Hence, his average mileage was \[\frac{d}{0.02 (d - 40)} = 55.\] Multiplying both sides by $0.02 (d - 40)$, we get \[d = 55 \cdot 0.02 \cdot (d - 40) = 1.1d - 44.\] Then $0.1d = 44$, so $d = \boxed{440}$. The answer is $\mathrm{(C)}\\qquad$

Video Solution

https://youtu.be/HQmkIPpuIEc

~savannahsolver

See Also

2011 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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_15&oldid=158263"