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

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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>
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Then <math>0.1d = 44</math>, so <math>d = \boxed{440}</math>. The answer is $\mathrm{(C)}\$
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Then <math>0.1d = 44</math>, so <math>d = \boxed{440}</math>. The answer is <math>mathrm{(C)}</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 $mathrm{(C)}$

Video Solution

https://youtu.be/HQmkIPpuIEc

~savannahsolver

See Also

2011 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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All AMC 10 Problems and Solutions

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