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

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== Problem 15 ==
 
== Problem 15 ==
  
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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?
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Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first <math>40</math> miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of <math>0.02</math> gallons per mile. On the whole trip he averaged <math>55</math> miles per gallon. How long was the trip in miles?
  
 
<math>
 
<math>
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</math>
 
</math>
  
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== Solution ==
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== Solution 1 ==
  
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We know that <math>\frac{\text{total miles}}{\text{total gas}}=55</math>. Let <math>x</math> be the distance the car traveled during the time it ran on gasoline, then the amount of gas used is <math>0.02x</math>. The total distance traveled is <math>40+x</math>, so we get <math>\frac{40+x}{0.02x}=55</math>. Solving this equation, we get <math>x=400</math>, so the total distance is <math>400 + 40 = \boxed{440 \ \mathbf{(C)}}</math>.
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We know that <math>\frac{\text{total miles}}{\text{total gas}}=55</math>. Let <math>x</math> be the distance in miles the car traveled during the time it ran on gasoline, then the amount of gas used is <math>0.02x</math>. The total distance traveled is <math>40+x</math>, so we get <math>\frac{40+x}{0.02x}=55</math>. Solving this equation, we get <math>x=400</math>, so the total distance is <math>400 + 40 = \boxed{440 \ \mathbf{(C)}}</math>.
  
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Solution 2  
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== Solution 2 ==
  
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The answer has to be divisble by 55, and the only answer that is divisible by 55 is C. so it's C.
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Let <math>d</math> 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 <math>d - 40</math> miles, for a total of <math>0.02 (d - 40)</math> gallons. Hence, his average mileage was
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<cmath>\frac{d}{0.02 (d - 40)} = 55.</cmath>
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Multiplying both sides by <math>0.02 (d - 40)</math>, we get
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<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 <math>(C)</math>.
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== Solution 3 (Answer choices) ==
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Since Roy had 0 gasoline usage during the first 40 miles, and we know his usage is 0.02 gallons per mile after that, we can look at all 5 individual answers and simply run them through.
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If Roy drove (A) 140 miles would mean he spent 140 - 40 = 100 miles on gasoline, so he used 100 * 0.02 = 2 gallons. <math>2/140</math> = 70 miles per gallon which does not equal 55, so (A) is incorrect. Similarly, (B) 240 gives:
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240 - 40 = 200 miles on gasoline, 200 * 0.02 gallons = 4 gallons, <math>4/240</math> = 60 m/g which does not equal 55.
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Plugging in (C) 440, we get:
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440 - 40 = 400, 400 * 0.02 = 8, <math>8/440</math> = 55 miles per gallon, so answer choice <math>\boxed{440 \ \mathbf{(C)}}</math> is correct.
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"And we're done."
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*Richard Rusczyk outro*
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~@BeepTheSheep954
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==Video Solution==
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https://youtu.be/HQmkIPpuIEc
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 +
~savannahsolver
  
 
== See Also ==
 
== See Also ==
 
{{AMC10 box|year=2011|ab=A|num-b=14|num-a=16}}
 
{{AMC10 box|year=2011|ab=A|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:18, 16 July 2024

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 in miles 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 $(C)$.


Solution 3 (Answer choices)

Since Roy had 0 gasoline usage during the first 40 miles, and we know his usage is 0.02 gallons per mile after that, we can look at all 5 individual answers and simply run them through.

If Roy drove (A) 140 miles would mean he spent 140 - 40 = 100 miles on gasoline, so he used 100 * 0.02 = 2 gallons. $2/140$ = 70 miles per gallon which does not equal 55, so (A) is incorrect. Similarly, (B) 240 gives: 240 - 40 = 200 miles on gasoline, 200 * 0.02 gallons = 4 gallons, $4/240$ = 60 m/g which does not equal 55.

Plugging in (C) 440, we get: 440 - 40 = 400, 400 * 0.02 = 8, $8/440$ = 55 miles per gallon, so answer choice $\boxed{440 \ \mathbf{(C)}}$ is correct.

"And we're done."

  • Richard Rusczyk outro*

~@BeepTheSheep954

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

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