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Difference between revisions of "2009 AMC 10B Problems/Problem 8"

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== Problem ==
 
== Problem ==
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gas was hijacked expensive
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In a certain year the price of gasoline rose by <math>20\%</math> during January, fell by <math>20\%</math> during February, rose by <math>25\%</math> during March, and fell by <math>x\%</math> during April. The price of gasoline at the end of April was the same as it had been at the beginning of January. To the nearest integer, what is <math>x</math>
  
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<math>\mathrm{(A)}\ 10000\qquad
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<math>\mathrm{(A)}\ 12\qquad
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\mathrm{(B)}\ 1700000\qquad
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\mathrm{(B)}\ 17\qquad
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\mathrm{(C)}\ 20000002\qquad
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\mathrm{(C)}\ 20\qquad
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\mathrm{(D)}\ 25000040000\qquad
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\mathrm{(D)}\ 25\qquad
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\mathrm{(E)}\ 350</math>
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\mathrm{(E)}\ 35</math>
  
 
== Solution ==
 
== Solution ==
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Let <math>p</math> be the price at the beginning of January.  The price at the end of March was <math>(1.2)(0.8)(1.25)p = 1.2p.</math> Because the price at the of April was <math>p</math>, the price decreased by <math>0.2p</math> during April, and the percent decrease was
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Let <math>p</math> be the price at the beginning of January.  The price at the end of March was <math>(1.2)(0.8)(1.25)p = 1.2p.</math> Because the price at the end of April was <math>p</math>, the price decreased by <math>0.2p</math> during April, and the percent decrease was
 
<cmath>x = 100 \cdot \frac{0.2p}{1.2p} = \frac {100}{6} \approx 16.7.</cmath>
 
<cmath>x = 100 \cdot \frac{0.2p}{1.2p} = \frac {100}{6} \approx 16.7.</cmath>
 
So to the nearest integer <math>x</math> is <math>\boxed{17}</math>.  The answer is <math>\mathrm{(B)}</math>.
 
So to the nearest integer <math>x</math> is <math>\boxed{17}</math>.  The answer is <math>\mathrm{(B)}</math>.
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== Solution 2 ==
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Without loss of generality, we can assume the price at the beginning of January was <math>\$100</math>.
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When it rose by <math>20\%</math>, it became <math>\$120</math>, when it fell by <math>20\%</math>, it became <math>\$96</math>, and when it rose by <math>25\%</math>, it became <math>\$120</math> again.
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In order for the price at the end of April to be the same as it was at the beginning of January (<math>\$100</math>), the price must decrease by <math>\$20</math>.
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20 is <math>\frac{1}{6}th</math> of 120, and <math>\frac{1}{6} \approx 0.167 \approx 17\%</math> So to the nearest integer, <math>x = 17</math> and the answer is <math>\boxed{\textbf{(B) } 17}</math>. ~azc1027
  
 
== See also ==
 
== See also ==

Latest revision as of 17:15, 14 June 2023

The following problem is from both the 2009 AMC 10B #8 and 2009 AMC 12B #7, so both problems redirect to this page.

Problem

In a certain year the price of gasoline rose by $20\%$ during January, fell by $20\%$ during February, rose by $25\%$ during March, and fell by $x\%$ during April. The price of gasoline at the end of April was the same as it had been at the beginning of January. To the nearest integer, what is $x$

$\mathrm{(A)}\ 12\qquad \mathrm{(B)}\ 17\qquad \mathrm{(C)}\ 20\qquad \mathrm{(D)}\ 25\qquad \mathrm{(E)}\ 35$

Solution

Let $p$ be the price at the beginning of January. The price at the end of March was $(1.2)(0.8)(1.25)p = 1.2p.$ Because the price at the end of April was $p$, the price decreased by $0.2p$ during April, and the percent decrease was \[x = 100 \cdot \frac{0.2p}{1.2p} = \frac {100}{6} \approx 16.7.\] So to the nearest integer $x$ is $\boxed{17}$. The answer is $\mathrm{(B)}$.

Solution 2

Without loss of generality, we can assume the price at the beginning of January was $$100$.

When it rose by $20\%$, it became $$120$, when it fell by $20\%$, it became $$96$, and when it rose by $25\%$, it became $$120$ again.

In order for the price at the end of April to be the same as it was at the beginning of January ($$100$), the price must decrease by $$20$.

20 is $\frac{1}{6}th$ of 120, and $\frac{1}{6} \approx 0.167 \approx 17\%$ So to the nearest integer, $x = 17$ and the answer is $\boxed{\textbf{(B) } 17}$. ~azc1027

See also

2009 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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
2009 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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

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