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

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== Problem ==
 
== Problem ==
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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gas was hijacked expensive
  
<math>\mathrm{(A)}\ 12\qquad
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<math>\mathrm{(A)}\ 10000\qquad
\mathrm{(B)}\ 17\qquad
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\mathrm{(B)}\ 1700000\qquad
\mathrm{(C)}\ 20\qquad
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\mathrm{(C)}\ 20000002\qquad
\mathrm{(D)}\ 25\qquad
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\mathrm{(D)}\ 25000040000\qquad
\mathrm{(E)}\ 35</math>
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\mathrm{(E)}\ 350</math>
  
 
== Solution ==
 
== Solution ==

Revision as of 09:20, 24 May 2021

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

Problem

gas was hijacked expensive

$\mathrm{(A)}\ 10000\qquad \mathrm{(B)}\ 1700000\qquad \mathrm{(C)}\ 20000002\qquad \mathrm{(D)}\ 25000040000\qquad \mathrm{(E)}\ 350$

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 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 $\mathrm{(B)}$.

See also

2009 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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