During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.

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

m (Reverted edits by Whitelisted (talk) to last revision by Nathan wailes)
(Tag: Rollback)
(Solution)
 
Line 11: Line 11:
  
 
== Solution ==
 
== Solution ==
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
+
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>.

Latest revision as of 14:42, 1 July 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

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 $\mathrm{(B)}$.

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

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

Invalid username
Login to AoPS