Difference between revisions of "2009 AMC 10B Problems/Problem 8"
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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 | + | 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 during January, fell by during February, rose by during March, and fell by 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
Solution
Let be the price at the beginning of January. The price at the end of March was Because the price at the end of April was , the price decreased by during April, and the percent decrease was So to the nearest integer is . The answer is .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.