Difference between revisions of "2009 AMC 10B Problems/Problem 8"
(→Solution 2) |
|||
| Line 16: | Line 16: | ||
== Solution 2 == | == Solution 2 == | ||
| − | Without loss of generality, we can assume the price at the beginning of January was <math>\$100</math>. 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. In order for the price to be the same as it was at the beginning of January (<math>\$100</math>), the price must decrease by <math>\$20</math>. 20 is <math>\frac{1}{6}</math>th 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 | + | Without loss of generality, we can assume the price at the beginning of January was <math>\$100</math>. |
| − | + | ||
| + | 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. | ||
| + | |||
| + | In order for the price to be the same as it was at the beginning of January (<math>\$100</math>), the price must decrease by <math>\$20</math>. | ||
| + | |||
| + | 20 is <math>\frac{1}{6}</math>th 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 == | ||
{{AMC10 box|year=2009|ab=B|num-b=7|num-a=9}} | {{AMC10 box|year=2009|ab=B|num-b=7|num-a=9}} | ||
{{AMC12 box|year=2009|ab=B|num-b=6|num-a=8}} | {{AMC12 box|year=2009|ab=B|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:11, 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.
Contents
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
![\[x = 100 \cdot \frac{0.2p}{1.2p} = \frac {100}{6} \approx 16.7.\]](http://latex.artofproblemsolving.com/f/7/a/f7a133f61c84005792fb54e3b863469ff7dc7de6.png)
So to the nearest integer is 
. The answer is 
.
Solution 2
Without loss of generality, we can assume the price at the beginning of January was 
.
When it rose by , it became 
, when it fell by , it became 
, and when it rose by , it became again.
In order for the price to be the same as it was at the beginning of January (), the price must decrease by 
.
20 is 
th of 120, and 
So to the nearest integer, 
and the answer is 
. ~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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
