Difference between revisions of "2007 AMC 8 Problems/Problem 6"
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so the percent decrease is <math>\frac{34}{41}</math> which is about <math> \boxed{\textbf{(E)}\ 80\%} </math> | so the percent decrease is <math>\frac{34}{41}</math> which is about <math> \boxed{\textbf{(E)}\ 80\%} </math> | ||
| + | |||
| + | ==Video Solution by SpreadTheMathLove== | ||
| + | https://www.youtube.com/watch?v=omFpSGMWhFc | ||
| + | |||
| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/UtFO7ICGLgI | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2007|num-b=5|num-a=7}} | {{AMC8 box|year=2007|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 16:35, 28 October 2024
Contents
Problem
The average cost of a long-distance call in the USA in 
was
cents per minute, and the average cost of a long-distance
call in the USA in 
was 
cents per minute. Find the
approximate percent decrease in the cost per minute of a long-
distance call.
Solution
The percent decrease is (the amount of decrease)/(original amount)
the amount of decrease is 
so the percent decrease is 
which is about 
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=omFpSGMWhFc
Video Solution by WhyMath
See Also
| 2007 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
