Difference between revisions of "2007 AMC 8 Problems/Problem 6"
(Created page with '== Problem == The average cost of a long-distance call in the USA in <math>1985</math> was <math>41</math> cents per minute, and the average cost of a long-distance call in the …') |
|||
(7 intermediate revisions by 6 users not shown) | |||
Line 15: | Line 15: | ||
the amount of decrease is <math>41 - 7 = 34</math> | the amount of decrease is <math>41 - 7 = 34</math> | ||
− | so the percent decrease is <math>\frac{34}{41}</math> which is about <math>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== | ||
+ | {{AMC8 box|year=2007|num-b=5|num-a=7}} | ||
+ | {{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.