Difference between revisions of "2007 AMC 8 Problems/Problem 6"

(One intermediate revision by one other user 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> \boxed{\textbf{(E)}\ 80%} </math>
+
so the percent decrease is <math>\frac{34}{41}</math> which is about <math> \boxed{\textbf{(E)}\ 80\%} </math>
  
 
==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}}

Revision as of 01:24, 5 July 2013

Problem

The average cost of a long-distance call in the USA in $1985$ was $41$ cents per minute, and the average cost of a long-distance call in the USA in $2005$ was $7$ cents per minute. Find the approximate percent decrease in the cost per minute of a long- distance call.

$\mathrm{(A)}\ 7 \qquad\mathrm{(B)}\ 17 \qquad\mathrm{(C)}\ 34 \qquad\mathrm{(D)}\ 41 \qquad\mathrm{(E)}\ 80$

Solution

The percent decrease is (the amount of decrease)/(original amount)

the amount of decrease is $41 - 7 = 34$

so the percent decrease is $\frac{34}{41}$ which is about $\boxed{\textbf{(E)}\ 80\%}$

See Also

2007 AMC 8 (ProblemsAnswer KeyResources)
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. AMC logo.png