Difference between revisions of "2002 AMC 8 Problems/Problem 10"
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/* AMC8 2002 #8, 9, 10 Problem */ | /* AMC8 2002 #8, 9, 10 Problem */ | ||
size(3inch, 1.5inch); | size(3inch, 1.5inch); | ||
| − | for ( int y = 0; y & | + | for ( int y = 0; y <= 5; ++y ) |
{ | { | ||
draw((0,y)--(18,y)); | draw((0,y)--(18,y)); | ||
| Line 59: | Line 59: | ||
= \frac{0.72+0.72+0.24+0.65}{43}\\ | = \frac{0.72+0.72+0.24+0.65}{43}\\ | ||
= \frac{2.33}{43} \approx \boxed{\text{(E)}\ 5.5\ \text{cents}}</cmath> | = \frac{2.33}{43} \approx \boxed{\text{(E)}\ 5.5\ \text{cents}}</cmath> | ||
| + | |||
| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/61m1NMsMCRI | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2002|num-b=9|num-a=11}} | {{AMC8 box|year=2002|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 13:30, 29 October 2024
Problem
Problems 8,9 and 10 use the data found in the accompanying paragraph and table:
Juan organizes the stamps in his collection by country and by the decade in which they were issued. The prices he paid for them at a stamp shop were: Brazil and France, 6 cents each, Peru 4 cents each, and Spain 5 cents each. (Brazil and Peru are South American countries and France and Spain are in Europe.)
![[asy] /* AMC8 2002 #8, 9, 10 Problem */ size(3inch, 1.5inch); for ( int y = 0; y <= 5; ++y ) { draw((0,y)--(18,y)); } draw((0,0)--(0,5)); draw((6,0)--(6,5)); draw((9,0)--(9,5)); draw((12,0)--(12,5)); draw((15,0)--(15,5)); draw((18,0)--(18,5)); draw(scale(0.8)*"50s", (7.5,4.5)); draw(scale(0.8)*"4", (7.5,3.5)); draw(scale(0.8)*"8", (7.5,2.5)); draw(scale(0.8)*"6", (7.5,1.5)); draw(scale(0.8)*"3", (7.5,0.5)); draw(scale(0.8)*"60s", (10.5,4.5)); draw(scale(0.8)*"7", (10.5,3.5)); draw(scale(0.8)*"4", (10.5,2.5)); draw(scale(0.8)*"4", (10.5,1.5)); draw(scale(0.8)*"9", (10.5,0.5)); draw(scale(0.8)*"70s", (13.5,4.5)); draw(scale(0.8)*"12", (13.5,3.5)); draw(scale(0.8)*"12", (13.5,2.5)); draw(scale(0.8)*"6", (13.5,1.5)); draw(scale(0.8)*"13", (13.5,0.5)); draw(scale(0.8)*"80s", (16.5,4.5)); draw(scale(0.8)*"8", (16.5,3.5)); draw(scale(0.8)*"15", (16.5,2.5)); draw(scale(0.8)*"10", (16.5,1.5)); draw(scale(0.8)*"9", (16.5,0.5)); label(scale(0.8)*"Country", (3,4.5)); label(scale(0.8)*"Brazil", (3,3.5)); label(scale(0.8)*"France", (3,2.5)); label(scale(0.8)*"Peru", (3,1.5)); label(scale(0.8)*"Spain", (3,0.5)); label(scale(0.9)*"Juan's Stamp Collection", (9,0), S); label(scale(0.9)*"Number of Stamps by Decade", (9,5), N);[/asy]](http://latex.artofproblemsolving.com/f/6/7/f679bb00d960506089f9cd632a80bd218347de4e.png)
The average price of his '70s stamps is closest to
Solution
The price of all the stamps in the '70s together over the total number of stamps is equal to the average price.
![\[\frac{(12)(0.06)+(12)(0.06)+(6)(0.04)+(13)(0.05)}{12+12+6+13}\\ = \frac{0.72+0.72+0.24+0.65}{43}\\ = \frac{2.33}{43} \approx \boxed{\text{(E)}\ 5.5\ \text{cents}}\]](http://latex.artofproblemsolving.com/5/1/1/51154c56d06f4969528699d870649d0d24f811c6.png)
Video Solution by WhyMath
See Also
| 2002 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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. 
