Difference between revisions of "2002 AMC 8 Problems/Problem 10"

 
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The price of all the stamps in the '70s together over the total number of stamps is equal to the average price.
 
The price of all the stamps in the '70s together over the total number of stamps is equal to the average price.
  
<cmath>\frac{(12)(0.06)+(12)(0.06)+(6)(0.04)+(13)(0.05)}{12+12+6+13}\\
+
<cmath>
= \frac{0.72+0.72+0.24+0.65}{43}\\
+
\begin{align*}
= \frac{2.33}{43} \approx \boxed{\text{(E)}\ 5.5\ \text{cents}}</cmath>
+
\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}}
 +
\end{align*}
 +
</cmath>
  
 
==Video Solution by WhyMath==
 
==Video Solution by WhyMath==
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{{AMC8 box|year=2002|num-b=9|num-a=11}}
 
{{AMC8 box|year=2002|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
[[Category:Introductory Algebra Problems]]

Latest revision as of 17:50, 25 June 2025

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 &lt;= 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]

The average price of his '70s stamps is closest to

$\text{(A)}\ 3.5 \text{ cents} \qquad \text{(B)}\ 4 \text{ cents} \qquad \text{(C)}\ 4.5 \text{ cents} \qquad \text{(D)}\ 5 \text{ cents} \qquad \text{(E)}\ 5.5 \text{ cents}$

Solution

The price of all the stamps in the '70s together over the total number of stamps is equal to the average price.

\begin{align*} \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}} \end{align*}

Video Solution by WhyMath

https://youtu.be/61m1NMsMCRI

See Also

2002 AMC 8 (ProblemsAnswer KeyResources)
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png