2002 AMC 8 Problems/Problem 10

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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]

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.

\[\frac{(12)(0.06)+(12)(0.06)+(6)(0.04)+(13)(0.05)}{12+12+6+13}\\ = \frac{0.72+0.72+0.24+0.65}{43}\\ = \frac{2.33}{43} \approx \boxed{\text{(E)}\ 5.4\ \text{cents}}\]

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

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