Difference between revisions of "2002 AMC 8 Problems/Problem 24"

Latest revision as of 09:15, 14 June 2024

Problem

Miki has a dozen oranges of the same size and a dozen pears of the same size. Miki uses her juicer to extract 8 ounces of pear juice from 3 pears and 8 ounces of orange juice from 2 oranges. She makes a pear-orange juice blend from an equal number of pears and oranges. What percent of the blend is pear juice?

$\text{(A)}\ 30\qquad\text{(B)}\ 40\qquad\text{(C)}\ 50\qquad\text{(D)}\ 60\qquad\text{(E)}\ 70$

Solution 1

A pear gives $8/3$ ounces of juice per pear. An orange gives $8/2=4$ ounces of juice per orange. If the pear-orange juice blend used one pear and one orange each, the percentage of pear juice would be

$\frac{8/3}{8/3+4} \times 100 = \frac{8}{8+12} \times 100 = \boxed{\text{(B)}\ 40}$

Solution 2

Since it doesn't matter how many pears and oranges there are, you can make the number of them whatever you like. In this case, we could use $6$ , because it's the LCM of $2$ and $3$ . Then for the $6$ pears, there are $6/3*8=16$ ounces of pear juice. For the 6 oranges, there are $6/2*8=24$ ounces of orange juice. Since we are looking for the percent of pear juice, we need to do $16/(16+24)=16/40$ . Simplifying, we get $2/5$ . Hence the answer is $\boxed{\text{(B)}\ 40}$ .

Video Solution

https://www.youtube.com/watch?v=TarSWoN3ne4 ~David

2002 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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.

@@ Line 13: / Line 13: @@
 ==Video Solution==
-https://youtu.be/KqOE0AvrRs8 Soo, DRMS, NM
-https://www.youtube.com/watch?v=TarSWoN3ne4
+https://www.youtube.com/watch?v=TarSWoN3ne4  ~David
 ==See Also==
 {{AMC8 box|year=2002|num-b=23|num-a=25}}
 {{MAA Notice}}

Page

Toolbox

Search