Difference between revisions of "2002 AMC 8 Problems/Problem 24"
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==Solution 2== | ==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 <math>6</math>, because it's the LCM of <math>2</math> and <math>3</math>. Then for the <math>6</math> pears, there are <math>6/3*8=16</math> ounces of pear juice. For the 6 oranges, there are <math>6/2*8=24</math> ounces of orange juice. Since we are looking for the percent of pear juice, we need to do <math>16/(16+24)=16/40</math>. Simplifying, we get <math>2/5</math>. Hence the answer is <math>\boxed{\text{(B)}\ 40}</math>. | 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 <math>6</math>, because it's the LCM of <math>2</math> and <math>3</math>. Then for the <math>6</math> pears, there are <math>6/3*8=16</math> ounces of pear juice. For the 6 oranges, there are <math>6/2*8=24</math> ounces of orange juice. Since we are looking for the percent of pear juice, we need to do <math>16/(16+24)=16/40</math>. Simplifying, we get <math>2/5</math>. Hence the answer is <math>\boxed{\text{(B)}\ 40}</math>. | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/KqOE0AvrRs8 Soo, DRMS, NM | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2002|num-b=23|num-a=25}} | {{AMC8 box|year=2002|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 11:41, 27 March 2022
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?
Solution 1
A pear gives 
ounces of juice per pear. An orange gives 
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
![\[\frac{8/3}{8/3+4} \times 100 = \frac{8}{8+12} \times 100 = \boxed{\text{(B)}\ 40}\]](http://latex.artofproblemsolving.com/4/5/9/45978beb64182b3338319ac23975cd8e57986031.png)
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 
, because it's the LCM of 
and 
. Then for the pears, there are 
ounces of pear juice. For the 6 oranges, there are 
ounces of orange juice. Since we are looking for the percent of pear juice, we need to do 
. Simplifying, we get 
. Hence the answer is 
.
Video Solution
https://youtu.be/KqOE0AvrRs8 Soo, DRMS, NM
See Also
| 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. 
