Difference between revisions of "2004 AMC 8 Problems/Problem 16"

(Created page with "==Problem== Two <math>600</math> mL pitchers contain orange juice. One pitcher is <math>1/3</math> full and the other pitcher is <math>2/5</math> full. Water is added to fill eac...")
 
 
(One intermediate revision by one other user not shown)
Line 6: Line 6:
 
==Solution==
 
==Solution==
 
The first pitcher contains <math>600 \cdot \frac13 = 200</math> mL of orange juice. The second pitcher contains <math>600 \cdot \frac25 = 240</math> mL of orange juice. In the large pitcher, there is a total of <math>200+240=440</math> mL of orange juice and <math>600+600=1200</math> mL of fluids, giving a fraction of <math>\frac{440}{1200} = \boxed{\textbf{(C)}\ \frac{11}{30}}</math>.
 
The first pitcher contains <math>600 \cdot \frac13 = 200</math> mL of orange juice. The second pitcher contains <math>600 \cdot \frac25 = 240</math> mL of orange juice. In the large pitcher, there is a total of <math>200+240=440</math> mL of orange juice and <math>600+600=1200</math> mL of fluids, giving a fraction of <math>\frac{440}{1200} = \boxed{\textbf{(C)}\ \frac{11}{30}}</math>.
 +
 +
==Video Solution==
 +
https://youtu.be/Dit3DvLEVJY Soo, DRMS, NM
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2004|num-b=15|num-a=17}}
 
{{AMC8 box|year=2004|num-b=15|num-a=17}}
 +
{{MAA Notice}}

Latest revision as of 01:50, 25 March 2022

Problem

Two $600$ mL pitchers contain orange juice. One pitcher is $1/3$ full and the other pitcher is $2/5$ full. Water is added to fill each pitcher completely, then both pitchers are poured into one large container. What fraction of the mixture in the large container is orange juice?

$\textbf{(A)}\ \frac18 \qquad \textbf{(B)}\ \frac{3}{16} \qquad \textbf{(C)}\ \frac{11}{30} \qquad \textbf{(D)}\ \frac{11}{19}\qquad \textbf{(E)}\ \frac{11}{15}$

Solution

The first pitcher contains $600 \cdot \frac13 = 200$ mL of orange juice. The second pitcher contains $600 \cdot \frac25 = 240$ mL of orange juice. In the large pitcher, there is a total of $200+240=440$ mL of orange juice and $600+600=1200$ mL of fluids, giving a fraction of $\frac{440}{1200} = \boxed{\textbf{(C)}\ \frac{11}{30}}$.

Video Solution

https://youtu.be/Dit3DvLEVJY Soo, DRMS, NM

See Also

2004 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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. AMC logo.png

Invalid username
Login to AoPS