2004 AMC 8 Problem 16

Attempt Problem 16 of the 2004 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

16.

Two 600600 mL pitchers contain orange juice. One pitcher is 13\frac 13 full and the other pitcher is 25\frac 25 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?
18\dfrac{1}{8}
316\dfrac{3}{16}
1130\dfrac{11}{30}
1119\dfrac{11}{19}
1115\dfrac{11}{15}
Answer: C
Concepts:mixturefraction
Small hint:
Compute the orange juice in each pitcher before water is added.
Big hint:
After filling both pitchers, the total mixture volume is 12001200 mL.
Solution:
The first pitcher contains 600⋅13=200600 \cdot \dfrac{1}{3} = 200 mL of orange juice. The second one has 600⋅25=240600 \cdot \dfrac{2}{5} = 240 mL.
The large container then has 200+240=440200 + 240 = 440 mL of orange juice. The total amount of mixture is 2⋅600=12002 \cdot 600 = 1200 mL.
Then the fraction of orange juice is 4401200=1130.\dfrac{440}{1200} = \dfrac{11}{30}.
Thus, C is the correct answer.
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