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Difference between revisions of "1999 AMC 8 Problems/Problem 22"

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Substituting <math>l</math> from the second equation back into the first gives us <math>3f=8r</math>. So each fish is worth <math>\frac{8}{3}</math> bags of rice, or <math>2 \frac{2}{3}\Rightarrow \boxed{D}</math>.
 
Substituting <math>l</math> from the second equation back into the first gives us <math>3f=8r</math>. So each fish is worth <math>\frac{8}{3}</math> bags of rice, or <math>2 \frac{2}{3}\Rightarrow \boxed{D}</math>.
  
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==See also==
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==Video Solution==
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https://youtu.be/JXQoEzh9eM4 Soo, DRMS, NM
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== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
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https://www.youtube.com/watch?v=Nm0zn-aXyl0
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==See Also==
 
{{AMC8 box|year=1999|num-b=21|num-a=23}}
 
{{AMC8 box|year=1999|num-b=21|num-a=23}}
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{{MAA Notice}}

Latest revision as of 21:28, 24 January 2024

Problem

In a far-off land three fish can be traded for two loaves of bread and a loaf of bread can be traded for four bags of rice. How many bags of rice is one fish worth?

$\text{(A)}\ \frac{3}{8} \qquad \text{(B)}\ \frac{1}{2} \qquad \text{(C)}\ \frac{3}{4} \qquad \text{(D)}\ 2\frac{2}{3} \qquad \text{(E)}\ 3\frac{1}{3}$

Solution

Let $f$ represent one fish, $l$ a loaf of bread, and $r$ a bag of rice. Then: $3f=2l$, $l=4r$

Substituting $l$ from the second equation back into the first gives us $3f=8r$. So each fish is worth $\frac{8}{3}$ bags of rice, or $2 \frac{2}{3}\Rightarrow \boxed{D}$.

Video Solution

https://youtu.be/JXQoEzh9eM4 Soo, DRMS, NM


Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=Nm0zn-aXyl0

See Also

1999 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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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