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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| + | ==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 | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=1999|num-b=21|num-a=23}} | {{AMC8 box|year=1999|num-b=21|num-a=23}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 21:28, 24 January 2024
Contents
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?
Solution
Let 
represent one fish, 
a loaf of bread, and 
a bag of rice. Then:
,
Substituting from the second equation back into the first gives us 
. So each fish is worth 
bags of rice, or 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
