Difference between revisions of "1999 AMC 8 Problems/Problem 22"
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+ | ==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? | ||
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+ | <math>\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}</math> | ||
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+ | ==Solution== | ||
+ | |||
Let <math>f</math> represent one fish, <math>l</math> a loaf of bread, and <math>r</math> a bag of rice. Then: | Let <math>f</math> represent one fish, <math>l</math> a loaf of bread, and <math>r</math> a bag of rice. Then: | ||
<math>3f=2l</math>, | <math>3f=2l</math>, | ||
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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>. | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC8 box|year=1999|num-b=21|num-a=23}} | ||
+ | {{MAA Notice}} |
Latest revision as of 23:34, 4 July 2013
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 .
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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