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

(Formatting: added problem and see also box)
m
Line 13: Line 13:
 
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==
+
==See Also==
 
{{AMC8 box|year=1999|num-b=21|num-a=23}}
 
{{AMC8 box|year=1999|num-b=21|num-a=23}}

Revision as of 14:04, 23 December 2012

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 $2 \frac{2}{3}\Rightarrow \boxed{D}$.

See Also

1999 AMC 8 (ProblemsAnswer KeyResources)
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