Difference between revisions of "2000 AMC 10 Problems/Problem 3"
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| − | + | ==Problem== | |
| − | <math>x | + | Each day, Jenny ate <math>20\%</math> of the jellybeans that were in her jar at the beginning of that day. At the end of the second day, <math>32</math> remained. How many jellybeans were in the jar originally? |
| + | |||
| + | <math>\mathrm{(A)}\ 40 \qquad\mathrm{(B)}\ 50 \qquad\mathrm{(C)}\ 55 \qquad\mathrm{(D)}\ 60 \qquad\mathrm{(E)}\ 75</math> | ||
| + | |||
| + | ==Solution== | ||
| + | |||
| + | <math>x(.8)(.8)=32</math> | ||
| + | |||
| + | <math>x(.64)=32</math> | ||
<math>x \cdot \frac{32}{50}=32</math> | <math>x \cdot \frac{32}{50}=32</math> | ||
| − | <math>x=50</math> | + | <math>x=50</math> |
| + | |||
| + | <math>\boxed{\text{B}}</math> | ||
| + | |||
| + | ==See Also== | ||
| − | + | {{AMC10 box|year=2000|num-b=2|num-a=4}} | |
Revision as of 18:58, 8 January 2009
Problem
Each day, Jenny ate 
of the jellybeans that were in her jar at the beginning of that day. At the end of the second day, 
remained. How many jellybeans were in the jar originally?
Solution
See Also
| 2000 AMC 10 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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 AMC 10 Problems and Solutions | ||
