Difference between revisions of "2002 AMC 8 Problems/Problem 13"
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==Solution== | ==Solution== | ||
| − | + | 1^3=1 | |
| + | 2^3=8 | ||
| + | 8*125=1000 | ||
| + | \boxed{\text{(E)}\ 1000}$. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2002|num-b=12|num-a=14}} | {{AMC8 box|year=2002|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 13:49, 26 October 2016
Problem
For his birthday, Bert gets a box that holds 125 jellybeans when filled to capacity. A few weeks later, Carrie gets a larger box full of jellybeans. Her box is twice as high, twice as wide and twice as long as Bert's. Approximately, how many jellybeans did Carrie get?
Solution
1^3=1 2^3=8 8*125=1000 \boxed{\text{(E)}\ 1000}$.
See Also
| 2002 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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. 
