Difference between revisions of "2002 AMC 8 Problems/Problem 13"
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==Solution== | ==Solution== | ||
| − | + | Since the volume ratio is equal to the sides ratio cubed, then the ratio of the larger box's volume to the smaller one is 2 cubed. | |
| − | 2^3=8 | + | |
| − | 8 | + | <math>2^3=8</math> |
| − | \boxed{\text{(E)}\ 1000} | + | |
| + | Now multiply 125 (the number of jellybeans that Bert's box can hold) by 8. | ||
| + | |||
| + | <math>8\cdot125= | ||
| + | \boxed{\text{(E)}\ 1000}</math>. | ||
| + | |||
| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/JnXzlMhq6pI | ||
==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}} | ||
Latest revision as of 13:31, 29 October 2024
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
Since the volume ratio is equal to the sides ratio cubed, then the ratio of the larger box's volume to the smaller one is 2 cubed.
Now multiply 125 (the number of jellybeans that Bert's box can hold) by 8.
.
Video Solution by WhyMath
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. 
