Difference between revisions of "2002 AMC 8 Problems/Problem 13"

(Solution)
(Solution)
Line 6: Line 6:
  
 
==Solution==
 
==Solution==
1^3=1
+
<math>1^3=1
 
2^3=8
 
2^3=8
 
8*125=1000
 
8*125=1000
\boxed{\text{(E)}\ 1000}$.
+
\boxed{\text{(E)}\ 1000}</math>.
  
 
==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 22:25, 30 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?

$\text{(A)}\ 250\qquad\text{(B)}\ 500\qquad\text{(C)}\ 625\qquad\text{(D)}\ 750\qquad\text{(E)}\ 1000$

Solution

$1^3=1 2^3=8 8*125=1000 \boxed{\text{(E)}\ 1000}$.

See Also

2002 AMC 8 (ProblemsAnswer KeyResources)
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. AMC logo.png