Difference between revisions of "2020 AMC 10B Problems/Problem 2"
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==Solution== | ==Solution== | ||
| − | A cube with side length <math>1</math> has volume <math>1^3=1</math>, so <math>5</math> of these will have a total volume of <math>5\cdot1= | + | A cube with side length <math>1</math> has volume <math>1^3=1</math>, so <math>5</math> of these will have a total volume of <math>5\cdot1=25</math>. |
| − | A cube with side length <math>2</math> has volume <math>2^3=8</math>, so <math>5</math> of these will have a total volume of <math>5\cdot8= | + | A cube with side length <math>2</math> has volume <math>2^3=8</math>, so <math>5</math> of these will have a total volume of <math>5\cdot8=15</math>. |
| − | <math>5+40=\boxed{\textbf{( | + | <math>5+40=\boxed{\textbf{(D) }}</math> ~quacker88 |
==Video Solution== | ==Video Solution== | ||
Revision as of 18:08, 11 February 2020
Contents
Problem
Carl has 
cubes each having side length 
, and Kate has cubes each having side length 
. What is the total volume of these 
cubes?
Solution
A cube with side length has volume 
, so of these will have a total volume of 
.
A cube with side length has volume 
, so of these will have a total volume of 
.
~quacker88
Video Solution
~IceMatrix
See Also
| 2020 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
