Difference between revisions of "2020 AMC 10B Problems/Problem 2"
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<math>\textbf{(A)}\ 24 \qquad\textbf{(B)}\ 25 \qquad\textbf{(C)}\ 28 \qquad\textbf{(D)}\ 40 \qquad\textbf{(E)}\ 45</math> | <math>\textbf{(A)}\ 24 \qquad\textbf{(B)}\ 25 \qquad\textbf{(C)}\ 28 \qquad\textbf{(D)}\ 40 \qquad\textbf{(E)}\ 45</math> | ||
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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=5</math>. | 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=5</math>. | ||
Revision as of 04:40, 17 May 2022
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 1
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
Solution 2
The total volume of Carl's cubes is 5. This is because to find the volume of a cube or a rectangular prism, you have to multiply the height by the length by the width. So in this question, it would be 1*1*1. This is equal to 1. Since Carl has 5 cubes, you will have to multiply 1 by 5, to account for all the 5 cubes.
Next, to find the total volume of Kate's cubes you have to do the same thing. Except, this time, the height, the width, and the length, are all 2, so it will be 2^3. This is equal to 8. Now you have to multiply by 5 to account for all the 5 blocks. This is 40. So the total volume of Kate's cubes are 40.
Lastly, to find the total of both Carl's and Kate's cubes, all you have to do it add the total volume of these people. This is going to be 5+40. And this is 45. - BrightPorcupine
Video Solution
Check It Out! :) Education, the study of everything https://www.youtube.com/watch?v=ExEfaIOqt_w
~IceMatrix
~savannahsolver
https://www.youtube.com/watch?v=GNPAgQ8fSP0&t=1s
~AlexExplains
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. 
