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> | ||
| − | ==Solution== | + | ==Solution 1== |
| − | 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=5</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=40</math>. |
| − | <math>5+40=\boxed{\textbf{( | + | <math>5+40=\boxed{\textbf{(E) }45}</math>. |
| − | ==Video Solution== | + | ~quacker88 |
| + | |||
| + | ==Solution 2 (more in depth about Solution 1)== | ||
| + | |||
| + | The total volume of Carl's cubes is <math>5</math>. 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 <math>1 \times 1 \times 1</math>. This is equal to <math>1</math>. Since Carl has <math>5</math> cubes, you will have to multiply <math>1</math> by <math>5</math>, to account for all the <math>5</math> 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 <math>2</math>, so it will be <math>2 \times 2 \times 2 = 8.</math> Now you have to multiply by <math>5</math> to account for all the <math>5</math> blocks. This is <math>40</math>. So the total volume of Kate's cubes is <math>40</math>. | ||
| + | |||
| + | Lastly, to find the total of Carl's and Kate's cubes, you must add the total volume of their cubes together. This is going to be <math>5+40=\boxed{\textbf{(E)} ~45}</math>. | ||
| + | |||
| + | ~BrightPorcupine | ||
| + | |||
| + | ~MrThinker | ||
| + | |||
| + | ~<B+ | ||
| + | |||
| + | ==Video Solution by Education, the study of everything== | ||
| + | https://www.youtube.com/watch?v=ExEfaIOqt_w | ||
| + | |||
| + | ~Education, the study of everything | ||
| + | |||
| + | ==Video Solution by TheBeautyofMath== | ||
https://youtu.be/Gkm5rU5MlOU | https://youtu.be/Gkm5rU5MlOU | ||
| − | ~ | + | ==Video Solution by WhyMath== |
| + | https://youtu.be/FcPO4EXDwzc | ||
| + | |||
| + | ~savannahsolver | ||
| + | |||
| + | ==Video Solution by AlexExplains== | ||
| + | https://www.youtube.com/watch?v=GNPAgQ8fSP0&t=1s | ||
| + | |||
| + | ~AlexExplains | ||
==See Also== | ==See Also== | ||
Latest revision as of 07:26, 15 August 2024
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 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 (more in depth about Solution 1)
The total volume of Carl's cubes is . 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 
. This is equal to . Since Carl has cubes, you will have to multiply by , to account for all the 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 , so it will be 
Now you have to multiply by to account for all the blocks. This is 
. So the total volume of Kate's cubes is .
Lastly, to find the total of Carl's and Kate's cubes, you must add the total volume of their cubes together. This is going to be 
.
~BrightPorcupine
~MrThinker
~<B+
Video Solution by Education, the study of everything
https://www.youtube.com/watch?v=ExEfaIOqt_w
~Education, the study of everything
Video Solution by TheBeautyofMath
Video Solution by WhyMath
~savannahsolver
Video Solution by AlexExplains
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. 
