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Difference between revisions of "2020 AMC 10B Problems/Problem 2"

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==Problem==
 
==Problem==
  
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What is the value of  
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Carl has <math>5</math> cubes each having side length <math>1</math>, and Kate has <math>5</math> cubes each having side length <math>2</math>. What is the total volume of these <math>10</math> cubes?
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<cmath>1-(-2)-3-(-4)-5-(-6)?</cmath>
 
  
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<math>\textbf{(A)}\ -20 \qquad\textbf{(B)}\ -3 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 21</math>
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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>
  
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==Solution==
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==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=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>.
  
 
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>.
 
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>.
  
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<math>5+40=\boxed{\textbf{(E) }21}</math> ~quacker88
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<math>5+40=\boxed{\textbf{(E) }45}</math>.
 +
 
 +
~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
 +
 
 +
==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==
  
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{{AMC12 box|year=2020|ab=B|num-b=1|num-a=3}}
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{{AMC10 box|year=2020|ab=B|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 07:26, 15 August 2024

Problem

Carl has $5$ cubes each having side length $1$, and Kate has $5$ cubes each having side length $2$. What is the total volume of these $10$ cubes?

$\textbf{(A)}\ 24 \qquad\textbf{(B)}\ 25 \qquad\textbf{(C)}\ 28 \qquad\textbf{(D)}\ 40 \qquad\textbf{(E)}\ 45$

Solution 1

A cube with side length $1$ has volume $1^3=1$, so $5$ of these will have a total volume of $5\cdot1=5$.

A cube with side length $2$ has volume $2^3=8$, so $5$ of these will have a total volume of $5\cdot8=40$.

$5+40=\boxed{\textbf{(E) }45}$.

~quacker88

Solution 2 (more in depth about Solution 1)

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 $1 \times 1 \times 1$. 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 $2 \times 2 \times 2 = 8.$ 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 is $40$.

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 $5+40=\boxed{\textbf{(E)} ~45}$.

~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

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

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

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