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

(Solution)
(Solution 2)
 
(13 intermediate revisions by 6 users not shown)
Line 5: Line 5:
 
<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=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>.
  
<math>5+40=\boxed{\textbf{(E) }45}</math> ~quacker88
+
<math>5+40=\boxed{\textbf{(E) }45}</math>.
 +
 
 +
~quacker88
  
 
==Solution 2==
 
==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.
+
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 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.
+
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 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.
+
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>.
  
==Video Solution==
+
~BrightPorcupine
  
Check It Out! :)
+
~MrThinker
Education, the study of everything  
+
 
 +
==Video Solution by Education, the study of everything==
 
https://www.youtube.com/watch?v=ExEfaIOqt_w
 
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
  
~IceMatrix
+
==Video Solution by WhyMath==
 
 
 
https://youtu.be/FcPO4EXDwzc
 
https://youtu.be/FcPO4EXDwzc
  
 
~savannahsolver
 
~savannahsolver
 +
 +
==Video Solution by AlexExplains==
 +
https://www.youtube.com/watch?v=GNPAgQ8fSP0&t=1s
 +
 +
~AlexExplains
  
 
==See Also==
 
==See Also==

Latest revision as of 15:14, 12 July 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

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

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 (ProblemsAnswer KeyResources)
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. AMC logo.png