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

m (Solution)
m (Solution 2)
 
(18 intermediate revisions by 9 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) }}</math> ~quacker88
+
<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
  
~IceMatrix
+
==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

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 (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