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

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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) }21}</math> ~quacker88
+
<math>5+40=\boxed{\textbf{(E) }45}</math> ~quacker88
 +
 
 +
==Video Solution==
 +
https://youtu.be/Gkm5rU5MlOU
 +
 
 +
~IceMatrix
 +
 
 +
https://youtu.be/FcPO4EXDwzc
 +
 
 +
~savannahsolver
  
 
==See Also==
 
==See Also==

Revision as of 16:59, 16 June 2020

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

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

Video Solution

https://youtu.be/Gkm5rU5MlOU

~IceMatrix

https://youtu.be/FcPO4EXDwzc

~savannahsolver

See Also

2020 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AMC 10 Problems and Solutions

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