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

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<math>5+40=\boxed{\textbf{(E) }21}</math> ~quacker88
 
<math>5+40=\boxed{\textbf{(E) }21}</math> ~quacker88
  
==See Also==
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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>.
  
{{AMC10 box|year=2020|ab=B|num-b=1|num-a=3}}
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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>.
{{MAA Notice}}
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<math>5+40=\boxed{\textbf{(E) }45}</math>

Revision as of 17:16, 7 February 2020

Problem

What is the value of \[1-(-2)-3-(-4)-5-(-6)?\]

$\textbf{(A)}\ -20 \qquad\textbf{(B)}\ -3 \qquad\textbf{(C)}\  3 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 21$

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 $5\cdot8=40$.

$5+40=\boxed{\textbf{(E) }21}$ ~quacker88

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 $5\cdot8=40$.

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