2020 AMC 10B Problems/Problem 2

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