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Difference between revisions of "2013 AMC 8 Problems/Problem 18"
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==Problem==
 
==Problem==
  
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Isabella uses one-foot cubical blocks to build a rectangular fort that is 12 feet long, 10 feet wide, and 5 feet high. The floor and the four walls are all one foot thick. How many blocks does the fort contain?
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Isabella uses one-foot cubical blocks to build a rectangular fort that is <math>12</math> feet long, <math>10</math> feet wide, and <math>5</math> feet high. The floor and the four walls are all one foot thick. How many blocks does the fort contain?
  
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\documentclass{article}
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<asy>import three;
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\usepackage[pdftex]{graphicx}
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currentprojection=orthographic(-8,15,15);
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\begin{document}
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triple A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P;
 +
A = (0,0,0);
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B = (0,10,0);
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C = (12,10,0);
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D = (12,0,0);
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E = (0,0,5);
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F = (0,10,5);
 +
G = (12,10,5);
 +
H = (12,0,5);
 +
I = (1,1,1);
 +
J = (1,9,1);
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K = (11,9,1);
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L = (11,1,1);
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M = (1,1,5);
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N = (1,9,5);
 +
O = (11,9,5);
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P = (11,1,5);
 +
//outside box far
 +
draw(surface(A--B--C--D--cycle),white,nolight);
 +
draw(A--B--C--D--cycle);
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draw(surface(E--A--D--H--cycle),white,nolight);
 +
draw(E--A--D--H--cycle);
 +
draw(surface(D--C--G--H--cycle),white,nolight);
 +
draw(D--C--G--H--cycle);
 +
//inside box far
 +
draw(surface(I--J--K--L--cycle),white,nolight);
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draw(I--J--K--L--cycle);
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draw(surface(I--L--P--M--cycle),white,nolight);
 +
draw(I--L--P--M--cycle);
 +
draw(surface(L--K--O--P--cycle),white,nolight);
 +
draw(L--K--O--P--cycle);
 +
//inside box near
 +
draw(surface(I--J--N--M--cycle),white,nolight);
 +
draw(I--J--N--M--cycle);
 +
draw(surface(J--K--O--N--cycle),white,nolight);
 +
draw(J--K--O--N--cycle);
 +
//outside box near
 +
draw(surface(A--B--F--E--cycle),white,nolight);
 +
draw(A--B--F--E--cycle);
 +
draw(surface(B--C--G--F--cycle),white,nolight);
 +
draw(B--C--G--F--cycle);
 +
//top
 +
draw(surface(E--H--P--M--cycle),white,nolight);
 +
draw(surface(E--M--N--F--cycle),white,nolight);
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draw(surface(F--N--O--G--cycle),white,nolight);
 +
draw(surface(O--G--H--P--cycle),white,nolight);
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draw(M--N--O--P--cycle);
 +
draw(E--F--G--H--cycle);
  
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This is my first image.
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label("10",(A--B),SE);
 +
label("12",(C--B),SW);
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label("5",(F--B),W);</asy>
  
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\includegraphics{box.png}
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<math>\textbf{(A)}\ 204 \qquad \textbf{(B)}\ 280 \qquad \textbf{(C)}\ 320 \qquad \textbf{(D)}\ 340 \qquad \textbf{(E)}\ 600</math>
 +
 
 +
==Solution 1==
 +
 
 +
There are <math>10 \cdot 12 = 120</math> cubes on the base of the box. Then, for each of the 4 layers above the bottom (as since each cube is 1 foot by 1 foot by 1 foot and the box is 5 feet tall, there are 4 feet left), there are <math>9 + 11 + 9 + 11 = 40</math> cubes. Hence, the answer is <math>120 + 4 \cdot 40 = \boxed{\textbf{(B)}\ 280}</math>.
 +
 
 +
==Solution 2 (Complementary Counting)==
 +
 
 +
We can just calculate the volume of the prism that was cut out of the original <math>12\times 10\times 5</math> box. Each interior side of the fort will be <math>2</math> feet shorter than each side of the outside. Since the floor is <math>1</math> foot, the height will be <math>4</math> feet. So the volume of the interior box is <math>10\times 8\times 4=320\text{ ft}^3</math>.
  
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That's a cool picture up above.
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The volume of the original box is <math>12\times 10\times 5=600\text{ ft}^3</math>. Therefore, the number of blocks contained in the fort is <math>600-320=\boxed{\textbf{(B)}\ 280}</math>
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\end{document}
 
  
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/FDgcLW4frg8?t=5261
  
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<math>\textbf{(A)}\ 204 \qquad \textbf{(B)}\ 280 \qquad \textbf{(C)}\ 320 \qquad \textbf{(D)}\ 340 \qquad \textbf{(E)}\ 600</math>
+
~ pi_is_3.14
  
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==Solution==
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==Video Solution 2==
 +
https://youtu.be/WlMUXUloTFM Soo, DRMS, NM
  
−
There are <math>10 \cdot 12 = 120</math> cubes on the base of the box. Then, for each of the 4 layers above the bottom (as since each cube is 1 foot by 1 foot by 1 foot and the box is 5 feet tall, there are 4 feet left), there are <math>9 + 11 + 9 + 11 = 40</math> cubes. Hence, the answer is <math>120 + 4 \cdot 40 = \boxed{\textbf{(B)}\ 280}</math>.
+
==Video Solution 3==
 +
https://youtu.be/zFNf8WxBdoY ~savannahsolver
  
 
==See Also==
 
==See Also==
 +
(Other problems)
 
{{AMC8 box|year=2013|num-b=17|num-a=19}}
 
{{AMC8 box|year=2013|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:30, 18 December 2023

Problem

Isabella uses one-foot cubical blocks to build a rectangular fort that is $12$ feet long, $10$ feet wide, and $5$ feet high. The floor and the four walls are all one foot thick. How many blocks does the fort contain?

[asy]import three; currentprojection=orthographic(-8,15,15); triple A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P; A = (0,0,0); B = (0,10,0); C = (12,10,0); D = (12,0,0); E = (0,0,5); F = (0,10,5); G = (12,10,5); H = (12,0,5); I = (1,1,1); J = (1,9,1); K = (11,9,1); L = (11,1,1); M = (1,1,5); N = (1,9,5); O = (11,9,5); P = (11,1,5); //outside box far draw(surface(A--B--C--D--cycle),white,nolight); draw(A--B--C--D--cycle); draw(surface(E--A--D--H--cycle),white,nolight); draw(E--A--D--H--cycle); draw(surface(D--C--G--H--cycle),white,nolight); draw(D--C--G--H--cycle); //inside box far draw(surface(I--J--K--L--cycle),white,nolight); draw(I--J--K--L--cycle); draw(surface(I--L--P--M--cycle),white,nolight); draw(I--L--P--M--cycle); draw(surface(L--K--O--P--cycle),white,nolight); draw(L--K--O--P--cycle); //inside box near draw(surface(I--J--N--M--cycle),white,nolight); draw(I--J--N--M--cycle); draw(surface(J--K--O--N--cycle),white,nolight); draw(J--K--O--N--cycle); //outside box near draw(surface(A--B--F--E--cycle),white,nolight); draw(A--B--F--E--cycle); draw(surface(B--C--G--F--cycle),white,nolight); draw(B--C--G--F--cycle); //top draw(surface(E--H--P--M--cycle),white,nolight); draw(surface(E--M--N--F--cycle),white,nolight); draw(surface(F--N--O--G--cycle),white,nolight); draw(surface(O--G--H--P--cycle),white,nolight); draw(M--N--O--P--cycle); draw(E--F--G--H--cycle); label("10",(A--B),SE); label("12",(C--B),SW); label("5",(F--B),W);[/asy]

$\textbf{(A)}\ 204 \qquad \textbf{(B)}\ 280 \qquad \textbf{(C)}\ 320 \qquad \textbf{(D)}\ 340 \qquad \textbf{(E)}\ 600$

Solution 1

There are $10 \cdot 12 = 120$ cubes on the base of the box. Then, for each of the 4 layers above the bottom (as since each cube is 1 foot by 1 foot by 1 foot and the box is 5 feet tall, there are 4 feet left), there are $9 + 11 + 9 + 11 = 40$ cubes. Hence, the answer is $120 + 4 \cdot 40 = \boxed{\textbf{(B)}\ 280}$.

Solution 2 (Complementary Counting)

We can just calculate the volume of the prism that was cut out of the original $12\times 10\times 5$ box. Each interior side of the fort will be $2$ feet shorter than each side of the outside. Since the floor is $1$ foot, the height will be $4$ feet. So the volume of the interior box is $10\times 8\times 4=320\text{ ft}^3$.

The volume of the original box is $12\times 10\times 5=600\text{ ft}^3$. Therefore, the number of blocks contained in the fort is $600-320=\boxed{\textbf{(B)}\ 280}$

Video Solution by OmegaLearn

https://youtu.be/FDgcLW4frg8?t=5261

~ pi_is_3.14

Video Solution 2

https://youtu.be/WlMUXUloTFM Soo, DRMS, NM

Video Solution 3

https://youtu.be/zFNf8WxBdoY ~savannahsolver

See Also

(Other problems)

2013 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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