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Difference between revisions of "2018 AMC 10B Problems/Problem 4"

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==Solution 2==
 
==Solution 2==
 
If you find the GCD of <math>24</math>, <math>48</math>, and <math>72</math> you get your first number, <math>12</math>. After this, do <math>48 \div 12</math> and <math>72 \div 12</math> to get <math>4</math> and <math>6</math>, the other 2 numbers. When you add up your <math>3</math> numbers, you get <math>22</math> which is <math>\boxed{B}</math>.
 
If you find the GCD of <math>24</math>, <math>48</math>, and <math>72</math> you get your first number, <math>12</math>. After this, do <math>48 \div 12</math> and <math>72 \div 12</math> to get <math>4</math> and <math>6</math>, the other 2 numbers. When you add up your <math>3</math> numbers, you get <math>22</math> which is <math>\boxed{B}</math>.
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==Solution 3==
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Without a loss of generality, <math>XY=24</math>, <math>XZ=48</math>, <math>YZ=72</math>.
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Multiplying the three gives <math>(XYZ)^2=2^{10}\cdot3^4</math> <math>\Rightarrow</math> <math>XYZ=2^5\cdot3^2</math>. <math>\frac{XYZ}{XY}=12</math>
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<math>\frac{XYZ}{XZ}=6</math> and <math>\frac{XYZ}{YZ}=4</math>
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<math>\therefore X+Y+Z=4+6+12=</math> <math>\boxed{\textbf{B } 22}</math>
  
 
==See Also==
 
==See Also==

Revision as of 02:32, 13 February 2019

Problem

A three-dimensional rectangular box with dimensions $X$, $Y$, and $Z$ has faces whose surface areas are $24$, $24$, $48$, $48$, $72$, and $72$ square units. What is $X$ + $Y$ + $Z$?

$\textbf{(A) }18 \qquad \textbf{(B) }22 \qquad \textbf{(C) }24 \qquad \textbf{(D) }30 \qquad \textbf{(E) }36 \qquad$

Solution 1

Let $X$ be the length of the shortest dimension and $Z$ be the length of the longest dimension. Thus, $XY = 24$, $YZ = 72$, and $XZ = 48$. Divide the first two equations to get $\frac{Z}{X} = 3$. Then, multiply by the last equation to get $Z^2 = 144$, giving $Z = 12$. Following, $X = 4$ and $Y = 6$.

The final answer is $4 + 6 + 12 = 22$. $\boxed{B}$

Solution 2

If you find the GCD of $24$, $48$, and $72$ you get your first number, $12$. After this, do $48 \div 12$ and $72 \div 12$ to get $4$ and $6$, the other 2 numbers. When you add up your $3$ numbers, you get $22$ which is $\boxed{B}$.

Solution 3

Without a loss of generality, $XY=24$, $XZ=48$, $YZ=72$. Multiplying the three gives $(XYZ)^2=2^{10}\cdot3^4$ $\Rightarrow$ $XYZ=2^5\cdot3^2$. $\frac{XYZ}{XY}=12$ $\frac{XYZ}{XZ}=6$ and $\frac{XYZ}{YZ}=4$

$\therefore X+Y+Z=4+6+12=$ $\boxed{\textbf{B } 22}$

See Also

2018 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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

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