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

Revision as of 22:30, 6 May 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

Simply use guess and check to find that the dimensions are $4$ by $6$ by $12$ . Therefore, the answer is $4 + 6 + 12 = 22$ . $\boxed{B}$

Solution 3

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 4

Since the surface areas of the faces are the product of two of the dimensions. Therefore, $XY=24$ , $XZ=48$ , and $YZ=72$ . You can multiply $XY \times XZ \times YZ$ , which simplifies to ${XYZ}^2=24 \times 48 \times 72$ which means that the volume $XYZ$ equals $\sqrt{24*48*72}=\sqrt{24^2*12^2}=24*12=288$ . The individual dimensions, $X$ , $Y$ , and $Z$ can be found by doing $\frac{XYZ}{XY}$ , $\frac{XYZ}{YZ}$ , and $\frac{XYZ}{XZ}$ , which yields $Z=12$ , $Y=6$ , and $X=4$ . Adding this up, we have that $X+Y+Z=22$ which is $\boxed{B}$

@@ Line 24: / Line 24: @@
 ==Solution 3==
 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>.
+==Solution 4==
+Since the surface areas of the faces are the product of two of the dimensions. Therefore, <math>XY=24</math>, <math>XZ=48</math>, and <math>YZ=72</math>. You can multiply <math>XY \times XZ \times YZ</math>, which simplifies to <math>{XYZ}^2=24 \times 48 \times 72</math> which means that the volume<math>XYZ</math> equals<math>\sqrt{24*48*72}=\sqrt{24^2*12^2}=24*12=288</math>. The individual dimensions, <math>X</math>, <math>Y</math>, and <math>Z</math> can be found by doing <math>\frac{XYZ}{XY}</math>, <math>\frac{XYZ}{YZ}</math>, and <math>\frac{XYZ}{XZ}</math>, which yields <math>Z=12</math>, <math>Y=6</math>, and <math>X=4</math>. Adding this up, we have that <math>X+Y+Z=22</math> which is <math>\boxed{B}</math>
 ==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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