Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2: and →‎Solution 3)
(→‎Video Solution)
 
(5 intermediate revisions by 4 users not shown)
Line 19: Line 19:
  
 
==Solution 2==
 
==Solution 2==
 +
 +
Simply guess and check to find that the dimensions are <math>4</math> by <math>6</math> by <math>12</math>. Therefore, the answer is <math>4 + 6 + 12 = 22</math>. <math>\boxed{B}</math>
 +
 +
==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>.
 
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>
 +
- Solution by smartninja2000
  
−
==Solution 3==
+
==Video Solution (HOW TO THINK CRITICALLY!!!)==
−
Without a loss of generality, <math>XY=24</math>, <math>XZ=48</math>, <math>YZ=72</math>.
+
https://youtu.be/FFrcsd9TzdM
−
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>
+
 
−
<math>\frac{XYZ}{XZ}=6</math> and <math>\frac{XYZ}{YZ}=4</math>
+
~Education, the Study of Everything
 +
 
 +
 
 +
==Video Solution==
 +
https://youtu.be/iIgNr9XUneg
  
−
<math>\therefore X+Y+Z=4+6+12=</math> <math>\boxed{\textbf{B } 22}</math>
+
~savannahsolver
  
 
==See Also==
 
==See Also==

Latest revision as of 12:30, 28 May 2023

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 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 $\boxed{B}$ - Solution by smartninja2000

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/FFrcsd9TzdM

~Education, the Study of Everything


Video Solution

https://youtu.be/iIgNr9XUneg

~savannahsolver

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_4&oldid=193626"