Difference between revisions of "2018 AMC 10B Problems/Problem 4"
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==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 | ||
+ | |||
+ | ==Video Solution (HOW TO THINK CRITICALLY!!!)== | ||
+ | https://youtu.be/FFrcsd9TzdM | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/iIgNr9XUneg | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Latest revision as of 13:30, 28 May 2023
Contents
Problem
A three-dimensional rectangular box with dimensions ,
, and
has faces whose surface areas are
,
,
,
,
, and
square units. What is
+
+
?
Solution 1
Let be the length of the shortest dimension and
be the length of the longest dimension. Thus,
,
, and
.
Divide the first two equations to get
. Then, multiply by the last equation to get
, giving
. Following,
and
.
The final answer is .
Solution 2
Simply guess and check to find that the dimensions are by
by
. Therefore, the answer is
.
Solution 3
If you find the GCD of ,
, and
you get your first number,
. After this, do
and
to get
and
, the other 2 numbers. When you add up your
numbers, you get
which is
.
Solution 4
Since the surface areas of the faces are the product of two of the dimensions. Therefore, ,
, and
. You can multiply
, which simplifies to
which means that the volume
equals
. The individual dimensions,
,
, and
can be found by doing
,
, and
, which yields
,
, and
. Adding this up, we have that
which is
- Solution by smartninja2000
Video Solution (HOW TO THINK CRITICALLY!!!)
~Education, the Study of Everything
Video Solution
~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.