Difference between revisions of "2018 AMC 10B Problems/Problem 4"
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Let <math>X</math> be the length of the shortest dimension and <math>Z</math> be the length of the longest dimension. Thus, <math>XY = 24</math>, <math>YZ = 72</math>, and <math>XZ = 48</math>. | Let <math>X</math> be the length of the shortest dimension and <math>Z</math> be the length of the longest dimension. Thus, <math>XY = 24</math>, <math>YZ = 72</math>, and <math>XZ = 48</math>. | ||
− | Divide the first | + | Divide the first two equations to get <math>\frac{Z}{X} = 3</math>. Then, multiply by the last equation to get <math>Z^2 = 144</math>, giving <math>Z = 12</math>. Following, <math>X = 4</math> and <math>Y = 6</math>. |
The final answer is <math>4 + 6 + 12 = 22</math>. <math>\boxed{B}</math> | The final answer is <math>4 + 6 + 12 = 22</math>. <math>\boxed{B}</math> |
Revision as of 21:52, 4 March 2018
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 use guess and check to find that the dimensions are by
by
. Therefore, the answer is
.
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.