Difference between revisions of "2018 AMC 10B Problems/Problem 4"
(→Solution) |
|||
Line 11: | Line 11: | ||
</math> | </math> | ||
− | ==Solution== | + | ==Solution 1== |
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>. | ||
Line 17: | Line 17: | ||
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> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Simply use 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> |
Revision as of 15:26, 16 February 2018
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 to 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 .