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

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Divide the first to 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>. | Divide the first to 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 <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 14:48, 16 February 2018

## Problem

A three-dimensional rectangular box with dimensions , , and has faces whose surface areas are , , , , , and square units. What is + + ?

## Solution

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 .