2018 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 4 Easy

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? (2018 AMC 10B Problem, Question#4)

  • A.

    18

  • B.

    22

  • C.

    24

  • D.

    30

  • E.

    36

Answer:B

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, mutiply 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.

Simply use guess and check to find that the dimensions are 4 by 6 by 12.Therefore, the answer is 4+6+12=22.

If you find the GCD of 24，48~and 72 you get your first number，12. After this, do 48\div12 and 72\div12 to get 4 and 6, the other 2 numbers.When you add up your 3 numbers, you get 22 which is \rm B.

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\times48\times72 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 \rm B.

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