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Listing down the combinations
XY - 24 - 2*12, 3*8, 4*6
YZ - 48 - 2*14, 3*16, 4*12, 6*8
ZX - 72 - 2*36, 3*24, 4*18, 6*12, 8*9

Assuming z is 9, x is 8, then y = 3 which doesn't validate with YZ.

Using similar logic only sides 12, 4,6 tally with all combinations.

Hence answer is 22

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Bunuel
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?

A. 18
B. 22
C. 24
D. 30
E. 36

possible combinations
x*y=24
x*z=48
y*z=72
x=24/y
z*24/x=72
z=3x
3*x^2=48 ; x=4
y=6,z=12
sum ; 22
IMO B
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Given,
XY=24--1
XZ=48--2
ZY=72--3

Multiply the above 3 we get

(XYZ)^2=24x48x72=(288)^2

Therefore XYZ=288
Using eqn 1,2 and 3

Z=12
Y=6
X=4

Therefore X+Y+Z=22

Hence B
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Bunuel
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?

A. 18
B. 22
C. 24
D. 30
E. 36

XY = 24
YZ = 48
XZ = 72

Instead of solving the equations, I will pick the greatest number 72 and look at its similarly sized factors 8 and 9. Since 9 is neither a factor of 24 nor of 48, so the two 3s must be distributed between the two factors. The 8 could be divided as 4 and 2 to give 4*3 and 2*3 i.e. 12 and 6 as factors of 72.
Let's try whether X and Z work as 6 and 12.

Now we see that 6 * 4 = 24 and 12 * 4 = 48 which means that Y = 4.

Hence we get the three dimensions as 4, 6 and 12 and so their sum is 22.

Answer (B)

Check this video for a discussion on rectangular solids: https://www.youtube.com/watch?v=kqVgiZRd6yM
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