Compare :
Quantity A : The volume of a rectangular solid with sides of x, 2y and 3z
Quantity B : The volume of a rectangular solid with sides of 3x, y and z
Options:
A) Quantity A > Quantity B
B) Quantity A < Quantity B
C) Quantity A = Quantity B
D) Cannot be determined
My solution:
Volume = l * b * h
Quantity A : 6xyz
Quantity B : 3xyz
Now even if all x,y,z are 1, A > B
If x,y,z are fractions, still A > B
Hence, Option A) Quantity A > Quantity B
Is this solution correct ? Can l, b or h be taken as 0 for volume to be 0?
As Hellen states in their answer, breaking up your solution into cases is not necessary. In fact, $6xyz > 3xyz$ is true for any $x,y,z >0$.
To further justify this answer, think about it this way: $2 > 1 \Rightarrow 2x>x$ if $x>0$. So for your case, $6>3 \Rightarrow 6xyz > 3xyz$, as long as $xyz >0$, which it is.
One final way to think about it is to take the difference between $6xyz$ and $3xyz$.
$$6xyz-3xyz = 3xyz, \text{which is always positive. }$$