A set of 125 small 1 x 1 x 1 cubes are arranged to form a 5 x 5 x 5 cube. How many cubes (including the original) can be formed with vertices chosen from these points such that their faces are parallel to those of the original cube?

I can't even understand what they are asking. which one is original cube?

Hi Dhruv. The wording seems ambiguous. I will try to answer using what I interpreted:
The original cube here should refer to the 5x5x5 cube that is formed. I believe 'vertices chosen from these points' means that at least one of the vertices of this original cube should be common to the cubes we are trying to form:
Take the left-bottom vertice. We can form a 1x1 cube that has a common vertice. This is equivalent to the 1x1x1 cube present in that corner of the larger cube.
We can form a 2x2x2 cube too. This will be equivalent to the 8 cubes present in the left-bottom of the larger cube.
Similarly, we can form 3x3x3 and 4x4x4 cubes, which gives us 4 cubes with 1 vertice common. We can do this for each of the 8 vertices, giving us 32 different cubes.
If we try to have more than 1 vertice common, one side of the formed cube will be 5 units, giving us the original cube itself, hence, 33 cubes.
It has not been given that we need to form the cubes only inside. We could form similar cubes outside the contours of the original cube, increasing the number of cases. Also, it has not been given that we can only use the 125 cubes. Then we can form infinitely many cubes. It is better to altogether avoid such vague worded questions. Hope this helps.