Assume the length of the original card board is x.
Then the volume of the expressed as 5 * (x -10) * (x - 10).
Hence 5 * (x -10) * (x - 10) = 40
(x-10)^2 = 8

x-10 = sqrt(8) or x-10 = -sqrt(8)

x = 10 + 2*sqrt(2) or x = 10 - 2*sqrt(2)

C is your answer.

Two squares of each 5 inches are taken out from the corner.

Therefore minimum length of the original cardboard is 5+5 = 10

You can eliminate all the options except C and D.

The cube or cuboid formed has a volume of 40. We no idea weather it is cuboid or cube. I think this info is missing. However, if we assume it to be a cube:

cube root (40) = 2cuberoot5

10+2cuberoot5

I dont understand why are the options in sqroot.

Morgoth wrote:Two squares of each 5 inches are taken out from the corner.

Therefore minimum length of the original cardboard is 5+5 = 10

You can eliminate all the options except C and D.

Great start, but remember that B is also > 10. We can narrow it down to C and D because we know that the length is 5 + 5 + length of the base of the box, so the answer should be in the form 10 + something.

Now we go 1 step further.

We know that the height of our open box is 5, since we lifted a flap of length 5 to form the outside of the box.

We also know that the base of the box is a square, since we started with a square and cut away the same amount from each dimension.

So, we know that our box has height 5 and length and width x.

Accordingly:

5(x^2) = 40

x^2 = 8

x = root8 = 2root2

So, the length of a side of the original square is:

5 (that we lifted up) + 2root2 (what's left on the base) + 5 (that we lifted up)

= 10 + 2root2

choose (C).

Note that letting x be the base of the box instead of the original length of the cardboard makes the math simpler.