Mr.X can do this with 5 pieces.
On May 1st, he gives the landlord 1 inch piece.
On the 2nd day, he gives the landlord a 2 inch piece, and takes the 1 inch back.
On the 3rd, gives the landlord the 2 inch piece. (now she has 2+1)= 3 inches for 3 days.
On the 4th day, he gives the landlord the 4 inch piece, but takes back the 2 and 1 inch pieces.
In the same way he can do it with 1+2+4+8+16 = 31 inches for 31 days, with 5 pieces.

Yes, as 'eagleeye' said, 5-pieces is the answer.
Beauty of math is illustrated in this problem.
If Mr X cuts the silver bar of 31 inches in to the pieces with following sizes (in inches), it will result in minimum number of pieces- solution:
1, 2, 4, 8, 16
On day-1 Mr X gives 1-inch piece to the landlord
On day-2 he gives 2-inch piece and takes back the 1-inch piece
On day-3 he gives 1-inch piece in addition to the already given 2-inch piece, thus making it 3-inches in total
On day-4 he gives 4-inch piece and takes back the 1-inch and 2-inch pieces
On day-5 he gives 1-inch piece in addition to the already given 4-inch piece, thus making it 5-inches in total
This will continue till day-31.
Hence a minimum of five pieces is enough
Math logic:
Any number up to 2n-1 can be represented by the combinations of the 'n' numbers: 2^0,2^1,2^2,...2n-1.
For example, if 7 is considered, all numbers from 1 to 7 can be represented as combinations of 1,2, and 4:
1 = 1; 2 = 2; 3 = 1+2; 4 = 4; 5 = 1+4; 6 = 2+4; 7 = 1+2+4

If 9 is considered, all numbers from 1 to 9 can be represented as combinations of 1,2,4 and 2:
1 = 1; 2 = 2; 3 = 1+2; 4 = 4; 5 = 1+4; 6 = 2+4; 7 = 1+2+4; 8 = 2+2+4; 9 = 1+2+2+4;

If 15 is considered, all numbers from 1 to 15 can be represented as combinations of 1,2,4 and 8:
1 = 1; 2 = 2; 3 = 1+2; 4 = 4; 5 = 1+4; 6 = 2+4; 7 = 1+2+4; 8 = 8; 9 = 1+8; 10 = 2+8;
11 = 1+2+8; 12 = 4+8; 13 = 1+4+8; 14 = 2+4+8; 15 = 1+2+4+8

Answer (A)