a = 205, l =495, d=10
l=a+(n-1)d
495=205+(n-1)10
290=(n-1)10
29=(n-1)
n=30

option b is correct

Sometimes using brute force is the best helper out there. Unit digit 5 means the numbers which are going to have 5 in the last position(from left to right).

So, let's consider a batch of 10 number from the range, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210. Only 205 meet our criteria, and there will 30 batches in total, 1 number from each batch is going to meet the criteria.

Now, Qnt A = 30, Qnt B = 31. So, the Answer is B

Positive integer between 200 to 500 with the units digit being 5.
The list starts with 205 and goes on till 295.
there are a total of 10 numbers in the column with 2 in the hundreds place and 5 in the units.

There will be equal number of integers in the list with 3 and 4 in the hundreds place and 5 in the units place.

therefore 10 x 3=30

Solution: Option B

(In such case I prefer doing the column x rows concept to work out the solutions:
205 305 405
215 - -
225 - -
235 - -
245 - -
255 - -
265 - -
275 - -
285 - -
295 - -

Columns x rows- 3 x 10=30 )

Multiples of 5 end with 0 or 5 alternating...Question has even number range, so we are to sure get an equal number of zero-ending and five-ending multiples...there are (500-200)/5 multiples of 5. Half the 60 end with zero, the other half with 5, so 30.

Multiples of 5 end with 0 or 5 alternating...Question has even number range, so we are to sure get an equal number of zero-ending and five-ending multiples...there are (500-200)/5 multiples of 5. Half the 60 end with zero, the other half with 5, so 30 for A...hence B.

Thank you so much. The best

What a clever answer! kudos for you!