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Bunuel
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The question deals with prime numbers. So we write down all the prime numbers between 1 and 200.
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199.
For an one time analysis we could see that except 2,5,7 we can't find such combination when the difference between two prime numbers is 3.

Statement I
The units digit of x is 5.
Insufficient information.

Statement II
x, x - 3 and x + 2 are all prime numbers
We could see the only combination is 2,5,7
Sufficient

IMO B

I agree with this approach, though no need to write out the prime number from 1-200. Once we find the 2,5, and 7 combination, we know that 2, 5, 7 is the only combination that fits the sequence regarding prime numbers. If we try X=7, we would get 4, 7, 9. Plugging in other numbers for X would result in an even number on the lower end (thus non-prime). 2, 5, and 7 being the only combination, B IMO

I have written the prime numbers from 2 to 200 as an extra check and make one acquainted with the prime numbers. But yes, the approach is to start from the lowest one i.e. 2.

Another approach I could find now is the difference between the numbers x-3, x and x+2 are 3 and 2.
We know that the difference between an odd number and the next odd number is always 2,also same for even numbers.
So here if x-3 is even then x-1 is also even,next x & x+2 are odd numbers. Which can be quite the case with x=5.
But if x-3 is odd then x-1 is also odd, next both x and x+2 have to be constitutive even numbers but we know only 2 is the even prime. So none of other cases are possible.

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