prime number

[sanjib]

If n = p + r, where n, p, and r are positive integers and n is odd, does p equal 2?
(1) p and r are prime numbers.
(2) r ≠ 2
IMO B
OA is C

[ssmiles08]

If n = p + r, where n, p, and r are positive integers and n is odd, does p equal 2?
(1) p and r are prime numbers.
(2) r ≠ 2
IMO B
OA is C

From the question stem, n = odd. the only possibility of n = O would be if it was E + O or O+ E.

1) insufficient. the only even prime number is 2. but that doesn't mean p could equal 2. r could also equal 2.

for ex; 2+3 = 5 (n)----> p+r
or for ex; 3+2 = 5---> p+r

2) is obviously Insufficient. p = 2, r = 5, or p = 4 r = 5

together: both are prime numbers and r is not equal to 2.

which means p = 2 and r = Odd number.

sufficient. (C)

[Stuart@KaplanGMAT]

If n = p + r, where n, p, and r are positive integers and n is odd, does p equal 2?
(1) p and r are prime numbers.
(2) r ≠ 2
IMO B
OA is C

If you think the answer is B, then what you've almost certainly done is assumed the information in (1) when assessing the sufficiency of (2).

Remember, for B to be correct, (2) must be sufficient only in conjunction with the original information, not in conjunction with (1).

Knowing that r ≠ 2 by itself does not preclude r from being any other even number (or p from being an even number other than 2); only by assuming that p and r are prime can you draw that conclusion. Therefore, you were using information from both (1) and (2) to solve, which leads to C as the correct choice.

[sanjib]

Thanks.
It was exactly what you mentioned ,I assumed the st#1.

[ghacker]

If n = p + r, where n, p, and r are positive integers and n is odd, does p equal 2?
(1) p and r are prime numbers.
(2) r ≠ 2

Answer is C

n = p+ r , if n is odd ---------> p or r must be even or a multiple of 2

Clearly statement I and II are insufficient

together we know that P= 2

Hence C