Interesting GMATFix Problem-11

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arora007: Community Manager; Posts: 1048; Joined: Mon Aug 17, 2009 3:26 amFollowed by: 27 members; Location: India; Thanked: 51 times; GMAT Score:670

by arora007 » Tue Sep 21, 2010 12:52 pm

X^2 + bx + c = 0 has two integer solutions for the values of x. b is an integer constant and c is a prime number constant. Is |x| >2 ?
1) b is odd
2) c is even

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Source: Beat The GMAT — Data Sufficiency | Tue Sep 21, 2010 12:52 pm

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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members; Location: Milpitas, CA; Thanked: 447 times

by Rahul@gurome » Tue Sep 21, 2010 11:00 pm

Solution:
Let the integer solutions of x be m , n.
So m+n = -b
and mn = c.
Since c is prime, one of m or n has to be 1 or -1 and the other has to be the prime number.
If lxl > 2, it means either x > 2 or x < -2.
So we need to know if both m, n > 2 or both m, n < -2.
Consider first (1) alone.
b is odd.
So m+n = - odd number.
Since one of m and n is 1 or -1, the other has to be even, because only even + odd gives odd.
Also it has to be prime.
The only possibility is -2 or 2.
So the only possible values of m, n are 1, -1, -2, 2.
Now l1l = 1< 2, l2l =2 ,l-1l = 1 < 2, l-2l = 2 .
So lml or lnl is not > 2.
Or lxl is not > 2
So (1) alone is sufficient to answer the question.
Next consider (2) alone.
If c is even c has to be 2 because it is given that is a prime.
Or mn is 2.
So one of m or n is 1 and the other is 2 or one of m, n is -1 and the other is -2.
Now l1l < 2, l2l = 2, l-1l = 1 < 2, l-2l = 2.
So lml or lnl is not > 2.
Or lxl is not > 2.
So (2) alone is also sufficient to answer the question.

The correct answer is (D).

Last edited by Rahul@gurome on Wed Sep 22, 2010 6:12 am, edited 1 time in total.

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arora007: Community Manager; Posts: 1048; Joined: Mon Aug 17, 2009 3:26 amFollowed by: 27 members; Location: India; Thanked: 51 times; GMAT Score:670

by arora007 » Wed Sep 22, 2010 2:44 am

OA is D, even i had selected B. see GMATFix is such a trap!!

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