Bunuel wrote:
5. If x^2 + 2x -15 = -m, where m is an integer from -10 and 10, inclusive, what is the probability that m is greater than zero?
A. 2/7
B. 1/3
C. 7/20
D. 2/5
E. 3/7
GyanOne wrote:
@Bunuel, for Q5, shouldn't the original question also say that we are only looking for solutions where x is an integer?
For this to be satisfied, in x^2 + 2x -15 = -m, 4 - 4(m-15)>0 and 4-4(m-15) must be a perfect square
=> 4 (16-m) must be a perfect square and >0
=> 16-m must be a perfect square and >0
The only values that satisfy this for -10<=m<=10 are m=-9,0,7 of which only 7 is positive
=> probability = 1/3
I agree. Proper answer to 5 as written is 10/21
If x is not explicitly constrained or if x is not constrained by the equation given, then we have to respect all real values of x, integer or not. Since x is not constrained to integers only, we have to respect the possibility that x is not an integer.
Since -10 <= m <= 10 and the equation is x^2 + 2x - 15 = -m, we will always have x^2 + 2x + c = 0 where -25 <= c <= -5. Using quadratic formula (just to prove it, though you won't need to know it for the GMAT), we need the calculate the discriminant, which is b^2 - 4ac, to determine whether or not there are real values for x. Here we get 2^2 - 4*1*c. Since c is always negative, we will always have a positive number for the discriminant. This means that we will always have a positive number underneath the square root of the quadratic formula and, therefore, 2 real outcomes for x.
There are 21 integer values of m (10 positive, 10 negative, and 0) and 10 of them are positive, so 10/21.
Good question though if we constrain x to only integer values.