Hi,
The correct answer is E) but I thought the answer was D).
So, the reason why answer has to be E) is because there are possibility that either p or q can be negative to make it zero??
23. If x is an integer, is (x + p)(x + q) an even integer?
(1) q is an even integer.
(2) p is an even integer.
ds 500 test 7 #23
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Often on DS picking #'s can help you make sense of the statements. On DS questions that ask you a yes/no question, you have to remember that the statment is only sufficient if it gives you always yes or always no, not both.
For statment 1, let's say x=2, q=2, and p=2.
that gives us (2+2)(2+2)=4*4=16.
But we could also say that q=2, p=2 and x=1.
That gives us (1+2)(1+2)=3*3=9.
So the statement is insuffiecient because the product could be either even or odd.
The same numbers can be used for statement 2, so it's insufficient as well.
And since we can use the same in both statmetns, then together they don't tell us anything new. E is the answer.
For statment 1, let's say x=2, q=2, and p=2.
that gives us (2+2)(2+2)=4*4=16.
But we could also say that q=2, p=2 and x=1.
That gives us (1+2)(1+2)=3*3=9.
So the statement is insuffiecient because the product could be either even or odd.
The same numbers can be used for statement 2, so it's insufficient as well.
And since we can use the same in both statmetns, then together they don't tell us anything new. E is the answer.
Matt McIver
Princeton Review Instructor
Princeton Review Instructor
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23. If x is an integer, is (x + p)(x + q) an even integer?
(1) q is an even integer.
(2) p is an even integer.
Even*Odd = Even
Even*Even = Even
Odd*Odd = Odd
Odd + Even = Odd
Even+Even = Even
Odd + Odd = Even
Keeping these in mind Statement I alone is insufficient ans so is Statement II
If put together even then they are not sufficient. Hence E
(1) q is an even integer.
(2) p is an even integer.
Even*Odd = Even
Even*Even = Even
Odd*Odd = Odd
Odd + Even = Odd
Even+Even = Even
Odd + Odd = Even
Keeping these in mind Statement I alone is insufficient ans so is Statement II
If put together even then they are not sufficient. Hence E