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If x and y are consecutive odd integers such that x < y,

Expert replies
by BTGmoderatorDC » Sun Aug 05, 2018 10:10 pm

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If x and y are consecutive odd integers such that x < y, what is the value of y + x?

(1) The product of xy is negative.

(2) The sum x + y is the square of an integer.

OA A

Source: Magoosh
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Source: — Data Sufficiency |

by Jay@ManhattanReview » Sun Aug 05, 2018 11:23 pm
BTGmoderatorDC wrote:If x and y are consecutive odd integers such that x < y, what is the value of y + x?

(1) The product of xy is negative.

(2) The sum x + y is the square of an integer.

OA A

Source: Magoosh
Given: x and y are consecutive odd integers such that x < y
Question: What is the value of y + x?

Let's take each statement one by one.

Some examples of the consecutive odd integers x and y are -3, -5; -1, 1; 1, 3; 101, 103, etc.

(1) The product of xy is negative.

Since the product of xy is negative, one of the numbers must be positive and other must be negative. There is only one possibility, i.e., x = -1 and y = 1. Thus, y + x = 1 - 1 = 0. Sufficient.

(2) The sum x + y is the square of an integer.

There can be few examples.

Case 1: If x = - 1 and y = 1, then y + x = 1 - 1 = 0, square of an integer 0.
Case 2: If x = 1 and y = 3, then y + x = 3 + 1 = 4, square of an integer 2. No unique value of y + x. Insufficient.

The correct answer: A

Hope this helps!

-Jay
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by Jay@ManhattanReview » Sun Aug 05, 2018 11:37 pm
BTGmoderatorDC wrote:If x and y are consecutive odd integers such that x < y, what is the value of y + x?

(1) The product of xy is negative.

(2) The sum x + y is the square of an integer.

OA A

Source: Magoosh
Another approach for those who are the fan of Algebra.

Given: x and y are consecutive odd integers such that x < y
Question: What is the value of y + x?

Some examples of the consecutive odd integers x and y are -3, -5; -1, 1; 1, 3; 101, 103, etc.

Say the consecutive odd integers are (2n - 1) and (2n + 1). I did not choose expressions such as (2n + 1) and (2n + 3) since, upon multiplication, they would render three terms, while that with (2n - 1) and (2n + 1) would render only two terms [(2n - 1)*(2n + 1) = 4n^2 - 1], easier to manage.

Let's take each statement one by one.

(1) The product of xy is negative.

=> (2n - 1)*(2n + 1) < 0
4n^2 - 1 < 0
n^2 < 1/4

=> -1/2 < n < 1/2
=> n = 0; since n is an integer

Thus, the consecutive odd integers are x = 2n - 1 = 2*0 - 1 = -1 and y = 2n + 1 = 2*0 + 1 = 1

Thus, y + x = 1 - 1 = 0. Sufficient.

(2) The sum x + y is the square of an integer.

x + y = (2n - 1) + (2n + 1) = 4n

n itself can take may value such as 0, 1, 4, etc, making 4n, not a unique value. Insufficient.

The correct answer: A

Hope this helps!

-Jay
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by Brent@GMATPrepNow » Mon Aug 06, 2018 8:00 am
BTGmoderatorDC wrote:If x and y are consecutive odd integers such that x < y, what is the value of y + x?

(1) The product of xy is negative.
(2) The sum x + y is the square of an integer.
Target question: What is the value of y + x?

Given: x and y are consecutive odd integers such that x < y

Statement 1: The product of xy is negative
In order for the product xy to be NEGATIVE, it must be the case that one value is POSITIVE and one value of NEGATIVE.
Since x and y are CONSECUTIVE ODD integers, it must be the case that x = -1 and y = 1
This is the ONLY way to satisfy statement 1.
If x = -1 and y = 1, then x + y = (-1) + 1 = 0
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: The sum x + y is the square of an integer.
There are several values of x and y that satisfy statement 2. Here are two:
Case a: x = 1 and y = 3. Notice that x + y = 1 + 3 = 4, and 4 IS the square of an integer (4 = 2²) In this case, the answer to the target question is x + y = 4
Case b: x = 7 and y = 9. Notice that x + y = 7 + 9 = 16, and 16 IS the square of an integer (16 = 4²) In this case, the answer to the target question is x + y = 16
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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