Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

If x and y are positive integers, is x a prime number?

(1)\(|x - 2| \lt 2 - y\). The left hand side of the inequality is an absolute value, so the least value of LHS is zero, thus \(0 \lt 2 - y\), thus \(y \lt 2\) (if \(y\) is more than or equal to 2, then \(y-2 \le{0}\) and it cannot be greater than \(|x - 2|\)). Next, since given that \(y\) is a positive integer, then \(y=1\).

So, we have that: \(|x - 2| \lt 1\), which implies that \(-1 \lt x-2 \lt 1\), or \(1 \lt x \lt 3\), thus \(x=2=prime\). Sufficient.

(2)\(x + y - 3 = |1-y|\). Since \(y\) is a positive integer, then \(1-y \le {0}\), thus \(|1-y|=-(1-y)\). So, we have that \(x + y - 3 = -(1-y)\), which gives \(x=2=prime\). Sufficient.

(if \(y\) is more than or equal to 2, then \(y-2 \le{0}\)

Since \(y\) is a positive integer, then \(1-y \le {0}\),

I have not understood the above two things. Could you please help me on this? How can we get these equation?

I tried to explain this in my solution.

1. (1) \(|x - 2| \lt 2 - y\). The left hand side of the inequality is an absolute value, so the least value of LHS is zero, thus \(0 \lt 2 - y\), thus \(y \lt 2\).

2. For (2): we know that y is a positive integer 1, 2, 3, ... so 1 - y must be less than or equal to 0.
_________________

Bunuel, from what I understand about absolute values, the number/expression within the absolute number sign could either be positive or negative (just like the way the square root of a number could either be positive or negative), so I didn't understand how the least value in the LHS is zero. Please explain? Thanks

Bunuel, from what I understand about absolute values, the number/expression within the absolute number sign could either be positive or negative (just like the way the square root of a number could either be positive or negative), so I didn't understand how the least value in the LHS is zero. Please explain? Thanks

The point is that |some expression| >= 0. So, when we have |some expression| = x, it means that x must be more than or equal to 0 too.

Also, the square root of a number cannot be negative \(\sqrt{4}=2\), not +/-2.

When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3; \(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).

If \(x\) and \(y\) are positive integers, is \(x\) a prime number?

(1) \(|x - 2| \lt 2 - y\)

(2) \(x + y - 3 = |1-y|\)

I've solved this problem in a little different way.

1) |x - 2| < 2 - y It means -(2 - y) < x - 2 < 2 - y => y - 2 < x - 2 < 2 - y => y < x < 4 - y

Try pick up numbers. It's said that x and y are positive integers. Let's say y is 1, then x could be 2 or 3. y can't be 2, 3 or any other integer, because it violates the inequality. In both cases, whether x is 2 or 3, x is a prime number. Therefore, first statement is sufficient.

2) x + y - 3 = |1 - y| It means x + y - 3 = 1 - y or x + y - 3 = -(1 - y), x + y - 3 = y - 1 x = 4 - 2y or x = 2 . 2 is a prime number. Therefore, second statement is sufficient too.

Thus, correct answer is D: both statements are sufficient.

I think this the explanation isn't clear enough, please elaborate. for option 1 my logic is since x is a positive integer, lx-2l>0, therefore x-2<2-y x+y<4 so, X can be 1 or 2, cannot determine can someone correct me, if my logic is wrong?

I think this the explanation isn't clear enough, please elaborate. for option 1 my logic is since x is a positive integer, lx-2l>0, therefore x-2<2-y x+y<4 so, X can be 1 or 2, cannot determine can someone correct me, if my logic is wrong?

(1) \(|x - 2| \lt 2 - y\). The left hand side of the inequality is an absolute value, so the least value of LHS is zero, thus \(0 \lt 2 - y\), thus \(y \lt 2\).
_________________

If x and y are positive integers, is x a prime number?

(1)\(|x - 2| \lt 2 - y\). The left hand side of the inequality is an absolute value, so the least value of LHS is zero, thus \(0 \lt 2 - y\), thus \(y \lt 2\) (if \(y\) is more than or equal to 2, then \(y-2 \le{0}\) and it cannot be greater than \(|x - 2|\)). Next, since given that \(y\) is a positive integer, then \(y=1\).

So, we have that: \(|x - 2| \lt 1\), which implies that \(-1 \lt x-2 \lt 1\), or \(1 \lt x \lt 3\), thus \(x=2=prime\). Sufficient.

(2)\(x + y - 3 = |1-y|\). Since \(y\) is a positive integer, then \(1-y \le {0}\), thus \(|1-y|=-(1-y)\). So, we have that \(x + y - 3 = -(1-y)\), which gives \(x=2=prime\). Sufficient.

Answer: D

Thanks for the solution. But, i can`t understand one point. In statement 1, you mentioned if y is more than or equal to 2, then y−2≤0y−2≤0 and it cannot be greater than |x−2||x−2|. Could you please explain this point in more detail?

If x and y are positive integers, is x a prime number?

(1)\(|x - 2| \lt 2 - y\). The left hand side of the inequality is an absolute value, so the least value of LHS is zero, thus \(0 \lt 2 - y\), thus \(y \lt 2\) (if \(y\) is more than or equal to 2, then \(y-2 \le{0}\) and it cannot be greater than \(|x - 2|\)). Next, since given that \(y\) is a positive integer, then \(y=1\).

So, we have that: \(|x - 2| \lt 1\), which implies that \(-1 \lt x-2 \lt 1\), or \(1 \lt x \lt 3\), thus \(x=2=prime\). Sufficient.

(2)\(x + y - 3 = |1-y|\). Since \(y\) is a positive integer, then \(1-y \le {0}\), thus \(|1-y|=-(1-y)\). So, we have that \(x + y - 3 = -(1-y)\), which gives \(x=2=prime\). Sufficient.

Answer: D

Thanks for the solution. But, i can`t understand one point. In statement 1, you mentioned if y is more than or equal to 2, then y−2≤0y−2≤0 and it cannot be greater than |x−2||x−2|. Could you please explain this point in more detail?

Thanks in advance.

If y is greater than 2 then the right hand side of the inequality (2-y) becomes negative. The left hand side if the inequality is an absolute value (|x - 2|) which cannot be negative, thus the inequality (\(|x - 2| \lt 2 - y\)) will not hold true. Please re-read the discussion above.
_________________