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# M02-25

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16 Sep 2014, 00:18
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36% (02:49) correct 64% (02:27) wrong based on 315 sessions

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

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

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

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16 Sep 2014, 00:18
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Official Solution:

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 $$2-y \le{0}$$ 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.

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27 Nov 2014, 12:44
[quote="Bunuel"]Official Solution:

(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?
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28 Nov 2014, 05:33
1
Raihanuddin wrote:
Bunuel wrote:
Official Solution:

(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.
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28 Nov 2014, 06:30
positive integer means we should not consider zero??
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28 Nov 2014, 08:25
sudd1 wrote:
positive integer means we should not consider zero??

Positive integers are integers which are to the right of 0 on the number line: 1, 2, 3, 4, ...
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20 Apr 2015, 23:00
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
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21 Apr 2015, 05:25
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Kigunda wrote:
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

I think this is explained here: m02-183573.html#p1448632

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 $$x=\sqrt{9}=3$$.

Theory on Number Properties: math-number-theory-88376.html
Tips on Numper Properties: number-properties-tips-and-hints-174996.html

All DS Number Properties Problems to practice: search.php?search_id=tag&tag_id=38
All PS Number Properties Problems to practice: search.php?search_id=tag&tag_id=59

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20 Nov 2015, 19:38
Can someone help me understand this part in detail

2)x+y−3=|1−y|. Since y is a positive integer, then 1−y≤0, thus |1−y|=−(1−y).

why we are not considering the |1−y|=(1−y) ??
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21 Nov 2015, 02:46
sajib2126 wrote:
Can someone help me understand this part in detail

2)x+y−3=|1−y|. Since y is a positive integer, then 1−y≤0, thus |1−y|=−(1−y).

why we are not considering the |1−y|=(1−y) ??

Let me ask you if y is a positive integer what are the possible values of 1 - y ?
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23 Dec 2015, 09:43
1
2
Bunuel wrote:
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.
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12 Dec 2016, 20:04
Bunuel wrote:
Official Solution:

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.

Thanks for the solution. But, i cant 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?

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13 Dec 2016, 01:55
jahidhassan wrote:
Bunuel wrote:
Official Solution:

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.

Thanks for the solution. But, i cant 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 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.
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27 Jun 2017, 20:50
Hi Bunuel,

So from statement 1 what I understand is that |x−2|<2−y will only happen if x being positive will be equal to 2. Since it is absolute value and will be positive so for RHS Y cannot be >2 since then RHS<LHS (-ve value). so y has to be only 1 (positive integer) and hence x has to be 2 as no other value will make it less than RHS (1).

Also alternatively, now since x−2| < 1 (value with y=1), -1<x-2<1 and hence 1<x<3 so x=2 which is prime.

Kindly confirm if the above approach for statement 1 is correct.
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27 Jun 2017, 23:08
kreel11 wrote:
Hi Bunuel,

So from statement 1 what I understand is that |x−2|<2−y will only happen if x being positive will be equal to 2. Since it is absolute value and will be positive so for RHS Y cannot be >2 since then RHS<LHS (-ve value). so y has to be only 1 (positive integer) and hence x has to be 2 as no other value will make it less than RHS (1).

Also alternatively, now since x−2| < 1 (value with y=1), -1<x-2<1 and hence 1<x<3 so x=2 which is prime.

Kindly confirm if the above approach for statement 1 is correct.

______________
Yes, that's correct.
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25 Dec 2017, 09:37
1
Hi Bunuel

Please let me know if my approach makes sense

Given: x, y -> +ve
x -> prime?

(1) |x-2| < 2-y
Opening up the Absolute Value:
when +ve
x-2 < 2-y
x+y < 4
OR when -ve
-x+2 < 2-y
-x<-y
x>y

Test Cases:
x=2
y=1
for x+y < 4 or = 3 [Given x, y > 0 & x>y, so total has to be 3]
and so x is prime

Sufficient

(2) x+y-3 = |1-y|
when +ve
x+y-3 = 1-y
x+2y = 4
Testing Cases:
x = 2, y = 1: Only option that works

OR when -ve
x+y-3 = -1+y
x=2 -> prime

Sufficient

Thanks
Prathamesh
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25 Dec 2017, 11:26
ppnimkar wrote:
Hi Bunuel

Please let me know if my approach makes sense

Given: x, y -> +ve
x -> prime?

(1) |x-2| < 2-y
Opening up the Absolute Value:
when +ve
x-2 < 2-y
x+y < 4
OR when -ve
-x+2 < 2-y
-x<-y
x>y

Test Cases:
x=2
y=1
for x+y < 4 or = 3 [Given x, y > 0 & x>y, so total has to be 3]
and so x is prime

Sufficient

(2) x+y-3 = |1-y|
when +ve
x+y-3 = 1-y
x+2y = 4
Testing Cases:
x = 2, y = 1: Only option that works

OR when -ve
x+y-3 = -1+y
x=2 -> prime

Sufficient

Thanks
Prathamesh

I guess when you test values for (1), you consider x + y < 4 and x > y as simultaneous equations, and that is why you test only x = 2 and y = 1, whereas x + y < 4 and x > y are two different cases (true in certain conditions).
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31 Mar 2019, 02:13
Bunuel wrote:
Official Solution:

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 $$2-y \le{0}$$ 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.

HI Bunuel

For S1:-

Since x & y > 0
then the expression |x - 2|< 2 - y can be simplified as follows:-
x -2 < 2 - y (if x>0 , then |x| = x)
x < 4 - y

Now, for y to be a positive integer y should be < 4 (i.e. y can be 1, 2 and 3)

Hence,

if y = 1,
x < 4 - y
x < 4 -1
x < 3 so x can be 1 or 2 ( 1 is NOT A PRIME & 2 IS A PRIME )

if y = 2

x < 4 - 2
x < 2 so x= 1 (1 is NOT A PRIME)

if y = 3

x < 4 - 3
x < 1 (this is not possible since x is a positive integer)

Can you please let me know where am i going wrong??

THANKS!
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01 Apr 2019, 05:41
JIAA wrote:
Bunuel wrote:
Official Solution:

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 $$2-y \le{0}$$ 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.

HI Bunuel

For S1:-

Since x & y > 0
then the expression |x - 2|< 2 - y can be simplified as follows:-
x -2 < 2 - y (if x>0 , then |x| = x)
x < 4 - y

Now, for y to be a positive integer y should be < 4 (i.e. y can be 1, 2 and 3)

Hence,

if y = 1,
x < 4 - y
x < 4 -1
x < 3 so x can be 1 or 2 ( 1 is NOT A PRIME & 2 IS A PRIME )

if y = 2

x < 4 - 2
x < 2 so x= 1 (1 is NOT A PRIME)

if y = 3

x < 4 - 3
x < 1 (this is not possible since x is a positive integer)

Can you please let me know where am i going wrong??

THANKS!

As you correctly wrote |x| = x, when x >= 0 (and |x| = -x, when x < 0).

So, |x - 2| = x - 2, when x - 2 >= 0, or which is the same, when x >= 2. In this case, so when x >= 2, x - 2 < 2 - y.
If y = 1, then x < 3 --> since we are looking for x >= 2, then x = 2.
If y = 2, then x < 2 --> since we are looking for x >= 2, then here we have no solution.
If y = 3, then x < 1 --> since we are looking for x >= 2, then here we have no solution.

Next, the second case would be when |x - 2| = -(x - 2) (|x| = -x, when x < 0). This would happen when x - 2 < 0, or which is the same, when x < 2. In this case, so when x < 2, -(x - 2) < 2 - y.
If y = 1, then x > 1 --> since we are looking for x < 2, then here we have no solution.
If y = 2, then x > 2 --> since we are looking for x > 2, then here we have no solution.
If y = 3, then x > 3 --> since we are looking for x > 2, then here we have no solution.

As you can see we have only one solution: x = 2 = prime.

Hope it's clear.
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01 Apr 2019, 06:59
Bunuel wrote:
JIAA wrote:
Bunuel wrote:
Official Solution:

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 $$2-y \le{0}$$ 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.

HI Bunuel

For S1:-

Since x & y > 0
then the expression |x - 2|< 2 - y can be simplified as follows:-
x -2 < 2 - y (if x>0 , then |x| = x)
x < 4 - y

Now, for y to be a positive integer y should be < 4 (i.e. y can be 1, 2 and 3)

Hence,

if y = 1,
x < 4 - y
x < 4 -1
x < 3 so x can be 1 or 2 ( 1 is NOT A PRIME & 2 IS A PRIME )

if y = 2

x < 4 - 2
x < 2 so x= 1 (1 is NOT A PRIME)

if y = 3

x < 4 - 3
x < 1 (this is not possible since x is a positive integer)

Can you please let me know where am i going wrong??

THANKS!

As you correctly wrote |x| = x, when x >= 0 (and |x| = -x, when x < 0).

So, |x - 2| = x - 2, when x - 2 >= 0, or which is the same, when x >= 2. In this case, so when x >= 2, x - 2 < 2 - y.
If y = 1, then x < 3 --> since we are looking for x >= 2, then x = 2.
If y = 2, then x < 2 --> since we are looking for x >= 2, then here we have no solution.
If y = 3, then x < 1 --> since we are looking for x >= 2, then here we have no solution.

Next, the second case would be when |x - 2| = -(x - 2) (|x| = -x, when x < 0). This would happen when x - 2 < 0, or which is the same, when x < 2. In this case, so when x < 2, -(x - 2) < 2 - y.
If y = 1, then x > 1 --> since we are looking for x < 2, then here we have no solution.
If y = 2, then x > 2 --> since we are looking for x > 2, then here we have no solution.
If y = 3, then x > 3 --> since we are looking for x > 2, then here we have no solution.

As you can see we have only one solution: x = 2 = prime.

Hope it's clear.

THANKS Bunuel

Just one more thing !

It's explicitly stated in Q stem that x is positive i.e. x> 0, shouldn't we consider second case ONLY when we don't know if x is positive or negative??

Why are we even considering second case here (i.e.|x| = -x, when x < 0)??

Re: M02-25   [#permalink] 01 Apr 2019, 06:59

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# M02-25

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