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5. What is the value of y?
(1) 3|x^2 -4| = y - 2
(2) |3 - y| = 11

(1) As we are asked to find the value of y, from this statement we can conclude only that y>=2, as LHS is absolute value which is never negative, hence RHS als can not be negative. Not sufficient.

(2) |3 - y| = 11:

y<3 --> 3-y=11 --> y=-8
y>=3 --> -3+y=11 --> y=14

Two values for y. Not sufficient.

(1)+(2) y>=2, hence y=14. Sufficient.

Answer: C.
Can we not solve the first statement showing positive and negative values when the mod is removed and hence prove the statement is not sufficient??

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Bunuel
7. |x+2|=|y+2| what is the value of x+y?
(1) xy<0
(2) x>2 y<2

This one is quite interesting.

First note that |x+2|=|y+2| can take only two possible forms:

A. x+2=y+2 --> x=y. This will occur if and only x and y are both >= than -2 OR both <= than -2. In that case x=y. Which means that their product will always be positive or zero when x=y=0.
B. x+2=-y-2 --> x+y=-4. This will occur when either x or y is less then -2 and the other is more than -2.

When we have scenario A, xy will be nonnegative only. Hence if xy is negative we have scenario B and x+y=-4. Also note that vise-versa is not right. Meaning that we can have scenario B and xy may be positive as well as negative.

(1) xy<0 --> We have scenario B, hence x+y=-4. Sufficient.

(2) x>2 and y<2, x is not equal to y, we don't have scenario A, hence we have scenario B, hence x+y=-4. Sufficient.

Answer: D.
Shouldn't statement 2 be x>-2 and y<-2 to become true??

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Bunuel
7. |x+2|=|y+2| what is the value of x+y?
(1) xy<0
(2) x>2 y<2

This one is quite interesting.

First note that |x+2|=|y+2| can take only two possible forms:

A. x+2=y+2 --> x=y. This will occur if and only x and y are both >= than -2 OR both <= than -2. In that case x=y. Which means that their product will always be positive or zero when x=y=0.
B. x+2=-y-2 --> x+y=-4. This will occur when either x or y is less then -2 and the other is more than -2.

When we have scenario A, xy will be nonnegative only. Hence if xy is negative we have scenario B and x+y=-4. Also note that vise-versa is not right. Meaning that we can have scenario B and xy may be positive as well as negative.

(1) xy<0 --> We have scenario B, hence x+y=-4. Sufficient.

(2) x>2 and y<2, x is not equal to y, we don't have scenario A, hence we have scenario B, hence x+y=-4. Sufficient.

Answer: D.

Hi Bunuel,
I tried using the number line to simulate A scenario and could get it something like below and this somehow looks fine.

--y/x-------- (-2)-----------

----------(-2)-------------x/y

I am not able to understand B scenario from the number line as we can have any distance between x and y. its just that the distance of x from -2 and y from -2 should be same and in opposite sides. How does the -4 come into the picture here ?

-------x<----------d------->(-2)<---------d-------->y---------


now this d here can be anything right ?
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Bunuel
10. If n is not equal to 0, is |n| < 4 ?
(1) n^2 > 16
(2) 1/|n| > n

Question basically asks is -4<n<4 true.

(1) n^2>16 --> n>4 or n<-4, the answer to the question is NO. Sufficient.

(2) 1/|n| > n, this is true for all negative values of n, hence we can not answer the question. Not sufficient.

Answer: A.

Can you explain why in statement n is true for negative values of n? Can we instead square both sides and solve for n?
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|n| is always non-negative.
for any negative number n, |n| will always be positive
hence, 1/|n| will always be >n
so, we cannot say with confidence that |n| <4
coz n can be -99999999999 ( which does not satisfy |n| <4) , and n can be -1 (which does satisfy |n| <4)
hence, St2 is not sufficient.

ritikapatni
Bunuel
10. If n is not equal to 0, is |n| < 4 ?
(1) n^2 > 16
(2) 1/|n| > n

Question basically asks is -4<n<4 true.

(1) n^2>16 --> n>4 or n<-4, the answer to the question is NO. Sufficient.

(2) 1/|n| > n, this is true for all negative values of n, hence we can not answer the question. Not sufficient.

Answer: A.

Can you explain why in statement n is true for negative values of n? Can we instead square both sides and solve for n?
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Bunuel
7. |x+2|=|y+2| what is the value of x+y?
(1) xy<0
(2) x>2 y<2

This one is quite interesting.

First note that |x+2|=|y+2| can take only two possible forms:

A. x+2=y+2 --> x=y. This will occur if and only x and y are both >= than -2 OR both <= than -2. In that case x=y. Which means that their product will always be positive or zero when x=y=0.
B. x+2=-y-2 --> x+y=-4. This will occur when either x or y is less then -2 and the other is more than -2.

When we have scenario A, xy will be nonnegative only. Hence if xy is negative we have scenario B and x+y=-4. Also note that vise-versa is not right. Meaning that we can have scenario B and xy may be positive as well as negative.

(1) xy<0 --> We have scenario B, hence x+y=-4. Sufficient.

(2) x>2 and y<2, x is not equal to y, we don't have scenario A, hence we have scenario B, hence x+y=-4. Sufficient.

Answer: D.

Hi Bunuel,

when I solve this by squaring method-
the eq becomes --> x+y= 4(y-x)/(x-y)

Now, with this equation I find info of statement 1 irrelevant.
Can you please tell why xy<0 will be relevant.
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Bunuel
7. |x+2|=|y+2| what is the value of x+y?
(1) xy<0
(2) x>2 y<2

This one is quite interesting.

First note that |x+2|=|y+2| can take only two possible forms:

A. x+2=y+2 --> x=y. This will occur if and only x and y are both >= than -2 OR both <= than -2. In that case x=y. Which means that their product will always be positive or zero when x=y=0.
B. x+2=-y-2 --> x+y=-4. This will occur when either x or y is less then -2 and the other is more than -2.

When we have scenario A, xy will be nonnegative only. Hence if xy is negative we have scenario B and x+y=-4. Also note that vise-versa is not right. Meaning that we can have scenario B and xy may be positive as well as negative.

(1) xy<0 --> We have scenario B, hence x+y=-4. Sufficient.

(2) x>2 and y<2, x is not equal to y, we don't have scenario A, hence we have scenario B, hence x+y=-4. Sufficient.

Answer: D.


Sir I have a question, what if in the question statement itself I square both the sides (LHS and RHS)
It'll be like:
x^2 + 4 + 4x = y^2 + 4 + 4y
x^2 - y^2 = 4y - 4x
(x-y) (x+y) = -4 ( x-y)
Therefore (x+y) = -4

Now both of the options tell me that x and y are not equal to zero, hence solved. Is this approach correct?
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Bunuel
7. |x+2|=|y+2| what is the value of x+y?
(1) xy<0
(2) x>2 y<2

This one is quite interesting.

First note that |x+2|=|y+2| can take only two possible forms:

A. x+2=y+2 --> x=y. This will occur if and only x and y are both >= than -2 OR both <= than -2. In that case x=y. Which means that their product will always be positive or zero when x=y=0.
B. x+2=-y-2 --> x+y=-4. This will occur when either x or y is less then -2 and the other is more than -2.

When we have scenario A, xy will be nonnegative only. Hence if xy is negative we have scenario B and x+y=-4. Also note that vise-versa is not right. Meaning that we can have scenario B and xy may be positive as well as negative.

(1) xy<0 --> We have scenario B, hence x+y=-4. Sufficient.

(2) x>2 and y<2, x is not equal to y, we don't have scenario A, hence we have scenario B, hence x+y=-4. Sufficient.

Answer: D.


Sir I have a question, what if in the question statement itself I square both the sides (LHS and RHS)
It'll be like:
x^2 + 4 + 4x = y^2 + 4 + 4y
x^2 - y^2 = 4y - 4x
(x-y) (x+y) = -4 ( x-y)
Therefore (x+y) = -4

Now both of the options tell me that x and y are not equal to zero, hence solved. Is this approach correct?

You cannot reduce by x - y there because x - y could be 0, and we cannot divide by 0.

You can check that approach here: https://gmatclub.com/forum/inequality-a ... l#p1111747

Hope it helps.
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Bunuel
7. |x+2|=|y+2| what is the value of x+y?
(1) xy<0
(2) x>2 y<2

This one is quite interesting.

First note that |x+2|=|y+2| can take only two possible forms:

A. x+2=y+2 --> x=y. This will occur if and only x and y are both >= than -2 OR both <= than -2. In that case x=y. Which means that their product will always be positive or zero when x=y=0.
B. x+2=-y-2 --> x+y=-4. This will occur when either x or y is less then -2 and the other is more than -2.

When we have scenario A, xy will be nonnegative only. Hence if xy is negative we have scenario B and x+y=-4. Also note that vise-versa is not right. Meaning that we can have scenario B and xy may be positive as well as negative.

(1) xy<0 --> We have scenario B, hence x+y=-4. Sufficient.

(2) x>2 and y<2, x is not equal to y, we don't have scenario A, hence we have scenario B, hence x+y=-4. Sufficient.

Answer: D.


Hi Bunuel,

I have query on Statement (2). Since X>2 we know |X+2| will open with positive sign, while Y<2 means that Y can be > -2 or Y can be <-2 , and hence we don't know the way the bracket for |Y+2| will open. Could you please let me know how we are sure that X>2 & Y<2 will result in case B from statement 2?

Regards,
Deepak Patel
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Bunuel
13. Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

Last one.

Is |x-1| < 1? Basically the question asks is 0<x<2 true?

(1) (x-1)^2 <= 1 --> x^2-2x<=0 --> x(x-2)<=0 --> 0<=x<=2. x is in the range (0,2) inclusive. This is the trick here. x can be 0 or 2! Else it would be sufficient. So not sufficient.

(2) x^2 - 1 > 0 --> x<-1 or x>1. Not sufficient.

(1)+(2) Intersection of the ranges from 1 and 2 is 1<x<=2. Again 2 is included in the range, thus as x can be 2, we cannot say for sure that 0<x<2 is true. Not sufficient.

Answer: E.

(x-1)^2 <= 1 --> x^2-2x<=0 --> x(x-2)<=0 --> 0<=x<=2 . I didn't understand this part. how is x>= 0?
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Bunuel
13. Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

Last one.

Is |x-1| < 1? Basically the question asks is 0<x<2 true?

(1) (x-1)^2 <= 1 --> x^2-2x<=0 --> x(x-2)<=0 --> 0<=x<=2. x is in the range (0,2) inclusive. This is the trick here. x can be 0 or 2! Else it would be sufficient. So not sufficient.

(2) x^2 - 1 > 0 --> x<-1 or x>1. Not sufficient.

(1)+(2) Intersection of the ranges from 1 and 2 is 1<x<=2. Again 2 is included in the range, thus as x can be 2, we cannot say for sure that 0<x<2 is true. Not sufficient.

Answer: E.

(x-1)^2 <= 1 --> x^2-2x<=0 --> x(x-2)<=0 --> 0<=x<=2 . I didn't understand this part. how is x>= 0?

This is explained in detail here: Solving Quadratic Inequalities: Graphic Approach


You could also approach \((x-1)^2 \leq 1\) in another way:
Take the square root: \(|x-1| \leq 1\) (recall that \(\sqrt{a^2}=|a|\));
Get rid of the modulus sign: \(-1 \leq x-1 \leq 1\);
Add 1 to all three parts of the inequality: \(- \leq x \leq 2\).

9. Inequalities



For more check Ultimate GMAT Quantitative Megathread



Hope it helps.
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Bunuel
13. Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

Last one.

Is |x-1| < 1? Basically the question asks is 0<x<2 true?

(1) (x-1)^2 <= 1 --> x^2-2x<=0 --> x(x-2)<=0 --> 0<=x<=2. x is in the range (0,2) inclusive. This is the trick here. x can be 0 or 2! Else it would be sufficient. So not sufficient.

(2) x^2 - 1 > 0 --> x<-1 or x>1. Not sufficient.

(1)+(2) Intersection of the ranges from 1 and 2 is 1<x<=2. Again 2 is included in the range, thus as x can be 2, we cannot say for sure that 0<x<2 is true. Not sufficient.

Answer: E.


With choice 1, if x were 2 wouldnt |2-1| < 1 = |1|<1 = 1<1? and If x were 0 |0-1| < 1, |-1| < 1, 1<1? Which answers the question if |x-1| < 1.
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Bunuel
13. Is |x-1| < 1?
(1) (x-1)^2 <= 1
(2) x^2 - 1 > 0

Last one.

Is |x-1| < 1? Basically the question asks is 0<x<2 true?

(1) (x-1)^2 <= 1 --> x^2-2x<=0 --> x(x-2)<=0 --> 0<=x<=2. x is in the range (0,2) inclusive. This is the trick here. x can be 0 or 2! Else it would be sufficient. So not sufficient.

(2) x^2 - 1 > 0 --> x<-1 or x>1. Not sufficient.

(1)+(2) Intersection of the ranges from 1 and 2 is 1<x<=2. Again 2 is included in the range, thus as x can be 2, we cannot say for sure that 0<x<2 is true. Not sufficient.

Answer: E.


With choice 1, if x were 2 wouldnt |2-1| < 1 = |1|<1 = 1<1? and If x were 0 |0-1| < 1, |-1| < 1, 1<1? Which answers the question if |x-1| < 1.

(1) is \((x-1)^2 \leq 1\), notice that there is the less than or equal sign (\(\leq\)) not the less than sign (\(<\)). So, both 2 and 0 satisfy the given inequality:

    \((x-1)^2 \leq 1\)

Take the square root:

    \(|x-1| \leq 1\)

Get rid of the modulus:

    \(-1 \leq x-1 \leq 1\)

Add -1 to all three parts:

    \(-2 \leq x \leq 0\)

Hope it helps.
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Hi Bunuel,

If we solve both the equation in 1 & 2 as follow-

1) (x + y)^2 = 9a
x^2 + y^2 + 2xy = 9a
x^2 + y^2 = 9a - 2xy

2) (x – y)^2 = a
x^2 + y^2 - 2xy = a
x^2 + y^2 = a + 2xy

Equating both the equations give us
9a - 2xy = a + 2xy
8a = 4xy
2a = xy

Putting this in (1) & (2)

x^2 + y^2 = 9a - 2xy = 9a - 4a = 5a

x^2 + y^2 = a + 2xy = a + 4a = 5a

This providers the solution to the question, hence should'nt the answer be C?
Please let me know if I miscalculated anything here. Thanks!

Bunuel
3. Is x^2 + y^2 > 4a?
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a

(1) (x + y)^2 = 9a --> x^2+2xy+y^2=9a. Clearly insufficient.

(2) (x – y)^2 = a --> x^2-2xy+y^2=a. Clearly insufficient.

(1)+(2) Add them up 2(x^2+y^2)=10a --> x^2+y^2=5a. Also insufficient as x,y, and a could be 0 and x^2 + y^2 > 4a won't be true, as LHS and RHS would be in that case equal to zero. Not sufficient.

Answer: E.
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Hi Bunuel,

If we solve both the equation in 1 & 2 as follow-

1) (x + y)^2 = 9a
x^2 + y^2 + 2xy = 9a
x^2 + y^2 = 9a - 2xy

2) (x – y)^2 = a
x^2 + y^2 - 2xy = a
x^2 + y^2 = a + 2xy

Equating both the equations give us
9a - 2xy = a + 2xy
8a = 4xy
2a = xy

Putting this in (1) & (2)

x^2 + y^2 = 9a - 2xy = 9a - 4a = 5a

x^2 + y^2 = a + 2xy = a + 4a = 5a

This providers the solution to the question, hence should'nt the answer be C?
Please let me know if I miscalculated anything here. Thanks!

Bunuel
3. Is x^2 + y^2 > 4a?
(1) (x + y)^2 = 9a
(2) (x – y)^2 = a

(1) (x + y)^2 = 9a --> x^2+2xy+y^2=9a. Clearly insufficient.

(2) (x – y)^2 = a --> x^2-2xy+y^2=a. Clearly insufficient.

(1)+(2) Add them up 2(x^2+y^2)=10a --> x^2+y^2=5a. Also insufficient as x,y, and a could be 0 and x^2 + y^2 > 4a won't be true, as LHS and RHS would be in that case equal to zero. Not sufficient.

Answer: E.

No. The answer is E. I believe your doubt has already been addressed in this thread.
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