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Re: Inequality and absolute value questions from my collection
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05 Apr 2018, 12:24
Aditi10 wrote: Bunuel wrote: 4. Are x and y both positive? (1) 2x2y=1 (2) x/y>1
(1) 2x2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x2y=1 > y=x1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.
(2) x/y>1 > x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.
(1)+(2) Again it can be done with different approaches. You should just find the one which is the less timeconsuming and comfortable for you personally.
One of the approaches: \(2x2y=1\) > \(x=y+\frac{1}{2}\) \(\frac{x}{y}>1\) > \(\frac{xy}{y}>0\) > substitute x > \(\frac{1}{y}>0\) > \(y\) is positive, and as \(x=y+\frac{1}{2}\), \(x\) is positive too. Sufficient.
Answer: C. Hi Bunuel, I solved it as below (1) 2x2y=1 > xy= 1/2. This means x>y by 1/2 but x can be 1/2 and y=0 so this is insufficient (2) x/y=1 > x>y which is no different from (1) so insufficient againBut I didn't follow your approach on combining the 2 statements, how did you get x to be substituted by 1/y? The fact that x is positive is already proved by (1) isn't it? Please help out. The red parts are not correct. \(\frac{x}{y}>1\) does not mean that \(x>y\). If both \(x\) and \(y\) are positive, then \(x>y\), BUT if both are negative, then \(x<y\). What you are actually doing when writing \(x>y\) from \(\frac{x}{y}>1\) is multiplying both parts of inequality by \(y\): never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero. So from (2) \(\frac{x}{y}>1\), we can only deduce that \(x\) and \(y\) have the same sigh (either both positive or both negative). Also, from (1) we can only say that x > y (because x = y + 1/2 = y + positive number) but we cannot say that x is positive. For example, consider x = 2 and y = 2.5. Finally, when we substitute \(x = y + \frac{1}{2}\) into \(\frac{x}{y}>1\) we'll get: \(\frac{y + \frac{1}{2}}{y}>1\); \(1 + \frac{1}{2y}>1\); \(\frac{1}{2y}>0\); \(y > 0\). If \(y > 0\), then \(x > y > 0\). Hope it helps.
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Re: Inequality and absolute value questions from my collection
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17 Apr 2018, 09:29
Hi This is really an interesting question, I want to share an alternative approach. Concept xa=b, it means distance of x from a = bit can be logically deducted that , if xa=ya, it means either x= y or the a and y are equidistant from a, hence average of x & y = a , or x+y = 2ait can be proved but if visualization can save time. 7. x+2=y+2 what is the value of x+y? x+2=y+2, it means either x and y are equal or are equidistant from 2. (1) xy<0 it means one is positive and one is negative, hence they are not equal. so x and y are equidistant from 2, hence sum of x and y = 4. SUFFICIENT. (2) x>2 y<2 x not equal to y, hence their sum is 4. SUFFICIENT. Hence Answer D Bunuel wrote: 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=y2 > 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 viseversa 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.
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Re: Inequality and absolute value questions from my collection
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17 Apr 2018, 23:32
Bunuel wrote: 13. Is x1 < 1? (1) (x1)^2 <= 1 (2) x^2  1 > 0
Last one.
Is x1 < 1? Basically the question asks is 0<x<2 true?
(1) (x1)^2 <= 1 > x^22x<=0 > x(x2)<=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. Still not clear on this one. Can you please explain why is 1 insufficient I am not able to eliminate 1 also why is not C sufficient? Thanks in Advance



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Re: Inequality and absolute value questions from my collection
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18 Apr 2018, 01:15
Is x1 <1? Since Absolute value function is always non negative, we can square both sides, We get is (x1)^2<1? Statement 1 is (x1)^2< = 1 If (x1)^2<1: answer to question is yes (x1)^2= 1: hence answer is No, So Statement 1 is NOT SUFFICIENT. Statement2: Question stem (x1)^2<1? Can be reduced to is x(x2)<0 Or is 0<x<2 Now St2: x^21>0 gives x >1 or x <1 Now x can be 2, answer to question stem is No Or 1.5 answer to question stem is yes. Hence NOT SUFFICIENT. Combining both Statement 1 & 2, we get X can be 2, answer to question stem NO Or 1.5 answer to question stem is yes. Hence Answer E Buttercup3 wrote: Bunuel wrote: 13. Is x1 < 1? (1) (x1)^2 <= 1 (2) x^2  1 > 0
Last one.
Is x1 < 1? Basically the question asks is 0<x<2 true?
(1) (x1)^2 <= 1 > x^22x<=0 > x(x2)<=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. Still not clear on this one. Can you please explain why is 1 insufficient I am not able to eliminate 1 also why is not C sufficient? Thanks in Advance
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Re: Inequality and absolute value questions from my collection
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27 Jun 2018, 07:14
Bunuel wrote: 2. If y is an integer and y = x + x, is y = 0?
Notice that from \(y=x+x\) it follows that y cannot be negative: If \(x>0\), then \(y=x+x=2x=2*positive=positive\); If \(x\leq{0}\) (when x is negative or zero) then \(y=x+x=0\).
(1) \(x<0\) > \(y=x+x=x+x=0\). Sufficient.
(2) \(y<1\). We found out above that y cannot be negative and we are given that y is an integer, hence \(y=0\). Sufficient.
Answer: D. I forgot the instruction by the time, I started assessing option B. Y is an integer.
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Re: Inequality and absolute value questions from my collection
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27 Jun 2018, 07:33
Bunuel wrote: 5. What is the value of y? (1) 3x^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 > 3y=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. I have done a silly mistake while solving this problem, in my mind I assumed that the statement B is an inequality rather than equal sign.
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12 Jul 2018, 00:47
sriharimurthy wrote: Quote: 1. If 6*x*y = x^2*y + 9*y, what is the value of xy? (1) y – x = 3 (2) x^3< 0 \(6*x*y = x^2*y + 9*y\) \(6*x = x^2 + 9 = 0\) \(x^2  6*x + 9 = 0\) \((x3)^2 = 0\) \(x = 3\) St. (1) : y  x = 3
y = 6 Sufficient. St. (2) : x^3 < 0
Invalid statement. Does not give us value of y. Insufficient. Answer : Asolving the equation gives either y=0 or x=3 (1) y=x+3 so if y=0, x=3, xy=0 if x=3 y =6, xy=18 insufficient (2) x^3<0 means x<0, so y=0 and xy=0 sufficient (B)



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Re: Inequality and absolute value questions from my collection
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15 Jul 2018, 03:00
Bunuel wrote: 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=y2 > 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 viseversa 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 Bunnuel, I didnt quite understand the part where you said "B. x+2=y2 > x+y=4. This will occur when either x or y is less than 2 and the other is more than 2."why can't x & y be 2 & 2 here?



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Re: Inequality and absolute value questions from my collection
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15 Jul 2018, 06:56
blueviper wrote: Bunuel wrote: 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=y2 > 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 viseversa 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 Bunnuel, I didnt quite understand the part where you said "B. x+2=y2 > x+y=4. This will occur when either x or y is less than 2 and the other is more than 2."why can't x & y be 2 & 2 here? You might find the following solution simpler: https://gmatclub.com/forum/inequalitya ... l#p1111747Hope it helps.
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Re: Inequality and absolute value questions from my collection
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31 Jul 2018, 14:56
Bunuel wrote: 8. a*b#0. Is a/b=a/b? (1) a*b=a*b (2) a/b=a/b a/b=a/b is true if and only a and b have the same sign, meaning a/b is positive.
(1) a*b=a*b, means a and b are both positive or both negative, as LHS is never negative (well in this case LHS is positive as neither a nor b equals to zero). Hence a/b is positive in any case. Hence a/b=a/b. Sufficient.
(2) a/b=a/b, from this we can not conclude whether they have the same sign or not. Not sufficient.
Answer: A. Can someone clarify whether a/b=a\b exists as a property?



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31 Jul 2018, 20:49
arunjohn43 wrote: Bunuel wrote: 8. a*b#0. Is a/b=a/b? (1) a*b=a*b (2) a/b=a/b a/b=a/b is true if and only a and b have the same sign, meaning a/b is positive.
(1) a*b=a*b, means a and b are both positive or both negative, as LHS is never negative (well in this case LHS is positive as neither a nor b equals to zero). Hence a/b is positive in any case. Hence a/b=a/b. Sufficient.
(2) a/b=a/b, from this we can not conclude whether they have the same sign or not. Not sufficient.
Answer: A. Can someone clarify whether a/b=a\b exists as a property? Yes, \(\frac{a}{b}=\frac{a}{b}\) and \(ab=a*b\) are generally true.
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Re: Inequality and absolute value questions from my collection
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17 Aug 2018, 23:00
In case anybody struggles with #4. I found it easier to think about (1+2) as 2 cases: Stmt 1: Step 1) 2x2y =1 Step 2) 2(xy) = 1 Step 3) xy = 1/2 At this point its easy to think of cases when either x or y are negative or both are positive, for example x = 0 y = (1/2) or x = 1 and y = 1/2, Hence, NS  BCE
Stmt 2: x/y > 1 From this statement it should be clear that both x and y are either negative or positive. NS  CE
Now Stmt 1 and 2 combined: consider 2 cases from stmt 2 split: case 1  both positive: x/y > 1 => (multiply both sides by "y", since "y" > 0 we keep the sign) x > y, we can still satisfy the condition from stmt 1. Example x = 1 and y = 1/2
case 2  both negative: x/y > 1 => (multiply both sides by "y", since "y" < 0 we flip the sign) x < y, no we can no longer satisfy the condition from stmt 1. x  y = 1/2. (x  y) has to be positive. We cannot subtract something bigger (y) from something smaller (x) and get positive number. Hence, case 2 does not work. (1+2) combined Sufficient  Answer is C



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14 Sep 2018, 11:36
Bunuel wrote: 2. If y is an integer and y = x + x, is y = 0?
Notice that from \(y=x+x\) it follows that y cannot be negative: If\(x>0\), then \(y=x+x=2x=2*positive=positive\); If \(x\leq{0}\) (when x is negative or zero) then \(y=x+x=0\).
(1) \(x<0\) > \(y=x+x=x+x=0\). Sufficient.
(2) \(y<1\). We found out above that y cannot be negative and we are given that y is an integer, hence \(y=0\). Sufficient.
Answer: D. Hi Bunuel, Shouldn't it be like this if x=x if \(x\geq 0\)then y=x+x which means y=2x and if f x=  x if x<0 then y=x+x which means y=0 Probus.
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15 Sep 2018, 03:53
Probus wrote: Bunuel wrote: 2. If y is an integer and y = x + x, is y = 0?
Notice that from \(y=x+x\) it follows that y cannot be negative: If\(x>0\), then \(y=x+x=2x=2*positive=positive\); If \(x\leq{0}\) (when x is negative or zero) then \(y=x+x=0\).
(1) \(x<0\) > \(y=x+x=x+x=0\). Sufficient.
(2) \(y<1\). We found out above that y cannot be negative and we are given that y is an integer, hence \(y=0\). Sufficient.
Answer: D. Hi Bunuel, Shouldn't it be like this if x=x if \(x\geq 0\)then y=x+x which means y=2x and if f x=  x if x<0 then y=x+x which means y=0 Probus. Both are correct. If x = 0, then 0 = 0 as well 0 = 0 = 0.
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Re: Inequality and absolute value questions from my collection
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16 Nov 2018, 11:02
Bunuel wrote: 9. Is n<0? (1) n=n (2) n^2=16
(1) n=n, means that either n is negative OR n equals to zero. We are asked whether n is negative so we can not be sure. Not sufficient.
(2) n^2=16 > n=4 or n=4. Not sufficient.
(1)+(2) n is negative OR n equals to zero from (1), n is 4 or 4 from (2). > n=4, hence it's negative, sufficient.
Answer: C. Is (1) even a correct clue? How can an absolute value be negative? Please help me understand.



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18 Nov 2018, 01:45
PB1712989 wrote: Bunuel wrote: 9. Is n<0? (1) n=n (2) n^2=16
(1) n=n, means that either n is negative OR n equals to zero. We are asked whether n is negative so we can not be sure. Not sufficient.
(2) n^2=16 > n=4 or n=4. Not sufficient.
(1)+(2) n is negative OR n equals to zero from (1), n is 4 or 4 from (2). > n=4, hence it's negative, sufficient.
Answer: C. Is (1) even a correct clue? How can an absolute value be negative? Please help me understand. We have n=n. Now, if n is negative or 0 then n is positive or 0, so the absolute value, which is on the right hand side, does not equal to negative number. All is correct there.
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Re: Inequality and absolute value questions from my collection
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25 Nov 2018, 06:30
4. Are x and y both positive? (1) 2x2y=1 (2) x/y>1
(1) 2x2y=1. Well this one is clearly insufficient. You can do it with number plugging OR consider the following: x and y both positive means that point (x,y) is in the I quadrant. 2x2y=1 > y=x1/2, we know it's an equation of a line and basically question asks whether this line (all (x,y) points of this line) is only in I quadrant. It's just not possible. Not sufficient.
(2) x/y>1 > x and y have the same sign. But we don't know whether they are both positive or both negative. Not sufficient.
(1)+(2) Again it can be done with different approaches. You should just find the one which is the less timeconsuming and comfortable for you personally.
One of the approaches: 2x−2y=12x−2y=1 > x=y+12x=y+12 xy>1xy>1 > x−yy>0x−yy>0 > substitute x > 1y>01y>0 > yy is positive, and as x=y+12x=y+12, xx is positive too. Sufficient.
Hello Bunuel,
Please help me in my understanding of statement 2.
x/y >1 we can write it as xy>y^2 Therefore, y(xy)>0 y(yx)<0 (sign flips on converting xy to yx) so we have 2 points on number line now 0 and x. hence, y lies between 0 and x >(0<y<x) this gives us both x and y are more than zero. so my answer was B. can we take variables on number line like I did?



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06 Dec 2018, 18:47
Hi Bunuelwas not able to find your answer for 1st Question. Can you post it. Also, I was able to get x = 3 from question stem, however statement II states that X^3 = Negative, I dont get it ? how can statement II clash with stem ? X^3 = Negative, definitely means X = Negative as per statement II, can you please help



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06 Dec 2018, 21:36
hero_with_1000_faces wrote: Hi Bunuelwas not able to find your answer for 1st Question. Can you post it. Also, I was able to get x = 3 from question stem, however statement II states that X^3 = Negative, I dont get it ? how can statement II clash with stem ? X^3 = Negative, definitely means X = Negative as per statement II, can you please help Check here: https://gmatclub.com/forum/inequalitya ... ml#p653690
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Inequality and absolute value questions from my collection
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25 Jan 2019, 09:55
Bunuel wrote: 5. What is the value of y? (1) 3x^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 > 3y=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. Dear Bunuel I understood your solution and I agree with it, but when I solved it in a different manner I got a different answer. Can you please tell me where went wrong? I solved it this way: (1) 3x^2  4 = y2 x^2  4 = (y2)/3 Case 1: x^2  4 = (y2)/3 Case 2: x^2  4 =  (y2)/3 Solving Case 1: x^2 = (y2)/3 + 4 x^2 = (y2+12)/3 x^2 = (y+10)/3 This means, (y+10)/3 >= 0 Therefore, y >= 10 Solving Case 2: x^2 = 4  (y2)/3 x^2 = (12y+2)/3 x^2 = (14y)/3 This means, (14y)/3 >= 0 Therefore, y <= 14 From Case 1 and 2 we get 10 <= y <= 14 (2) 3  y = 11: y<3 > 3y=11 > y=8 y>=3 > 3+y=11 > y=14 Two values for y. Not sufficient. (1)+(2) 10 <= y <= 14, y=8 or y=14. Still two values of y. Therefore, Not Sufficient. Option E




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