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Re: Inequality and absolute value questions from my collection
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16 Jul 2010, 07:21
ramadossvinodh wrote: Hi Bunuel, I forgot to thank you for these great questions and solutions. You are the BEST!!!!
If y is an integer and y = x + x, is y = 0? (1) x < 0 (2) y < 1
(2) y<1, as we concluded y is never negative, and we are given that y is an integer, hence . Sufficient.
In the above problem, it is very obvious that Y is not equal to 0 from the 2nd statement. Do you think GMAT questions will have this much obvious statements? The trick here is to conclude that \(y\) can not be negative, the rest is relatively easy. And yes, I think you can see such questions on GMAT.
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20 Jul 2010, 01:34
lagomez wrote: 10. If n is not equal to 0, is n < 4 ? (1) n^2 > 16 (2) 1/n > n
answer A because in number 2 n can be negative or a fraction Doubt :The question is is N<4 ,from statement 2 we can see that n can only be a fraction or its value lies between 0 and 1 like say 1/2 .1/3 etc.same for negative fractions for example for n =1/2 , 1/(1/2)=2 which is greater than n=1/2 Also for n=1/2,1/(1/2)=2 which is greater than n=1/2 for negative integers also its true ,say we take n=3 , 1/n=1/3 which is greater than n=3 But point to note is that both negative nos and fractions are less than 4 ,so this statement is equally sufficient ,correct me if i am wrong !!!!



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20 Jul 2010, 01:47
amlan009 wrote: lagomez wrote: 10. If n is not equal to 0, is n < 4 ? (1) n^2 > 16 (2) 1/n > n
answer A because in number 2 n can be negative or a fraction Doubt :The question is is N<4 ,from statement 2 we can see that n can only be a fraction or its value lies between 0 and 1 like say 1/2 .1/3 etc.same for negative fractions for example for n =1/2 , 1/(1/2)=2 which is greater than n=1/2 Also for n=1/2,1/(1/2)=2 which is greater than n=1/2 for negative integers also its true ,say we take n=3 , 1/n=1/3 which is greater than n=3 But point to note is that both negative nos and fractions are less than 4 ,so this statement is equally sufficient ,correct me if i am wrong !!!! \(\frac{1}{n}>n\) holds true for ALL negative values of \(n\), as if \(n<0\) then \(LHS=positive>RHS=negative\). Hence we don't know whether \(4<n<4\) is true. That's why statement (2) is not sufficient. The complete range of \(n\) for which \(\frac{1}{n}>n\) holds true is \(n<1\). P.S. OA's and solutions for all question are given in my posts on pages 2 and 3. Hope it's clear.
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31 Oct 2010, 23:55
Dear Bunuel,
I don't quite understand your explanation on question 4, could you please explain to me again?. When I solved both statement 1 and 2, the answer can be + or , therefore, how the answer comes up with only 1 sign?.
Thanks New member



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01 Nov 2010, 00:04
sumitra wrote: Dear Bunuel,
I don't quite understand your explanation on question 4, could you please explain to me again?. When I solved both statement 1 and 2, the answer can be + or , therefore, how the answer comes up with only 1 sign?.
Thanks New member Welcome to GMATClub ! 4. Are x and y both positive? (1) 2x2y=1 (2) x/y>1 (1) 2(xy)=1 OR (xy)=0.5 Does not need both to be positive or not, EG. x=5, y=4.5 ... x=4.5, y=5 (2) x/y>1 Again not sufficient. EG. x=5, y=2 ... x=5,y=2 (1+2) x/y>1 does mean that the sign of x & y has to be the same since their ratio is greater than 0 Now (xy)=0.5 is also true which means x is greater than y Either x&y are both negative or both positive If x and y are both negative, and we know x is greater than y, then (x/y) will be less than 1 (x=3, y=4 .. x/y=3/4<1) which contradicts statement 1 Hence x & y must both be positive Sufficient Answer is (C) : Both statements together but neither alone is sufficient
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01 Nov 2010, 06:46
sumitra wrote: Dear Bunuel,
I don't quite understand your explanation on question 4, could you please explain to me again?. When I solved both statement 1 and 2, the answer can be + or , therefore, how the answer comes up with only 1 sign?.
Thanks New member Hi, and welcome to Gmat Club. I guess you understand why each statement alone is not sufficient. As for (1)+(2): From (1): \(2x2y=1\) > \(x=y+\frac{1}{2}\) From (2) \(\frac{x}{y}>1\), we can only deduce that x and y have the same sigh (either both positive or both negative). When we consider two statement together: From (2): \(\frac{x}{y}>1\) > \(\frac{x}{y}1>0\) > \(\frac{xy}{y}>0\) > substitute \(x\) from (1) > \(\frac{y+\frac{1}{2}y}{y}>0\)> \(\frac{1}{2y}>0\) ( we can drop 2 as it won't affect anything here and write as I wrote \(\frac{1}{y}>0\), but basically it's the same) > \(\frac{1}{2y}>0\) means \(y\) is positive, and from (2) we know that if y is positive x must also be positive. OR: as \(y\) is positive and as from (1) \(x=y+\frac{1}{2}\), \(x=positive+\frac{1}{2}=positive\), hence \(x\) is positive too. Hope it helps.
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Re: Inequality and absolute value questions from my collection
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19 Dec 2010, 02:12
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 I am not sure if the answer is C.From my point of view it needs to be E.Reason is even if we combine both I and II ,the answer is insufficient. Consider putting the values as 1) x=1 y=1/2 ,now x>y and xy=1/2.In this case both x and y are positive Now2) x=1 and y=1/2,now again x>y and xy=1/2.In this case both are negative. Hence insufficient to deduce whther x and y are both positive. Please share your inputs on the same.



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19 Dec 2010, 02:27
Eshika 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 I am not sure if the answer is C.From my point of view it needs to be E.Reason is even if we combine both I and II ,the answer is insufficient. Consider putting the values as 1) x=1 y=1/2 ,now x>y and xy=1/2.In this case both x and y are positive Now2) x=1 and y=1/2,now again x>y and xy=1/2.In this case both are negative. Hence insufficient to deduce whther x and y are both positive. Please share your inputs on the same. OA for this question is C, not E.The red part in your reasoning is not correct: if x=1 and y=1/2 then y>x. Here is the logic for C:From (2) \(\frac{x}{y}>1\), we can only deduce that x and y have the same sigh (either both positive or both negative). When we consider two statement together: From (1): \(2x2y=1\) > \(x=y+\frac{1}{2}\) From (2): \(\frac{x}{y}>1\) > \(\frac{x}{y}1>0\) > \(\frac{xy}{y}>0\) > substitute \(x\) from (1) > \(\frac{y+\frac{1}{2}y}{y}>0\)> \(\frac{1}{2y}>0\) ( we can drop 2 as it won't affect anything here and write as I wrote \(\frac{1}{y}>0\), but basically it's the same) > \(\frac{1}{2y}>0\) means \(y\) is positive, and from (2) we know that if y is positive x must also be positive. OR: as \(y\) is positive and as from (1) \(x=y+\frac{1}{2}\), \(x=positive+\frac{1}{2}=positive\), hence \(x\) is positive too. Does it make sense now?
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24 Jan 2012, 08:01
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=2. 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. (...)
Bunuel, how did you figure out that x+2=y+2 can take only two possible forms? I did it by evaualting four scenarios when I combine these possibilities: x+2>0, x+2<0, y+2>0, y+2<'0 Is there a faster method to do it? I tried by using the method proposed by "walker", but I think that it doesn't work when there are more than one variable. Please your help. Thanks!
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Re: Inequality and absolute value questions from my collection
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24 Jan 2012, 12:47
metallicafan 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=2. 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. (...)
Bunuel, how did you figure out that x+2=y+2 can take only two possible forms? I did it by evaualting four scenarios when I combine these possibilities: x+2>0, x+2<0, y+2>0, y+2<'0 Is there a faster method to do it? I tried by using the method proposed by "walker", but I think that it doesn't work when there are more than one variable. Please your help. Thanks! We have \(x+2=y+2\). If both absolute values expand with + or  sign we'll get: \(x+2=y+2\) (notice that \((x+2)=(y+2)\) is exactly the same); If they will expand with different signs we'll get: \((x+2)=y+2\) (notice that \(x+2=(y+2)\) is exactly the same). So only two forms. Hope it's clear.
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23 Feb 2012, 10:41
Quote: 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.
In the first statement from 2x2y=1 > we can sat xy=1/2 So it cud be 8.58 or 0.25  (0.25) HOw can we say both x and y are positive? similarily statement 2 x/y>1 =>x>y how can we be sure x and y have the same sign we can have 8>7 or 8>8 Bunel can you pls xplain..or am i missing sumtin fundamental?



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23 Feb 2012, 11:00
shankar245 wrote: Quote: 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.
In the first statement from 2x2y=1 > we can sat xy=1/2 So it cud be 8.58 or 0.25  (0.25) HOw can we say both x and y are positive? similarily statement 2 x/y>1 =>x>y how can we be sure x and y have the same sign we can have 8>7 or 8>8 Bunel can you pls xplain..or am i missing sumtin fundamental? For (1) we are not saying that x and y are positive, we saying that if they were both positive then it would mean that the line y=x1/2 is only in I quadrant, which is not possible. So from (1) we cannot say that x and y are both positive. Or consider following numbers: x=2, y=1.5 and x=2, y=2.5. For (2). First of all you can not multiply x/y>1 by y and write x>y, since you don't know the sign of y: if it's positive then x>y but if it's negative then when multiplying by negative number you should flip the sing of the inequality and write x<y. Never multiply or divide inequality by a variable (or by an expression with variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequality.Next, since x/y>1 then x/y>0, which means that x and y are either both positive or both negative, hence they have the same sign. Hope it's clear.
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25 Feb 2012, 06:51
Quote: 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. when we have x+y=4 Acc to the explanation , we have This will occur when either x or y is less then 2 and the other is more than 2so x can be be 3,4,5..(X<2) and y can be 1,0,1...(y>2) a solution could that 22=4 in this case it does not satisfy ? May be this is trivial but your clarification will help me correct my assumption Thanks in advance.



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Re: Inequality and absolute value questions from my collection
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25 Feb 2012, 08:06
shankar245 wrote: Quote: 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. when we have x+y=4 Acc to the explanation , we have This will occur when either x or y is less then 2 and the other is more than 2so x can be be 3,4,5..(X<2) and y can be 1,0,1...(y>2) a solution could that 22=4 in this case it does not satisfy ? May be this is trivial but your clarification will help me correct my assumption Thanks in advance. When we consider ranges to expand an absolute value we should put equal sign (=) in either of the range. For our question we put equal sign for the first range (case A) when we are analyzing the case when x and y are both \(\geq{2}\) than 2 OR both \(\leq{2}\). So the scenario when \(x=y=2\) (\(x+y=22=4\)) is included in case A. Hope it's clear.
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Re: Inequality and absolute value questions from my collection
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09 Aug 2012, 16:06
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 Bunuel, I am getting E and just cannot understand D. Please see my solution below  I used number picking. A. xy<0, x=+ and y= For this condition choosing different values of x and y (x=2,y=6: x=3, y=7)satisfies the given condition in modulus. Hence x=y can be different value or x= and y=+  This condition doesn't satisfy the modulus condiotion B x>2 and y<2  As per the above stmt 1  condition 1, there can be various values for x and y, hence x+y is different. Hence E. I know I am going wrong some where, please help. thanks jay



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Re: Inequality and absolute value questions from my collection
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09 Aug 2012, 16:21
jayaddula 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 Bunuel, I am getting E and just cannot understand D. Please see my solution below  I used number picking. A. xy<0, x=+ and y= For this condition choosing different values of x and y (x=2,y=6: x=3, y=7)satisfies the given condition in modulus. Hence x=y can be different value or x= and y=+  This condition doesn't satisfy the modulus condiotion B x>2 and y<2  As per the above stmt 1  condition 1, there can be various values for x and y, hence x+y is different. Hence E. I know I am going wrong some where, please help. thanks jay In your example, both pairs give the same value for x+y: 26=4 and 37=4. We can solve this question in another way: 7. x+2=y+2 what is the value of x+y?Square both sides: \(x^2+4x+4=y^2+4y+4\) > \(x^2y^2+4x4y=0\) > \((x+y)(xy)+4(xy)=0\) > \((xy)(x+y+4)=0\) > either \(x=y\) or \(x+y=4\). (1) xy<0 > the first case is not possible, since if \(x=y\), then \(xy=x^2\geq{0}\), not \(<0\) as given in this statement, hence we have the second case: \(x+y=4\). Sufficient. (2) x>2 and y<2. This statement implies that \(x\neq{y}\), therefore \(x+y=4\). Sufficient. Answer: D. Hope it's clear.
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Re: Inequality and absolute value questions from my collection
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25 Aug 2012, 06:26
Bunuel wrote: SOLUTIONS:
1. If 6*x*y = x^2*y + 9*y, what is the value of xy? (1) y – x = 3 (2) x^3< 0
First let's simplify given expression \(6*x*y = x^2*y + 9*y\):
\(y*(x^26x+9)=0\) > \(y*(x3)^2=0\). Note here that we CAN NOT reduce this expression by \(y\), as some of you did. Remember we are asked to determine the value of \(xy\), and when reducing by \(y\) you are assuming that \(y\) doesn't equal to \(0\). We don't know that.
Next: we can conclude that either \(x=3\) or/and \(y=0\). Which means that \(xy\) equals to 0, when y=0 and x any value (including 3), OR \(xy=3*y\) when y is not equal to zero, and x=3.
(1) \(yx=3\). If y is not 0, x must be 3 and yx to be 3, y must be 6. In this case \(xy=18\). But if y=0 then x=3 and \(xy=0\). Two possible scenarios. Not sufficient.
OR:
\(yx=3\) > \(x=y3\) > \(y*(x3)^2=y*(y33)^2=y(y6)^2=0\) > either \(y=0\) or \(y=6\) > if \(y=0\), then \(x=3\) and \(xy=0\) \(or\) if \(y=6\), then \(x=3\) and \(xy=18\). Two different answers. Not sufficient.
(2) \(x^3<0\). x is negative, hence x is not equals to 3, hence y must be 0. So, xy=0. Sufficient.
Answer: B.
The answer to this one is C right? B alone is not sufficient.



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Re: Inequality and absolute value questions from my collection
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26 Aug 2012, 00:23
dvinoth86 wrote: Bunuel wrote: SOLUTIONS:
1. If 6*x*y = x^2*y + 9*y, what is the value of xy? (1) y – x = 3 (2) x^3< 0
First let's simplify given expression \(6*x*y = x^2*y + 9*y\):
\(y*(x^26x+9)=0\) > \(y*(x3)^2=0\). Note here that we CAN NOT reduce this expression by \(y\), as some of you did. Remember we are asked to determine the value of \(xy\), and when reducing by \(y\) you are assuming that \(y\) doesn't equal to \(0\). We don't know that.
Next: we can conclude that either \(x=3\) or/and \(y=0\). Which means that \(xy\) equals to 0, when y=0 and x any value (including 3), OR \(xy=3*y\) when y is not equal to zero, and x=3.
(1) \(yx=3\). If y is not 0, x must be 3 and yx to be 3, y must be 6. In this case \(xy=18\). But if y=0 then x=3 and \(xy=0\). Two possible scenarios. Not sufficient.
OR:
\(yx=3\) > \(x=y3\) > \(y*(x3)^2=y*(y33)^2=y(y6)^2=0\) > either \(y=0\) or \(y=6\) > if \(y=0\), then \(x=3\) and \(xy=0\) \(or\) if \(y=6\), then \(x=3\) and \(xy=18\). Two different answers. Not sufficient.
(2) \(x^3<0\). x is negative, hence x is not equals to 3, hence y must be 0. So, xy=0. Sufficient.
Answer: B.
The answer to this one is C right? B alone is not sufficient. Don't forget that \(y*(x3)^2=0.\) B alone is definitely sufficient.
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Re: Inequality and absolute value questions from my collection
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03 Oct 2012, 04:39
Hi Bunuel, I have read all the responses to Q4. But I am still confused why C is the answer. Here is how I solved it.
4. Are x and y both positive? (1) 2x2y=1 (2) x/y>1
1) Insufficient . First reduced equation to xy=0.5 . Plugged in 2 positive and 2 negative values. I chose x=3, y=2.5 => 32.5=0.5 works.. then chose x= 1, y=1.5 => 1  (1.5)= 0.5. works again. So x and y can be both +ve and ve. So 1) is Insufficient. 2) x/y>1. Just tells us that both x and y have same sign. both are ve or both are +ve. So Insufficient.
Now Combining, Picking the same values used in 1) x=3, y=2.5. both signs positive and xy=0.5. works then chose x= 1, y=1.5 => 1  (1.5)= 0.5 both signes negative. works as well. So we still don't know if both signs are +ve or ve. So my answer is E.
Could you please take a look at my solution and tell me where I am going wrong? That would be a big help. Thanks a ton!




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