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If x is not equal to y, is (xy)/(x+y) > 1? [#permalink]
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28 Jul 2011, 04:17
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If x is not equal to y, is (xy)/(x+y) > 1? (1) x > 0 (2) y < 0
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Re: Inequalities Help  OG Question [#permalink]
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28 Jul 2011, 04:20
I originally came up with B  but I'm obviously wrong I think I've violated a rule of inequalities here, so, at the risk of looking stupid, I'll write out my train of thought: I rephrased the given into: 1 is x  y > x + y ? (I think that is where I am fundamentally wrong  are you allowed to crossmultiply inequalities?) then 2 is x > x + 2y ? then 3 is 0 > 2y or is Y < 0 ? I immediately spotted "b" as answering the given, so I selected it... I'm gonna go have a look in the MGMAT inequalities guide, but I thought I'd post the question up here anyway for other peoples' benefit
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Re: Inequalities Help  OG Question [#permalink]
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28 Jul 2011, 05:21
Its E. My method was to multiply top and bottom with (x+y)/(x+y) = 1 You get (x^2  y^2) / (x+y)^2
In this case it doesn't matter if you know x is positive and y is negative because it all squares out to positive. So you can't tell is top or bottom is bigger so maybe less than 1 or bigger than 1.



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Re: Inequalities Help  OG Question [#permalink]
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28 Jul 2011, 08:19
MackyCee wrote: From OG11... Q139) if "x" is not equal to "y", is xy / x+y > 1? (1) X > 0 (2) Y < 0 Statement 1: Even if x> 0 we don't know what is Y hence not sufficient Statement 2: Even if y< 0 we don't know what is X hence not sufficient 1 and 2 together: let x be 1 and y be 4 satisfying statement 1 and 2. Then 1(4)/14 = 5 which is less than 0 now let us take x = 4 and y = 1 satisfying 1 and 2 again. Then we have 4(1)/41 = 5/3 which is greater than 0. Hence Not sufficient and IMO should be E



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Re: Inequalities Help  OG Question [#permalink]
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28 Jul 2011, 09:26
MackyCee wrote: I originally came up with B  but I'm obviously wrong
I think I've violated a rule of inequalities here, so, at the risk of looking stupid, I'll write out my train of thought:
I rephrased the given into:
1 is x  y > x + y ? (I think that is where I am fundamentally wrong  are you allowed to crossmultiply inequalities?)
then
2 is x > x + 2y ?
then
3 is 0 > 2y or is Y < 0 ?
I immediately spotted "b" as answering the given, so I selected it...
I'm gonna go have a look in the MGMAT inequalities guide, but I thought I'd post the question up here anyway for other peoples' benefit The most important thing to remember about inequalities is that you have to reverse the inequality if you multiply or divide by a negative. In this case you can't tell what the sign of x + y will be, so cross multiplying is dangerous  you have to consider two scenarios.



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Re: Inequalities Help  OG Question [#permalink]
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28 Jul 2011, 23:14
MackyCee wrote: From OG11... Q139) if "x" is not equal to "y", is xy / x+y > 1? (1) X > 0 (2) Y < 0 We know that we cannot multiply an inequality by an unknown if we don't know whether the unknown is a positive value or a negative value. The inequality behaves differently in the two cases. If the unknown is positive, the inequality stays as it is. If the unknown is negative, the inequality flips sign. Therefore, xy > x+y is not an option. Next, '>1' is much more complicated that '>0' where we just need to consider whether the variables are positive or negative. Therefore, bring whatever is on the right hand side to the left hand side. Question: Is \(\frac{(x  y)}{(x + y)} > 1\) Is \(\frac{(x  y)}{(x + y)}  1 > 0\) Is \(\frac{2y}{(x + y)} > 0\) For this expression to be positive, y/(x+y) should be negative. Therefore, either 'y should be negative and (x+y) should be positive' or 'y should be positive and (x+y) should be negative'. Statement 1: X > 0 No info about y so not sufficient. Statement 2: Y < 0 No info about x so not sufficient. Both together, we know that y is negative. We now need to know the sign of (x+y). Just knowing that x is positive doesn't tell us the sign of (x+y) because we don't know which one out of x and y has greater absolute value. If absolute value of x is greater than that of y, x+y is positive. If absolute value of x is less than that of y, x+y is negative. We do not know the sign of x+y so we still cannot say whether \(\frac{2y}{(x + y)}\) is positive. Not sufficient. Answer E.
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Re: Inequalities Help  OG Question [#permalink]
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29 Jul 2011, 07:12
hey karishma , can u tel the exact solution for this problem??? Is x > 0 ? (1) /x  3/ < 5 (2) /x + 2/ > 5



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Re: Inequalities Help  OG Question [#permalink]
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30 Jul 2011, 21:47
sushantarora wrote: hey karishma , can u tel the exact solution for this problem??? Is x > 0 ? (1) /x  3/ < 5 (2) /x + 2/ > 5 Put it in a new post.
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Re: Inequalities Help  OG Question [#permalink]
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30 Jul 2011, 22:42
sushantarora wrote: hey karishma , can u tel the exact solution for this problem??? Is x > 0 ? (1) /x  3/ < 5 (2) /x + 2/ > 5 As suggested, every new question should be put in a new post. Just pm me the link and I will reply to it. Is x > 0? Statement 1: x  3 < 5 This translates to 2 < x < 8 (check out the explanation at http://www.veritasprep.com/blog/2011/01 ... edoredid/) x could be positive or negative. Not sufficient. Statement 2: x + 2 > 5 This translates to x < 7 or x > 3. x could be positive or negative. Not sufficient. Both together, we get that 3 < x < 8. These will be the only values satisfying both conditions. If this is not intuitive, draw the two inequalities on the number line. x should lie between 2 and 8 and it should be either less than 7 or greater than 3. The only overlapping region is from 3 to 8. x will be positive so sufficient. Answer C.
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Re: Inequalities Help  OG Question [#permalink]
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31 Jul 2011, 12:21
hey,
i simplified to is 0>y^2+xy is 0>y(y+x)
Statement 1: X > 0 X positive but Y negative gives 1 solution and Y positive gives different solution so NS Statement 2: Y < 0
Y big negative and X small positive gives NO. Y little negative and X big positive gives yes.
combining: from S2 you can conclude that the answer varies on the values of X and Y. again X big positive and Y little negative the answer is 0>y(y+x) i.e 0> 1(1+8)
X little positive and Y big negative answer is 0<y(y+x) i.e 0< 8(8+3).
hence E



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Re: Inequalities Help  OG Question [#permalink]
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31 Jul 2011, 23:29
can someone help please:
in the second Q.(the modulus) i recieved 2 solutions for S1:
3<x<8 and 2<x
1. do i have to combine the solution of S1 to 2<x<8.
for S2 I got: x>3 and 7>x
How do i combine both statements to 3<x<8?



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Re: Inequalities Help  OG Question [#permalink]
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01 Aug 2011, 02:05
dimri10 wrote: can someone help please:
in the second Q.(the modulus) i recieved 2 solutions for S1:
3<x<8 and 2<x
1. do i have to combine the solution of S1 to 2<x<8.
How did you get this? Even if you want to use the two step algebra: Step 1: Assume x >= 3 (x  3) < 5 x < 8 So 3 <= x < 8 Step 2: Assume x < 3 and (x  3) < 5 gives x > 2 2 < x < 3 So basically x can lie from 2 to 3 and from 3 to 8 i.e. it can lie from 2 to 8. OR Just say that if x  3 < 5, then 5 < x  3 < 5. Adding 3 to the inequality, 2 < x < 8. dimri10 wrote: for S2 I got: x>3 [highlight]and[/highlight] 7>x
How do i combine both statements to 3<x<8? The highlighted part has to be 'OR'. When you have x + 2 > 5, then x+2 > 5 OR x+2 < 5 i.e. x > 3 OR x < 7 To satisfy the following two conditions: 1. 2 < x < 8 2. x>3 OR 7>x Attachment:
Ques3.jpg [ 4.45 KiB  Viewed 1691 times ]
From the figure, which are the only values of x satisfying both the conditions? Between 3 and 8, right? So 3 < x < 8
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Re: Inequalities Help  OG Question [#permalink]
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04 Aug 2011, 08:34
i am sorry karishma. just got confused. the number system approach is way better to solve such questions. i have seen some other posts in that issue (yours as well as Bunuel) and now it's a cake. 11 out of 10.



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