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If x and y are nonzero integers, is (x^(1) + y^(1))^(1) > [#permalink]
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04 Feb 2011, 17:29
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If x and y are nonzero integers, is \((x^{1}+y^{1})^{1}> (x^{1}*y^{1})^{1}\)? (1) x = 2y (2) x + y > 0 I need to clarify a doubt in these types of questions. Firstly apologies guys for giving away too much info but it is important to clarify a doubt. If you solve the above equations you'll get 1/x+y > 1 ...so my question is why cant you solve it one step further and say is x+y < 1. Now if i follow the initial approach i get 1/3y > 1 for 1st statement so why cant we say 3y< 1. I got my ans wrong because of this . Please can some clarify if i am doing some thing conceptually wrong. Apologies again for giving out too much.
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Last edited by Bunuel on 21 Oct 2014, 07:12, edited 4 times in total.
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If x and y are nonzero integers, is (x^(1) + y^(1))^(1) > [#permalink]
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ajit257 wrote: If x and y are nonzero integers, is (x1 + y1)1 > [(x1)(y1)]1 ?
(1) x = 2y
(2) x + y > 0
I need to clarify a doubt in these types of questions. Firstly apologies guys for giving away too much info but it is important to clarify a doubt. If you solve the above equations you'll get 1/x+y > 1 ...so my question is why cant you solve it one step further and say is x+y < 1. Now if i follow the initial approach i get 1/3y > 1 for 1st statement so why cant we say 3y< 1. I got my ans wrong because of this . Please can some clarify if i am doing some thing conceptually wrong.
Apologies again for giving out too much. First of all the question should be: If x and y are nonzero integers, is (x^(1) + y^(1))^(1) > (x^(1)*y^(1))^(1) ?Is \((x^{1}+y^{1})^{1}> (x^{1}*y^{1})^{1}\)? > is \((\frac{1}{x}+\frac{1}{y})^{1}>(\frac{1}{xy})^{1}\)? > is \((\frac{x+y}{xy})^{1}>xy\) > is \(\frac{xy}{x+y}>xy\)? Now, from this point you cannot divide both parts of the inequality by \(xy\) and write \(\frac{1}{x+y}>1\) (as you did), because you don't know whether \(xy\) is positive or negative: if \(xy>0\) then you should write \(\frac{1}{x+y}>1\) BUT if \(xy<0\) then you should flip the sign and write \(\frac{1}{x+y}<1\). But even if you knew that \(xy>0\) then the next step of writing \(x+y<1\) from \(\frac{1}{x+y}>1\) would still be incorrect for the same exact reason: you don't k now whether \(x+y\) is positive or negative, hence you can not muliply both sides of the inequality by \(x+y\). 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.Thus the question is boiled down to: is \(\frac{xy}{x+y}>xy\)? Actually we can manipulate further but there is no need. (1) x = 2y > question becomes: is \(\frac{2y^2}{3y}>2y^2\)? Now, as we know that \(y\) is nonzero then \(2y^2>0\) and we can divide both parts by it > is \(\frac{1}{3y}>1\)? As \(y\) is an integer (no matter positive or negative) then the answer to this question is always NO (if it's a positive integer then \(\frac{1}{3y}<1\) and if it's a negative integer then again: \(\frac{1}{3y}<0<1\)). Sufficient. (2) x + y > 0 > if \(x=y=1\) then the answer will be NO but if \(x=3\) and \(y=1\) then the answer will be YES. Not sufficient. Answer: A.
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Re: If x and y are nonzero integers, is (x^(1) + y^(1))^(1) > [#permalink]
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Re: If x and y are nonzero integers, is (x^(1) + y^(1))^(1) > [#permalink]
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05 Feb 2011, 07:37
apologies AmrithS...i forgot to structure the question after posting it... Thanks a ton ! Bunuel ....a major concept cleared.
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Re: If x and y are nonzero integers, is (x^(1) + y^(1))^(1) > [#permalink]
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Re: If x and y are nonzero integers, is (x^(1) + y^(1))^(1) > [#permalink]
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abilash10 wrote: If x and y are nonzero integers, is (x^1 + y^1)^1 > [(x^1)(y^1)]^1 ? (1) x = 2y (2) x + y > 0 I'm not quite satisfied with the official answer for this question from MGMAT The original question can be reduced to Is \(\frac{xy}{(x+y)}\) > \(xy\) ? Statement 1\(2y^2\)/3y >\(2y^2\) \(2y^2\) (13y)/3y >0 eq(2) If Y is positive, the answer will be + /+ >Negative () If Y is negative, the answer will be + +/ >Negative () Sufficient Statement 2x+y>0 If x = + & y = + , then \(\frac{xy}{(x+y)}\) > \(xy\) will be false If x =  & y = + , then \(\frac{xy}{(x+y)}\) > \(xy\) will be True Thus Insufficient
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Re: If x and y are nonzero integers, is (x^(1) + y^(1))^(1) > [#permalink]
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28 Jul 2013, 21:29
The original question can be reduced to Is \frac{xy}{(x+y)} > xy ?
why can i not cancel out xy on both sides ? x & y are both nonzero. if i do that, i'm left with : is (x+y)^1 >1.
in which case, statement 2 gives me the answer, since it has a positive denominator



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Re: If x and y are nonzero integers, is (x^(1) + y^(1))^(1) > [#permalink]
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28 Jul 2013, 21:31
damn.. i jus realised that they may be nonzero, but xy cud still be negative.. my bad.. thanks...



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Re: If x and y are nonzero integers, is (x^(1) + y^(1))^(1) > [#permalink]
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30 Jan 2015, 16:32
Hi Bunuel  in your explanation above, how do you get from (1/x + 1/y)^(1) to ((x+y)/xy)^(1)?



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Re: If x and y are nonzero integers, is (x^(1) + y^(1))^(1) > [#permalink]
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If x and y are nonzero integers, is (x^(1) + y^(1))^(1) > [#permalink]
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16 Mar 2015, 03:35
I solved it this way: from the original inequality > is xy/y+x > xy > is xy((1/x+y) 1) > 0? to satisfy the inequality either xy > 0 and (1/x+y)1 > 0 or xy < 0 and (1/x+y)1 < 0. Notice that 1/x+y is a proper fraction. So unless x+y = 1 this expression (1/x+y) is going to be negative. If that expression is negative we want xy to be negative as well. statement 1. x=2y > x/y=2/1 x and y have the same sign. If x and y are negative ((1/x+y) 1) is surely negative and xy is positive. Thus the overall expression is not > 0. If x and y are positive ((1/x+y) 1) is also negative and xy is positive. Thus the overall expression is not > 0. sufficient statement 2. x+y >0 we don't know the exact values of x and y. Assume that x is negative and y is positive and x<y then x+y > 0 still ((1/x+y) 1) is negative and xy is negative too, making the overall expression > 0. Assume that both x and y are positive and you end up with the same scenario as statement 1. Assume that x+y=1 and the overall expression becomes 0. not sufficient. answer A.
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Re: If x and y are nonzero integers, is (x^(1) + y^(1))^(1) > [#permalink]
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28 Jan 2017, 07:01
Bunuel wrote: ajit257 wrote: If x and y are nonzero integers, is (x1 + y1)1 > [(x1)(y1)]1 ?
(1) x = 2y
(2) x + y > 0
I need to clarify a doubt in these types of questions. Firstly apologies guys for giving away too much info but it is important to clarify a doubt. If you solve the above equations you'll get 1/x+y > 1 ...so my question is why cant you solve it one step further and say is x+y < 1. Now if i follow the initial approach i get 1/3y > 1 for 1st statement so why cant we say 3y< 1. I got my ans wrong because of this . Please can some clarify if i am doing some thing conceptually wrong.
Apologies again for giving out too much. First of all the question should be: If x and y are nonzero integers, is (x^(1) + y^(1))^(1) > (x^(1)*y^(1))^(1) ?Is \((x^{1}+y^{1})^{1}> (x^{1}*y^{1})^{1}\)? > is \((\frac{1}{x}+\frac{1}{y})^{1}>(\frac{1}{xy})^{1}\)? > is \((\frac{x+y}{xy})^{1}>xy\) > is \(\frac{xy}{x+y}>xy\)? Now, from this point you cannot divide both parts of the inequality by \(xy\) and write \(\frac{1}{x+y}>1\) (as you did), because you don't know whether \(xy\) is positive or negative: if \(xy>0\) then you should write \(\frac{1}{x+y}>1\) BUT if \(xy<0\) then you should flip the sign and write \(\frac{1}{x+y}<1\). But even if you knew that \(xy>0\) then the next step of writing \(x+y<1\) from \(\frac{1}{x+y}>1\) would still be incorrect for the same exact reason: you don't k now whether \(x+y\) is positive or negative, hence you can not muliply both sides of the inequality by \(x+y\). 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.Thus the question is boiled down to: is \(\frac{xy}{x+y}>xy\)? Actually we can manipulate further but there is no need. (1) x = 2y > question becomes: is \(\frac{2y^2}{3y}>2y^2\)? Now, as we know that \(y\) is nonzero then \(2y^2>0\) and we can divide both parts by it > is \(\frac{1}{3y}>1\)? As \(y\) is an integer (no matter positive or negative) then the answer to this question is always NO ( if it's a positive integer then \(\frac{1}{3y}<1\) and if it's a negative integer then again: \(\frac{1}{3y}<0<1\) ). Sufficient. (2) x + y > 0 > if \(x=y=1\) then the answer will be NO but if \(x=3\) and \(y=1\) then the answer will be YES. Not sufficient. Answer: A. Dear Bunuel, Could you elaborate on the highlighted sentence as I am confused with the sign
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Re: If x and y are nonzero integers, is (x^(1) + y^(1))^(1) > [#permalink]
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28 Jan 2017, 07:59
ziyuenlau wrote: Bunuel wrote: ajit257 wrote: If x and y are nonzero integers, is (x1 + y1)1 > [(x1)(y1)]1 ?
(1) x = 2y
(2) x + y > 0
I need to clarify a doubt in these types of questions. Firstly apologies guys for giving away too much info but it is important to clarify a doubt. If you solve the above equations you'll get 1/x+y > 1 ...so my question is why cant you solve it one step further and say is x+y < 1. Now if i follow the initial approach i get 1/3y > 1 for 1st statement so why cant we say 3y< 1. I got my ans wrong because of this . Please can some clarify if i am doing some thing conceptually wrong.
Apologies again for giving out too much. First of all the question should be: If x and y are nonzero integers, is (x^(1) + y^(1))^(1) > (x^(1)*y^(1))^(1) ?Is \((x^{1}+y^{1})^{1}> (x^{1}*y^{1})^{1}\)? > is \((\frac{1}{x}+\frac{1}{y})^{1}>(\frac{1}{xy})^{1}\)? > is \((\frac{x+y}{xy})^{1}>xy\) > is \(\frac{xy}{x+y}>xy\)? Now, from this point you cannot divide both parts of the inequality by \(xy\) and write \(\frac{1}{x+y}>1\) (as you did), because you don't know whether \(xy\) is positive or negative: if \(xy>0\) then you should write \(\frac{1}{x+y}>1\) BUT if \(xy<0\) then you should flip the sign and write \(\frac{1}{x+y}<1\). But even if you knew that \(xy>0\) then the next step of writing \(x+y<1\) from \(\frac{1}{x+y}>1\) would still be incorrect for the same exact reason: you don't k now whether \(x+y\) is positive or negative, hence you can not muliply both sides of the inequality by \(x+y\). 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.Thus the question is boiled down to: is \(\frac{xy}{x+y}>xy\)? Actually we can manipulate further but there is no need. (1) x = 2y > question becomes: is \(\frac{2y^2}{3y}>2y^2\)? Now, as we know that \(y\) is nonzero then \(2y^2>0\) and we can divide both parts by it > is \(\frac{1}{3y}>1\)? As \(y\) is an integer (no matter positive or negative) then the answer to this question is always NO ( if it's a positive integer then \(\frac{1}{3y}<1\) and if it's a negative integer then again: \(\frac{1}{3y}<0<1\) ). Sufficient. (2) x + y > 0 > if \(x=y=1\) then the answer will be NO but if \(x=3\) and \(y=1\) then the answer will be YES. Not sufficient. Answer: A. Dear Bunuel, Could you elaborate on the highlighted sentence as I am confused with the sign We know that y is an integer. Now, for integer y, the answer to the question whether \(\frac{1}{3y}>1\) will always be NO (no matter whether y is negative or positive). If y > 0, then \(\frac{1}{3y}<1\) and if y < 0, then again \(\frac{1}{3y}<1\). Hope it's clear.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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