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If x and y are integers and xy does not equal 0, is xy < 0?
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15 Oct 2014, 16:49
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Tough and Tricky questions: Inequalities. If x and y are integers and xy does not equal 0, is xy < 0? (1) \(y = x^4 – x^3\) (2) \(12y^2 – y^2x + x^2y^2 > 0\)
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Re: If x and y are integers and xy does not equal 0, is xy < 0?
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15 Oct 2014, 18:54
Bunuel wrote: Tough and Tricky questions: Statistics. If x and y are integers and xy does not equal 0, is xy < 0? (1) y = x^4 – x^3 (2) 12y^2 – y^2x + x^2y^2 > 0 I got the answer correct in 2 Mins 20 Sec, however, I don't know if my method is correct. Feedback's are welcome if I have missed or omitted anything from my method *************************************************************************************************************************************************************************************** Now Statement 1 > \(Y=X^4X^3\)
From this statement, you can also gather that \(Y's\) value depends upon \(X\). So, let take into considerations all possible signs & values of \(X\).
Assume \(X\) is positive (+ve), then...
\(Y=X^4X^3\) \(Y=X^3*(X1)\) \(Y=(+ve)*(+ve  1) = +ve * +ve\) But, according to the main statement \(X\) is not equal to zero, therefore \(X\) could be equal to 1. If \(X=1\), then, \(Y=0\) > Substitute \(X=1\) in statement 1 \(Y=X^4X^3 = (1)^4(1)^3 =11= Zero\)
Therefore, \(XY = +ve\) when \(X>1\) and \(XY = 0\) when \(X=1\)
Hence, Statement 1 is insufficient, Eliminate A, D
Now, Statement 2 > \(12Y^2Y^2X+X^2Y^2 > 0\)
Rearranging & Simplifying > \(Y^2(12X+X^2) > 0\) \(Y^2(X^2X12) > 0\) \(Y^2((X4)(X+3)) > 0\)
Now, since main stem of the question mentions \(X\) is not equal to zero Lets again assume \(X\) is positive \((+ve)\) > (It doesnt matter whether Y is +ve or ve since eqn 2 contains \(Y^2\)),
Then... in the simplified statement 2, we again fall into the same scenario where \(XY = +ve\) when \(X>4\), \(XY = 0\) when \(X=4\) and \(XY = ve\) when \(X < 4\)
Hence, Statement 2 is insufficient, Eliminate B
Now, Combining both the statements doesn't provide us any additional information.
Therefore, both statements together are insufficient, Eliminate option C
Quote: Only option E remains, select option E!



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Re: If x and y are integers and xy does not equal 0, is xy < 0?
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15 Oct 2014, 22:38
Bunuel wrote: Tough and Tricky questions: Statistics. If x and y are integers and xy does not equal 0, is xy < 0? (1) y = x^4 – x^3 (2) 12y^2 – y^2x + x^2y^2 > 0 Ans is E. Statement 1 : y = x^4 – x^3 if X = 1, then y = x^4 – x^3 X^4 X^3 = 0 and since XY does not equal zero X cannot be equal to 1.if X= 1, then y = x^4 – x^3 X^4 X^3 = 2 ; X is ve and Y is +ve therefore XY < 0 If X = 2, then y = x^4 – x^3 X^4 X^3 = 8 ; X is +ve and Y is +ve therefore XY > 0 Therefore Statement 1 is insufficient.Statement 2 : 12y^2 – y^2x + x^2y^2 > 0= y^2 ( 12 X + X^2 ) > 0 = y^2 ( X^2 X 12 ) > 0 = y^2 { (X4) (X+3) } > 0When X=4 then XY = 0...... Therefore X cannot be 4 When X = 5 then y^2 { (X4) (X+3) } = y^2 * 8 ; X is +ve and Y can be +ve or  ve When X = 4 then y^2 { (X4) (X+3) } = y^2 * 8 ; X is ve and Y can be +ve or  ve Statement 2 is insufficient.Taken together both are still insufficient.Hence answer is E



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Re: If x and y are integers and xy does not equal 0, is xy < 0?
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18 Oct 2014, 01:11
Bunuel wrote: Tough and Tricky questions: Statistics. If x and y are integers and xy does not equal 0, is xy < 0? (1) y = x^4 – x^3 (2) 12y^2 – y^2x + x^2y^2 > 0 E. 1) y = x^3(x1) x=2, y=8 ; xy=16 (+ve) x=1, y=2 ; xy=2 (ve) so inconclusive. 2) y^2(x^2x12) > 0 y^2(x4)(x+3) > 0 y^2 is always positive so (x4)(x+3) > 0 x>4 and x<3 in one case xy is +ve and in other xy is ve (1)+(2) > still inconclusive
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Re: If x and y are integers and xy does not equal 0, is xy < 0?
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18 Dec 2015, 08:40
thefibonacci wrote: Bunuel wrote: Tough and Tricky questions: Statistics. If x and y are integers and xy does not equal 0, is xy < 0? (1) y = x^4 – x^3 (2) 12y^2 – y^2x + x^2y^2 > 0 E. 1) y = x^3(x1) x=2, y=8 ; xy=16 (+ve) x=1, y=2 ; xy=2 (ve) so inconclusive. 2) y^2(x^2x12) > 0 y^2(x4)(x+3) > 0 y^2 is always positive so (x4)(x+3) > 0 x>4 and x<3 in one case xy is +ve and in other xy is ve (1)+(2) > still inconclusive Hey "Thefibonacci" shouldnt it be x>4 and x>3? instead of x<3 ?



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Re: If x and y are integers and xy does not equal 0, is xy < 0?
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19 Dec 2015, 05:11
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. If x and y are integers and xy does not equal 0, is xy < 0? (1) y = x^4 – x^3 (2) 12y^2 – y^2x + x^2y^2 > 0 In the original condition, there are 2 variables(x,y), which should match with the number equations. So, you need 2 equations. But, for 1) 1 equation, for 2) 1 equation, which is likely to make C the answer. In 1) & 2), from y=x^3(x1) and y^2(x^2x12)>0, y^2(x4)(x+3)>0, y^2 is always a positive numver. When you divide the 2 equations, (x4)(x+3)>0 becomes x<3, 4<x. When x=4, y=x^3(x1)>0 > xy<0, which is yes. When x=5, y=x^3(x1)>0 >xy>0, which is no and not sufficient. There fore the answer is E. > For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: If x and y are integers and xy does not equal 0, is xy < 0?
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19 Dec 2015, 05:32
nikdiablo129 wrote: thefibonacci wrote: Bunuel wrote: Tough and Tricky questions: Statistics. If x and y are integers and xy does not equal 0, is xy < 0? (1) y = x^4 – x^3 (2) 12y^2 – y^2x + x^2y^2 > 0 E. 1) y = x^3(x1) x=2, y=8 ; xy=16 (+ve) x=1, y=2 ; xy=2 (ve) so inconclusive. 2) y^2(x^2x12) > 0 y^2(x4)(x+3) > 0 y^2 is always positive so (x4)(x+3) > 0 x>4 and x<3 in one case xy is +ve and in other xy is ve (1)+(2) > still inconclusive Hey "Thefibonacci" shouldnt it be x>4 and x>3? instead of x<3 ? Hi, it will be x<3 that is it takes the values 4,5,6,7 and so on.. anything greater than 3, say 2 will make the eq y^2(24)(2+3)=y^2* 6*1 thus ,making it <0..
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Re: If x and y are integers and xy does not equal 0, is xy < 0?
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05 Aug 2018, 21:55
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Re: If x and y are integers and xy does not equal 0, is xy < 0?
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