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Re: Is x  y > x  z? [#permalink]
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24 Jun 2012, 12:39
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kuttingchai wrote: Voted for E
Solved it following way
(A) Y > Z
Case 1 :
[X =4][Y = 3][Z = 2][0][Z = 2][Y=3][X=4]
Case 2 :
[Y = 3][Z = 2][X =1][0][X=1][Z = 2][Y=3]
when x>0
XY = 13 = 2 and XZ = 12 = 1 therefore XY > XZ
XY = 43 = 1 and XZ = 42 = 2 therefore XY < XZ
when x<0
XY = 1+3 = 2 and XZ = 1+2 = 1 therefore XY > XZ
XY = 4+3 = 1 and XZ = 4+2 = 2 therefore XY < XZ
not sufficient
(B) X<0 dont know about Y and Z, therefore nit sufficient
(C) X < 0 and Y > Z
when x<0
XY = 1+3 = 2 and XZ = 1+2 = 1 therefore XY > XZ
XY = 4+3 = 1 and XZ = 4+2 = 2 therefore XY < XZ [not sufficent]
therefore E First, remember what do we use the absolute value for ? We use it to express the distance between two real numbers on the number line. So, x  y means the distance between x and y, and x = x  0 means the distance between x and 0. Also, x + 3 = x  (3) expresses the distance between x and 3. Obviously, x  y = y  x (it is the same distance). So, the question can be rephrased as "is the distance between x and y greater that the distance between x and z ?" or "is x closer to z than to y?" (1) y > z means that the distance from y to 0 is greater than the distance from z to 0. This in fact is not important in this case. But certainly y and z are distinct. Regardless of wheather y > z or y < z (both cases are possible, try to draw the number line and visualize the points), we can place x closer to either y or z. So, (1) is not sufficient. (2) Is evidently not sufficient. Take a point x on the number line at the left of 0, then you can place y and z as you please, each one can be closer or farther from x. Also, now you can consider y = z, in which case the two distances are equal. Evidently, considering (1) and (2) together won't help either. For example, put x at the left of 0, y and z on either side of x such that y < x and z > x, both negative. You can play with each distance and put either y or z closer to x. Therefore, answer, E.
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Re: Is x  y > x  z? [#permalink]
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27 Jun 2012, 07:52
Friends, Am new here. Just started to prepare. Please let me know if my thought process on this problem is right.
XY > XZ?
1) Y > Z.  > Y = +Y OR Y  > Z = +Z OR Z Basis the above, XY = XY OR X+Y OR YX OR YX XZ = XZ OR X+Z OR ZX OR ZX Hence we are not sure which one out of the above matches to give a certain answer.
2) X<0 This by itself is not sufficient because we dont know the signs of Y and Z
1 + 2 , we know X is negative and if we consider the negative value of X in XY = XY OR X+Y OR YX OR YX & XZ = XZ OR X+Z OR ZX OR ZX , still we are unclear which one will certainly answer the question. Hence E.
( pl direct me correctly , if Iam wrong)



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Re: Is x  y > x  z? [#permalink]
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17 Jan 2013, 04:32
Hussain15 wrote: Is x  y>x  z?
(1) y>z (2) x < 0 My approach is to use the distance perspective... xz represents the distance bet. x and z xy represents the distance bet. x and y 1. y > z <0zy> or <y0z> or <yz0> or <z0y> But we don't know where x lies... INSUFFICIENT 2. x < 0 <x0> But we dont know where y and z lie... INSUFFICIENT Combine: scenario1: <x0zy> xz < xy scenario2: <yx0z> xz > xy and many more scenarios... BUT two shows that info is NOT SUFFICIENT ANsweR: E
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Re: Is x  y > x  z? [#permalink]
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28 Mar 2013, 02:20
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Can anyone solve the following problem in a completely algebraic way? I mean, without plugging any numbers. Is xz > xy ? 1. z > y 2. 0 > x \(xz > xy = (xz)^2>(xy)^2 = x^2+z^22xz>x^2+y^22xy\) \(2xy2xz>y^2z^2\) \(2x(yz)>(y+z)(yz)\) 1. z > y\(z^2>y^2\) \(z^2y^2>0=y^2z^2<0=(yz)(y+z)<0\) So one between (yz) (y+z) is negative, the other is positive, but we cannot tell which one is +ve or ve. \(2x(yz)>(y+z)(yz)\) the second part is ve ((yz)(y+z)<0) we cannot say anything about 2x(yz) Not sufficient 2. 0 > x x<0 so the first term is ve (2x) but we cannot say anything about the other part (...(yz)>(y+z)(yz)) (1)+(2) Still not sufficient, let me explain. Here are the combinations ( remeber that one between (yz) (y+z) is negative and the other positive) \(2x(yz)>(y+z)(yz)\) Case one (yz)ve: 2x<0 (yz)<0 (y+z)>0 veNumber*veNumber>+veNumber*veNumber +ve>ve always true. Case two (y+z)ve: 2x<0 (yz)>0 (y+z)<0 veNumber*+veNumber>veNumber*+veNumber veNumber>veNumber we cannot say if this is true, since we have no numerical value E Do I deserve a Kudos for this?
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Re: Is x  y > x  z? [#permalink]
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10 May 2013, 17:36
Hussain15 wrote: Is x  y>x  z?
(1) y>z (2) x < 0 Absolute Value of a number describe the distance of a number from 0 in a number line irrespective of a positive or negative number. Similarly xy gives the distance between two numbers irrespective of their sign. Pick some numbers and you can reach this conclusion. 1)y > z > just tells you y is further from z on the number line. It doesn't tell you where x is > INSUFFICIENT. 2) x < 0. Again doesn't help because we are talking about distance between numbers. We don't any relation between x and y, hence INSUFFICIENT. Combining 1 and 2 also doesn't give much information. We just know y is farther from z and x is negative. This can be interpreted in multiple ways. Hence E.



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Re: Is x  y > x  z? [#permalink]
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13 May 2013, 04:36
square both sides , question becomes is 2x(y+z)<0 , this is possible in 2 cases
a) xve and y+z +ve b) x +ve and y+z ve
from 1
y^2  x2 >0 , i.e. (yx)(y+x) > 0 ... no idea about x on its own , (y+x) we can never tell ... insuff
from 2
x is ve .... no idea about ( y+x) ...... insuff
both
x is ve so assessing whether (y+z) is +ve or not .... we cant tell.........E



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Re: Is x  y > x  z? [#permalink]
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14 May 2013, 00:25
+1 Bunuel.....Nice explanation..I went to the extent of taking examples for each case and solving it..This graphical approach is awesome!!!!
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Re: Is x  y > x  z? [#permalink]
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30 Jun 2013, 11:26
Is x  y>x  z?
first, let's square both sides:
Is (xy)^2 > (xz)^2 Is (xy)*(xy) > (xz)*(xz) Is x^2 2xy + y^2 > x^2  2xz + z^2 Is 2xz  2xy > z^2  y^2? Is 2(xzxy) > z^2  y^2? Is 2x(zy) > (z+y)*(zy)?
1. z > y We can square both sides to factor out in an attempt to understand more about x and y [as we see in the stem  (z+y)*(zy)]
z^2  y^2 > 0 (z + y)*(z  y) > 0
So, (z+y)*(zy) may be both positive or both negative for the product to be greater than zero. However, this tells us nothing about the left hand side. INSUFFICIENT
2. 0 > x
This tells us the the left hand side of 2x(zy) > (z+y)*(zy) is positive, however, it tells us nothing about the right hand side. INSUFFICIENT
1+2) We know that the RHS of 2x(zy) > (z+y)*(zy) is positive and we also know that 2x is positive as well. However, there is one problem. We have established that (z + y)*(z  y) is positive which means BOTH (z + y)*(z  y) are positive or BOTH (z + y)*(z  y) are negative. We don't know if they are positive or negative though, which means that 2x(zy) could be positive or negative. This problem could also be tested by picking numbers. INSUFFICIENT
(E)



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Re: Is x  y > x  z? [#permalink]
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23 Jul 2013, 13:55
Is x  y>x  z?
(1) y>z
I. 4>3 II. 4>3 III. 4>3 IV. 4>3
This tells us nothing about x. INSUFFICIENT
(2) x < 0 This tells us nothing about y and z. INSUFFICIENT
1+2) y>z and x < 0 If the absolute value of y is greater than the absolute value of z and we are told that x < 0 (i.e. x is negative)
x  y>x  z 2  (4) > 2  3 2 > 5 2 > 5 Invalid
x  y>x  z 2  4>23 6>5 6>5 Valid
Even knowing that x is negative, we can't sufficently narrow down the signs of x and y to know if x  y>x  z holds true or not. INSUFFICIENT
(E)



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Re: Is x  y > x  z? [#permalink]
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Is x  y > x  z? [#permalink]
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31 Aug 2015, 07:31
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 equations ensures a solution. Is x  y > x  z? (1) y > z (2) x < 0 In the original condition there are 3 variables (x,y,z) and since we need to match the number of variables and equations, we need 3 equations. Since there is 1 each in 1) and 2), we need 1 more equation and thus E isl likely the answer and it turns out that E actually is the answer. In a number line, absolute number depicts the distance between two points. Looking back at the question, Is x  y > x  z? (1) y0 > z0 (2) x < 0 asks if the distance from y to 0 is greater than the distance from z to 0. Essentially is asks if, assuming x is negative, the distance from x to y is greater then the distance from x to z. if x=1, y=5, z=2, then the answer is yes, but if y=5, z=2, x=4 then the answer is no and therefore it is not sufficient. Therefore the answer is E. However, keep in mind that these direct substitutions are not the key methods but comparing the number of variables and equations in the original condition are.
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Re: Is x  y > x  z? [#permalink]
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