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If xy 0, is x > y? (1) 4x = 3y (2) |y - x| = x - y [#permalink]
 12 Nov 2009, 09:45
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If xy ≠ 0, is x > y?
(1) 4x = 3y
(2) |y - x| = x - y
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Economist wrote: If xy ≠ 0, is x > y?
(1) 4x = 3y
(2) |y - x| = x - y I'm getting B 1. 4x = 3y 4/3x = y if x = 3 y = 4 so yes if x = -3 y = - 4 so no not sufficient 2. |y - x| = x - y y - x = x - y 2x = 2y x=y so answer is no -y + x = x-y 0=0 answer is no sufficient
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Senior Manager
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Economist wrote: If xy ≠ 0, is x > y?
(1) 4x = 3y
(2) |y - x| = x - y B 1. 4x=3y if we take x=1 then y = 4/3 and we get x>y as No if we take x=-3 then y= -4 then x>y is Yes hence insuff 2. Considering x and y have different value,mod value will always be +ve so x-y = some +ve number.hence x>y. Suff
Last edited by kp1811 on 12 Nov 2009, 10:12, edited 1 time in total.
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kp1811 wrote: Economist wrote: If xy ≠ 0, is x > y?
(1) 4x = 3y
(2) |y - x| = x - y B 1. 4x=3y if we take x=1 then y = 4/3 and we get x>y as No if we take x=-3 then y= -4 then x>y is Yes hence insuff 2. mod value will always be +ve so x-y = some +ve number.hence x>y. Suff i got the answer but where is my logic off on statement 2? what if x = 3 and y = 3?
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Economist wrote: If xy ≠ 0, is x > y?
(1) 4x = 3y
(2) |y - x| = x - y This is a good one. +1 Economist for it. (1) x=\frac{3}{4}y, well this one is relatively easy. This statement only tells that x and y have the same sign When they are both positive then x>y, BUT when they are both negative y>x. Not sufficient. (2) |y - x| = x - y, LHS is absolute value which is never negative, hence RHS must also be >=0 -- > x-y>=0, --> y-x<=0. Hence LHS |y - x|= -y+x. |y - x|= -y+x=x-y --> 0=0. Which means that |y - x| = x - y is always true, for any values of x and y not violating the condition y-x<=0 or y<=x. Basically this statement is saying that x can be more or equal to y. Not sufficient. (1)+(2) x=\frac{3}{4}y and y<=x. From (1) x is not equal to y (baring in mind that xy ≠ 0), hence x is more then y. Sufficient.
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This is a good question, simple with a small trap
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Bunuel wrote: Economist wrote: If xy ≠ 0, is x > y?
(1) 4x = 3y
(2) |y - x| = x - y This is a good one. +1 Economist for it. (1) x=\frac{3}{4}y, well this one is relatively easy. This statement only tells that x and y have the same sign When the are both positive then x>y, BUT when they are both negative y>x. Not sufficient. (2) |y - x| = x - y, LHS is absolute value which is never negative, hence RHS must also be >=0 -- > x-y>=0, --> y-x<=0. Hence LHS |y - x|= -y+x. |y - x|= -y+x=x-y --> 0=0. Which means that |y - x| = x - y is always true, for any values of x and y not violating the condition y-x<=0 or y<=x. Basically this statement is saying that x can be more or equal to y. Not sufficient. (1)+(2) x=\frac{3}{4y} and y<=x. From (1) x is not equal to y(baring in mind that xy ≠ 0), hence it's more then y. Sufficient. okay so my answer is wrong..can't figure out what i did wrong on statement 2
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Bunuel wrote: lagomez wrote:
okay so my answer is wrong..can't figure out what i did wrong on statement 2
What you missed is that (2) is true not only when x>y but also when x=y. Hence we can not say for sure that x>y, as x >=y. Nor sufficient. so what is the best way to approach a statement like this? i have an absolute value and an equal sign but the question is asking for greater than or less than that always throws me off because had it been |y-x| > x-y would have been easier but that equal sign throws me off
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lagomez wrote:
2. |y - x| = x - y
y - x = x - y
2x = 2y
x=y so answer is no
-y + x = x-y
0=0
answer is no
sufficient
One more thing: the red part is not correct. |y - x| = x - y. Look at the LHS it's absolute value, it's never negative, hence RHS or x-y is never negative, x-y>=0 or y-x<=0. If so then only one possibility for |y - x|, it must be -y+x. So we'll have: Condition: y-x<=0, which is the same as x>=yAnd: |y - x| = x - y--> -y+x=x-y, --> 0=0. The above means that |y - x| = x - y is always true for x>=y. But as we concluded earlier it's not enough. We need to be sure that x>y. And x>=y leaves the possibility that they are equal, which is removed with the statement (1) when considering together. Hence C.
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If xy ≠ 0, is x > y?
(1) 4x = 3y
(2) |y - x| = x - y
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Manager
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st 1) y=1, x=3/4 - false y=-1, x=-3/4 - true Not sufficient st 2) |y-x| =x-y--> x>y Sufficient coz |-x| = -(-x) = x -- basically a negative numbers should be negated again to get the result for absolute value all the cases as cay that x>y but fails when x=y combining we can answer the question C <didnt consider x=y option, so answered as B  >
Last edited by chix475ntu on 11 Feb 2010, 10:06, edited 1 time in total.
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OA is C both combined.
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chix475ntu wrote: st 1) y=1, x=3/4 - false
y=-1, x=-3/4 - true
Not sufficient
st 2)
|y-x| =x-y--> x>y
Sufficient
coz |-x| = -(-x) = x -- basically a negative numbers should be negated again to get the result for absolute value
B B alone is not sufficient as we dont know whether x=y? So A gives information that x is not = y. So Final ans is C This was nice question...even I would have done the same mistake in the exam..Its really important to think of all the possibilities on the G day.
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can any one discuss more on this..iam weak in inequalities....espl with stmt 2.....
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dmetla wrote: If xy ≠ 0, is x > y?
(1) 4x = 3y
(2) |y - x| = x - y Neither of x or y is zero as xy≠0. 1) Now x=(3/4)y so y > x means statement 1 alone is sufficient to answer the question. 2) Using some values for x & y we can find if x>y then statement |y - x| = x - y is true and if y>x then statement |y - x| = x - y is not true. Hence statement 2 alone is sufficient to answer question. Either statement is sufficient to answer question so answer is D.
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Given: xy\neq0. Question: is x>y true? (1) 4x = 3y --> if x and y are both positive, then x<y BUT if they are both negative then x>y. Not sufficient. But from this statement we can grasp an important property we'll use while evaluating statements together: as xy\neq0 then from statement (1) x\neq{y}. (2) |y-x|=x-y. Now as LHS is absolute value, which is never negative, RHS must also be \geq0 --> so x-y\geq0 --> then |y-x|=-(y-x), and we'll get -(y-x)=x-y --> 0=0, which is true. This means that equation |y-x|=x-y holds true when x-y\geq0 or, which is the same, when x\geq{y}. But this not enough as x=y is still possible. Not sufficient. (1)+(2) From (1) we got that x\neq{y} and from (2) x\geq{y}, hence x>y. Sufficient. Answer: C. Hope it helps.
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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. NEW!!!
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If xy ≠ 0, is x > y?
(1) 4x = 3y
(2) |y - x| = x - y
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