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If y >= 0, what is the value of x? [#permalink]
 03 Aug 2009, 15:00
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0% (00:00) correct
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If y >= 0, what is the value of x? (1) |x - 3| >= y (2) |x - 3| <= -y
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I choose C .
according to 1 and 2: y<=absolute value" x-3 "<=-y
and y>=0 (y is 0 or positive); that is to say -y is negative
==> only when y=0, the inequation is available. so when y=0, x=3.
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what is the OA? Please post it. Think it is D...
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Last edited by Bunuel on 09 Sep 2009, 17:25, edited 1 time in total.
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y >= 0
1.) |x-3| >= y
=> -y >= (x-3) >= y
=> 3-y >= x >= 3+y
Now, if y = 0 , then, 3 >= x >= 3 => x can only be 3
but, if y = 1 , then, 2 >= x >= 4 => which is not possible for any value of x.
=> x can only be 3 when y = 0 ..sufficient..
2.) |x-3| <= -y
=> y <= (x-3) <= -y
=> 3+y <= x <= 3-y
when y = 0 ; then, 3 <= x <= 3 => x can only be 3..
when y = 1 ; then 4 <= x <= 2, which is not possible for any value of x.
=> x can only be 3 when y = 0 ,, sufficient..
Hence, Ans is D.
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gmate2010 wrote: y >= 0
1.) |x-3| >= y
=> -y >= (x-3) >= y
=> 3-y >= x >= 3+y
Now, if y = 0 , then, 3 >= x >= 3 => x can only be 3
but, if y = 1 , then, 2 >= x >= 4 => which is not possible for any value of x.
=> x can only be 3 when y = 0 ..sufficient..
2.) |x-3| <= -y
=> y <= (x-3) <= -y
=> 3+y <= x <= 3-y
when y = 0 ; then, 3 <= x <= 3 => x can only be 3..
when y = 1 ; then 4 <= x <= 2, which is not possible for any value of x.
=> x can only be 3 when y = 0 ,, sufficient..
Hence, Ans is D. Just plug numbers for statement (1) if 6=3 y=1 |x-3|=3 > 1 or x=-2 y=2 |x-3|=5>2 not sufficient. Answer B
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If y>=0 specified in the stem. Further mod of anything is positive. So, 0>=y according to 1. both y>=0 or y<=0, means y=0. So mod(x-3)>=0. Then x-3>=0 or x-3<=0 So, x>=3 or x<=3. Thus, x=3. Same thing for 2. So, D. gmate2010 wrote: y >= 0
1.) |x-3| >= y
=> -y >= (x-3) >= y
=> 3-y >= x >= 3+y
Now, if y = 0 , then, 3 >= x >= 3 => x can only be 3
but, if y = 1 , then, 2 >= x >= 4 => which is not possible for any value of x.
=> x can only be 3 when y = 0 ..sufficient..
2.) |x-3| <= -y
=> y <= (x-3) <= -y
=> 3+y <= x <= 3-y
when y = 0 ; then, 3 <= x <= 3 => x can only be 3..
when y = 1 ; then 4 <= x <= 2, which is not possible for any value of x.
=> x can only be 3 when y = 0 ,, sufficient..
Hence, Ans is D.
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tusharvk
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I will go with B. x can be anything and value of x can not be determined from A
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My 2 cents on answer D Rules for inequalities 1)|a| <b if - b<a<b 2)|a|>b if a<-b or a>b apply rule 2 and derive the equation just as gmate2010 and tusharvk have done.
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O.K. as I see there has been kind of confusion about this question. I'm pretty sure answer is B. I looked through the logic of people who think answer should be D and here is were they are making mistake: (1) First of all: To check |x - 3|>=0 is sufficient or not just plug numbers A. x=5 y=1>0 and B. x=8 y=2>0; you'll see that both fits in |x - 3|>=0, y>=0. So ABSOLUTELY NO WAY (1) CAN BE SUFFICIENT. |x - 3|>=y means that: x - 3>=y>=0 when x-3>0 --> x>3 OR (not and) -x+3>=y>=0 when x-3<0 --> x<3 Generally speaking |x - 3|>=y>=0 means that |x - 3|, an absolute value, is at least some positive number. So, there's no way you'll get a unique value for x. INSUFFICIENT. (2) |x-3|<=-y, y>=0, equation says that |x-3|less or equals to zero, but |x-3| never negative (|x-3|>=0), so only solution is if |x-3|=0=y --> x-3=0 --> x=3 SUFFICIENT In other words: (-y) is 0 or less, and the absolute value (|x-3|) must be at (0) or below this value. But absolute value (in this case |x-3|) can not be less than zero, so it must be 0. So answer is B.
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PLEASE READ AND FOLLOW: 11 Rules for Posting!!!
RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders
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. NEW!!! ,11 Mixed Questions NEW!!!, 12 Fresh Meat NEW!!!
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!!!, 11 New DS set. NEW!!!
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Bunuel wrote: O.K. as I see there has been kind of confusion about this question. I'm pretty sure answer is B.
I looked through the logic of people who think answer should be D and here is were they are making mistake:
(1) First of all:
To check |x - 3|>=0 is sufficient or not just plug numbers A. x=5 y=1>0 and B. x=8 y=2>0; you'll see that both fits in |x - 3|>=0, y>=0. So ABSOLUTELY NO WAY (1) CAN BE SUFFICIENT.
|x - 3|>=y means that:
x - 3>=y>=0 when x-3>0 --> x>3
OR (not and)
-x+3>=y>=0 when x-3<0 --> x<3
Generally speaking |x - 3|>=y>=0 means that |x - 3|, an absolute value, is at least some positive number. So, there's no way you'll get a unique value for x. INSUFFICIENT.
(2) |x-3|<=-y, y>=0, equation says that |x-3|less or equals to zero, but |x-3| never negative (|x-3|>=0), so only solution is if |x-3|=0=y --> x-3=0 --> x=3 SUFFICIENT
In other words:
(-y) is 0 or less, and the absolute value (|x-3|) must be at (0) or below this value. But absolute value (in this case |x-3|) can not be less than zero, so it must be 0.
So answer is B. Got it finally! Thank you for clear explanation, Bunuel. B it is. Of course, A is not suff...
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