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29 Apr 2016, 02:44
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
63% (02:11) correct
37%
(02:06)
wrong
based on 326
sessions
History
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Time
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Not Attempted Yet
If xy + x = z, is |x + y| > z?
(1) x < 0
(2) y > 0
(1) x < 0
(2) y > 0
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
29 Apr 2016, 03:32
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Answer: C
Re-writing xy+x=z as x(y+1)=z --(a)
Now looking at the question we can definitely say
|x+y|>z if a) z is -ve as mod of any number is always +ve
OR b) it can be proved that z is positive and > x+y
Taking i) x< 0 our modified equation (a) suggests that sign of z depends on sign of (y+1) so it can be both +ve or -ve. There is no data on the actual numeric value of x ,y or z. so. Not sufficient
Taking ii) y>0 our modified equation (a) suggests that sign of z depends on sign of x so it can be both +ve or -ve. There is no data on the actual numeric value of x ,y or z. so. Not sufficient
Combining i and ii and looking at eq -(a)
x< 0 and y>0 => (y+1)>0. Hence x(y+1) <0 => z <0. From our earlier inference if z<0 then |x+y|>z. Sufficient.
Re-writing xy+x=z as x(y+1)=z --(a)
Now looking at the question we can definitely say
|x+y|>z if a) z is -ve as mod of any number is always +ve
OR b) it can be proved that z is positive and > x+y
Taking i) x< 0 our modified equation (a) suggests that sign of z depends on sign of (y+1) so it can be both +ve or -ve. There is no data on the actual numeric value of x ,y or z. so. Not sufficient
Taking ii) y>0 our modified equation (a) suggests that sign of z depends on sign of x so it can be both +ve or -ve. There is no data on the actual numeric value of x ,y or z. so. Not sufficient
Combining i and ii and looking at eq -(a)
x< 0 and y>0 => (y+1)>0. Hence x(y+1) <0 => z <0. From our earlier inference if z<0 then |x+y|>z. Sufficient.
21 Sep 2017, 07:34
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neither statement can give us much
but if we combine both here is the interesting thing we can find: whether |x+y|> xy+x
if x is negative and y is positive, then xy will be negative and plus negative will always be negative, and if we sum negative and negative the result will be negative so, xy+x is negative if we combine statements 1) and 2)
|x+y| though will be positive no matter what, so the answer will be a definite yes, so, the answer is C (both 1 &2 are sufficient together)
but if we combine both here is the interesting thing we can find: whether |x+y|> xy+x
if x is negative and y is positive, then xy will be negative and plus negative will always be negative, and if we sum negative and negative the result will be negative so, xy+x is negative if we combine statements 1) and 2)
|x+y| though will be positive no matter what, so the answer will be a definite yes, so, the answer is C (both 1 &2 are sufficient together)
