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13 Feb 2023, 00:28
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E
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:
50% (01:41) correct
50%
(01:46)
wrong
based on 28
sessions
History
Date
Time
Result
Not Attempted Yet
Is \(x + y >0\)?
(1) \(x - y = 1\)
(2) \(\frac{x}{y} < 1\)
(1) \(x - y = 1\)
(2) \(\frac{x}{y} < 1\)
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
13 Feb 2023, 00:47
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TBT
Question
Is \(x + y >0\)
Statement 1
(1) \(x - y = 1\)
Case 1: x and y are positive
---------- 0 ------------ y ---------- x ----------
\(x + y >0\) - Yes !
Case 2: x and y negative positive
---------- y ------------ x ---------- 0 ----------
\(x + y >0\) - No!
Statement 1 is not sufficient.
Statement 2
(2) \(\frac{x}{y} < 1\)
Case 1: x and y are positive
x < y
Is \(x + y >0\) - Yes !
Case 2: x and y are negative
|x| < |y|
x > y
Is \(x + y >0\) - No!
Case 2: Either x is positive and y is negative or vice versa.
Is \(x + y >0\) - Maybe
Eliminate B
Combined
When the statements are combined, we can have the following scenarios
Case 1: x and y are negative
-------- y ---------- x --------- 0 --------
In this case : Is \(x + y >0\) - No!
Ex.
- x = -3
- y = -4
x + y = -7
Case 2: x is positive and y is negative
---- y -- 0 -------- x ---
In this case: Is \(x + y >0\) - Yes!
Ex.
- x = 0.9
- y = -0.1
x + y = 0.8
Option E
13 Feb 2023, 01:57
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Is \(x + y >0\)?
(1) \(x - y = 1\).
One number (x) is 1 greater than another (y). This is clearly insufficient to say whether their sum is greater than 1.
OR - \(x = y + 1\), so the question becomes:
(2) \(\frac{x}{y} < 1\).
If x = 0, then y can be any nonzero number so x + y can be positive or negative. Not sufficient.
(1)+(2) From (1) \(x = y + 1\) so from (2) we get:
We still cannot answer the question whether \(y > -\frac{1}{2}\) (from (1)). For example:
Answer: E.
(1) \(x - y = 1\).
One number (x) is 1 greater than another (y). This is clearly insufficient to say whether their sum is greater than 1.
OR - \(x = y + 1\), so the question becomes:
- Is \((y + 1) + y > 0\);
Is \(y > -\frac{1}{2}\).
(2) \(\frac{x}{y} < 1\).
If x = 0, then y can be any nonzero number so x + y can be positive or negative. Not sufficient.
(1)+(2) From (1) \(x = y + 1\) so from (2) we get:
- \(\frac{(y + 1)}{y} < 1;\\
1 + \frac{1}{y} < 1;\\
\frac{1}{y} < 0;\\
y < 0.\)
We still cannot answer the question whether \(y > -\frac{1}{2}\) (from (1)). For example:
- If \((y = -10) < -\frac{1}{2} < 0\) and \(x = y + 1 = -9\), then \(x + y < 0\).
If \(-\frac{1}{2} < (y = -\frac{1}{4}) < 0\) and \(x = y + 1 = \frac{3}{4}\), then \(x + y > 0\).
Answer: E.
