If x neq -y, then is (x-y)/(x + y) > 1? x > 0 \(\mathrm{y}...

GMAT Data Sufficiency : (DS) Questions

Source: Official Guide

Data Sufficiency

DS-Basics

HARD

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Notes

Post a Query

If \(\mathrm{x} \neq -\mathrm{y}\), then is \(\frac{\mathrm{x}-\mathrm{y}}{\mathrm{x} + \mathrm{y}} > 1\)?

\(\mathrm{x} > 0\)
\(\mathrm{y} < 0\)

Statement (1) ALONE is sufficient but statement (2) ALONE is not sufficient.

Statement (2) ALONE is sufficient but statement (1) ALONE is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are not sufficient.

Solution

Understanding the Question

We need to determine whether the fraction \(\frac{\mathrm{x-y}}{\mathrm{x+y}}\) is greater than 1, given that \(\mathrm{x} \neq -\mathrm{y}\).

What We Need to Determine

The question asks: Is \(\frac{\mathrm{x-y}}{\mathrm{x+y}} > 1\)?

This is a yes/no question. We need to determine if we can definitively answer "yes" or "no" to whether this inequality holds true.

Key Insight

For a fraction to be greater than 1, its numerator must exceed its denominator AND both must have the same sign. But here's the crucial catch: when the denominator \(\mathrm{(x+y)}\) is negative, the entire fraction becomes negative and cannot possibly be greater than 1. This makes the sign of \(\mathrm{(x+y)}\) our "master switch."

The given constraint \(\mathrm{x} \neq -\mathrm{y}\) ensures that \(\mathrm{x + y} \neq 0\), so division by zero is not a concern.

Analyzing Statement 1

Statement 1: \(\mathrm{x} > 0\)

This tells us that x is positive, but we know nothing about y.

Let's test different scenarios to see what happens:


Scenario 1: Small negative y (Example: \(\mathrm{x} = 3, \mathrm{y} = -1\))
  
  \(\mathrm{x + y} = 3 + (-1) = 2\) (positive denominator)
  \(\mathrm{x - y} = 3 - (-1) = 4\) (positive numerator)
  The fraction: \(\frac{4}{2} = 2\), which IS > 1 ✓
  

Scenario 2: Large negative y (Example: \(\mathrm{x} = 1, \mathrm{y} = -3\))
  
  \(\mathrm{x + y} = 1 + (-3) = -2\) (negative denominator)
  \(\mathrm{x - y} = 1 - (-3) = 4\) (positive numerator)
  The fraction: \(\frac{4}{-2} = -2\), which is NOT > 1 ✗
  



Since we get different answers ("yes" in Scenario 1, "no" in Scenario 2), Statement 1 alone is NOT sufficient.

[STOP - Not Sufficient!] This eliminates choices A and D.

Analyzing Statement 2

Now let's forget Statement 1 completely and analyze Statement 2 independently.

Statement 2: \(\mathrm{y} < 0\)

This tells us that y is negative, but we know nothing about x.

Testing different scenarios:


Scenario 1: Large positive x (Example: \(\mathrm{x} = 3, \mathrm{y} = -1\))
  
  \(\mathrm{x + y} = 3 + (-1) = 2\) (positive denominator)
  \(\mathrm{x - y} = 3 - (-1) = 4\) (positive numerator)
  The fraction: \(\frac{4}{2} = 2\), which IS > 1 ✓
  

Scenario 2: Small positive x (Example: \(\mathrm{x} = 1, \mathrm{y} = -3\))
  
  \(\mathrm{x + y} = 1 + (-3) = -2\) (negative denominator)
  \(\mathrm{x - y} = 1 - (-3) = 4\) (positive numerator)
  The fraction: \(\frac{4}{-2} = -2\), which is NOT > 1 ✗
  



Again, we get different answers to our yes/no question, so Statement 2 alone is NOT sufficient.

[STOP - Not Sufficient!] This eliminates choices B and D (already eliminated).

Combining Statements

Let's see what happens when we use BOTH statements together.

From both statements, we know:

\(\mathrm{x} > 0\) (x is positive)
\(\mathrm{y} < 0\) (y is negative)


The Critical Factor: Relative Magnitudes

Even with both pieces of information, the key question remains: Is x larger than \(|\mathrm{y}|\) or smaller?

This determines the sign of our denominator \(\mathrm{(x+y)}\):

When \(\mathrm{x} > |\mathrm{y}|\): The positive x "wins," so \(\mathrm{(x+y)} > 0\)
When \(\mathrm{x} < |\mathrm{y}|\): The negative y "wins," so \(\mathrm{(x+y)} < 0\)


Let's verify with concrete examples:

Case 1: \(\mathrm{x} > |\mathrm{y}|\) (Example: \(\mathrm{x} = 3, \mathrm{y} = -1\))

\(\mathrm{x + y} = 3 + (-1) = 2 > 0\)
\(\mathrm{x - y} = 3 - (-1) = 4 > 0\)
Fraction: \(\frac{4}{2} = 2 > 1\) ✓
Answer to question: YES


Case 2: \(\mathrm{x} < |\mathrm{y}|\) (Example: \(\mathrm{x} = 1, \mathrm{y} = -3\))

\(\mathrm{x + y} = 1 + (-3) = -2 < 0\)
\(\mathrm{x - y} = 1 - (-3) = 4 > 0\)
Fraction: \(\frac{4}{-2} = -2 < 1\) ✗
Answer to question: NO


Since we can construct examples that give different answers to our yes/no question even with both statements, they are NOT sufficient together.

[STOP - Not Sufficient!] This eliminates choice C.

The Answer: E

Even with both statements combined, we cannot definitively answer whether \(\frac{\mathrm{x-y}}{\mathrm{x+y}} > 1\) because the answer depends on the relative magnitudes of x and \(|\mathrm{y}|\), which we cannot determine.

Answer Choice E: "The statements together are not sufficient."

Answer Choices Explained

Statement (1) ALONE is sufficient but statement (2) ALONE is not sufficient.

Statement (2) ALONE is sufficient but statement (1) ALONE is not sufficient.

BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

EACH statement ALONE is sufficient.

Statements (1) and (2) TOGETHER are not sufficient.

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Understanding the Question We need to determine whether the fraction \(\frac{\mathrm{x-y}}{\mathrm{x+y}}\) is greater than 1, given that \(\mathrm{x} \neq -\mathrm{y}\). What We Need to Determine The question asks: Is \(\frac{\mathrm{x-y}}{\mathrm{x+y}} > 1\)? This is a yes/no question. We need to determine if we can definitively answer "yes" or "no" to whether this inequality holds true. Key Insight For a fraction to be greater than 1, its numerator must exceed its denominator AND both must have the same sign. But here's the crucial catch: when the denominator \(\mathrm{(x+y)}\) is negative, the entire fraction becomes negative and cannot possibly be greater than 1. This makes the sign of \(\mathrm{(x+y)}\) our "master switch." The given constraint \(\mathrm{x} \neq -\mathrm{y}\) ensures that \(\mathrm{x + y} \neq 0\), so division by zero is not a concern. Analyzing Statement 1 Statement 1: \(\mathrm{x} > 0\) This tells us that x is positive, but we know nothing about y. Let's test different scenarios to see what happens: Scenario 1: Small negative y (Example: \(\mathrm{x} = 3, \mathrm{y} = -1\)) \(\mathrm{x + y} = 3 + (-1) = 2\) (positive denominator) \(\mathrm{x - y} = 3 - (-1) = 4\) (positive numerator) The fraction: \(\frac{4}{2} = 2\), which IS > 1 ✓ Scenario 2: Large negative y (Example: \(\mathrm{x} = 1, \mathrm{y} = -3\)) \(\mathrm{x + y} = 1 + (-3) = -2\) (negative denominator) \(\mathrm{x - y} = 1 - (-3) = 4\) (positive numerator) The fraction: \(\frac{4}{-2} = -2\), which is NOT > 1 ✗ Since we get different answers ("yes" in Scenario 1, "no" in Scenario 2), Statement 1 alone is NOT sufficient. [STOP - Not Sufficient!] This eliminates choices A and D. Analyzing Statement 2 Now let's forget Statement 1 completely and analyze Statement 2 independently. Statement 2: \(\mathrm{y} < 0\) This tells us that y is negative, but we know nothing about x. Testing different scenarios: Scenario 1: Large positive x (Example: \(\mathrm{x} = 3, \mathrm{y} = -1\)) \(\mathrm{x + y} = 3 + (-1) = 2\) (positive denominator) \(\mathrm{x - y} = 3 - (-1) = 4\) (positive numerator) The fraction: \(\frac{4}{2} = 2\), which IS > 1 ✓ Scenario 2: Small positive x (Example: \(\mathrm{x} = 1, \mathrm{y} = -3\)) \(\mathrm{x + y} = 1 + (-3) = -2\) (negative denominator) \(\mathrm{x - y} = 1 - (-3) = 4\) (positive numerator) The fraction: \(\frac{4}{-2} = -2\), which is NOT > 1 ✗ Again, we get different answers to our yes/no question, so Statement 2 alone is NOT sufficient. [STOP - Not Sufficient!] This eliminates choices B and D (already eliminated). Combining Statements Let's see what happens when we use BOTH statements together. From both statements, we know: \(\mathrm{x} > 0\) (x is positive) \(\mathrm{y} < 0\) (y is negative) The Critical Factor: Relative Magnitudes Even with both pieces of information, the key question remains: Is x larger than \(|\mathrm{y}|\) or smaller? This determines the sign of our denominator \(\mathrm{(x+y)}\): When \(\mathrm{x} > |\mathrm{y}|\): The positive x "wins," so \(\mathrm{(x+y)} > 0\) When \(\mathrm{x} < |\mathrm{y}|\): The negative y "wins," so \(\mathrm{(x+y)} < 0\) Let's verify with concrete examples: Case 1: \(\mathrm{x} > |\mathrm{y}|\) (Example: \(\mathrm{x} = 3, \mathrm{y} = -1\)) \(\mathrm{x + y} = 3 + (-1) = 2 > 0\) \(\mathrm{x - y} = 3 - (-1) = 4 > 0\) Fraction: \(\frac{4}{2} = 2 > 1\) ✓ Answer to question: YES Case 2: \(\mathrm{x} < |\mathrm{y}|\) (Example: \(\mathrm{x} = 1, \mathrm{y} = -3\)) \(\mathrm{x + y} = 1 + (-3) = -2 < 0\) \(\mathrm{x - y} = 1 - (-3) = 4 > 0\) Fraction: \(\frac{4}{-2} = -2 < 1\) ✗ Answer to question: NO Since we can construct examples that give different answers to our yes/no question even with both statements, they are NOT sufficient together. [STOP - Not Sufficient!] This eliminates choice C. The Answer: E Even with both statements combined, we cannot definitively answer whether \(\frac{\mathrm{x-y}}{\mathrm{x+y}} > 1\) because the answer depends on the relative magnitudes of x and \(|\mathrm{y}|\), which we cannot determine. Answer Choice E: "The statements together are not sufficient."