Question

If \( \frac{x+y}{z}>0 \), is \(x<0\)?
(1) \(x<y\)
(2) \(z<0\)

• A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
• B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
• C. BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient.
• D. EACH statement ALONE is sufficient to answer the question asked.
• E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data are needed.

Hint:

When combining inequalities, look for opportunities to link them. Here, manipulating one inequality (\(x<y\)) to include a term from the other inequality (\(x+y\)) was the key to solving the problem.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We are given an inequality and asked to determine the sign of \(x\).
The condition \( \frac{x+y}{z}>0 \) implies that the numerator (\(x+y\)) and the denominator (\(z\)) must have the same sign. This leads to two possible cases:

Case A: \(x+y>0\) and \(z>0\)
Case B: \(x+y<0\) and \(z<0\)
The question is: Is \(x<0\)?
Step 2: Detailed Explanation:
Analyze Statement (1): \(x<y\).
This statement alone is not sufficient.

Example 1 (Answer "Yes"): Let \(y=5\) and \(x=-2\). Then \(x<y\) is true. Let \(z=1\). Then \(\frac{x+y}{z} = \frac{-2+5}{1} = 3>0\). Here, \(x = -2\), so \(x<0\). The answer is "Yes".

Example 2 (Answer "No"): Let \(y=5\) and \(x=2\). Then \(x<y\) is true. Let \(z=1\). Then \(\frac{x+y}{z} = \frac{2+5}{1} = 7>0\). Here, \(x = 2\), so \(x\) is not less than 0. The answer is "No".

Statement (1) is not sufficient.
Analyze Statement (2): \(z<0\).
If \(z<0\), we must be in Case B from our initial analysis. This means we must have \(x+y<0\). The question now becomes: If \(x+y<0\), is it necessary that \(x<0\)?

Example 1 (Answer "Yes"): Let \(y=-1\) and \(x=-2\). Then \(x+y = -3<0\). Here, \(x = -2\), so \(x<0\). The answer is "Yes".

Example 2 (Answer "No"): Let \(y=-5\) and \(x=2\). Then \(x+y = -3<0\). Here, \(x = 2\), so \(x\) is not less than 0. The answer is "No".

Statement (2) is not sufficient.
Analyze Both Statements Together:
From statement (2), we know \(z<0\), which implies \(x+y<0\). From statement (1), we know \(x<y\). We have a system of two inequalities: 1) \(x+y<0\) 2) \(x<y\) Let's manipulate the second inequality. We can add \(x\) to both sides: \(x + x<y + x\) \(2x<x + y\) Now we can combine this with the first inequality: \(2x<x+y\) and \(x+y<0\). This gives us a transitive relation: \(2x<0\). If \(2x<0\), then dividing by 2 gives \(x<0\). The answer to the question "Is \(x<0\)" is definitively "Yes". Therefore, the statements together are sufficient.
Step 3: Final Answer:
Neither statement alone is sufficient, but both statements together are sufficient.