If k is a real number such that k \neq -1, is \frac{|k - 3|}{k + 1} 1 (2) k^2 - 2k - 3 > 0

If k is a real number such that k \neq -1, is \frac{|k - 3|}{k + 1} < 1? (1) |k| > 1 (2) k… — GMAT

If $k$ is a real number such that $k \neq -1$ , is $\frac{|k - 3|}{k + 1} < 1$ ?

(1) $|k| > 1$
(2) $k^2 - 2k - 3 > 0$

A
Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
EACH statement ALONE is sufficient.Answer
E
Statements (1) and (2) TOGETHER are NOT sufficient.

Answer

EACH statement ALONE is sufficient.

Rephrasing the question stem shows that

\frac{|k - 3|}{k + 1} < 1

holds whenever

k < -1

k > 1

. Statement (1) gives

k < -1

k > 1

, yielding a definitive YES. Statement (2) gives

k < -1

k > 3

, which is a subset of

k < -1

k > 1

, also yielding a definitive YES. Thus, each statement alone is sufficient.

Step-by-Step Solution

Rephrase the question stem target inequality

The target inequality

\frac{|k - 3|}{k + 1} < 1

is satisfied if and only if

k < -1

k > 1

Analyze by cases depending on the sign of the denominator

k + 1

.
- Case 1:

k + 1 < 0 \implies k < -1

. The denominator is negative and the numerator

|k - 3| \ge 0

, so

\frac{|k - 3|}{k + 1} \le 0 < 1

is always true for all

k < -1

.
- Case 2:

k + 1 > 0 \implies k > -1

. Multiplying by

k + 1 > 0

yields

|k - 3| < k + 1

, which expands to

-(k + 1) < k - 3 < k + 1

. The left boundary

-k - 1 < k - 3

reduces to

2k > 2 \implies k > 1

. The right boundary

k - 3 < k + 1 \implies -3 < 1

is universally true. Thus for

k > -1

, the condition holds when

k > 1

.
Combining both cases, the inequality holds whenever

k < -1

k > 1

Evaluate Statement (1)

Statement (1) alone is SUFFICIENT.

Statement (1) states

|k| > 1

, which unwraps to

k < -1

k > 1

. This matches the target condition exactly. For every value of

k

satisfying this statement, the answer to the stem question is a definitive YES.

Evaluate Statement (2)

Statement (2) alone is SUFFICIENT.

Statement (2) states

k^2 - 2k - 3 > 0

, which factors as

(k - 3)(k + 1) > 0

. Solving gives

k < -1

k > 3

.
- If

k < -1

, it falls into the left target region

k < -1

.
- If

k > 3

, it falls into the right target region

k > 1

.
In all cases allowed by Statement (2), the answer to the stem question is a definitive YES.

Key Concept

Inequalities with absolute values and variables in the denominator require case analysis based on the denominator's sign.

Question

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Step-by-Step Solution

Key Concept

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