Question Stem Simplification and Target Rephrasing Practice Questions — GMAT

Question 1Question →

If $x$ is a nonzero real number, is $\frac{x^2 - 4x}{x} > 0$ ?

(1) $x > 5$
(2) $x^2 > 16$

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.

Show answer & explanation

Answer: Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Answer

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

Simplifying the target expression

\frac{x^2 - 4x}{x}

for

x \neq 0

yields

x - 4

. Thus, the question target rephrases to 'Is

x > 4

?'. Statement (1) states

x > 5

, which guarantees

x > 4

, yielding a definitive 'Yes' answer. Statement (2) allows

x > 4

or

x < -4

, yielding both 'Yes' and 'No' outcomes. Therefore, Statement (1) alone is sufficient, but Statement (2) alone is not sufficient.

Step-by-Step Solution

1

Simplify the question stem expression

Since

x \neq 0

, factor out

x

from the numerator:

\frac{x(x - 4)}{x} = x - 4

.

Simplifying the expression reveals the underlying condition required by the question stem.

2

Rephrase the target question

The original question 'Is

\frac{x^2 - 4x}{x} > 0

?' simplifies to 'Is

x - 4 > 0

?', which is equivalent to 'Is

x > 4

?'.

Target rephrasing turns a fraction inequality into a simple comparison.

3

Evaluate Statement (1):

x > 5

If

x > 5

, then

x

is strictly greater than 4. The answer to 'Is

x > 4

?' is a definitive YES.

Since Statement (1) provides a conclusive 'Yes', Statement (1) alone is sufficient.

4

Evaluate Statement (2):

x^2 > 16

Taking the square root gives

|x| > 4

, meaning

x > 4

or

x < -4

.
- If

x = 5

, then

x > 4

(YES).
- If

x = -5

, then

x < 4

(NO).

Because Statement (2) allows both 'Yes' and 'No' answers, it is not sufficient.

Key Concept

Question Stem Simplification and Target Rephrasing

Question 2Question →

For all non-zero real numbers $a$ and $b$ with $|a| \neq |b|$ , the Data Sufficiency Yes/No target question "Is $\frac{a^2 - b^2}{ab} > 0$ ?" is algebraically equivalent to asking whether $a$ and $b$ have the same sign when $|a| > |b|$ , or different signs when $|a| < |b|$ .

True

False

Show answer & explanation

Answer: True

Answer

True. The rephrased question accurately state the necessary and sufficient conditions for the fraction

\frac{a^2 - b^2}{ab}

to be positive.

The quotient

\frac{a^2 - b^2}{ab}

is positive when both

a^2 - b^2

and

ab

are positive, or when both are negative. If

a

and

b

have the same sign (

ab > 0

), then

a^2 - b^2 > 0

, which simplifies to

a^2 > b^2

or

|a| > |b|

. If

a

and

b

have different signs (

ab < 0

), then

a^2 - b^2 < 0

, which simplifies to

a^2 < b^2

or

|a| < |b|

. Hence, the statement correctly describes the target simplification.

Step-by-Step Solution

1

Set up the quotient inequality condition

The fraction

\frac{a^2 - b^2}{ab} > 0

holds when numerator

a^2 - b^2

and denominator

ab

have matching signs.

For any real fraction

\frac{N}{D} > 0

, either (

N > 0

and

D > 0

) or (

N < 0

and

D < 0

).

2

Analyze Case 1 where denominator

ab > 0

a

and

b

have the same sign. In this case,

\frac{a^2 - b^2}{ab} > 0 \implies a^2 - b^2 > 0 \implies |a| > |b|

.

When

ab > 0

, multiplying both sides of the inequality by

ab

preserves the inequality sign.

3

Analyze Case 2 where denominator

ab < 0

a

and

b

have different signs. In this case,

\frac{a^2 - b^2}{ab} > 0 \implies a^2 - b^2 < 0 \implies |a| < |b|

.

When

ab < 0

, multiplying both sides of the inequality by

ab

flips the inequality direction.

4

Combine the cases to evaluate the target rephrasing

The target question simplifies to: "Do

a

and

b

have the same sign with

|a| > |b|

, or opposite signs with

|a| < |b|

?"

Both cases together form the complete rephrased condition for the original statement to be true.

Key Concept

Algebraic Rephrasing of Quotients involving Absolute Values and Variable Signs

Question 3Question →

For all real numbers $x$ and $y$ such that $xy \neq 0$ and $|x| \neq |y|$ , the Data Sufficiency Yes/No target question "Is $\frac{x^2 y - x y^2}{x^3 y - x y^3} > 0$ ?" is algebraically equivalent to asking "Is $x + y > 0$ ?".

True

False

Show answer & explanation

Answer: True

Answer

The statement is True because algebraic factorization and cancellation simplify the target inequality directly to asking whether

x + y > 0

.

The statement is True. Factoring the given expression yields

\frac{xy(x - y)}{xy(x - y)(x + y)}

, which simplifies to

\frac{1}{x + y}

. The inequality

\frac{1}{x + y} > 0

is satisfied if and only if

x + y > 0

.

Step-by-Step Solution

1

Factor the numerator of the rational expression in the target question.

x^2 y - x y^2 = xy(x - y)

Extract the greatest common monomial factor

xy

.

2

Factor the denominator of the rational expression.

x^3 y - x y^3 = xy(x^2 - y^2) = xy(x - y)(x + y)

Extract

xy

and then apply the difference of squares formula

x^2 - y^2 = (x - y)(x + y)

.

3

Cancel common factors between the numerator and denominator.

\frac{xy(x - y)}{xy(x - y)(x + y)} = \frac{1}{x + y}

Since

xy \neq 0

and

|x| \neq |y|

,

xy(x - y) \neq 0

, allowing safe cancellation.

4

Rephrase the target inequality

\frac{1}{x + y} > 0

.

\frac{1}{x + y} > 0 \iff x + y > 0

A unit fraction is positive if and only if its denominator is strictly positive.

Key Concept

Target stem simplification via algebraic factoring and difference of squares

Question 4Question →

In Data Sufficiency, simplifying the question target "Is $x^2 < x$ ?" yields the algebraically equivalent question target "Is $0 < x < 1$ ?" for all real numbers $x$ .

True

False

Show answer & explanation

Answer: True

Answer

The statement is true because solving the inequality

x^2 < x

yields the exact range

0 < x < 1

.

Rephrasing the question stem target

x^2 < x

algebraically leads directly to

x(x - 1) < 0

, which holds true if and only if

x

is strictly between

0

and

1

. Therefore, the simplified target 'Is

0 < x < 1

?' is completely equivalent.

Step-by-Step Solution

1

Rearrange the inequality to set one side to zero.

x^2 - x < 0

Standard algebraic procedure for solving non-linear inequalities requires comparing a factored expression to zero.

2

Factor the quadratic expression.

x(x - 1) < 0

Factoring isolates the roots (

x = 0

and

x = 1

) that define the boundary intervals on the real number line.

3

Determine the interval where the product of the factors is negative.

The product is negative between the roots, which corresponds to

0 < x < 1

.

When

x < 0

, both factors are negative (product is positive). When

x > 1

, both factors are positive (product is positive). Only when

0 < x < 1

is

x

positive and

(x - 1)

negative.

Key Concept

Simplifying Data Sufficiency question stems by finding equivalent inequality ranges reduces complex targets to direct boundary checks.

Estimated Time:45s

Question 5Question →

For all real numbers $x$ such that $x \neq 0$ and $x \neq 1$ , the Data Sufficiency Yes/No target question "Is $\frac{x^2 - x}{|x - 1|} > 0$ ?" is algebraically equivalent to asking "Is $x > 0$ ?".

True

False

Show answer & explanation

Answer: False

Answer

The statement is False.

The statement is False because simplifying the target inequality

\frac{x(x - 1)}{|x - 1|} > 0

yields

x < 0

or

x > 1

, which is not equivalent to

x > 0

.

Step-by-Step Solution

1

Factor the numerator of the expression in the target inequality.

\frac{x^2 - x}{|x - 1|} = \frac{x(x - 1)}{|x - 1|}

.

Factoring allows for analyzing the individual signs of the linear factors.

2

Analyze the denominator to simplify the inequality.

Since

|x - 1| > 0

for all

x \neq 1

, multiplying both sides of

\frac{x(x - 1)}{|x - 1|} > 0

by

|x - 1|

gives the equivalent inequality

x(x - 1) > 0

.

Multiplying an inequality by a strictly positive quantity preserves the direction of the inequality sign.

3

Solve the quadratic inequality

x(x - 1) > 0

.

The product

x(x - 1)

is positive when both factors have the same sign, yielding the solution set

x < 0

or

x > 1

.

A product of two real terms is positive when both terms are positive or both terms are negative.

4

Compare the rephrased target (

x < 0

or

x > 1

) with the proposed target (

x > 0

).

The range

0 < x < 1

makes

x > 0

true but makes

x(x - 1) > 0

false. Thus, the targets are not equivalent.

Two target questions are algebraically equivalent if and only if they yield identical truth values for all values in the domain.

Key Concept

Question Stem Simplification and Target Rephrasing

Question 6Question →

For all non-zero real numbers $m$ and $n$ with $m \neq n$ , the Data Sufficiency target question "Is $\frac{m^2 + n^2}{2mn} > 1$ ?" is algebraically equivalent to the simplified target question "Are $m$ and $n$ of opposite signs?".

True

False

Show answer & explanation

Answer: False

Answer

False. The target question "Is

\frac{m^2 + n^2}{2mn} > 1

?" simplifies to "Do

m

and

n

have the same sign?", not opposite signs.

The statement is false. Rephrasing

\frac{m^2 + n^2}{2mn} > 1

yields

\frac{(m-n)^2}{2mn} > 0

. Because

(m-n)^2 > 0

when

m \neq n

, the ratio is positive if and only if

mn > 0

, which requires

m

and

n

to have the same sign, not opposite signs.

Step-by-Step Solution

1

Subtract 1 from both sides of the inequality to consolidate into a single rational expression.

\frac{m^2 + n^2}{2mn} - 1 > 0 \implies \frac{m^2 - 2mn + n^2}{2mn} > 0

Consolidating terms onto one side avoids cross-multiplying by

2mn

, which has an unknown sign.

2

Factor the numerator quadratic expression.

\frac{(m - n)^2}{2mn} > 0

Recognize that

m^2 - 2mn + n^2

is a perfect square binomial

(m - n)^2

.

3

Determine the sign conditions of the numerator and denominator.

Since

m \neq n

,

(m - n)^2 > 0

is strictly positive. Therefore,

\frac{(m - n)^2}{2mn} > 0

holds if and only if

2mn > 0 \implies mn > 0

.

A fraction with a positive numerator is strictly greater than zero if and only if its denominator is also strictly positive.

4

Translate the condition

mn > 0

into verbal sign properties.

mn > 0

implies that

m

and

n

share the SAME non-zero sign (both positive or both negative).

Comparing this condition to the statement in the prompt reveals that the claim of 'opposite signs' is incorrect.

Key Concept

Question Stem Simplification and Target Rephrasing

Question 7Question →

In Data Sufficiency analysis, for any non-zero real numbers $x$ and $y$ with $|x| \neq |y|$ , the target question 'Is $\frac{x^2 - y^2}{xy} > 0$ ?' is algebraically equivalent to the rephrased target question 'Do $x$ and $y$ have opposite signs if and only if $|x| < |y|$ ?'

True

False

Show answer & explanation

Answer: True

Answer

The statement is TRUE.

The quotient

\frac{x^2 - y^2}{xy}

is positive in two distinct cases: when

x, y

have the same sign and

|x| > |y|

(in which case both 'opposite signs' and '

|x| < |y|

' are false, making their biconditional true), or when

x, y

have opposite signs and

|x| < |y|

(in which case both components are true, making their biconditional true). In cases where the quotient is negative, the biconditional evaluates to false. Thus, the rephrased target question is logically and algebraically identical to the original target question.

Step-by-Step Solution

1

Analyze the sign conditions required for the fraction to be positive.

\frac{x^2 - y^2}{xy} > 0

requires numerator

x^2 - y^2

and denominator

xy

to have the exact same algebraic sign.

A rational expression

\frac{A}{B}

is strictly positive if and only if

A > 0, B > 0

or

A < 0, B < 0

.

2

Evaluate Case 1: Both numerator and denominator are positive (

xy > 0

and

x^2 - y^2 > 0

).

xy > 0 \implies x, y

have SAME signs (Opposite signs = False).

x^2 - y^2 > 0 \implies |x| > |y|

(

|x| < |y|

= False).

The rephrased question tests 'Opposite signs

\iff |x| < |y|

'. Here, False

\iff

False evaluates to TRUE.

3

Evaluate Case 2: Both numerator and denominator are negative (

xy < 0

and

x^2 - y^2 < 0

).

xy < 0 \implies x, y

have OPPOSITE signs (Opposite signs = True).

x^2 - y^2 < 0 \implies |x| < |y|

(

|x| < |y|

= True).

The rephrased question tests 'Opposite signs

\iff |x| < |y|

'. Here, True

\iff

True evaluates to TRUE.

4

Evaluate cases where the fraction is negative to ensure complete equivalence.

If

xy > 0

and

x^2 - y^2 < 0

, we get False

\iff

True (FALSE). If

xy < 0

and

x^2 - y^2 > 0

, we get True

\iff

False (FALSE).

The biconditional statement yields TRUE in exactly the same scenarios where

\frac{x^2 - y^2}{xy} > 0

, establishing strict algebraic equivalence.

Key Concept

Target Stem Rephrasing via Quotient Sign Decomposition and Logical Biconditional Equivalence

Question 8Question →

For all non-zero real numbers $x$ and $y$ such that $x \neq y$ , the Data Sufficiency Yes/No target question "Is $\frac{x}{y} + \frac{y}{x} > 2$ ?" is algebraically equivalent to the simplified target question "Is $xy > 0$ ?"

True

False

Show answer & explanation

Answer: True

Answer

True

The statement is True because rewriting the inequality as

\frac{(x - y)^2}{xy} > 0

shows that since

(x - y)^2 > 0

for all

x \neq y

, the inequality holds if and only if

xy > 0

.

Step-by-Step Solution

1

Combine the fractions on the left-hand side of the target inequality

\frac{x^2 + y^2}{xy} > 2

Finding a common denominator

xy

combines the two reciprocal ratios into a single rational expression.

2

Move all terms to one side to compare against zero

\frac{x^2 + y^2}{xy} - 2 > 0 \implies \frac{x^2 - 2xy + y^2}{xy} > 0

Comparing rational expressions directly to zero simplifies testing for positive or negative regions.

3

Factor the numerator into a perfect square

\frac{(x - y)^2}{xy} > 0

Recognizing

x^2 - 2xy + y^2

as

(x - y)^2

isolates a term with a guaranteed non-negative property.

4

Evaluate sign conditions under the problem constraints

Because

x \neq y

,

(x - y)^2

is strictly positive. Thus,

\frac{(x - y)^2}{xy} > 0

holds if and only if

xy > 0

.

A quotient with a strictly positive numerator is positive if and only if its denominator is positive.

Key Concept

Question Stem Simplification via Rational Expression Factoring

Question 9Question →

If $p$ and $q$ are non-zero real numbers such that $p \neq q$ , is $\frac{p^3 q - p q^3}{p - q} > 0$ ?

(1) $pq > 0$
(2) $p + q > 0$

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.

Show answer & explanation

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

Answer

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

Factoring the numerator gives

p^3 q - p q^3 = pq(p-q)(p+q)

. Since

p \neq q

, dividing by

(p-q)

yields

pq(p+q)

. To determine whether

pq(p+q) > 0

, we need the sign of the product

pq(p+q)

. Statement (1) tells us

pq > 0

, which is insufficient by itself because

p+q

could be positive or negative. Statement (2) tells us

p+q > 0

, which is insufficient by itself because

pq

could be positive or negative. Together, both terms

pq

and

p+q

are positive, so their product must be positive, providing a definitive 'Yes'.

Step-by-Step Solution

1

Rephrase the target question stem algebraically by factoring the numerator.

\frac{p^3 q - p q^3}{p - q} = \frac{p q (p^2 - q^2)}{p - q} = \frac{p q (p - q)(p + q)}{p - q} = pq(p + q)

. The target question simplifies to: 'Is

pq(p+q) > 0

?'

Simplifying the stem isolates the specific product whose sign determines sufficiency.

2

Evaluate Statement (1):

pq > 0

.

If

p = 1, q = 1

, then

pq = 1 > 0

and

pq(p+q) = 1(2) = 2 > 0

(Yes). If

p = -1, q = -1

, then

pq = 1 > 0

and

pq(p+q) = 1(-2) = -2 < 0

(No).

Knowing

pq > 0

leaves the sign of

p+q

unknown, so Statement (1) alone is NOT sufficient.

3

Evaluate Statement (2):

p + q > 0

.

If

p = 2, q = 1

, then

p+q = 3 > 0

and

pq(p+q) = 2(3) = 6 > 0

(Yes). If

p = 3, q = -1

, then

p+q = 2 > 0

and

pq(p+q) = (-3)(2) = -6 < 0

(No).

Knowing

p+q > 0

leaves the sign of

pq

unknown, so Statement (2) alone is NOT sufficient.

4

Evaluate Statement (1) and Statement (2) together.

From Statement (1),

pq > 0

. From Statement (2),

p+q > 0

. Therefore,

pq(p+q) > 0

as it is the product of two strictly positive real numbers.

Combining both statements provides a definitive 'Yes' answer to the simplified target question.

Key Concept

Question Stem Simplification and Target Rephrasing

Question 10Question →

If $x$ and $y$ are positive real numbers, is $\frac{1}{x} < \frac{1}{y}$ ?

(1) $x^2 > y^2$
(2) $x - y > 0$

Which of the following describes the sufficiency of the statements to answer the question?

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.

Show answer & explanation

Answer: EACH statement ALONE is sufficient.

Answer

EACH statement ALONE is sufficient.

Rephrasing the question stem

\frac{1}{x} < \frac{1}{y}

for positive numbers

x

and

y

yields

x > y

. Statement (1) states

x^2 > y^2

, which for positive numbers implies

x > y

(sufficient). Statement (2) states

x - y > 0

, which simplifies directly to

x > y

(sufficient). Therefore, each statement alone is sufficient.

Step-by-Step Solution

1

Rephrase the question stem target inequality.

Since

x

and

y

are both positive, multiplying

\frac{1}{x} < \frac{1}{y}

by

xy

preserves the inequality sign, yielding

y < x

, which is equivalent to

x > y

. The target question simplifies to: 'Is

x > y

?'

Simplifying the target question stem allows direct evaluation of each statement.

2

Evaluate Statement (1):

x^2 > y^2

.

Since

x

and

y

are positive real numbers, taking the positive square root of both sides of

x^2 > y^2

yields

x > y

. This gives a definitive 'Yes' to the rephrased question.

Statement (1) alone provides sufficient information.

3

Evaluate Statement (2):

x - y > 0

.

Adding

y

to both sides of

x - y > 0

gives

x > y

. This gives a definitive 'Yes' to the rephrased question.

Statement (2) alone provides sufficient information.

Key Concept

Question Stem Simplification and Target Rephrasing

Question 11Question →

For any non-zero real numbers $x$ and $y$ , the Data Sufficiency target question 'Is $\frac{x}{y} > 0$ ?' is algebraically equivalent to the rephrased target question 'Do $x$ and $y$ have the same sign?'

True

False

Show answer & explanation

Answer: True

Answer

True. The target question simplifies directly to asking whether

x

and

y

share the same sign.

The target question asks if the ratio

\frac{x}{y}

is positive. In arithmetic, the quotient of two real numbers is positive if and only if both numbers have the same sign (both positive or both negative). Thus, rephrasing the target to 'Do

x

and

y

have the same sign?' is a completely equivalent simplification.

Step-by-Step Solution

1

Analyze the condition for a fraction to be positive

For

\frac{x}{y} > 0

, either (

x > 0

and

y > 0

) or (

x < 0

and

y < 0

).

Dividing two numbers with identical signs yields a positive quotient, while dividing numbers with opposite signs yields a negative quotient.

2

Compare the condition to the proposed rephrased question

The condition 'both positive or both negative' is the definition of

x

and

y

having the same sign.

Rephrasing a Data Sufficiency stem requires finding a target condition that is true under the exact same conditions as the original question.

Key Concept

Question Stem Rephrasing for Quotient Inequalities

Question 12Question →

If $p$ and $q$ are distinct non-zero real numbers, is $\frac{p^3 - q^3}{p - q} > p^2 + q^2$ ?

(1) $\frac{p}{q} + \frac{q}{p} > 2$

(2) $|p + q| > |p - q|$

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.

Show answer & explanation

Answer: EACH statement ALONE is sufficient.

Answer

EACH statement ALONE is sufficient.

Rephrasing the question stem is the key strategy. Factoring the numerator gives

\frac{(p - q)(p^2 + pq + q^2)}{p - q} = p^2 + pq + q^2

. Subtracting

p^2 + q^2

from both sides simplifies the target question to 'Is

pq > 0

?'. Statement (1) simplifies to

\frac{p^2 + q^2}{pq} > 2

; since

p^2 + q^2 > 0

, this inequality requires

pq > 0

, answering 'Yes'. Statement (2) simplifies by squaring both sides to

4pq > 0 \iff pq > 0

, also answering 'Yes'. Thus, each statement alone is sufficient.

Step-by-Step Solution

1

Rephrase the question stem using algebraic factoring.

Since

p \neq q

, factor

p^3 - q^3 = (p - q)(p^2 + pq + q^2)

. The target inequality

\frac{(p - q)(p^2 + pq + q^2)}{p - q} > p^2 + q^2

simplifies directly to

p^2 + pq + q^2 > p^2 + q^2

, which further simplifies to 'Is

pq > 0

?'

Simplifying the target question stem upfront converts a complex cubic rational expression into a simple condition about whether

p

and

q

have the same sign.

2

Evaluate Statement (1):

\frac{p}{q} + \frac{q}{p} > 2

.

Combine fractions over a common denominator:

\frac{p^2 + q^2}{pq} > 2

. Since

p

and

q

are distinct non-zero real numbers,

(p - q)^2 > 0 \implies p^2 + q^2 > 2pq

. For

\frac{p^2 + q^2}{pq} > 2

to hold,

pq

must be positive (if

pq < 0

, the fraction would be negative). Thus,

pq > 0

must be true.

Statement (1) yields a definitive 'Yes' to the rephrased target question 'Is

pq > 0

?'. Therefore, Statement (1) alone is sufficient.

3

Evaluate Statement (2):

|p + q| > |p - q|

.

Square both non-negative sides:

(p + q)^2 > (p - q)^2 \implies p^2 + 2pq + q^2 > p^2 - 2pq + q^2 \implies 4pq > 0 \implies pq > 0

.

Statement (2) also yields a definitive 'Yes' to the target question 'Is

pq > 0

?'. Therefore, Statement (2) alone is sufficient.

Key Concept

Question Stem Simplification and Target Rephrasing

Question 13Question →

If $x$ and $y$ are real numbers such that $x \neq y$ and $x + y \neq 0$ , is $\frac{x^3 - y^3}{x^2 - y^2} > x + y$ ?

(1) $x > y > 0$

(2) $|x + y| < |x - y|$

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.

Show answer & explanation

Answer: Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Answer

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

Factoring the numerator and denominator of the target expression transforms the question 'Is

\frac{x^3 - y^3}{x^2 - y^2} > x + y

?' into the simplified target 'Is

\frac{xy}{x + y} < 0

?'. Under Statement (1), both

x

and

y

are positive, so

xy > 0

and

x + y > 0

, making

\frac{xy}{x + y} > 0

. This conclusively answers the target question with a definitive 'No', proving Statement (1) is sufficient alone. Statement (2) simplifies to

xy < 0

, which allows

x + y

to be positive or negative, resulting in both 'Yes' and 'No' possibilities, rendering Statement (2) insufficient.

Step-by-Step Solution

1

Rephrase the question target algebraically before evaluating the statements.

Simplifying

\frac{x^3 - y^3}{x^2 - y^2}

by factoring numerator and denominator yields

\frac{(x - y)(x^2 + xy + y^2)}{(x - y)(x + y)} = \frac{x^2 + xy + y^2}{x + y}

. The inequality

\frac{x^2 + xy + y^2}{x + y} > x + y

simplifies to

\frac{x^2 + xy + y^2 - (x + y)^2}{x + y} > 0

, which reduces to

\frac{-xy}{x + y} > 0

, or equivalently

\frac{xy}{x + y} < 0

.

Target rephrasing reveals that the question is asking whether

xy

and

x + y

have opposite signs.

2

Evaluate Statement (1):

x > y > 0

.

Since

x > 0

and

y > 0

,

xy > 0

and

x + y > 0

. Consequently,

\frac{xy}{x + y} > 0

. This yields a definitive 'No' to the rephrased question 'Is

\frac{xy}{x + y} < 0

?'. A definitive 'No' is sufficient.

A statement that guarantees a definitive 'No' answer to a Yes/No question is sufficient.

3

Evaluate Statement (2):

|x + y| < |x - y|

.

Squaring both sides gives

(x + y)^2 < (x - y)^2 \implies x^2 + 2xy + y^2 < x^2 - 2xy + y^2 \implies 4xy < 0 \implies xy < 0

. However, knowing

xy < 0

leaves the sign of

x + y

undetermined. For example, if

x = 3, y = -1

, then

x + y = 2 > 0

and

\frac{xy}{x+y} = -1.5 < 0

(Yes). If

x = 1, y = -3

, then

x + y = -2 < 0

and

\frac{xy}{x+y} = 1.5 > 0

(No).

Since the outcome can be either 'Yes' or 'No', Statement (2) alone is not sufficient.

Key Concept

Data Sufficiency Target Simplification and Yes/No Decision Logic

Question 14Question →

For all distinct non-zero real numbers $x$ and $y$ , the Data Sufficiency Yes/No target question "Is $\frac{x|y| - y|x|}{x - y} > 0$ ?" is algebraically equivalent to the simplified target question "Is $xy < 0$ ?"

True

False

Show answer & explanation

Answer: True

Answer

The statement is true. Simplifying the target question by considering the relative signs of

x

and

y

proves that the expression is strictly positive if and only if

x

and

y

have opposite signs (

xy < 0

).

Rephrasing a Data Sufficiency target question requires finding a simpler, logically equivalent condition. Testing the two distinct operational domains (

xy > 0

vs.

xy < 0

) demonstrates that the given algebraic fraction yields a positive result if and only if

x

and

y

carry opposite signs (

xy < 0

). Therefore, the proposed rephrasing is completely accurate and equivalent.

Step-by-Step Solution

1

Analyze the expression when

x

and

y

have the same sign (

xy > 0

).

If

x > 0, y > 0

, then

|x| = x

and

|y| = y

, so

x|y| - y|x| = xy - yx = 0

. If

x < 0, y < 0

, then

|x| = -x

and

|y| = -y

, so

x|y| - y|x| = x(-y) - y(-x) = -xy + xy = 0

. Thus, when

xy > 0

,

\frac{x|y| - y|x|}{x - y} = 0

, yielding a 'No' to the question 'Is the expression

> 0

?'.

Determining the behavior of absolute value expressions under identical signs simplifies the numerator to zero.

2

Analyze the expression when

x

and

y

have opposite signs (

xy < 0

).

Case A: If

x > 0

and

y < 0

, then

|x| = x

and

|y| = -y

. The numerator is

x(-y) - y(x) = -2xy > 0

. The denominator is

x - y > 0

. The quotient is

\frac{\text{positive}}{\text{positive}} > 0

.
Case B: If

x < 0

and

y > 0

, then

|x| = -x

and

|y| = y

. The numerator is

x(y) - y(-x) = 2xy < 0

. The denominator is

x - y < 0

. The quotient is

\frac{\text{negative}}{\text{negative}} > 0

.

Evaluating absolute values under opposite signs demonstrates that the numerator and denominator always have matching signs.

3

Compare the conditions for a 'Yes' answer.

The target expression is strictly positive if and only if

x

and

y

have opposite signs, which is defined by the inequality

xy < 0

.

Establishing biconditional equivalence confirms that rephrasing the target stem to 'Is

xy < 0

?' preserves all logical outcomes.

Key Concept

Data Sufficiency target rephrasing using piecewise definition of absolute value and sign analysis.

Question 15Question →

If $x$ and $y$ are positive real numbers, is $\frac{x + y}{2} - \sqrt{xy} > 1$ ?

(1) $(\sqrt{x} + \sqrt{y})^2 = 9$ and $xy = 4$
(2) $x + y = 8$ and $xy = 9$

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.

Show answer & explanation

Answer: EACH statement ALONE is sufficient.

Answer

EACH statement ALONE is sufficient.

Rephrasing the question stem shows that

\frac{x + y}{2} - \sqrt{xy} > 1

is equivalent to

(\sqrt{x} - \sqrt{y})^2 > 2

. Statement (1) establishes that

(\sqrt{x} - \sqrt{y})^2 = 1

, giving a definitive 'No' to the target question. Statement (2) establishes that

(\sqrt{x} - \sqrt{y})^2 = 2

, which also gives a definitive 'No' since 2 is not strictly greater than 2. Thus, each statement alone is sufficient.

Step-by-Step Solution

1

Rephrase the target question stem algebraically.

The target question 'Is

\frac{x + y}{2} - \sqrt{xy} > 1

?' simplifies to 'Is

x + y - 2\sqrt{xy} > 2

?', which is equivalent to 'Is

(\sqrt{x} - \sqrt{y})^2 > 2

?'

Since

x

and

y

are positive real numbers,

x + y - 2\sqrt{xy}

can be rewritten as the perfect square

(\sqrt{x} - \sqrt{y})^2

.

2

Evaluate Statement (1) using the rephrased target.

From

(\sqrt{x} + \sqrt{y})^2 = 9

and

xy = 4

, we have

\sqrt{xy} = 2

. Using the algebraic identity

(\sqrt{x} - \sqrt{y})^2 = (\sqrt{x} + \sqrt{y})^2 - 4\sqrt{xy}

, we find

(\sqrt{x} - \sqrt{y})^2 = 9 - 4(2) = 1

. Asking 'Is

1 > 2

?' yields a definitive 'No'.

A statement that yields a definitive 'No' to a Yes/No question is sufficient.

3

Evaluate Statement (2) using the rephrased target.

From

x + y = 8

and

xy = 9

, we have

\sqrt{xy} = 3

. Expanding

(\sqrt{x} - \sqrt{y})^2 = x + y - 2\sqrt{xy}

gives

8 - 2(3) = 2

. Asking 'Is

2 > 2

?' yields a definitive 'No'.

Since 2 is not strictly greater than 2, Statement (2) provides a definitive 'No' and is therefore sufficient.

4

Conclude the final sufficiency decision.

Because each statement independently provides a definitive 'No' answer, each statement alone is sufficient.

Both statements satisfy the Data Sufficiency requirements independently.

Key Concept

Question Stem Simplification and Target Rephrasing

Estimated Time:2m 0s

Question 16Question →

For all real numbers $a$ and $b$ with $a \neq b$ , the Data Sufficiency Yes/No target question "Is $\frac{a^2 - b^2}{(a - b)^2} > 1$ ?" is algebraically equivalent to the simplified target question "Is $a > b$ ?"

True

False

Show answer & explanation

Answer: False

Answer

The statement is False because rephrasing the target inequality yields

\frac{2b}{a-b} > 0

, which requires analyzing the signs of both

b

and

a-b

, rather than evaluating whether

a > b

alone.

The statement is False. Correct simplification of

\frac{a^2 - b^2}{(a - b)^2} > 1

leads to

\frac{2b}{a-b} > 0

. This inequality requires

2b

and

a-b

to have identical signs, which holds either when

b > 0

and

a > b

or when

b < 0

and

a < b

. Because

a > b

can be true while

b < 0

(making the original inequality false), the proposed rephrasing is invalid.

Step-by-Step Solution

1

Factor the algebraic expression in the numerator of the target question.

The numerator

a^2 - b^2

factors as

(a-b)(a+b)

, yielding

\frac{(a-b)(a+b)}{(a-b)^2} > 1

.

Factoring allows simplification of common terms between the numerator and denominator.

2

Simplify the fraction by canceling common non-zero terms.

Since

a \neq b

,

a - b \neq 0

, so

\frac{a+b}{a-b} > 1

.

Canceling

(a-b)

is valid as long as

a \neq b

.

3

Compare the fraction to zero by subtracting 1 from both sides.

\frac{a+b}{a-b} - 1 > 0 \implies \frac{(a+b) - (a-b)}{a-b} > 0 \implies \frac{2b}{a-b} > 0.

Subtracting 1 avoids multiplying by a variable expression

(a-b)

whose sign is unknown.

4

Determine the conditions under which

\frac{2b}{a-b} > 0

.

The quotient is positive when

2b

and

a-b

have the same sign: Case 1 (

b > 0

and

a > b

) OR Case 2 (

b < 0

and

a < b

).

A quotient is strictly positive if and only if its numerator and denominator share the same sign.

5

Test whether "Is

a > b

?" is equivalent to the derived condition using a counterexample.

If

a = 2

and

b = -1

, then

a > b

is true (

2 > -1

). However,

\frac{2(-1)}{2 - (-1)} = -\frac{2}{3} \ngtr 0

.

Finding a scenario where

a > b

is true but the original inequality fails proves the two target questions are not algebraically equivalent.

Key Concept

Question Stem Simplification and Target Rephrasing

Question 17Question →

If $x$ and $y$ are non-zero real numbers, is $x^3 y + x y^3 < 2 x^2 y^2$ ?

(1) $|x + y| < |x - y|$
(2) $x^2 y^3 > 0$

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.

Show answer & explanation

Answer: Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Answer

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

The correct response is the option stating that Statement (1) alone is sufficient, but Statement (2) alone is not sufficient. Rephrasing the stem

x^3 y + x y^3 < 2 x^2 y^2

by factoring yields

xy(x - y)^2 < 0

. Because

(x - y)^2 > 0

whenever

x \neq y

, the inequality holds if and only if

xy < 0

. Statement (1) reduces to

(x+y)^2 < (x-y)^2

, which simplifies directly to

xy < 0

, giving a definitive YES. Statement (2) reduces to

y > 0

, which leaves the sign of

x

(and thus

xy

) unknown.

Step-by-Step Solution

1

Rephrase the target question stem algebraically.

The target inequality

x^3 y + x y^3 < 2 x^2 y^2

can be rearranged as

x^3 y + x y^3 - 2 x^2 y^2 < 0

. Factoring out

xy

yields

xy(x^2 - 2xy + y^2) < 0

, which is

xy(x - y)^2 < 0

.

Simplifying the stem isolates the core algebraic condition needed to answer the question.

2

Analyze the conditions under which

xy(x - y)^2 < 0

holds.

Since

x

and

y

are non-zero, if

x = y

,

xy(x-y)^2 = 0

, so the inequality is false (NO). If

x \neq y

,

(x - y)^2 > 0

, so the sign of

xy(x - y)^2

is determined entirely by

xy

. If

xy < 0

, then

x \neq y

and

xy(x - y)^2 < 0

(YES). If

xy > 0

, then

xy(x - y)^2 \ge 0

(NO). Thus, the target question simplifies to: 'Is

xy < 0

?'

A perfect square of a real number is non-negative, so sign evaluation reduces to checking

xy < 0

.

3

Evaluate Statement (1):

|x + y| < |x - y|

.

Squaring both sides of

|x + y| < |x - y|

gives

(x + y)^2 < (x - y)^2 \implies x^2 + 2xy + y^2 < x^2 - 2xy + y^2 \implies 4xy < 0 \implies xy < 0

.

Since Statement (1) directly proves

xy < 0

, it gives a definitive YES to the rephrased target question, making Statement (1) sufficient.

4

Evaluate Statement (2):

x^2 y^3 > 0

.

Since

x \neq 0

,

x^2 > 0

. Thus

x^2 y^3 > 0

simplifies to

y^3 > 0

, which means

y > 0

. However,

x

can still be positive (yielding

xy > 0

) or negative (yielding

xy < 0

).

Knowing only that

y > 0

leaves the sign of

xy

undetermined, so Statement (2) is not sufficient.

Key Concept

Data Sufficiency Target Simplification and Algebraic Rephrasing

Question 18Question →

If $x$ and $y$ are real numbers with $x \neq 0$ , is $\frac{x^4 - y^4}{x^3 + x y^2} > x - y$ ?

(1) $y = 2x$
(2) $xy > 0$

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.

Show answer & explanation

Answer: Statements (1) and (2) TOGETHER are NOT sufficient.

Answer

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

Simplifying the numerator

(x^4 - y^4) = (x^2 - y^2)(x^2 + y^2)

and denominator

x^3 + xy^2 = x(x^2 + y^2)

reduces the left side of the inequality to

\frac{x^2 - y^2}{x}

. Rephrasing the question

\frac{x^2 - y^2}{x} > x - y

shows that it is equivalent to asking whether

\frac{y(x - y)}{x} > 0

.

Under Statement (1), substituting

y = 2x

transforms the target into asking 'Is

x < 0

?'. Because Statement (1) gives no information about the sign of

x

, it is insufficient.

Under Statement (2),

xy > 0

means

x

and

y

share the same sign, but testing values shows both 'Yes' and 'No' are possible (

x=2, y=1

yields Yes, whereas

x=1, y=2

yields No). Thus Statement (2) is insufficient.

Combining both statements gives

y = 2x

and

xy = 2x^2 > 0

, which holds true for all non-zero real numbers regardless of sign. Because

x

can still be positive or negative, the question 'Is

x < 0

?' cannot be answered. Hence, both statements together are NOT sufficient, making the option stating that statements (1) and (2) together are not sufficient the correct choice.

Step-by-Step Solution

1

Simplify the algebraic expression on the left-hand side of the target inequality.

\frac{x^4 - y^4}{x^3 + xy^2} = \frac{(x^2 - y^2)(x^2 + y^2)}{x(x^2 + y^2)} = \frac{x^2 - y^2}{x}

since

x^2 + y^2 > 0

for all non-zero real

x

.

Factoring the numerator as a difference of squares and factoring out

x

in the denominator allows canceling the non-zero common factor

(x^2 + y^2)

.

2

Rephrase the target inequality in terms of simplified components.

The target question 'Is

\frac{x^2 - y^2}{x} > x - y

?' subtracts to

\frac{x^2 - y^2 - x(x - y)}{x} > 0 \implies \frac{y(x - y)}{x} > 0

.

Simplifying the target question establishes the precise algebraic condition that must be evaluated by the statements.

3

Evaluate Statement (1):

y = 2x

.

Substituting

y = 2x

into the simplified target yields

\frac{2x(x - 2x)}{x} > 0 \implies \frac{-2x^2}{x} > 0 \implies -2x > 0 \implies x < 0

.

Statement (1) reduces the question to 'Is

x < 0

?'. Without information about the sign of

x

, this statement alone is NOT sufficient.

4

Evaluate Statement (2):

xy > 0

.

If

x = 2, y = 1

,

\frac{1(2-1)}{2} = 0.5 > 0

(Yes). If

x = 1, y = 2

,

\frac{2(1-2)}{1} = -2 \ngtr 0

(No).

Knowing that

x

and

y

have the same sign does not determine relative magnitudes, so Statement (2) alone is NOT sufficient.

5

Evaluate Statements (1) and (2) together.

Combining

y = 2x

and

xy > 0

gives

x(2x) = 2x^2 > 0

, which is true for all non-zero real

x

. If

x = 1, y = 2

, the target gives 'No'; if

x = -1, y = -2

, the target gives 'Yes'.

Combining both statements still does not determine whether

x < 0

or

x > 0

, so both statements together are NOT sufficient.

Key Concept

Question Stem Simplification and Target Rephrasing

Question 19Question →

If $m$ and $n$ are non-zero real numbers such that $m^2 \neq n^2$ , is $\frac{m^3 - m^2 n + m n^2 - n^3}{m^2 - n^2} < 0$ ?

(1) $m + n = -5$
(2) $m - n = 3$

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.

Show answer & explanation

Answer: Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

Answer

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

Simplifying the target expression by factoring gives

\frac{(m - n)(m^2 + n^2)}{(m - n)(m + n)} = \frac{m^2 + n^2}{m + n}

. Because

m^2 + n^2 > 0

for all non-zero real numbers, the quotient is negative if and only if

m + n < 0

. Statement (1) gives

m + n = -5 < 0

, providing a definitive 'Yes' answer. Statement (2) gives

m - n = 3

, which allows

m + n

to be either positive or negative depending on the specific values of

m

and

n

. Thus, Statement (1) alone is sufficient.

Step-by-Step Solution

1

Rephrase the target question stem algebraically

The target expression

\frac{m^3 - m^2 n + m n^2 - n^3}{m^2 - n^2}

simplifies to

\frac{m^2 + n^2}{m + n}

.

Factor the numerator by grouping:

m^2(m - n) + n^2(m - n) = (m - n)(m^2 + n^2)

. Factor the denominator:

m^2 - n^2 = (m - n)(m + n)

. Since

m^2 \neq n^2

,

m - n \neq 0

, so cancel

(m - n)

.

2

Determine the condition for the simplified expression to be negative

The rephrased target question is 'Is

m + n < 0

?'

Since

m

and

n

are non-zero real numbers,

m^2 + n^2 > 0

always. Therefore, the sign of

\frac{m^2 + n^2}{m + n}

depends solely on the denominator

m + n

.

3

Evaluate Statement (1):

m + n = -5

Statement (1) is SUFFICIENT.

Statement (1) directly tells us

m + n = -5

, which is less than

0

. This yields a definitive 'Yes' to the rephrased question 'Is

m + n < 0

?'.

4

Evaluate Statement (2):

m - n = 3

Statement (2) is NOT SUFFICIENT.

Knowing

m - n = 3

gives

m = n + 3

, so

m + n = 2n + 3

. If

n = 0.5

,

m + n = 4 > 0

('No'). If

n = -5

,

m + n = -7 < 0

('Yes'). Since

m + n

can be positive or negative, Statement (2) is insufficient.

Key Concept

Question Stem Simplification in Data Sufficiency

Estimated Time:2m 0s

Question 20Question →

For all positive real numbers $p$ and $q$ , the Data Sufficiency Yes/No target question "Is $\frac{p^2 + q^2}{2pq} > 1$ ?" is algebraically equivalent to the simplified target question "Is $p \neq q$ ?"

True

False

Show answer & explanation

Answer: True

Answer

True

The statement is true because clearing the positive denominator

2pq

and rearranging terms converts the target inequality

\frac{p^2 + q^2}{2pq} > 1

into

(p - q)^2 > 0

. A real squared quantity is strictly positive if and only if its base is non-zero, making the target equivalent to "Is

p \neq q

?"

Step-by-Step Solution

1

Multiply both sides of the target inequality

\frac{p^2 + q^2}{2pq} > 1

by

2pq

.

p^2 + q^2 > 2pq

, with the direction of the inequality preserved since

p > 0

and

q > 0

implies

2pq > 0

.

Clearing positive denominators simplifies fractional inequalities without changing the inequality sign.

2

Subtract

2pq

from both sides of the inequality.

p^2 - 2pq + q^2 > 0

.

Grouping all non-zero terms on one side sets up a recognizable quadratic form.

3

Factor the quadratic expression as a perfect square.

(p - q)^2 > 0

.

Recognizing

p^2 - 2pq + q^2

as the expansion of

(p - q)^2

isolates the core variable relationship.

4

Analyze the conditions required for

(p - q)^2 > 0

to hold for real numbers.

The square of any real number is always non-negative, and it is strictly greater than zero if and only if the base is non-zero (

p - q \neq 0

, which means

p \neq q

).

Simplifying the logical condition reveals the exact target question needed for Data Sufficiency evaluation.

Key Concept

Target Rephrasing via Perfect Square Inequalities

Question Stem Simplification and Target Rephrasing

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