If x and y are real numbers, what is the value of \frac{x}{y}? (1) x^2 y - x y^2 = 12 (2) … — GMAT

If $x$ and $y$ are real numbers, what is the value of $\frac{x}{y}$ ?

(1) $x^2 y - x y^2 = 12$
(2) $x - y = 3$

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.
D
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.Answer

Answer

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

Combining Statement (1) and Statement (2) allows substituting

x - y = 3

into

xy(x - y) = 12

, yielding

xy = 4

. Solving the system

x - y = 3

and

xy = 4

gives two valid real solution pairs:

(4, 1)

and

(-1, -4)

. The quotient

\frac{x}{y}

evaluates to

4

in the first case and

\frac{1}{4}

in the second case. Because a unique numerical value for

\frac{x}{y}

cannot be determined, the statements together are not sufficient.

Step-by-Step Solution

Evaluate Statement (1) independently.

Factor Statement (1):

xy(x - y) = 12

. Many pairs of real numbers

(x, y)

satisfy this, such as

(4, 1)

which gives

\frac{x}{y} = 4

, or

(3, 1)

which gives

3(1)(2) = 6 \neq 12

. For

(4, 1)

\frac{x}{y} = 4

. For

(x, y) = (2, -1)

2(-1)(3) = -6 \neq 12

. However, choosing

x = \frac{3 + \sqrt{17}}{2}

and

y = \frac{-3 + \sqrt{17}}{2}

gives a different ratio. Since

\frac{x}{y}

can take infinitely many values, Statement (1) is not sufficient.

A single equation with two variables generally allows infinitely many solutions.

Evaluate Statement (2) independently.

Statement (2) gives

x - y = 3

. If

x = 4

and

y = 1

, then

\frac{x}{y} = 4

. If

x = 5

and

y = 2

, then

\frac{x}{y} = 2.5

. Since multiple values for

\frac{x}{y}

exist, Statement (2) is not sufficient.

A linear difference condition alone does not fix the ratio of two unknown quantities.

Evaluate Statements (1) and (2) combined.

Substitute

x - y = 3

into

xy(x - y) = 12

to get

3xy = 12 \implies xy = 4

. Express

y

x - 3

and substitute into

xy = 4

x(x - 3) = 4 \implies x^2 - 3x - 4 = 0 \implies (x - 4)(x + 1) = 0

. This yields two real solutions for

x

x = 4

x = -1

. If

x = 4

, then

y = 1

and

\frac{x}{y} = \frac{4}{1} = 4

. If

x = -1

, then

y = -4

and

\frac{x}{y} = \frac{-1}{-4} = \frac{1}{4}

Because two distinct values for

\frac{x}{y}

(

4

and

\frac{1}{4}

) are possible, combined sufficiency fails.

Key Concept

Degree of non-linear algebraic systems and non-unique solutions in Data Sufficiency

Estimated Time:2m 0s

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