An agricultural facility blends two fertilizer solutions, Solution X (which is 20\% nitrog… — GMAT

An agricultural facility blends two fertilizer solutions, Solution $X$ (which is $20\%$ nitrogen by volume) and Solution $Y$ (which contains nitrogen and inert liquid). When Solution $X$ and Solution $Y$ are mixed, the resulting blend has an overall nitrogen concentration of $32\%$ by volume. What is the percentage of nitrogen by volume in Solution $Y$ ?

(1) The volume of Solution $Y$ in the blend is $50\%$ greater than the volume of Solution $X$ .
(2) The blend contains a total of $160$ liters of Solution $X$ .

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.Answer
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.
E
Statements (1) and (2) TOGETHER are NOT sufficient.

Answer

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

The question asks for a relative percentage concentration, which depends strictly on the volumetric ratio of the two component solutions. Statement (1) specifies that Solution Y has a volume 50% greater than Solution X, fixing the ratio of Solution Y to Solution X at 3 to 2. Substituting this ratio into the weighted average concentration formula yields a single unique solution for the nitrogen concentration of Solution Y. Thus, Statement (1) alone is sufficient. Statement (2) provides an absolute volume for Solution X without establishing the volume or ratio of Solution Y, leaving the mixture concentration under-determined and thus insufficient.

Step-by-Step Solution

Formulate the weighted average equation from the question stem.

Let

V_X

and

V_Y

be the volumes of Solution

X

and Solution

Y

, respectively, and let

y\%

be the nitrogen concentration of Solution

Y

. The total nitrogen content equation is

0.20 V_X + \frac{y}{100} V_Y = 0.32 (V_X + V_Y)

The total volume of nitrogen in the final mixture is the sum of the nitrogen contributed by each individual solution.

Simplify the target equation to isolate

y

in terms of the volume ratio

\frac{V_Y}{V_X}

Dividing the equation by

V_X

yields

0.20 + \frac{y}{100} \left(\frac{V_Y}{V_X}\right) = 0.32 \left(1 + \frac{V_Y}{V_X}\right)

Knowing the ratio of

V_Y

V_X

is both necessary and sufficient to determine

y

, making absolute volume values unnecessary.

Evaluate Statement (1): The volume of Solution

Y

50\%

greater than the volume of Solution

X

V_Y = 1.5 V_X \implies \frac{V_Y}{V_X} = 1.5 = \frac{3}{2}

. Substituting

\frac{3}{2}

into the simplified equation gives

0.20 + \frac{y}{100}(1.5) = 0.32(2.5) \implies 0.20 + 0.015 y = 0.80 \implies 0.015 y = 0.60 \implies y = 40\%

Statement (1) provides a unique numerical value for

y

, so Statement (1) alone is SUFFICIENT.

Evaluate Statement (2): The blend contains a total of

160

liters of Solution

X

V_X = 160

liters, but

V_Y

and total volume

V_X + V_Y

remain unknown.

Without knowing

V_Y

or the ratio

\frac{V_Y}{V_X}

y

cannot be uniquely determined. Statement (2) alone is NOT SUFFICIENT.

Key Concept

Weighted Averages and Volumetric Ratios in Data Sufficiency

Estimated Time:2m 0s

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