A pharmaceutical facility uses two liquid compounding lines, Line A and Line B, to produce… — GRE General Test

A pharmaceutical facility uses two liquid compounding lines, Line $A$ and Line $B$ , to produce a standardized saline solution. Line $A$ operates at a constant flow rate of $120$ liters per hour, producing a solution with active compound and purified water in a ratio of $1 : 3$ by volume. Line $B$ operates at a constant flow rate of $180$ liters per hour, producing a solution with active compound and purified water in a ratio of $2 : 3$ by volume. If Line $A$ runs for $4$ hours and Line $B$ runs for $5$ hours, and all output is collected into a single storage tank, what is the ratio of active compound to purified water in the storage tank?

$8 : 15$ Answer
B
$8 : 23$
C
$13 : 27$
D
$1 : 2$
E
$38 : 31$

Answer

The ratio of active compound to purified water in the storage tank is

8 : 15

The correct answer is derived by first converting the flow rates and operating hours into total volumes (

480

L for Line A and

900

L for Line B). Using the part-to-whole fraction conversion, Line A contributes

120

L of active compound and

360

L of water, while Line B contributes

360

L of active compound and

540

L of water. Adding these quantities yields

480

L of active compound and

900

L of water, which simplifies to the ratio

8 : 15

Step-by-Step Solution

Calculate total output volume and component quantities from Line A

Total Line A Volume =

120 \text{ L/hr} \times 4 \text{ hrs} = 480 \text{ liters}

. Active Compound =

480 \times \frac{1}{1+3} = 120 \text{ liters}

. Purified Water =

480 - 120 = 360 \text{ liters}

Line A produces a total volume of 480 liters, and a

1 : 3

ratio means active compound constitutes

\frac{1}{4}

of the total volume.

Calculate total output volume and component quantities from Line B

Total Line B Volume =

180 \text{ L/hr} \times 5 \text{ hrs} = 900 \text{ liters}

. Active Compound =

900 \times \frac{2}{2+3} = 360 \text{ liters}

. Purified Water =

900 - 360 = 540 \text{ liters}

Line B produces a total volume of 900 liters, and a

2 : 3

ratio means active compound constitutes

\frac{2}{5}

of the total volume.

Sum the components from both lines in the storage tank

Total Active Compound =

120 + 360 = 480 \text{ liters}

. Total Purified Water =

360 + 540 = 900 \text{ liters}

Combining the output of both lines adds the respective volumes of active compound and purified water.

Compute and simplify the final ratio of active compound to purified water

Ratio =

\frac{480}{900} = \frac{8}{15}

, expressed as

8 : 15

Dividing both numerator and denominator by their greatest common divisor, 60, yields the simplest integer ratio.

Key Concept

Weighted mixture ratios combining multi-stage rates and part-to-whole fraction conversions

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