A laboratory container initially holds 120 liters of a chemical solution that is 40\% acid… — GMAT

A laboratory container initially holds $120$ liters of a chemical solution that is $40\%$ acid by volume. A chemist removes $x$ liters of the solution and replaces it with an equal volume of pure acid. Next, the chemist removes $x$ liters of the resulting mixture and replaces it with an equal volume of pure water. If the final concentration of acid in the container is $40\%$ by volume, what is the value of $x$ ?

A
$20$
B
$30$
$40$ Answer
D
$50$
E
$60$

Answer

The value of

x

40

The initial amount of acid is

120 \times 0.40 = 48

liters. After replacing

x

liters with pure acid, the amount of acid becomes

48 - 0.40x + x = 48 + 0.60x

liters. Removing

x

liters of this intermediate solution removes a fraction

\frac{x}{120}

of its acid, leaving

(48 + 0.60x)\left(1 - \frac{x}{120}\right)

liters. Adding pure water does not add acid. Setting this equal to the final acid quantity of

48

liters yields the equation

(48 + 0.60x)(120 - x) = 5760

, which simplifies to

24x - 0.60x^2 = 0

. Solving for non-zero

x

gives

x = 40

Step-by-Step Solution

Calculate initial acid quantity and acid quantity after the first replacement.

Initial acid =

120 \times 0.40 = 48

liters. Removing

x

liters of

40\%

solution removes

0.40x

liters of acid. Adding

x

liters of pure acid gives a new acid quantity of

48 - 0.40x + x = 48 + 0.60x

liters.

Pure acid contains

100\%

acid, so replacing

x

liters adds

1.00x

acid while removing

0.40x

acid.

Determine the acid quantity after the second replacement.

The solution concentration after Step 1 is

\frac{48 + 0.60x}{120}

. Removing

x

liters removes a fraction

\frac{x}{120}

of the solution, leaving a fraction

\left(1 - \frac{x}{120}\right)

. Replacing with pure water adds

0

acid, so final acid quantity is

(48 + 0.60x)\left(1 - \frac{x}{120}\right)

Pure water contributes

0\%

acid to the total solute count.

Set up and solve the equation for the final acid amount.

(48 + 0.60x)\left(\frac{120 - x}{120}\right) = 48 \implies (48 + 0.60x)(120 - x) = 5760

. Expanding gives

5760 - 48x + 72x - 0.60x^2 = 5760 \implies 24x - 0.60x^2 = 0

. Since

x > 0

0.60x = 24 \implies x = 40

The final concentration is given as

40\%

, which corresponds to

48

liters of acid in

120

liters total volume.

Key Concept

Multi-stage mixture dilution and replacement equations using solute mass balance.

Estimated Time:2m 0s

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