A chemical laboratory manufactures two liquid formulations: Formulation Alpha, which conta… — GMAT

A chemical laboratory manufactures two liquid formulations: Formulation Alpha, which contains $10\%$ active solute by mass, and Formulation Beta, which contains $40\%$ active solute by mass. A technician needs to prepare a $100\text{-kg}$ mixture containing $40\%$ active solute by mass by combining Formulation Alpha, Formulation Beta, and a pure solute additive ( $100\%$ active solute by mass). The ratio of the mass of Formulation Alpha used to the mass of Formulation Beta used must be $2 : 1$ .

Which of the following correctly identifies the mass of Formulation Alpha used and the mass of Pure Additive used, respectively?

Formulation Alpha: 50 kg; Pure Additive: 25 kgAnswer
B
Formulation Alpha: 25 kg; Pure Additive: 50 kg
C
Formulation Alpha: 40 kg; Pure Additive: 20 kg
D
Formulation Alpha: 50 kg; Pure Additive: 10 kg
E
Formulation Alpha: 60 kg; Pure Additive: 25 kg

Answer

Formulation Alpha: 50 kg; Pure Additive: 25 kg

The correct combination specifies

50\text{ kg}

for Formulation Alpha and

25\text{ kg}

for Pure Additive. This leaves

25\text{ kg}

for Formulation Beta, maintaining the

2:1

ratio of Alpha to Beta (

50 : 25

). The total mass is

50 + 25 + 25 = 100\text{ kg}

, and the total active solute mass is

0.10(50) + 0.40(25) + 1.00(25) = 5 + 10 + 25 = 40\text{ kg}

, exactly

40\%

of the total mass.

Step-by-Step Solution

Define variables and establish system equations from the given ratios and constraints.

Let

A

B

, and

P

be the masses (in kg) of Formulation Alpha, Formulation Beta, and Pure Additive, respectively. Given

A : B = 2 : 1

, we have

A = 2B

. The total mass equation is

A + B + P = 100

Relating the component masses allows substituting

A = 2B

into the total mass condition:

2B + B + P = 100 \implies 3B + P = 100

Set up the solute mass balance equation.

Solute content equation:

0.10A + 0.40B + 1.00P = 0.40 \times 100 = 40\text{ kg}

Substituting

A = 2B

gives

0.10(2B) + 0.40B + P = 40 \implies 0.60B + P = 40

Solve the system of equations for

B

A

, and

P

Subtracting

(0.60B + P = 40)

from

(3B + P = 100)

yields

(3B - 0.60B) = 100 - 40 \implies 2.4B = 60 \implies B = 25\text{ kg}

With

B = 25\text{ kg}

, calculate

A = 2(25) = 50\text{ kg}

and

P = 100 - 3(25) = 25\text{ kg}

Key Concept

Weighted Mixture Ratios and Simultaneous Mass Balance Equations

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

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