A container originally holds a liquid mixture in which \frac{2}{5} of the total volume is … — GRE General Test

A container originally holds a liquid mixture in which $\frac{2}{5}$ of the total volume is pure juice. After $6$ liters of pure juice are added to the container, pure juice makes up $\frac{1}{2}$ of the new total volume. How many liters of mixture were originally in the container?

A
14
B
20
C
24
30Answer
E
36

Answer

30 liters

The original mixture volume

V

contains

\frac{2}{5}V

liters of pure juice. When

6

liters of pure juice are added, the new volume of juice becomes

\frac{2}{5}V + 6

and the new total volume becomes

V + 6

. Setting

\frac{2}{5}V + 6 = \frac{1}{2}(V + 6)

leads to

\frac{1}{10}V = 3

, giving

V = 30

liters.

Step-by-Step Solution

Define variables for the original volume and pure juice volume.

Let

V

represent the original volume of the mixture in liters. The original volume of pure juice is

\frac{2}{5}V

Expressing the initial quantity of juice in terms of the unknown total volume establishes the baseline algebraic relationship.

Set up an equation incorporating the added quantity of pure juice.

New juice volume

= \frac{2}{5}V + 6

, and new total volume

= V + 6

. The equation is

\frac{2}{5}V + 6 = \frac{1}{2}(V + 6)

Adding pure juice increases both the pure juice component and the total volume of the mixture by

6

liters.

Solve the linear equation for

V

Expanding the right side gives

\frac{2}{5}V + 6 = \frac{1}{2}V + 3

. Subtracting

\frac{2}{5}V

from both sides gives

6 - 3 = \left(\frac{1}{2} - \frac{2}{5}\right)V \Rightarrow 3 = \frac{1}{10}V \Rightarrow V = 30

Finding a common denominator for the fractions

\frac{1}{2} = \frac{5}{10}

and

\frac{2}{5} = \frac{4}{10}

allows solving for

V

directly.

Key Concept

Solving linear equations involving fractions of whole quantities and mixture relationships.

Estimated Time:1m 0s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic