Vessel A contains 60 liters of a beverage mixture that is 75\% fruit concentrate by volume… — GMAT

Vessel A contains $60$ liters of a beverage mixture that is $75\%$ fruit concentrate by volume, and Vessel B contains $90$ liters of a beverage mixture that is $25\%$ fruit concentrate by volume. First, $10$ liters of pure fruit concentrate are added to Vessel B. Next, $x$ liters of the mixture in Vessel A are removed and replaced with $x$ liters of pure water. If the concentration of fruit concentrate in Vessel A is now equal to the concentration of fruit concentrate in Vessel B, what is the value of $x$ ?

A
$10$
B
$20$
C
$25.5$
$34$ Answer
E
$40$

Answer

The value of

x

34

To find

x

, first determine the resulting concentration in Vessel B. Vessel B initially contains

0.25 \times 90 = 22.5

liters of concentrate. Adding

10

liters of pure concentrate gives

32.5

liters of concentrate in

100

liters of total mixture, which is a concentration of

32.5\%

. In Vessel A, there are initially

0.75 \times 60 = 45

liters of concentrate. Removing

x

liters of the mixture removes

0.75x

liters of concentrate. When replaced with

x

liters of pure water, the total volume remains

60

liters, so the new concentration is

\frac{45 - 0.75x}{60}

. Setting

\frac{45 - 0.75x}{60} = 0.325

gives

45 - 0.75x = 19.5

, which simplifies to

0.75x = 25.5

, yielding

x = 34

Step-by-Step Solution

Calculate the new concentration of fruit concentrate in Vessel B after adding

10

liters of pure concentrate.

Initial concentrate in B =

0.25 \times 90 = 22.5

liters. Adding

10

liters of pure concentrate yields

22.5 + 10 = 32.5

liters of concentrate in a total volume of

90 + 10 = 100

liters. Thus, the new concentration in B is

\frac{32.5}{100} = 32.5\%

(or

0.325

Adding pure concentrate increases both the solute amount and the total solution volume.

Express the concentration of fruit concentrate in Vessel A after removing

x

liters of mixture and replacing it with

x

liters of pure water.

Initial concentrate in A =

0.75 \times 60 = 45

liters. Removing

x

liters removes

0.75x

liters of concentrate. Replacing with

x

liters of pure water keeps the total volume at

60

liters. The remaining concentrate is

45 - 0.75x

, so the new concentration in A is

\frac{45 - 0.75x}{60}

The replacement maintains constant total volume while diluting the concentrate.

Equate the concentrations of Vessel A and Vessel B to solve for

x

\frac{45 - 0.75x}{60} = 0.325 \implies 45 - 0.75x = 19.5 \implies 0.75x = 25.5 \implies x = \frac{25.5}{0.75} = 34

Setting the two concentration expressions equal gives a single linear equation in terms of

x

Key Concept

Dilution and multi-vessel mixture balance

Question

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Step-by-Step Solution

Key Concept

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