A water storage tank is initially filled to \frac{4}{5} of its total capacity with water. … — GRE General Test

A water storage tank is initially filled to $\frac{4}{5}$ of its total capacity with water. First, $\frac{1}{4}$ of the water in the tank is drained for irrigation. Next, an amount of water equal to $\frac{1}{3}$ of the remaining water in the tank is added back. Finally, $\frac{3}{8}$ of the water currently in the tank is removed for domestic use. Which of the following statements about the volume of water in the tank must be true? Select all such statements.

After the second operation, the volume of water in the tank is equal to the initial volume of water before any operations.Answer
The final volume of water in the tank is equal to $\frac{5}{8}$ of the initial volume of water.Answer
The total volume of water removed from the tank during the first and third operations combined is equal to $\frac{1}{2}$ of the total capacity of the tank.Answer
D
The volume of water removed in the first step is equal to $\frac{1}{4}$ of the total capacity of the tank.
E
The final volume of water in the tank is equal to $\frac{2}{5}$ of the total capacity of the tank.

Answer

The correct statements are: (1) After the second operation, the volume of water in the tank is equal to the initial volume of water before any operations; (2) The final volume of water in the tank is equal to 5/8 of the initial volume of water; and (3) The total volume of water removed from the tank during the first and third operations combined is equal to 1/2 of the total capacity of the tank.

The statement regarding the volume after the second operation is correct because adding 1/3 of the remaining 3/5 capacity adds 1/5 capacity, returning the total volume to 4/5 capacity. The statement comparing the final volume to the initial volume is correct because (1/2) divided by (4/5) equals 5/8. The statement regarding total water removed in the first and third operations is correct because 1/5 capacity plus 3/10 capacity equals 1/2 capacity.

Step-by-Step Solution

Define total tank capacity as

T

and find initial water volume.

Initial volume

V_0 = \frac{4}{5}T

The problem states the tank is initially

\frac{4}{5}

full.

Calculate water remaining and removed after Operation 1.

Removed =

\frac{1}{4} \times \frac{4}{5}T = \frac{1}{5}T

. Remaining

V_1 = \frac{4}{5}T - \frac{1}{5}T = \frac{3}{5}T

Draining

\frac{1}{4}

of existing water leaves

\frac{3}{4}

of the existing water.

Calculate water remaining after Operation 2.

Added =

\frac{1}{3} \times \frac{3}{5}T = \frac{1}{5}T

. New volume

V_2 = \frac{3}{5}T + \frac{1}{5}T = \frac{4}{5}T

Adding

\frac{1}{3}

of the remaining volume increases it by a factor of

1 + \frac{1}{3} = \frac{4}{3}

Calculate water remaining and removed after Operation 3.

Removed =

\frac{3}{8} \times \frac{4}{5}T = \frac{3}{10}T

. Final volume

V_3 = \frac{4}{5}T \times \left(1 - \frac{3}{8}\right) = \frac{4}{5}T \times \frac{5}{8} = \frac{1}{2}T

Removing

\frac{3}{8}

of current water leaves

\frac{5}{8}

of that volume.

Evaluate each given statement against calculated values.

V_2 = V_0 = \frac{4}{5}T

(True).

\frac{V_3}{V_0} = \frac{\frac{1}{2}T}{\frac{4}{5}T} = \frac{5}{8}

(True). Total removed =

\frac{1}{5}T + \frac{3}{10}T = \frac{1}{2}T

(True).

Direct comparison with calculated step outcomes confirms these three statements are valid.

Key Concept

Sequential Fraction Multiplication and Part-to-Whole Relationships

Estimated Time:1m 45s

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