A storage tank is initially \frac{2}{3} full of liquid. After \frac{1}{4} of the liquid in… — GRE General Test

A storage tank is initially $\frac{2}{3}$ full of liquid. After $\frac{1}{4}$ of the liquid inside the tank is drained out, $15$ gallons of liquid are added to the tank. If the tank is then $\frac{5}{6}$ full, what is the total capacity of the tank, in gallons?

A
30
B
36
45Answer
D
60
E
90

Answer

45 gallons

The correct answer of 45 gallons is found by calculating the liquid remaining after draining

\frac{1}{4}

of the initial

\frac{2}{3} C

, which leaves

\frac{1}{2} C

. Adding 15 gallons brings the total volume to

\frac{5}{6} C

. Solving

\frac{5}{6} C - \frac{1}{2} C = 15

gives

\frac{1}{3} C = 15

, so

C = 45

Step-by-Step Solution

Determine the fraction of liquid in the tank after draining.

The initial amount of liquid is

\frac{2}{3} C

, where

C

is the total capacity. Draining

\frac{1}{4}

of this liquid leaves

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

of the initial liquid. Thus, the remaining liquid is

\frac{3}{4} \times \frac{2}{3} C = \frac{1}{2} C

The fraction drained applies to the existing liquid volume, not the total container capacity.

Set up an equation incorporating the added liquid and final tank fraction.

\frac{1}{2} C + 15 = \frac{5}{6} C

Adding

15

gallons to the remaining liquid yields a volume equal to

\frac{5}{6}

of the total capacity.

Solve for the total capacity

C

Subtracting

\frac{1}{2} C

(which is

\frac{3}{6} C

) from both sides gives

15 = \frac{5}{6} C - \frac{3}{6} C = \frac{2}{6} C = \frac{1}{3} C

. Multiplying both sides by

3

yields

C = 45

Isolating

C

determines the overall volume of the tank.

Key Concept

Multi-step operations with fractions and fractional parts of a whole

Estimated Time:1m 30s

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