Each of the three tanks holds n liters of water. After some amount of

Two approaches that I can think of

Approach 1 : Algebraic

Volume of water contained in three tanks = \(n + n + n = 3n\)

After distribution :

Assume that the common ratio is \(x\) is

Tank 1: Tank 2 : Tank 3 = \(x\) : \(6x\) : \(8x\)

\(x + 6x + 8x = 15x\)

As the volume remains the same among the three tanks

15x = 3n

n = 5x --- (1)

The initial volume of water in Tank 2 = \(n = 5x\)
The final volume of water in Tank 2 = \(6x \)
Volume of water poured = 6x - 5x = \(x\)

From (1)

\(x = \frac{n}{5}\)

Approach 2 : Assume Values

Assume that three tanks combined hold 15 gallons of water. Hence, each tank holds 15/3 = 5 gallons of water.

Hence, n = 5

After distribution Tank 2 hold 6 gallons of water.

Amount of water poured into tank 2 = 6 - 5 = 1 gallon

Amount of water poured in terms of n ⇒ \(\frac{n}{5}\)

Option C