Two fish tanks, with 12 gallons & 3 gallons of water. The first fish tank is filled at rate of 1/3 gallons/minute and the other at a rate 1 ½ of the first. After how much time will both have equal amount of water if hoses are turned on simultaneously?

Jun 8, 2017

The rate of filling the 1st tank $\frac{1}{3}$ gallon/min

The rate of filling the 2nd tank $1 \frac{1}{2} o f \frac{1}{3} = \frac{3}{2} o f \frac{1}{3} = \frac{1}{2}$ gallon/min

Let after t min the two tanks have equal amount of water,if hoses are turned on simultaneously. Initially two fish tanks, with 12 gallons & 3 gallons of water respectively.

So by the problem

$3 + \frac{t}{2} = 12 + \frac{t}{3}$

$\implies \frac{t}{2} - \frac{t}{3} = 12 - 3$

$\implies \frac{3 t - 2 t}{6} = 9$

$\implies \frac{t}{6} = 9$

$\implies t = 9 \times 6 = 54 \min$

Jun 8, 2017

After $54$ minutes both tanks will have same amount of water.

Explanation:

Let $t$ be time in minutes after which both tanks will have same amount of water.
Rate of filling water in first tank is $\frac{1}{3}$ gallons per minute.
Rate of filling water in second tank is $1 \frac{1}{2} \cdot \frac{1}{3} = \frac{\cancel{3}}{2} \cdot \frac{1}{\cancel{3}} = \frac{1}{2}$ gallons per minute.

By given condition , $12 + \frac{1}{3} t = 3 + \frac{1}{2} t \mathmr{and} \frac{1}{2} t - \frac{1}{3} t = 12 - 3$ or

$\frac{1}{6} t = 9 \mathmr{and} t = 6 \cdot 9 = 54$ minutes.

After $54$ minutes both tanks will have same amount of water.[Ans]