Expert: Josh - 10/22/2009

Question
Marco is draining his two aquariums. The tank are the same size. One tank has 7in. Of water in it. It is being drained at a rate of 1in./min. The other tank has 5in. Of water in it. It is being drained at a rate of 0.5in./min. After how many min will the tanks contain the same amount of water?

Answer
Ivory,

The amount of water which has been drained at any point in time is given by: R*T, where R is the drainage rate and T is the elapsed time. For one of the tanks, if we begin with X inches of water, then the amount of water left in the tank at time "t" is given by D(t)=X-R*t.

For the first tank, we can model the situation with the equation: D1(t)=X1-R1*t.
Here, D1(t) is the depth of water left in the first tank at time "t",
     X1 is the initial depth of the water in the first tank (when t=0),
     R1 is the drainage rate for the first tank measured in inch/min.

Similarly, we can model the situation with the second tank using D2(t)=X2-R2*t.
D2, X2, R2 are the corresponding values for the second tank.

Data: we have been told basically that X1=7, R1=1, X2=5 and R2=0.5.

Our task is to find the moment in time "t" when the depth of both tanks are the same. i.e., we want to solve D1(t)=D2(t).

Using the expressions for D1 and D2 we obtained earlier, we start with
X1-R1*t = X2-R2*t, the values of X1,X2,R1,R2 are known.

Substituting X1=7, R1=1, X2=5 and R2=0.5 into the above equation, we have
7-t = 5-0.5t. From this point on, finding "t" is just an algebra exercise.

For instance, we may subtract 5 from both sides (of the equation) to get 2-t = 0.5t. Then, add t to both sides to get 2 = 0.5*t. Finally, dividing both sides by 0.5, we get t=4 (hour).

As an exercise, check to see if this answer is correct. i.e., plugging this "t" value into the expression of D1(t) and D2(t) to check the water depths for both tanks.