# Math Help - Rational equation to solve mixture problem

1. ## Rational equation to solve mixture problem

A pollutant has leaked into a water tank containing 2000 gallons of water. The tank now is 3% pollutant. How much clean water should be added to the tank to decrease the concentration of the pollutant to a safe level of 0.5%?

TIA!

2. Originally Posted by eagertolearn
A pollutant has leaked into a water tank containing 2000 gallons of water. The tank now is 3% pollutant. How much clean water should be added to the tank to decrease the concentration of the pollutant to a safe level of 0.5%?

TIA!
Let the present volume of liquid present in the fluid be Vi.
3% is pollutant.
After adding fresh water, let the vol. of the tank be Vf and pollutant drops to 0.5%. Since the amount of pollutant remains same before and after adding water, we will take a material balance of the pollutant.

therefore,
(3/100)*Vi = (0.5/100)*Vf

We need to know amount of clean water added, i.e. Vf-Vi

Now, initially we had 2000 gallons of water. after addition of pollutant it becomes 97%.
therefore, 97% of Vi = 2000 gallons.

First solve for vVi, then Vf and finally Vf - Vi to get the answer.

3. Hello eagertolearn
Originally Posted by eagertolearn
A pollutant has leaked into a water tank containing 2000 gallons of water. The tank now is 3% pollutant. How much clean water should be added to the tank to decrease the concentration of the pollutant to a safe level of 0.5%?

TIA!
Can I suggest an alternative method, using ratios?

Initially, there are 2000 gallons of water in the tank, which is 97% of the total amount. The remaining 3% is pollutant. So the ratio of water to pollutant is $97:3$. The amount of pollutant in the tank is therefore:
$\frac{3}{97}\times 2000= 61.856$ gallons
After adding extra water, this has to be 0.5% of the total, the other 99.5% being water. So now the ratio of water to pollutant is $99.5:0.5=199:1$. Since there are still $61.856$ gallons of pollutant in the tank, the amount of water must now be:
$61.856\times199 = 12309$ gallons
Therefore, we must add $12309 - 2000 = 10309$ gallons of water.