How much pure water must be added to lower the concentration by 2%

Question

The concentration of salt in a 6000-gallon sea world aquarium is a bit high at 4% salt. How much pure water must be added to lower the concentration by 2%?

What I tried: I tried to draw this out and I believe that the equations are:

6000+x=y .04+0=.02

I am having a hart time figuring out the equations.

Answer 1

First of all let's work out how much of the water is currently salt - $6000 \cdot 4\% = 240$ gallons of salt.

Say we add $x$ gallons of pure water. The new total amount of water is $6000 + x$, so the proportion of salt is $240 / (6000 + x) = 2\%$ (or whatever percentage you decide is what you're aiming for, depending how you read the question. Could be $3.92\%$ :)). Can you take it from here?

Answer 2

As noted by @saulspatz the natural meaning of your question is a reduction to $3.92$%.

So you require

$$6000\times4=(6000+v)\times3.92$$ $v\approx 122.45$ gallons.