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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.

Grace Hey
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  • Hint: the amount of salt you have doesn't change when you add water. Should say: I'm not quite sure what it means to "lower the concentration by $2%$.". Do you mean that the final mix should be $2%$ salt instead of $4%$ salt? That's a big change. – lulu Feb 10 '20 at 18:41
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    It appears that you are trying to lower the concentration to $2%$ rather than by $2%$. I would think the new concentration should be $.98\cdot.04=3.92%$. – saulspatz Feb 10 '20 at 18:56

2 Answers2

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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.

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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?

Izaak van Dongen
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    I'm not sure that this is the right interpretation of the problem. Please see my comment on the original post. – saulspatz Feb 10 '20 at 18:58