Finding original concentration after adding water to a salt solution

Key takeaway: Salt-solution concentration changes predictably when 200 g of water is added, and the mass of salt stays 45 g throughout the dilution process.

Identify the dilution relationship

The salt mass remains 45 g before and after the 200 g of water is added. What changes is the total mass of the solution, so the concentration drops because the denominator gets larger while the numerator stays fixed.

Let the original mass of the solution be $m$ grams. Then the original concentration is

$\frac{45}{m}$

as a decimal fraction. After adding 200 g of water, the new mass becomes $m+200$, so the new concentration is

$\frac{45}{m+200}.$

The problem says this concentration is 8% lower than before, so the decrease is

$\frac{45}{m}-\frac{45}{m+200}=0.08.$

Build the equation from the concentration drop

Because the salt amount is fixed at 45 g, the only way the concentration can decrease by 0.08 is through the change in total solution mass. Substituting the two concentration expressions gives

$\frac{45}{m}-\frac{45}{m+200}=0.08.$

Multiply both sides by $m(m+200)$:

$45(m+200)-45m=0.08m(m+200).$

The left side simplifies immediately to

$9000=0.08m^2+16m.$

Move everything to one side:

$0.08m^2+16m-9000=0.$

To avoid decimals, multiply by 100:

$8m^2+1600m-900000=0.$

Divide by 8:

$m^2+200m-112500=0.$

Solve for the original solution mass

Use the quadratic formula:

$m=\frac{-200\pm\sqrt{200^2-4(1)(-112500)}}{2}.$

So

=\frac{-200\pm\sqrt{490000}}{2} =\frac{-200\pm700}{2}.$$ The positive solution is $$m=250.$$ So the original concentration is $$\frac{45}{250}=0.18.$$ Since the answer is requested as a decimal, the original concentration is **0.18**. ## Common mistake to avoid Do not treat the 8% drop as 8 percentage points of salt mass. The 45 g of salt never changes; only the total mass changes. Also, be careful not to add 200 g directly to the numerator. The added water affects the whole solution, not the amount of salt. ## Check the result If the original solution weighs 250 g, then its concentration is 0.18. After adding 200 g of water, the total becomes 450 g, and the new concentration is $45/450=0.10$. The drop is $0.18-0.10=0.08$, which matches the condition exactly. --- **Pitfalls the pros know** 👇 The equation usually breaks down when the 8% decrease is interpreted too loosely. In this dilution problem, the 45 g of salt stays constant, so the concentration must be written as salt divided by total solution mass, not as a subtraction on the salt amount itself. Another common error is using 200 g as if it were added to the salt mass. The 200 g is pure water, so it only increases the denominator. If you set the original mass of solution to a variable, remember that both the original and the diluted concentrations must be expressed with that same unknown. Skipping that consistency leads to an equation that looks reasonable but cannot give the correct decimal answer. **What if the problem changes?** If the problem were changed to: “If 150 g of water is added to a solution containing 45 g of salt, the concentration decreases by 6%. Find the original concentration as a decimal,” the setup would be the same type of dilution equation. Let the original solution mass be $m$. Then the concentrations are $45/m$ and $45/(m+150)$, and the condition becomes $45/m - 45/(m+150) = 0.06$. Solving that equation gives the new original concentration. This variant is useful because it tests whether you can reuse the same ratio model when the water amount and percent decrease both change, while the salt mass remains fixed. `Tags`: [solution concentration formula](/qa/seo/Math), [dilution equation setup](/qa/seo/Math), [mass percent concentration](/qa/seo/Math)