Mixture Problems | Naruhodo Tutoring

Problem Setup

c₁·V₁ + c₂·V₂ = c_mix·(V₁+V₂)

Find (unknown)

V₁ (liters)

c₁ (%)

V₂ (liters)

c₂ (%)

Target conc. c_mix (%)

Equation: c₁·V₁ + c₂·V₂ = c_mix·(V₁+V₂). Leave the unknown field blank or set the dropdown.

Examples

c₁·V₁ + 0·x = c_mix·(V₁+x) [dilute with water]

Add to existing solution

Existing volume (liters)

Existing conc. (%)

Desired concentration c_mix (%)

Add water (0%) to dilute; find how much water x to add.

Examples

Solution

Enter values above and press Solve to see the unknown quantity, the mixture equation, and a mixture table.

Answer

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Mixture Equation (substituted)

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Mixture Table

Solution	Volume (L)	Conc. (%)	Amount (L)

Mixture Equation

c₁V₁ + c₂V₂ = c_mix(V₁ + V₂)

The key idea: the amount of pure substance in each component adds up to the amount in the final mixture.

Amount = Volume × Concentration (as a decimal). So 4 liters of a 20% solution contains 4 × 0.20 = 0.8 liters of pure substance.

Depending on which value is unknown, solve the equation for V₁, V₂, c₁, or c₂ by isolating the variable algebraically.

The concentrations must be in the same units (both as percents or both as decimals). Do not mix them.

Mixture Table

Solution	Volume	Conc.	Amount
Solution 1	V₁	c₁	c₁ · V₁
Solution 2	V₂	c₂	c₂ · V₂
Mixture	V₁ + V₂	c_mix	c_mix(V₁+V₂)

Set up the table first — it makes the equation obvious.
Adding water: use concentration 0%. Adding pure: use 100%.
The mixture concentration must be between the two component concentrations.
Volume is always additive: V_total = V₁ + V₂.

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