Lesson 50 of 100 | Quantitative Aptitude / Mixtures / मिश्रण
Alligation for two-component mixtures
Outcome
Find mixing ratios and actual quantities by alligation, and reject impossible targets.
Concept and assumptions
Assume homogeneous liquids, additive volumes in litres, and no reaction or solute loss. Use percentage by volume (% v/v) throughout; water is 0%. Alligation rearranges the weighted-average equation.
Let lower concentration be a%, higher concentration b%, and target m%, with a < b. If x L of the lower solution and y L of the higher solution are mixed:
ax + by = m(x + y).
The percentage factors of 100 cancel. Rearranging gives:
(m − a)x = (b − m)y.
Thus:
Lower-concentration quantity : higher-concentration quantity = (b − m):(m − a).
The lower solution’s shortfall balances the higher solution’s excess. Differences are percentage points; label the component order.
For positive amounts of both solutions, a < m < b. At an endpoint, only that endpoint solution can be used; the other amount is zero. Outside the interval, the target is impossible with nonnegative amounts. If a = b, every mixture has that concentration and the difference formula cannot determine a ratio.
Ratios give relative amounts, not litres. Divide a specified total by total ratio parts. For a known component, divide its quantity by its own ratio part.
Worked examples
Example 1 — Finding the ratio. Mix 10% and 40% solutions to obtain 22%.
Since 10 < 22 < 40, both quantities can be positive. Lower:higher = (40 − 22):(22 − 10) = 18:12 = 3:2. Check: (3 × 10 + 2 × 40)/(3 + 2) = 110/5 = 22%.
Example 2 — A fixed total. Prepare 80 L of 27% solution from 15% and 45% solutions.
Lower:higher = (45 − 27):(27 − 15) = 18:12 = 3:2. Total parts = 5; one part = 80/5 = 16 L. Use 3 × 16 = 48 L of 15% and 2 × 16 = 32 L of 45%. Check: solute = 48 × 0.15 + 32 × 0.45 = 7.2 + 14.4 = 21.6 L. Concentration = 100 × 21.6/80 = 27%.
Example 3 — A known starting quantity. How much 50% solution should be added to 24 L of 18% solution to obtain 30%?
Lower:higher = (50 − 30):(30 − 18) = 20:12 = 5:3. The existing 24 L represents 5 parts, so one part = 24/5 = 4.8 L. Add 3 × 4.8 = 14.4 L of 50% solution. Check: solute = 24 × 0.18 + 14.4 × 0.50 = 4.32 + 7.2 = 11.52 L. Total volume = 24 + 14.4 = 38.4 L; 100 × 11.52/38.4 = 30%.
Common mistakes
Reversing components; confusing ratio parts with litres; scaling a known component using total parts; accepting out-of-range targets; mixing incompatible units.
Practice questions
- In what ratio should 12% and 36% solutions be mixed to obtain 20%? State lower:higher.
- Prepare 60 L of 18% solution using 8% and 32% solutions. Find both volumes.
- Prepare 21 L of 24% solution using water and 40% solution. Find both volumes.
- Can 15% and 35% solutions produce a nonzero quantity of 40% solution by mixing alone? Explain.
Worked answers
- Lower:higher = (36 − 20):(20 − 12) = 16:8 = 2:1. Check: (2 × 12 + 36)/3 = 20%.
- Lower:higher = 14:10 = 7:5. One part = 60/(7 + 5) = 5 L. Use 35 L of 8% and 25 L of 32%.
- Water:40% solution = (40 − 24):(24 − 0) = 16:24 = 2:3. One part = 21/5 = 4.2 L. Use 8.4 L water and 12.6 L of 40% solution.
- No. Both concentrations are at most 35%, so their weighted mean cannot exceed 35%. Algebraically, 15x + 35y = 40(x + y) gives 25x + 5y = 0. With x, y ≥ 0, only x = y = 0 works, which is not a nonzero mixture.