Mixture and Alligation made clear.
A mixture’s concentration or unit price is a weighted average. Alligation is a compact way to recover the mixing ratio when two ingredients and a target average are known.
Key concepts in Mixture and Alligation
Weighted mean
Mean = total value / total quantity. For concentrations use solute amount divided by total mixture.
Alligation
For lower value L, higher value H and target M, low:high = (H−M):(M−L), provided L < M < H.
Replacement
After removing and replacing x from volume V, the original substance remaining after n well-mixed operations is original × (1−x/V)^n.
Units
Concentrations must use the same basis; keep litres, kilograms and prices per unit consistent.
Alligation map: 20% and 50% to make 30%
| Ingredient | Opposite difference | Required parts |
|---|---|---|
| 20% solution | 50−30 = 20 | 2 |
| 50% solution | 30−20 = 10 | 1 |
A reliable solving method
- Identify quantities and what each percentage or price measures.
- Find total solute or total cost before averaging.
- Use alligation only for two values on opposite sides of the target.
- For repeated replacement, recalculate through the remaining-fraction factor.
Step-by-step solved examples
Mix 2 kg at ₹40/kg with 3 kg at ₹60/kg. Find mean price.
- Total cost = 2×40 + 3×60 = ₹260.
- Total mass = 5 kg.
- Mean = 260/5 = ₹52/kg.
Mix 20% and 50% solutions to obtain 30%. Find their ratio.
- Low:high = (50−30):(30−20).
- Ratio = 20:10 = 2:1.
- Check: (2×20+1×50)/3 = 30%.
A 10 L milk container has 2 L removed and replaced with water twice. Milk left?
- Each operation leaves 1−2/10 = 0.8 of the milk.
- After two well-mixed operations: 10×0.8².
- Milk remaining = 6.4 L; water = 3.6 L.
Mixture and Alligation: practice questions
Choose one answer before opening the explanation. Write down your reasoning as well as the option; a correct guess is different from a method you can repeat.
Equal quantities at ₹30 and ₹50 per kg: mean?
Show answer and explanation — question 1
Answer B: ₹40
Equal weights give (30+50)/2 = ₹40.
Mix 10% and 40% to make 20%. Low:high?
Show answer and explanation — question 2
Answer C: 2:1
(40−20):(20−10) = 20:10 = 2:1.
4 L of 25% solution contains how much solute?
Show answer and explanation — question 3
Answer B: 1 L
4×0.25 = 1 L on a volume basis.
Add 2 L water to 2 L of 40% solution. New concentration?
Show answer and explanation — question 4
Answer B: 20%
Solute remains 0.8 L in 4 L: 20%.
Remove 5 L from 20 L pure milk and replace once. Milk left?
Show answer and explanation — question 5
Answer C: 15 L
One removal leaves 20−5 = 15 L milk.
Can 20% and 50% solutions alone make 60%?
Show answer and explanation — question 6
Answer D: No
A weighted average cannot exceed the larger ingredient concentration.
Frequently asked questions
When is the simple average valid?
When both ingredients have equal quantities on the same measurement basis.
Does adding water remove solute?
No. It increases total volume, so the same solute becomes less concentrated.
Identify quantities and what each percentage or price measures. For repeated replacement, recalculate through the remaining-fraction factor.