Mixture and Alligation
This summary simplifies the key points of the Mixture and Alligation PDF, including core concepts, formulas, worked examples, a cheatsheet, and all practice questions with clear answers.
Core Concepts
Alligation: A rule to find the ratio in which two or more ingredients at given prices must be mixed to obtain a mixture of a desired price.
Mean Price: The cost price per unit of the mixture.
Key Formula (Alligation Rule)
If two ingredients A and B cost $x$ and $y$ per unit, and the mean price is $M$:
$$ \text{Required Ratio} = \frac{M - Y}{X - M} = \frac{Y - M}{M - X} $$Where:
- $X$ = Cost of cheaper ingredient
- $Y$ = Cost of dearer ingredient
- $M$ = Mean price (desired price of the mixture)
Worked Examples
Example 1: In what ratio must rice costing Rs. 8.50 per kg be mixed with rice costing Rs. 13 per kg so that the mixture is worth Rs. 10 per kg?
- $X = 8.5$, $Y = 13$, $M = 10$
- Ratio = $\frac{13 - 10}{10 - 8.5} = \frac{3}{1.5} = 2:1$
Example 2: A grocer mixes sugar costing Rs. 60/kg and Rs. 65/kg to sell at Rs. 68.20/kg with 10% profit. Find the ratio.
- Let cost price of mixture = $M$
- Selling price = Rs. 68.20, Profit = 10%
- Cost price = $68.20 / 1.10 = 62$
- Ratio = $\frac{65 - 62}{62 - 60} = \frac{3}{2} = 3:2$
Example 3: 729 litres of a mixture (milk:water = 7:2). How much water to add so that new ratio is 7:3?
- Milk = $729 \times \frac{7}{9} = 567$ L
- Water = $729 \times \frac{2}{9} = 162$ L
- Let added water = $x$
- $\frac{567}{162 + x} = \frac{7}{3}$ ⇒ $567 \times 3 = 7 \times (162 + x)$ ⇒ $1701 = 1134 + 7x$ ⇒ $7x = 567$ ⇒ $x = 81$ L
Example 4: Three types of rice at Rs. 1.27, Rs. 1.29, Rs. 1.32/kg are mixed to sell at Rs. 1.30/kg. Find the ratio.
- Use alligation for each pair:
- (1.32 - 1.30):(1.30 - 1.29) = 0.02:0.01 = 2:1
- For third, sum accordingly.
- Final ratio: $2:1:5$
Example 5: Mix 4 types of rice (Rs. 95, 60, 90, 50/kg) to get mixture worth Rs. 80/kg.
- Use alligation stepwise.
- Final ratio: $4:4:5:1$
Replacement Concept Formula:
When a container of capacity $a$ units has $b$ units replaced $k$ times:
$$ \text{Quantity left} = a \left(1 - \frac{b}{a}\right)^k $$Example 6: 8 L drawn from a cask full of wine and replaced with water, repeated 4 times. Wine left to water is 16:65. Find original volume.
- Let original = $x$
- After 4 times: $x \left(1 - \frac{8}{x}\right)^4 = \frac{16}{81}x$
- Solve: $x = 24$ L
Cheatsheet
- Alligation Ratio: $\frac{\text{Higher price} - \text{Mean price}}{\text{Mean price} - \text{Lower price}}$
- Replacement after k operations: $a \left(1 - \frac{b}{a}\right)^k$
- For mixtures with more than 2 ingredients: Apply alligation stepwise for each pair.
Practice Questions with Answers
| Q# | Question (Summary) | Answer |
|---|---|---|
| 1 | 330 L mixture, water 24%. Sold 80 L, added 60 L milk & 26 L water. Final % water? | 25.59% (B) |
| 2 | Paint maker: 806, 930, 992 barrels of 3 qualities. Min buckets of equal size? | 44 (D) |
| 3 | Vessels A (4:3) & B (2:3) milk:water. Mix to get 1:1? | 7:5 (A) |
| 4 | Milkman sells milk+water at Rs. 9/L, cost of pure milk Rs. 10/L, profit 20%. Ratio milk:water? | 3:1 (A) |
| 5 | Stainless steel: Cr:Steel 2:11 & 5:21. Mix so Cr:Steel is 7:32? | 1:2 (C) |
| 6 | 60 L solution, 80% acid. Water added to make 60% acid? | 20 L (B) |
| 7 | Mixture alcohol:water 4:3. Add 5 L water, becomes 4:5. Alcohol in mixture? | 10 L (D) |
| 8 | Can with A:B = 7:5. Drain 9 L, fill with B, becomes 7:9. Initial A? | 21 L (C) |
Quick Application Tips
- Alligation is best for price/ratio mixture problems.
- Replacement formula is key for repeated removal/addition scenarios.
- For more than two components, apply alligation stepwise.
Summary Table of Practice Questions
| Q# | Key Concept | Answer |
|---|---|---|
| 1 | Successive mixing & % calculation | 25.59% |
| 2 | HCF for equal buckets | 44 |
| 3 | Mixing ratios for desired composition | 7:5 |
| 4 | Profit & mixture ratio | 3:1 |
| 5 | Mixing alloys for target ratio | 1:2 |
| 6 | Dilution to required % | 20 L |
| 7 | Ratio adjustment by addition | 10 L |
| 8 | Replacement & ratio | 21 L |
Conclusion: Mastering mixture and alligation involves understanding the alligation formula, the replacement concept, and practicing a variety of ratio and percentage problems. Use the above examples, formulas, and practice questions for quick revision and exam success.