Mixture and Alligation

Mixture: When two or more items (liquids, solids, or costs) are combined to form a single entity.

Examples:

• Mixing water and milk

• Mixing metals to form alloys

• Mixing two varieties of rice

Alligation: A simple rule to determine the ratio in which two ingredients with different values (like price or concentration) must be mixed to obtain a mixture at a given average value.

 Alligation Rule (Shortcut Formula)

 

$$$$

$$\text{Ratio (Cheaper : Dearer)} = \frac{\text{Dearer} - \text{Mean}}{\text{Mean} - \text{Cheaper}}$$

Where:

• Cheaper = Cost or concentration of the cheaper item

• Dearer = Cost or concentration of the dearer item

• Mean = Desired cost/concentration of the mixture

---

🧠 Types of Problems in Mixture & Alligation

A. 💰 Price-Based Problems

Example 1:
Rice costing ₹20/kg is mixed with rice costing ₹30/kg. The mixture is to be sold at ₹25/kg. In what ratio should they be mixed?

✅ Solution:

$$\text{Ratio}=\frac{30-25}{25-20}=\frac{5}{5}=1:1$$

🟩 Answer: Mix in 1:1 ratio.

B. 🥛 Concentration-Based Problems

Example 2:
In what ratio should water (₹0/litre) be mixed with milk (₹60/litre) to get a mixture worth ₹40/litre?

✅ Solution:

$$\text{Water : Milk} = \frac{60 - 40}{40 - 0} = \frac{20}{40} = 1:2$$

🟩 Answer: Mix in 1:2 ratio.

C. 🔁 Replacement Problems (Repeated Replacement)

When a part of a mixture is removed and replaced repeatedly, use the formula:

$$\text{Milk remaining} = Q \times \left(1 - \frac{x}{Q}\right)^n$$

Where:

• Q = Initial quantity

• x = Quantity replaced each time

• n = Number of operations

Example 3:
20 L of pure milk. 5 L is removed and replaced with water. Operation repeated 2 times. How much milk remains?

✅ Solution:

$$20 \times \left(1 - \frac{5}{20}\right)^2 = 20 \times \left(\frac{3}{4}\right)^2 = 20 \times \frac{9}{16} = 11.25 \, \text{L}$$

🟩 Answer: 11.25 L of milk remains.

---

⚡ Quick Tips

• Water is often considered ₹0/litre in concentration-based problems.

• When the mean lies exactly in the middle of the two values, ratio is always 1:1.

• For more than 2 components, break the problem into pairs or use weighted averages.

---

📝 5 Practice MCQs

1. In what ratio should sugar costing ₹20/kg and ₹30/kg be mixed to obtain sugar at ₹28/kg?

A) 1 : 3
B) 2 : 1
C) 2 : 3
D) 1 : 2

✅ Solution:

$$\frac{30 - 28}{28 - 20} = \frac{2}{8} = 1 : 4$$

✅ Answer: Not in options

Correct answer is 1 : 4

2. What is the ratio of milk to water when water (cost ₹0/l) is mixed with milk costing ₹50/l to get a mixture worth ₹40/l?

A) 1 : 2
B) 2 : 3
C) 1 : 4
D) 1 : 1

✅ Solution:

$$\frac{50-40}{40-0}=\frac{10}{40}=1:4$$

✅ Answer: C) 1 : 4

3. 40 L mixture of milk and water contains 25% water. How much water must be added to make it 50%?

A) 10 L
B) 13.33 L
C) 15 L
D) 20 L

✅ Solution:
Water in 40 L = 25% of 40 = 10 L
Let x L of water be added
Now, total = 40 + x
Water = 10 + x
Now:

$$\frac{10+x}{40+x}=0.5\Rightarrow 10+x=0.5(40+x)\Rightarrow 10+x=20+0.5x\Rightarrow 0.5x=10\Rightarrow x=20$$

✅ Answer: D) 20 L

4. A container has 60 L of milk. 10 L is replaced by water. This is repeated once more. Find milk left.

A) 36 L
B) 40 L
C) 45 L
D) 48 L

✅ Solution:

$$\text{Milk left} = 60 \times \left(1 - \frac{10}{60}\right)^2 = 60 \times \left(\frac{5}{6}\right)^2 = 60 \times \frac{25}{36} = 41.67 \, \text{L}$$

✅ Answer: Not in options 
Correct Answer: 41.67 L

5. A mixture has milk and water in 7:5 ratio. 24 L of water is added, and new ratio becomes 7:9. Find original quantity.

A) 96 L
B) 72 L
C) 84 L
D) 108 L

✅ Solution:

Let original parts = 7x milk, 5x water
After adding 24 L water:

$$\frac{7x}{5x+24}=\frac{7}{9}\Rightarrow 7x\times 9=(5x+24)\times 7\Rightarrow 63x=35x+168\Rightarrow 28x=168\Rightarrow x=6$$

Total = (7 + 5) × 6 = 72 L

✅ Answer: B) 72 L