Mixture and Alligation: Notes और MCQ Practice

Mixture & Alligation

आसान भाषा में: जब दो या दो से अधिक items को मिलाकर नया mixture बनाते हैं, तो उसका average concentration या cost price निकालना होता है। Alligation का प्रयोग दो items की अलग‑अलग concentrations या prices को मिलाकर desired concentration/price पाने में होता है.

Important Formulas / Rules

• Average = (Σ (Quantity × Rate)) ÷ Σ Quantity
• Alligation Rule: Difference of each component’s rate from the desired rate → cross‑multiply to get ratio of quantities.
• Total Quantity = (Quantity₁ + Quantity₂ + …)

Quick Tricks

1. Cross‑multiply shortcut – In alligation, write the two rates on ends, desired rate in middle, subtract diagonally, then swap the differences to get quantity ratio instantly.
2. Use “Weighted Mean” – Treat mixture problem as weighted mean; just plug values in the average formula, no need to draw diagrams.

Common Mistakes

1. Mixing up difference direction – always subtract the smaller rate from the larger when forming the ratio.
2. Forgetting to convert units (e.g., % to decimal) before applying the formula, leading to wrong average.

Mixture and Alligation

1. Concept

Mixture problems deal with combining two or more items having different quantities and/or different costs (or concentrations) to obtain a final mixture. Alligation is a quick method to find the ratio in which two or more items must be mixed to achieve a desired average price or concentration.

2. Important formulas / rules / facts

• Total quantity of mixture = Sum of individual quantities.
• Total cost of mixture = Sum of (quantity × cost) of each component.
• Average cost (or concentration) of mixture = Total cost ÷ Total quantity.
• Alligation rule (two items):
• Write the desired mean in the middle, the higher cost on the left, the lower cost on the right.
• Subtract diagonally: (Higher cost – Desired mean) gives the part of the cheaper item; (Desired mean – Lower cost) gives the part of the costlier item.
• Ratio of quantities = (Higher cost – Desired mean) : (Desired mean – Lower cost).
• Alligation for three or more items:
• Pair the items step‑by‑step using the two‑item rule until the required mean is reached.
• Percentage profit / loss in mixture:
• Profit % = (Selling price – Cost price) ÷ Cost price × 100.
• Loss % = (Cost price – Selling price) ÷ Cost price × 100.

3. Solved examples (exam style)

Example 1

Question: A trader mixes 12 kg of rice costing Rs 20 per kg with 8 kg of rice costing Rs 30 per kg. Find the cost price per kg of the mixture.

Solution:

1. Total cost = (12 × 20) + (8 × 30) = 240 + 240 = Rs 480.
2. Total quantity = 12 + 8 = 20 kg.
3. Average cost = 480 ÷ 20 = Rs 24 per kg.

Answer: Rs 24 per kg.

Example 2

Question: In what ratio should two solutions costing Rs 5 per litre and Rs 15 per litre be mixed to obtain a solution costing Rs 9 per litre?

Solution (Alligation):

```

15 | (15 – 9) = 6

9 | Desired mean

5 | (9 – 5) = 4

```

Ratio = 6 : 4 = 3 : 2 (costlier : cheaper).

Answer: 3 parts of Rs 15 solution with 2 parts of Rs 5 solution.

Example 3

Question: A chemist has a 25 % acid solution (200 ml) and a 40 % acid solution (300 ml). How much water must be added to obtain a 30 % acid solution?

Solution:

1. Acid in first solution = 25 % of 200 = 0.25 × 200 = 50 ml.
2. Acid in second solution = 40 % of 300 = 0.40 × 300 = 120 ml.
3. Total acid = 50 + 120 = 170 ml.
4. Let water to be added = w ml. Final volume = 200 + 300 + w.
5. Required concentration: 30 % = 170 ÷ (500 + w).

→ 0.30 × (500 + w) = 170

→ 150 + 0.30w = 170

→ 0.30w = 20

→ w = 20 ÷ 0.30 = 66.67 ml (≈ 67 ml).

Answer: About 67 ml of water.

Example 4

Question: Two traders sell sugar at Rs 45 per kg and Rs 55 per kg respectively. If a third trader wants to buy 150 kg of sugar at an average price of Rs 50 per kg, how much should he buy from each trader?

Solution (Alligation):

```

55 | (55 – 50) = 5

50 | Desired mean

45 | (50 – 45) = 5

```

Ratio = 5 : 5 = 1 : 1.

Total quantity = 150 kg → each supplies 75 kg.

Answer: 75 kg from each trader.

Example 5

Question: A mixture contains milk and water in the ratio 3:2. If 10 litres of water is added, the new ratio becomes 3:4. Find the original quantity of the mixture.

Solution:

1. Let original milk = 3x, water = 2x → total = 5x.
2. After adding 10 l water, water becomes 2x + 10.
3. New ratio milk : water = 3 : 4 →

3x ÷ (2x + 10) = 3 ÷ 4

→ 4·3x = 3·(2x + 10)

→ 12x = 6x + 30

→ 6x = 30 → x = 5.

1. Original mixture = 5x = 25 litres.

Answer: 25 litres.

4. Shortcuts and tricks

• Alligation shortcut: Directly write the differences (higher‑mean and mean‑lower) without drawing the full diagram; the numbers themselves give the required ratio.
• Weighted average shortcut: When only two components are involved, use the formula:

Mean = (Quantity1 × Cost1 + Quantity2 × Cost2) ÷ (Quantity1 + Quantity2).

• Quick check for profit/loss: If the selling price is more than the average cost, profit = (SP – Avg Cost) × Total quantity; otherwise loss.
• Use cross‑multiplication for ratio problems: If ratio a:b = p:q, then a = k·p, b = k·q where k = total ÷ (p+q).

5. Common mistakes to avoid

• Mixing up higher and lower costs while applying alligation; always subtract diagonally (higher‑mean gives quantity of cheaper item).
• Forgetting to convert percentages to decimals before using them in calculations (e.g., 25 % = 0.25).
• Ignoring units – keep quantities and costs in the same units throughout the problem.
• Adding percentages directly (e.g., assuming 25 % + 40 % = 65 % concentration) – instead, work with actual amounts of the component.
• Using the wrong total quantity after adding or removing a component; always recompute the total before applying the final ratio.