Average, Mixture & Alligation

Average, Mixture & Alligation

🟢 Lite

Key Formula/Rule

Average = Sum of all values ÷ Number of values. For mixture/alligation: when you mix two groups, the final average sits somewhere between the two individual averages — weighted toward whichever group is larger.

Memory Trick

Average is just “fair share” — if everyone got the same score, what would it be? Total ÷ Count. For alligation, think of mixing juice: 1 litre of 30% juice + 1 litre of 60% juice = 2 litres of 45% juice. The concentration just averages out when quantities are equal.

1-Sentence Summary

Averages measure the central tendency of a group; alligation solves “if I mix these two things, what do I get?” — and conversely, “what ratio do I need to get a target concentration?“

30-Second Example

Q: The average of 4, 8, 12, and 16 is? A: 10 — Sum = 40, Count = 4, so 40 ÷ 4 = 10.

🟡 Standard

Concept

Average is the most intuitive statistic. Imagine 5 students score 60, 70, 80, 90, and 100. Their average is (60+70+80+90+100) ÷ 5 = 400 ÷ 5 = 80. Every student “effectively” scored 80 — it’s the fair-share number. The formula Average = Sum ÷ Count works for any group of numbers, whether they’re ages, prices, speeds, or anything else.

Weighted Average comes into play when groups have different sizes. If Group A (10 students) averages 60 and Group B (20 students) averages 80, you can’t just average 60 and 80 to get 70. The 20 students from Group B count twice as much. The combined average = (10×60 + 20×80) ÷ (10+20) = (600 + 1600) ÷ 30 = 2200 ÷ 30 = 73.3. This is the single most important formula in this topic.

Alligation solves a specific practical problem: if you have rice costing ₹30/kg and rice costing ₹50/kg, what ratio must you mix them to get a mean price of ₹40/kg? The alligation rule says:

• Difference from mean for cheaper: 40 − 30 = 10 parts cheap
• Difference from mean for expensive: 50 − 40 = 10 parts expensive
• Ratio = 10 : 10 = 1 : 1

When quantities are equal, the mean price is the simple average. When quantities differ, the alligation rule automatically accounts for the weight difference.

Key Formulas

Formula	Use
Average = (Sum of values) / (Number of values)	Finding the central value
Sum = Average × Number of values	Reversing the average
Missing number = Avg × n − (sum of rest)	Finding a missing value in a group
Weighted Avg = (n₁×avg₁ + n₂×avg₂) / (n₁+n₂)	Combining two groups
Alligation ratio = (d − m) : (m − c)	Finding mixing ratio (d=dear, c=cheap, m=mean)

Worked Example

Q: A merchant mixes 20 kg of tea costing ₹200/kg with 30 kg of tea costing ₹300/kg. What is the price of the mixture per kg?

Step 1: Total cost of cheap tea = 20 × 200 = ₹4000 Step 2: Total cost of expensive tea = 30 × 300 = ₹9000 Step 3: Total mixture = 20 + 30 = 50 kg Step 4: Price per kg = (4000 + 9000) ÷ 50 = 13000 ÷ 50 = ₹260/kg

Answer: ₹260 per kg

Common Errors

• Simply averaging two averages (60+80)/2 = 70 when groups have different sizes → use weighted average formula
• Mixing up which price goes on which side of the alligation diagram → cheap on left, dear on right, mean in the middle
• Forgetting to multiply by quantity in alligation problems → always check: is it ratio by weight, or ratio by value?

🔴 Extended

Full Concept

Why Weighted Average Formula Works — The Proof

When combining two groups, you’re really just adding all their individual values together. Group 1 has n₁ items with average avg₁, contributing a total of n₁ × avg₁ to the grand sum. Group 2 contributes n₂ × avg₂. The total sum is (n₁×avg₁ + n₂×avg₂) over (n₁ + n₂) items, giving the weighted average formula.

This isn’t just a formula — it explains why the simple average of two averages (60 and 80) is wrong when group sizes differ. If 10 people average 60 and 30 people average 80, the 30 people from Group B drag the combined average up toward 80. You can’t treat a group of 10 and a group of 30 identically.

Alligation — The Graphical Method Derivation

The alligation cross is a visual shortcut. Place the cheaper concentration on the left, the dearer on the right, and the desired mean in the middle:

Cheaper (c)     Mean (m)     Dearer (d)
    c -------|------ m ------|------- d

The difference d − m tells you how many parts of the cheaper you need. The difference m − c tells you how many parts of the dearer you need. Why? Because the area of each rectangle in the diagram is proportional to the quantity needed — heavier weights are farther from the mean.

More formally: In the mixture, if you need x kg at ₹c/kg and y kg at ₹d/kg to get mean ₹m/kg:

• Total cost = cx + dy
• Total quantity = x + y
• Mean price: m = (cx + dy)/(x + y)
• Solving: mx + my = cx + dy → m(x − y) = (c − m)x + (d − m)y

For the special case where x + y = 1 (unit mixture): the ratio x : y = (d − m) : (m − c).

When Values Repeat in the Set

If a number appears multiple times in a group, treat it as repeated entries. The average of {5, 5, 5, 7, 7, 8} is (5+5+5+7+7+8) ÷ 6 = 37 ÷ 6 = 6.17. You can also think: the sum of the five 5s and 7s… no shortcut here, just count each occurrence.

Advanced Mixture Problems

Alloy problems: A jeweller has gold of 80% purity and gold of 95% purity. How much of each must be mixed to get 100g of 88% pure gold?

• Using alligation: cheaper = 80%, dearer = 95%, mean = 88%
• Ratio = (95 − 88) : (88 − 80) = 7 : 8
• Total parts = 15, so each part = 100/15 g ≈ 6.67 g
• 80% gold needed = 7 × 6.67 ≈ 46.67 g
• 95% gold needed = 8 × 6.67 ≈ 53.33 g

Spirit-water problems: A vessel contains 40 litres of a mixture with 30% spirit. How much water must be added to make it 20% spirit?

• Spirit quantity stays constant: 30% of 40 = 12 litres of spirit
• If 12 litres = 20% of new total, then new total = 12 ÷ 0.2 = 60 litres
• Water added = 60 − 40 = 20 litres

Replacement problems: A vessel has 60 litres of milk. 12 litres are removed and replaced with water. This is done twice. How much milk remains?

• After 1st replacement: milk = 60 × (1 − 12/60) = 60 × 48/60 = 48 litres
• After 2nd replacement: milk = 48 × (48/60) = 48 × 0.8 = 38.4 litres
• Formula: Final = Initial × (1 − r/n)ⁿ where r = replaced amount, n = total

Multiple Approaches

Finding a missing number when average is known:

• Standard: Total = Average × n, Missing = Total − (sum of known numbers)
• Shortcut: The missing number always equals (given average × total count) − (sum of all other numbers)

Alligation when one component is pure water (0%):

• If mixing water (0%) with 50% juice to get 30% juice, ratio = (50 − 30) : (30 − 0) = 20 : 30 = 2 : 3
• Water always goes on the 0% side of the diagram

CUET-Level Problems

Q1: The average weight of 30 students is 45 kg. One student weighing 60 kg leaves and is replaced by a new student. The new average becomes 44 kg. What is the weight of the new student? Working: Total initial weight = 30 × 45 = 1350 kg. After one leaves, weight = 1350 − 60 = 1290 kg. New total for 30 students = 30 × 44 = 1320 kg. New student’s weight = 1320 − 1290 = 30 kg. Answer: 30 kg

Q2: In what ratio must rice at ₹40/kg be mixed with rice at ₹60/kg so that the mixture costs ₹50/kg? Working: Alligation: cheaper 40, dearer 60, mean 50. Difference right = 60 − 50 = 10 parts dearer. Difference left = 50 − 40 = 10 parts cheaper. Ratio = 10 : 10 = 1 : 1. Answer: 1 : 1

Tricky Cases

• Mean is outside the range — if you want a mean price of ₹25 between ₹20 and ₹30, that’s fine. But if someone asks for mean price of ₹35 between ₹20 and ₹30, it’s impossible — the mean must always lie between the two extremes.
• Replacement without mixing: Removing some mixture and replacing with pure water repeatedly — always use (1 − r/n)ⁿ formula
• Average of the same number repeated: Adding numbers equal to the average doesn’t change the average — useful in sequential addition problems
• Weighted average when n₁ = n₂: Falls back to simple average (avg₁ + avg₂)/2 — but only when group sizes are equal

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