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Mixture & Alligation

SBI Clerk · Quantitative Aptitude

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Core Concept

Read this first — the foundation of the topic

→CORE CONCEPT

Alligation is a quick method to find the ratio of mixing two items when we know their individual values and the mean (average) value of the mixture. It works on the principle that gain of one quantity equals loss of another quantity

💡KEY RULES

The ratio of quantities is inversely proportional to their differences from the mean 2. Cheaper quantity × Its difference from mean = Dearer quantity × Its difference from mean 3. For replacement problems, use: Final strength = Initial strength × (1 - R/C)^n where R = replaced quantity, C = total capacity, n = number of operations

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Formula Block

Memorise — at least one formula appears in every paper

• Basic Alligation Formula: Ratio = (Mean - Cheaper value) : (Dearer value - Mean)

• For percentages: If x% and y% solutions are mixed to get z%, then Ratio = (z-x) : (y-z)

• Weighted Average: Mean = (A×a + B×b) / (A+B) where A,B are quantities and a,b are their values

• Replacement Formula: New concentration = Old × (1 - fraction removed)^number of operations

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Exam Patterns

What examiners ask — read before attempting PYQs

→Common types include

mixing two different priced items, alcohol-water mixtures, milk-water problems, successive mixing, and replacement of mixtures

⚡SHORTCUT TRICK - CROSS METHOD

Draw a cross diagram. Put mean in center, cheaper value top-left, dearer value bottom-left. Differences go on right side. This gives the mixing ratio directly

✏️WORKED EXAMPLE 1

1

Apply alligation formula Cheaper tea = Rs 64, Dearer tea = Rs 86, Mean = Rs 68

2

Find differences Difference of cheaper from mean = 68 - 64 = 4 Difference of dearer from mean = 86 - 68 = 18

3

Write ratio Ratio = 4:18 = 2:9 So tea worth Rs 64 and Rs 86 should be mixed in ratio 2:9

✏️WORKED EXAMPLE 2

1

Find initial milk quantity Initial milk = 60 × 80% = 48 litres

2

Set up equation for final state Let x litres be removed and replaced with water Milk removed = x × 80% = 0.8x Final milk = 48 - 0.8x

3

Apply condition Final milk percentage = 60% (48 - 0.8x)/60 = 60/100 48 - 0.8x = 36 0.8x = 12 x = 15 litres SHORTCUT FOR SUCCESSIVE OPERATIONS: When same fraction is removed repeatedly, use: Final = Initial × (1-f)^n where f = fraction removed, n = number of times. COMMON MISTAKE #1: Students often confuse the direction of ratio in alligation

💡Remember

the ratio is OPPOSITE to what you might think. If mean is closer to cheaper value, you need MORE of the cheaper item, not less. Always write (Mean - Cheaper) : (Dearer - Mean) for the ratio of Cheaper : Dearer.

🔑 Key Points

Alligation finds mixing ratios when individual values and mean value are known
Formula: Ratio = (Mean - Cheaper) : (Dearer - Mean) for quantities cheaper:dearer
Cross method: Put mean in center, write differences on opposite sides to get ratio
For replacement: Final concentration = Initial × (1 - removed fraction)^operations
In successive operations, multiply the reduction factor repeatedly
Weighted average formula: Mean = (Q1×V1 + Q2×V2)/(Q1+Q2)
The mixing ratio is inversely proportional to distance from mean value
When mean is closer to cheaper value, more cheaper quantity is needed
For percentage mixtures, treat percentages as values in alligation formula
Always check if final answer makes logical sense with given constraints

📌 Exam Facts

Alligation rule: Cheaper quantity : Dearer quantity = (Mean - Cheaper value) : (Dearer value - Mean)
For milk-water problems, concentration typically reduces by factor (1 - R/C) per operation
When equal quantities of two mixtures are combined, resultant concentration is arithmetic mean of both
In successive mixing, final ratio = Product of individual operation ratios
Standard milk percentage in problems: Pure milk = 100%, typical mixtures = 80%, 75%, 60%
Replacement formula for n operations: Final = Initial × (1 - removed/total)^n
Cost price mixing follows same alligation rules as percentage mixing
Two mixtures of same concentration when mixed give same concentration

Questions Asked in Previous Exams

Real questions from SSC papers — 2015 to 2024 · Showing 4 of 13

Exam Q 12021Previous Year Pattern

Two containers have sugar solutions of 20% and 50% concentration respectively. In what ratio should they be mixed to get a 30% solution?

Exam Q 22021Previous Year Pattern

A shopkeeper mixes two types of rice costing ₹40 per kg and ₹60 per kg in the ratio 3:2. What is the cost price of the mixture per kg?

Exam Q 32021Previous Year Pattern

A container has 60 litres of a mixture of alcohol and water in the ratio 7:5. How much pure alcohol must be added to make the ratio 3:1?

Exam Q 42021Previous Year Pattern

A goldsmith mixes gold of 18 carats and 24 carats in the ratio 1:2. What is the carat value of the resulting mixture?

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🚀 60-Second Revision

Formula: Ratio = (Mean - Cheaper) : (Dearer - Mean) - remember the cross pattern
Replacement: New concentration = Old × (1 - fraction removed)^number of times
Trap: Ratio direction is opposite to intuition - more cheaper when mean is closer to cheaper
Quick check: Final ratio quantities should be inversely related to their distance from mean
For successive operations, multiply the (1 - removed fraction) repeatedly
Cross method saves time: Draw cross, put mean center, differences give ratio directly
Always verify answer makes sense: percentages between 0-100, positive quantities

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