Mixture and Alligation for SSC CGL — Alligation Rule, Replacement Formula & Weighted Average

Concept

Mixture and alligation is fundamentally about answering one question: when you blend two things of different quality (price, concentration, speed, anything measurable), what ratio produces a target quality?

Think of it this way. You have cheap tea at ₹30/kg and premium tea at ₹50/kg. You want a blend that costs ₹38/kg. Common sense says: the blend should contain more cheap tea than premium (since ₹38 is closer to ₹30 than to ₹50). Alligation gives you the exact ratio — fast, without any algebra.

The word "alligation" comes from the Latin word for "binding together." It's a rule that binds two quantities at different values to produce a weighted mean.

Here's the analogy that makes this stick. Imagine two workers filling a tank — one fast, one slow. The combined fill rate is a weighted average of their individual rates. Alligation is just the arithmetic of that blending process, applied to prices, concentrations, or any per-unit measure.

There are three situations you'll see in SSC CGL:

Type 1 — Simple Mixing: Two ingredients mixed in some ratio. Find the resulting quality (price, concentration).

Type 2 — Reverse Mixing: Target quality is given. Find the ratio in which two ingredients should be mixed. This is the classic alligation cross.

Type 3 — Repeated Replacement: A fixed volume is removed and replaced repeatedly. Find remaining quantity after n replacements.

The good news: all three have direct formulas. Once you identify the type, the problem reduces to a 30-second arithmetic exercise. The bad news: SSC CGL papers often disguise the type with profit/loss or percentage language — the skill is recognition, not calculation.

One more thing worth noting before we go deep. Alligation is not limited to liquids. Any time two groups with different per-unit values are merged, alligation applies. Profit percentage, speed, batting average, interest rate — all valid. If you see "two groups combined → average of the whole," reach for alligation.

Deep Dive

The Alligation Cross Rule

This is the core tool. Set it up like this:

Cheaper value (C)          Dearer value (D)
         \                   /
          \                 /
           Mean value (M)
          /                 \
         /                   \
   (D - M)                (M - C)
   [quantity of C]        [quantity of D]

In ratio form:

$\text{Quantity of Cheaper} : \text{Quantity of Dearer} = (D - M) : (M - C)$

The logic is a direct consequence of the weighted average formula. If you mix a units of value C with b units of value D to get mean M:

$M = \frac{aC + bD}{a + b}$

Rearranging: $aM + bM = aC + bD$, so $a(M - C) = b(D - M)$, giving $a : b = (D - M) : (M - C)$.

No need to re-derive this in the exam. Just remember: each ingredient's quantity is proportional to the other ingredient's deviation from the mean.

Important convention: Water, which costs ₹0, is always treated as the "cheaper" ingredient when mixing with any liquid. Never forget to plug in ₹0 for water — it's the most common error.

Weighted Average (Type 1 — Forward Direction)

When the ratio is given and you need the mean value:

$M = \frac{r_1 \times C + r_2 \times D}{r_1 + r_2}$

where $r_1$ and $r_2$ are the quantities (or ratio parts) of the cheaper and dearer ingredients respectively.

This is just the weighted average formula. Apply it directly when ratio is known.

Repeated Replacement Formula

This is the formula SSC loves to test. A container has V litres of a liquid. Every round, x litres are removed and replaced with a different liquid. After n rounds:

$\text{Original liquid remaining} = V \times \left(1 - \frac{x}{V}\right)^n$

Derivation intuition: After the first removal, the fraction remaining is $\frac{V - x}{V}$. On the second removal, you take out the same fraction of what's left — so the fraction remaining compounds as $\left(\frac{V-x}{V}\right)^2$. After n rounds, it's the $n$th power.

The fraction $\frac{V - x}{V}$ is sometimes written as $\frac{V-x}{V}$ or simplified. Work with whatever form makes calculation cleaner.

Common shortcut for the replacement formula: When $x/V$ is a unit fraction (like $1/10$, $1/5$), the computation is fast. When it's not clean, convert to a fraction first.

Selling Price + Profit Percentage Hidden Inside

Some SSC CGL problems don't give you the mean cost directly. Instead, they say "sold at ₹X with Y% profit." You must first back-calculate the mean cost:

$\text{Mean cost} = \frac{\text{Selling Price}}{1 + \frac{\text{Profit \%}}{100}}$

Only then apply alligation. This two-step setup catches a lot of test-takers who rush to apply the cross without extracting the mean cost.

Mixing More Than Two Ingredients

SSC CGL rarely asks for three-ingredient mixtures directly, but they do appear. The approach: fix one pair, compute their resultant, then alligation of that resultant with the third. Or use the direct weighted average with three terms. Either way, it's two applications of the same formula, not a new technique.

Concentration and Volume Problems

When dealing with percentages of a component (e.g., "40% alcohol solution"):

• Treat the percentage as the "value" of each solution
• The "mean value" is the target concentration percentage
• Apply alligation exactly as you would with prices

The rule generalises because both prices and concentrations are per-unit measures. The underlying algebra is identical.

Memory Tricks & Shortcuts

Fast-Solving Framework

When you see a mixture/alligation problem in the exam hall, run this decision tree:

Step 1 — Identify what's being mixed. Is it price? Concentration? Speed? Label the two values as Cheaper (C) and Dearer (D).

Step 2 — Is the ratio given or unknown?

• Ratio given → use weighted average to find mean. Done.
• Ratio unknown, mean given → use alligation cross. Subtract diagonally. Done.

Step 3 — Is there a hidden mean? Check if the problem gives SP + profit% instead of cost directly. If yes, back-calculate CP before touching alligation.

Step 4 — Is there repeated replacement? If a fixed volume is removed and refilled n times, use $V \times \left(\frac{V-x}{V}\right)^n$ directly. No alligation needed.

Step 5 — Verify. Use the reverse weighted average check. Five seconds. Always worth it.

One timing note: a clean alligation cross problem should take under 45 seconds. A replacement problem with clean fractions should take under 60 seconds. If you're past 90 seconds on either, something is wrong — re-read the question for what type it actually is.

Solved PYQs

Why this question: The classic water-milk setup. Every SSC aspirant encounters this, yet many get the direction of the ratio wrong (water:milk vs milk:water). Know the convention cold.

Solving path: Water costs ₹0. Alligation cross: cheaper = ₹0 (water), dearer = ₹12 (milk), mean = ₹8. Quantity of water : quantity of milk = (12 - 8) : (8 - 0) = 4 : 8 = 1 : 2. Answer: water to milk = 1 : 2.

Why this question: The repeated replacement formula. SSC sets this up regularly. The key is recognising it as a powers problem, not an alligation cross problem.

Solving path: Formula: $40 \times \left(1 - \frac\{4\}\{40\}\right)^2 = 40 \times \left(\frac\{9\}\{10\}\right)^2 = 40 \times \frac\{81\}\{100\} = 32.4$ litres.

Why this question: Forward direction — ratio given, find mean. Tests whether you know the weighted average formula and can avoid alligation (which doesn't apply here).

Solving path: Weighted average = $\frac{3 \times 20 + 2 \times 30}{3 + 2} = \frac{60 + 60}\{5\} = \frac\{120\}\{5\} = ₹24$ per kg.

Why this question: The profit-percentage trap. If you alligation on ₹68.20, you get a wrong ratio. The paper is testing whether you remember to extract CP first.

Solving path: CP of mixture = $\frac{68.20}{1.10} = ₹62$. Now alligation: cheaper = ₹60, dearer = ₹65, mean = ₹62. Ratio = (65 - 62) : (62 - 60) = 3 : 2.

Why this question: Removal-and-replacement but with a pre-existing ratio in the container. Tests whether you track individual components through each replacement step.

Solving path: Original: milk = 15L, water = 5L (ratio 3:1 in 20L). Remove 10L (maintaining 3:1 ratio): milk removed = 7.5L, water removed = 2.5L. Remaining: milk = 7.5L, water = 2.5L. Add 10L pure milk: milk = 17.5L, water = 2.5L. Ratio = 17.5 : 2.5 = 7 : 1.

Common Mistakes

• Flipping the cross diagonals. The quantity of the cheaper ingredient comes from (D - M), not (M - C). If you subtract the wrong way, your ratio is inverted. Always write C top-left, D top-right, subtract from D to M for the left quantity, M to C for the right quantity.

• Forgetting water costs ₹0. This sounds obvious, but under time pressure, aspirants set up alligation with water having some positive value they invent. Water is ₹0. Full stop.

• Applying alligation when the ratio is already given. If the problem tells you the mixing ratio, use weighted average. Alligation is for finding the ratio — running it when the ratio is known wastes 30 seconds and often confuses.

• Skipping the SP → CP conversion. When profit percentage is embedded in a mixing problem, the mean value for alligation is the cost price of the mixture, not the selling price. Back-calculate CP before setting up the cross.

• Using the replacement formula with the wrong base. After the first replacement, the base for the second removal is still the original total volume V, not the remaining liquid volume. The formula $V \times \left(\frac{V-x}{V}\right)^n$ handles this correctly — don't re-derive it each time and introduce an error.

• Misreading "ratio of milk to water" vs "ratio of water to milk." The alligation cross gives you cheaper : dearer. If water is cheaper, the cross gives water : milk. If the question asks for milk : water, you must invert. Read the question's exact phrasing in the last 5 seconds before marking.

Related Topics

• ratio and proportion
• percentage
• profit and loss
• simple interest
• time speed distance
• averages

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