A barista blends two coffees to hit a target price. A gasoline station mixes fuel grades. A nurse dilutes a solution to the right strength. A sari-sari store owner stretches pure milk with water. All of these are the same math problem in different clothes: combine two things of different values to land on a target average. There is a beautifully quick method for this — called alligation — that turns the whole thing into simple subtraction. This chapter teaches it thoroughly, along with the closely related "repeated dilution" pattern the exam loves.
Mixture problems describe combining two ingredients at different prices, strengths, or concentrations, and ask for the ratio needed to reach a target average — or how much of each to use for a given total. Alligation is a shortcut that reads the ratio straight off two subtractions.
On the Civil Service Exam, mixture and alligation questions appear in Numerical Reasoning, usually 1 to 3 items, and they build directly on Ratio & Proportion and on weighted Averages. Because the alligation shortcut is so fast, these become easy marks once you've drilled it.
The difficulty is intermediate. The math is light; the exam tests whether you set the alligation ratio the right way around, remember it gives parts (not quantities), treat water as zero value, and handle repeated replacement with a power. Master those four points and mixtures are quick, satisfying points.
After completing this lesson you will be able to:
You should be comfortable with:
A quick refresher on the key intuition: a mixture's value always lands between the values of its two ingredients, and it sits closer to the ingredient you use more of. Mix a lot of cheap coffee with a little expensive coffee and the blend is cheap-ish; mix them equally and the blend is right in the middle. Alligation is just a fast way to read how much of each is needed to hit an exact in-between target. Keep this "between, and closer to the bigger share" picture in mind.
Mixing to a target is everywhere in Filipino life and work:
Alligation is the professional's fast tool for all of these — and the exam rewards you for knowing it.
We build from the alligation rule to the trickier repeated-dilution pattern.
To mix two ingredients with values A (cheaper/weaker) and B (dearer/stronger) to reach a target mean M, the ratio of the amounts is:
Quantity of A : Quantity of B = (B − M) : (M − A).
Read it as "each ingredient's distance from the target, but crossed over." The ingredient farther from the target is needed in the smaller amount, and vice versa.
Mix a 20%-sugar solution with a 40%-sugar solution to get 30% sugar. Ratio = (40 − 30) : (30 − 20) = 10 : 10 = 1 : 1 (equal parts).
Blend coffee at ₱200/kg with coffee at ₱320/kg to get a ₱260/kg blend. Ratio = (320 − 260) : (260 − 200) = 60 : 60 = 1 : 1.
Analogy: Picture a see-saw with the target M as the balance point. Ingredient A sits at distance (M − A) on one side, B at distance (B − M) on the other. To balance, the heavier share goes on the side with the shorter arm. The alligation "cross" is exactly this balancing.
It feels backwards that A's amount uses B's distance. Here's why: to balance the see-saw, the ingredient closer to the target can be used in bulk (it barely pulls the average), while the ingredient far from the target must be used sparingly (a little pulls the average a lot). So each amount is proportional to the other ingredient's distance — hence the cross.
If the target sits nearer one ingredient, the ratio comes out lopsided — more of the closer ingredient.
Mix milk at ₱40/liter with water (₱0/liter) to make a mixture worth ₱30/liter. Ratio (milk : water) = (40 − 30) : (30 − 0)... careful with labels. Taking A = water (0) and B = milk (40), target 30: ratio A : B = (40 − 30) : (30 − 0) = 10 : 30 = 1 : 3, i.e. water : milk = 1 : 3 — three parts milk to one part water, since ₱30 is close to milk's ₱40.
Water (or any free ingredient) has value 0. Plugging 0 into the rule handles all "dilute with water" problems.
Alligation gives parts, not final amounts. To get real quantities, sum the parts and scale to the total (the same shares method from Ratio & Proportion).
Using milk : water = 3 : 1, how much of each makes 20 liters of mixture? Total parts = 3 + 1 = 4; one part = 20 ÷ 4 = 5 L. Milk = 3 × 5 = 15 L; water = 1 × 5 = 5 L.
A common variant asks how much water to add (or evaporate) to change a concentration. The trick: the amount of the solute (salt, sugar, acid, pure milk) doesn't change when you add or remove water — so find that fixed amount, then solve for the new total.
40 liters of a 20%-acid solution — how much water must be added to dilute it to 15% acid? Acid = 0.20 × 40 = 8 liters (this stays constant). We need 8 to be 15% of the new volume: new volume = 8 ÷ 0.15 ≈ 53.33 L. Water to add = 53.33 − 40 ≈ 13.33 liters.
A favorite exam pattern: some of a mixture is removed and replaced with a pure substance (usually water), repeatedly. After each replacement, the remaining amount of the original substance follows a power rule:
Remaining original = Original amount × (1 − replaced fraction)^(number of replacements).
A 10-liter container of pure milk: 2 liters are removed and replaced with water, done 3 times. How much pure milk remains? Replaced fraction each time = 2 ÷ 10 = 1/5. Remaining milk = 10 × (1 − 1/5)³ = 10 × (4/5)³ = 10 × 64/125 = 5.12 liters.
The power (the number of replacements) is the crucial part — using the fraction only once, instead of cubing it here, is the classic error.
Everything here is a rearranged weighted-average equation. Alligation is simply faster to compute under time pressure. If you ever forget the cross, you can always fall back on the weighted-average setup: (amount of A × value of A + amount of B × value of B) ÷ total amount = target. Both give the same answer.
With the toolkit built, let's picture, tabulate, and drill.
| Concept | Formula |
|---|---|
| Alligation ratio | A : B = (B − M) : (M − A) |
| Water / free ingredient | value = 0 |
| Parts → quantity | one part = total ÷ (sum of parts) |
| Keep solute constant | solute amount is unchanged by adding/removing water |
| Repeated dilution | remaining = original × (1 − f)^n |
| Weighted-average check | (aA + bB) ÷ (a + b) = M |
Two reminders. (1) The alligation ratio is A : B = (B − M) : (M − A) — each ingredient paired with the other's distance (the cross). (2) It gives parts; scale to the real total. And for repeated dilution, the exponent is the number of times the replacement happens.
| When the problem says… | Do this… |
|---|---|
| "in what ratio to mix… to get an average/target" | alligation: (B − M) : (M − A) |
| "mix with water / a free ingredient" | set that value to 0 |
| "how much of each to make N liters/kg" | find the ratio, then scale to N |
| "how much water to add to change the strength" | keep the solute constant, solve for new total |
| "removed and replaced, repeated k times" | remaining = original × (1 − f)^k |
| "blend to sell at a certain price/profit" | alligation with prices |
Step 1 — Identify the two values (A, B) and the target M. Convert percentages to numbers; set water/free ingredients to 0. ↓ Step 2 — Apply the alligation cross: ratio A : B = (B − M) : (M − A). Simplify. ↓ Step 3 — If a total quantity is asked, scale the parts to that total. ↓ Step 4 — For "add/remove" problems, hold the solute constant and solve for the new total. For "repeated replacement," use (1 − f)^n. ↓ Step 5 — Sanity-check: the target must lie between A and B, and the larger share goes to the ingredient nearer the target.
Why Step 2's direction matters most: setting the ratio backward (straight instead of crossed) is the defining mistake. Remember each amount pairs with the other ingredient's distance from the target.
Example 1. In what ratio should a 10%-salt solution be mixed with a 30%-salt solution to get 25% salt? Solution: (30 − 25) : (25 − 10) = 5 : 15 = 1 : 3 (weak : strong). Check: the target 25 is closer to 30, so more of the strong solution — matches 1 : 3. ✓ Difficulty: ★★☆☆☆
Example 2. Rice at ₱45/kg and rice at ₱55/kg are blended to sell at ₱48/kg. Find the mixing ratio. Solution: (55 − 48) : (48 − 45) = 7 : 3 (cheap : dear). Difficulty: ★★☆☆☆
Example 3 (water, value 0). In what ratio must water be mixed with milk costing ₱24/liter to sell the mixture at ₱20/liter? Thinking: Water = 0; target 20. Solution: water : milk = (24 − 20) : (20 − 0) = 4 : 20 = 1 : 5. Difficulty: ★★★☆☆
Example 4 (scale to quantity). Using the 1 : 3 ratio from Example 1, how much of each solution makes 24 liters of the 25% mixture? Solution: Parts = 1 + 3 = 4; one part = 24 ÷ 4 = 6 L. Weak = 6 L; strong = 18 L. Difficulty: ★★★☆☆
Example 5 (blend price). A grocer mixes 8 kg of ₱60/kg coffee with 12 kg of ₱90/kg coffee. What is the blend's price per kg? Thinking: This gives amounts, so use the weighted average. Solution: (8 × 60 + 12 × 90) ÷ 20 = (480 + 1,080) ÷ 20 = 1,560 ÷ 20 = ₱78/kg. Difficulty: ★★★☆☆
Example 6 (how much water to add). A 50-liter solution is 12% acid. How much water must be added to dilute it to 10% acid? Thinking: Acid stays constant. Solution: Acid = 0.12 × 50 = 6 L. Need 6 = 10% of new volume → new volume = 6 ÷ 0.10 = 60 L. Water to add = 60 − 50 = 10 liters. Difficulty: ★★★★☆
Example 7 (evaporate water). 40 liters of a 15%-salt solution — how much water must evaporate to make it 20% salt? Thinking: Salt stays constant. Solution: Salt = 0.15 × 40 = 6 L. Need 6 = 20% of new volume → new volume = 6 ÷ 0.20 = 30 L. Water evaporated = 40 − 30 = 10 liters. Difficulty: ★★★★☆
Example 8 (repeated dilution). A 20-liter can of pure milk has 4 liters removed and replaced with water, done twice. How much pure milk remains? Solution: Replaced fraction = 4 ÷ 20 = 1/5. Remaining = 20 × (4/5)² = 20 × 16/25 = 12.8 liters. Difficulty: ★★★★☆
Example 9 (profit via alligation). A trader wants an average cost of ₱52/kg by mixing tea at ₱40/kg and ₱70/kg. In what ratio should he mix them? Solution: (70 − 52) : (52 − 40) = 18 : 12 = 3 : 2 (cheap : dear). Difficulty: ★★★★☆
Example 10 (find the total from parts). Water is mixed with 15 liters of pure juice so the mixture is 75% juice. How much water is added? Thinking: Juice (15 L) is the solute and stays constant; it must be 75% of the new total. Solution: New total = 15 ÷ 0.75 = 20 L. Water added = 20 − 15 = 5 liters. Difficulty: ★★★★☆
Example 11 (three replacements). A 25-liter tank of pure milk has 5 liters replaced with water three times. How much pure milk remains, to two decimals? Solution: Fraction = 5 ÷ 25 = 1/5. Remaining = 25 × (4/5)³ = 25 × 64/125 = 12.8 liters. Difficulty: ★★★★★
Example 12 (reverse: find the mix from a known blend). A ₱78/kg blend was made from ₱60/kg and ₱90/kg coffee. In what ratio were they mixed? Solution: (90 − 78) : (78 − 60) = 12 : 18 = 2 : 3 (cheap : dear). (Consistent with Example 5, where 8 : 12 = 2 : 3.) Difficulty: ★★★★☆
Mixture and alligation problems combine two things of different value to hit a target average. The alligation rule reads the mixing ratio straight off two subtractions: A : B = (B − M) : (M − A) — each ingredient paired with the other's distance from the target (the cross). Set water (or any free ingredient) to 0, remember the ratio gives parts you must scale to the real total, and for "add or evaporate water" problems, keep the solute constant and solve for the new total. For repeated replacement, the original substance shrinks by (1 − replaced fraction) raised to the number of replacements. It's all weighted averages underneath, just faster. Cross correctly, treat water as zero, scale the parts, and power up the repeats — and mixtures become quick, dependable marks.
| Item | Key point |
|---|---|
| Alligation | A : B = (B − M) : (M − A) (cross) |
| Water / free item | value = 0 |
| Parts → quantity | one part = total ÷ sum of parts |
| Add/remove water | solute stays constant |
| Repeated replacement | original × (1 − f)^n |
| Check | target between A and B, nearer the bigger share |
Which way does the alligation ratio go? The amount of the cheaper/weaker ingredient A is proportional to (B − M), and the dearer/stronger B is proportional to (M − A) — each takes the other's distance from the target. Drawing the cross keeps it straight.
Why is water's value zero? Water (or any free filler) contributes no price or strength, so it enters the formula as 0. That turns "dilute with water" into a simple alligation.
Does alligation give me the actual amounts? No — it gives a ratio (parts). To get real quantities, sum the parts and scale to the total the question asks for.
How do I handle "add water to change the concentration"? Keep the dissolved substance (solute) constant, since water doesn't change its amount. Find the solute, then divide by the new percentage to get the new total, and subtract to find the water added.
What changes when replacement happens several times? Each replacement multiplies the remaining original substance by (1 − replaced fraction). For n replacements, raise that factor to the nth power — cube it for three times, square it for two.
Tick them all and mixtures and alligation become fast, reliable points — and you'll blend, dilute, and price like a pro in real life.
Put it to the test with 1,338 practice questions on this topic.