Ratio, Proportion and Mixtures

Overview

Ratio, Proportion and Mixtures appears about 1.8 times per paper across the last four AFCAT solved papers, placing it in the high yield band of Numerical Ability.

Ratio, proportion and mixture problems together account for roughly 1.75 marks per AFCAT paper across the last four solved sittings. That puts the topic in the second tier of Numerical Ability — smaller than percentages and time-speed-distance, but larger than probability or permutation. Because the question pool is small and the methods are predictable, this is one of the highest-return chapters in the entire paper for a candidate who drills the right shortcuts.

Almost every item in this chapter reduces to one of four tools. The first is the chained-ratio rule, which combines two ratios that share a middle term. The second is the alligation cross, which turns any two-component mixing problem — milk and water, two grades of rice, two interest rates, two driving speeds — into a single subtraction. The third is the partnership formula, which says profit shares track the product of investment and time. The fourth is the successive-replacement formula, which gives the remaining quantity after repeated draw-and-refill cycles in closed form.

This page builds the full toolkit. You get the ratio identities, the proportion rules, a step-by-step method for chaining ratios, the alligation derivation with its cross-diagram visual, the partnership ratio table, the successive-replacement formula with two worked applications, twelve worked examples spanning every standard pattern, and a list of AFCAT trap wordings to memorise before exam day.

Why ratio and mixtures is a high-yield AFCAT topic

Across the last four solved AFCAT papers, ratio-proportion-and-mixture items averaged about 1.75 per paper. That sits behind percentages (~3.0) and time-speed-distance (~2.5), but the per-item return is among the best in Numerical Ability. Three reasons.

• The methods are short. Alligation collapses a three-equation system into a single cross subtraction. Chained ratios reduce to one scaling step. Successive replacement is a one-formula plug-in.
• The patterns recycle. AFCAT keeps returning to milk-and-water alligation, two-grade rice mixing, two-investment partnership and ratio-of-ages. Once you have done ten of each variety, the eleventh is recognised in under thirty seconds.
• The arithmetic is clean. Question setters pick numbers that simplify — ratios such as 2:3, 3:5, 4:5, 5:7. Decimal grind is rare. Speed gains come almost entirely from method, not from calculation power.

For a paper worth 300 marks scored at +3/−1, two correct items in this topic produce six marks of gain. Two attempted-wrong items produce two marks of loss. The expected value of attempting any item where you know the method is therefore strongly positive. That is the bet this page is built to win.

Ratio basics — notation, equivalence, simplification

A ratio compares two quantities of the same kind. The notation a : b is read ‘a to b’ and is equivalent to the fraction a/b. The first quantity is the antecedent, the second is the consequent.

Simplest form. A ratio is in simplest form when the highest common factor of the two parts is 1. 18 : 24 reduces to 3 : 4 after dividing both parts by their HCF of 6. Always simplify before chaining or comparing.

Equivalent ratios. Two ratios are equivalent when they reduce to the same simplest form. 6 : 8, 9 : 12 and 15 : 20 are all equivalent to 3 : 4. AFCAT will sometimes ask you to spot the equivalent ratio from a list; reduce each option and pick the match.

Proportion — direct, inverse and the rule of three

A proportion is a statement that two ratios are equal: a : b = c : d, written also as a/b = c/d. Cross-multiplying gives ad = bc, the fundamental proportion identity. The four quantities are called the first, second, third and fourth proportionals; a and d are the extremes, b and c are the means.

• Direct proportion. Two quantities are in direct proportion when an increase in one causes a proportional increase in the other. Cost is in direct proportion to quantity bought; distance is in direct proportion to time at constant speed.
• Inverse proportion. Two quantities are in inverse proportion when an increase in one causes a proportional decrease in the other. Speed and time are inversely proportional at constant distance; number of workers and time taken are inversely proportional at constant work.
• Rule of three. Given three quantities and asked for the fourth in a proportion, write the known proportion, cross-multiply, and divide. If 5 kg of rice costs ₹250, the cost of 12 kg is found from 5 : 250 :: 12 : x → x = 250 × 12 / 5 = ₹600.

Mean proportional. The mean proportional between a and b is √(ab). The mean proportional between 4 and 25 is √100 = 10. Third proportional. The third proportional to a and b is b²/a. The third proportional to 4 and 12 is 144/4 = 36.

Chained ratios — combining a:b and b:c via the middle term

The single most common ratio-arithmetic task on AFCAT is combining two ratios into one. The method has three steps.

1. Identify the common term. If you are given a : b and b : c, the common term is b. If you are given a : b and c : d with no common term, you cannot chain them — recheck the question.
2. Scale each ratio so the common term has the same numerical value. Take the LCM of the two values of b. Multiply the first ratio so the b-value equals the LCM. Multiply the second ratio so its b-value also equals the LCM.
3. Read off the chained ratio. Once both b-values match, the chained ratio is (scaled a) : (matched b) : (scaled c).

Worked walk-through. Given a : b = 2 : 3 and b : c = 4 : 5, find a : b : c.

• b-values are 3 and 4. LCM = 12.
• Scale a : b = 2 : 3 by 4 → 8 : 12. Scale b : c = 4 : 5 by 3 → 12 : 15.
• Chained: a : b : c = 8 : 12 : 15.

Three-link chain. Given a : b = 2 : 3, b : c = 4 : 5 and c : d = 6 : 7, chain in stages. First combine a : b : c using the method above to get 8 : 12 : 15. Then match the c-values (15 and 6) at their LCM 30. Scale 8 : 12 : 15 by 2 → 16 : 24 : 30. Scale c : d = 6 : 7 by 5 → 30 : 35. Final: a : b : c : d = 16 : 24 : 30 : 35.

Adding to all parts versus multiplying — what actually changes

AFCAT loves the wording ‘Each member receives an extra ₹50’ or ‘Five workers leave and the ratio changes to …’. The trap is to assume the ratio is preserved. It is not.

• Multiplying every part by the same non-zero number preserves the ratio. 2 : 3 multiplied by 5 gives 10 : 15, which reduces back to 2 : 3.
• Adding the same number to every part changes the ratio. 2 : 3 plus 1 gives 3 : 4, a different ratio.
• Subtracting the same number from every part changes the ratio. 5 : 6 minus 1 gives 4 : 5, again different.

Worked. ‘Two numbers are in the ratio 5 : 7. If 4 is added to each, the new ratio becomes 3 : 4. Find the original numbers.’

1. Let the original numbers be 5x and 7x.
2. After adding 4: (5x + 4) / (7x + 4) = 3 / 4.
3. Cross-multiply: 4(5x + 4) = 3(7x + 4) → 20x + 16 = 21x + 12 → x = 4.
4. Original numbers: 20 and 28.

This is the canonical setup. Whenever the problem describes a ratio, then a uniform addition or subtraction, then a new ratio — let the parts be ax and bx, write the algebraic equation, solve.

Partition problems — splitting N in a given ratio

To split a quantity N into parts in the ratio a : b : c, the parts are (a/(a+b+c)) × N, (b/(a+b+c)) × N, (c/(a+b+c)) × N. The sum of the fractional weights is 1, which guarantees the parts add back to N.

Worked. Split ₹1800 between three children in the ratio 2 : 3 : 4. Sum of parts = 9. Each unit = 200. Shares are 400, 600, 800.

Worked with a twist. ‘₹6300 is divided among A, B and C such that A gets half of what B gets, and B gets two-thirds of what C gets. Find each share.’

1. Let C = 6 units. Then B = (2/3) × 6 = 4 units. A = (1/2) × 4 = 2 units.
2. Ratio A : B : C = 2 : 4 : 6 = 1 : 2 : 3. Sum = 6.
3. One unit = 6300 / 6 = 1050. Shares: A = 1050, B = 2100, C = 3150.

The trick is to pick the unit that produces the cleanest whole numbers for every named part. Start with the most-divided variable — here, C — and work backwards.

Age problems via ratios — present, past, future

Age problems on AFCAT almost always reduce to a ratio equation across two time points. The standard setup uses one variable for the common multiplier.

Worked. ‘The present ages of A and B are in the ratio 5 : 7. Four years ago the ratio was 3 : 5. Find their present ages.’

1. Let present ages be 5x and 7x.
2. Four years ago: (5x − 4) / (7x − 4) = 3 / 5.
3. Cross-multiply: 5(5x − 4) = 3(7x − 4) → 25x − 20 = 21x − 12 → 4x = 8 → x = 2.
4. Present ages: A = 10, B = 14.

Future-tense variant. ‘Six years hence the ratio of ages of A and B will be 7 : 9, and twelve years ago the ratio was 1 : 3. Find their present ages.’ Let present ages be A and B. Then (A + 6) / (B + 6) = 7 / 9 and (A − 12) / (B − 12) = 1 / 3. Two equations, two unknowns. Solve to get A = 22, B = 30.

Sum-given variant. ‘The sum of the present ages of a father and son is 60. Six years ago the father was five times as old as the son. Find present ages.’ Let father = F, son = S. F + S = 60 and F − 6 = 5(S − 6) → F − 6 = 5S − 30 → F = 5S − 24. Substitute: 5S − 24 + S = 60 → 6S = 84 → S = 14, F = 46.

Partnership and investment — profit shares track (investment × time)

In a partnership, each partner contributes capital for a period of time. The profit at the end is divided in the ratio of each partner’s (capital × time) product.

Worked. ‘A invests ₹12,000 for 8 months. B invests ₹9,000 for 12 months. Profit at year-end is ₹6,300. Find each share.’

1. Ratio of (C × T): A = 12000 × 8 = 96000. B = 9000 × 12 = 108000.
2. Ratio = 96 : 108 = 8 : 9. Sum = 17.
3. A’s share = 6300 × 8 / 17 = ₹2964.71. B’s share = 6300 × 9 / 17 = ₹3335.29.

Worked with changing capital. ‘A invests ₹10,000 for 12 months. After 6 months he adds ₹2,000 more. B invests ₹15,000 for 9 months. Find profit ratio.’ A’s contribution = 10000 × 6 + 12000 × 6 = 60000 + 72000 = 132000. B’s contribution = 15000 × 9 = 135000. Ratio = 132 : 135 = 44 : 45.

Alligation rule — derivation, cross diagram, applications

Alligation is a one-step shortcut for any two-component mixture problem in which you know two component values and the mean of the mixture. It applies to mixing two grades of rice, mixing milk and water, mixing two solutions of different strength, blending two interest rates, averaging two driving speeds — anywhere a weighted average is in play.

Statement. Two ingredients are mixed. Ingredient one has value (price, strength, rate) a, ingredient two has value b, and the mixture has mean value m, with a < m < b. Then the ratio of quantities mixed is:

Quantity of cheaper : Quantity of dearer = (b − m) : (m − a)

Derivation. Let q₁ units of ingredient one and q₂ units of ingredient two be mixed. Total value = q₁ × a + q₂ × b. Total quantity = q₁ + q₂. Mean value m = (q₁a + q₂b) / (q₁ + q₂). Cross-multiply: m(q₁ + q₂) = q₁a + q₂b → q₁(m − a) = q₂(b − m) → q₁ / q₂ = (b − m) / (m − a). That is the alligation rule.

Cross diagram. Write the cheaper value top-left, the dearer value top-right, the mean in the middle, and take the cross differences:

Read the bottom row as ratio: quantity of cheaper : quantity of dearer = (b − m) : (m − a).

Two-component mixture problems — milk-water, dal-of-two-grades

The classic AFCAT two-component mixture problem fixes two ingredient values and asks for the mixing ratio that produces a stated mean.

Worked. ‘In what ratio must water be added to milk worth ₹24 per litre so that the resulting mixture is worth ₹20 per litre?’

1. Water has value 0 per litre (treat as the cheaper ingredient). Milk has value 24. Mean = 20.
2. Cross differences: (24 − 20) and (20 − 0) → 4 and 20.
3. Ratio water : milk = 4 : 20 = 1 : 5.

Worked with a target quantity. ‘Rice at ₹30 per kg is mixed with rice at ₹45 per kg to obtain 60 kg of mixture costing ₹36 per kg. How much of each is used?’

1. Cross differences: (45 − 36) and (36 − 30) → 9 and 6.
2. Ratio cheaper : dearer = 9 : 6 = 3 : 2. Sum = 5.
3. Cheaper (₹30 rice) = 60 × 3 / 5 = 36 kg. Dearer (₹45 rice) = 60 × 2 / 5 = 24 kg.

Worked with strength. ‘A 60-litre solution contains milk and water in the ratio 7 : 3. How much water must be added to make the ratio 7 : 5?’

1. Milk in original = 60 × 7/10 = 42 L. Water = 18 L.
2. Milk stays 42 L. New ratio demands water : milk = 5 : 7, so water = 42 × 5 / 7 = 30 L.
3. Water to add = 30 − 18 = 12 L.

Successive replacement — the (1 − x/V)ⁿ rule

A vessel contains V litres of a pure liquid (milk). At each step, x litres of the contents are drawn off and replaced with water. The process is repeated n times. The amount of milk remaining after n operations is:

Milk after n operations = V × (1 − x/V)ⁿ

Derivation. After the first draw-and-refill, the vessel still contains V litres of liquid, of which milk is V − x = V(1 − x/V). The fraction of milk in the vessel is therefore (1 − x/V). At the second draw, the x litres drawn off contain milk in the same fraction; what remains is V × (1 − x/V) × (1 − x/V) = V × (1 − x/V)². Induct over n.

Water present after n operations. Water = V − milk = V × [1 − (1 − x/V)ⁿ].

Application 1. ‘A drum has 50 L of milk. 5 L is drawn off and replaced with water; the process is repeated three times. Find the milk left.’ Milk = 50 × (1 − 5/50)³ = 50 × (0.9)³ = 50 × 0.729 = 36.45 L.

Application 2. ‘A cask contains 64 L of milk. 8 L is drawn off and replaced with water; repeated twice. Find the ratio of milk to water in the final mixture.’ Milk = 64 × (1 − 8/64)² = 64 × (7/8)² = 64 × 49/64 = 49 L. Water = 64 − 49 = 15 L. Ratio = 49 : 15.

Reverse-direction question. Sometimes the question gives the final milk-to-water ratio and asks for n. Set up V × (1 − x/V)ⁿ equal to the milk portion of the final ratio scaled to V litres, take logs (or test n = 2, 3, 4 mentally). For AFCAT, n is almost always 2 or 3 — guess-and-check is fastest.

Mean price method — when alligation will not fit

When three or more components are being mixed, alligation does not apply directly. Use the mean-price method instead.

Method. Mean price of mixture = (Total cost of all components) / (Total quantity of all components). Solve for the missing variable from the resulting equation.

Worked. ‘A trader mixes 20 kg of rice at ₹30 per kg, 25 kg at ₹40 per kg and 15 kg at ₹50 per kg. Find the mean price per kg of the mixture.’

1. Total cost = 20 × 30 + 25 × 40 + 15 × 50 = 600 + 1000 + 750 = ₹2350.
2. Total quantity = 20 + 25 + 15 = 60 kg.
3. Mean price = 2350 / 60 = ₹39.17 per kg.

Worked — unknown quantity. ‘How many kilograms of rice at ₹35 per kg must be mixed with 50 kg of rice at ₹50 per kg so that the mixture is worth ₹40 per kg?’

1. Let x kg of cheaper rice be added.
2. (35x + 50 × 50) / (x + 50) = 40.
3. 35x + 2500 = 40x + 2000 → 5x = 500 → x = 100 kg.

Both alligation and mean-price give the same answer for two-component problems. Alligation is faster; mean-price scales to any number of components. Pick the right tool by counting how many components are in the mix.

AFCAT trap patterns to memorise

• Adding a constant to all parts. The ratio changes. Multiplying preserves the ratio; adding does not. Always set up the algebra rather than treating addition as scaling.
• Alligation denominator order. Cheaper : dearer = (b − m) : (m − a). The first term is the difference between the dearer value and the mean, sitting on the cheaper side of the cross. Reverse it and your answer flips.
• False ‘free’ water in milk problems. Water has value 0, not ‘missing’. Plug 0 into the alligation cross. AFCAT will sometimes test this by phrasing as ‘milk diluted with water’ rather than ‘milk mixed with water of zero cost’.
• Partnership with mid-year capital change. Each leg of the year is a separate (capital × time) term. Sum the legs before forming the ratio with the other partner.
• Successive replacement — wrong n. n is the number of draw-and-refill cycles, not the number of litres drawn. ‘Repeated three times’ means n = 3 cycles total, not n = 2.
• Conserved ingredient overlooked. When pure water is added to a milk-water mix, only the water amount changes. Set the unchanged quantity as the anchor and solve.
• Three-way ratio split with non-uniform descriptions. ‘A gets half of B; B gets two-thirds of C.’ Start with the most-divided variable and back-compute. Do not try to write a single equation in three unknowns.
• Age problem trap on direction of time. ‘Four years ago’ subtracts from current age. ‘Four years hence’ adds. AFCAT will mix both phrases in one stem.

Time budget per item

AFCAT averages 72 seconds per question across the 100-question, 120-minute paper. For ratio-and-mixture items, target the following bands.

If any single item passes 90 seconds, mark it for review and move on. The opportunity cost of one stuck item is two faster items lost later in the paper. With ratio-and-mixture in particular, the method should announce itself within the first ten seconds of reading — if it has not, the chance that grinding will rescue the item is low.

Worked AFCAT-style examples

Exam-day strategy

1. For chained ratios, identify the common middle term first and scale to the LCM. If the two middle values are coprime, the scaling factors are simply the values themselves.
2. Default to alligation for any two-component mixing problem — milk-water, rice grades, interest rates, speeds, solution strengths. It collapses three lines of algebra into one cross subtraction.
3. When water (cost 0) or a pure ingredient (strength 100%) is one component, plug it straight into the alligation cross. Do not waste time on a separate setup.
4. For successive replacement, memorise V × (1 − x/V)ⁿ. n is the number of cycles, not the number of litres drawn.
5. Partnership share = ratio of (capital × time). For mid-year capital changes, split each partner’s contribution into legs and sum before forming the ratio.
6. On age problems, let present ages be ax and bx (the ratio multiplier), then apply the second-time-point ratio to find x.
7. When a quantity is added uniformly to all parts of a ratio, the ratio changes — always set up the algebra rather than rescaling.
8. Aim for 45–60 seconds per ratio item and 60–75 seconds per mixture item. Past 90 seconds, mark and move.

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