Unit 4: Ratio, Proportion, Alligation and Mixture

I. Orientation

Ratio, proportion, alligation and mixture problems compare quantities, relate corresponding values, or determine how different components combine to produce a required result. Their governing principle is equivalence through comparison: a ratio compares quantities, a proportion equates two ratios, alligation determines the mixing relationship between unlike-priced or unlike-strength materials, and replacement tracks the changing composition of a mixture.

• Defining properties: A ratio must compare quantities expressed in the same unit; (a:b) means (a/b).
• Order convention: In (a:b), (a) is the antecedent and (b) is the consequent; reversing them changes the meaning.
• Proportionality: If (a:b=c:d), then (ad=bc), provided the quantities are appropriately comparable.
• Common scale: Ratios remain unchanged when both terms are multiplied or divided by the same non-zero number.
• Mixture assumption: Unless stated otherwise, components are completely mixed and volume or quantity is additive.
• Rate convention: Partnership shares depend on both invested capital and the duration for which it remains invested.
• Alligation condition: The required mean value must lie between the values of the two components being mixed.

II. Ratio and Proportion — Fundamental comparison tools

A. Concept of ratio and proportion

The concept of ratio and proportion provides the algebraic foundation for comparing quantities and solving unknown relationships.

• Ratio definition: The ratio of (a) to (b) is written (a:b) or (a/b), where (b\neq0). For example, (12\text{ kg}:18\text{ kg}=2:3).
• Simplification: Divide both terms by their greatest common divisor. Thus, (45:60=3:4).
• Equivalent ratios: Multiplying both terms by the same number preserves the ratio: (3:5=12:20).
• Compound ratio: The compound ratio of (a:b) and (c:d) is (ac:bd). For (2:3) and (4:5), it is (8:15).
• Proportion definition: Four quantities are proportional when (a:b=c:d), also written (a:b::c:d).
• Cross multiplication: In (a/b=c/d), (ad=bc). If (x:7=9:21), then (21x=63), so (x=3).
• Direct proportion: If (x) increases in the same ratio as (y), then (x/y=k). For 5 notebooks costing ₹100, 8 notebooks cost (100(8/5)=₹160).
• Inverse proportion: If one quantity increases while the other decreases proportionally, then (xy=k). For a fixed task, workers and days satisfy (w_1d_1=w_2d_2).
• Part-to-whole conversion: If quantities are in the ratio (2:3:5), total parts are (10); their shares are (2/10,3/10,) and (5/10) of the total.

B. Applications and limitations

Ratio methods are reliable only when the quantities and conditions being compared are correctly identified.

• Unit consistency: Convert (2\text{ m}:50\text{ cm}) to (200:50=4:1), not (2:50).
• Difference information: If two numbers are in ratio (5:7) and differ by 18, their difference is (2) parts; one part is (9), giving (45) and (63).
• Fractional comparison: If (A:B=3:4), then (A/(A+B)=3/7), not (3/4).
• Limitation: Direct and inverse proportion must not be confused; cost may be directly proportional to quantity, while time for a fixed task is inversely proportional to workers.

III. Problems based on ages — Translating time into ratios

Age problems apply ratio and linear-equation principles to quantities that change by equal amounts over time.

A. Problems based on ages

The central rule is that every person’s age changes by the same number of years when the same time interval passes.

• Present-age representation: If the present ages of two people are in ratio (a:b), write them as (ax) and (bx), where (x) is one year-unit.
• Age difference: The difference remains constant. If ages are (5x) and (3x), their difference is (2x) both now and in the past or future.
• Future ages: After (t) years, ages become (ax+t) and (bx+t).
• Past ages: (t) years ago, ages were (ax-t) and (bx-t).
• Ratio equation: A statement such as “after 6 years, the ratio is (4:5)” becomes
  [
  \frac{ax+6}{bx+6}=\frac45.
  ]
• Worked example: The present ages of A and B are in ratio (3:5), and their difference is 12 years. Let ages be (3x,5x). Then (2x=12), so (x=6); their ages are 18 and 30.
• Parent-child condition: If a father is four times his son’s age and their age difference is 30, write ages (4x,x). Then (3x=30), giving 40 and 10.
• Verification: Substitute the calculated ages into both the ratio and the stated time condition; this detects sign errors involving “ago” and “after.”

IV. Problems based on partnership — Sharing profit fairly

Partnership problems distribute profit according to the value and duration of each partner’s investment.

A. Problems based on partnership

The governing formula is that a partner’s profit share is proportional to capital multiplied by time.

• Profit-share ratio: For partners A and B,
  [
  \text{A's share}:\text{B's share}=C_A T_A:C_B T_B,
  ]
  where (C) is capital and (T) is investment duration.
• Equal duration: If A invests ₹40,000 and B ₹60,000 for the same period, their profit ratio is (40,000:60,000=2:3).
• Unequal duration: A invests ₹50,000 for 12 months and B ₹80,000 for 9 months. Their capital-time products are (600,000) and (720,000), so the ratio is (5:6).
• Worked example: A invests ₹30,000 for 12 months; B joins with ₹40,000 after 3 months and remains for 9 months. Products are (30,000\times12=360,000) and (40,000\times9=360,000), so profit is shared equally.
• Withdrawals: If capital changes, divide the investment period into intervals. A’s contribution is the sum of each capital amount multiplied by the months it was maintained.
• Working partner salary: If a partner receives a fixed salary or commission before profit division, subtract that amount first; divide the remaining profit in the capital-time ratio.
• Losses: Unless a different agreement is given, losses are distributed using the same ratio as profits.

V. Alligation and Mixture — Combining quantities and values

Alligation is a shortcut for finding the ratio in which two components of different values or concentrations must be mixed to obtain a desired mean value.

A. Concept-based questions on alligation and mixtures

Alligation depends on balancing the excess of the higher component against the deficit of the lower component.

• Terms: Let (L) be the lower value, (H) the higher value, and (M) the desired mean, with (L<M<H).
• Alligation rule:
  [
  \text{Quantity of lower}:\text{quantity of higher}
  =(H-M):(M-L).
  ]
• Reason: Each unit of lower material is short by (M-L), while each unit of higher material exceeds the mean by (H-M); total shortage must equal total excess.
• Value example: Rice costing ₹40/kg and ₹60/kg is mixed to obtain rice costing ₹48/kg:
  [
  \text{cheaper}:\text{costlier}=(60-48):(48-40)=12:8=3:2.
  ]
• Weighted-average formula: For quantities (q_1,q_2) and values (v_1,v_2),
  [
  M=\frac{q_1v_1+q_2v_2}{q_1+q_2}.
  ]
  Here (M) is the average value per unit.
• Concentration mixtures: For water and alcohol, use the amount of pure alcohol. If (x) litres of 30% solution are mixed with (y) litres of 50% solution, pure alcohol is (0.30x+0.50y).
• Percentage interpretation: A 20% acid solution contains (20) units of acid in every (100) units of solution; the remaining (80) units are solvent.
• Feasibility: A mean of ₹70/kg cannot be formed only from materials costing ₹40/kg and ₹60/kg; the target must lie between component values.
• Mixture quantity: Once the ratio is known, multiply its parts by the required total. A (3:2) ratio for 25 kg gives 15 kg and 10 kg.

B. Applications and limitations

Mixture equations are necessary when more than two components, repeated additions, or changing quantities make simple alligation insufficient.

• Two-component use: Alligation directly handles two known values and one target mean.
• Several components: Use the weighted-average equation:
  [
  \text{Mean}=\frac{\sum q_iv_i}{\sum q_i},
  ]
  where (q_i) and (v_i) denote the quantity and value of component (i).
• Profit and selling price: If a mixture costs ₹50/kg and is sold at ₹60/kg, profit per kilogram is ₹10; the profit percentage is (10/50\times100=20\%).
• Limitation: Do not average percentages without weighting. Equal volumes of 20% and 40% solution give 30%, but unequal volumes require the weighted formula.

VI. Replacement-based questions — Tracking changing composition

Replacement problems examine a mixture from which a quantity is removed and replaced repeatedly, causing the original component to decrease geometrically.

A. Replacement-based questions

The standard replacement model assumes that each removal takes out the same fraction of every component because the mixture is uniform.

• Basic notation: Let (V) be the original volume, (r) the volume removed and replaced each time, and (n) the number of operations.
• Remaining fraction: After one operation, the fraction of original liquid left is (1-r/V). After (n) operations:
  [
  \text{original fraction remaining}=\left(1-\frac rV\right)^n.
  ]
• Original quantity remaining: If the vessel initially contains (V) litres of a component, then
  [
  V\left(1-\frac rV\right)^n
  ]
  litres of that original component remain.
• Worked example: A 20-litre vessel is full of milk. Four litres are removed and replaced with water twice. Milk remaining is
  [
  20\left(1-\frac4{20}\right)^2
  =20\left(\frac45\right)^2=12.8\text{ litres}.
  ]
  Water present is (20-12.8=7.2) litres.
• Repeated replacement: The exponent (n) counts complete removal-and-replacement operations, not merely the number of times the vessel is disturbed.
• Different replacement liquid: If milk is removed and water is added, the original milk follows the geometric formula; the added water is the complement of the remaining milk when the vessel stays full.
• Uniformity assumption: Stirring or complete mixing is essential; otherwise the removed portion may not have the same composition as the vessel.
• Changing vessel volume: If the vessel is not restored to the same total volume after each operation, the standard formula does not apply directly; calculate each stage separately.
• Equivalent percentage form: Removing (r/V) of the mixture removes that same fraction of the original component. Replacing 25% each time leaves 75% after one operation and (75^n\%) after (n) operations.