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Quantitative Aptitude

Mixture & Alligation

Profit Loss & Discount9 / 51Percentage
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Concepts (2)

Alligation Rule finds the ratio to mix two ingredients of different costs to achieve a desired mean cost, using a simple cross method for quick calculation of quantities.

Core Formula

The Alligation Rule (or Cross Method) is a powerful technique to determine the ratio in which two ingredients (or items) of different prices (or values) must be mixed to obtain a mixture of a desired mean price (or value).

Let C be the price/value of the cheaper ingredient. Let D be the price/value of the dearer ingredient. Let M be the mean price/value of the mixture.

The ratio of the quantity of the cheaper ingredient (Q_C) to the quantity of the dearer ingredient (Q_D) is given by:

Q_C / Q_D = (D - M) / (M - C)

Visually, the cross method is set up as follows:

          Dearer (D)
         /        \
        /          \
       M (Mean)     (D - M)  <-- Quantity of Cheaper
      /          \
     /            \
    Cheaper (C)  (M - C)  <-- Quantity of Dearer

The ratio of Cheaper quantity to Dearer quantity is (D - M) : (M - C).

Worked Example 1

Two varieties of rice costing Rs. 30/kg and Rs. 45/kg are mixed. In what ratio should they be mixed so that the mixture costs Rs. 36/kg?

Solution:

  1. Identify values: Cheaper (C) = Rs. 30/kg, Dearer (D) = Rs. 45/kg, Mean (M) = Rs. 36/kg.
  2. Apply the cross method: 
          45 (Dearer)
     /        \
    /          \
   36 (Mean)     (45 - 36) = 9  (Quantity of Cheaper)
  /          \
 /            \
30 (Cheaper)  (36 - 30) = 6  (Quantity of Dearer)
  3. Form the ratio: Quantity of Cheaper : Quantity of Dearer = 9 : 6.
  4. Simplify the ratio: 9 : 6 = 3 : 2. So, the two varieties of rice must be mixed in the ratio 3:2.

Worked Example 2

In what ratio must water be mixed with milk costing Rs. 60 per litre to obtain a mixture worth Rs. 45 per litre?

Solution:

  1. Identify values: Water's cost (C) = Rs. 0/litre (cheaper ingredient), Milk's cost (D) = Rs. 60/litre, Mean (M) = Rs. 45/litre.
  2. Apply the cross method: 
          60 (Milk)
     /        \
    /          \
   45 (Mean)     (60 - 45) = 15  (Quantity of Water)
  /          \
 /            \
0 (Water)    (45 - 0) = 45  (Quantity of Milk)
  3. Form the ratio: Quantity of Water : Quantity of Milk = 15 : 45.
  4. Simplify the ratio: 15 : 45 = 1 : 3. So, water and milk must be mixed in the ratio 1:3.

Shortcuts & Tricks

  1. Direct Setup: Always place the mean value in the center. Dearer value on top-left, cheaper on bottom-left. Subtract diagonally and place the result on the opposite side to get the respective quantity ratio.
  2. Zero Value: Remember that water, air, or any 'free' ingredient has a value of 0 for calculation purposes in such problems.
  3. Unit Consistency: Ensure all values (prices, percentages, etc.) are in the same units before applying the rule.
  4. Quick Check: The mean price (M) must always lie between the cheaper (C) and dearer (D) prices. If it doesn't, recheck your setup or calculations.

Common Mistakes

  1. Incorrect Placement: Swapping (D-M) and (M-C) results in an inverted ratio. Always remember (D-M) corresponds to the quantity of the cheaper item, and (M-C) to the dearer.
  2. Ignoring Zero Value: For ingredients like water, not taking its cost as 0 can lead to incorrect calculations.
  3. Units Mismatch: Mixing percentages with absolute values or different units (e.g., Rs/kg and Rs/litre) without proper conversion.
  4. Applying to Wrong Problems: Alligation is specifically for finding mixing ratios to achieve a mean value. It's not a universal solution for all mixture problems, especially those involving successive replacements without a mean value context.

Derivation (brief)

The Alligation Rule is fundamentally a simplified application of the weighted average concept. If we mix Q_C units of an ingredient costing C per unit with Q_D units of an ingredient costing D per unit, the total cost of the mixture is Q_C * C + Q_D * D. The total quantity is Q_C + Q_D. The mean cost M of the mixture is therefore: M = (Q_C * C + Q_D * D) / (Q_C + Q_D) Rearranging this equation to find the ratio: M * (Q_C + Q_D) = Q_C * C + Q_D * D M * Q_C + M * Q_D = Q_C * C + Q_D * D Group terms with Q_C and Q_D: M * Q_D - Q_D * D = Q_C * C - M * Q_C Factor out Q_D and Q_C: Q_D * (M - D) = Q_C * (C - M) To express this as a positive ratio (since M-D would be negative and C-M would be negative, as M is between C and D): Q_D * (D - M) = Q_C * (M - C) Finally, the ratio of quantities: Q_C / Q_D = (D - M) / (M - C) This mathematical derivation directly validates the visual cross-method.

Advanced Examples

  1. Problem with Profit/Loss: A grocer mixes two varieties of pulses costing Rs. 15/kg and Rs. 20/kg respectively. He sells the mixture at Rs. 21/kg, making a profit of 20%. In what ratio did he mix the two varieties? Solution: First, calculate the cost price (CP) of the mixture. The selling price (SP) is Rs. 21/kg and profit is 20%. CP of mixture = SP / (1 + Profit%) = 21 / (1 + 0.20) = 21 / 1.2 = Rs. 17.5/kg. Now, apply alligation with C = Rs. 15/kg, D = Rs. 20/kg, and M = Rs. 17.5/kg:

          20 (Dearer)
     /        \
    /          \
   17.5 (Mean)   (20 - 17.5) = 2.5  (Quantity of Cheaper)
  /          \
 /            \
15 (Cheaper)  (17.5 - 15) = 2.5  (Quantity of Dearer)

    Ratio of Cheaper to Dearer = 2.5 : 2.5 = 1 : 1.

  2. Percentage-based Problem: In what ratio must a solution of 30% alcohol be mixed with a solution of 50% alcohol to get a solution of 45% alcohol? Solution: Here, the 'value' is the percentage concentration. C = 30% alcohol (cheaper/lower concentration) D = 50% alcohol (dearer/higher concentration) M = 45% alcohol (mean/desired concentration)

          50 (Dearer)
     /        \
    /          \
   45 (Mean)     (50 - 45) = 5  (Quantity of 30% solution)
  /          \
 /            \
30 (Cheaper)  (45 - 30) = 15  (Quantity of 50% solution)

    Ratio of 30% solution to 50% solution = 5 : 15 = 1 : 3.

Variation Types

Alligation is versatile and can be applied to various scenarios:

  1. Finding the Mean Value: When quantities and individual values are known, you can use the alligation diagram to visualize the weighted average, or simply calculate it directly.
  2. Finding one Quantity: If the ratio, one quantity, and all values are given, you can find the other quantity.
  3. Profit/Loss Scenarios: As demonstrated above, where the mean cost price needs to be derived from selling price and profit/loss percentage.
  4. Percentage/Concentration Problems: Mixing solutions with different concentrations (e.g., alcohol, acid, sugar solutions).
  5. Average Speed/Performance Problems: Can be used when two different speeds/performances are maintained for different durations/quantities, leading to an overall average.
  6. Population/Group Averages: For instance, if the average age of men and women in a group is known, along with the overall average age, alligation can determine the ratio of men to women.

Time-Saving Methods

  1. Mental Math: Practice performing the diagonal subtractions mentally. This saves precious seconds in the exam.
  2. Immediate Simplification: As soon as you get the ratio, simplify it to its lowest terms. Don't wait for a separate step.
  3. Recognize Zero: For ingredients like water or any 'free' component, instantly place 0 as its value. This makes one subtraction trivial.
  4. Contextual Sanity Check: Always quickly verify that the mean value you're working with (or have calculated) falls between the two individual values. This is a powerful self-correction mechanism.
  5. Pattern Recognition: With sufficient practice, you'll start recognizing common ratios and problem structures, allowing for even faster problem-solving.
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Start Lesson

Master mixture replacement: calculate final concentration after removing a part and refilling with another liquid, often successively. Use specific formulas and ratio methods for speed.

Core Formula

When a certain quantity of a pure liquid is replaced by another liquid (e.g., water), and this process is repeated 'n' times, the quantity of the original liquid remaining in the mixture can be found using the formula:

Final quantity of original liquid = Initial quantity * (1 - (Quantity removed / Initial quantity))^n

Where:

  • Initial quantity = Total volume of the pure liquid initially.
  • Quantity removed = Volume of mixture removed in each operation.
  • n = Number of times the operation (remove and refill) is repeated.

The concentration of the original liquid after 'n' operations will be (Final quantity of original liquid / Initial quantity) * 100%.

Worked Example 1

A container contains 80 litres of milk. 8 litres of milk are removed and replaced with water. What is the quantity of milk in the container now?

Solution: Initial quantity of milk = 80 litres Quantity removed = 8 litres Number of operations (n) = 1

Using the formula: Final quantity of milk = 80 * (1 - (8 / 80))^1 = 80 * (1 - 1/10) = 80 * (9/10) = 72 litres

So, there are 72 litres of milk in the container.

Worked Example 2

From a 100-litre barrel of pure milk, 10 litres are removed and replaced with water. This process is repeated one more time. What is the final quantity of milk in the barrel?

Solution: Initial quantity of milk = 100 litres Quantity removed = 10 litres Number of operations (n) = 2 (initial removal + 1 repeat)

Using the formula: Final quantity of milk = 100 * (1 - (10 / 100))^2 = 100 * (1 - 1/10)^2 = 100 * (9/10)^2 = 100 * (81/100) = 81 litres

So, the final quantity of milk in the barrel is 81 litres.

Shortcuts & Tricks

  1. Ratio Method for Successive Replacement: If x litres are removed from V litres, the fraction remaining is (V-x)/V. After n operations, the fraction of original liquid remaining is ((V-x)/V)^n. Multiply this fraction by the initial quantity V to get the final quantity.
    • Example 2 shortcut: (100-10)/100 = 90/100 = 9/10. After 2 operations, (9/10)^2 = 81/100. Final milk = 100 * (81/100) = 81 litres.
  2. Percentage Approach: If 10% is removed, 90% remains. After n operations, it's (0.9)^n of the original. This is particularly useful when numbers are easy percentages.

Common Mistakes

  1. Incorrect 'n' value: Students often confuse the number of times the process is repeated with the total number of operations. If it says "removed and replaced, and this process is repeated n more times", then the total n in the formula becomes 1 + n. Read carefully!
  2. Miscalculating the remaining quantity: Ensure (Quantity removed / Initial quantity) is correctly subtracted from 1 before raising to the power n.
  3. Assuming constant total volume: The core formula assumes the quantity removed is equal to the quantity refilled, thus maintaining a constant total volume. If different quantities are removed and added, the total volume changes, and a step-by-step calculation is needed.

Derivation (brief)

Let V be the initial volume of pure liquid. Let x be the volume removed and replaced with water in each step.

Step 1: Quantity of pure liquid removed = x Quantity of pure liquid remaining = V - x Fraction of pure liquid remaining = (V - x) / V = 1 - (x/V)

Step 2: Now, the mixture has V - x of pure liquid and x of water. Total volume is V. When x litres of mixture are removed, the quantity of pure liquid removed from this x litres will be x * ( (V-x) / V ). (Since (V-x)/V is the concentration of pure liquid in the mixture).

Quantity of pure liquid remaining after 2nd removal = (V - x) - x * ( (V-x) / V ) = (V - x) * (1 - x/V) = V * (1 - x/V) * (1 - x/V) = V * (1 - x/V)^2

Generalizing for 'n' operations, the quantity of pure liquid remaining is V * (1 - x/V)^n.

Advanced Examples

Problem: A 90-litre container is full of milk. 9 litres of milk are removed and replaced with water. Then, 10 litres of the mixture are removed and replaced with water. What is the final quantity of milk?

Solution:

  • Step 1: (Remove 9L milk, add 9L water) Initial milk = 90L Quantity removed = 9L Milk remaining after 1st step = 90 * (1 - 9/90)^1 = 90 * (9/10) = 81 litres. Total volume remains 90L. The mixture now has 81L milk and 9L water.

  • Step 2: (Remove 10L mixture, add 10L water) Current milk = 81L Total volume = 90L Concentration of milk in mixture = 81/90 = 9/10 Quantity of milk removed in this step = 10L * (9/10) = 9 litres. Milk remaining after 2nd step = 81 - 9 = 72 litres. Total volume remains 90L.

    Final quantity of milk = 72 litres.

    Note: The core formula V * (1 - x/V)^n cannot be directly applied here because the Quantity removed (x) is different in the second step (10L vs 9L). You must calculate step-by-step.

Variation Types

  1. Mixing two different solutions: Problems where two different mixtures (e.g., Milk:Water 3:1 and Milk:Water 2:3) are combined. Use weighted averages or alligation to find the final ratio/concentration.
  2. Multiple ingredients: Mixtures involving three or more components (e.g., milk, water, syrup). Track each component's quantity separately or focus on the target component.
  3. Changing total volume: As seen in the advanced example, if the quantity removed is not equal to the quantity added, the total volume changes, requiring step-by-step calculation rather than the direct formula.

Time-Saving Methods

  • Fractional approach: Always think in terms of fractions. If 1/10th is removed, 9/10th remains. This simplifies calculations, especially with successive replacements. (9/10)^n is easier to compute mentally or quickly than (1 - x/V)^n.
  • Mental Math for small 'n': For n=2 or n=3, practice calculating (fraction)^2 or (fraction)^3 quickly. For example, (9/10)^2 = 81/100, (4/5)^3 = 64/125.
  • Focus on the question: Sometimes, only the ratio of milk to water is asked, not the absolute quantity. Calculate the remaining fraction of milk and then deduce the water fraction (1 - milk fraction).
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