Alligation or Mixture: Formulas, Shortcuts and Solved Examples — MCQs with Answers (Set 1)

Start with x units of pure liquid in the container. Each time, y units of the current mixture are drawn off and replaced with water. The fraction of liquid drawn off each time is y/x. So the fraction that stays pure after each operation is (1 – y/x). After one operation, pure liquid = x(1 – y/x). After n such operations, pure liquid = x(1 – y/x)^n.

Water has no purchase cost, so it carries no cost price of its own. In an alligation problem it is treated as the cheaper ingredient, with cost price Rs 0 per unit. This lets the usual alligation rule, (dearer price – mean price) : (mean price – cheaper price), apply directly. Here the cheaper price is simply set to 0.

Let the cheaper ingredient cost c per unit, and the dearer one cost d per unit. Let the mean price of the mixture be m per unit. The rule of alligation gives cheaper quantity : dearer quantity = (d – m) : (m – c). Here (d – m) measures how far the dearer price sits above the mean. And (m – c) measures how far the mean sits above the cheaper price. A bigger gap on one side means less of that ingredient is required.

When two ingredients are mixed, the mixture itself has a cost price. This cost price, for one unit of the mixture, is called the mean price. The mean price always lies between the cheaper ingredient’s price and the dearer ingredient’s price. It can never fall outside that range.

Milk fraction in vessel A = 1/(1+4) = 0.2. Milk fraction in vessel B = 1/(1+1) = 0.5. Target milk fraction = 2/(2+3) = 0.4. By alligation, ratio of A to B = (B’s fraction – target) : (target – A’s fraction) = (0.5-0.4) : (0.4-0.2) = 0.1 : 0.2. Writing this over a common denominator and reducing gives the whole-number ratio 1 : 2. Check: mixing 1 part of A with 2 parts of B gives milk fraction (1×0.2 + 2×0.5) / (1+2) = 0.4, matching the target 0.4.

Milk fraction in vessel A = 2/(2+5) = 0.29. Milk fraction in vessel B = 4/(4+1) = 0.8. Target milk fraction = 3/(3+2) = 0.6. By alligation, ratio of A to B = (B’s fraction – target) : (target – A’s fraction) = (0.8-0.6) : (0.6-0.29) = 0.2 : 0.31. Writing this over a common denominator and reducing gives the whole-number ratio 7 : 11. Check: mixing 7 parts of A with 11 parts of B gives milk fraction (7×0.29 + 11×0.8) / (7+11) = 0.6, matching the target 0.6.

Milk fraction in vessel A = 1/(1+4) = 0.2. Milk fraction in vessel B = 4/(4+5) = 0.44. Target milk fraction = 2/(2+3) = 0.4. By alligation, ratio of A to B = (B’s fraction – target) : (target – A’s fraction) = (0.44-0.4) : (0.4-0.2) = 0.04 : 0.2. Writing this over a common denominator and reducing gives the whole-number ratio 2 : 9. Check: mixing 2 parts of A with 9 parts of B gives milk fraction (2×0.2 + 9×0.44) / (2+9) = 0.4, matching the target 0.4.

Milk fraction in vessel A = 2/(2+5) = 0.29. Milk fraction in vessel B = 5/(5+2) = 0.71. Target milk fraction = 3/(3+2) = 0.6. By alligation, ratio of A to B = (B’s fraction – target) : (target – A’s fraction) = (0.71-0.6) : (0.6-0.29) = 0.11 : 0.31. Writing this over a common denominator and reducing gives the whole-number ratio 4 : 11. Check: mixing 4 parts of A with 11 parts of B gives milk fraction (4×0.29 + 11×0.71) / (4+11) = 0.6, matching the target 0.6.

Milk fraction in vessel A = 5/(5+1) = 0.83. Milk fraction in vessel B = 5/(5+4) = 0.56. Target milk fraction = 3/(3+2) = 0.6. By alligation, ratio of A to B = (target – B’s fraction) : (A’s fraction – target) = (0.6-0.56) : (0.83-0.6) = 0.04 : 0.23. Writing this over a common denominator and reducing gives the whole-number ratio 4 : 21. Check: mixing 4 parts of A with 21 parts of B gives milk fraction (4×0.83 + 21×0.56) / (4+21) = 0.6, matching the target 0.6.

Milk fraction in vessel A = 1/(1+2) = 0.33. Milk fraction in vessel B = 5/(5+3) = 0.62. Target milk fraction = 1/(1+1) = 0.5. By alligation, ratio of A to B = (B’s fraction – target) : (target – A’s fraction) = (0.62-0.5) : (0.5-0.33) = 0.12 : 0.17. Writing this over a common denominator and reducing gives the whole-number ratio 3 : 4. Check: mixing 3 parts of A with 4 parts of B gives milk fraction (3×0.33 + 4×0.62) / (3+4) = 0.5, matching the target 0.5.

Milk fraction in vessel A = 5/(5+4) = 0.56. Milk fraction in vessel B = 2/(2+5) = 0.29. Target milk fraction = 1/(1+2) = 0.33. By alligation, ratio of A to B = (target – B’s fraction) : (A’s fraction – target) = (0.33-0.29) : (0.56-0.33) = 0.05 : 0.22. Writing this over a common denominator and reducing gives the whole-number ratio 3 : 14. Check: mixing 3 parts of A with 14 parts of B gives milk fraction (3×0.56 + 14×0.29) / (3+14) = 0.33, matching the target 0.33.

Milk fraction in vessel A = 3/(3+5) = 0.38. Milk fraction in vessel B = 5/(5+2) = 0.71. Target milk fraction = 2/(2+1) = 0.67. By alligation, ratio of A to B = (B’s fraction – target) : (target – A’s fraction) = (0.71-0.67) : (0.67-0.38) = 0.05 : 0.29. Writing this over a common denominator and reducing gives the whole-number ratio 8 : 49. Check: mixing 8 parts of A with 49 parts of B gives milk fraction (8×0.38 + 49×0.71) / (8+49) = 0.67, matching the target 0.67.

Milk fraction in vessel A = 5/(5+2) = 0.71. Milk fraction in vessel B = 1/(1+4) = 0.2. Target milk fraction = 1/(1+2) = 0.33. By alligation, ratio of A to B = (target – B’s fraction) : (A’s fraction – target) = (0.33-0.2) : (0.71-0.33) = 0.13 : 0.38. Writing this over a common denominator and reducing gives the whole-number ratio 7 : 20. Check: mixing 7 parts of A with 20 parts of B gives milk fraction (7×0.71 + 20×0.2) / (7+20) = 0.33, matching the target 0.33.

Milk fraction in vessel A = 1/(1+4) = 0.2. Milk fraction in vessel B = 2/(2+3) = 0.4. Target milk fraction = 1/(1+2) = 0.33. By alligation, ratio of A to B = (B’s fraction – target) : (target – A’s fraction) = (0.4-0.33) : (0.33-0.2) = 0.07 : 0.13. Writing this over a common denominator and reducing gives the whole-number ratio 1 : 2. Check: mixing 1 part of A with 2 parts of B gives milk fraction (1×0.2 + 2×0.4) / (1+2) = 0.33, matching the target 0.33.

Let the container start with x = 72 litres of pure milk. Each time, 24 litres of the mixture are drawn off and replaced with water. Fraction drawn off each time = 24/72 = 0.33. So the fraction of pure milk remaining after each operation = 1 – 0.33 = 0.67. This is repeated 2 times. After operation 1, pure milk = 72 x 0.67 = 48 litres. After operation 2, pure milk = 48 x 0.67 = 32 litres. By the repeated-dilution formula, pure milk left = 72 x (0.67)^2 = 32 litres.

Let the container start with x = 375 litres of pure milk. Each time, 300 litres of the mixture are drawn off and replaced with water. Fraction drawn off each time = 300/375 = 0.8. So the fraction of pure milk remaining after each operation = 1 – 0.8 = 0.2. This is repeated 2 times. After operation 1, pure milk = 375 x 0.2 = 75 litres. After operation 2, pure milk = 75 x 0.2 = 15 litres. By the repeated-dilution formula, pure milk left = 375 x (0.2)^2 = 15 litres.

Let the container start with x = 150 litres of pure milk. Each time, 120 litres of the mixture are drawn off and replaced with water. Fraction drawn off each time = 120/150 = 0.8. So the fraction of pure milk remaining after each operation = 1 – 0.8 = 0.2. This is repeated 2 times. After operation 1, pure milk = 150 x 0.2 = 30 litres. After operation 2, pure milk = 30 x 0.2 = 6 litres. By the repeated-dilution formula, pure milk left = 150 x (0.2)^2 = 6 litres.

Let the container start with x = 144 litres of pure milk. Each time, 24 litres of the mixture are drawn off and replaced with water. Fraction drawn off each time = 24/144 = 0.17. So the fraction of pure milk remaining after each operation = 1 – 0.17 = 0.83. This is repeated 2 times. After operation 1, pure milk = 144 x 0.83 = 120 litres. After operation 2, pure milk = 120 x 0.83 = 100 litres. By the repeated-dilution formula, pure milk left = 144 x (0.83)^2 = 100 litres.

Let the container start with x = 100 litres of pure milk. Each time, 20 litres of the mixture are drawn off and replaced with water. Fraction drawn off each time = 20/100 = 0.2. So the fraction of pure milk remaining after each operation = 1 – 0.2 = 0.8. This is repeated 2 times. After operation 1, pure milk = 100 x 0.8 = 80 litres. After operation 2, pure milk = 80 x 0.8 = 64 litres. By the repeated-dilution formula, pure milk left = 100 x (0.8)^2 = 64 litres.

Let the container start with x = 144 litres of pure milk. Each time, 72 litres of the mixture are drawn off and replaced with water. Fraction drawn off each time = 72/144 = 0.5. So the fraction of pure milk remaining after each operation = 1 – 0.5 = 0.5. This is repeated 2 times. After operation 1, pure milk = 144 x 0.5 = 72 litres. After operation 2, pure milk = 72 x 0.5 = 36 litres. By the repeated-dilution formula, pure milk left = 144 x (0.5)^2 = 36 litres.

Let the container start with x = 250 litres of pure milk. Each time, 200 litres of the mixture are drawn off and replaced with water. Fraction drawn off each time = 200/250 = 0.8. So the fraction of pure milk remaining after each operation = 1 – 0.8 = 0.2. This is repeated 3 times. After operation 1, pure milk = 250 x 0.2 = 50 litres. After operation 2, pure milk = 50 x 0.2 = 10 litres. After operation 3, pure milk = 10 x 0.2 = 2 litres. By the repeated-dilution formula, pure milk left = 250 x (0.2)^3 = 2 litres.

Let the container start with x = 405 litres of pure milk. Each time, 135 litres of the mixture are drawn off and replaced with water. Fraction drawn off each time = 135/405 = 0.33. So the fraction of pure milk remaining after each operation = 1 – 0.33 = 0.67. This is repeated 3 times. After operation 1, pure milk = 405 x 0.67 = 270 litres. After operation 2, pure milk = 270 x 0.67 = 180 litres. After operation 3, pure milk = 180 x 0.67 = 120 litres. By the repeated-dilution formula, pure milk left = 405 x (0.67)^3 = 120 litres.

Let the container start with x = 81 litres of pure milk. Each time, 27 litres of the mixture are drawn off and replaced with water. Fraction drawn off each time = 27/81 = 0.33. So the fraction of pure milk remaining after each operation = 1 – 0.33 = 0.67. This is repeated 3 times. After operation 1, pure milk = 81 x 0.67 = 54 litres. After operation 2, pure milk = 54 x 0.67 = 36 litres. After operation 3, pure milk = 36 x 0.67 = 24 litres. By the repeated-dilution formula, pure milk left = 81 x (0.67)^3 = 24 litres.

Let the container start with x = 120 litres of pure milk. Each time, 60 litres of the mixture are drawn off and replaced with water. Fraction drawn off each time = 60/120 = 0.5. So the fraction of pure milk remaining after each operation = 1 – 0.5 = 0.5. This is repeated 3 times. After operation 1, pure milk = 120 x 0.5 = 60 litres. After operation 2, pure milk = 60 x 0.5 = 30 litres. After operation 3, pure milk = 30 x 0.5 = 15 litres. By the repeated-dilution formula, pure milk left = 120 x (0.5)^3 = 15 litres.

Let the container start with x = 216 litres of pure milk. Each time, 36 litres of the mixture are drawn off and replaced with water. Fraction drawn off each time = 36/216 = 0.17. So the fraction of pure milk remaining after each operation = 1 – 0.17 = 0.83. This is repeated 3 times. After operation 1, pure milk = 216 x 0.83 = 180 litres. After operation 2, pure milk = 180 x 0.83 = 150 litres. After operation 3, pure milk = 150 x 0.83 = 125 litres. By the repeated-dilution formula, pure milk left = 216 x (0.83)^3 = 125 litres.