UTKALPRESS
ARITHMETIC HANDBOOK — ALLIGATION & MIXTURE
1. What is a Mixture?
A mixture is formed when two or more substances are combined together. The substances may have different prices, quantities, strengths, concentrations or qualities.
Examples include mixing rice of different prices, milk and water, solutions of different concentrations, metals of different purities, or chemicals of different strengths.
Important Terms
- Quantity: Amount of each substance used.
- Cost Price: Price per unit of a substance.
- Mean Price: Price per unit of the final mixture.
- Concentration: Amount of a particular component in a mixture.
- Alligation: A shortcut method for finding the ratio in which two substances should be mixed.
2. Weighted Average Formula
If two substances having prices P₁ and P₂ are mixed in quantities Q₁ and Q₂, the average price of the mixture is:
Mean Price =
(P₁Q₁ + P₂Q₂) / (Q₁ + Q₂)
Example 1: Rice costing ₹40/kg and ₹60/kg are mixed in equal
quantities. Find the average price.
Mean Price = (40 + 60)/2
= ₹50/kg
Mean Price = (40 + 60)/2
= ₹50/kg
3. Basic Alligation Rule
Suppose a cheaper substance and an expensive substance are mixed to obtain a mixture having a mean price between their prices.
Cheaper Price = C
↓
Mean Price = M
↑
Dearer Price = D
``` Ratio = (D − M) : (M − C) ```
↓
Mean Price = M
↑
Dearer Price = D
``` Ratio = (D − M) : (M − C) ```
Quantity of Cheaper : Quantity of Dearer
= (Dearer − Mean) : (Mean − Cheaper)
= (D − M) : (M − C)
= (Dearer − Mean) : (Mean − Cheaper)
= (D − M) : (M − C)
Example 2: Rice costing ₹30/kg and ₹50/kg are mixed to obtain
a mixture costing ₹40/kg. Find the ratio.
Cheaper : Dearer
= (50 − 40) : (40 − 30)
= 10 : 10
= 1 : 1
Cheaper : Dearer
= (50 − 40) : (40 − 30)
= 10 : 10
= 1 : 1
Golden Rule: In alligation, subtract the mean price
diagonally from the two given prices. The resulting differences
give the required quantity ratio.
UTKALPRESS
ALLIGATION — RATIO & APPLICATIONS
4. Direct Alligation Formula
If C < M < D:
Cheaper : Dearer
= (D − M) : (M − C)
Cheaper : Dearer
= (D − M) : (M − C)
Example: Tea costing ₹120/kg and ₹180/kg are mixed to obtain
a mixture costing ₹150/kg. Find the ratio.
Ratio = (180 − 150) : (150 − 120)
= 30 : 30
= 1 : 1
Ratio = (180 − 150) : (150 − 120)
= 30 : 30
= 1 : 1
5. Finding Quantity of Each Substance
Example: Rice costing ₹25/kg and ₹40/kg are mixed to obtain
a mixture costing ₹31/kg. If the total mixture is 90 kg, find the
quantity of each type.
``` Ratio of cheaper : dearer
= (40 − 31) : (31 − 25)
= 9 : 6
= 3 : 2
Total parts = 3 + 2 = 5
Cheaper rice = 90 × 3/5 = 54 kg
Dearer rice = 90 × 2/5 = 36 kg ```
``` Ratio of cheaper : dearer
= (40 − 31) : (31 − 25)
= 9 : 6
= 3 : 2
Total parts = 3 + 2 = 5
Cheaper rice = 90 × 3/5 = 54 kg
Dearer rice = 90 × 2/5 = 36 kg ```
6. Finding Mean Price
When quantities are known, use the weighted average formula.
Example: 20 kg of rice at ₹30/kg is mixed with 30 kg of rice
at ₹50/kg. Find the average price.
``` Mean Price = (20×30 + 30×50)/(20+30)
= (600 + 1500)/50
= 2100/50
= ₹42/kg ```
``` Mean Price = (20×30 + 30×50)/(20+30)
= (600 + 1500)/50
= 2100/50
= ₹42/kg ```
7. Three or More Components
For three or more substances, the weighted average method is often the safest approach.
Mean =
(Q₁P₁ + Q₂P₂ + Q₃P₃ + ...)
/ (Q₁ + Q₂ + Q₃ + ...)
Example: 10 kg at ₹20/kg, 20 kg at ₹30/kg and 30 kg at
₹40/kg are mixed. Find the average price.
Mean = (10×20 + 20×30 + 30×40)/60
= (200 + 600 + 1200)/60
= 2000/60
= ₹33⅓/kg
Mean = (10×20 + 20×30 + 30×40)/60
= (200 + 600 + 1200)/60
= 2000/60
= ₹33⅓/kg
8. Concentration Mixtures
Alligation can also be applied to percentages and concentrations. The same difference method is used.
Example: Solutions of 20% and 50% concentration are mixed
to obtain a 30% solution. Find the ratio.
Ratio = (50 − 30) : (30 − 20)
= 20 : 10
= 2 : 1
Ratio = (50 − 30) : (30 − 20)
= 20 : 10
= 2 : 1
Important: The mean concentration must lie between the two
concentrations being mixed.
UTKALPRESS
MIXTURE — REPLACEMENT & SPECIAL PROBLEMS
9. Replacement of a Liquid
In replacement problems, a certain quantity of liquid is removed from a container and replaced with another liquid, usually water.
If a vessel contains V litres and x litres are removed and replaced
each time, then after n operations:
Quantity of original liquid remaining = V × (1 − x/V)n
Quantity of original liquid remaining = V × (1 − x/V)n
Example: A vessel contains 20 litres of milk. 5 litres are
removed and replaced with water. The process is repeated twice.
Find the quantity of milk left.
Milk remaining = 20 × (1 − 5/20)²
= 20 × (3/4)²
= 20 × 9/16
= 11.25 litres
Milk remaining = 20 × (1 − 5/20)²
= 20 × (3/4)²
= 20 × 9/16
= 11.25 litres
10. Quantity Removed in Repeated Replacement
Fraction of original substance remaining after n replacements:
(1 − x/V)n
Fraction replaced:
1 − (1 − x/V)n
(1 − x/V)n
Fraction replaced:
1 − (1 − x/V)n
Example: A tank contains 100 litres of pure milk.
20 litres are removed and replaced with water. This is done
three times. What fraction of the original milk remains?
Fraction remaining = (1 − 20/100)³
= (4/5)³
= 64/125
Fraction remaining = (1 − 20/100)³
= (4/5)³
= 64/125
11. Mixture of Two Liquids
If one liquid is present in quantity x and another in quantity y, then the fraction of the first liquid in the mixture is:
Fraction of first liquid = x/(x+y)
Percentage of first liquid = [x/(x+y)] × 100
Percentage of first liquid = [x/(x+y)] × 100
Example: A mixture contains 30 litres of milk and 10 litres
of water. Find the percentage of milk.
Total = 30 + 10 = 40 litres
Milk% = 30/40 × 100
= 75%
Total = 30 + 10 = 40 litres
Milk% = 30/40 × 100
= 75%
12. Finding the Amount to Add
Sometimes a question asks how much of a cheaper or more expensive substance should be added to an existing quantity to obtain a desired mean price.
Example: Rice costing ₹60/kg is available. How much rice
costing ₹40/kg should be mixed with 20 kg of it to obtain a mixture
costing ₹50/kg?
By alligation:
Cheaper : Dearer
= (60 − 50) : (50 − 40)
= 10 : 10
= 1 : 1
Therefore, if dearer rice = 20 kg, cheaper rice required = 20 kg.
By alligation:
Cheaper : Dearer
= (60 − 50) : (50 − 40)
= 10 : 10
= 1 : 1
Therefore, if dearer rice = 20 kg, cheaper rice required = 20 kg.
13. Important Cases
| Given | Required | Method |
|---|---|---|
| Two prices + mean | Quantity ratio | Alligation |
| Quantities + prices | Mean price | Weighted average |
| Concentrations + mean concentration | Ratio | Alligation |
| Repeated replacement | Original liquid left | Replacement formula |
Common Mistake: In the alligation ratio,
the difference is taken crosswise. Do not use
(M−C):(D−M) for the cheaper:dearer order without checking the
corresponding quantities.
UTKALPRESS
ALLIGATION & MIXTURE — FORMULA & PRACTICE
14. Complete Formula Sheet
| Concept | Formula |
|---|---|
| Weighted Mean | Σ(PQ) / ΣQ |
| Alligation Ratio | Cheaper : Dearer = (D−M):(M−C) |
| Mixture Quantity | Sum of individual quantities |
| Mixture Cost | Σ(CP × Quantity) |
| Mean Price | Total Cost / Total Quantity |
| Fraction of Component | Component Quantity / Total Quantity |
| Percentage of Component | Component/Total × 100 |
| Replacement Remaining | V(1−x/V)n |
| Fraction Replaced | 1−(1−x/V)n |
15. Solved Revision Examples
Example 1: Two varieties of rice cost ₹35/kg and ₹55/kg.
In what ratio should they be mixed to get rice worth ₹40/kg?
Ratio = (55−40):(40−35)
= 15:5
= 3:1
Ratio = (55−40):(40−35)
= 15:5
= 3:1
Example 2: 15 kg of sugar at ₹40/kg is mixed with 25 kg
at ₹60/kg. Find the mean price.
Mean = (15×40 + 25×60)/40
= (600 + 1500)/40
= ₹52.50/kg
Mean = (15×40 + 25×60)/40
= (600 + 1500)/40
= ₹52.50/kg
Example 3: A 30% solution is mixed with a 60% solution to
obtain a 40% solution. Find the ratio.
Ratio = (60−40):(40−30)
= 20:10
= 2:1
Ratio = (60−40):(40−30)
= 20:10
= 2:1
16. Practice Questions
- Rice costing ₹20/kg and ₹30/kg are mixed to obtain a mixture costing ₹24/kg. Find the ratio. ```
- Tea costing ₹80/kg and ₹120/kg are mixed in the ratio 3:2. Find the average price.
- In what ratio should milk at ₹50/litre be mixed with milk at ₹70/litre to obtain a mixture worth ₹55/litre?
- 20 kg of rice at ₹40/kg is mixed with 30 kg at ₹60/kg. Find the average price.
- Solutions of 10% and 40% concentration are mixed to obtain a 25% solution. Find the ratio.
- A mixture contains 45 litres of milk and 15 litres of water. Find the percentage of milk.
- A vessel contains 30 litres of milk. 6 litres are removed and replaced with water. The process is repeated twice. Find the quantity of original milk remaining.
- Rice costing ₹25/kg is mixed with rice costing ₹45/kg in the ratio 2:3. Find the mean price.
- A 20% solution is mixed with an 80% solution in the ratio 3:1. Find the resulting concentration.
- A vessel contains 40 litres of milk. 10 litres are removed and replaced with water three times. Find the quantity of milk left. ```
17. Answers
1. 3 : 2
2. ₹96/kg
3. 3 : 1
4. ₹52/kg
5. 1 : 1
6. 75%
7. 15.36 litres
8. ₹37/kg
9. 35%
10. 16.875 litres
18. Quick Exam Shortcuts
Shortcut 1:
For two quantities, use weighted average:
Mean = Total Cost ÷ Total Quantity.
``` Shortcut 2: For two prices and a desired mean, use:
Cheaper : Dearer = (Dearer−Mean):(Mean−Cheaper).
Shortcut 3: The mean price or concentration must normally lie between the two values being mixed.
Shortcut 4: In repeated replacement, remember:
Original remaining = V(1−x/V)n.
Shortcut 5: Reduce the alligation ratio to its lowest terms before calculating individual quantities. ```
Mean = Total Cost ÷ Total Quantity.
``` Shortcut 2: For two prices and a desired mean, use:
Cheaper : Dearer = (Dearer−Mean):(Mean−Cheaper).
Shortcut 3: The mean price or concentration must normally lie between the two values being mixed.
Shortcut 4: In repeated replacement, remember:
Original remaining = V(1−x/V)n.
Shortcut 5: Reduce the alligation ratio to its lowest terms before calculating individual quantities. ```
Final Rule:
Alligation is mainly a shortcut for finding the mixing ratio,
while the weighted-average method is the general method for
finding the average value of a mixture.
