10. Alligation
10. Alligation problems as a part of Arithmetic problems
Alligation is a method used in arithmetic to solve mixture problems by finding the ratio in which two or more ingredients with different properties (e.g., concentrations, costs, or qualities) are mixed to achieve a desired average property. It is based on the principle that the weighted average of the components equals the target average. The alligation rule states that the ratio of the quantities of two ingredients is inversely proportional to the differences between their individual properties and the desired mean.
Below are 10 alligation problems with step-by-step solutions and detailed explanations. Each problem includes the problem statement, solution using the alligation method, verification, and an explanation of the concept. These problems avoid alcohol-related scenarios, as per your preference, and focus on practical applications like cost, concentration, and quality.
Problem 1
A grocer wants to mix two types of rice: one costing $5 per kg and another costing $8 per kg, to make a blend costing $6 per kg. In what ratio should they be mixed?
Step-by-Step Solution:
- Identify the given values:
- Cost of first rice () = $5/kg
c_1 - Cost of second rice () = $8/kg
c_2 - Desired mean cost (( m )) = $6/kg
- Apply the alligation rule: The ratio of quantities is given by:
\text{Ratio} = \frac{c_2 - m}{m - c_1}- Difference between second cost and mean:
8 - 6 = 2 - Difference between mean and first cost:
6 - 5 = 1 - Ratio of first rice to second rice: ( 2 : 1 )
- Interpret the ratio: Mix 2 parts of $5 rice with 1 part of $8 rice.
- Verify: Assume 2 kg of $5 rice and 1 kg of $8 rice:
- Total weight =kg
2 + 1 = 3 - Total cost =
5 \cdot 2 + 8 \cdot 1 = 10 + 8 = 18 - Cost per kg =, which matches the target.
\frac{18}{3} = 6
- Answer: The ratio is 2:1 (2 parts $5 rice to 1 part $8 rice).
Explanation: Alligation finds the ratio by comparing how far each ingredient’s cost is from the mean. The $5 rice is $1 below the mean, and the $8 rice is $2 above, so the ratio is 2:1 to balance the costs. This method is quicker than setting up equations for simple mixture problems.
Problem 2
How many liters of a 10% sugar solution must be mixed with a 30% sugar solution to make 20 liters of a 15% sugar solution?
Step-by-Step Solution:
- Identify the given values:
- Concentration of first solution () = 10%
c_1 - Concentration of second solution () = 30%
c_2 - Desired mean concentration (( m )) = 15%
- Total volume = 20 liters
- Apply the alligation rule:
\text{Ratio} = \frac{c_2 - m}{m - c_1} = \frac{30 - 15}{15 - 10} = \frac{15}{5} = 3:1- Ratio of 10% solution to 30% solution = 3:1
- Calculate quantities: Total parts =. Total volume = 20 liters.
3 + 1 = 4- Volume of 10% solution:liters
\frac{3}{4} \cdot 20 = 15 - Volume of 30% solution:liters
\frac{1}{4} \cdot 20 = 5
- Verify: Total volume =liters. Total sugar:
15 + 5 = 20- From 10% solution:
0.1 \cdot 15 = 1.5 - From 30% solution:
0.3 \cdot 5 = 1.5 - Total sugar =
1.5 + 1.5 = 3 - Concentration =
\frac{3}{20} = 0.15 = 15\%
- Answer: 15 liters of 10% solution and 5 liters of 30% solution.
Explanation: The 15% target is closer to 10% than 30%, so more of the 10% solution is needed. Alligation gives the ratio based on the concentration differences, and we scale it to the total volume.
Problem 3
A shop mixes tea costing $12 per kg with tea costing $18 per kg to make a blend costing $15 per kg. If 30 kg of the blend is needed, how much of each type is used?
Step-by-Step Solution:
- Identify the given values:
- Cost of first tea () = $12/kg
c_1 - Cost of second tea () = $18/kg
c_2 - Desired mean cost (( m )) = $15/kg
- Total weight = 30 kg
- Apply the alligation rule:
\text{Ratio} = \frac{18 - 15}{15 - 12} = \frac{3}{3} = 1:1- Ratio of $12 tea to $18 tea = 1:1
- Calculate quantities: Total parts =. Total weight = 30 kg.
1 + 1 = 2- Weight of $12 tea:kg
\frac{1}{2} \cdot 30 = 15 - Weight of $18 tea:kg
\frac{1}{2} \cdot 30 = 15
- Verify: Total weight =kg. Total cost:
15 + 15 = 30- From $12 tea:
12 \cdot 15 = 180 - From $18 tea:
18 \cdot 15 = 270 - Total cost =
180 + 270 = 450 - Cost per kg =
\frac{450}{30} = 15
- Answer: 15 kg of $12 tea and 15 kg of $18 tea.
Explanation: The mean cost ($15) is exactly halfway between $12 and $18, so equal quantities are needed. Alligation simplifies this to a 1:1 ratio, scaled to 30 kg.
Problem 4
How much water must be added to 8 liters of a 40% salt solution to make a 20% salt solution?
Step-by-Step Solution:
- Identify the given values:
- Concentration of salt solution () = 40%
c_1 - Concentration of water () = 0%
c_2 - Desired mean concentration (( m )) = 20%
- Apply the alligation rule:
\text{Ratio} = \frac{0 - 20}{20 - 40} = \frac{-20}{-20} = 1:1- Ratio of salt solution to water = 1:1
- Calculate quantities: Let the volume of salt solution = 8 liters (given). Since the ratio is 1:1, volume of water = 8 liters.
- Verify: Total volume =liters. Total salt:
8 + 8 = 16- From 40% solution:
0.4 \cdot 8 = 3.2 - From water:
0 \cdot 8 = 0 - Total salt = 3.2 liters
- Concentration =
\frac{3.2}{16} = 0.2 = 20\%
- Answer: 8 liters of water must be added.
Explanation: Water has 0% salt, diluting the 40% solution to 20%. The 1:1 ratio means equal volumes of the original solution and water, doubling the volume while halving the concentration.
Problem 5
A jeweler mixes gold that is 80% pure with gold that is 95% pure to make a 90% pure alloy. In what ratio should they be mixed?
Step-by-Step Solution:
- Identify the given values:
- Purity of first gold () = 80%
c_1 - Purity of second gold () = 95%
c_2 - Desired mean purity (( m )) = 90%
- Apply the alligation rule:
\text{Ratio} = \frac{95 - 90}{90 - 80} = \frac{5}{10} = 1:2- Ratio of 80% gold to 95% gold = 1:2
- Interpret the ratio: Mix 1 part of 80% gold with 2 parts of 95% gold.
- Verify: Assume 1 gram of 80% gold and 2 grams of 95% gold:
- Total weight =grams
1 + 2 = 3 - Pure gold =
0.8 \cdot 1 + 0.95 \cdot 2 = 0.8 + 1.9 = 2.7 - Purity =
\frac{2.7}{3} = 0.9 = 90\%
- Answer: The ratio is 1:2 (1 part 80% gold to 2 parts 95% gold).
Explanation: The 90% purity is closer to 95% than 80%, so more of the higher-purity gold is needed. Alligation gives the ratio based on purity differences.
Problem 6
A farmer mixes two types of fertilizer: one with 15% nitrogen and another with 25% nitrogen, to make 100 kg of a 18% nitrogen fertilizer. How much of each is used?
Step-by-Step Solution:
- Identify the given values:
- Nitrogen in first fertilizer () = 15%
c_1 - Nitrogen in second fertilizer () = 25%
c_2 - Desired mean nitrogen (( m )) = 18%
- Total weight = 100 kg
- Apply the alligation rule:
\text{Ratio} = \frac{25 - 18}{18 - 15} = \frac{7}{3} = 7:3- Ratio of 15% fertilizer to 25% fertilizer = 7:3
- Calculate quantities: Total parts =. Total weight = 100 kg.
7 + 3 = 10- Weight of 15% fertilizer:kg
\frac{7}{10} \cdot 100 = 70 - Weight of 25% fertilizer:kg
\frac{3}{10} \cdot 100 = 30
- Verify: Total weight =kg. Total nitrogen:
70 + 30 = 100- From 15% fertilizer:
0.15 \cdot 70 = 10.5 - From 25% fertilizer:
0.25 \cdot 30 = 7.5 - Total nitrogen =
10.5 + 7.5 = 18 - Nitrogen percentage =
\frac{18}{100} = 18\%
- Answer: 70 kg of 15% fertilizer and 30 kg of 25% fertilizer.
Explanation: The 18% target is closer to 15%, so more of the 15% fertilizer is used. Alligation scales the ratio to the total weight.
Problem 7
How many liters of a 50% vinegar solution must be mixed with a 20% vinegar solution to make 60 liters of a 35% vinegar solution?
Step-by-Step Solution:
- Identify the given values:
- Concentration of first solution () = 50%
c_1 - Concentration of second solution () = 20%
c_2 - Desired mean concentration (( m )) = 35%
- Total volume = 60 liters
- Apply the alligation rule:
\text{Ratio} = \frac{20 - 35}{35 - 50} = \frac{-15}{-15} = 1:1- Ratio of 50% solution to 20% solution = 1:1
- Calculate quantities: Total parts =. Total volume = 60 liters.
1 + 1 = 2- Volume of 50% solution:liters
\frac{1}{2} \cdot 60 = 30 - Volume of 20% solution:liters
\frac{1}{2} \cdot 60 = 30
- Verify: Total volume =liters. Total vinegar:
30 + 30 = 60- From 50% solution:
0.5 \cdot 30 = 15 - From 20% solution:
0.2 \cdot 30 = 6 - Total vinegar =
15 + 6 = 21 - Concentration =
\frac{21}{60} = 0.35 = 35\%
- Answer: 30 liters of 50% solution and 30 liters of 20% solution.
Explanation: The 35% target is halfway between 50% and 20%, so equal volumes are needed. Alligation confirms this with a 1:1 ratio.
Problem 8
A baker mixes flour costing $3 per kg with flour costing $5 per kg to make a blend costing $4.20 per kg. In what ratio should they be mixed?
Step-by-Step Solution:
- Identify the given values:
- Cost of first flour () = $3/kg
c_1 - Cost of second flour () = $5/kg
c_2 - Desired mean cost (( m )) = $4.20/kg
- Apply the alligation rule:
\text{Ratio} = \frac{5 - 4.2}{4.2 - 3} = \frac{0.8}{1.2} = \frac{8}{12} = 2:3- Ratio of $3 flour to $5 flour = 2:3
- Interpret the ratio: Mix 2 parts of $3 flour with 3 parts of $5 flour.
- Verify: Assume 2 kg of $3 flour and 3 kg of $5 flour:
- Total weight =kg
2 + 3 = 5 - Total cost =
3 \cdot 2 + 5 \cdot 3 = 6 + 15 = 21 - Cost per kg =, which matches.
\frac{21}{5} = 4.2
- Answer: The ratio is 2:3 (2 parts $3 flour to 3 parts $5 flour).
Explanation: The $4.20 cost is closer to $5, so more of the $5 flour is needed. Alligation gives the ratio based on cost differences.
Problem 9
How much pure acid (100% acid) must be added to 10 liters of a 20% acid solution to make a 50% acid solution?
Step-by-Step Solution:
- Identify the given values:
- Concentration of acid solution () = 20%
c_1 - Concentration of pure acid () = 100%
c_2 - Desired mean concentration (( m )) = 50%
- Apply the alligation rule:
\text{Ratio} = \frac{100 - 50}{50 - 20} = \frac{50}{30} = 5:3- Ratio of 20% solution to pure acid = 5:3
- Calculate quantities: Let the volume of 20% solution = 10 liters (given). Ratio is 5 parts 20% solution to 3 parts pure acid.
- If 5 parts = 10 liters, 1 part =liters
\frac{10}{5} = 2 - Pure acid =liters
3 \cdot 2 = 6
- Verify: Total volume =liters. Total acid:
10 + 6 = 16- From 20% solution:
0.2 \cdot 10 = 2 - From pure acid:
1.0 \cdot 6 = 6 - Total acid =
2 + 6 = 8 - Concentration =
\frac{8}{16} = 0.5 = 50\%
- Answer: 6 liters of pure acid must be added.
Explanation: Pure acid increases the concentration significantly, so less is needed compared to the 20% solution. Alligation scales the ratio to the given volume.
Problem 10
A store mixes candy costing $2.50 per kg with candy costing $4 per kg to make 25 kg of a mixture costing $3.25 per kg. How much of each is used?
Step-by-Step Solution:
- Identify the given values:
- Cost of first candy () = $2.50/kg
c_1 - Cost of second candy () = $4/kg
c_2 - Desired mean cost (( m )) = $3.25/kg
- Total weight = 25 kg
- Apply the alligation rule:
\text{Ratio} = \frac{4 - 3.25}{3.25 - 2.5} = \frac{0.75}{0.75} = 1:1- Ratio of $2.50 candy to $4 candy = 1:1
- Calculate quantities: Total parts =. Total weight = 25 kg.
1 + 1 = 2- Weight of $2.50 candy:kg
\frac{1}{2} \cdot 25 = 12.5 - Weight of $4 candy:kg
\frac{1}{2} \cdot 25 = 12.5
- Verify: Total weight =kg. Total cost:
12.5 + 12.5 = 25- From $2.50 candy:
2.5 \cdot 12.5 = 31.25 - From $4 candy:
4 \cdot 12.5 = 50 - Total cost =
31.25 + 50 = 81.25 - Cost per kg =
\frac{81.25}{25} = 3.25
- Answer: 12.5 kg of $2.50 candy and 12.5 kg of $4 candy.
Explanation: The $3.25 cost is halfway between $2.50 and $4, so equal weights are needed. Alligation confirms this with a 1:1 ratio.
General Notes on Alligation
- Key Concept: Alligation is a shortcut for mixture problems where two ingredients are combined to achieve a mean value. It uses the formula:
\text{Ratio of ingredient 1 to ingredient 2} = \frac{c_2 - m}{m - c_1}whereare the properties (e.g., cost, concentration), and ( m ) is the desired mean.c_1, c_2 - Key Steps:
- Identify the properties of the ingredients and the desired mean.
- Calculate the differences from the mean.
- Form the ratio and scale to the total quantity if needed.
- Verify by checking the weighted average.
- Advantages: Alligation is faster than solving equations for two-component mixtures, especially when only the ratio or quantities are needed.
- Limitations: Alligation works best for two ingredients. For more complex mixtures or when additional constraints (e.g., total cost) are involved, equations may be needed.
- Verification: Always verify by calculating the total property (e.g., cost, concentration) to ensure the mean is achieved.
These problems demonstrate alligation’s utility in various contexts (cost, concentration, purity) and highlight its efficiency. If you need more problems, variations, or further clarification, let me know!
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