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Mixtures are an important part of daily life, from blending different varieties of tea or milk to creating alloys or solutions with specified concentrations. A mixture combines two or more substances, each possibly having a different quantity, price, or concentration. The value or composition of the mixture depends on how these substances are combined.
Understanding mixture problems is crucial in competitive exams because they test your ability to analyze and calculate quantities, ratios, and averages quickly and accurately. To solve mixture problems effectively, you need to be comfortable with ratios, percentages, and basic arithmetic operations like addition, multiplication, and division.
We will start by reviewing these foundational concepts before moving on to practical strategies for solving various types of mixture questions.
Ratios are fundamental in describing how two or more substances combine. A ratio compares the relative size of two quantities, showing how many parts of one substance there are for a set number of parts of another.
For example, if a mixture consists of two liquids, A and B, in the ratio 3:2, it means for every 3 litres of liquid A, there are 2 litres of liquid B.
This ratio helps us determine proportions, total amounts, or costs related to the mixture. Ratios are especially useful when you are given the cost or concentration of individual substances and want to find out how much of each to mix to get a desired overall result.
In the above bar, the green section represents 3 parts and the yellow section 2 parts of the mixture, making a total of 5 parts.
The alligation method is a handy shortcut for solving mixture problems involving blending two substances with different concentrations or prices to achieve a desired quality or cost.
It helps find the ratio in which the two substances should be mixed without setting up complex equations.
graph TD A[Write down Price/Concentration of Component 1 (C₁)] --> B[Write down Price/Concentration of Component 2 (C₂)] B --> C[Write desired Price/Concentration of mixture (Cₘ)] C --> D[Calculate difference |C₂ - Cₘ|] C --> E[Calculate difference |Cₘ - C₁|] D --> F[Form ratio \(A_1 : A_2 = \frac{|C_2 - C_m|}{|C_m - C_1|}\)] E --> F F --> G[Use ratio to find quantities of each component]Here, \(C_1\) and \(C_2\) are the concentrations or prices of the two components, and \(C_m\) is the desired concentration or price of the mixture. \(A_1\) and \(A_2\) are the amounts of the two substances to mix.
Step 1: Identify the prices of the two types of milk: \(C_1 = 45\) and \(C_2 = 55\).
Step 2: Desired price of mixture, \(C_m = 50\).
Step 3: Apply the alligation formula:
\[ Amount_1 : Amount_2 = \frac{|C_2 - C_m|}{|C_m - C_1|} = \frac{|55 - 50|}{|50 - 45|} = \frac{5}{5} = 1 : 1 \]
Step 4: The ratio is 1:1, meaning equal parts of both milks.
Step 5: Total quantity = 50 litres, so quantity of each = \( \frac{50}{1+1} = 25 \) litres.
Answer: Mix 25 litres of milk at INR 45/litre and 25 litres at INR 55/litre to get 50 litres costing INR 50/litre.
Step 1: Cost of metals: \(C_1 = 120\), \(C_2 = 150\).
Step 2: Desired cost \(C_m = 135\).
Step 3: Calculate the ratio using alligation formula:
\[ Amount_1 : Amount_2 = \frac{|150 - 135|}{|135 - 120|} = \frac{15}{15} = 1 : 1 \]
Step 4: Total mixture = 20 kg; so amounts of each metal = \( \frac{20}{2} = 10 \) kg.
Answer: Mix 10 kg of each metal costing INR 120 and INR 150 per kg to make 20 kg alloy costing INR 135 per kg.
Step 1: Concentrations: \(C_1 = 20\%\), \(C_2 = 50\%\), desired \(C_m = 40\%\).
Step 2: Use alligation to find ratio:
\[ Amount_1 : Amount_2 = \frac{|50 - 40|}{|40 - 20|} = \frac{10}{20} = 1 : 2 \]
Step 3: Let quantity of 20% solution = \(x\) litres, and 50% solution = \(2x\) litres.
Step 4: Total solution = 30 litres, so:
\[ x + 2x = 30 \implies 3x = 30 \implies x = 10 \]
Answer: Mix 10 litres of 20% solution and 20 litres of 50% solution.
Step 1: Calculate effective prices after discounts:
Price A = \(80 - 10\%\) of 80 = \(80 - 8 = 72\) INR/kg
Price B = \(100 - 20\%\) of 100 = \(100 - 20 = 80\) INR/kg
Step 2: Desired mixture price \(C_m = 90\) INR/kg.
Step 3: Since 90 is more than both discounted prices, this suggests a mistake in interpretation. However, the problem implies mixing to get a priced mixture of 90 after discounts which need reconfirmation.
Step 4: Let's check carefully: The discounts made the effective prices lower than the desired price 90, so no mixture is possible at 90 with these discounted prices. If the target price is before discount, apply a different approach.
Correct Approach: Assume mixture price after discount is 90 INR/kg.
But since discounted prices are 72 and 80 only, mixture can't be priced at 90.
Alternative scenario: Possibly prices before discount should be used for mixing, then discounts applied.
For demonstration, assume mixture price before discount is 90.
Step 5: Using alligation with prices before discount:
\[ Amount_1 : Amount_2 = \frac{|100 - 90|}{|90 - 80|} = \frac{10}{10} = 1 : 1 \]
Answer: They should be mixed in equal ratio before discount, then discounts apply individually.
Note: Carefully examine problem wording in exams to clarify whether prices are before or after discounts.
Step 1: Known data:
- Cost 1 (\(C_1\)) = 100 INR/kg, quantity \(A_1 = 10\) kg
- Cost 2 (\(C_2\)) = 120 INR/kg, quantity \(A_2 = x\) kg (unknown)
- Cost 3 (\(C_3\)) = 150 INR/kg, quantity \(A_3 = y\) kg (unknown)
- Total weight = 30 kg
- Mixture cost \(C_m = 130\) INR/kg
Step 2: The total quantity equation:
\[ 10 + x + y = 30 \implies x + y = 20 \]
Step 3: Let's combine the two unknown varieties 2 and 3 into one "combined" part.
We can write the total cost equation as weighted average:
\[ 130 \times 30 = 100 \times 10 + 120x + 150y \]
\[ 3900 = 1000 + 120x + 150y \]
\[ 120x + 150y = 2900 \]
Step 4: Use \( y = 20 - x \), substitute:
\[ 120x + 150(20 - x) = 2900 \]
\[ 120x + 3000 - 150x = 2900 \]
\[ -30x = -100 \implies x = \frac{100}{30} = \frac{10}{3} \approx 3.33 \text{ kg} \]
\[ y = 20 - x = 20 - \frac{10}{3} = \frac{60 - 10}{3} = \frac{50}{3} \approx 16.67 \text{ kg} \]
Answer: Use approximately 3.33 kg of the INR 120/kg tea and 16.67 kg of the INR 150/kg tea with 10 kg of the INR 100/kg tea to get 30 kg costing INR 130/kg.
When to use: To correctly apply the alligation method and avoid sign errors.
When to use: Selecting the right approach for mixture problems.
When to use: At the start of any mixture problem involving different unit measures.
When to use: To confirm correctness before finalizing your answer.
When to use: When prices given include discounts or separate markups that affect final values.
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