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Mixtures & Solutions

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Introduction

Mixture problems are very common in aptitude exams. They involve combining two or more ingredients (like milk & water, sugar solutions, metals, etc.) in a given ratio, and sometimes replacing or adding a part of the mixture.

These problems may look confusing, but with the ratio method, they become simple and quick to solve.

Pattern: Mixtures & Solutions

Pattern: Mixtures & Solutions

The key idea:

If two ingredients are mixed in the ratio a : b, then in the total mixture:

Quantity of first ingredient = (a ÷ (a + b)) × Total
Quantity of second ingredient = (b ÷ (a + b)) × Total

Always divide total into ratio parts before solving further.

Step-by-Step Example

Question

A container has 60 litres of a mixture of milk and water in the ratio 2 : 1. Find the quantity of milk and water.

Solution

  1. Step 1: Write the ratio and total.

    Ratio = 2 : 1 → Total = 60 litres

  2. Step 2: Find total ratio parts.

    2 + 1 = 3 parts

  3. Step 3: Find each part value.

    1 part = 60 ÷ 3 = 20 litres

  4. Step 4: Allocate parts to each ingredient.

    Milk = 2 parts = 2 × 20 = 40 litres
    Water = 1 part = 1 × 20 = 20 litres

  5. Step 5: Final Answer.

    Milk = 40 litres, Water = 20 litres

  6. Step 6: Quick Check.

    Ratio check: 40 : 20 = 2 : 1 ✅ Sum check: 40 + 20 = 60 (matches total) ✅

Question

A container has 40 litres of pure milk. If 8 litres of milk is removed and replaced with water, what is the ratio of milk to water now?

Solution

  1. Step 1: Initial mixture.

    Initially, 40 litres = all milk (milk = 40, water = 0)

  2. Step 2: Remove 8 litres milk.

    Milk left = 40 - 8 = 32 litres

  3. Step 3: Replace with water.

    Add 8 litres water → Mixture = 32 milk + 8 water

  4. Step 4: Write ratio.

    Ratio = 32 : 8 = 4 : 1

  5. Step 5: Final Answer.

    Milk : Water = 4 : 1

  6. Step 6: Quick Check.

    Total = 32 + 8 = 40 litres (same as initial) ✅ Ratio simplified correctly = 4 : 1 ✅

Quick Variations

When two mixtures are mixed: Calculate each part separately, then add totals and simplify ratio.

Replacement type: Remove and replace method → reduce first, then add new part.

Concentration type: In solutions, percentage concentration can be treated like ratio parts.

Trick to Always Use

  • Step 1: Add ratio parts (a + b).
  • Step 2: Each part = Total ÷ (a + b).
  • Step 3: Multiply by each ratio term to get quantities.
  • Step 4: For replacement, subtract first → then add new.
  • Step 5: Always verify by checking total and ratio again.

Summary

Mixture questions are solved using the ratio method:

  • Formula: Part = (ratio term ÷ sum of terms) × total
  • Replacement: Subtract removed → Add new part
  • Check: Ratio must match, total must remain consistent

Once you practice, mixture questions become very quick to solve in exams.

Practice

(1/5)
1. A container has 60 liters of milk and water in the ratio 2 : 1. Find the quantity of water.
easy
A. 15 liters
B. 20 liters
C. 25 liters
D. 30 liters

Solution

  1. Step 1: Calculate total ratio parts

    Total ratio = 2 + 1 = 3 parts.

  2. Step 2: Find value of one part

    1 part = 60 ÷ 3 = 20 liters.

  3. Step 3: Compute required term

    Water = 1 part = 20 liters.

  4. Final Answer:

    20 liters → Option B

  5. Quick Check:

    Milk = 40 liters, Water = 20 liters → ratio 40:20 = 2:1 ✅
Hint: Divide total into ratio parts, then multiply by required term.
Common Mistakes: Taking 1/2 of 60 instead of using ratio parts.
2. A mixture of 90 liters contains milk and water in the ratio 4 : 5. How much milk is in the mixture?
easy
A. 36 liters
B. 40 liters
C. 45 liters
D. 50 liters

Solution

  1. Step 1: Compute total ratio parts

    Total ratio = 4 + 5 = 9 parts.

  2. Step 2: Find one part value

    1 part = 90 ÷ 9 = 10 liters.

  3. Step 3: Multiply by required part

    Milk = 4 parts = 4 × 10 = 36 liters.

  4. Final Answer:

    36 liters → Option A

  5. Quick Check:

    Milk:Water = 36:54 = 4:5 ✅
Hint: Use (ratio term ÷ total parts) × total mixture.
Common Mistakes: Dividing by only one ratio term instead of sum of terms.
3. A 72-liter mixture has water and alcohol in the ratio 5 : 7. How much alcohol does it contain?
medium
A. 30 liters
B. 36 liters
C. 42 liters
D. 48 liters

Solution

  1. Step 1: Add ratio parts

    Total ratio = 5 + 7 = 12 parts.

  2. Step 2: Determine one part

    1 part = 72 ÷ 12 = 6 liters.

  3. Step 3: Calculate required quantity

    Alcohol = 7 parts = 7 × 6 = 42 liters.

  4. Final Answer:

    42 liters → Option C

  5. Quick Check:

    Water = 30, Alcohol = 42 → 30:42 = 5:7 ✅
Hint: Find value of one part, then multiply by ratio part.
Common Mistakes: Using 7/12 of 72 incorrectly as 72 ÷ 7.
4. A 100-liter solution contains sugar and water in the ratio 3 : 7. If 20 liters of water is added, what is the new ratio?
medium
A. 1 : 3
B. 3 : 8
C. 3 : 10
D. 3 : 1

Solution

  1. Step 1: Calculate initial quantities

    Sugar = (3/10) × 100 = 30 liters; Water = 70 liters.

  2. Step 2: Add the given amount to water

    New water quantity = 70 + 20 = 90 liters.

  3. Step 3: Form the new ratio

    New ratio = 30 : 90 = 1 : 3.

  4. Final Answer:

    1 : 3 → Option A

  5. Quick Check:

    Dividing both by 30 gives 1 : 3 → correct ✅
Hint: Convert ratio into quantities first, then apply the addition.
Common Mistakes: Forgetting to add before simplifying the ratio.
5. In a mixture of 80 liters, the ratio of acid to water is 7 : 9. If 16 liters of water is added, what is the new ratio?
hard
A. 7 : 10
B. 7 : 11
C. 35 : 61
D. 7 : 13

Solution

  1. Step 1: Compute initial quantities

    Total ratio = 7 + 9 = 16; 1 part = 80 ÷ 16 = 5 liters. Acid = 7 × 5 = 35 liters; Water = 9 × 5 = 45 liters.

  2. Step 2: Add water

    Water = 45 + 16 = 61 liters.

  3. Step 3: Form the new ratio

    New ratio = 35 : 61.

  4. Final Answer:

    35 : 61 → Option C

  5. Quick Check:

    No further simplification possible → ratio is exact and correct ✅
Hint: Always compute exact new quantities, then write the precise ratio.
Common Mistakes: Trying to force simplification when none exists.