Mixtures & Alligation

1. Repeated Replacement of a Liquid

What is the question?

These questions usually say:

“A container has milk/water. Some quantity is removed and replaced with another liquid. This is repeated several times. How much of the original liquid remains?”

Idea

Every time you remove some liquid, you remove the same fraction of the original liquid that is present at that moment.

If:

Total quantity = V
Quantity removed = r

then the fraction remaining after one operation is:

(V - r) / V

So after n operations:

Original liquid remaining
= V × [(V-r)/V]^n

Example

Question: A vessel contains 40 L of milk. 10 L is removed and replaced with water. The same process is repeated once more. How much milk remains?

Step 1: Find the fraction of milk remaining each time

Total = 40 L
Removed = 10 L

Remaining fraction = (40 - 10)/40
                   = 30/40
                   = 3/4

Step 2: The operation happens twice

Milk remaining
= 40 × (3/4)²

Step 3: Calculate

= 40 × 9/16
= 22.5 L

Answer: 22.5 L of milk

Shortcut

If the same amount is removed and replaced every time:

Original remaining
= Initial × (1 - removed/total)^number of times

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2. Basic Mixture / Weighted Average

What is the question?

These questions usually say:

“A quantity of two materials with different prices is mixed. Find the ratio in which they should be mixed to get a particular average price.”

For example:

Sugar A = ₹24/kg
Sugar B = ₹30/kg

Required price = ₹27/kg

Idea

The required price lies between the two prices.

Use the differences:

Cheaper      Required      Dearer
  ₹24  -------- ₹27 -------- ₹30
         3             3

The quantities are in the opposite ratio of the differences.

Formula

If:

Cheaper price = C
Dearer price = D
Mean price    = M

then:

Cheaper : Dearer
= (D - M) : (M - C)

Example

Question: Sugar costing ₹24/kg and ₹30/kg is mixed to get a mixture costing ₹27/kg. Find the ratio in which they should be mixed.

Step 1: Find the differences

30 - 27 = 3

27 - 24 = 3

Step 2: Take the opposite ratio

Cheaper : Dearer
= 3 : 3
= 1 : 1

Answer: 1 : 1

Visual

₹24                  ₹30
  \                    /
   \                  /
    \       ₹27      /
     \              /
       3        3

Ratio = 3 : 3
      = 1 : 1

Remember

Difference crosswise gives the mixing ratio.

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3. Mixture Ratio / Percentage Problems

What is the question?

These questions give the ratio or percentage of ingredients in a mixture and ask:

“How much of each ingredient is present?”

or:

“What is the ratio of one ingredient to another?”

Idea

Treat the ratio as parts.

For example:

Milk : Water = 3 : 2

means:

Milk   = 3 parts
Water  = 2 parts

Total  = 5 parts

Example

Question: A mixture contains milk and water in the ratio 3:2. If the total mixture is 25 L, find the amount of milk and water.

Step 1: Add the parts

3 + 2 = 5 parts

Step 2: Find one part

25 ÷ 5 = 5 L

Step 3: Find each quantity

Milk = 3 × 5
     = 15 L

Water = 2 × 5
      = 10 L

Answer:

Milk  = 15 L
Water = 10 L

Shortcut

Ratio → Add parts → Find 1 part → Multiply

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4. Adding Water / Dilution

What is the question?

These questions usually say:

“A solution contains a certain amount of pure substance. How much water should be added to reduce its concentration to a particular percentage?”

Idea

The amount of the original substance stays the same when only water is added.

Only the total volume increases.

Example

Question: A 20 L solution contains 40% alcohol. How much water should be added to make the solution 25% alcohol?

Step 1: Find the amount of alcohol

Alcohol = 40% of 20

        = 8 L

The 8 L of alcohol does not change.

Step 2: Find the new total volume

We want alcohol to be 25% of the final mixture:

25% of final volume = 8 L

Therefore:

Final volume = 8 / 0.25
             = 32 L

Step 3: Find water added

Water added
= Final volume - Original volume

= 32 - 20

= 12 L

Answer: 12 L of water

Remember

When only water is added:

Amount of substance = SAME
Total mixture       = INCREASES
Concentration       = DECREASES

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5. Removing and Replacing Liquid

What is the question?

These questions are similar to repeated replacement, but may ask:

“A container has milk and water. Some mixture is removed and replaced with milk/water. Find the new ratio.”

The important point is that the removed liquid contains the ingredients in the same ratio as the current mixture.

Example

Question: A 40 L mixture contains milk and water in the ratio 3:1. 8 L of the mixture is removed and replaced with water. Find the new amount of milk.

Step 1: Find the amount of milk initially

Milk : Water = 3 : 1

Total parts = 4

Milk = 3/4 × 40
     = 30 L

Step 2: Find the fraction of mixture removed

Removed = 8 L
Total = 40 L

Fraction removed = 8/40
                 = 1/5

So 1/5 of the milk is removed.

Milk removed = 30 × 1/5
            = 6 L

Step 3: Find milk remaining

Milk remaining = 30 - 6
               = 24 L

The replacement is water, so no new milk is added.

Answer: 24 L of milk

Shortcut

When r litres are removed from a mixture of total V:

Fraction of every ingredient removed
= r/V

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6. Mixing Two Mixtures

What is the question?

These questions give two mixtures with different concentrations and ask:

“In what ratio should the two mixtures be combined to obtain a required concentration?”

Idea

This is another alligation problem.

Suppose:

Mixture A = 20% acid
Mixture B = 50% acid
Required = 30% acid

Place them like this:

20%              50%
  \                /
   \              /
      30%
   /              \
10                 10

The ratio is:

Mixture A : Mixture B
= (50 - 30) : (30 - 20)
= 20 : 10
= 2 : 1

Example

Question: A 20% acid solution and a 50% acid solution are mixed to obtain a 30% acid solution. Find the ratio in which they should be mixed.

Step 1: Find the differences

50 - 30 = 20

30 - 20 = 10

Step 2: Take the cross differences

20% solution : 50% solution
= 20 : 10
= 2 : 1

Answer: 2:1

Remember

Required concentration must lie between the two given concentrations.

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7. Replacement + Final Ratio

What is the question?

These are slightly more advanced questions where you are told:

“A vessel contains milk and water in a certain ratio. Some mixture is removed and replaced with water. Find the final ratio.”

These are common enough to know for aptitude exams.

Example

Question: A vessel contains 40 L of milk and 20 L of water. 12 L of the mixture is removed and replaced with 12 L of water. Find the new ratio of milk to water.

Step 1: Find the initial total

Milk  = 40 L
Water = 20 L

Total = 60 L

Step 2: Find the fraction removed

Removed = 12 L
Total = 60 L

Fraction = 12/60
         = 1/5

So 1/5 of both milk and water is removed.

Step 3: Find milk remaining

Milk removed = 40 × 1/5
            = 8 L

Milk remaining = 40 - 8
               = 32 L

Step 4: Find water remaining

Water removed = 20 × 1/5
             = 4 L

Water remaining = 20 - 4
                = 16 L

Now 12 L of water is added:

Water = 16 + 12
      = 28 L

Step 5: Find the ratio

Milk : Water
= 32 : 28
= 8 : 7

Answer: 8:7

Visual

BEFORE

Milk  = 40 L
Water = 20 L
        |
        | Remove 12 L mixture
        ↓
Milk  = 32 L
Water = 16 L
        |
        | Add 12 L water
        ↓
Milk  = 32 L
Water = 28 L

Final ratio = 32 : 28
            = 8 : 7