AFCAT Numerical Ability · Mixture and Alligation

Adding One Liquid to a Mixture

Split a mixture into its liquids, add one of them, and find the new ratio, the amount to add for a target ratio, or the amount there was to begin with.

All Mixture and Alligation notes

Why this matters

Nearly half the chapter's questions are here, including AFCAT past-year questions from 2015 and 2018. Each one adds a single liquid, so the other liquid never changes. Marks are lost when the working lets it change too, or when the answer given is the new total instead of the litres added.

Concept 1 of 5: Splitting a mixture, then adding to it

A mixture of milk and water in a ratio is a total cut into equal parts. Find the litres of each liquid first; only then add the new litres to the liquid they belong to.

Definition

Milk : water $= m : w$ in a total $T$ : the milk is $\dfrac{m}{m + w}T$ and the water is $\dfrac{w}{m + w}T$ .
Given one liquid's amount instead of the total, that amount divided by its term is one part.
Add the new litres to their own liquid, then write the new ratio in its simplest form.

Milk in a mixture

\text{milk} = \dfrac{m}{m + w} \times T

Worked example

A 48-litre mixture has milk and water in the ratio

5 : 3

. 6 litres of water are added. Find the new ratio of milk to water.

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

AFCAT · 2015 · AFCAT (II) 2015 — Numerical Ability · 03-208Easy

Example 1 · Mixture and Alligation · Adding to a Mixture

In a mixture of 35 L, the ratio of milk to water is 4 : 1. If 7 L of water is added to the mixture, the ratio of milk to water in the new mixture will be

Add litres to litres, not to ratio terms

A 40-litre mixture with milk and water

3 : 1

has 30 litres of milk and 10 of water. With 10 more litres of water it is

30 : 20 = 3 : 2

. Adding the 10 to the ratio term gives

3 : 11

, which is meaningless.

Drill 6 more on splitting a mixture, then adding to it

Concept 2 of 5: Adding water to reach a ratio: the milk is fixed

Pouring in water does not change the milk. So the milk decides how much water the new ratio needs; the litres to add are what is missing.

Definition

Find the milk $M$ and the water $W$ now.
For the new ratio $r : s$ , the water must become $\dfrac{s}{r} \times M$ .
The water to add is that amount minus $W$ .
The same works for any pair: alcohol and water, juice and water, two oils. Hold fixed whichever liquid is not poured in.

Water to add

\text{add} = \dfrac{s}{r} \times M - W

Worked example

A 90-litre mixture has milk and water in the ratio

4 : 1

. How much water must be added to make the ratio

3 : 1

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

AFCAT · 2018 · AFCAT (II) 2018 — Numerical Ability · 03-203Easy

Example 2 · Mixture and Alligation · Adding to a Mixture

729 mL of a mixture contains milk and water in the ratio 7 : 2. How much more water must be added to get a new mixture containing milk and water in the ratio 7 : 3?

Do not split the old total in the new ratio

To turn 40 litres of a

3 : 1

milk and water mixture into

1 : 1

, halving 40 gives 20 litres of water and an answer of 10. But that quietly cuts the milk to 20. The milk stays 30, so the water must reach 30: add 20 litres.

Answer with the litres added

The question asks how much water to add, not how much water there will be. Subtract the water already in the mixture before you choose an option.

Drill 18 more on adding water to reach a ratio: the milk is fixed

Concept 3 of 5: Adding milk to reach a ratio: the water is fixed

Now the liquid poured in is the milk (or acid, syrup, the first oil). The water is what stays, so the water decides how much milk the new ratio needs.

Definition

Find the milk $M$ and the water $W$ now; the water stays $W$ .
For the new ratio $r : s$ , the milk must become $\dfrac{r}{s} \times W$ .
The milk to add is that amount minus $M$ .
'Reverse the ratio' means swap the terms: $2 : 5$ reversed is $5 : 2$ .

Milk to add

\text{add} = \dfrac{r}{s} \times W - M

Worked example

A 60-litre mixture has acid and water in the ratio

2 : 3

. How much acid must be added to make the ratio

1 : 1

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionModerate

Example 3 · Mixture and Alligation · Adding to a Mixture

36

litre mixture contains milk and water in ratio

7:2

. How many litres of milk must be added to make the ratio

4:1

Adding milk, scale from the water

When milk is poured in, the milk is the liquid that changes. Working from the old milk (as if it were fixed) answers a different question. Start from the water.

A reversed ratio is a fraction of the water, not of the new total

To reverse

3 : 5

by adding milk, the milk must become

\tfrac 53

of the water. It is not

\tfrac 58

of anything, because the new total is not yet known.

Drill 19 more on adding milk to reach a ratio: the water is fixed

Concept 4 of 5: Unknown amounts: when the unchanged liquid keeps its term

Milk and water

4 : 3

become

4 : 5

after 10 litres of water are added. The milk's term is 4 both times, so the parts are the same size, and the water grew by 2 parts. Those 2 parts are the 10 litres.

Definition

Write the amounts as $4k$ and $3k$ : $k$ is the size of one part.
If the liquid that was not added has the same term in both ratios, the other term grew by a number of parts. Litres added divided by those extra parts is $k$ .
The equation route gives the same $k$ : $\dfrac{4k}{3k + 10} = \dfrac 45$ .
Then answer what is asked: the milk, the water, or the whole mixture, before or after.

One part

k = \dfrac{\text{litres added}}{\text{extra parts}}

Worked example

A mixture has milk and water in the ratio

4 : 3

. After 10 litres of water are added the ratio is

4 : 5

. How much milk is there?

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionModerate

Example 4 · Mixture and Alligation · Adding to a Mixture

A mixture of milk and water is in the ratio 5:2. If 14 liters of water is added, the ratio becomes 5:4. Find the original quantity of milk.

k is not the answer

Finding one part is half the work. With

k = 6

in a

2 : 3

mixture the milk is 12, the water 18 and the mixture 30 litres; the options usually hold all three. Read which one is asked.

Drill 12 more on unknown amounts: when the unchanged liquid keeps its term

Concept 5 of 5: When the unchanged liquid's term differs: rescale first

If milk and water go from

2 : 1

4 : 5

as water is added, the milk did not change but its term did. Write the old ratio with the same milk term,

4 : 2

, and the parts are the same size again.

Definition

Find the liquid that was not added. Multiply each ratio so that liquid has the same term in both (use the LCM of its two terms).
Now the other term's growth counts the extra parts, and litres added divided by extra parts is one part.
'Equal amounts' or $1 : 1$ is a ratio too: write it as $5 : 5$ or $3 : 3$ to match.
The equation route still works when the rescaling is awkward: $\dfrac{2k}{k + 12} = \dfrac 45$ .

One part after rescaling

k = \dfrac{\text{litres added}}{\text{extra parts after matching the fixed term}}

Worked example

Milk and water are in the ratio

2 : 1

. After 12 litres of water are added the ratio is

4 : 5

. How much milk is there?

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionModerate

Example 5 · Mixture and Alligation · Adding to a Mixture

In a mixture, milk and water are in the ratio

9:4

. If

10

litres of water is added, the ratio becomes

3:2

. The initial quantity of mixture was:

Match the liquid that did not change

Milk and water

4 : 1

become

2 : 3

after water is added. The milk is fixed, so write

2 : 3

4 : 6

: the water grew by 5 parts. Reading the water's 1 and 3 as 2 extra parts compares parts of different sizes.

Drill 5 more on when the unchanged liquid's term differs: rescale first

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

Splitting a mixture, then adding to it
Milk in a mixture
$\text{milk} = \dfrac{m}{m + w} \times T$
Adding water to reach a ratio: the milk is fixed
Water to add
$\text{add} = \dfrac{s}{r} \times M - W$
Adding milk to reach a ratio: the water is fixed
Milk to add
$\text{add} = \dfrac{r}{s} \times W - M$
Unknown amounts: when the unchanged liquid keeps its term
One part
$k = \dfrac{\text{litres added}}{\text{extra parts}}$
When the unchanged liquid's term differs: rescale first
One part after rescaling
$k = \dfrac{\text{litres added}}{\text{extra parts after matching the fixed term}}$

Watch out for (7)

Add litres to litres, not to ratio terms
→ Splitting a mixture, then adding to it
Do not split the old total in the new ratio
→ Adding water to reach a ratio: the milk is fixed
Answer with the litres added
→ Adding water to reach a ratio: the milk is fixed
Adding milk, scale from the water
→ Adding milk to reach a ratio: the water is fixed
A reversed ratio is a fraction of the water, not of the new total
→ Adding milk to reach a ratio: the water is fixed
k is not the answer
→ Unknown amounts: when the unchanged liquid keeps its term
Match the liquid that did not change
→ When the unchanged liquid's term differs: rescale first

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