AFCAT Numerical Ability · Percentage

Mixtures and Solutions in Percentages

A solution's strength is the percentage of one component: add water or the pure component, change a ratio, mix two strengths, or draw some off and replace it.

All Percentage notes

Why this matters

These questions all turn on one idea: find the component that does not change, and let the percentage follow the total. Nearly a quarter of the page is HARD, mostly replacements and 'how much must be added', where the slip is to apply the new percentage to the old total.

Concept 1 of 5: Strength of a solution when water is added or removed

60 L of a 15% salt solution holds 9 L of salt. Adding water does not change the salt; it only makes the total bigger, so the percentage falls.

Definition

Amount of a component $=$ its percentage $\times$ the total.
Adding water or letting water evaporate leaves the other component unchanged. New percentage $= \dfrac{\text{unchanged amount}}{\text{new total}} \times 100$ .
'How much must evaporate to reach a target': the unchanged amount must be the target percentage of the new total.
A ratio converts to a percentage: copper and zinc in the ratio $3 : 5$ is $\tfrac 38 = 37.5\%$ copper.

New strength

\text{new } \% = \dfrac{\text{unchanged amount}}{\text{new total}} \times 100

Worked example

15 L of water is added to 60 L of a 15% salt solution. What percentage of the new solution is salt?

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Practice questionModerate

Example 1 · Percentage · Mixtures and Solutions

An 80-litre mixture contains

25

% alcohol. How many litres of water must evaporate so that alcohol becomes

40

% of the mixture?

Keep the right thing fixed

Adding water fixes the salt, not the water. Working with the water's percentage gives the right answer only if you remember the water itself changed.

Drill 10 more on strength of a solution when water is added or removed

Concept 2 of 5: How much of the pure component to add

Adding pure alcohol to a mixture changes the alcohol and the total, but not the water. So the water is the fixed amount: it must be the right percentage of the new total.

Definition

Adding pure A: the other component is fixed. New total $= \dfrac{\text{other}}{100 - \text{target}\%} \times 100$ ; subtract the old total.
Or solve directly: $\dfrac{a + x}{T + x} = \dfrac{\text{target}}{100}$ .
An impurity works the same way: adding pure glycerine leaves the impurity fixed, and it must become the target share.

Pure component to add

\dfrac{a + x}{T + x} = \dfrac{\text{target}}{100}

Worked example

A 75 L solution is 12% alcohol. How much pure alcohol must be added to make it 34% alcohol?

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The same idea in a practice question:

Practice questionModerate

Example 2 · Percentage · Mixtures and Solutions

Ninety litres of solution are

40

% acid. How many litres of acid must be added to make it

50

% acid?

The total grows too

To lift 75 L from 12% to 34% alcohol, adding 22% of 75 L is not enough: what you add also enlarges the total. Fix the water and solve.

Drill 5 more on how much of the pure component to add

Concept 3 of 5: Changing a ratio by adding one component

72 L of milk and water in the ratio

5 : 4

is 40 L of milk and 32 L of water. To make the ratio

5 : 6

, only the water changes; the milk's 40 L stays and fixes what one part is.

Definition

Split the mixture into its components first.
The component you do not add is fixed. Use the new ratio to find what the other must become, then subtract what is there.
Equal amounts means simply adding until the two match.

Amount to add

\text{added} = \text{fixed} \times \dfrac{\text{new term of the other}}{\text{new term of the fixed}} - \text{other now}

Worked example

72 L of a mixture has milk and water in the ratio

5 : 4

. How much water must be added to make the ratio

5 : 6

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Practice questionModerate

Example 3 · Percentage · Mixtures and Solutions

A mixture of

120

litres contains milk and water in the ratio

3:2

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

1:1

Find the parts before adding

In 72 L at

5 : 4

, one part is 8 L. Adding the change in ratio terms directly (2 parts) only works if you scale by that 8 L, which gives the 16 L.

Drill 5 more on changing a ratio by adding one component

Concept 4 of 5: Mixing two strengths, and alligation

Mix 6 L of 50% acid with 4 L of 75% acid and the acid adds up: 3 L plus 3 L in 10 L, so 60%. Going the other way, to hit a target strength, the two quantities are in the ratio of their distances from the target.

Definition

Mixing: strength of the mixture $= \dfrac{q_1 s_1 + q_2 s_2}{q_1 + q_2}$ , a weighted average.
Alligation: to get strength $m$ from $s_1 < m < s_2$ , mix them in the ratio $q_1 : q_2 = (s_2 - m) : (m - s_1)$ .
Water counts as 0% (or a price of ₹0): to sell milk worth ₹32 a litre at ₹28, water : milk $= (32 - 28) : (28 - 0) = 1 : 7$ .
Two alloys mixed in equal weights give the plain average of their strengths.

Alligation

\dfrac{q_1}{q_2} = \dfrac{s_2 - m}{m - s_1}

Worked example

6 L of a 50% acid solution is mixed with 4 L of a 75% acid solution. Find the strength of the mixture.

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Example 4 · Percentage · Mixtures and Solutions

To prepare

300

L of

30

% chemical, how much of

25

% and

40

% solutions are needed?

The ratio is crossed

To get 32% from 20% and 50%, the 20% solution takes the larger share,

50 - 32 = 18

parts, because the target is nearer to it. Pairing each strength with its own distance reverses the answer.

Drill 6 more on mixing two strengths, and alligation

Concept 5 of 5: Drawing off and replacing

Draw 10 L from a 40 L can of milk and the can keeps

\tfrac{30}{40} = 0.75

of its milk. Top it up with water and repeat: the milk is multiplied by 0.75 again.

Definition

Removing a share of a well-mixed solution removes that share of each component.
Drawn off and replaced with water $n$ times: the original liquid left $=$ volume $\times \left(1 - \tfrac{\text{drawn}}{\text{volume}}\right)^n$ .
Replaced with the pure component instead: subtract that component's share in what was drawn, then add back the full amount poured in.
Given the strength before and after one replacement with water, the fraction kept is new strength $\div$ old strength.

Repeated replacement with water

\text{left} = V\left(1 - \dfrac{x}{V}\right)^n

Worked example

A can holds 40 L of milk. 10 L is drawn off and replaced with water, and this is done twice. How much milk is left?

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Example 5 · Percentage · Mixtures and Solutions

Twenty litres of a mixture (

45

% acid) are replaced by water, and the acid percentage falls to

36

%. Find the original volume of the mixture.

The second draw takes less milk

After the first round the can is only 75% milk, so the second 10 L takes out 7.5 L of milk, not 10. That is why the factor is squared rather than doubled.

Drill 4 more on drawing off and replacing

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)

Strength of a solution when water is added or removed
New strength
$\text{new } \% = \dfrac{\text{unchanged amount}}{\text{new total}} \times 100$
How much of the pure component to add
Pure component to add
$\dfrac{a + x}{T + x} = \dfrac{\text{target}}{100}$
Changing a ratio by adding one component
Amount to add
$\text{added} = \text{fixed} \times \dfrac{\text{new term of the other}}{\text{new term of the fixed}} - \text{other now}$
Mixing two strengths, and alligation
Alligation
$\dfrac{q_1}{q_2} = \dfrac{s_2 - m}{m - s_1}$
Drawing off and replacing
Repeated replacement with water
$\text{left} = V\left(1 - \dfrac{x}{V}\right)^n$

Watch out for (5)

Keep the right thing fixed
→ Strength of a solution when water is added or removed
The total grows too
→ How much of the pure component to add
Find the parts before adding
→ Changing a ratio by adding one component
The ratio is crossed
→ Mixing two strengths, and alligation
The second draw takes less milk
→ Drawing off and replacing

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