MathsCBSE Class 9 · How Quantities Combine: Understanding Data

Mixtures, Concentration and Percentage Shares: Class 9

Updated 17 October 2026 · Fact-checked

Concentration is the amount of the pure ingredient divided by the total amount of the mixture. In any mixing problem the amount of pure ingredient adds up, while the percentages do not. So write: (ingredient in part 1) + (ingredient in part 2) = (ingredient in the final mixture), and solve. The same idea gives percentage shares: part divided by total, times 100.

Notice that the orange juice amounts add up (0.4 + 0.9 = 1.3 L), but the percentages do not: the mixture is 26%, not 25% or 50%.

Illustration drawn for QuestionBuddy.

Understand Mixtures, Concentration and Percentage Shares

The concentration of a mixture is $\frac{amount of the ingredient}{total amount of the mixture}$ , usually written as a percentage. A 30% acid solution has 30 mL of acid in every 100 mL. The key fact is this: quantities add; percentages do not. If you pour 2 L of 20% juice into 3 L of 30% juice, the juice content is $0.4 + 0.9 = 1.3$ L in 5 L, a concentration of $26%$ , not $25%$ .

Dilution. Adding water changes the total but not the amount of the pure ingredient. To bring 6 L of 30% acid to 20% we need $1.8$ L of acid to make up 20% of the final volume: $\frac{1.8}{V} = 0.2$ , so $V = 9$ L, and 3 L of water is added.

Mixing two grades. To get 60 L of a 20% syrup from 40% and 10% syrups, let $x$ L be the 40% syrup. Then $0.4 x + 0.1 (60 - x) = 0.2 \times 60 = 12$ , so $0.3 x = 6$ and $x = 20$ L; the other part is 40 L. The result lies between the two strengths and nearer the one used in larger quantity.

Alloys and carats. The purity of gold is measured in carats, where 24 carat is pure gold. An 18 carat alloy is therefore $\frac{18}{24} = 75%$ gold. If $x$ g of pure gold is added to 30 g of 18 carat alloy to make 22 carat, the gold in the new alloy is $22.5 + x$ out of $30 + x$ g, so $\frac{22.5 + x}{30 + x} = \frac{22}{24} = \frac{11}{12}$ , and $x = 60$ g.

Repeated replacement. If a quantity is drawn out and replaced by water, each draw removes some of the pure ingredient and some water, in the current proportions. From 10 L of pure milk, remove 2 L and add 2 L of water: milk left $= 8$ L. Remove 2 L of the mixture again: the 2 L drawn contains $2 \times 0.8 = 1.6$ L of milk, leaving $6.4$ L of milk, a concentration of $64%$ , not $60%$ .

Percentage shares and stacked bars. The percentage share of a category is $\frac{category}{total} \times 100$ . A *stacked bar chart* shows the counts, so bars for groups of different sizes have different heights. A *100% stacked bar chart* scales every bar to the same height, which lets you compare shares between groups of different size. To combine groups, add the counts and recompute the shares from the totals; never average the percentages unless the groups are equal in size.

Exam approach. Pick a single quantity to track (the pure ingredient), write it for each part, equate, solve, and finally check with a quick percentage calculation.

Mixtures, Concentration and Percentage Shares: formulas to remember

Concentration: $c = \frac{pure ingredient}{total mixture} \times 100%$; Same units in numerator and denominator.
Mixing rule: $c_{1} V_{1} + c_{2} V_{2} = c (V_{1} + V_{2})$; Ingredient quantities add; percentages do not.
Dilution: $ingredient is unchanged when water is added$; Final volume $= \frac{ingredient}{new concentration}$ .
Gold purity: $purity = \frac{carat}{24} \times 100%$; 24 carat is pure gold.
Repeated replacement: $left after one draw = V (1 - \frac{r}{V})$; $r$ drawn out and replaced by water, $V$ the container volume.
Percentage share: $share = \frac{part}{total} \times 100%$; Shares of all categories add to 100%.

How to solve Mixtures, Concentration and Percentage Shares questions

Track the pure ingredient through the process.

1Name the ingredient you will track (acid, gold, milk, a category).
2Write the amount of that ingredient in each part: concentration x quantity.
3Add the ingredient amounts and the quantities of the mixture.
4Form an equation: ingredient total = final concentration x final quantity, and solve.
5For repeated replacement, multiply by the remaining fraction after each draw.
6For shares, compute part/total for each group, and recompute from combined counts for the whole.
7Verify: recompute the concentration from your answer.

Mixtures, Concentration and Percentage Shares in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Dilution

Suppose a cleaner is sold as 6 litres of a 30% concentrated solution, and a label says to dilute it to 20% before use.

How much water must be added?

Try it first, then show the solutionSolution

Concentrate in the 6 L: $0.30 \times 6 = 1.8$ L.
Final solution $V$ with the same $1.8$ L at 20%: $V = \frac{1.8}{0.2} = 9$ L.
Water added $= 9 - 6 = 3$ L.

Answer: 3 litres of water.

Memory hook: Water changes the total, never the ingredient.

Tricks to remember Mixtures, Concentration and Percentage Shares

Short memory aids. Say each one aloud once, then use it on a question the same day.

Track the pure part
Choose the pure ingredient and write its amount on each side of the equation.
Use it when: All mixture questions.
Quantities add, percentages do not
Add litres or grams, then convert to a percentage once at the end.
Use it when: Avoiding the average-of-percentages error.
Carat over 24
Divide carat by 24 to get the gold fraction.
Use it when: Gold and alloy questions.

Mixtures, Concentration and Percentage Shares: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Mixing two solutions · 3 marks · Short answerBoard level

2 L of a 20% juice concentrate is mixed with 3 L of a 30% juice concentrate. Find the percentage of juice in the mixture.

Try it first, then show the solutionSolution

Juice in the first part $= 0.20 \times 2 = 0.4$ L. (1)
Juice in the second part $= 0.30 \times 3 = 0.9$ L. (1)
Concentration $= \frac{0.4 + 0.9}{2 + 3} = \frac{1.3}{5} = 0.26 = 26%$ . (1)

Answer: The mixture is 26% juice.

Example 2: Dilution · 3 marks · Short answerBoard level

How many litres of water must be added to 6 L of a 30% acid solution to make it a 20% solution?

Try it first, then show the solutionSolution

Acid in the solution $= 0.30 \times 6 = 1.8$ L, unchanged by adding water. (1)
Let the final volume be $V$ : $\frac{1.8}{V} = 0.20$ , so $V = 9$ L. (1)
Water to add $= 9 - 6 = 3$ L. (1)

Answer: 3 litres of water must be added.

Example 3: Two syrups · 3 marks · Short answerHarder board level

How many litres of 40% syrup and 10% syrup must be mixed to make 60 L of 20% syrup?

Try it first, then show the solutionSolution

Let $x$ L be the 40% syrup, so $(60 - x)$ L is the 10% syrup. (½)
Syrup content: $0.40 x + 0.10 (60 - x) = 0.20 \times 60$ . (1)
$0.30 x + 6 = 12$ , so $x = 20$ . (1)
40% syrup 20 L and 10% syrup 40 L; check $8 + 4 = 12$ L of pure syrup in 60 L. (½)

Answer: 20 L of 40% syrup and 40 L of 10% syrup.

Example 4: Raising gold purity · 5 marks · Long answerHarder board level

30 g of an 18 carat gold alloy is melted with some pure gold to make 22 carat gold. Find the mass of pure gold added.

Try it first, then show the solutionSolution

Purity of 18 carat $= \frac{18}{24} = 75%$ ; gold in the alloy $= 0.75 \times 30 = 22.5$ g. (1)
Required purity $= \frac{22}{24} = \frac{11}{12}$ . (1)
Let $x$ g of pure gold be added: $\frac{22.5 + x}{30 + x} = \frac{11}{12}$ . (1)
$12 (22.5 + x) = 11 (30 + x)$ , so $270 + 12 x = 330 + 11 x$ . (1)
$x = 60$ g. Check: $\frac{82.5}{90} = \frac{11}{12}$ . (1)

Answer: 60 g of pure gold must be added.

Example 5: Shares across two classes · 4 marks · Case-basedHOTS / competency

Case: a survey asks how students travel to school. Class A has 40 students: 16 by bus, 12 on foot, 8 by cycle, 4 by car. Class B has 60 students: 21 by bus, 18 on foot, 12 by cycle, 9 by car. (i) Find the percentage shares for each class. (ii) Find the shares for the two classes together. (iii) Which class has the larger share travelling by car? (iv) Why would a 100% stacked bar chart suit this comparison better than a stacked bar chart of counts?

Try it first, then show the solutionSolution

(i) Class A: bus $40%$ , foot $30%$ , cycle $20%$ , car $10%$ . Class B: bus $35%$ , foot $30%$ , cycle $20%$ , car $15%$ . (1)
(ii) Totals out of 100: bus $37$ , foot $30$ , cycle $20$ , car $13$ , so shares $37%, 30%, 20%, 13%$ . (1)
(iii) Class B (15% against 10%). (1)
(iv) The classes differ in size (40 and 60), so counts give bars of different height; the 100% chart scales both bars to the same height, so the shares can be compared directly. (1)

Answer: (i) A: 40/30/20/10%; B: 35/30/20/15%; (ii) 37/30/20/13%; (iii) Class B; (iv) it removes the effect of the different class sizes.

Where students lose marks in Mixtures, Concentration and Percentage Shares

The slips examiners see most often, and what to write instead.

Mistake: Averaging percentages
Ignoring the amounts of each part.
Wrong working
$\frac{20 + 30}{2} = 25%$ .
Right working
$\frac{0.4 + 0.9}{5} = 26%$ .
Do this: Convert each percentage to a quantity first.
Usually costs: 2 of 3 marks.
Mistake: Changing the ingredient on dilution
Forgetting that water does not change the amount of acid.
Wrong working
New acid $= 0.20 \times 6$ .
Right working
Acid stays $0.30 \times 6 = 1.8$ L; the final volume is $9$ L.
Do this: Write "ingredient unchanged" beside the equation.
Usually costs: The whole answer.
Mistake: Reporting the final volume as the water added
Stopping at $V = 9$ .
Wrong working
Water added $= 9$ L.
Right working
Water added $= 9 - 6 = 3$ L.
Do this: Re-read what the question asks for.
Usually costs: 1 mark.

Exam tips

Always name the ingredient you are tracking.
Draw a tiny table: quantity, concentration, ingredient for each part and the mixture.
Check the answer by recomputing the concentration.
For charts, say whether bars show counts or percentage shares before comparing.

Practice questions on Mixtures, Concentration and Percentage Shares

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
5 L of a 40% solution is mixed with 5 L of water. The concentration of the mixture is
(a) 10%
(b) 20%
(c) 25%
(d) 40%
Hint
The pure amount is $2$ L in a total of $10$ L.
Answer
(b) 20%. Pure amount $= 0.4 \times 5 = 2$ L in $10$ L gives $20%$ .
Q2 · 3 marks · Short answerBoard level
In a month a shop sells 120 units of item A, 180 units of item B and 100 units of item C. Find the percentage share of each item and verify that the shares add to 100%.
Hint
Total sales are $120 + 180 + 100$ .
Answer
Total $= 400$ . A: $\frac{120}{400} = 30%$ ; B: $\frac{180}{400} = 45%$ ; C: $\frac{100}{400} = 25%$ . Sum $= 30 + 45 + 25 = 100%$ .
Q3 · 3 marks · Short answerHarder board level
After 10 innings a batter's average is 42 runs. How many runs must be scored in the 11th innings to raise the average to 45?
Hint
Total after 11 innings at an average of 45, minus total after 10.
Answer
Required total $= 11 \times 45 = 495$ ; present total $= 10 \times 42 = 420$ ; runs needed $= 495 - 420 = 75$ .
Q4 · 5 marks · Long answerHarder board level
80 g of an alloy contains 75% silver. How much copper must be added to make an alloy with 60% silver? What is the percentage of copper in the final alloy, and why is it larger than the original?
Hint
Silver stays the same; only the total changes.
Answer
Silver $= 0.75 \times 80 = 60$ g. Final mass $= \frac{60}{0.60} = 100$ g, so copper added $= 100 - 80 = 20$ g. Copper in the final alloy $= 100 - 60 = 40$ g, that is $40%$ (originally $25%$ ). It is larger because the copper has increased while the silver is unchanged, so the total grows and the silver share falls.
Q5 · 4 marks · Case-basedHOTS / competency
Case: a 10 L can holds pure milk. A dairy worker removes 2 L of milk and tops the can up with 2 L of water. Then she removes 2 L of the mixture and tops up again with water. (i) How much milk is left after the first step? (ii) How much milk is in the 2 L drawn in the second step? (iii) What is the final milk concentration? (iv) Why is it not 60%?
Hint
After step 1 the mixture is 80% milk; the second draw removes milk and water together.
Answer
(i) $10 - 2 = 8$ L of milk. (ii) The mixture is $80%$ milk, so $2 \times 0.8 = 1.6$ L. (iii) Milk left $= 8 - 1.6 = 6.4$ L in 10 L, a concentration of $64%$ . (iv) The second draw removes some water as well as milk, so it removes only $1.6$ L of milk, not $2$ L; the milk left is therefore more than $6$ L.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Concentration formula?Show answer
Pure ingredient over total mixture, times 100%.
2.What stays the same in dilution?Show answer
The amount of the pure ingredient.
3.Purity of 18 carat gold?Show answer
$\frac{18}{24} = 75%$ .
4.Why use a 100% stacked bar chart?Show answer
To compare shares of groups of different sizes.
5.Do shares of all categories add to?Show answer
100%.

Mixtures, Concentration and Percentage Shares: frequently asked questions

Why do I add quantities and not percentages?

Percentages are ratios to different totals, so they cannot be added directly. Quantities of the pure ingredient can.

Is 24 carat gold usable for jewellery?

It is pure gold and is soft, so jewellery usually uses lower carats such as 22 or 18, which are alloys with other metals.

How do I combine percentage shares of two groups?

Add the counts of each category across the groups and recompute the percentages from the combined totals.