Index Numbers Worked Examples (Medium)

Q: Do I divide by the number of items or by the total weight?

By the total weight, ∑w\sum w∑w. Dividing by the number of items only works when every weight is equal, which is the special case of no weightage.

Q: How do I find a missing weight from a given composite?

Set the composite equal to ∑Iw∑w\frac{\sum I w}{\sum w}∑w∑Iw, multiply both sides by the denominator, then collect the terms in the unknown and solve. Always substitute your answer back to confirm the composite.

What these examples cover

These medium Index Numbers examples take the step that most SPM questions turn on: combining several price indices into one composite index using weights, written $\bar{I}=\frac{\sum I w}{\sum w}$ . The three questions ask you to compute a composite index from a small table, compare the same items with and without weightage so you can see what weights actually do, and then work backwards from a known composite to a missing weight.

Each uses small, clean numbers so the arithmetic never hides the method. Cover the solution, attempt the question in full on paper, then check line by line.

When your answer differs, find the exact step where the two solutions part company, that line is where the learning is.

Worked examples

Work through all three in order. Attempt each fully before you read the matching solution, and watch how the same formula, $\bar{I}=\frac{\sum I w}{\sum w}$ , is read forwards to find a composite and backwards to find a missing quantity.

Q1[4 marks]

A household basket contains three items A, B and C. Their price indices for 2023 based on 2020 are $110$ , $120$ and $130$ , with weights $2$ , $3$ and $5$ respectively.

Find the composite index for 2023 based on 2020.

Show worked solution

The composite index is the weighted mean of the separate indices. Write the formula, where each index $I$ is multiplied by its weight $w$ :

\bar{I}=\frac{\sum I w}{\sum w}

Multiply each index by its weight in the numerator, and add the weights in the denominator:

\bar{I}=\frac{110(2)+120(3)+130(5)}{2+3+5}

Work the products, then the two sums:

\bar{I}=\frac{220+360+650}{10}=\frac{1230}{10}=123

Answer

The composite index is $123$ . It lies between the smallest index $110$ and the largest $130$ , as a weighted mean must, and leans toward $130$ because item C carries the heaviest weight.

Q2[4 marks]

Two items X and Y have price indices $118$ and $132$ for 2024 based on 2019. (a) Find the composite index if the two items are equally weighted (no weightage).

(b) Find the composite index if X and Y are given weights $3$ and $1$ . Explain briefly why the two answers differ.

Show worked solution

(a) With no weightage, the composite index is just the simple mean of the two indices:

\bar{I}=\frac{118+132}{2}=\frac{250}{2}=125

(b) With weights $3$ and $1$ , multiply each index by its weight and divide by the total weight:

\bar{I}=\frac{118(3)+132(1)}{3+1}=\frac{354+132}{4}=\frac{486}{4}=121.5

Answer

Without weightage the composite is $125$ ; with weights $3$ and $1$ it is $121.5$ . The heavier weight of $3$ sits on item X, whose index $118$ is the lower of the two, so the weighted composite is pulled down toward X.

Q3[4 marks]

A basket contains three items P, Q and R with price indices $105$ , $115$ and $125$ and weights $3$ , $w$ and $2$ respectively. The composite index is $114$ .

Find the value of $w$ .

Show worked solution

Read the composite formula backwards: put the known composite on the left and the weighted sum on the right, leaving $w$ as the unknown.

114=\frac{105(3)+115w+125(2)}{3+w+2}

Simplify the parts that do not involve $w$ : the numerator becomes $315+115w+250=565+115w$ , and the denominator is $5+w$ :

114=\frac{565+115w}{5+w}

Multiply both sides by $5+w$ to clear the fraction, then collect the $w$ terms:

114(5+w)=565+115w

570+114w=565+115w

570-565=115w-114w \;\Rightarrow\; 5=w

Answer

The weight is $w=5$ . Check by substituting back: $\frac{105(3)+115(5)+125(2)}{3+5+2}=\frac{315+575+250}{10}=\frac{1140}{10}=114$ , the given composite.

Q4[4 marks]

The price of a particular textbook was RM24 in 2021 and rose to RM28.80 in 2024. (a) Find the price index of the textbook for 2024 based on 2021.

(b) If the price keeps rising at the same rate every three years, find the price of the textbook in 2027.

Show worked solution

Part (a) asks for a single price index, so apply the basic index formula, which compares the current price $P_1$ with the base-year price $P_0$ :

I=\frac{P_1}{P_0}\times100

Substitute $P_1=28.80$ for 2024 and $P_0=24$ for 2021:

I=\frac{28.80}{24}\times100=120

Part (b): since the price grows at the same rate every three years, the index for the next three-year period is again $120$ . Apply it to the 2024 price to reach 2027:

P_{2027}=28.80\times\frac{120}{100}=34.56

Answer

The price index for 2024 based on 2021 is $120$ (a 20% rise). Continuing at the same rate gives a 2027 price of RM34.56, each three-year jump multiplies the price by $1.2$ , and $24\times1.2\times1.2=34.56$ confirms it.

Q5[4 marks]

The index number of a certain toy for 2022 based on 2020 is $120$ . The index number of the same toy for 2024 based on 2022 is $125$ .

Find the index number of the toy for 2024 based on 2020.

Show worked solution

Three years are linked here, 2020, 2022 and 2024, so use the chain rule for index numbers: multiply the two given indices and divide by 100, since the middle index is itself a percentage of the first base.

I_{2020,2024}=\frac{I_{2020,2022}\times I_{2022,2024}}{100}

Substitute the two given index numbers:

I_{2020,2024}=\frac{120\times125}{100}

I_{2020,2024}=\frac{15000}{100}=150

Answer

The index number for 2024 based on 2020 is $150$ , a 50% rise overall. That is more than simply adding the two separate rises of 20% and 25%, because the second rise applies to an already-higher 2022 price, chaining multiplies the growth rather than adding it.

Q6[4 marks]

The price indices of three household items P, Q and R for 2024 based on 2022 are $108$ , $115$ and $122$ , with weights $1$ , $2$ and $2$ respectively. (a) Calculate the composite index for 2024 based on 2022.

(b) A family spent RM350 in total on these three items in 2022. Estimate their total spending on the same items in 2024.

Show worked solution

Part (a) is the familiar weighted mean:

\bar{I}=\frac{108(1)+115(2)+122(2)}{1+2+2}

\bar{I}=\frac{108+230+244}{5}=\frac{582}{5}=116.4

Part (b): the composite index compares total spending the same way a single price index compares one price, so total spending in 2024 equals total spending in 2022 scaled by $\frac{\bar{I}}{100}$ :

\text{Spending}_{2024}=350\times\frac{116.4}{100}=407.4

Answer

The composite index is $116.4$ , so the family's total spending rises to about RM407.40 in 2024, an increase of about 16.4%, matching the index.

Q7[4 marks]

In 2020 a shop sold 5 boxes of notebooks at RM12 per box, and 2 boxes of pens at RM20 per box. Taking the amount spent on each item in 2020 as its weightage, and given that the price indices for 2023 based on 2020 are $110$ for notebooks and $125$ for pens, find the composite index for 2023 based on 2020.

Show worked solution

The weightage here is not given directly, it is the amount spent on each item in 2020, so find each weight first as price times quantity:

w_{\text{notebooks}}=5(12)=60 \qquad w_{\text{pens}}=2(20)=40

Now substitute these weights and the given price indices into the composite index formula:

\bar{I}=\frac{110(60)+125(40)}{60+40}

\bar{I}=\frac{6600+5000}{100}=\frac{11600}{100}=116

Answer

The composite index is $116$ . It falls between the two separate indices $110$ and $125$ as it should, and sits closer to $110$ because notebooks carry the larger weight of $60$ .

Q8[4 marks]

A workshop produced $2400$ units of a toy car in 2020 and $2760$ units in 2023. (a) Find the quantity index for 2023 based on 2020.

(b) If the workshop wants the quantity index for 2025 based on 2020 to reach $150$ , find the number of units it must produce in 2025.

Show worked solution

An index number can compare quantities produced just as easily as prices, the same formula applies, with $Q$ in place of $P$ :

I=\frac{Q_1}{Q_0}\times100

Part (a): substitute the 2023 and 2020 production figures:

I=\frac{2760}{2400}\times100=115

Part (b): use the same formula with the target index $150$ and solve for the unknown 2025 production $Q_{2025}$ :

150=\frac{Q_{2025}}{2400}\times100 \;\Rightarrow\; Q_{2025}=\frac{150\times2400}{100}=3600

Answer

The 2023 quantity index is $115$ ; to reach an index of $150$ by 2025 the workshop must produce $3600$ units, one and a half times the 2020 output, matching an index of $150$ .

Across these three you have read the composite formula both ways: forwards to blend indices into one number, and backwards to recover a missing weight. The discipline is identical each time, multiply every index by its weight, sum the numerator and the weights separately, and keep the fraction intact until the final step.

Key method points

These three examples rehearse the composite-index skills that carry most Index Numbers questions worth several marks. Keep the following in mind as you practise more.

The composite index is a weighted mean, $\bar{I}=\frac{\sum I w}{\sum w}$ : each index times its weight on top, the total weight underneath.
A composite index always lies between the smallest and largest separate index, and leans toward the indices with the heaviest weights.
With no weightage, every item counts equally, so the composite is the simple mean of the indices.
To find a missing weight or index, put the known composite equal to the fraction and solve, multiply out by the denominator before collecting terms.
Check a composite by confirming it sits between the extreme indices, and check a back-solved weight by substituting it into the full formula.
Because marking is analytic, writing $\bar{I}=\frac{\sum I w}{\sum w}$ with a clear substitution earns method marks even before the final value.

How a teacher helps

The composite index trips students in two predictable places: dividing by the number of items instead of the total weight, and losing a term when they multiply out to solve for a missing weight. In a one-to-one lesson our teacher catches the exact line where that slip appears and shows the tidy layout that prevents it.

Because our teachers are experienced, you work with someone who explains why the denominator is the sum of the weights, not the count. Lessons are taught in English, while SPM papers are set in both Malay and English, so the notation reads the same to you in either version.

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Frequently asked questions

What is the difference between an index number and a composite index?

An index number compares one item's price with its base-year price. A composite index combines several such indices into a single number using weights: $\bar{I}=\frac{\sum I w}{\sum w}$ .

Do I divide by the number of items or by the total weight?

By the total weight, $\sum w$ . Dividing by the number of items only works when every weight is equal, which is the special case of no weightage.

What does the weight actually represent?

It reflects how important an item is in the basket, how much of the spending it accounts for. A larger weight pulls the composite index closer to that item's own index.

How do I find a missing weight from a given composite?

Set the composite equal to $\frac{\sum I w}{\sum w}$ , multiply both sides by the denominator, then collect the terms in the unknown and solve. Always substitute your answer back to confirm the composite.

Source:SRC-DSKP-EN

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

Index Numbers, Worked Examples (medium)

What these examples cover

Worked examples

Key method points

How a teacher helps

Frequently asked questions

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