Additional Mathematics notes, chapter 10
Index Number
Textbook Additional Mathematics Form 4 KSSM (DLP), pages 270–287
What this chapter covers
- Define and describe index numbers, and calculate an index number I = Q1/Q0 × 100 (10.1.1)
- Find the index number, the base value or the current value, and relate indices with different base years (10.1.2)
- Solve problems involving index numbers (10.1.3)
- Determine and interpret the composite index using weightages (10.2.1)
- Solve problems involving index numbers and composite index (10.2.2)
10.1Index numbers
Index number: a statistical measure of how a quantity (price, output, number of people and so on) in one period compares with the same quantity in a base year. The base year has the value 100.
It is the change expressed as a ratio, multiplied by 100. The percentage sign is not written.
I = Q1Q0 × 100, where Q0 is the price or quantity in the base year and Q1 is the price or quantity in the year you want to compare.
| Index | Meaning |
|---|---|
| More than 100 | An increase compared with the base year (I = 125 means +25%) |
| Exactly 100 | No change (this is the base year itself, or the same value) |
| Less than 100 | A decrease compared with the base year (I = 80 means −20%) |
Example 1 (textbook p. 274): find the index
A watch cost RM500 in 2017 and RM550 in 2018. Index for 2018 based on 2017?I = 550 ÷ 500 × 100 = 110. The price increased by 10%.
Example 2 (textbook p. 274): find the base value
Registrations in 2017 were 893, and the 2017 index based on 2010 is 156.39.156.39 = 893 ÷ Q0 × 100, so Q0 = 893 × 100 ÷ 156.39 = 571.
Which year is the base? "Index in 2018 based on 2017" puts 2017 in the denominator. Swapping the years gives the wrong answer.
The price of rice was RM2.50 in 2015 and RM3.00 in 2020. Find the index for 2020 based on 2015.
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I = 3.00 ÷ 2.50 × 100 = 120, an increase of 20%.
Changing the base year
When two indices share the same base year, divide them to get the index between the other two years.
I2/0 = I2/1 × I1/0 ÷ 100. In words: index of year 2 on year 0 = (index of year 2 on year 1 × index of year 1 on year 0) ÷ 100.
Example 3 (textbook p. 275): same base year
The price index of a bicycle in 2018 based on 2010 is 176, and in 2018 based on 2015 is 110. Find the index in 2015 based on 2010.Divide the two equations: Q2015/Q2010 = 176/110, so I = 176 ÷ 110 × 100 = 160.
| Given | Find | Working |
|---|---|---|
| 2016/2015 = 130, 2015/2012 = 120 | 2016/2012 | 130 × 120 ÷ 100 = 156 |
| 2019/2011 = 225, 2013/2011 = 150 | 2019/2013 | 225 ÷ 150 × 100 = 150 |
Alternative method (textbook p. 275): I2015/2010 = I2018/2010 ÷ I2018/2015 × 100, the same result.
The notation "2013 (2011 = 100)" means the 2013 index is based on 2011.
Solving problems
Follow the textbook's four steps: understand the problem, plan the strategy, carry it out, and conclude by checking.
Example 4 (textbook p. 276): repeated increase
17 721 visitors in 2016; 10% more in 2018; the same rate again from 2018 to 2020.2018: 17 721 × 110 ÷ 100 = 19 493. 2020: 19 493 × 1.10 = 21 442 (check: 21 442 ÷ 19 493 × 100 ≈ 110).
Two increases of 10% make 21% overall, not 20%. The index from 2016 to 2020 is 110 × 110 ÷ 100 = 121.
The index of a price in 2019 based on 2017 is 120, and in 2021 based on 2019 is 110. Find the index in 2021 based on 2017.
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120 × 110 ÷ 100 = 132.
10.2Composite index
Composite index: the average of several indices that takes into account how important each item is. The importance is the weightage (w). It can be given as numbers, ratios, percentages, or read from a bar chart or pie chart.
I = ∑Iiwi∑wi
Example 6 (textbook p. 280): weightage from a pie chart
Expenditure index 2017 based on 2011: water 135 (17%), electricity 140 (58%), fuel 125 (25%).I = (135×17 + 140×58 + 125×25) ÷ 100 = 13 540 ÷ 100 = 135.4.
Example 5 (textbook p. 280): no weightage given
Price indices of three fruits in 2018 based on 2010: 175, 120 and 160.I = (175 + 120 + 160) ÷ 3 = 151.67.
If no weightages are given, use the same weightage (1) for every item. The composite index is then the ordinary mean.
Example 7 (textbook p. 281): a complete problem
Steel materials (price 2010 → 2018, per tonne, weightage): iron 2 025 → 3 424 (72%), chromium 8 431 → 9 512 (18%), nickel 62 235 → 50 916 (10%).
| Material | Index Ii | wi | Iiwi |
|---|---|---|---|
| Iron | 169.09 | 72 | 12 174.48 |
| Chromium | 112.82 | 18 | 2 030.76 |
| Nickel | 81.81 | 10 | 818.10 |
| Total | 100 | 15 023.34 |
Do not average the indices without weights when weightages are given. The item with the largest weightage should pull the answer most.
Indices 110, 120 and 140 have weightages 2, 3 and 5. Find the composite index.
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(110×2 + 120×3 + 140×5) ÷ 10 = 1 280 ÷ 10 = 128.
Unknown weightage or index: put the symbol in the formula, multiply both sides by ∑wi and solve the equation. Using a "reduced by 10%" change: new index = old index × 0.9.