DoRevision

Proportion and Rates of Change

Proportion, rates and growth are one idea: find the fixed link between two quantities. It is a constant ratio, a rate, or a repeated multiplier, and once you have it every answer follows.

⏱️ 20 min 🎯 16 activities
Best used for
Intervention Mock preparation Cover lesson

Get the method right under pressure

Free interactive practice on the steps that lose marks under exam pressure.

Start revising free

What you'll cover

Two quantities, one fixed link

This whole topic looks like several separate skills: sharing in a ratio, working with percentages, direct and inverse proportion, speed and density, compound interest. They are one idea wearing different clothes: two quantities are linked by a fixed rule, and the job is to find that rule and use it. There are three shapes that rule can take. A constant ratio. In direct proportion, one quantity is always a fixed number of times the other, written as y equals a constant times x. Double one and you double the other. An inverse link. In inverse proportion, one quantity is a fixed number divided by the other, written as y equals a constant divided by x. Double one and you halve the other, and the two multiplied together always give the same number. A rate. A rate is simply a proportion between two DIFFERENT quantities. Speed is distance per unit of time, density is mass per unit of volume, pressure is force per unit of area. Each is a fixed link with units attached. And a repeated multiplier. Compound growth or decay is the same percentage change applied again and again, so a total is multiplied by the same factor each step, never added to by the same amount. Carry one question through everything that follows: what is the fixed link between the two quantities, and as the first rises does the second rise or fall?

Words for how things change together

Five terms, each defined by what it is. They name the fixed links this topic is built on.

Tap the two that are direct proportion

Tap the TWO situations that are direct proportion, where one quantity rises as the other rises.

  • The more hours you work, the more pay you earn
  • The more identical items you buy, the greater the total cost
  • The more workers on a job, the less time it takes
  • The faster you drive, the less time a fixed journey takes

Find the value in direct proportion

y is directly proportional to x. When x is 4, y is 20. First find the constant by dividing, then use it. Work out y when x is 7.

Direct against inverse

Both are fixed links between two quantities. The difference is whether they move the same way or opposite ways.

Match each rate to what it measures

  • Speed
  • Density
  • Pressure
  • Rate of pay
  • Flow rate
  • distance travelled per unit of time
  • mass per unit of volume
  • force per unit of area
  • money earned per hour worked
  • volume passing per unit of time

Two things true of inverse proportion

Two quantities are in inverse proportion. Select the TWO statements that are true.

  • If one of them doubles, the other halves
  • The two quantities multiplied together give a constant
  • They both increase together
  • Their sum stays the same

The compound growth trap

A savings amount grows by 10 per cent each year. A student says that after three years it has grown by 30 per cent in total. Why is that wrong?

  • The growth compounds: each year multiplies by 1.1, so after three years the amount is multiplied by 1.1 three times, which is about 33 per cent more, not 30
  • The real growth is less than 30 per cent because some interest is lost
  • It is right, because 10 plus 10 plus 10 is 30
  • The amount does not change at all over three years

Answering a proportion or rate question

A method for the whole topic. Decide which fixed link it is. Do the two quantities rise together, which is direct proportion, or does one rise as the other falls, which is inverse? Or is it a rate, one quantity per another, or growth that repeats? Find the constant. For a proportion, use the pair of values you are given to find the constant. For a rate, that constant is the rate itself. For compound growth, it is the multiplier. Use it. Substitute the value you are asked about and work out the answer, keeping the units if it is a rate. Two habits cost marks. The first is treating compound growth as repeated addition. Adding the same amount each year is a different, simpler pattern; compound growth multiplies by the same factor each time, on a total that keeps getting bigger. The second is mixing up direct and inverse: always check whether the second quantity should go up or down as the first goes up. At Higher you may be asked to set up the relationship yourself, writing y equals a constant times x for direct proportion or y equals a constant divided by x for inverse, then finding the constant from a given pair of values.

Grow it year on year

A sum of 200 pounds grows by 10 per cent each year, so each year it is multiplied by 1.1. After 1 year it is 220 pounds. Work out its value after 2 years, in pounds.

Order how to solve a proportion problem

Put the steps of solving a proportion problem into a sensible order.

  • Decide whether the two quantities are in direct or inverse proportion
  • Use the pair of values you are given to find the constant
  • Write the relationship, such as y equals the constant times x
  • Substitute the new value you are asked about
  • Work out the answer, and add the correct units if it is a rate

Build the proportion rule

This is the rule that separates the two kinds of proportion. Assemble it.

Why compound is not just adding

Here is why growth that repeats is not the same as adding the same amount each time. Suppose 100 pounds grows by 10 per cent a year. If you simply added the same amount each year, you would add 10 pounds every year: 110, then 120, then 130. That is what many people expect, and it is wrong, because after the first year the 10 per cent is taken of a larger amount. What really happens is that each year the total is multiplied by 1.1. Year one: 100 times 1.1 is 110. Year two: 110 times 1.1 is 121, not 120, because this time the growth is 10 per cent of 110, which is 11, not 10. Year three: 121 times 1.1 is 133 pounds and 10 pence, not 130. The gap looks small over three years, but it widens fast, because each year the growth is worked out on a bigger number than the last. That is the whole idea: the same repeated multiply, applied to a total that keeps getting larger. So whenever growth or decay repeats, reach for a multiplier and apply it again and again. Adding the same amount each time is a different, simpler pattern, and it is not what this kind of growth means.

Complete the proportion facts

When two quantities rise and fall together in a fixed ratio, they are in _____. When one rises as the other falls, they are in _____. A measure of one quantity per unit of another, such as speed or density, is a _____. The fixed number that links the two quantities in a proportion is the _____.

direct proportion inverse proportion rate constant of proportionality compound growth ratio percentage multiplier

Three proportion answers to sharpen

Three student answers about proportion and rates. In each case the reasoning is what needs work.

  • A student sees that more taps fill a tank faster and calls it direct proportion. What is the mistake?
  • A student works out compound growth by adding the same percentage on again and again. How would you correct it?
  • A student gives a speed as just the number 60, with no units. Why does that lose marks?

Explain how proportion and rates work

A friend muddles direct and inverse proportion and thinks compound growth is just adding a percentage each year. Explain how proportion and rates really work.

  • Explain what direct proportion is, and what happens to one quantity when the other doubles
  • Explain what inverse proportion is, and how it differs from direct proportion
  • Explain what a rate is, using speed or density as an example, and why the units matter
  • Explain why compound growth multiplies rather than adds, and what that does over several years
  • Finish with the one question to ask at the start of any such problem, about the fixed link between the two quantities