Direct and Inverse Proportion Questions, Worksheets and Revision

Direct and Inverse Proportion Questions, Worksheets and Revision

GCSE 6 - 7GCSE 4 - 5KS3AQAEdexcelOCRWJECAQA 2022Edexcel 2022OCR 2022WJEC 2022

Direct and Inverse Proportion

In maths, we say that two quantities are proportional if as one changes, the other changes in a specific way. There are two types of proportionality that you need to be familiar with, direct and inverse proportion. Make sure you are happy with the following topics before continuing.

Level 4-5 GCSE KS3 AQA Edexcel OCR WJEC

Direct Proportionality – The Basics

If two quantities are directly proportional, then as one increases the other also increases at the same rate (proportionally), e.g. as one doubles, the other one also doubles.

Example: Jeremy uses \textcolor{blue}{400\text{ g}} of flour to make \textcolor{red}{8} muffins. How much flour would he need to make \textcolor{purple}{30} muffins?

Step 1: Divide the amount of flour by 8 to find the value of flour for 1 muffin.

\textcolor{blue}{400} \div \textcolor{red}{8} = 50 g of flour.

Step 2: Multiply the amount of flour needed for 1 muffin by the \textcolor{purple}{30} muffins needed.

\textcolor{purple}{30} \times 50 = 1500 g of flour.

Level 4-5 GCSE KS3 AQA Edexcel OCR WJEC

Inverse Proportionality – The Basics

If two quantities are inversely proportional, then as one increases the other decreases at the same rate (proportionally), e.g. as one doubles, the other one halves.

Example: \textcolor{blue}{6} builders can build \textcolor{red}{10} houses in 30 months. How long would it take \textcolor{purple}{18} builders to build the same number of houses?

Step 1: Multiply the number of months by the number of builders to get the time for 1 builder.

\textcolor{red}{10} houses would take 1 builder \textcolor{blue}{6} \times 30 = 180 months

Step 2: Divide the time it would take for 1 builder by the \textcolor{purple}{18} builders.

\textcolor{red}{10} houses would take \textcolor{purple}{18} builders 180 \div \textcolor{purple}{18} = 10 months.

Level 4-5 GCSE KS3 AQA Edexcel OCR WJEC
Level 6-7 GCSE AQA Edexcel OCR WJEC
Direct Proportionality Table Algebra

Direct Proportionality – with Algebra

If two quantities, y and x are directly proportional, we can write

y \textcolor{blue}{\propto} x

This reads as ‘y is directly proportional to x‘, where ‘\textcolor{blue}{\propto}‘ is the proportionality symbol. We can turn it into an equation by replacing \propto with = \textcolor{orange}k:

y = \textcolor{orange}{k}x

\textcolor{orange}{k} tells us how y and x are related, and is called the ‘constant of proportionality‘. Other examples of direct proportionality for y can be seen in the table.

Direct Proportionality Table Algebra
Level 6-7 GCSE AQA Edexcel OCR WJEC
Inverse Proportionality Table Algebra

Inverse Proportionality – with Algebra

If two quantities, y and x are inversely proportional, we can write

y \textcolor{blue}{\propto} \dfrac{1}{x}

This reads as ‘y is inversely proportional to x‘, where ‘\textcolor{blue}{\propto}‘ is the same proportionality symbol as before. We again turn this into an equation by replacing ‘\textcolor{blue}{\propto}‘ by ‘= \textcolor{orange}{k}‘:

 

y = \dfrac{\textcolor{orange}{k}}{x}

Other examples of inverse proportionality for y can be seen in the table.

Inverse Proportionality Table Algebra
Level 6-7 GCSE AQA Edexcel OCR WJEC
Level 6-7 GCSE AQA Edexcel OCR WJEC

Proportionality graphs

Once we convert the proportion into an equation, we can plot a graph easily.

Proportionality Graphs

Note: If you are a foundation student, you won’t explicitly be asked to form an equation. However, you will be expected to know how to work with constants of proportionality and use these equation, so all of this content is still important to you.

Level 6-7GCSEAQAEdexcelOCRWJEC
Level 6-7 GCSE AQA Edexcel OCR WJEC

Example 1: Direct Proportion

y is directly proportional to x. When y = \textcolor{red}{24}, x = \textcolor{blue}{8}. Work out the value of y when x = \textcolor{blue}{2}.

[3 marks]

Step 1: We have that y is directly proportional to x, i.e. y \propto x. Then, expressing this as an equation we get y = kx

Step 2: In the question we’re given that when y = \textcolor{red}{24}, x = \textcolor{blue}{8}. Substituting these into the equation above, we get

\textcolor{red}{24} = \textcolor{blue}{8} \times k

k = \textcolor{red}{24} \div \textcolor{blue}{8} = \textcolor{orange}3

Thus, the proportionality equation becomes, y = \textcolor{darkorange}{3}x

Step 3: We can now use this to work out that when x = \textcolor{blue}{2} y = \textcolor{darkorange}{3} \times \textcolor{blue}{2} = 6

Level 6-7 GCSE

Example 2: Inverse Proportion

y is inversely proportional to x. When y = \textcolor{red}{2}, x = \textcolor{blue}{3}. Work out the value of y when x = \textcolor{blue}{18}.

[3 marks]

Step 1: We have that y is inversely proportional to x, so we write y \propto \frac{1}{x}. Then, expressing this as an equation we get

y = \dfrac{k}{x}

Step 2: In the question we’re given that when y = \textcolor{red}{2}, x = \textcolor{blue}{3}. Substituting these into the equation above, we get

\textcolor{red}{2} = \dfrac{k}{\textcolor{blue}{3}}

k = \textcolor{red}{2} \times \textcolor{blue}{3} = \textcolor{orange}{6}

Thus, the proportionality equation becomes  y = \dfrac{\textcolor{darkorange}{6}}{x}

Step 3: Therefore, when x = \textcolor{blue}{18},

y = \dfrac{\textcolor{darkorange}{6}}{\textcolor{blue}{18}} = \dfrac{1}{3}

Level 6-7 GCSE AQA Edexcel OCR WJEC

Example Questions

F is inversely proportional to the square of r (i.e. r^2), so

F \propto \dfrac{1}{r^2}

We can rewrite this equation using k to represent the constant of proportionality:

F = \dfrac{k}{r^2}

We know that when F = 50, r = 3, so we can work out k by substituting these known values into the equation:

50 = \dfrac{k}{3^2}=\dfrac{k}{9}

Since we want to work out the value of k, we need to rearrange the formula to make k the subject which we can do by multiplying each side by 9:

k = 50 \times 9 = 450

We can now rewrite the original equation with a value for the constant k:

F = \dfrac{450}{r^2} = \dfrac{450}{12^2} = 3.125

Since y is inversely proportional to x, we can write this as an equation, as follows:

y \propto \dfrac{1}{x}

We can rewrite this equation using k to represent the constant of proportionality:

y = \dfrac{k}{x}

 

From the table, we know that when x = 4, y = 7.5. If we substitute these values into our equation, we can work out the value of the constant k.

Since

y = \dfrac{k}{x}

then

7.5 = \dfrac{k}{4}

 

Since we want to work out the value of k, we need to rearrange the formula to make k the subject which we can do by multiplying each side by 4:

\begin{aligned} k &= 7.5 \times 4 \\ k &= 30 \end{aligned}

 

We can now rewrite the original equation with a value for the constant k:

y = \dfrac{30}{x}

 

Now that we have calculated the value of the constant k and rewritten the original proportionality equation, we can work out the value of y when  x = 60:

When x = 60, y= \dfrac{30}{60} = 0.5

 

We can also work out the value of x when  y = 12:

When y = 1212 = \dfrac{30}{x}

If we multiply both sides by x and divide both sides by 12, we have rearranged the equation making x is the subject:

x = \dfrac{30}{12} = 2.5

Therefore when y = 12, x = 2.5

 

So, the completed table should look like this:

 

x and y Table

a) Since M is directly proportional to h, we can write this as an equation, as follows:

M \propto h

We can rewrite this equation using k to represent the constant of proportionality:

M = kh

 

We know that when M = \pounds155.80, h = 9.5 hours, so we can work out k by substituting these known values into the equation:

155.80 = k \times 9.5

 

Since we want to work out the value of k, we need to rearrange the formula to make k the subject which we can do by dividing each side by 9.5:

k = \dfrac{155.80}{9.5} = 16.4

 

We can now rewrite the original equation with a value for the constant k:

M = 16.4h

 

 

b) Now that we have calculated the value of the constant k and rewritten the original proportionality equation, we can work out the value of h when  M = \pounds688.80.

 

Since

M = 16.4h

then

\pounds688.80 = 16.4h

If

16.4h = \pounds688.80

then

h = \pounds688.80 \div16.4 = 42

 

So it will take Sasha 42 hours to earn £688.80.

a)  x is directly proportional to y, so

x \propto y

We can therefore write this as an equation where k is a constant:

x = ky

 

We now need to work out the value of the constant k which we can do by substituting in the known values for x and y:

Since

x = ky

then

2 = k \times 8

 

If we make k the subject of the formula by dividing both sides by 8, we can work out the value of k:

\dfrac{2}{8} = k

 

Therefore k has a value of \frac{2}{8}, which can be simplified to \frac{1}{4}.

 

Now that we know the value of the constant k, we can write the equation connecting x and y:

Since

x = ky

then

x = \frac{1}{4}y

or

x = \frac{y}{4}

 

 

b)  We now have an equation connecting x and y, so we can work out the value of x when y is 32.

 

Since

x = \frac{y}{4}

then

x = \frac{32}{4}

 

32 \div4 = 8

So

x = 8

 

 

c)  We now have an equation connecting x and y, so we can work out the value of y when x is 50.

 

Since

x = \frac{y}{4}

then

y = x \times 4

 

50 \times 4 = 200

So

y = 200

a)  Since the time taken (t) is inversely proportional to the square of the number of staff on duty (s), we can write a basic equation as follows:

t \propto \dfrac{1}{s^2}

This formula can be rewritten using k as a constant that connects t and s:

t = \dfrac{k}{s^2}

 

We can now work out the value of k if we substitute in the known values for t and s:

Since

t = \dfrac{k}{s^2}

then

20 = \dfrac{k}{4^2}

so

20 = \dfrac{k}{16}

 

To work out the value of k, we need to rearrange the formula, making k the subject.  We can do by multiplying both sides by 16:

20 \times 16 = k

So

k = 320

 

Since we now know the exact value of k, we can rewrite the proportionality formula:

Since

t = \dfrac{k}{s^2}

then

t = \dfrac{320}{s^2}

 

 

b)  It would be too easy to assume that if you double the staff, then the time taken would be halved!  Sadly, this is not the case.

 

If the number of staff is doubled, then there would be 8 staff members on duty instead of 4.  We know from part a) that k = 320 and we know that s = 8.  If we substitute these values into our formula, it should look like this:

Since

t = \dfrac{k}{s^2}

then

t = \dfrac{320}{8^2}

so

t = \dfrac{320}{64}

 

320 \div 64 = 5 minutes

 

Therefore, if you double the number of staff the orders take 5 minutes to be received instead of 20 minutes, so the orders are received 4 times faster.

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