Problem Solving with Equations (Cambridge (CIE) IGCSE International Maths)

Revision Note

Written by: Amber

Reviewed by: Dan Finlay

Equations & Problem Solving

What is problem solving?

Problem solving means you are given a specific situation (real life or constructed) and you need to form and solve equations to find answers to the questions asked
The equations can be any type that are in the course, including
- Linear, e.g. $2 (x + 4) = 3 x$
- Quadratic, e.g. $x^{2} - 7 x + 12 = 0$
Answers must always be given in context
- This means related to the situation using words, phrases and units from the question
  - For example, "The population density is 225 people per square km" (not just $x = 225$ )

What type of algebra can come up in a problem solving question?

Many questions will require you to solve quadratic equations
- You need to be able to spot these
  - This may require bringing all the terms to one side to get "= 0"
- You are often free to choose which quadratic method to use to solve them
- If you get two solutions, you may need to justify which solution is correct
You may be asked to use algebra in other settings such as geometry or numbers
- $P$ % is $\frac{P}{100}$ as a decimal
- If the ratio $x : (x + 2)$ is equivalent to 5 : 8, then $\frac{x}{x + 2} = \frac{5}{8}$
You may have unfamiliar equations to solve
- such as multiplying both sides by $x$ in $\frac{12}{x} = 7 - x$ to form a quadratic

Examiner Tips and Tricks

If part (a) asks you to prove an equation and part (b) uses that equation, you can still do part (b) without having done part (a)!
- This means you won't lose all the marks if you can't do part (a)

Worked Example

The net of a cube is shown below.

Let $x$ cm be the side length of the cube.

Find an expression in terms of $x$ for

(a) the perimeter of the net shown.

Count the number of edges around the outside of the net

14 edges

Multiply this number by the side length, $x$

The perimeter of the net shown is $14 x$ cm

(b) the area of the net shown.

Count the number of faces (squares) in the net

6 faces

Multiply this number by the area of one face, $x \times x = x^{2}$

The area of the net shown is $6 x^{2}$ cm²

When the net is folded into a cube, the difference between the volume of the cube and the surface area of the net is eight times the perimeter of the net.

First find an expression for the volume of the cube

This will have dimensions $x \times x \times x$

The volume is $x^{3}$

Then find an expression for the difference between the volume and the surface area in part (b)

Subtract the surface area from the volume

$x^{3} - 6 x^{2}$

Set this difference equal to 8 times the perimeter of the net

Use the answer in part (a)

$x^{3} - 6 x^{2} = 8 \times 14 x$

Simplify $8 \times 14$ and bring all the terms to one side

$\begin{array}{rcl} x^{3} - 6 x^{2} & = & 112 x \\ x^{3} - 6 x^{2} - 112 x & = & 0 \end{array}$

You are nearly at the correct equation given in the question

Cancel both sides by $x$ (as $x$ cannot be zero)

To show this, you can factorise out an $x$ first

$x (x^{2} - 6 x - 112) = 0$

$x^{2} - 6 x - 112 = 0$

(d) Hence, find the exact volume of the cube when the net is folded.

Hence means use the previous results

The volume of the cube is $x^{3}$ which involves an unknown, $x$

To find $x$ , solve the equation in part (c), for example using the quadratic formula

$\begin{array}{rcl} x & = & \frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \times 1 \times (- 112)}}{2} \\ = & \frac{6 \pm \sqrt{484}}{2} \\ = & \frac{6 \pm 22}{2} \end{array}$

Find the two possible answers

14 or -8

The side length $x$ cannot be a negative number

$x = 14$

Substitute this into $x^{3}$ to find the volume

$14^{3}$

The question asks for the answer to be exact, so do not round

The volume is 2744 cm³

Last updated: 1 July 2024

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