O LevelMathsCambridge (CIE)Revision NotesAlgebra & SequencesForming & Solving EquationsProblem Solving with Equations

Problem Solving with Equations (Cambridge (CIE) O Level Maths): Revision Note

Exam code: 4024

Amber

Written by: Amber

Reviewed by: Dan Finlay

Updated on

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 open parentheses x plus 4 close parentheses equals 3 x

    • Quadratic, e.g. x squared minus 7 x plus 12 equals 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 equals 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 P over 100 as a decimal

    • If the ratio x space colon space open parentheses x plus 2 close parentheses is equivalent to 5 : 8, then fraction numerator x over denominator x plus 2 end fraction equals 5 over 8

  • You may have unfamiliar equations to solve

    • such as multiplying both sides by x in 12 over x equals 7 minus 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.

Cross shaped net of a cube

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 cross times x equals x squared

The area of the net shown is 6 x squared 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.

(c) Show that x squared minus 6 x minus 112 equals 0.

First find an expression for the volume of the cube

This will have dimensions x cross times x cross times x

The volume is x cubed

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 cubed minus 6 x squared

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

Use the answer in part (a)

x cubed minus 6 x squared equals 8 cross times 14 x

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

table row cell x cubed minus 6 x squared end cell equals cell 112 x end cell row cell x cubed minus 6 x squared minus 112 x end cell equals 0 end table

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

up diagonal strike x open parentheses x squared minus 6 x minus 112 close parentheses equals 0

x squared minus 6 x minus 112 equals 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 cubed which involves an unknown, x

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

table row x equals cell fraction numerator 6 plus-or-minus square root of open parentheses negative 6 close parentheses squared minus 4 cross times 1 cross times open parentheses negative 112 close parentheses end root over denominator 2 end fraction end cell row blank equals cell fraction numerator 6 plus-or-minus square root of 484 over denominator 2 end fraction end cell row blank equals cell fraction numerator 6 plus-or-minus 22 over denominator 2 end fraction end cell end table

Find the two possible answers

14 or -8

The side length x cannot be a negative number

x equals 14

Substitute this into x cubed to find the volume

14 cubed

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

The volume is 2744 cm3

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    Author
    Reviewer
    Amber

    Author: Amber

    Expertise: Maths Content Creator

    Amber gained a first class degree in Mathematics & Meteorology from the University of Reading before training to become a teacher. She is passionate about teaching, having spent 8 years teaching GCSE and A Level Mathematics both in the UK and internationally. Amber loves creating bright and informative resources to help students reach their potential.

    Dan Finlay

    Reviewer: Dan Finlay

    Expertise: Maths Subject Lead

    Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.