Secondary Worksheets & Practice Questions

15 Algebra Questions And Practice Problems (KS3 & KS4): Harder GCSE Exam Style Questions Included

April 2, 2024 | 2 min read

Beki Christian

Algebra questions involve using letters or symbols to represent unknown values or values that can change. Here you will find 15 algebra questions to test your knowledge and show you the different ways that algebra can be used to solve a problem to find an unknown value or to make generalisations.

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Algebra in KS3 and KS4

There are many topics and techniques within algebra. In KS3 we learn to write and manipulate basic algebraic expressions and linear equations. In KS4 we develop these techniques to allow us to deal with more complicated algebra problems such as ones that involve quadratic equations or a system of equations.

How to solve algebraic questions

When you are presented with an algebraic problem it is important to establish what you are being asked to do. Here are some of the key terms along with what they mean:

Solve the equation – find out the value of the unknown
Substitute – put the values you have been given into the algebraic expression
Simplify – collect together like terms to make the expression or equation look simpler
Expand brackets – multiply out the brackets
Factorise – put into brackets
Make x the subject – rewrite the equation in the form x =…..

Download this 15 Algebra Questions And Practice Problems (KS3 & KS4) Worksheet

Help your students prepare for their Maths GCSE with this free Algebra worksheet of 15 multiple choice questions and answers.

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Algebra in KS2

The ideas of writing and simplifying expressions, solving equations and substitution are introduced in KS2. Here are some example KS2 algebra questions:

Algebra questions KS2

c+d

2c+d

2c+2d

2c-2d

2 chocolate bars would cost 2 lots of c, or 2c, and 2 drinks would cost 2 lots of d, or 2d.

10m

6m+4

6m+6

6m-4

We need to collect together like terms here so 4m + 2m = 6m and 5 – 1 = 4 (watch out for the negative).

Algebra questions KS3

In KS3 we learn a variety of different algebra techniques to answer algebra questions and to practise problem solving with algebra. These include:

Simplifying algebraic expressions
Expanding brackets and factoring
Forming algebraic equations from word problems
Solving algebraic equations and inequalities
Substituting into expressions
Changing the subject of an equation
Working with real life graphs and straight line graphs
Sequences

A slide guiding pupils through the equation of a straight line from Third Space Learning's online intervention. — A slide guiding pupils through the equation of a straight line from Third Space Learning’s online intervention.

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Algebra questions KS3: basic algebra

8a-2b

12a-b

12a+5b

10a+b

basic algebra question 1b

£24

£67

£48

£27

In this question, n is 7 so we can substitute 7 into the formula.

$Charge = £20 + 4 \times 7$

$Charge = £48$

$4x$ and $6x$

$4$ and $x-6$

$2$ and $2x-3$

$2x$ and $2x-3$

There are two ways of attempting this question. We know that $area \;of \;a \;rectangle = length × width$ so we could multiply each pair together to see which pair makes 4x − 6.

\begin{aligned} &4x × 6x \quad \quad\quad\; \;24x^2 \\ &4 × (x − 6) \quad \quad \quad4x − 24\\ &2 × (2x − 3) \quad \quad \;\;4x − 6\\ &2x × (2x − 3) \quad \quad 4x^2 − 6x\\ \end{aligned}

Alternatively, if we factorise 4x − 6 we get 2(2x − 3) meaning the sides could be 2 and 2x − 3.

C=\frac{5(F-32)}{9}

C=\frac{5F-32}{9}

C=\frac{5F}{9}-32

C=5F-\frac{32}{9}

\begin{aligned} F&=\frac{9C}{5}+32 \hspace{3cm} &\text{subtract 32}\\\\ F-32&=\frac{9C}{5} &\text{multiply by 5}\\\\ 5(F-32)&=9C &\text{divide by 9}\\\\ \frac{5(F-32)}{9}&=C \end{aligned}

Algebra questions KS3: forming and solving equations

20^{\circ}

26^{\circ}

8^{\circ}

34^{\circ}

The angles in a triangle add up to $180^{\circ}$ therefore we can write

4x+2x-10+3x-8=180

Now we have an equation we can solve.

\begin{aligned} 9x-18&=180 \hspace{3cm} &\text{add } 18\\ 9x&=198 &\text{divide by } 9\\ x&=22^{\circ} \end{aligned}

The angles are :

\begin{aligned} 4\times22&=88^{\circ}\\ 2\times22 -10&=34^{\circ}\\ 3\times22 -8&=58^{\circ} \end{aligned}

The smallest angle is $34^{\circ}$ .

70

42

22

35

To solve this we need to write an equation.

Let Jamie’s age now be $x$ . Then Jamie’s dad’s age is $4x$ .

In $14$ years time Jamie’s age will be $x + 14$ and Jamie’s dad’s age will be $4x + 14$ .

Since we know Jamie’s dad’s age will be two times Jamie’s age, we can write

$4x+14=2(x+14)$

Now we have an equation we can solve.

\begin{aligned} 4x+14&=2(x+14) \hspace{3cm} &\text{expand the brackets}\\ 4x+14&=2x+28 \hspace{3cm} &\text{subtract 2x}\\ 2x+14&=28 \hspace{3cm} &\text{subtract 14}\\ 2x&=14 \hspace{3cm} &\text{expand the brackets}\\ x&=7 \hspace{3cm} \end{aligned}

Jamie is currently $7$ years old meaning his dad is $28$ years old. The sum of their ages is $35$ .

Algebra questions KS3: graphs

y=2x-1

y=2x+1

y=4x-2

y=2x+5

At the point (2, 5), x is 2 and y is 5. We can check which equation works when we substitute in these values:

\begin{aligned} y&=2x−1 \quad \quad \quad 5=2×2−1 \quad \quad \text{False}\\ y&=2x+1 \quad \quad 5=2×2+1 \quad \quad \text{True}\\ y&=4x−2 \quad \quad \quad 5=4×2−2 \quad \quad \text{False}\\ y&=2x+5 \quad \quad 5=2×2+5 \quad \quad \text{False} \end{aligned}

Algebra questions KS4

Algebra is studied extensively in the GCSE and IGCSE curriculum.

In KS4 we build on the techniques learnt in KS3. Topics include:

Expanding and factorising polynomials
Solving quadratic equations
Solving simultaneous equations
Inequalities
Algebraic fractions
Further work on graphs
Functions

One of the GCSE algebra slides on Third Space Learning's online intervention, guiding students through factorising quadratics. — One of the GCSE algebra slides on Third Space Learning’s online intervention, guiding students through factorising quadratics.

Algebra questions KS4: algebraic manipulation

\frac{1}{2}(a-b)

ab

b^2

b - 4a

\begin{aligned} &\frac{1}{2}(a−b) \quad \quad \quad 4 \\\\ &ab \quad \quad \quad \quad-15 \\\\ &b^2 \quad \quad \quad \quad \quad 9 \\\\ &b−4a \quad \quad \quad -23 \end{aligned}

30x^3-49x^2+4x+3

30x^3-46x^2-8x-3

30x^3-3

30x^3+3

Volume of a cuboid = length \times width \times height

Volume $= (5x+1)(2x-3)(3x-1)$

Volume $= (10x^2+2x-15x-3)(3x-1)$

Volume $= (10x^2-13x-3)(3x-1)$

Volume $= 30x^3-39x^2-9x-10x^2+13x+3$

Volume $= 30x^3-49x^2+4x+3$

Algebra questions KS4: forming and solving equations

16cm

24cm

12cm

9cm

The area of a triangle is $area = \frac{1}{2} × b × h.$

If we fill in what we know we get:

\begin{aligned} 24&=\frac{1}{2}\times 6 \times (3x-1) \hspace{3cm} &\text{simplify}\\\\ 24&=3(3x-1) &\text{multiply out the brackets}\\\\ 24&=9x-3 &\text{add 3}\\\\ 27&=9x &\text{divide by 9}\\\\ x&=3 \end{aligned}

Since $x = 3$ , the side lengths are $6m, 8cm$ and $10cm$ .

The perimeter is $6 + 8 + 10 = 24cm$ .

$x=-2$ or $x=15$

$x=-3$ or $x=5$

$x=-5$ or $x=3$

$x=-15$ or $x=2$

We can make this a bit easier by getting rid of the fraction involving x. We do this by multiplying each term by x.

\begin{aligned} x+2-\frac{15}{x}&=0 \hspace{3cm} &\text{multiply by x}\\\\ x^{2}+2x-15&=0 &\text{factorise}\\\\ (x+5)(x-3)&=0 &\text{solve}\\\\ &x=-5 \text{ or } x=3 \end{aligned}

£10

£8

£12

£6

We can write simultaneous equations to solve this.

$2a+3c=48$ (Equation 1)
$3a+c=44$ (Equation 2)

Multiply equation 2 by 3 to make the coefficients of c equal: $9a+3c=132$ (Equation 3)

Subtract equation 1 from equation 3:
$7a= 84$
$a=12$

Substitute a into equation 3:
$3\times12+c=44$
$36+c=44$
$c=8$

The cost of an adult ticket is $£12$ and a child ticket is $£8$ .

Algebra questions KS4: graphs

3y=x+7

y=2x-2

y=\frac{1}{2}x-9

2y=3x+8

For two lines to be parallel, their gradient must be equal.

If we rearrange $2y=x+7$ to make $y$ the subject we get $y=\frac{1}{2}x+\frac{7}{2}.$

The gradient is $\frac{1}{2}$

1

-2

5

0

To find the minimum value we need to complete the square.

\begin{aligned} f(x)&=x^2+4x+5\\ f(x)&=(x+2)^2-4+5\\ f(x)&=(x+2)^2+1 \end{aligned}

The minimum value is 1. This occurs when $(x+2)$ is 0.

y=5x+5

y=3x+4

y=\frac{-3}{4}x+3

y=\frac{-3}{4}x+\frac{21}{4}

To work out the gradient of the line we need to work out the gradient of the normal.

We know that the normal goes through the points (0, 0) and (3, 4) so we can calculate the gradient: $\frac{4-0}{3-0}=\frac{4}{3}.$

The gradient of the tangent will be $\frac{-3}{4}.$

We can now use $y=mx+c$ . We know the tangent goes through the point (3, 4) and that it’s gradient is $\frac{-3}{4}$ .

Therefore

\begin{aligned} 4&=\frac{-3}{4} \times 3 +c\\\\ 4&=\frac{-9}{4}+c\\\\ 4+\frac{9}{4}&=c\\\\ \frac{25}{4}&=c \end{aligned}

y=\frac{-3}{4}x+\frac{25}{4} \text{ or } 4y=-3x+25

Looking for more algebra questions and resources?

Third Space Learning’s free GCSE maths resource library contains detailed lessons with step-by-step instructions on how to solve algebra problems, as well as worksheets with algebra practice questions and more GCSE exam questions.

Take a look at the Algebra lessons today – more are added every week.

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