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

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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.

## Algebra in KS3 and KS4

Algebra questions ks2, you may also like:, algebra questions ks3: forming and solving equations, algebra questions ks3: graphs, algebra questions ks4: algebraic manipulation, algebra questions ks4: forming and solving equations, algebra questions ks4: graphs, looking for more algebra questions and resources, looking for more ks3 and ks4 maths questions.

Free GCSE maths revision resources for schools As part of the Third Space Learning offer to schools, the personalised online GCSE maths tuition can be supplemented by hundreds of free GCSE maths revision resources including: – GCSE maths past papers – GCSE maths worksheets – GCSE maths questions – GCSE maths topic list

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 =…..

Remember, when working with algebra, we must still apply BODMAS / BIDMAS. i.e. B rackets, I ndices (powers, exponents, square roots), D ivision, M ultiplication, A ddition, S ubtraction.

When working with algebraic expressions and equations we must consider carefully which operations to deal with first.

## 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.

## 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:

1. A chocolate bar costs c pence and a drink costs d pence. Write down an expression for the cost of 2 chocolate bars and 2 drinks.

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

2. Simplify the expression 4m+5+2m-1

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

- Year 6 Maths Test
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## Algebra questions KS3: basic algebra

1. In this pyramid, you add two adjacent blocks to find the value of the block above.

What expression will be in the top box?

2. Brian is a window cleaner. He uses the following formula to calculate the amount to charge his customers:

Charge = £20 + 4n

Where n is the number of windows a house has.

If a house has 7 windows, how much would Brian charge?

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

Charge = £20 + 4 × 7

Charge = £48

3. The area of a rectangle is 4x-6.

Which of the following pairs could be the length and width of the rectangle?

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.

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

4. The formula for changing degrees Celsius to degrees Fahrenheit is

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

Rearrange this formula to make C the subject.

5. Work out the size of the smallest angle.

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

Now we have an equation we can solve.

The angles are :

The smallest angle is 34^{\circ} .

6. Jamie’s dad is 4 times older than Jamie. In 14 years time, Jamie’s dad will be twice the age of Jamie.

What is the sum of Jamie’s age now and Jamie’s dad’s age now?

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)

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

7. Which of the following lines passes through the point (2, 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:

## 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

8. Which of the following expressions has the smallest value when a=5 and b=-3?

9. Find an expression in terms of x for the volume of this cuboid

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

10. The area of this triangle is 24cm^2 .

Work out the perimeter of the triangle.

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

If we fill in what we know we get:

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

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

11. Solve the equation x+2-\frac{15}{x}=0

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.

12. At a theme park the Jones family purchased 2 adult tickets and 3 child tickets for £48 . The Evans family purchased 3 adult tickets and 1 child ticket for £44 .

Calculate the cost of one child ticket.

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: 3×12+c=44 36+c=44 c=8

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

13. Which of these lines is parallel to the line 2y = x + 7

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}

14. Find the minimum value of the function f(x) = x^2+4x+5 .

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

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

15. The diagram shows the circle x^2+y^2=25 . The line is a tangent to the circle at the point (3,4) . Work out the equation of the line.

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} .

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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## Course: Algebra 1 > Unit 2

- Why we do the same thing to both sides: Variable on both sides
- Intro to equations with variables on both sides
- Equations with variables on both sides: 20-7x=6x-6

## Equations with variables on both sides

- Equation with variables on both sides: fractions
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- Your answer should be
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## Algebra through Puzzles

What's inside.

- Introduction
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- Arithmetic Logic and Magic
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- Equations and Unknowns
- Manipulating Exponents
- Algebra in Motion
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## Rational Functions

Piecewise functions, community wiki.

Browse through thousands of Algebra wikis written by our community of experts.

## Expressions and Variables

- What Makes A Good Problem?
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## Non-Linear Inequalities

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## Quadratic Equations (Parabolas)

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## Square Roots (Radicals)

- Simplifying Radicals
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## Arithmetic and Geometric Progressions

- Arithmetic Progressions
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## SAT Math : Algebra

Study concepts, example questions & explanations for sat math, all sat math resources, example questions, example question #92 : how to find the solution to an equation.

If 6 x = 42 and xk = 2, what is the value of k ?

Solve the first equation for x by dividing both sides of the equation by 6; the result is 7. Solve the second equation for k by dividing both sides of the equation by x, which we now know is 7. The result is 2/7.

## Example Question #93 : How To Find The Solution To An Equation

If 4 x + 5 = 13 x + 4 – x – 9, then x = ?

Start by combining like terms.

4 x + 5 = 13 x + 4 – x – 9

4 x + 5 = 12 x – 5

## Example Question #94 : How To Find The Solution To An Equation

If 3 – 3 x < 20, which of the following could not be a value of x ?

First we solve for x .

Subtracting 3 from both sides gives us –3 x < 17.

Dividing by –3 gives us x > –17/3.

–6 is less than –17/3.

## Example Question #1 : Algebra

Let x be a number. Increasing x by twenty percent yields that same result as decreasing the product of four and x by five. What is x?

The problem tells us that increasing x by twenty percent gives us the same thing that we would get if we decreased the product of four and x by five. We need to find expressions for these two situations, and then we can set them equal and solve for x.

Let's find an expression for increasing x by twenty percent. We could represent this as x + 20%x = x + 0.2x = 1.2x = 6x/5.

Let's find an expression for decreasing the product of four and x by five. First, we must find the product of four and x, which can be written as 4x. Then we must decrease this by five, so we must subtract five from 4x, which could be written as 4x - 5.

Now we must set the two expressions equal to one another.

6x/5 = 4x - 5

Subtract 6x/5 from both sides. We can rewrite 4x as 20x/5 so that it has a common denominator with 6x/5.

0 = 20x/5 - 6x/5 - 5 = 14x/5 - 5

0 = 14x/5 - 5

Now we can add five to both sides.

Now we can multiply both sides by 5/14, which is the reciprocal of 14/5.

5(5/14) = (14x/5)(5/14) = x

The answer is 25/14.

## Example Question #96 : How To Find The Solution To An Equation

If 4 xs = v, v = ks , and sv ≠ 0, which of the following is equal to k ?

This question gives two equalities and one inequality. The inequality ( sv ≠ 0 ) simply says that neither s nor v is 0. The two equalities tell us that 4 xs and ks are both equal to v, which means that 4 xs and ks must be equal to each other--that is, 4 xs = ks . Dividing both sides by s gives 4 x = k , which is our solution.

## Example Question #97 : How To Find The Solution To An Equation

If 2x 2 (5-x)(3x+2) = 0, then what is the sum of all of the possible values of x?

Since we are told that 2x 2 (5-x)(3x+2) = 0, in order to find x, we must let each of the factors of our equation equal zero. The equation is already factored for us, which means that our factors are 2x 2 , (5-x), and (3x+2). We must let each of these equal zero separately, and these will give us the possible values of x that satisfy the equation.

Let's look at the factor 2x 2 and set it equal to zero.

Then, let's look at the factor 5-x.

Add x to both sides

Finally, we set the last factor equal to zero.

Subtract two from both sides

Divide both sides by three.

This means that the possible values of x are 0, 5, or -2/3. The question asks us to find the sum of these values.

0 + 5 + -2/3

Remember to find a common denominator of 3.

15/3 + -2/3 = 13/3.

The answer is 13/3.

## Example Question #98 : How To Find The Solution To An Equation

If bx + c = e – ax, then what is x?

To solve for x:

bx + c = e – ax

bx + ax = e – c

x(b+a) = e-c

x = (e-c) / (b+a)

√( x 2 -7) = 3

To solve, remove the radical by squaring both sides

(√( x 2 -7)) 2 = 3 2

## Example Question #100 : How To Find The Solution To An Equation

(√3x) 2 = 9 2

x = 81/3 = 27

## Example Question #1 : Linear / Rational / Variable Equations

√(8y) + 18 = 4

First, simplify the equation:

√(8y) = -14

Then square both sides

(√8y) 2 = -14 2

y = 196/8 = 24.5

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Algebra has been developed over thousands of years in several different countries. The earliest methods for solving mathematical problems with one or more unknown quantities come from ancient Egypt.

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