15 Algebra Questions And Practice Problems (KS3 & KS4): Harder GCSE Exam Style Questions Included (2023)

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

Charge = £20 + 4 × 7

Charge = £48

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.

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} .

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 .

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}

Volume of a cuboid = length × width × 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

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 .

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}

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 .

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}

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.

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