Daily Maths Revision – Week 2 Walkthrough

If you’re struggling with this week’s questions, this walkthrough should help you get the skills you need.

If it’s a formula you need, here’s the one that is relevant this week. You aren’t given this so you will need to remember it. It can be helpful to write the things you need to remember and stick them somewhere visible like in the bathroom!

Gradient

$\text{gradient} = \frac{\text{change in }y}{\text{change in }x}$

This week, we’re exploring:

Gradient
Spheres
Upper and Lower Quartiles
Reverse Percentages
Factorising into Double Brackets

Question 1: Gradient

The gradient of a line gives information about how steep the line is. It is the change in the $y$ -coordinate divided by the change in the $x$ -coordinate. If we have an equation in the form $y=mx+c$ , the gradient is the coefficient of $x$ ( $m$ ).

Let’s find the gradient of the line that passes through the points with coordinates (3, 4) and (7, 2).

Make sure you are consistent with which coordinates you use first – you are finding the change not the difference so your answer could be negative.

Change in $y$ : $4-2=2$
Change in $x$ : $3-7=\text{-}4$

$\begin{aligned} \text{gradient} &= \frac{2}{\text-4} \\ &=\text{-}\boldsymbol{0.5} \end{aligned}$

Question 2: Spheres

You are given the formula for both the surface area and the volume of a sphere. So finding these values is mostly about substitution. Just make sure you are using the radius – the distance from the centre to the surface of the sphere.

$\begin{aligned} \text{Volume} &= \frac{4}{3}\pi r^3 \\ \text{Surface area} &= 4\pi r^2 \end{aligned}$

Let’s find the volume and surface area of this sphere. We will give both answers to 1 decimal place.

14cm is the diameter and we want the radius. So, we simply divide it by 2.

$\begin{aligned} r &= 14\div2\\&=7\text{cm}\end{aligned}$

$\begin{aligned} V &= \frac{4}{3}\pi r^3 \\&= \frac{4}{3} \times \pi \times 7^3 \\&= \frac{1372}{3}\pi \\&= 1436.8 \text{cm}^3 \text{ (1d.p.)} \end{aligned}$

$\begin{aligned} \text{S.A.} &= 4\pi r^2 \\&= 4 \times \pi \times 7^2 \\&= 196\pi \\&= 615.8\text{cm}^2 \text{ (1d.p.)}\end{aligned}$

Your calculator will, at first, give your answer in terms of $\pi$ . It’s always a good idea to write this version down as well as a rounded version

Question 3: Upper and Lower Quartiles

The upper and lower quartile, along with the median, split the data into quarters. When finding the quartiles, it’s important you start by ordering the data.

Find the upper and lower quartiles of the data below.

4.7 3.2 4.5 5.1 2.7 5.6

Let’s start by ordering the numbers, starting with the smallest:

2.7 3.2 4.5 4.7 5.1 5.6
First, we find the median. Now we’ve ordered the data, it’s just the central number. In this case, there are two central numbers, 4.5 and 4.7, and so the median is the midpoint of 4.5 and 4.7:

Median = (4.5 + 4.7) ÷ 2
= 4.6

This splits the data in half:

2.7 3.2 4.5 | 4.7 5.1 5.6

The lower quartile is the median of the lower half of the data (3.2) and the upper quartile is the median of the upper half of the data (5.1).

Lower quartile = 3.2
Upper quartile = 5.1

Question 4: Reverse Percentages

These questions are sometimes called “original amount problems”. Essentially, we need to work backwards to find a value before a percentage change, when given the new amount and the percentage it changed by.

In a sale, the price of a cooker is reduced by 12%. The cooker now costs £333.96. Find the original price of the cooker.

There are two ways to approach this – the first method is based on finding what percentage of the original price the new price is.

The original price will always be 100%. If we reduce this by 12%, we have 100 – 12 = 88%. This is equal to the new amount. Now we know this, we can use the unitary method to find what value 100% is.

88% = 333.96
÷ 88 ÷ 88
1% = 3.795
× 100 × 100
100% = 379.5

So, the original amount (100%) is £379.50.

We could also use multipliers. To reduce something by 12%, we can multiply it by a certain number called a multiplier. To find the multiplier for a percentage decrease, you subtract the percentage from 100 and then divide by 100:

(100 – 12) ÷ 100 = 0.88

So, if we call the original price $r$ , then:

333.96 = 0.88 × $r$
$r$ = 333.96 ÷ 0.88
= £379.50

Question 5: Factorising into Double Brackets

Factorising is the opposite of expanding. It’s the easiest way to solve a quadratic equation, although you do also get questions that are just on factorising.

This week, we are looking at questions in the form $x^2 + bx + c$ . These will factorise into two brackets.

When this is the case, we look for two numbers that multiply together to give $c$ and add together to give $b$ . These are the two numbers that will go into our brackets.

Let’s factorise $\boldsymbol{x^2+7x+10}$ .

In this case, $b=7$ and $c=10$ . So, we are looking for two numbers that multiply to give 10 and add to give 7. You might be able to spot it straight away but, if you can’t, write out the factors of 10:

1 10
2 5

Now, we can see that the numbers we want are 2 and 5. We write these into our brackets:

$x^2+7x+10=\boldsymbol{(x+2)(x+5)}$

It doesn’t matter which way round the two brackets are; $(x+5)(x+2)$ would also be correct for this question.

Now, let’s factorise $\boldsymbol{x^2-5x-14}$ .

As you can see, we have negative numbers this time. This doesn’t change the process though; we are still looking for two numbers that multiply together to give -14 and add together to give -5. Make sure you include the signs when you find the values of $b$ and $c$ .

Let’s start by writing out the factors of -14. Since $c$ is negative, the signs of our two numbers must be different.

1          -14
2          -7
7          -2
14        -1

The only pair that add together to give -5 is 2 and -7. So, these are what we write in our brackets:

$x^2-5x-14=\boldsymbol{(x+2)(x-7)}$

Don’t forget to read even more of our blogs here and you can find our main Daily Maths Revision Page here! You can also subscribe to Beyond for access to thousands of secondary teaching resources. You can sign up for a free account here and take a look around at our free resources before you subscribe too.

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