# Daily Maths Revision – Week 8 Walkthrough

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

This week, we’re exploring:

### Question 1: Box Plots

A box plot, sometimes called a box and whisker diagram, is a way of displaying data. Five vertical lines are marked; at the minimum value, the lower quartile, the median, the upper quartile and the maximum value.

For a reminder on how to find the upper and lower quartile, take a look at week 2. We will use the same set of values in the example below.

Draw a box plot for the following data.

2.7     3.2     4.5     4.7     5.1     5.6

First, we need to identify the median and the quartiles.

median = (4.5 + 4.7) ÷ 2
= 4.6
lower quartile = 3.2
upper quartile = 5.1

Now, we mark these three values and the maximum, 5.6, and minimum, 2.7, on a scale:

Then, we use horizontal lines to connect the central 3 lines to make a box and the maximum and minimum to the centre of the box:

This type of diagram is sometimes called a box and whisker plot – the centre area is the box and the lines either side are the whiskers.

### Question 2: Simultaneous Equations

Solving simultaneous equations is a tricky bit of algebra. However, there is a series of steps you can follow.

Make sure you set this work out neatly – it’ll be easier for both you and whoever is marking your work.

Let’s solve the simultaneous equations below.

\begin{aligned} \boldsymbol{6x-5y}&\boldsymbol{=9} \\ \boldsymbol{7x+3y}&\boldsymbol{=37} \end{aligned}

To be able to solve these equations, we need to eliminate one of the variables. We do this by multiplying the equations to get a pair of coefficients that are the same, ignoring the sign. In this case, we will get the $y$-coefficients to be the same.

The lowest common multiple of 3 and 5 is 15. So, we will multiply both equations to get a common $y$-coefficient of 15, again ignoring the sign.

\begin{aligned} 18x - 15y &= 27 \\ 35x + 15y &= 185 \end{aligned}

Now, we can add the two equations together. This will eliminate the $y$-variable. Make sure you are careful with negatives.

\begin{aligned} 18x &-15y &&=27 \\ 35x&+15y&&=185 \\ \hline 53x& && =212 \end{aligned}

Then, we solve the equation to find the value of $x$.

\begin{aligned} 53x &= 212\\ &= 4 \end{aligned}

Finally, we substitute this value of $x$ into one of the equations to find the value of $y$:

\begin{aligned} 7 \times 4 + 3y&=37\\ 28+3y&=37\\3y&=9\\y&=3 \end{aligned}

So, the solution to our simultaneous equations is $\boldsymbol{x=4}$ and $\boldsymbol{y=4}$.

Sometimes, you will get equations where the sign in the middle is the same. In this case, the process is almost identical – you just subtract the equations instead of adding them.

### Question 3: The Difference of Two Squares

In week 2, we looked at factorising expressions of the form $x^2+bx+c$. The difference of two squares is an extension of this – the expressions are of the form $x^2 - d^2$.

When factorising $x^2+bx+c$, we looked for a pair of numbers that multiplied together to give $c$ and added together to give $b$.

When the expression is in the form $x^2 - d^2$, we are looking for a pair of numbers that multiply together to give $- d^2$ and add together to give 0.

Let’s factorise $\boldsymbol{x^2 - 196}$.

We want to find a pair of numbers that multiply together to give -196 and add together to give 0. The only such pair is +14 and -14.

$x^2 - 196 = \boldsymbol{(x+14)(x-14)}$

Expressions of this form will always factorise into this form. It’s just a case of spotting that it is a difference of two squares! Knowing the first 15 square numbers can really help with this.

### Question 4: Circle Theorems

These are the two circle theorems that are relevant this week.

Given that AC is a diameter of the circle, find the size of angles ABC and BCD.

ABCD is a cyclic quadrilateral so the opposite angles will add up to 180°.

BCD = 180 – 102
= 78°

If AC is a diameter, then ABC is the angle in a semicircle – which is a right angle.

ABC = 90°

### Question 5: Two-Way Tables

Two-way tables are an excellent way to organise information. We carefully split up the information into its component parts – it’s all about setting it out in a clear, usable way.

100 people were asked to choose an activity (swimming or running) and a dessert (cake or pie). One person is chosen at random. Given that they chose swimming, find the probability that they also chose cake.

We are given that the person chose swimming. That means we are choosing from 70 people – those that chose swimming. Of these 70 people, 64 of them chose cake.

So, the probability that someone chose cake, given they chose swimming, is $\frac{64}{70}$.

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