# Daily Maths Revision – Week 19 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: Simplifying Ratio

Simplifying ratios is a similar method to simplifying fractions. We start by identifying a common factor for each part of the ratio and then divide both parts by this factor. If we use the highest common factor, we only need to do this process once but if not, we will need to repeat the process until there is no common factor other than 1.

Let’s simplify 60:96:84.

First, we will use the most efficient method – we will find the highest common factor of 60, 96 and 84. We can see that these three numbers are all in the 12 times table. So, we can divide each part of the ratio by 12.

60 ÷ 12 = 5
96 ÷ 12 = 8
84 ÷ 12 = 7

So, 60:96:84 simplifies to 5:8:7.

However, you may not be able to spot that all 3 parts are a multiple of 12. If this is the case, start by dividing by a factor you can spot. For example, they are all divisible by 2 so we can simplify to 30:48:42.

If using this method, at every stage you need to check if there are still common factors. All these numbers are still even, so we can divide each part by 2 again to get 15:24:21.

These numbers are all multiples of 3, so we divide each part by 3 to get 5:8:7. These numbers have no common factors, except 1, so we’re done.

### Question 2: Pie Charts

A pie chart shows the proportion of a group that share a certain characteristic. Usually, it doesn’t give information about the frequency of each characteristic – although this information can be added.

The table below shows the ages of some members of a tennis club. Draw an accurate pie chart to show this information.

To be able to draw the pie chart accurately, we need to start by finding the angle that will represent each section. First, we find the total frequency: 12 + 16 + 24 + 38 = 90

We know that there are 360° in a circle, so divide this by 90:      360 ÷ 90 = 4.

This means that each person is represented by 4°. To find the angle, we multiply each frequency by 4:

12 × 4 = 48°
16 × 4 = 64°
24 × 4 = 96°
38 × 4 = 152°

To check these answers, add them together. If they are correct, you should get 360°.

Now, we draw each angle accurately. Start with the vertical and then using a protractor complete each section in turn. When you’ve done this, label the sections and double check your angles.

Sometimes, you will need to use the frequency of a particular section to find the relationship between the frequency and angle.

### Question 3: Lowest Common Multiple and Highest Common Factor

This week, question 3 is a mixture of worded lowest common multiple and highest common factor questions. The biggest challenge with these questions is identifying if you need to find the lowest common multiple or the highest common factor.

A red light flashes every 6 seconds, a yellow light flashes every 10 seconds and a blue light every 9 seconds. They all flash together and a stopwatch is started. After how many seconds will all 3 lights flash together again?

First, we need to decide if we are looking for the lowest common multiple or the highest common factor. Let’s look at the red light – this will flash at 0 seconds (when the stopwatch starts) then at 6 seconds, 12 seconds, 18 seconds and so on. These are all the multiples of 6 so we are looking for the lowest common multiple.

We can do this by prime factorisation or by listing the multiples, depending on the size of the numbers. With numbers as small as this, we can easily write out the multiples until we identify the first number in all three lists. Write out the first 15:

Multiples of 6
6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, …

Multiples of 9
9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99, 108, 117, 126, 135, …

Multiples of 10
10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 140, 150, …

We can see that the first number that appears in all three lists is 90, so the lowest common multiple of 6, 9 and 10 is 90. The three lights will flash together again after 90 seconds.

James has two lengths of ribbon. One length of ribbon is 28cm long and the other length is 42cm long. James wants to cut lengths of ribbon into shorter lengths that are of equal length. James doesn’t want any ribbon left over. What is the longest possible length for each of the shorter lengths of ribbon?

In this case, we are dividing the ribbon into smaller parts, so we are looking for factors. More specifically, we are looking for the highest common factor of 28 and 42.

As before, we could use prime factorisation to do this, but it is probably more efficient to write out the factors.

Factors of 28:   1 and 28
2 and 14
4 and 7

Factors of 42:   1 and 42
2 and 21
3 and 14
6 and 7

As we can see, the highest common factor is 14 so James can cut the ribbons into 14cm lengths.

### Question 4: Solving Quadratic Equations by Factorising

Factorising is one way to solve quadratic equations. It doesn’t work for all quadratics but, if the question is on a non-calculator paper, the chances are you can solve the quadratic by factorising it.

Solve 𝑥2 – 7𝑥 = 30.

For this method to work, we need to rearrange the equation so it’s equal to 0.

𝑥2 – 7𝑥 = 30
– 30                                             – 30
𝑥2 – 7𝑥 – 30 = 0

Now, we factorise the left-hand side. You may want to look back at week 9 if you think you need to revisit this.

To factorise, we look for a pair of numbers that multiply together to give -30 and add together to give -7. The only pair of numbers that fit both requirements is -10 and 3. We put these into our two brackets.

(𝑥 – 10)( 𝑥 + 3) = 0

For two numbers to multiply together to give 0, one of them must be zero.

𝑥 – 10 = 0       or        𝑥 + 3 = 0
𝑥 = 10                         𝑥 = -3

So, the two solutions to 𝑥2 – 7𝑥 = 30 are 𝑥 = 10 and 𝑥 = -3.

### Question 5: Listing the Elements in a Set

A large part of this is knowing your set notation. If you need a refresher on this, check out this blog.

Given that A = {2, 4, 6, 8, 10, 11} and B = {2, 3, 5, 11}, write down the elements in the set A B.

Remember, A ∪ B is everything that’s in A or in B or in both.

A ∪ B = {2, 3, 4, 5, 6, 8, 10, 11}

If we drew this as a Venn diagram, this would be all elements in either circle or the section in the middle.

Given that A = {2, 4, 6, 8, 10, 11} and B = {2, 3, 5, 11}, write down the elements in the set A B.

Remember, A ∩ B is every element that’s in both A and B.

A ∩ B = {2, 11}

If we drew this as a Venn diagram, this would be all elements in section in the middle.

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