Daily Maths Revision – Week 11 Walkthrough

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

This week, we’re exploring:

  1. Error Intervals
  2. Averages from a Grouped Table
  3. Enlargement
  4. Expanding and Simplifying Brackets
  5. Drawing Straight Line Graphs

Question 1: Error Intervals

An error interval uses an inequality to show all the numbers that a rounded number could have been.

Let’s say a number, 𝑥, is rounded to 1 decimal place and the result is 3.4.

To find the error interval, we need to think about the smallest number that would round up to 3.4 and the largest number that would round down to 3.4.

Let’s think about the smallest number first – this is called the lower bound. When rounding, we look at the number to the right of what we’re asked to round to – in this case that would be the number in the hundredth’s column. If this number is 4 or lower, we round down but if it’s 5 or higher, we round up. So, any number starting 3.34… would round down, but 3.35 would round up. So, our lower bound is 3.35. We can write 3.35 ≤ 𝑥.

Now, let’s think about the upper bound – the largest number that would round down to 3.4. As explained above, if the number in the hundredths is 4 or lower, we round down – 3.44 would round down to 3.4. However, this isn’t the largest number that rounds down: 3.445 would also round to 3.4, as would 3.449 and 3.4499. In fact, any number smaller than, but not equal to, 3.45 would round down. So, we use 3.45 as our upper bound and use inequality signs to show that it can’t equal 3.45. In other words, we write 𝑥 < 3.45

Putting these two together gives our error interval:
3.35 ≤ 𝑥 < 3.45

Question 2: Averages from a Grouped Table

When working with a grouped table, you could be asked to estimate the mean, find the modal class or find the class that contains the median. We are going to find each of these for the table below. This table shows the time spent revising in one week for 60 different students. 

Let’s start by finding the modal class. 

Remember, the mode is the most common result. Therefore, the modal class is the group with the highest frequency. In this case, the highest frequency is 24 so the modal class is the corresponding class: 10 < 𝑡 ≤ 15.

Next, let’s find the class containing the median.

This is the middle value when the data is in order. When data is in a frequency table, it is already ordered, so to find the central value we need to simply work out which of the values it is. To find which is the central value we add one to the total frequency (60) and then divide by 2. 

(60 + 1) ÷ 2 = 30.5
We are looking for the 30.5th value. Now, we simply add the frequencies up until we have passed this point:

So, the median is in the group 10 < 𝑡 ≤ 15

Finally, let’s estimate the mean. 

To find the mean, we add up all the data and divide by the total number of students. To add up the data in a frequency table, we need to multiply the times spent revising by their corresponding frequencies. Since the data is grouped, we don’t know the exact time. Instead, we use the midpoint as an estimate.

Now, we find the total of these values and divide it by the total number of students (60).

10 + 67.5 + 300 + 315 + 112.5 = 805
805 ÷ 60 = 13.4166….

So, the mean is 13.4 to 1 decimal place.

Question 3: Enlargement

When enlarging a shape, you will be given a scale factor and centre of enlargement. The scale factor tells us how much larger the shape needs to be, and the centre of enlargement tells us the point to enlarge it from.

Let’s enlarge the shape below by scale factor 3 with a centre (1, 2).  

First, we need to identify the centre of enlargement – the point (1, 2). Then, we will draw a straight line through each of this point and each of the vertices (corners). 

The enlarged triangle needs to be 3 times larger but also 3 times further away from the point (1, 2). So, we can find the vertices of the new shape by counting the distance from the point (1, 2).

For example, the right-angle is currently 1 square to the right of the centre of enlargement and 1 square up. So, the right-angle of the enlarged shape should be 3 squares to the right of the centre of enlargement and 3 squares up.

Once we have one point, we can complete the shape by enlarging each side by a factor of 3. The left-hand side will be 3 units and the base will be 6 units. Joining this together gives us our enlarged triangle.

Each vertex should sit on the rays we drew, if they don’t, then you need to check your answer.

If asked to describe a transformation, you can identify it as an enlargement if one of the shapes is larger than the other. You then need to specify that it’s an enlargement and give the scale factor and centre of enlargement.

Let’s describe the transformation that maps shape A onto shape B.

To identify the scale factor, we choose one side of shape A and compare it to the corresponding side of shape B.

We can see that the original shape has a height of 3 units and the enlarged shape has a height of 6 units. To find the scale factor, we divide the height of the enlarged shape by the height of the original shape. In this case, that’s 6 ÷ 3 = 2.

Finally, to find the centre of enlargement we draw a straight line through a vertex on Shape B and the corresponding vertex on shape A. Do this for at least 3 sets of vertices to make sure they all meet at the same point:

The point at which these lines meet is the centre of enlargement. In this case, it’s the point (1, 1). So, this transformation is an enlargement of scale factor 2 and centre (1, 1).

Question 4: Expanding and Simplifying Brackets

The trick to expanding and simplifying brackets is to be systematic. Expand each bracket in turn and then simplify the expanded expression. You may want to look back at week 1 to remind yourself how to expand single brackets. 

Let’s expand and simplify 3(5𝑥 – 7) – 4(2𝑥 – 3).

First, we will set up a grid for each of the sets of brackets. Remember to take the signs with the terms – we need to use -4 for the second bracket.

Now, we multiply the term in each row by the term in each column.

Now, we write out the white boxes as a single expression and simplify.

3(5𝑥 – 7) – 4(2𝑥 – 3) = 15𝑥 – 21 – 8𝑥 + 12
                                    = 7𝑥 – 9

Question 5: Drawing Straight Line Graphs

When asked to draw a straight line graph, you will sometimes be given a table to complete. If you aren’t, you should draw a table of values.

Let’s draw the line with equation 𝑦 = 2𝑥 – 1.

We can see the graph will be a straight line because it’s in the form 𝑦 = 𝑚𝑥 + 𝑐. If you are given a table, it will usually look something like this:

If you aren’t given the table, you should draw one with at least 3 values of 𝑥. We need to substitute each value for 𝑥 into the equation 𝑦 = 2𝑥 – 1 to find the corresponding value for 𝑦.

You may find it easier to draw a larger table. You can use one row to find the value of 2𝑥 and then a second row to subtract 1 to give the value of 𝑦.

Our coordinates are the top and bottom value of each column: (1, 1), (2, 3), (3, 5) and (4, 7). We can plot them on a set of axes and join them to make a straight line.

As you can see, the points make a nice straight line. If this isn’t the case, you need to check your table of values for a mistake.

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