Daily Maths Revision – Week 10 Walkthrough

This image has an empty alt attribute; its file name is 5-a-day-blog-images-Grade-3-5-Social-Media-Blog.jpg

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

The formulae for the volume or curved surface area of a cone are given in the exam (and in the booklet).

This week, we’re exploring:

  1. Bearings
  2. Standard Form Calculations
  3. Cones
  4. Simultaneous Equations
  5. Two-Way Tables

Question 1: Bearings

Bearings are a particular way of measuring angles. There are 3 rules you need to remember when working with bearings:

1. Bearings must be measured from north.
2. Bearings must be measured clockwise.
3. Bearings must be three digits.

You may be asked for the bearing of A from B. You will always measure the angle at the point the bearing is from.

Find the bearing of X from Y.

The angle we need is marked in orange. This is measure clockwise, from North at Y.

First, we use co-interior angles to find the acute angle at Y. Remember, co-interior angles add up to 180°.

180 – 107 = 73°

This isn’t our final answer. The angle needs to be measured clockwise from north. To find the bearing, we need to subtract 73° from 360°.

360 – 73 = 287°

Now, quickly run through the checklist. The angle has been measured clockwise from north and is 3-digits long. If our answer was 2-digits long, we would add a 0 to the front of the angle. This won’t change the angle but will mean the bearing is 3-digits.

Question 2: Standard Form Calculations

When adding and subtracting numbers in standard form, the most reliable method is to convert the numbers to ordinary numbers and calculate from there.

For multiplication or division, however, it’s best to leave the numbers in standard form and use index laws to find the solution.

Let’s find the value of 2.5 × 105 + 8.4 × 106.

As we are adding, we convert the numbers to ordinary numbers:

2.5 × 105 = 250 000
8.4 × 106 = 8 400 000

Now, we can add these two numbers together. Use column addition and make sure that you align the numbers carefully:

   2 5 0 0 0 0
8 4 0 0 0 0 0
8 6 5 0 0 0 0

Finally, we convert this number back into standard form.

8 650 000 = 8.65 × 106

Now, let’s find 2 × 107 × 5.7 × 103.

Multiplication is commutative so we can do it in any order. When multiplying with standard form, start by multiplying the numbers then deal with the powers.

2 × 5.7 = 11.4
107 × 103 = 1010

So, 2 × 107 × 5.7 × 103 = 11.4 × 1010

However, this number isn’t in standard form. To be in correct standard form, the first number must be between 1 and 10. To adjust this number, we divide 11.4 by 10 and, to keep it balanced, times 1010 by 10:

11.4 × 1010 = 11.4 ÷ 10 × 1010 × 10
                   = 1.14 × 1011

Question 3: Cones

You are given the formula for both the curved surface area and volume of a cone. So, finding these values is mostly about substitution. Just make sure you are using the radius for r, the slant height for l and the perpendicular height for h.

\text{volume} = \frac{1}{3}\pi r^2h
\text{curved surface area} = \pi rl

Let’s find the volume and total surface area of this cone. We will give both answers in terms of \boldsymbol{\pi} .

The slant height is 13cm and the perpendicular height is 12cm. 10cm is the diameter but we want the radius, so we simply divide it by 2.

\begin{aligned} r &= 10 \div 2 \\ &= 5\text{cm} \\ h &= 12\text{cm} \\ l &= 13\text{cm} \end{aligned}


\begin{aligned} V &= \frac{1}{3}\pi r^2h \\ &= \frac{1}{3}\pi \times 5^2 \times 12  \\ &= \boldsymbol{100\pi}\text{ cm}^2 \end{aligned}

Surface Area

We want to find the total surface area but the formula we are given only calculates the curved surface area. To find the total surface area, we need to add the base of the cone – a simple circle with the same radius as the cone.

\begin{aligned} \text{curved surface area} &= \pi r l \\ &= \pi \times 5 \times 13 \\ &= 65\pi \text{ cm}^2 \end{aligned}

\begin{aligned} \text{base of cone} &= \pi r^2 \\ &= \pi \times 5^2 \\ &= 25\pi \text {cm}^2\end{aligned}

\begin{aligned} \text{total surface area} &= 65\pi + 25\pi \\&= \boldsymbol{90\pi} \text{ cm}^2 \end{aligned}

Question 4: 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. 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 y -coefficient of 15. It can help to number the equations to help keep track of them:

\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 for x .

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

Finally, we substitute this 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=3}.

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 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). 30 people chose running. Of the 74 people who chose cake, 10 also chose running. Use this information to complete the two-way table below.

Start by adding any information that we are given in the question:

Now, we can calculate each of the missing numbers – it’s like solving a simple puzzle.

Total pie = 100 – 74
                 = 26

Pie and running = 30 – 10
                               = 20

Total swimming = 100 – 30
                               = 70

Cake and swimming = 74 – 10
                                        = 64

Let’s add these four numbers in and see what’s left:

The only missing piece of information is pie and swimming. We can find this by calculating 26 – 20 or 70 – 64. Both of these will give the answer 6.

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.

Leave a Reply