Daily Maths Revision – Week 3 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 are the ones that are relevant this week. You aren’t given these so you will need to remember them. It can be helpful to write the things you need to remember and stick them somewhere visible like in the bathroom!

Compound Interest
original amount × (multiplier)number of years

\text{speed} = \frac{distance}{time}

Trigonometry Ratios
\sin \theta = \frac{O}{H}        
\cos \theta = \frac{A}{H}        
\tan \theta = \frac{O}{A}      

You are given the trigonometric ratios and a formula for compound interest in the 2023 exams but it’s in a different format than your used to so it may be worth memorising them anyway.

This week, we’re exploring:

  1. Compound Interest
  2. Changing the Subject of a Formula
  3. Speed
  4. Product Rule for Counting
  5. Right-Angled Trigonometry

Question 1: Compound Interest

The key to calculating compound interest is percentage multipliers.

Let’s say a bank account containing £1000 earns compound interest of 1.5% a year. How much will be in the bank account after 4 years?

Each year, the amount will increase by 1.5%. If we are increasing by 1.5%, we are finding 101.5%. To find the multiplier, we divide this by 100:

(100 + 1.5) ÷ 100 = 1.015

If your calculator has a percentage button, you can use 101.5% for this calculation. However, not all calculators have this function, so we are going to use the method that always works.

After year 1, there will be 1000 × 1.015 (= £1015) in the account. In year 2, the additional £15 is also included when calculating the interest. So, after year 2, there will be 1015 × 1.015 (= £1030.23) in the account.

We can keep multiplying by 1.015 until we get to year 4.

Year 3: 1030.225 × 1.015 = £1045.68
Year 4: 1045.678… × 1.015 = £1061.36

Remember, money is always given to 2 decimal places. However, you shouldn’t round your answer until the end of your work as this could lead to a rounding error.

There is a quicker way. If you are good at remembering formulae, you can remember:

original amount × (multiplier)number of years

Let’s have a look at the same example with this formula.

Original amount= £1000
Percentage multiplier = 1.015
Number of years = 4

Total amount = 1000 × 1.0154
                        = £1061.36

Question 2: Changing the Subject of a Formula

To change the subject of a formula, we need to make sure we keep everything balanced. Just like when you are solving equations, whatever you do to one side you must do to the other.

Let’s rearrange the equation below to make \boldsymbol{x} the subject.


First, expand the brackets. It’s much easier to rearrange our formula without any brackets.

\begin{aligned} bx + ab &= a(m - x)\\bx+ab&=am-ax \end{aligned}

Now, we want to get all the terms that include an x on one side and any terms without an x on the other side.

\begin{aligned} &&bx+ab&=am-ax&&\\&+ax&&&&+ax\\&&bx+ab+ax&=am&&\\&-ab&&&&-ab\\&&bx+ax&=am-ab&&\end{aligned}

Next, we need to factorise the left-hand side; we can take out a common factor of x.

\begin{aligned} bx + ax &= am - ab\\x(b+a)&=am-ab \end{aligned}

Finally, we divide by the expression in the brackets:

\boldsymbol{x = \frac{am-ab}{b+a}}

Question 3: Speed

To find the speed of something, we divide the distance travelled by the time taken:

\text{speed} = \frac{\text{distance}}{\text{time}}

You can also use a formula triangle to help organise the formula – this is particularly helpful when you need to find the distance or time rather than the speed.

speed, distance, time

Let’s find the time taken for Jonathon to travel 500m at 6m/s.

In this case, we want to find the time taken so we cover this up in the formula triangle and use the other two variables. Since distance is over speed, we divide the distance by the speed:

time = distance ÷ speed
         = 500 ÷ 6
        = 83.33….
        = 83.3 seconds (1d.p.)

Make sure you include units in your answer – there is often a mark for this.

If the question asked for the distance travelled, we would use distance = speed × time. If the question asked for the speed, we would use
speed = distance ÷ time.

Question 4: Product Rule for Counting

The product rule for counting is an efficient way to find the total number of ways of choosing from different options.

Say there are three types of hot drink (tea, coffee and hot chocolate) and two types of cake (vanilla and chocolate). We could write out all the different ways to choose one hot drink and one cake:

tea and vanilla
tea and chocolate
coffee and vanilla
coffee and chocolate
hot chocolate and vanilla
hot chocolate and chocolate

Now, we can see there are 6 different combinations. However, if there were 30 hot drinks and 20 different cakes, writing out all the options would take too long! So, we use the product rule for counting.

There are 3 × 2 = 6 ways of choosing from 3 drinks and 2 cakes. Similarly, there are 30 × 20 = 600 ways of choosing from 30 drinks and 20 cakes.

A teacher chooses 2 different students from a class of 30 to complete a task. How many different ways can the students be chosen?

This question is a little more complicated because we are choosing 2 students from the same class. So, there are 30 ways to pick the first student but only 29 ways to pick the second student – you can’t pick the same student twice.

However, there are some combinations that are identical. For example, picking Bruce then Matilda is the same as picking Matilda then Bruce. So, we multiply 30 and 29 and then we divide by 2 to correct for this.

(30 × 29) ÷ 2 = 870 ÷ 2
                        = 435

Question 5: Right-Angled Trigonometry

Trigonometry is sometimes referred to as SOHCAHTOA – but SOHCAHTOA is a mnemonic that helps you remember one type of trigonometry, involving right-angled triangles. It gives the 3 trig ratios, sin, cos and tan. For example, Sinθ is the Opposite over the Hypotenuse – hence SOH.

We can also write these trig ratios using formula triangles.

To use these triangles, we simply cover up the value we are looking for.

Let’s find the length of the side labelled \boldsymbol{x} , giving your answer to 2 decimal places.

We start by labelling the sides. The longest side, which is the one opposite the right-angle, is the hypotenuse (H). The side opposite the angle we are given is the opposite (O). The side next to angle we are given is the adjacent (A).

We are given the adjacent and we are looking for the hypotenuse. So, we need to use the trig ratio that links A and H – that’s cos (CAH).

Now, we cover up the value we’re looking for – this leaves us with A above C. This means we divide the adjacent (A) by cos (C). In other words, if we call the angle \theta :

H = \frac{A}{\cos\theta}

We can substitute the values of A and \theta into this equation:

H = \frac{4.2}{\cos(35)}
      = \boldsymbol{5.13}cm (2d.p.)

Let’s try this question instead. Find the length of the side labelled \boldsymbol{x} , giving your answer to 1 decimal place.

Always start by labelling the sides. Just like before, identify the hypotenuse first, then the opposite and finally the adjacent.

This time, we are given the hypotenuse and we are trying to find the opposite. We need to use sin (SOH).

If we cover up O, we can see we multiply sin by the hypotenuse. 

O = H \times {\sin(\theta)}
      = 12 \times \sin(62)
      = \boldsymbol{10.6}cm (1d.p.)

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