Uncategorized

Daily Maths Revision – Week 1 Walkthrough

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

  1. Standard Form
  2. Simplifying Surds
  3. Negative Indices
  4. Proportion
  5. Changing the Subject of a Formula

Question 1: Standard Form

Standard form is a way to write very large or very small numbers more efficiently. We use powers of 10 to replace long lists of zeros.

Write 275 000 in standard form.

We look at the first significant figure, 2 – this represents 200 000. This is the same as 2 × 105.

So, 275 000 is equivalent to 2.75 × 105.

Write 0.00782 in standard form.

We look at the first significant figure again – this represents 0.007. This is the same as 7 × 10-3.

So, 0.00782 is equivalent to 7.82 × 10-3.

Question 2: Simplifying Surds

Surds are irrational numbers that can be written as square roots. To write them as simply as possible, we take out any factors that are square numbers.

Simplify \boldsymbol{\sqrt{50}}.

25 is both a factor of 50 and a square number, so we can write 50 as 25 × 2:

\sqrt{50} = \sqrt{25\times2}

We can then split this multiplication into two square roots and simplify:

\begin{aligned} \sqrt{25\times2} &= \sqrt{25} \times \sqrt{2} \\&=5\sqrt{2} \end{aligned}

We can use a similar technique when multiplying or dividing surds.

Simplify \boldsymbol{\frac{\sqrt{450}}{\sqrt{18}}}.

Let’s start by simplifying each of the surds. There are many ways of doing this, depending on what factors you can see. Here, we have split 450 into 9, 25 and 2.

\begin{aligned} \sqrt{450} &= \sqrt{9\times25\times2} \\ &= \sqrt{9} \times \sqrt{25} \times \sqrt{2} \\ &= 3 \times 5 \times \sqrt{2} \\&=15\sqrt{2} \end{aligned}

\begin{aligned} \sqrt{18} &= \sqrt{9\times2} \\ &= 3\sqrt{2} \end{aligned}

Now, we can simplify just like we do with fractions:

\begin{aligned} \frac{\sqrt{450}}{\sqrt{18}} &= \frac{15\sqrt{2}}{3\sqrt{2}} \\&= \boldsymbol{5} \end{aligned}

We can multiply and divide surds as normal but we can’t simply add and subtract them. However, we can simplify using addition and subtraction.

Simplify \boldsymbol{\sqrt{450} + \sqrt{18}}.

\begin{aligned} \sqrt{450} &=15\sqrt{2} \\ \sqrt{18} &= 3\sqrt{2} \end{aligned}

\begin{aligned} \sqrt{450} + \sqrt{18} &=15\sqrt{2} + 3\sqrt{2} \\&= \boldsymbol{18\sqrt{2}} \end{aligned}

Question 3: Negative Indices

Let’s start with a general rule and then we will look at how we apply that rule:

a^{\text{-}m} = \frac{1}{a^m}

This looks a bit complicated but it’s simpler than it looks. Here are some examples:

\begin{aligned} 9^{\text{-}1} &= \frac{1}{9} \\ 7^{\text{-}2} & = \frac{1}{49} \\ \left( \frac{2}{3} \right)^{\text{-}1} &= \frac{3}{2} \end{aligned}

As you can see with the last example, with a fraction, we find the reciprocal fraction (we swap the numerator and denominator).

Let’s find the value of \boldsymbol{\left( \frac{3}{5} \right)^{\text{-}3}}.

In this case, we will find the reciprocal fraction and then cube the result.

\begin{aligned} \left( \frac{3}{5} \right)^{\text{-}3} &= \left( \frac{5}{3} \right)^{3} \\ &= \frac{125}{27} \end{aligned}

Question 4: Proportion

Proportion questions are very logical – it’s all about keeping things balanced and doing the same thing to each side.

At a factory, 5 machines can make 3600 parts a day. Another 3 machines are bought. Given that all the machines work at the same rate, how many parts can the factory make in one day now?

Let’s set this out nice and neatly. It is sensible to include column headings to help organise your working. There are now 8 machines in total (the original 5 add the 3 new machines) so we want to know how many parts 8 machines can make.

Start by working out how many parts 1 machine makes then find how many parts 8 machines can make.

MachineNumber of Parts
53600
÷ 5÷ 5
1720
× 8× 8
85760

The factory can now make 5760 parts a day.

Question 5: 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 formula below to make \boldsymbol{x} the subject.

\boldsymbol{\frac{xy}{3}-am=b}

We want to get all the terms that include an x on one side and any terms without an x on the other side. We start by isolating the term that includes x .

\begin{aligned} &&\frac{xy}{3}-am&=b&&\\&+am&&&&+am \\ &&\frac{xy}{3}&=b+am\end{aligned}

Next, we multiply and divide the equation to get x on its own.

\begin{aligned} &&\frac{xy}{3}&=b+am&&\\&\times3&&&&\times3 \\ &&xy&=3(b+am) \\&\div y&&&&\div y \\ &&\boldsymbol{x}&=\boldsymbol{\frac{3(b+am)}{y}} \end{aligned}

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