Uncategorized

Daily Maths Revision – Week 4 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:

  1. Fractional Indices
  2. Standard Form
  3. The Equation of Parallel Lines
  4. Direct Proportion
  5. Rationalising Surds

Question 1: Fractional Indices

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

a^{\frac{1}{m}} = \sqrt[m]{a}

This looks a bit complicated but it’s simpler than it looks. If we have a number to the power of one half, we take the square root; if it’s to the power of one third, we take the cube root and so on.  Here are some examples:

\begin{aligned} 81^\frac{1}{2} &= 9 \\ 125^\frac{1}{3} &= 5 \\ \left(\frac{1}{32}\right)^\frac{1}{5}&=\frac{1}{2}\end{aligned}

On the last example, we take the fifth root. With a fraction, we take the fifth root of the numerator and the denominator.

Sometimes, you’ll have a fraction where the numerator isn’t 1.

Let’s find the value of \boldsymbol{25^\frac{3}{2}}.

In this case, we will find the square root and then we will cube the answer.

 \begin{aligned} 25^\frac{3}{2} &= \left(25^\frac{1}{2}\right)^3 \\&= 5^3 \\&=\boldsymbol{125} \end{aligned}

Question 2: 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 3: The Equation of Parallel Lines

Two (or more) lines are parallel if their gradients are the same. This means we can find the equation of a line that is parallel to another line and that passes through a set point.

Remember, the equation of any straight line can be written in the form y=mx+c where m is the gradient and c is the y-intercept.

Find the equation of the line that passes through the point (1, -3) and is parallel to the line with equation \boldsymbol{y=7x-2}

Since the equation of the line is in the form y=mx+c, we know the gradient is the coefficient of x, 7.

Now we have this gradient, we can substitute into the general form and

y=7x+c

To find the value of c, we substitute in the point we were given, (1, -8), and solve to find c.

\begin{aligned} \text{-}8 &= 7\times1 + c\\ \text{-}8 &=-7+c\\c&=\text{-}-7\\&=\text{-}15 \end{aligned}

Now we can use this to replace c in the equation we had above:

\boldsymbol{y=7x-15}

Question 4: Direct Proportion

Two (or more) values are directly proportional if they change at the same rate. If one doubles, the other also doubles.

For example, the time taken to fill some bottles and the number of bottles to fill. Assuming everything else stays the same, if there are twice as many bottles, it will take twice as long to fill them!

The easiest way to answer these questions is to use algebra to represent the relationship. If two values, A and B , are directly proportional, we can describe this by writing A \propto B . The sign in the middle, \propto , just means “is proportional to”.

When two values are directly proportional, we can also describe their relationship using an equation of the form A=kb where k is a constant called the constant of proportionality. To solve most proportion problems, you will need to find the value of k .

Let’s say \boldsymbol{A} is directly proportional to \boldsymbol{B} . When \boldsymbol{A=12} , \boldsymbol{B=14} . Find the value of \boldsymbol{A} when \boldsymbol{B=91} .

First, we’ll look at how to use algebra. In this case, A \propto B so A = kB for some value of k . To find that value, we substitute in A = 12 and B=14 .

\begin{aligned} A &= kb \\ 12 &= k \times 14 \\ k &= \frac{12}{14} \\ k&= \frac{6}{7} \end{aligned}

Putting this back into the equation gives:

\begin{aligned} A = \frac{6}{7}B \end{aligned}

Now, we can substitute B=91 into our equation.

\begin{aligned} A &= \frac{6}{7} \times 91 \\ &= \boldsymbol{78} \end{aligned}

Question 5: Rationalising Surds

Rationalising a fraction means finding an equivalent fraction that has an integer on the denominator. In other words, you want to manipulate the fraction so that any surds are on the top.

Let’s simplify:

\huge\boldsymbol{\frac{3}{\sqrt{5}}}

To eliminate the surd on the denominator, we multiply it by \sqrt5. However, if we only multiplied the denominator this would change the value of the fraction. So instead, we multiply both the numerator and denominator by the same surd:

\frac{3}{\sqrt5}\times\frac{\sqrt5}{\sqrt{5}} = \boldsymbol{\frac{3\sqrt5}{5}}

That’s it! You have no surds on the bottom of the fraction but you haven’t changed the value of the number.

If you have a fraction on the bottom that can be simplified, such as \sqrt{28} , it’s a good idea to simplify that first.

Let’s simplify:

\huge\boldsymbol{\frac{7}{\sqrt{28}}}

\begin{aligned} \frac{7}{\sqrt{28}} &= \frac{7}{\sqrt{4}\times\sqrt{7}} \\ &= \frac{7}{2\sqrt{7}} \end{aligned}

Now, we multiply by \sqrt{7}  as it’s only this part of the denominator we want to change:

\begin{aligned} \frac{7}{2\sqrt{7}} \times \frac{\sqrt7}{\sqrt7} &= \frac{7\sqrt{7}}{2\times7} \\ &= \frac{7\sqrt{7}}{14} \end{aligned}

Now you might be able to see that this fraction isn’t in its simplest form. To simplify it fully, we need to divide the numerator and denominator by 7:

\begin{aligned} \frac{7\sqrt{7}}{14} &= \boldsymbol{\frac{\sqrt{7}}{2}} \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