Daily Maths Revision – Week 4 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. Multiplying Fractions
  2. Laws of Indices
  3. The nth Term of Linear Sequences
  4. Averages from a Table
  5. The Area of Composite Shapes

Question 1: Multiplying Fractions

When multiplying fractions, we simply multiply the numerators (the top numbers) and then multiply the denominators (the bottom numbers). Then, if we need to, we simplify our answer.

Let’s calculate:

\huge{ \boldsymbol{\frac{3}{7} \times \frac{5}{6}}}

\begin{aligned} \frac{3}{7} \times \frac{5}{6} &= \frac{3\times5}{7\times6}\\ &= \frac{15}{42} \end{aligned}

We need to see if we can simplify our answer by checking for any common factors of 15 and 42. In this case, they are both divisible by 3 so we divide both the numerator and denominator by 3.

\begin{aligned} \frac{15}{42} &= \frac{5}{14} \end{aligned}

5 and 14 have no common factors so this fraction is in its simplest form.

Question 2: Laws of Indices

There are three laws of indices you need to know. Often, exam questions will require you to use a mixture of these so it’s really important you are familiar with them.

Let’s say we want to find \boldsymbol{x^7\times x^3} .

When we are multiplying, we add the powers:

\begin{aligned} x^7 \times x^3 &= x^{7+3}\\ &= \boldsymbol{x^{10}} \end{aligned}

Instead, let’s say we want to find \boldsymbol{6m^9 \div 2m^5} .

This time, we have both coefficients (the large numbers in front of the letters) and indices to think about. We are also dividing the indices instead of multiplying.

Let’s think about the coefficients first. We simply divide the first coefficient by the second:

6 \div 2 = 3

Now, let’s look at the indices. When we divide, we simply subtract the powers:

\begin{aligned} m^9 \div m^5 &= m^{9-5}\\ &= m^{4} \end{aligned}

Putting these together gives our final answer:

6m^9 \div 2m^5 = \boldsymbol{3m^4}

Finally, let’s say we want to find \boldsymbol{(2ab^6)^4} .

Raising a term to the power of four means that you multiply it by itself four times.


(2ab^6)^4 = 2ab^6 \times 2ab^6 \times 2ab^6 \times 2ab^6

Just like before, we can think about each component of the term separately:

2 \times 2 \times 2 \times 2 = 16
a \times a \times a \times a = a^4
b^6 \times b^6 \times b^6 \times b^6 = b^{24}

Alternatively, we could have just raised each component of the term to the power of 4. When we have indices to a power, we multiply the powers.

2^4 = 16
(a)^4 = a^4                 (Remember, a = a^1 ).
(b^6)^4 = b^{24}

We then put it all together to get our final answer:

(2ab6)4 = 16a4b24

Question 3: The \huge{\boldsymbol{n}} th Term of Linear Sequences

A linear sequence is one that goes up by the same amount each time. We use the n th term of a sequence to describe the different terms in relation to their position in the sequence, n . For the first term, n =1; for the second term, n =2 and so on.

Let’s say we want to find the first 3 terms of the sequence with \boldsymbol{n} th term \boldsymbol{3n+1} .

We simply substitute in n=1 , n=2 and n=3 .

n=1 :     3(1) + 1 = 4
n=2 :     3(2) + 1 = 7
n=3 :     3(3) + 1 = 10

So, the first 3 terms of the sequence are 4, 7 and 10.

What if we are given the start of a sequence and we want to find the n th term? This is a common question and the process is relatively simple to learn.

Let’s find the \boldsymbol{n} th term of the sequence that starts 2, 9, 16, 23, 30….

We start by finding the common difference of the terms. This simply means finding the amount we add each time:

We are adding 7 each time – this means the sequence is related to the 7 times table. The n th term will start with 7n .

Next, we find the 0th term; this is the term that would come before 2 in our sequence. This term gives us the amount to add or subtract from 7n to get our sequence. It works because, for this term, n =0 and 7n is also 0. This means the amount we add or subtract will be the term itself.

2 – 7 = -5, so -5 would come before 2 in our sequence.

Therefore, our complete n th term is \boldsymbol{7n-5} .

Question 4: Averages from a Table

There are three averages you could be asked to find from a table: mean, median and mode. We are going to find each of these for the table below. This table shows the number of exams in one week for 60 different students.

Let’s start by finding the mode.

Remember, the mode is the most common result. In other words, it’s the result with the highest frequency. In this case, the highest frequency is 24 so the mode is the number of exams corresponding to that frequency: 4.

Next, let’s find the median.

This is the middle value if 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 1 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 4.

Finally, let’s find the mean.

To find the mean, we add up all the data and divide by the total number of students. However, to be able to add up the data we need to multiply the number of exams by their corresponding frequency. 

Take the last row for example, there are 5 students that each sat 6 exams. In total, these 5 students sat 30 exams because 6 + 6 + 6 + 6 + 6 = 30. The most efficient way to calculate this value for every row is by multiplying the two columns together.

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

2 + 39 + 96 + 85 + 30 = 252
252 ÷ 60 = 4.2

So, the mean is 4.2.  

Question 5: The Area of Composite Shapes

The best way to approach the area of a composite shape is to split it into shapes that you can find the area of. If you have managed to do this but are struggling to find the area of one part, take a look at our revision blog on the area of 2D shapes. It covers all the 2-dimensional shapes you need for your GCSE.

Let’s find the area of the composite shape below.

This shape is made up of a rectangle and a right-angle triangle. To find the total area, we can find each separate area and add them together. There are many ways we could split the shape up, but we will split it like this:

Let’s find the area of the rectangle first. We simply multiply the base (15cm) by the height (8cm).

Area of the rectangle = 15 × 8
                                         = 120cm2

To find the area of a triangle, we multiply the base by the perpendicular height and divide by 2. However, we don’t know the base of the triangle. The base of the triangle plus the small bit that sticks out will be the same as the base of the rectangle. So, we simply subtract 2.5cm from 15cm to find the triangle base.

Base of triangle = 15 – 2.5
                              = 12.5cm

Area of triangle = (12.5 × 3) ÷ 2
                              = 37.5 ÷ 2
                              = 18.75cm2

So, the area of the composite shape is these two values added together.

Total area = 120 + 18.75
                    = 138.75cm2

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