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, this is the relevant one this week.
This week, we’re exploring:
- Similar Shapes
- The Equation of a Line Through Two Points
- Solving Single Linear Inequalities
- Simple Interest
Question 1: Similar Shapes
Two shapes are similar if one is an enlargement of the other. The key to finding missing sides is finding the scale factor.
The two shapes below are similar. The height of the smaller shape is 10cm. Find the height of the larger shape.
First, we compare the two lengths we’re given. By dividing the larger length by the smaller length, we can find the scale factor.
Scale factor for length = 6 ÷ 4
So, if we multiply any of the lengths on the smaller shape by 1.5, we will get the corresponding length on the larger shape.
Height of the larger shape = 10 × 1.5
Question 2: The Equation of a Line Through Two Points
Every straight line can be written in the form 𝑦 = 𝑚𝑥 + 𝑐 where 𝑚 is the gradient of the line and 𝑐 is the 𝑦-intercept. This week, we are given two points that a straight line passes through and we need to use this information to find the equation of the straight line.
A straight line passes through the points (2, 10) and (-3, -5). Work out the equation of the line in the form 𝑦 = 𝑚𝑥 + 𝑐.
The gradient of the line is the change in the 𝑦-coordinates divided by the change in the 𝑥-coordinates. So, using our two points we can find the gradient, 𝑚.
Since the line passes through the point (2, 10), we can substitute these coordinates in to find the value of 𝑐. We could also substitute in (-3, -5) but using the positive values, if possible, means we can reduce the risk of making an error with the negatives.
Putting this into our equation gives us our final answer: 𝑦 = 3𝑥 + 4.
Question 3: Solving Single Linear Inequalities
Solving linear inequalities is very similar to solving linear equations. We covered this back in week 2 so you may want to glance back at that week. We have to make sure the inequality stays balanced by doing the same to each side. When working with inequalities, it’s best to avoid dividing by negatives as this means we need to change the inequality sign. Instead, add the negative 𝑥-term to both sides.
Solve 3𝑥 – 2 ≤ 5 – 7𝑥.
As with linear equations, we start by getting all our 𝑥 values on one side. In this case, we will start by adding 7𝑥 to both sides.
Question 4: Simple Interest
Simple interest is when the same amount of interest is added each time. This amount of interest is based on the amount of money originally invested, as opposed to the amount of money currently in the account. This is different to compound interest which we covered back in week 8.
Manuel invests £300 in a bank account that pays 3.5% simple interest per annum. How much will he have in the account after 5 years?
First, we need to find 3.5% of £300. When we have a calculator, the easiest way to do this is by using multipliers. Take a look at week 3 if you need a reminder.
To find a multiplier, we simply divide the percentage by 100. We can then multiply £300 by this value to find the amount we need.
3.5% of 300 = 300 × (3.5 ÷ 100)
= 300 × 0.035
This is the amount of interest that is paid per annum (or each year). If Manuel leaves the money in the account for 5 years, he will make 5 × 10.50 = £52.50 in interest. To find the total amount he has in his account, we add this to the original investment.
300 + 5 × 10.50 = 300 + 52.5
Question 5: Pressure
The pressure exerted by an object is the force per unit area; usually N/cm2 or N/m2. In physics, you may use the unit Pa (Pascals) for pressure, but this isn’t necessary in maths.
You can also write this as a formula triangle:
Let’s find the force required to exert a pressure of 5N/m2 over an area of 2.5m2.
In this case, we want to find the force, so we cover this up in the formula triangle and use the other two variables. Since pressure and area are on the same level, we multiply them together.
force = pressure × area
= 5 × 2.5
Make sure you include units in your answer – there is often a mark for this.
If the question asked for the pressure, we would use pressure = force ÷ area. If the question asked for the area, we would use area = force ÷ pressure.
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