Following on from the last blog, in which I discussed 5 common mistakes in foundation tier GCSE maths, this time we’ll look at 5 common mistakes in higher tier GCSE maths.
Even if you’re studying for the higher tier exam, I’d still recommend going back and reading the foundation tier blog – as those questions could still come up in your exam. If you are studying for the foundation tier exam, do not read this blog as this content will not be relevant to you.
1. Index laws
Question:
Work out 4-2
Incorrect solution:
1/8
In this incorrect solution, we’ve correctly identified that taking a number to a negative power reciprocates it (in other words, flips it). This would turn 4 into ¼. However, the 2 in the power does not mean we multiply 4 by 2! The 2 means we square the number.
It’s a silly mistake to make, as I doubt students would ever make that mistake if the question were just “Work out 42”. For some reason when it’s -2 I see this mistake a lot. It’s potentially because there are two components to the index and that throws students off.
Correct solution:
42 = 16, so 4-2 = 1/16
To practice more questions like this, try our index laws worksheet.
2. Bounds
Question:
A = B ÷ C
B = 24 correct to 2 significant figures
C = 5.6 correct to 2 significant figures
Find an error interval for A, giving both bounds to 3 significant figures.
Incorrect solution:
Lower bound of B = 23.5
Upper bound of B = 24.5
Lower bound of C = 5.55
Upper bound of C = 5.65
Lower bound of A = Lower bound of B ÷ Lower bound of C
= 23.5 ÷ 5.55
= 4.234…
= 4.23 (to 3 significant figures)
Upper bound of A = Upper bound of B ÷ Lower bound of C
= 24.5 ÷ 5.65
= 4.336…
= 4.34 (to 3 significant figures)
So, error interval is
4.23 ≤ x < 4.34
The upper and lower bounds for B and C have been calculated perfectly. However, it’s when we used them to find the upper and lower bounds for A that the solution went wrong.
If we want to find the upper bound for A, we actually want to divide the upper bound of B by the lower bound of C.
Because A = B ÷ C, the smaller C is the larger A will be.
Let’s quickly prove that with an example.
If B = 12 and C = 4, then A = 12 ÷ 4 = 3
If I decrease C from 4 to 3, then A = 12 ÷ 3 = 4
So decreasing C increased A.
It also follows therefore that the larger C is the smaller A will be. So, to find the lower bound of A, we would need to divide the lower bound of B by the upper bound of C.
Correct solution:
Lower bound of B = 23.5
Upper bound of B = 24.5
Lower bound of C = 5.55
Upper bound of C = 5.65
Lower bound of A = Lower bound of B ÷ Upper bound of C
= 23.5 ÷ 5.65
= 4.159…
= 4.16 (to 3 significant figures)
Upper bound of A = Upper bound of B ÷ Lower bound of C
= 24.5 ÷ 5.55
= 4.414…
= 4.41 (to 3 significant figures)
So, error interval is
4.16 ≤ x < 4.41
This is an incredibly common, and – to be fair – very understandable mistake. To practice more questions like this, try our bounds worksheet.
3. Perimeter of sector
Question:
Find the perimeter of the sector, to 2 significant figures.
Incorrect solution:
Everything has been done perfectly here – the formula is correct, the numbers have been put in to the formula correctly and the final answer has been rounded to 2 significant figures correctly also. BUT… we have only found the arc length. The arc length is the part of the shape I’ve highlighted in red below.
The question asks for the perimeter, which is all the way around the shape. So we have missed out the two radii (highlighted in blue above). These need to be added on to the 19 cm we calculated earlier.
Correct solution:
To practice more questions like this, try our sectors and arcs worksheet.
4. Quadratic formula
Question:
Solve 3x2 – 10x + 4 = 0, giving your answers to 2 decimal places.
Incorrect solution:
a = 3, b = –10 and c = 4
If you type both of those into your calculator, you will get a Math Error. This is because we have tried to square root a negative number, which is impossible. If this happens to you in an exam, don’t panic! It’s almost certain that you’ve made one of the two following mistakes…
I’ve actually made two separate mistakes here! Both errors are extremely common. First of all, the quadratic formula stated is correct and so are the values of a, b and c.
The first mistake (highlighted in red) is that the formula starts with –b
b = –10, so –b = – – 10
Two minuses next to each other make a +, so –b = 10, not –10
The second mistake (highlighted in blue) is in the square root, we should put (–10)2 instead of –102. These two give us very different results.
(–10)2 = –10 × –10 = 100
–102 = –10 × 10 = –100
The first option is the one we want, because we want to square –10 i.e. to do –10 × –10.
The second option does not do that – it squares the 10 to get 100 and subtracts it. The reason for this is BIDMAS (you may know it as BODMAS, or the order of operations) – Indices (I) come before Subtraction (S).
Because we put the –10 in brackets in the first option, it did this first, as Brackets (B) comes before everything else.
Correct solution:
To practice more questions like this, try our quadratic formula worksheet.
5. Quadratic sequences
Question:
Find the nth term of the sequence 6, 7, 10, 15, …
Incorrect solution:
Find the difference of the differences
2 ÷ 2 = 1
So the nth term begins with n2. Now compare our sequence to this.
Now find the nth term of the green sequence:
The sequence has a difference of 2 and if there were a previous term it would be –7.
So the nth term is 2n – 7.
So our final answer is n2 + 2n – 7
It’s not obvious what the error is here. But, if we are checking our answers correctly, we will be able to spot that it is wrong. When you are finding the nth term of a quadratic sequence, you should always check your answer immediately after.
The way you check your answer is by finding the 1st, 2nd, 3rd and 4th terms of your sequence according to your nth term formula. These terms should match the original sequence if you are correct.
n = 1 12 + 2×1 – 7 = 1 + 2 – 7 = –4
We have a problem already. The first term of the sequence is 6, not –4. So our answer cannot be correct.
The mistake that was made in the solution was at this stage:
We subtracted our sequence from the n2 sequence. We should have subtracted the n2 sequence from our sequence.
Correct solution:
Find the difference of the differences
2 ÷ 2 = 1
So the nth term begins with n2. Now compare our sequence to this.
Now find the nth term of the green sequence:
The sequence has a difference of –2 and if there were a previous term it would be 7.
So the nth term is –2n + 7.
So our final answer is n2 – 2n + 7
Let’s check our answer to make sure it is correct this time:
n = 1 12 – 2×1 + 7 = 1 – 2 + 7 = 6 this matches the sequence
n = 2 22 – 2×2 + 7 = 4 – 4 + 7 = 7 this also matches
n = 3 32 – 2×3 + 7 = 9 – 6 + 7 = 10 this also matches
To practice more questions like this, try our quadratic sequences worksheet.
For more examples of this method, read my guide on how to find the nth term of a quadratic sequence.
I hope you’ve found this useful and that you won’t make these mistakes when you do your exams!
Also please check out a similar blog on 5 common mistakes in foundation GCSE maths, as if you are studying for the higher tier exam, foundation topics will still come up.
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