Curtis M. A Level Maths tutor, GCSE Maths tutor, GCSE Chemistry tutor...

Curtis M.

Currently unavailable: for regular students

Degree: Chemical Engineering (Masters) - Sheffield University

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About me

About Me:
I am currently studying and enjoying Chemical Engineering at the University of Sheffield.
As a training engineer maths is obviously pivotal to my studies, and not only do I understand pure maths, but I also understand how maths is applied into real life situations.
I have always had a passion for mathematics and I hope I can share at least some of this passion with you!

My Sessions
The content and pace of my sessions will be conducted around what you want. Whether you want me to prep you in all areas for a given exam, cover one particular topic that you are struggling with, or for me to supplement the work that is being done in the classroom, I am happy to fulfill any of these roles.
I believe that maths is best taught not by learning things by memory but by actually understanding the concepts involved. This is particularly critical if you intend to take maths further as it is far easier to build on content that you fully understand. I personally see the benefits of a good understanding of maths every day of my studies and I hope to be able to bring those benefits to you, and to push you to heights in maths you may never have thought you could reach!

So if you are looking for an enthusiastic GCSE or A-Level Maths tutor, then please do contact me, either by "Webmail" or by booking a "Meet the Tutor" session - all possible via this website. I look forward to helping you (or your child) succeed in maths!

Subjects offered

SubjectLevelMy prices
Maths A Level £20 /hr
Chemistry GCSE £18 /hr
Maths GCSE £18 /hr
Physics GCSE £18 /hr

Qualifications

QualificationLevelGrade
MathsA-LevelA
ChemistryA-LevelA
PhysicsA-LevelA
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Currently unavailable: for regular students

Questions Curtis has answered

How do you solve the #EdexcelMaths question?

Remember the uproar on Twitter about an Edexcel GCSE Maths paper, sparking the #EdexcelMaths trend on Twitter? Whether you do or don't, the question that caused the controversyis within the GCSE syllabus and it is important to be able to tackle it. The question is as follows:   Q) There are ...

Remember the uproar on Twitter about an Edexcel GCSE Maths paper, sparking the #EdexcelMaths trend on Twitter? Whether you do or don't, the question that caused the controversy is within the GCSE syllabus and it is important to be able to tackle it. The question is as follows:

 

Q) There are n sweets in a bag. 6 of the sweets are orange. The rest of the sweets are yellow.

Hannah takes at random a sweet from the bag. She eats the sweet.

The probability that Hannah eats two orange sweets is 1/3

Show that n2 - n - 90 = 0

 

It is easy to be thrown off with this question because of the surprising ending. You may perhaps have been expecting a probability question and then you were asked a quadratic equation question. The truth is, if we work through this logically, we can use both our probability skills and our algebra skills to crack this.

First we need to find the probability that Hannah eats two orange sweets in terms of n.

The probability that Hannah eats an orange sweet first is the number of orange sweets in the bag (6, from the question) divided by the total number of sweets in the bag (n, again, from the question). So the probability of the first sweet being orange is 6/n.

Now we need to determine the probability of Hannah picking an orange sweet second from the bag, if the first she picked was also orange. To figure this out, we again do the number of orange sweets in the bag (5, as Hannah ate one), divided by the total number of sweets in the bag (n-1, again, as Hannah ate one). So the probability of Hannah picking a second orange sweet from the bag is 5/(n-1).

To figure out the probability of Hannah picking 2 orange sweets from the bag we multiply those 2 probabilities together.

6/n * 5/(n-1) = 30/n(n-1) = 30/(n2-n)

And we were told in the question that this probability was actually equal to 1/3. Therefore:

30/(n2-n) = 1/3

All we have to do now is manipulate this equation so that we get the one asked for in the question. First we can multiply both sides by 3:

90/(n2-n) = 1

Then we can multiply both sides by (n2-n):

90 = n2-n

And if we collect all the terms on one side we get:

n2-n-90=0

Which is what we were asked to prove in the original question.

 

Hopefully you are able to follow this explanation, and if not, perhaps try running through it again step by step to try and understand it. There is no one skill used here that is particularly difficult - the challenge in this question is knowing how and when to apply the maths skills that you have, and this is a challenge that will keep facing you with maths if you intend to take it further.

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1 year ago

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