Matthew B. A Level Maths tutor, A Level Further Mathematics  tutor, A...

Matthew B.

Currently unavailable: for regular students

Degree: Mathematics (Masters) - Warwick University

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About me

About me:

Hello! My name is Matt and I'm a 4th year Maths student at the University of Warwick. I achieved all A*s at GCSE and 4 A*s at A-Level in Maths, Further Maths, Physics, and Chemistry. I am offering tutoring for A-Level MathsFurther Maths and Physics, as well as GSCE Maths and Science.

At school I informally tutored other students and used to volunteer after school to help younger pupils with their homework. I then worked for a year as a teaching assistant with children ranging from year 7 up to year 13. During this time I discovered my passion for teaching and gained a huge amount of experience working with students of all ability levels and ages. I am extremely patient, easy going, and friendly.

The sessions:

The tutorials will provide you with an opportunity to go through topics that you find particularly challenging, and it will be up to you to decide which areas to focus on. My main focus is to build a solid understanding of the simple concepts and then slowly apply that knowledge to more complex situations. 

I also hope to build the confidence of the student, so that eventually they feel comfortable explaining the idea to someone else (nothing tests understanding like explaining it yourself).

I believe that the key to productive learning is through a relaxed, unpressured environment. Therefore, I will try very hard to make the sessions as fun and engaging as possible, so that you actually look forward to learning Maths and Science, rather than dread it.

What next?

Feel free to send me a "WebMail" or book a "Meet the tutor session" if you think I might be the tutor for you!

Subjects offered

SubjectLevelMy prices
Further Mathematics A Level £20 /hr
Maths A Level £20 /hr
Physics A Level £20 /hr
Maths GCSE £18 /hr
Science GCSE £18 /hr

Qualifications

QualificationLevelGrade
MathematicsA-LevelA*
Further MathematicsA-LevelA*
PhysicsA-LevelA*
ChemistryA-LevelA*
Step IIUni Admissions Test3
Step IIIUni Admissions Test2
Disclosure and Barring Service

CRB/DBS Standard

17/08/2012

CRB/DBS Enhanced

No

Currently unavailable: for regular students

General Availability

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Please get in touch for more detailed availability

Ratings and reviews

5from 10 customer reviews

Colette (Parent) March 14 2016

Good solid tutoring with sound understanding

Colette (Parent) March 21 2016

Great attitude

Colette (Parent) February 29 2016

Very helpful

Colette (Parent) February 22 2016

Great Tutor.
See all reviews

Questions Matthew has answered

How do I add up the integers from 1 to 1000 without going insane?

Suppose you are asked by your teacher to sum all the integers from 1 up to 1000. You might be thinking they must hold some kind of grudge against you. If you try to calculate the sum by adding on each integer one at a time, you will have to perform 999 separate additions, some of which will be...

Suppose you are asked by your teacher to sum all the integers from 1 up to 1000. You might be thinking they must hold some kind of grudge against you. If you try to calculate the sum by adding on each integer one at a time, you will have to perform 999 separate additions, some of which will be quite long and tedious. This would take far longer than anyone can be bothered to spend adding up numbers.

There is, however, a quicker way. Let's give our sum a letter to represent its unknown value. Let's use "S" for "sum". Then:

S = 1 + 2 + 3 + ... + 999 + 1000

But we can rewrite this sum in reverse order. Starting with 1000 and ending with 1:

S = 1000 + 999 + 998 + ... + 2 + 1

We can now add together these two equations to give us:

S + S = (1+1000) + (2+999) + (3+998) + ... + (999+2) + (1000+1)

Simplifying both sides gives us:

2S = 1001 + 1001 + 1001 + ... + 1001 + 1001

The right hand side has 1000 separate terms, since our original sum contained 1000 numbers. So:

2S = 1000 x 1001 = 1001000

Dividing both sides by two we find that S = 500500. Therefore:

1 + 2 + 3 + ... + 999 + 1000 = 500500

This calculation is an example of a more general concept called an arithmetic series, where you sum a sequence of numbers which differ by adding on a fixed amount with each step.

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1 year ago

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