Hanumanth Srikar K. IB Further Mathematics  tutor, IB Maths tutor, GC...

Hanumanth Srikar K.

Currently unavailable: for new students

Degree: Maths and Economics (Bachelors) - LSE University

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About me

Hanumanth

EDUCATION

London School of Economics                                                                                      London, United Kingdom

Mathematics & Economics                                                                                                          Sep 2015 - Present

King Edward’s School                                                                                            Birmingham, United Kingdom

International Baccalaureate – 39 Points                                                                                 Sep 2013 – May 2015

HL Further Mathematics [6], HL Mathematics [6], HL Chemistry [6], SL Economics [7],

SL German [6], SL English [6], TOK & Mathematics Extended Essay [2]

GCSEs – 10 A*s including Mathematics [A*], English Language [A*]                                           Sep 2011 – Jun 2013

   Distinctions in Mathematics 2008 & 2012

‘Outstanding GCSEs’ Prize 2013

Awarded an Academic Scholarship and KES Fn. Bursary for entire 7 years

WORK EXPERIENCE

  Summer School Job                                                                                                  Birmingham, United Kingdom

Working as a teacher of a group of children at an 11+ summer school                                                         July 2015

Worked as part of a student-teacher team to deliver an educational week for 100 underprivileged children.

•   Planned and delivered stimulating lessons in English, Maths and problem solving for groups of 12 children.

•   Responsible for the care and safety of a form of children throughout the school week.

Role as Head of Mentoring at King Edward’s School                                           Birmingham, United Kingdom

  Restarted ‘Mentoring’ society helping children struggling in specific areas                                 Sept 2013 - June 2015

Organised and ran a society providing free tutoring for over 100 pupils by older boys in the school.

Taught a wide range of subjects to groups of students aged 11-16 for 2-3 hours a week.

  Volunteering for Birmingham Children’s Hospital                                               Birmingham, United Kingdom

  Raising funds for sensory equipment in the Pediatric intensive care unit                                     Sept 2013 - June 2014

•   Worked with a team of students to raise money for the intensive care unit at the hospital.

•   In charge of organising a ‘Balloon day’ event involving 800 children which raised over £6500 in a single day

  

INTERESTS AND ACHIEVEMENTS

2012 Duke of Edinburgh Bronze Award

2013 Appointed Year 12 Student representative

2014 Appointed Head of Mentoring

2015 Mentoring Prize

SKILLS

Languages: English (fluent), Telugu (fluent), German (proficient), Hindi (proficient), French (intermediate)

Maths: Distinction in British Maths Olympiad 2014, Merit in Intermediate Maths Olympiad 2012, Gold in UKMT Maths challenges 2008-2015

IT: Currently on a Python course, Proficient at LaTeX, Advanced Proficiency in MS Office

Sports: Represented KES Hockey, Cricket, Badminton teams in interschool competitions

Voluntary work: Religious committee volunteer at Balaji Temple UK since 2006

Other: Captain of U14 chess team 2010, won several individual prizes and team trophies across school career

  REFERENCES AVAILABLE ON REQUEST

Subjects offered

SubjectLevelMy prices
Chemistry GCSE £20 /hr
German GCSE £20 /hr
Maths GCSE £20 /hr
Economics IB £22 /hr
Further Mathematics IB £22 /hr
Maths IB £22 /hr

Qualifications

QualificationLevelGrade
MathematicsBaccalaureate6
Further MathematicsBaccalaureate6
ChemistryBaccalaureate6
EconomicsBaccalaureate7
GermanBaccalaureate6
EnglishBaccalaureate6
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Currently unavailable: for new students

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Ratings and reviews

5from 7 customer reviews

Valeriya (Parent) April 28 2016

we did not get to meet

Jamila (Parent) September 21 2016

Tessa (Student) August 21 2016

Tessa (Student) August 18 2016

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Questions Hanumanth Srikar has answered

Differentiation from first principles

Differentiaiton from principles requires the use of the following formula which is provided in the formula booklet: f'(x) = limh->0 ((f(x+h) - f(x))/(h)) Consider a function: f(x) = 6x2 Clearly we know that the function differentiates to: f'(x) = 12x  by using the process of multiplying th...

Differentiaiton from principles requires the use of the following formula which is provided in the formula booklet:

f'(x) = limh->0 ((f(x+h) - f(x))/(h))

Consider a function:

f(x) = 6x2

Clearly we know that the function differentiates to:

f'(x) = 12x 

by using the process of multiplying the coefficient by the power and then reducing the power by 1.

Using first principles however we must consider the formula mentioned previously.

f'(x) = limh->0 ((f(x+h) - f(x))/(h))

By computing the function for x+h and x we get:

f'(x) = limh->0 (6(x+h)2 - 6x2)/(h))

f'(x) = limh->0 (6(x2+2xh+h2) - 6x2)/(h))

f'(x) = limh->0 (6x2+12xh+6h2) - 6x2)/(h))

f'(x) = limh->0 (12xh+6h2)/(h))

We now cancel the h from above and below to get:

f'(x) = limh->0 12x+6h

Now consider the limit as h-> 0, clearly 12x remains unaffected but 6h will become 0 and is hence removed. Hence we are left with:

f'(x) = 12x

Which we know to be true from the trivial methods of differentiation considered earlier. 

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9 months ago

226 views

Factorising a Quadratic

Factorising a quadratic polynomial of the form, ax2 + bx + c = 0 can be done in many depends depending on the values we have for a, b and c.  Some simple polynomial could be done simply by recognition. For example:  x2 + 4x + 3 = 0 In this scenario we would use the assumption that the factor...

Factorising a quadratic polynomial of the form,

ax2 + bx + c = 0

can be done in many depends depending on the values we have for a, b and c. 

Some simple polynomial could be done simply by recognition. For example:

 x2 + 4x + 3 = 0

In this scenario we would use the assumption that the factorisation is of the form:

(dx+e)(fx+g) = 0

and we would consider the values of d, e, f and g.

As the coefficient of the xterm is 1, d and f would both be 1. Hence:

(x+e)(x+g) = 0

Now we need numbers for e and g such that:

e*g = 3 and,

e+g = 4

We find that e = 3 and g = 1 or visa versa. 

Hence we can factorise 

x2 + 4x + 3 = 0 

to (x+3)(x+1) = 0

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10 months ago

232 views
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