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Aditya Kumar Class 9 Tuition trainer in Delhi

Aditya Kumar

Mathematics

Mukherjee Nagar, Delhi, India - 110009.

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Referral Discount: Get ₹ 500 off when you make a payment to start classes. Get started by Booking a Demo.

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Education

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Overview

Initiative and high level of energy.
Tolerant and flexible with different situations.
Verbal communication skills, decision making, critical thinking, organizing and planning.

Languages Spoken

English

Hindi

Education

Sir MVIt 2017

Bachelor of Engineering (B.E.)

Address

Mukherjee Nagar, Delhi, India - 110009

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Teaches

Class 9 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE, State, ICSE

CBSE Subjects taught

Mathematics, Science

ICSE Subjects taught

Mathematics, Physics

Taught in School or College

Yes

State Syllabus Subjects taught

Science, Mathematics

Class 10 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE, State, ICSE

CBSE Subjects taught

Mathematics, Science

ICSE Subjects taught

Mathematics, Physics

Taught in School or College

Yes

State Syllabus Subjects taught

Science, Mathematics

Class 6 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 6 Tuition

1

Board

State, CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics

Class 7 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 7 Tuition

1

Board

State, CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics

Class 8 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 8 Tuition

1

Board

State, CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics

Reviews

No Reviews yet! Be the first one to Review

FAQs

1. Which school boards of Class 10 do you teach for?

CBSE, State, ICSE

2. Do you have any prior teaching experience?

Yes

3. Which classes do you teach?

I teach Class 10 Tuition, Class 6 Tuition, Class 7 Tuition, Class 8 Tuition and Class 9 Tuition Classes.

4. Do you provide a demo class?

Yes, I provide a free demo class.

5. How many years of experience do you have?

I have been teaching for less than a year.

Contact

Answers by Aditya Kumar (2)

Answered on 16/12/2017 Learn CBSE/Class 10/Mathematics +1 Tuition/Class IX-X Tuition

If the zeroes of polynomial x3 – ax2 + bx – c are in AP then show that 2a3 – 9ab + 27c = 0

zreors of polynomial means roots of polynomial if the roots of polynomial are in AP then the roots are A+d, A, A-d Sum of the roots = (-) Co-efficient of X^2/ Co-efficient of X^3 = A+d +A +A-d -(-a) = 3A a =3A A= a/3 since... ...more
zreors of polynomial means roots of polynomial if the roots of polynomial are in AP then the roots are A+d, A, A-d Sum of the roots = (-) Co-efficient of X^2/ Co-efficient of X^3 = A+d +A +A-d -(-a) = 3A a =3A A= a/3 since f(x)= x^3 - ax^2 +bx - c f(A)= A^3 - aA^2 +bA -c put A= a/3 (a/3)^3 -a(a/3)^2 +b(a/3) -c=0 (a^3)/27 - (a^3)/9 +ab/3 -c = 0 (a^3) -3(a^3) +9ab -27c =0 -2(a^3) +9ab - 27c = 0 2a^3 -9ab +27c = 0 hence PROVED
Answers 5 Comments
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Answered on 16/12/2017 Learn CBSE/Class 10/Mathematics +1 Tuition/Class IX-X Tuition

A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same... ...more
A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/hr more. Find the original speed of the train.

distance=360 km time difference b/w both speed = 48/60 hr let speed of train is x km/hr then time taken by train when speed is x km/hr = 360/x hr time taken by train when speed is x+5 km/hr = 360/(x+5) hr A/C to question 360/x = 360/(x+5) +48/60 after solving this equation you will get the original speed... ...more
distance=360 km time difference b/w both speed = 48/60 hr let speed of train is x km/hr then time taken by train when speed is x km/hr = 360/x hr time taken by train when speed is x+5 km/hr = 360/(x+5) hr A/C to question 360/x = 360/(x+5) +48/60 after solving this equation you will get the original speed of train i.e equal to 45 km/hr
Answers 2 Comments
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Contact

Teaches

Class 9 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE, State, ICSE

CBSE Subjects taught

Mathematics, Science

ICSE Subjects taught

Mathematics, Physics

Taught in School or College

Yes

State Syllabus Subjects taught

Science, Mathematics

Class 10 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE, State, ICSE

CBSE Subjects taught

Mathematics, Science

ICSE Subjects taught

Mathematics, Physics

Taught in School or College

Yes

State Syllabus Subjects taught

Science, Mathematics

Class 6 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 6 Tuition

1

Board

State, CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics

Class 7 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 7 Tuition

1

Board

State, CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics

Class 8 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 8 Tuition

1

Board

State, CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics

No Reviews yet! Be the first one to Review

Answers by Aditya Kumar (2)

Answered on 16/12/2017 Learn CBSE/Class 10/Mathematics +1 Tuition/Class IX-X Tuition

If the zeroes of polynomial x3 – ax2 + bx – c are in AP then show that 2a3 – 9ab + 27c = 0

zreors of polynomial means roots of polynomial if the roots of polynomial are in AP then the roots are A+d, A, A-d Sum of the roots = (-) Co-efficient of X^2/ Co-efficient of X^3 = A+d +A +A-d -(-a) = 3A a =3A A= a/3 since... ...more
zreors of polynomial means roots of polynomial if the roots of polynomial are in AP then the roots are A+d, A, A-d Sum of the roots = (-) Co-efficient of X^2/ Co-efficient of X^3 = A+d +A +A-d -(-a) = 3A a =3A A= a/3 since f(x)= x^3 - ax^2 +bx - c f(A)= A^3 - aA^2 +bA -c put A= a/3 (a/3)^3 -a(a/3)^2 +b(a/3) -c=0 (a^3)/27 - (a^3)/9 +ab/3 -c = 0 (a^3) -3(a^3) +9ab -27c =0 -2(a^3) +9ab - 27c = 0 2a^3 -9ab +27c = 0 hence PROVED
Answers 5 Comments
Dislike Bookmark

Answered on 16/12/2017 Learn CBSE/Class 10/Mathematics +1 Tuition/Class IX-X Tuition

A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same... ...more
A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/hr more. Find the original speed of the train.

distance=360 km time difference b/w both speed = 48/60 hr let speed of train is x km/hr then time taken by train when speed is x km/hr = 360/x hr time taken by train when speed is x+5 km/hr = 360/(x+5) hr A/C to question 360/x = 360/(x+5) +48/60 after solving this equation you will get the original speed... ...more
distance=360 km time difference b/w both speed = 48/60 hr let speed of train is x km/hr then time taken by train when speed is x km/hr = 360/x hr time taken by train when speed is x+5 km/hr = 360/(x+5) hr A/C to question 360/x = 360/(x+5) +48/60 after solving this equation you will get the original speed of train i.e equal to 45 km/hr
Answers 2 Comments
Dislike Bookmark

Contact

Aditya Kumar describes himself as Mathematics. He conducts classes in Class 10 Tuition, Class 6 Tuition and Class 7 Tuition. Aditya is located in Mukherjee Nagar, Delhi. Aditya takes at students Home and Regular Classes- at his Home. He has 1 years of teaching experience . Aditya has completed Bachelor of Engineering (B.E.) from Sir MVIt in 2017. He is well versed in English and Hindi.

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