https://www.urbanpro.com/delhi/rahul-vikrant # Rahul Vikrant

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Rahul Vikrant

Rohini Sector 16, Delhi, India - 110089

Rohini Sector 16, Delhi, India - 110089.

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I am an experienced, qualified teacher and tutor with over 4 years of experience in teaching Maths, across different boards including CBSE, ICSE and state boards. Passionate about solving Mathematical problems over the years I have helped thousands of students overcome their fear of Maths.

Hindi Mother Tongue (Native)

English Proficient

BRABU, Muzaffarpur 2016

Bachelor of Science (B.Sc.)

Rohini Sector 16, Delhi, India - 110089

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Class 12 Tuition

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 12 Tuition

5

Board

State, ISC/ICSE, CBSE

ISC/ICSE Subjects taught

Mathematics

CBSE Subjects taught

Mathematics

Taught in School or College

No

State Syllabus Subjects taught

Mathematics

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5.0 out of 5.0 1 review

5.0051

N

18 Jul, 2019

Nischal Kaushik attended Class 12 Tuition

"Fantastic teacher and friendt is really lucky to have a teacher like Rahul sir.Excellent knowledge about the subject. "

Reply by Rahul

Thank you Nishchal.

Have you attended any class with Rahul? Write a Review

1. Which school boards of of Class 11 do you teach for?

State, ISC/ICSE, CBSE

2. Have you ever taught in any School or College?

No

3. Which classes do you teach?

I teach Class 10 Tuition, Class 11 Tuition, Class 12 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 5 years.

Answered on 17 Jul CBSE/Class 10/Mathematics/UNIT I: Number systems/Real numbers/NCERT Solutions/Exercise 1.1

Let 'a' be any +ve integer. Also suppose there exists two +ve integers q and r as quotient and remainder respectively when 'a' is divided by 6. Then , By using Euclid's Division lemma, we get a = 6q + r .................( A ) , where 0 ≤ r < 6. Obviously, r = 0 or 1 or 2 or 3 or... ...more

Let 'a' be any **+ve** integer.

Also suppose there exists two +ve integers **q and** r as quotient and remainder respectively when 'a' is divided by **6. **

**Then , **

**By using**** Euclid's Division lemma, we get **

a = 6q + r .................( A ) , where **0 ≤ r < 6.**

Obviously, r = 0 or **1** or **2** or **3** or **4** or **5.**

**Case-(i) **When r = 0 , then from ( A ) we have

*a = 6q + 0 *

=> **a = 6q = 2×3q = 2m _{ }**

**Case-(ii) When r = 1 , then **

**a = 6q + 1 = 2(3q) + 1 = 2m + 1 = odd number, where m = 3q.**

**Case-(iii) When r = 2 then **

**a = 6q + 2 = 2(3q + 1) = 2m _{1 }**

**Case-(iv) When r = 3 then **

**a = 6q + 3 = 2(3q + 1) + 1 = 2m _{1 }**

*Case-(v) When r= 4 then*

*a = 6q + 4 = 2(3q + 2) = 2m _{2 }*

**Case -(vi) When r = 5 then **

**a = 6q + 5 = 2(3q + 2) + 1 = 2m _{2 }+ 1 = odd number, where m_{2 = 3q + 2.}**

**Clearly from above cases ( ii ) , ( iv ) and ( vi ) we conclude that any +ve odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5.**

* PROVED *

Like 1

Answers 3 Comments RahulDirections

x Class 12 Tuition 5.0

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 12 Tuition

5

Board

State, ISC/ICSE, CBSE

ISC/ICSE Subjects taught

Mathematics

CBSE Subjects taught

Mathematics

Taught in School or College

No

State Syllabus Subjects taught

Mathematics

Class 11 Tuition 5.0

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 11 Tuition

5

Board

State, ISC/ICSE, CBSE

ISC/ICSE Subjects taught

Mathematics

CBSE Subjects taught

Mathematics

Experience in School or College

I've experience of more than 5 years of teaching mathematics of 11 and 12.

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics

Class 10 Tuition 5.0

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 10 Tuition

5

Board

ICSE, CBSE, State

CBSE Subjects taught

Mathematics

ICSE Subjects taught

Mathematics

Taught in School or College

No

State Syllabus Subjects taught

Mathematics

Class 9 Tuition 5.0

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 9 Tuition

5

Board

ICSE, CBSE, State

CBSE Subjects taught

Mathematics

ICSE Subjects taught

Mathematics

Taught in School or College

No

State Syllabus Subjects taught

Mathematics

this is test message this is test message this is test message this is test message this is test message this is test message this is test message

5.0 out of 5.0 1 review

N

18 Jul, 2019

Nischal Kaushik attended Class 12 Tuition

"Fantastic teacher and friendt is really lucky to have a teacher like Rahul sir.Excellent knowledge about the subject. "

Reply by Rahul

Thank you Nishchal.

Have you attended any class with Rahul? Write a Review

Answered on 17 Jul CBSE/Class 10/Mathematics/UNIT I: Number systems/Real numbers/NCERT Solutions/Exercise 1.1

Let 'a' be any +ve integer. Also suppose there exists two +ve integers q and r as quotient and remainder respectively when 'a' is divided by 6. Then , By using Euclid's Division lemma, we get a = 6q + r .................( A ) , where 0 ≤ r < 6. Obviously, r = 0 or 1 or 2 or 3 or... ...more

Let 'a' be any **+ve** integer.

Also suppose there exists two +ve integers **q and** r as quotient and remainder respectively when 'a' is divided by **6. **

**Then , **

**By using**** Euclid's Division lemma, we get **

a = 6q + r .................( A ) , where **0 ≤ r < 6.**

Obviously, r = 0 or **1** or **2** or **3** or **4** or **5.**

**Case-(i) **When r = 0 , then from ( A ) we have

*a = 6q + 0 *

=> **a = 6q = 2×3q = 2m _{ }**

**Case-(ii) When r = 1 , then **

**a = 6q + 1 = 2(3q) + 1 = 2m + 1 = odd number, where m = 3q.**

**Case-(iii) When r = 2 then **

**a = 6q + 2 = 2(3q + 1) = 2m _{1 }**

**Case-(iv) When r = 3 then **

**a = 6q + 3 = 2(3q + 1) + 1 = 2m _{1 }**

*Case-(v) When r= 4 then*

*a = 6q + 4 = 2(3q + 2) = 2m _{2 }*

**Case -(vi) When r = 5 then **

**a = 6q + 5 = 2(3q + 2) + 1 = 2m _{2 }+ 1 = odd number, where m_{2 = 3q + 2.}**

**Clearly from above cases ( ii ) , ( iv ) and ( vi ) we conclude that any +ve odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5.**

* PROVED *

Like 1

Answers 3 Comments Share this Profile

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