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Anoop Class 12 Tuition trainer in Hyderabad
Referral Discount: Get ₹ 250 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

I am from iit-jee background from my schooling .
Cleared my iitjee, mains, eamcet and attempted bitsat,other entrance examinations.
Merit certified in apptitude .
Cleared many entrance examinations through maths.

Languages Spoken

Telugu

English Proficient

Education

osmania university 2018

Bachelor of Technology (B.Tech.)

Address

Jeedimetla, Hyderabad, India - 500067

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Teaches

Class 12 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 12 Tuition

1

Board

State

Taught in School or College

No

State Syllabus Subjects taught

Mathematics

Class 11 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 11 Tuition

1

Board

State, CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

No

State Syllabus Subjects taught

Mathematics

Class 10 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 10 Tuition

1

Board

CBSE, State

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

No

State Syllabus Subjects taught

Mathematics

Class 9 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 9 Tuition

1

Board

CBSE, State

CBSE Subjects taught

Mathematics

Taught in School or College

No

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

No

State Syllabus Subjects taught

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

No

State Syllabus Subjects taught

Mathematics

Class I-V Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class I-V Tuition

1

Fees

₹ 240.0 per hour

Board

State, CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

No

State Syllabus Subjects taught

Mathematics

Engineering Entrance Coaching classes
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Engineering Entrance Coaching classes

1

Engineering Entrance Exams

IIT JEE Coaching Classes, BITSAT Coaching Classes, EAMCET, GATE Coaching Classes

IITJEE Coaching

IIT JEE Mains Coaching, IIT JEE Integrated Coaching, IIT JEE Advanced Coaching, IIT JEE Crash Course, IIT JEE Foundation Course

Type of class

Regular Classes, Crash Course

IIT-JEE Subjects

Maths

Cricket Coaching classes

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Cricket Coaching classes

1

Reviews (7)

5 out of 5 7 reviews

Anoop https://p.urbanpro.com/tv-prod/member/photo/6919969-small.jpg Jeedimetla
5.0057
Anoop
V

Cricket Coaching

"One of the best cricket coach, very young but so much talent.He has perfect knowledge about the game. My son has really improved his game. "

Anoop
R

Class 11 Tuition

"The way sir explains the core concepts are undoubtedly great, the passion he has towards teaching is incredible, sometimes sir stays overtime to complete the topic, which shows his dedication levels in teaching, sir was very friendly and keen in his explanation.I really loved his way of teaching and I do suggested his name for my brother and his friends. "

Anoop
S

Class 8 Tuition

"The teaching was good. I was able to understand the teaching .After attending this classes I was able to score more marks in my exam. This classes was helped me a lot. "

Anoop
R

Class 12 Tuition

"1)Knowledgeable in his actions and disciplinary in his attitude 2)Knowledge gaining and conceptual clarity established with his teaching 3)Simple,humble and smart in character 4)Had good educational background and Great explanation skills 5)Multitalented and logical in his approach Thank you sir for sharing your wonderful knowledge with me and I hope you continue to inspire the minds. "

Have you attended any class with Anoop?

FAQs

1. Which school boards of Class 12 do you teach for?

State

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, Class 6 Tuition, Class 7 Tuition, Class 8 Tuition, Class 9 Tuition, Class I-V Tuition, Cricket Coaching and Engineering Entrance Coaching 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 1 year.

Answers by Anoop (32)

Answered on 02/11/2019 Learn CBSE/Class 6/Maths/Mensuration/NCERT Solutions/Exercise 10.1

(a) Perimeter = 4 × Side 30 = 4 × Side Side = (b) Perimeter = 3 × Side 30 = 3 × Side Side = (c) Perimeter = 6 × Side 30 = 6 × Side Side = read less ...more

(a) Perimeter = 4 × Side

30 = 4 × Side

Side =

(b) Perimeter = 3 × Side

30 = 3 × Side

Side = 

(c) Perimeter = 6 × Side

30 = 6 × Side

Side = 

read less

Answers 3 Comments
Dislike Bookmark

Answered on 02/11/2019 Learn CBSE/Class 6/Maths/Mensuration/NCERT Solutions/Exercise 10.1

Perimeter of triangle = Sum of the lengths of all sides of the triangle Perimeter = 10 + 14 + 15 = 39 cm
Answers 3 Comments
Dislike Bookmark

Answered on 22/10/2019 Learn CBSE/Class 6/Maths/Mensuration/NCERT Solutions/Exercise 10.1

Perimeter of regular pentagon = 5 × Length of side 100 = 5 × Side Side = 100 side=100/5 cm=20
Answers 2 Comments
Dislike Bookmark

Answered on 22/10/2019 Learn CBSE/Class 6/Maths/Mensuration/NCERT Solutions/Exercise 10.3

(a) Total area of the region = 100 × 144 = 14400 cm2 Area of one tile = 12 × 5 = 60 cm2 Number of tiles required = Therefore, 240 tiles are required. (b) Total area of the region = 70 × 36 = 2520 cm2 Area of one tile = 60 cm2 Number of tiles required = Therefore, 42 tiles are requir... ...more

(a) Total area of the region = 100 × 144 = 14400 cm2 Area of one tile = 12 × 5 = 60 cm2 Number of tiles required = Therefore, 240 tiles are required. (b) Total area of the region = 70 × 36 = 2520 cm2 Area of one tile = 60 cm2 Number of tiles required = Therefore, 42 tiles are required.

Answers 2 Comments
Dislike Bookmark

Teaches

Class 12 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 12 Tuition

1

Board

State

Taught in School or College

No

State Syllabus Subjects taught

Mathematics

Class 11 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 11 Tuition

1

Board

State, CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

No

State Syllabus Subjects taught

Mathematics

Class 10 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 10 Tuition

1

Board

CBSE, State

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

No

State Syllabus Subjects taught

Mathematics

Class 9 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 9 Tuition

1

Board

CBSE, State

CBSE Subjects taught

Mathematics

Taught in School or College

No

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

No

State Syllabus Subjects taught

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

No

State Syllabus Subjects taught

Mathematics

Class I-V Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class I-V Tuition

1

Fees

₹ 240.0 per hour

Board

State, CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

No

State Syllabus Subjects taught

Mathematics

Engineering Entrance Coaching classes
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Engineering Entrance Coaching classes

1

Engineering Entrance Exams

IIT JEE Coaching Classes, BITSAT Coaching Classes, EAMCET, GATE Coaching Classes

IITJEE Coaching

IIT JEE Mains Coaching, IIT JEE Integrated Coaching, IIT JEE Advanced Coaching, IIT JEE Crash Course, IIT JEE Foundation Course

Type of class

Regular Classes, Crash Course

IIT-JEE Subjects

Maths

Cricket Coaching classes

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Cricket Coaching classes

1

5 out of 5 7 reviews

Anoop
V

Cricket Coaching

"One of the best cricket coach, very young but so much talent.He has perfect knowledge about the game. My son has really improved his game. "

Anoop
R

Class 11 Tuition

"The way sir explains the core concepts are undoubtedly great, the passion he has towards teaching is incredible, sometimes sir stays overtime to complete the topic, which shows his dedication levels in teaching, sir was very friendly and keen in his explanation.I really loved his way of teaching and I do suggested his name for my brother and his friends. "

Anoop
S

Class 8 Tuition

"The teaching was good. I was able to understand the teaching .After attending this classes I was able to score more marks in my exam. This classes was helped me a lot. "

Anoop
R

Class 12 Tuition

"1)Knowledgeable in his actions and disciplinary in his attitude 2)Knowledge gaining and conceptual clarity established with his teaching 3)Simple,humble and smart in character 4)Had good educational background and Great explanation skills 5)Multitalented and logical in his approach Thank you sir for sharing your wonderful knowledge with me and I hope you continue to inspire the minds. "

Load More
Have you attended any class with Anoop?

Answers by Anoop (32)

Answered on 02/11/2019 Learn CBSE/Class 6/Maths/Mensuration/NCERT Solutions/Exercise 10.1

(a) Perimeter = 4 × Side 30 = 4 × Side Side = (b) Perimeter = 3 × Side 30 = 3 × Side Side = (c) Perimeter = 6 × Side 30 = 6 × Side Side = read less ...more

(a) Perimeter = 4 × Side

30 = 4 × Side

Side =

(b) Perimeter = 3 × Side

30 = 3 × Side

Side = 

(c) Perimeter = 6 × Side

30 = 6 × Side

Side = 

read less

Answers 3 Comments
Dislike Bookmark

Answered on 02/11/2019 Learn CBSE/Class 6/Maths/Mensuration/NCERT Solutions/Exercise 10.1

Perimeter of triangle = Sum of the lengths of all sides of the triangle Perimeter = 10 + 14 + 15 = 39 cm
Answers 3 Comments
Dislike Bookmark

Answered on 22/10/2019 Learn CBSE/Class 6/Maths/Mensuration/NCERT Solutions/Exercise 10.1

Perimeter of regular pentagon = 5 × Length of side 100 = 5 × Side Side = 100 side=100/5 cm=20
Answers 2 Comments
Dislike Bookmark

Answered on 22/10/2019 Learn CBSE/Class 6/Maths/Mensuration/NCERT Solutions/Exercise 10.3

(a) Total area of the region = 100 × 144 = 14400 cm2 Area of one tile = 12 × 5 = 60 cm2 Number of tiles required = Therefore, 240 tiles are required. (b) Total area of the region = 70 × 36 = 2520 cm2 Area of one tile = 60 cm2 Number of tiles required = Therefore, 42 tiles are requir... ...more

(a) Total area of the region = 100 × 144 = 14400 cm2 Area of one tile = 12 × 5 = 60 cm2 Number of tiles required = Therefore, 240 tiles are required. (b) Total area of the region = 70 × 36 = 2520 cm2 Area of one tile = 60 cm2 Number of tiles required = Therefore, 42 tiles are required.

Answers 2 Comments
Dislike Bookmark

Contact

Load More

Anoop describes himself as I completed by btech i am a trained and experienced teacher with merit certifications. He conducts classes in Class 10 Tuition, Class 11 Tuition and Class 12 Tuition. Anoop is located in Jeedimetla, Hyderabad. Anoop takes at students Home and Regular Classes- at his Home. He has 1 years of teaching experience . Anoop has completed Bachelor of Technology (B.Tech.) from osmania university in 2018. HeĀ is well versed in English and Telugu. Anoop has got 7 reviews till now with 100% positive feedback.

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