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Engineering Diploma Tuition > Engineering Diploma Tuition in Belgaum > Afroz Ahmed

Afroz Ahmed Engineering Diploma Tuition trainer in Belgaum

Afroz Ahmed

Tutor

Kanabargi, Belgaum, India - 590016.

1 Student

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Overview

I pursued my B.E in mechanical stream from Bangalore Institute Of Technology, Bangalore in 2015. For the past one year I have been taking home tuition for physics, maths and diploma engineering. Also, I work as an online tutor teaching Math and Physics to US students for the past 10 months. I am very passionate about teaching. I am very strong in maths n physics of PUC level. <br/> <br/>My teaching strength is that I study every minute detail of the topic to be taught or you can say that I do a deep research about everything related to the topic before beginning the class. <br/> <br/>Before taking a class, I prepare myself fully and get all the details about the topic to be taught. I myself question myself about the various topics and try to find its answers so that my students get a very clear idea and concept of that particular topic and thus I try to build the concepts of my students to great extent. <br/> <br/>I am not any stereo-type tutor who study the topic and then blindly teach them to students without realizing that whether they are understanding it or not. I enjoy studying and thereby enjoy it more while teaching or more precisely sharing my knowledge with the students. <br/> <br/>I assure you all that it will be a real fun in my company and surely you will get all your concepts clear by the end of the class.

Languages Spoken

Hindi

English

Education

VTU 2015

Bachelor of Engineering (B.E.)

Address

Kanabargi, Belgaum, India - 590016

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Teaches

Engineering Diploma Tuition
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Engineering Diploma Branch

Mechanical Engineering Diploma, Engineering Diploma 1st Year, Automobile Engineering Diploma

Engineering Diploma Subject

Basic Math, Basic Physics, Engineering Mathematics, Engineering Mechanics, Computer Fundamentals

Mechanical Engineering Diploma Subject

Thermal Engineering, Computer Programming, Strength of Materials, Fluid Mechanics and Machinery, Applied Mathematics, Theory of Machines & Mechanisms

Automobile Engineering Diploma Subject

Applied Math, Computer Programming, Theory of Machines & Mechanisms, Strength of Materials

Type of class

Crash Course, Regular Classes

Class strength catered to

Group Classes, One on one/ Private Tutions

Taught in School or College

No

Class 11 Tuition
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE

CBSE Subjects taught

Physics, English, Mathematics

Taught in School or College

No

Class 12 Tuition
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE

CBSE Subjects taught

Physics, English, Mathematics

Taught in School or College

No

BTech Tuition
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in BTech Tuition

1

BTech Mechanical subjects

Fluid Mechanics, Hydraulic and Pneumatic Control, Turbomachines, Mechanics of Machines, Finite Element Analysis, Heat & Mass Transfer, Kinematics of Machinery, Operations Research, Thermodynamics, Tribology, Computer Integrated Manufacturing, Strength of Materials, Automobile Engineering, Energy Engineering and Management

BTech Branch

BTech 1st Year Engineering, BTech Mechanical Engineering

Type of class

Crash Course

Class strength catered to

Group Classes, One on one/ Private Tutions

Taught in School or College

No

BTech 1st Year subjects

Engineering Mathematics (M1), Computer science, Basic Mechanical Engineering, Engineering Physics, Advanced Mathematics (M2)

BCA Tuition
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in BCA Tuition

1

BCA Subject

Mathematics

Type of class

Crash Course

Class strength catered to

Group Classes, One on one/ Private Tutions

Taught in School or College

No

Class 9 Tuition
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE, ICSE

CBSE Subjects taught

Science, English, Hindi, Mathematics

ICSE Subjects taught

Chemistry, English, Physics, Computer Application, Mathematics, English Literature, Hindi

Taught in School or College

No

Class 10 Tuition
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE, ICSE

CBSE Subjects taught

Science, English, Hindi, Mathematics

ICSE Subjects taught

Chemistry, English, Physics, Computer Application, Mathematics, English Literature, Hindi

Taught in School or College

No

Quantitative Aptitude Coaching
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Teaching Experience in detail in Quantitative Aptitude Coaching

I got great interest in aptitude during my college campus placement time where you have to clear aptitude test to get into next round. So I started solving many apti questions and gradually became an expert in it.

Reviews

No Reviews yet!

FAQs

1. Which Engineering Diploma branches do you tutor for?

Mechanical Engineering Diploma, Engineering Diploma 1st Year and Automobile Engineering Diploma

2. Do you have any prior teaching experience?

No

3. Which classes do you teach?

I teach BCA Tuition, BTech Tuition, Class 10 Tuition, Class 11 Tuition, Class 12 Tuition, Class 9 Tuition, Engineering Diploma Tuition and Quantitative Aptitude 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 Afroz Ahmed (8)

Answered on 09/12/2017 Learn CBSE/Class 9/Mathematics +1 Tuition/Class IX-X Tuition

Solve : 55x + 52y = 217; 52x + 55y = 217

First of all this is a question of simultaneous equations. But it has a trick under its sleeve. The main thing is to identify that little twist in this question. The RHS of both the equations have the same number. So, if we subtract one equation from the other, we can express x in terms of y. Lets... ...more
First of all this is a question of simultaneous equations. But it has a trick under its sleeve. The main thing is to identify that little twist in this question. The RHS of both the equations have the same number. So, if we subtract one equation from the other, we can express x in terms of y. Lets do it... 55x + 52y - 52x - 55y = 217-217 3x - 3y = 0 x = y. Great!! We got x = y. So substituting x in place of y in 1st equation we can get the value of x. Therefore, 55x + 52x = 217 107x = 217 x = 217/107 = 2.028 Also, y = x So y = 2.028 Therefore, the solution is (2.028,2.028)
Answers 8 Comments
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Answered on 06/12/2017 Learn CBSE/Class 10/Mathematics +1 Tuition/Class IX-X Tuition

If sin A=1/2, then find the value of cos A.

Use the trigonometric identity sin^2 A + cos^2 A = 1 Express the given in terms of cos A. So, cos A = sqrt(1-sin^2 A) sin A = 1/2 sin^2 A = (1/2)^2 = 1/4 So, cos A = sqrt(1 - 1/4) = sqrt(3/4) = sqrt(3)/ 2 Therefore, cos A = sqrt(3)/2
Answers 43 Comments
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Answered on 06/12/2017 Learn CBSE/Class 9/Mathematics +1 Tuition/Class IX-X Tuition

Factorise: (a – b)3 + (b – c)3 + (c – a)3

First factorize (a-b)^3 , (b-c)^3 and (c-a)^3 separately and then add them together to get the final result. (a-b)^3 = a^3 - b^3 - 3a^2b + 3ab^2 ................(1) (b-c)^3 = b^3 - c^3 - 3b^2c + 3bc^2 .................(2) (c-a)^3 = c^3 - a^3 - 3ac^2 + 3 a^2c ... ...more
First factorize (a-b)^3 , (b-c)^3 and (c-a)^3 separately and then add them together to get the final result. (a-b)^3 = a^3 - b^3 - 3a^2b + 3ab^2 ................(1) (b-c)^3 = b^3 - c^3 - 3b^2c + 3bc^2 .................(2) (c-a)^3 = c^3 - a^3 - 3ac^2 + 3 a^2c ..................(3) Now add (1), (2) and (3). This gives, a^3 - b^3 - 3a^2b + 3ab^2 + b^3 - c^3 - 3b^2c + 3bc^2 + c^3 - a^3 - 3ac^2 + 3 a^2c = -3a^2b + 3ab^2 - 3b^2c + 3bc^2 - 3ac^2 + 3 a^2c
Answers 15 Comments
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Answered on 17/11/2017 Learn Tuition/Class IX-X Tuition

What is Newton's second law?

Newton's second law of motion states that the force applied on an object is directly proportional to the rate of change of its momentum in the direction of the applied force. Momentum of an object is equal to the product of its mass and velocity. If the mass of the object is constant, then the force... ...more
Newton's second law of motion states that the force applied on an object is directly proportional to the rate of change of its momentum in the direction of the applied force. Momentum of an object is equal to the product of its mass and velocity. If the mass of the object is constant, then the force applied is equal to the acceleration of the object. Acceleration is defined as the rate of change of velocity.
Answers 206 Comments
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Answered on 17/09/2016 Learn Language/Hindi Language +1 Language/Hindi Language/Hindi Speaking

How to learn Hindi in 3 weeks?

1. Learn the grammar part. 2. Try to learn some words in Hindi which we use in our daily life. 3. Now try to frame small sentences in Hindi using the words and the grammar. By this way u will definitely enjoy learning Hindi and 3 weeks will be enough to learn basic Hindi language. If u need my help... ...more
1. Learn the grammar part. 2. Try to learn some words in Hindi which we use in our daily life. 3. Now try to frame small sentences in Hindi using the words and the grammar. By this way u will definitely enjoy learning Hindi and 3 weeks will be enough to learn basic Hindi language. If u need my help I am always there for u..
Answers 67 Comments
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Contact

View All

Teaches

Engineering Diploma Tuition
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Engineering Diploma Branch

Mechanical Engineering Diploma, Engineering Diploma 1st Year, Automobile Engineering Diploma

Engineering Diploma Subject

Basic Math, Basic Physics, Engineering Mathematics, Engineering Mechanics, Computer Fundamentals

Mechanical Engineering Diploma Subject

Thermal Engineering, Computer Programming, Strength of Materials, Fluid Mechanics and Machinery, Applied Mathematics, Theory of Machines & Mechanisms

Automobile Engineering Diploma Subject

Applied Math, Computer Programming, Theory of Machines & Mechanisms, Strength of Materials

Type of class

Crash Course, Regular Classes

Class strength catered to

Group Classes, One on one/ Private Tutions

Taught in School or College

No

Class 11 Tuition
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE

CBSE Subjects taught

Physics, English, Mathematics

Taught in School or College

No

Class 12 Tuition
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE

CBSE Subjects taught

Physics, English, Mathematics

Taught in School or College

No

BTech Tuition
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in BTech Tuition

1

BTech Mechanical subjects

Fluid Mechanics, Hydraulic and Pneumatic Control, Turbomachines, Mechanics of Machines, Finite Element Analysis, Heat & Mass Transfer, Kinematics of Machinery, Operations Research, Thermodynamics, Tribology, Computer Integrated Manufacturing, Strength of Materials, Automobile Engineering, Energy Engineering and Management

BTech Branch

BTech 1st Year Engineering, BTech Mechanical Engineering

Type of class

Crash Course

Class strength catered to

Group Classes, One on one/ Private Tutions

Taught in School or College

No

BTech 1st Year subjects

Engineering Mathematics (M1), Computer science, Basic Mechanical Engineering, Engineering Physics, Advanced Mathematics (M2)

BCA Tuition
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in BCA Tuition

1

BCA Subject

Mathematics

Type of class

Crash Course

Class strength catered to

Group Classes, One on one/ Private Tutions

Taught in School or College

No

Class 9 Tuition
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE, ICSE

CBSE Subjects taught

Science, English, Hindi, Mathematics

ICSE Subjects taught

Chemistry, English, Physics, Computer Application, Mathematics, English Literature, Hindi

Taught in School or College

No

Class 10 Tuition
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE, ICSE

CBSE Subjects taught

Science, English, Hindi, Mathematics

ICSE Subjects taught

Chemistry, English, Physics, Computer Application, Mathematics, English Literature, Hindi

Taught in School or College

No

Quantitative Aptitude Coaching
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Teaching Experience in detail in Quantitative Aptitude Coaching

I got great interest in aptitude during my college campus placement time where you have to clear aptitude test to get into next round. So I started solving many apti questions and gradually became an expert in it.

No Reviews yet!

Answers by Afroz Ahmed (8)

Answered on 09/12/2017 Learn CBSE/Class 9/Mathematics +1 Tuition/Class IX-X Tuition

Solve : 55x + 52y = 217; 52x + 55y = 217

First of all this is a question of simultaneous equations. But it has a trick under its sleeve. The main thing is to identify that little twist in this question. The RHS of both the equations have the same number. So, if we subtract one equation from the other, we can express x in terms of y. Lets... ...more
First of all this is a question of simultaneous equations. But it has a trick under its sleeve. The main thing is to identify that little twist in this question. The RHS of both the equations have the same number. So, if we subtract one equation from the other, we can express x in terms of y. Lets do it... 55x + 52y - 52x - 55y = 217-217 3x - 3y = 0 x = y. Great!! We got x = y. So substituting x in place of y in 1st equation we can get the value of x. Therefore, 55x + 52x = 217 107x = 217 x = 217/107 = 2.028 Also, y = x So y = 2.028 Therefore, the solution is (2.028,2.028)
Answers 8 Comments
Dislike Bookmark

Answered on 06/12/2017 Learn CBSE/Class 10/Mathematics +1 Tuition/Class IX-X Tuition

If sin A=1/2, then find the value of cos A.

Use the trigonometric identity sin^2 A + cos^2 A = 1 Express the given in terms of cos A. So, cos A = sqrt(1-sin^2 A) sin A = 1/2 sin^2 A = (1/2)^2 = 1/4 So, cos A = sqrt(1 - 1/4) = sqrt(3/4) = sqrt(3)/ 2 Therefore, cos A = sqrt(3)/2
Answers 43 Comments
Dislike Bookmark

Answered on 06/12/2017 Learn CBSE/Class 9/Mathematics +1 Tuition/Class IX-X Tuition

Factorise: (a – b)3 + (b – c)3 + (c – a)3

First factorize (a-b)^3 , (b-c)^3 and (c-a)^3 separately and then add them together to get the final result. (a-b)^3 = a^3 - b^3 - 3a^2b + 3ab^2 ................(1) (b-c)^3 = b^3 - c^3 - 3b^2c + 3bc^2 .................(2) (c-a)^3 = c^3 - a^3 - 3ac^2 + 3 a^2c ... ...more
First factorize (a-b)^3 , (b-c)^3 and (c-a)^3 separately and then add them together to get the final result. (a-b)^3 = a^3 - b^3 - 3a^2b + 3ab^2 ................(1) (b-c)^3 = b^3 - c^3 - 3b^2c + 3bc^2 .................(2) (c-a)^3 = c^3 - a^3 - 3ac^2 + 3 a^2c ..................(3) Now add (1), (2) and (3). This gives, a^3 - b^3 - 3a^2b + 3ab^2 + b^3 - c^3 - 3b^2c + 3bc^2 + c^3 - a^3 - 3ac^2 + 3 a^2c = -3a^2b + 3ab^2 - 3b^2c + 3bc^2 - 3ac^2 + 3 a^2c
Answers 15 Comments
Dislike Bookmark

Answered on 17/11/2017 Learn Tuition/Class IX-X Tuition

What is Newton's second law?

Newton's second law of motion states that the force applied on an object is directly proportional to the rate of change of its momentum in the direction of the applied force. Momentum of an object is equal to the product of its mass and velocity. If the mass of the object is constant, then the force... ...more
Newton's second law of motion states that the force applied on an object is directly proportional to the rate of change of its momentum in the direction of the applied force. Momentum of an object is equal to the product of its mass and velocity. If the mass of the object is constant, then the force applied is equal to the acceleration of the object. Acceleration is defined as the rate of change of velocity.
Answers 206 Comments
Dislike Bookmark

Answered on 17/09/2016 Learn Language/Hindi Language +1 Language/Hindi Language/Hindi Speaking

How to learn Hindi in 3 weeks?

1. Learn the grammar part. 2. Try to learn some words in Hindi which we use in our daily life. 3. Now try to frame small sentences in Hindi using the words and the grammar. By this way u will definitely enjoy learning Hindi and 3 weeks will be enough to learn basic Hindi language. If u need my help... ...more
1. Learn the grammar part. 2. Try to learn some words in Hindi which we use in our daily life. 3. Now try to frame small sentences in Hindi using the words and the grammar. By this way u will definitely enjoy learning Hindi and 3 weeks will be enough to learn basic Hindi language. If u need my help I am always there for u..
Answers 67 Comments
Dislike Bookmark

Contact

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Afroz Ahmed describes himself as Tutor. He conducts classes in BCA Tuition, BTech Tuition and Class 10 Tuition. Afroz Ahmed is located in Kanabargi, Belgaum. Afroz Ahmed takes Regular Classes- at his Home and Online Classes- via online medium. He has 1 years of teaching experience . Afroz Ahmed has completed Bachelor of Engineering (B.E.) from VTU in 2015. He is well versed in Hindi and English.

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