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Niraj Mishra C++ Language trainer in Pune

Niraj Mishra

Trainer

Wakad, Pune, India - 411027.

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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 a C/C++ developer working with a Product based MNC for the past 6 years. My strengths in Software Development lies in C/C++ and Data Structures. Apart from these, I have a fair knowledge of basic Java, Angular 2, Ionic frameworks etc. I have been conducting C/C++ training for new joiners for the past 3 years now and have been awarded for the training. Apart from this, I have created a study group among my friend circle and we conduct regular sessions on different technologies to keep ourselves at par with the industry. Apart from my professional expertise, I am very strong in Mathematics and Quantitative Aptitude. I have always scored more than 90% in mathematics and have recently scored a perfect 170 (out of 170) in GRE quant. I look forward to take tuition on C/C++/Data Structures and Quantitative Aptitude for GRE/GMAT/Bank PO etc.

Languages Spoken

English

Hindi

Education

Institute of Technical Education and Research 2010

Bachelor of Technology (B.Tech.)

Address

Wakad, Pune, India - 411027

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Teaches

C++ Language Classes

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in C++ Language Classes

3

Proficiency level taught

Basic C++, Advanced C++

Teaching Experience in detail in C++ Language Classes

I have been conducting C++ training for new hires in my company for the past 3 years and have won few awards s well. I have received a lot of encouraging feedback from the trainees and hence I have decided to take classes out of my company duties as well.

C Language Classes

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in C Language Classes

3

Teaching Experience in detail in C Language Classes

I have been conducting C training for new hires in my company for the past 3 years and have won few awards s well. I have received a lot of encouraging feedback from the trainees and hence I have decided to take classes out of my company duties as well.

GRE Coaching classes

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in GRE Coaching classes

2

Background

Working Professional

Experience in taking GRE exam

Yes

Awards and Recognition

No

Teaching Experience in detail in GRE Coaching classes

I appeared for my GRE exam around 2.5 years back. Having a strong Mathematics and Quantitative background, I scored a perfect 170 (out of 170) in GRE Quantitative section. After this, I started to teach lots of my friends who were applying for GRE. I think now is a good time to devote myself into it on a professional level. Note: I will only conduct the training for Quantitative section of GRE.

Quantitative Aptitude Coaching

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Reviews

No Reviews yet!

FAQs

1. For what proficiency level do you teach ?

Basic C++ and Advanced C++

2. Which classes do you teach?

I teach C Language, C++ Language, GRE Coaching and Quantitative Aptitude Classes.

3. Do you provide a demo class?

Yes, I provide a free demo class.

4. How many years of experience do you have?

I have been teaching for 3 years.

Contact

Answers by Niraj Mishra (4)

Answered on 24/08/2017 Learn Exam Coaching/Quantitative Aptitude

A number divided by 169 leaves 91 as a remainder. If the same number is divided by 13, what will be the remainder?

The number is of the form 169m+91 (m being the quotient) now when this number is divided by 13, the Remainder will be zero. Because, 169m will be completely divisible by 13 and so will be 91. As per Remainder theorem, Remainder=> Remainder + Remainder]/C So Remainder=> Remainder /13 => Remainder... ...more
The number is of the form 169m+91 (m being the quotient) now when this number is divided by 13, the Remainder will be zero. Because, 169m will be completely divisible by 13 and so will be 91. As per Remainder theorem, Remainder[(A+B)/C] => Remainder [Remainder[A/C] + Remainder[B/C]]/C So Remainder [(169m + 91)/13] => Remainder [Remainder(169m/13) + Remainder(91/13)]/13 => Remainder [0 + 0]/13 => Remainder = 0
Answers 4 Comments
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Answered on 24/08/2017 Learn Exam Coaching/Quantitative Aptitude

Find the remainder when 7 to the power 162 is divided by 800?

We will apply Remainder Theorem here. Remainder Theorem: Remainder of /P is x Remainder]/P e.g. Remainder of (34 x 36)/32 can be calculated as: Remainder -> 2 Remainder -> 4 so Remainder /32 -> Remainder /32 -> 8 Applying this theorem the current problem: (7^162)/800 can be visualized as... ...more
We will apply Remainder Theorem here. Remainder Theorem: Remainder of [M x N ]/P is [Remainder[M/P] x Remainder[N/P]]/P e.g. Remainder of (34 x 36)/32 can be calculated as: Remainder[34/32] -> 2 Remainder[36/32] -> 4 so Remainder [34 x 36]/32 -> Remainder [2 x 4]/32 -> 8 Applying this theorem the current problem: (7^162)/800 can be visualized as 7 x 7 x 7 x 7........162 times / 800 now, Remainder [7/800] -> 7 Remainder [7^2/800] -> 49 Remainder [7^3/800] -> 343 Remainder [7^4/800] -> Remainder [2401/800] -> 1 Now, when you see a Remainder of 1, your can now find a pattern. Remainders of the next powers of 7 will form a cycle of 7, 49, 343, 1, 7, 49. 343, 1........ This happens because, as per the Remainder Theorem, 7^5 can be written as 7^4 x 7 so Remainder[7^5/800] -> Remainder [(7^4 x 7)/800] ->Remainder[(1 x 7)/800] -> 7 And so on for 7^6, 7^7, 7^8........ Every time 7 has a power that is a multiple of 4, the remainder will be 1. 162 can be seen as (4*40) + 2 so Remainder [7^162/800] -> Remainder [(7^160) x (7^2)/800] -> Remainder [Remainder[(7^160)/800] x [Remainder[(7^2)/800]]/800-> Remainder [1 x 49]/800 -> So, Remainder will be 49
Answers 3 Comments
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Answered on 24/08/2017 Learn Exam Coaching/Quantitative Aptitude

Find the greatest number which when divided by 16,18 and 32 exactly?

The greatest number that divide 16, 18 and 32 is the HCF of three numbers. 16 = 2x2x2x2 18 = 2x3x3 32 = 2x2x2x2x2 HCF of three numbers = 2 So the greatest number is 2
Answers 2 Comments
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Answered on 24/08/2017 Learn Exam Coaching/Quantitative Aptitude

The average age of A, B and C is 26 years. If the average age of A and C is 29 years, what is the age of B in a year?

Average Age of A, B and C = 26 years Therefore, A+B+C = 26 x 3 = 78 years Average age of A and C = 29 years Therefore, A+C = 29 x 2 = 58 years Age of B = (Sum of A, B and C) - (sum of A and C) = 78 - 58 = 20 years
Answers 3 Comments
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Contact

Teaches

C++ Language Classes

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in C++ Language Classes

3

Proficiency level taught

Basic C++, Advanced C++

Teaching Experience in detail in C++ Language Classes

I have been conducting C++ training for new hires in my company for the past 3 years and have won few awards s well. I have received a lot of encouraging feedback from the trainees and hence I have decided to take classes out of my company duties as well.

C Language Classes

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in C Language Classes

3

Teaching Experience in detail in C Language Classes

I have been conducting C training for new hires in my company for the past 3 years and have won few awards s well. I have received a lot of encouraging feedback from the trainees and hence I have decided to take classes out of my company duties as well.

GRE Coaching classes

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in GRE Coaching classes

2

Background

Working Professional

Experience in taking GRE exam

Yes

Awards and Recognition

No

Teaching Experience in detail in GRE Coaching classes

I appeared for my GRE exam around 2.5 years back. Having a strong Mathematics and Quantitative background, I scored a perfect 170 (out of 170) in GRE Quantitative section. After this, I started to teach lots of my friends who were applying for GRE. I think now is a good time to devote myself into it on a professional level. Note: I will only conduct the training for Quantitative section of GRE.

Quantitative Aptitude Coaching

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

No Reviews yet!

Answers by Niraj Mishra (4)

Answered on 24/08/2017 Learn Exam Coaching/Quantitative Aptitude

A number divided by 169 leaves 91 as a remainder. If the same number is divided by 13, what will be the remainder?

The number is of the form 169m+91 (m being the quotient) now when this number is divided by 13, the Remainder will be zero. Because, 169m will be completely divisible by 13 and so will be 91. As per Remainder theorem, Remainder=> Remainder + Remainder]/C So Remainder=> Remainder /13 => Remainder... ...more
The number is of the form 169m+91 (m being the quotient) now when this number is divided by 13, the Remainder will be zero. Because, 169m will be completely divisible by 13 and so will be 91. As per Remainder theorem, Remainder[(A+B)/C] => Remainder [Remainder[A/C] + Remainder[B/C]]/C So Remainder [(169m + 91)/13] => Remainder [Remainder(169m/13) + Remainder(91/13)]/13 => Remainder [0 + 0]/13 => Remainder = 0
Answers 4 Comments
Dislike Bookmark

Answered on 24/08/2017 Learn Exam Coaching/Quantitative Aptitude

Find the remainder when 7 to the power 162 is divided by 800?

We will apply Remainder Theorem here. Remainder Theorem: Remainder of /P is x Remainder]/P e.g. Remainder of (34 x 36)/32 can be calculated as: Remainder -> 2 Remainder -> 4 so Remainder /32 -> Remainder /32 -> 8 Applying this theorem the current problem: (7^162)/800 can be visualized as... ...more
We will apply Remainder Theorem here. Remainder Theorem: Remainder of [M x N ]/P is [Remainder[M/P] x Remainder[N/P]]/P e.g. Remainder of (34 x 36)/32 can be calculated as: Remainder[34/32] -> 2 Remainder[36/32] -> 4 so Remainder [34 x 36]/32 -> Remainder [2 x 4]/32 -> 8 Applying this theorem the current problem: (7^162)/800 can be visualized as 7 x 7 x 7 x 7........162 times / 800 now, Remainder [7/800] -> 7 Remainder [7^2/800] -> 49 Remainder [7^3/800] -> 343 Remainder [7^4/800] -> Remainder [2401/800] -> 1 Now, when you see a Remainder of 1, your can now find a pattern. Remainders of the next powers of 7 will form a cycle of 7, 49, 343, 1, 7, 49. 343, 1........ This happens because, as per the Remainder Theorem, 7^5 can be written as 7^4 x 7 so Remainder[7^5/800] -> Remainder [(7^4 x 7)/800] ->Remainder[(1 x 7)/800] -> 7 And so on for 7^6, 7^7, 7^8........ Every time 7 has a power that is a multiple of 4, the remainder will be 1. 162 can be seen as (4*40) + 2 so Remainder [7^162/800] -> Remainder [(7^160) x (7^2)/800] -> Remainder [Remainder[(7^160)/800] x [Remainder[(7^2)/800]]/800-> Remainder [1 x 49]/800 -> So, Remainder will be 49
Answers 3 Comments
Dislike Bookmark

Answered on 24/08/2017 Learn Exam Coaching/Quantitative Aptitude

Find the greatest number which when divided by 16,18 and 32 exactly?

The greatest number that divide 16, 18 and 32 is the HCF of three numbers. 16 = 2x2x2x2 18 = 2x3x3 32 = 2x2x2x2x2 HCF of three numbers = 2 So the greatest number is 2
Answers 2 Comments
Dislike Bookmark

Answered on 24/08/2017 Learn Exam Coaching/Quantitative Aptitude

The average age of A, B and C is 26 years. If the average age of A and C is 29 years, what is the age of B in a year?

Average Age of A, B and C = 26 years Therefore, A+B+C = 26 x 3 = 78 years Average age of A and C = 29 years Therefore, A+C = 29 x 2 = 58 years Age of B = (Sum of A, B and C) - (sum of A and C) = 78 - 58 = 20 years
Answers 3 Comments
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Contact

Niraj Mishra describes himself as Trainer. He conducts classes in C Language, C++ Language and GRE Coaching. Niraj is located in Wakad, Pune. Niraj takes Regular Classes- at his Home and Online Classes- via online medium. He has 3 years of teaching experience . Niraj has completed Bachelor of Technology (B.Tech.) from Institute of Technical Education and Research in 2010. He is well versed in English and Hindi.

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