UrbanPro
Login Ask & Answer Success Stories Paid Courses Free Classes Tuition & Classes Fees Write a Review All Categories Help Center
Post your Learning Need Signup as a Tutor
✕

Popular Cities

Bangalore

Hyderabad

Delhi

Mumbai

Chennai

Pune

Kolkata

Gurgaon

Ahmedabad

Jaipur

Noida

✕
true
Sujeet Kumar BCA Tuition trainer in Bangalore

Sujeet Kumar

Mathematics Faculty

Kavery Nagar, Bangalore, India - 560087.

Verified 1 Student

Contact
Referral Discount: Get ₹ 250 off when you make a payment to start classes. Get started by Booking a Demo.

Details verified of Sujeet Kumar✕

Identity

Education

Know how UrbanPro verifies Tutor details

Identity is verified based on matching the details uploaded by the Tutor with government databases.

Overview

Mathematics Faculty with 8 year experience for competitive level exams like Aryabhatta, NTSE, NSTSE, JSTSE, NMTC, PRE-RMO etc including different boards like CBSE, ICSE and State boards. I take classes for SSC- CGL (Quant Section) and BCA also.

Languages Spoken

English

Hindi

Education

Banaras Hindu University, Varanasi 2013

Master of Science (M.Sc.)

Address

Kavery Nagar, Bangalore, India - 560087

Verified Info

ID Verified

Education Verified

Phone Verified

Email Verified

Taught Students from these Schools

A

Amity International School

Saket, Delhi

Amity International School

Sector-44 Noida-201303 , Noida

R

Ryan International School

Adhyatmic Nagar, Ghaziabad

A

Amity international school

Pushp Vihar, Delhi

K

Kendriya Vidayalya

Sector 9 Noida, Delhi

A

Amity International School

Vasundhra, Ghaziabad

A

Amity International School

Gurgaon HO, Delhi

K

Kendriya Vidyalaya

Atta Market, Noida

Report this Profile

✕
x

Is this listing inaccurate or duplicate? Any other problem?

Please tell us about the problem and we will fix it.

Please describe the problem that you see in this page.

Type the letters as shown below *

Please enter the letters as show below

Teaches

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

3

BCA Subject

Mathematics

Type of class

Regular Classes, Crash Course

Class strength catered to

One on one/ Private Tutions

Taught in School or College

Yes

Bank Clerical Exam Coaching classes
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Bank Clerical Exam Coaching classes

2

Target section catered to

Reasoning Ability, Quantitative aptitude/ Numerical ability

Class 9 Tuition
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 9 Tuition

3

Board

ICSE, CBSE, State, International Baccalaureate

IB Subjects taught

Mathematics

CBSE Subjects taught

Mathematics

ICSE Subjects taught

Mathematics

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics

Class 10 Tuition
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 10 Tuition

4

Board

ICSE, CBSE, State, International Baccalaureate

IB Subjects taught

Mathematics

CBSE Subjects taught

Mathematics

ICSE Subjects taught

Mathematics

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics

Class 11 Tuition
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 11 Tuition

3

CBSE Subjects taught

Mathematics

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics

Class 12 Tuition
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 12 Tuition

3

Board

CBSE, State

CBSE Subjects taught

Mathematics

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics

Quantitative Aptitude Coaching
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Quantitative Aptitude Coaching

2

Reviews

No Reviews yet!

FAQs

1. Which BCA subjects do you tutor for?

Mathematics

2. Do you have any prior teaching experience?

Yes

3. Which classes do you teach?

I teach BCA Tuition, Bank Clerical Exam Coaching, Class 10 Tuition, Class 11 Tuition, Class 12 Tuition, Class 9 Tuition, Quantitative Aptitude and RRB Exam 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 3 years.

Contact

Answers by Sujeet Kumar (3)

Answered on 09/12/2014 Learn Tuition/Class XI-XII Tuition (PUC)

Hi friends, please post the solution for following problem. If, (root X) + Y = 11 X + (root Y)... ...more
Hi friends, please post the solution for following problem. If, (root X) + Y = 11 X + (root Y) =7 Then X , Y = ?

root(x)+y=11.........................(1) root(y)+x=7...........................(2) y <=11 & x<=7 root(y)+x is equivalent to 0 (mod 7) or,x is equivalent to root(y) (mod 7) suppose, root(x)=k or,x=k^2 So, x is a perfect square integer and <=7 Now perfect square integer less than... ...more
root(x)+y=11.........................(1) root(y)+x=7...........................(2) y <=11 & x<=7 root(y)+x is equivalent to 0 (mod 7) or,x is equivalent to root(y) (mod 7) suppose, root(x)=k or,x=k^2 So, x is a perfect square integer and <=7 Now perfect square integer less than 7 is 4 and 1; and 7 is not a perfect square Thus, x=1 or 4 let, x = 1 then x is equivalent to -root(y) (mod 7) or 1 is equivalent to -6 (mod 7) For , x=1,y=36 {not possible} let, x=4 x is equivalent to -root(y) (mod 7) or 4 is equivalent to -3 (mod 7) For, x=4 ,y=9 {possible} Thus, one root of x=4 and that of y=9
Answers 85 Comments
Dislike Bookmark

Answered on 09/12/2014 Learn Tuition/Class XI-XII Tuition (PUC)

Hi friends, please post the solution for following problem. If, (root X) + Y = 11 X + (root Y)... ...more
Hi friends, please post the solution for following problem. If, (root X) + Y = 11 X + (root Y) =7 Then X , Y = ?

Sorry, here question marks is the root sign..
Answers 85 Comments
Dislike Bookmark

Answered on 09/12/2014 Learn Tuition/Class XI-XII Tuition (PUC)

Hi friends, please post the solution for following problem. If, (root X) + Y = 11 X + (root Y)... ...more
Hi friends, please post the solution for following problem. If, (root X) + Y = 11 X + (root Y) =7 Then X , Y = ?

?x+y=11............................(1) ?y+x=7.................................(2) y?11 & x?7 ?y+x?0 (mod 7) or,x?-?y (mod 7) Suppose, ?x=k or,x=k^2 & So, x will be a perfect square and ?7 Now perfect square integer less than 7 is 4 and 1.....& 7 is not a perfect square. Thus, x= either... ...more
?x+y=11............................(1) ?y+x=7.................................(2) y?11 & x?7 ?y+x?0 (mod 7) or,x?-?y (mod 7) Suppose, ?x=k or,x=k^2 & So, x will be a perfect square and ?7 Now perfect square integer less than 7 is 4 and 1.....& 7 is not a perfect square. Thus, x= either 1 or 4 let, x = 1 then x?-?y (mod 7) 1?-6 (mod 7) For , x=1,y=36 {not possible} let, x=4 x?-?y (mod 7) 4?-3 (mod 7) For, x=4 ,y=9 {possible} Thus, one root of x=4 and that of y=9
Answers 85 Comments
Dislike Bookmark

Contact

Teaches

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

3

BCA Subject

Mathematics

Type of class

Regular Classes, Crash Course

Class strength catered to

One on one/ Private Tutions

Taught in School or College

Yes

Bank Clerical Exam Coaching classes
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Bank Clerical Exam Coaching classes

2

Target section catered to

Reasoning Ability, Quantitative aptitude/ Numerical ability

Class 9 Tuition
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 9 Tuition

3

Board

ICSE, CBSE, State, International Baccalaureate

IB Subjects taught

Mathematics

CBSE Subjects taught

Mathematics

ICSE Subjects taught

Mathematics

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics

Class 10 Tuition
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 10 Tuition

4

Board

ICSE, CBSE, State, International Baccalaureate

IB Subjects taught

Mathematics

CBSE Subjects taught

Mathematics

ICSE Subjects taught

Mathematics

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics

Class 11 Tuition
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 11 Tuition

3

CBSE Subjects taught

Mathematics

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics

Class 12 Tuition
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 12 Tuition

3

Board

CBSE, State

CBSE Subjects taught

Mathematics

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics

Quantitative Aptitude Coaching
1 Student

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Quantitative Aptitude Coaching

2

No Reviews yet!

Answers by Sujeet Kumar (3)

Answered on 09/12/2014 Learn Tuition/Class XI-XII Tuition (PUC)

Hi friends, please post the solution for following problem. If, (root X) + Y = 11 X + (root Y)... ...more
Hi friends, please post the solution for following problem. If, (root X) + Y = 11 X + (root Y) =7 Then X , Y = ?

root(x)+y=11.........................(1) root(y)+x=7...........................(2) y <=11 & x<=7 root(y)+x is equivalent to 0 (mod 7) or,x is equivalent to root(y) (mod 7) suppose, root(x)=k or,x=k^2 So, x is a perfect square integer and <=7 Now perfect square integer less than... ...more
root(x)+y=11.........................(1) root(y)+x=7...........................(2) y <=11 & x<=7 root(y)+x is equivalent to 0 (mod 7) or,x is equivalent to root(y) (mod 7) suppose, root(x)=k or,x=k^2 So, x is a perfect square integer and <=7 Now perfect square integer less than 7 is 4 and 1; and 7 is not a perfect square Thus, x=1 or 4 let, x = 1 then x is equivalent to -root(y) (mod 7) or 1 is equivalent to -6 (mod 7) For , x=1,y=36 {not possible} let, x=4 x is equivalent to -root(y) (mod 7) or 4 is equivalent to -3 (mod 7) For, x=4 ,y=9 {possible} Thus, one root of x=4 and that of y=9
Answers 85 Comments
Dislike Bookmark

Answered on 09/12/2014 Learn Tuition/Class XI-XII Tuition (PUC)

Hi friends, please post the solution for following problem. If, (root X) + Y = 11 X + (root Y)... ...more
Hi friends, please post the solution for following problem. If, (root X) + Y = 11 X + (root Y) =7 Then X , Y = ?

Sorry, here question marks is the root sign..
Answers 85 Comments
Dislike Bookmark

Answered on 09/12/2014 Learn Tuition/Class XI-XII Tuition (PUC)

Hi friends, please post the solution for following problem. If, (root X) + Y = 11 X + (root Y)... ...more
Hi friends, please post the solution for following problem. If, (root X) + Y = 11 X + (root Y) =7 Then X , Y = ?

?x+y=11............................(1) ?y+x=7.................................(2) y?11 & x?7 ?y+x?0 (mod 7) or,x?-?y (mod 7) Suppose, ?x=k or,x=k^2 & So, x will be a perfect square and ?7 Now perfect square integer less than 7 is 4 and 1.....& 7 is not a perfect square. Thus, x= either... ...more
?x+y=11............................(1) ?y+x=7.................................(2) y?11 & x?7 ?y+x?0 (mod 7) or,x?-?y (mod 7) Suppose, ?x=k or,x=k^2 & So, x will be a perfect square and ?7 Now perfect square integer less than 7 is 4 and 1.....& 7 is not a perfect square. Thus, x= either 1 or 4 let, x = 1 then x?-?y (mod 7) 1?-6 (mod 7) For , x=1,y=36 {not possible} let, x=4 x?-?y (mod 7) 4?-3 (mod 7) For, x=4 ,y=9 {possible} Thus, one root of x=4 and that of y=9
Answers 85 Comments
Dislike Bookmark

Contact

Sujeet Kumar describes himself as Mathematics Faculty. He conducts classes in BCA Tuition, Bank Clerical Exam Coaching and Class 10 Tuition. Sujeet is located in Kavery Nagar, Bangalore. Sujeet takes at students Home, Regular Classes- at his Home and Online Classes- via online medium. He has 4 years of teaching experience . Sujeet has completed Master of Science (M.Sc.) from Banaras Hindu University, Varanasi in 2013. He is well versed in English and Hindi.

X
X

Post your Learning Need

Let us shortlist and give the best tutors and institutes.

or

Send Enquiry to Sujeet Kumar

Let Sujeet Kumar know you are interested in their class

Reply to 's review

Enter your reply*

1500/1500

Please enter your reply

Your reply should contain a minimum of 10 characters

Your reply has been successfully submitted.

This website uses cookies

We use cookies to improve user experience. Choose what cookies you allow us to use. You can read more about our Cookie Policy in our Privacy Policy

Accept All
Decline All

UrbanPro.com is India's largest network of most trusted tutors and institutes. Over 55 lakh students rely on UrbanPro.com, to fulfill their learning requirements across 1,000+ categories. Using UrbanPro.com, parents, and students can compare multiple Tutors and Institutes and choose the one that best suits their requirements. More than 7.5 lakh verified Tutors and Institutes are helping millions of students every day and growing their tutoring business on UrbanPro.com. Whether you are looking for a tutor to learn mathematics, a German language trainer to brush up your German language skills or an institute to upgrade your IT skills, we have got the best selection of Tutors and Training Institutes for you. Read more

ThinkVidya Learning Pvt Ltd © 2010-2024All Rights Reserved