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BCA Tuition > BCA Tuition in Bangalore > Sujeet Kumar

Sujeet Kumar BCA Tuition trainer in Bangalore

Sujeet Kumar

Mathematics Faculty

Kavery Nagar, Bangalore, India - 560087.

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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

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

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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

Crash Course, Regular Classes

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

State, CBSE, International Baccalaureate, ICSE

Subjects taught

Mathematics

Taught in School or College

Yes

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

State, CBSE, International Baccalaureate, ICSE

Subjects taught

Mathematics

Taught in School or College

Yes

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

Subjects taught

Mathematics

Taught in School or College

Yes

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

State, CBSE

Subjects taught

Mathematics

Taught in School or College

Yes

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)...

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)...

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)...

?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

Crash Course, Regular Classes

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

State, CBSE, International Baccalaureate, ICSE

Subjects taught

Mathematics

Taught in School or College

Yes

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

State, CBSE, International Baccalaureate, ICSE

Subjects taught

Mathematics

Taught in School or College

Yes

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

Subjects taught

Mathematics

Taught in School or College

Yes

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

State, CBSE

Subjects taught

Mathematics

Taught in School or College

Yes

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)...

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)...

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)...

?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.

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