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Sujeet Kumar BCA Tuition trainer in Bangalore

Sujeet Kumar

Mathematics Faculty

Kavery Nagar, Bangalore, India - 560087.

Verified 1 Student

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Referral Discount: Get ₹ 500 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

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Teaches

BCA Tuition

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

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

Quantitative aptitude/ Numerical ability, Reasoning Ability

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

3

Board

International Baccalaureate, CBSE, State, ICSE

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

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

International Baccalaureate, CBSE, State, ICSE

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

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

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

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! Be the first one to Review

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

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

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

Quantitative aptitude/ Numerical ability, Reasoning Ability

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

3

Board

International Baccalaureate, CBSE, State, ICSE

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

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

International Baccalaureate, CBSE, State, ICSE

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

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

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

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! Be the first one to Review

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.

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