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BCA Tuition > BCA Tuition in Bangalore > Sujeet Kumar

Sujeet Kumar BCA Tuition trainer in Bangalore

Sujeet Kumar

locationImg Kavery Nagar, Bangalore
3 yrs of Exp
students 1 student
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Details verified of Sujeet Kumar✕

Identity

Education

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

Mathematics Faculty

Online Classes
Student's home
Tutor's home
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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Taught Students from these Schools

K

Kendriya Vidayalya

Sector 9 Noida, Delhi

Amity International School

Sector-44 Noida-201303 , Noida

A

Amity International School

Saket, Delhi

K

Kendriya Vidyalaya

Atta Market, Noida

R

Ryan International School

Adhyatmic Nagar, Ghaziabad

A

Amity international school

Pushp Vihar, Delhi

A

Amity International School

Gurgaon HO, Delhi

A

Amity International School

Vasundhra, Ghaziabad

Teaches

BCA Tuition

Class Location

Online class via Zoom

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

Class Location

Online class via Zoom

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

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 9 Tuition

3

Board

ICSE, CBSE, State, International Baccalaureate

Subjects taught

Mathematics

Taught in School or College

Yes

Class 10 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 10 Tuition

4

Board

ICSE, CBSE, State, International Baccalaureate

Subjects taught

Mathematics

Taught in School or College

Yes

Class 11 Tuition

Class Location

Online class via Zoom

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

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 12 Tuition

3

Board

CBSE, State

Subjects taught

Mathematics

Taught in School or College

Yes

Quantitative Aptitude Coaching

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Quantitative Aptitude Coaching

2

Reviews

No Reviews yet!

Answers by Sujeet Kumar

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

Ask a Question

Post a Lesson

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)

Ask a Question

Post a Lesson

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)

Ask a Question

Post a Lesson

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
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Teaches

BCA Tuition

Class Location

Online class via Zoom

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

Class Location

Online class via Zoom

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

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 9 Tuition

3

Board

ICSE, CBSE, State, International Baccalaureate

Subjects taught

Mathematics

Taught in School or College

Yes

Class 10 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 10 Tuition

4

Board

ICSE, CBSE, State, International Baccalaureate

Subjects taught

Mathematics

Taught in School or College

Yes

Class 11 Tuition

Class Location

Online class via Zoom

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

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 12 Tuition

3

Board

CBSE, State

Subjects taught

Mathematics

Taught in School or College

Yes

Quantitative Aptitude Coaching

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Quantitative Aptitude Coaching

2

No Reviews yet!

Answers by Sujeet Kumar

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

Ask a Question

Post a Lesson

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)

Ask a Question

Post a Lesson

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)

Ask a Question

Post a Lesson

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
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