Olaichanditala, Kolkata, India - 712103.
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English
Hindi
Bengali
Korean
Calcutta University 2012
Bachelor of Science (B.Sc.)
WEST BENGAL UNIVERSITY OF TECHNOLOGY 2014
Master of Business Administration (M.B.A.)
ITTT 2019
TESOL
Olaichanditala, Kolkata, India - 712103
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Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class I-V Tuition
1
Board
International Baccalaureate, State, ICSE, CBSE
IB Subjects taught
English, Mathematics
CBSE Subjects taught
Mathematics, English
ICSE Subjects taught
Mathematics, English
Taught in School or College
Yes
State Syllabus Subjects taught
Mathematics, English
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 9 Tuition
1
Board
ICSE, CBSE, State
CBSE Subjects taught
English, Mathematics
ICSE Subjects taught
Mathematics, English, English Literature
Taught in School or College
Yes
State Syllabus Subjects taught
English, Mathematics
1. Which school boards of Class 1-5 do you teach for?
International Baccalaureate, State, ICSE and others
2. Have you ever taught in any School or College?
Yes
3. Which classes do you teach?
I teach Class 9 Tuition and Class I-V Tuition 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 1 year.
Answered on 28/11/2019 Learn CBSE/Class 9
Group the numbers 1st, 13in one group, 69 in another
. Now find a number whose square root is equal to or less than 13. The name is 3. So, 3 is both your divisor and quotient.
. Now 3*3=9. So, subtract nine from 13.The remainder in 4. Bring down the numbers from other pair and write it besides 4. Here the number is 69. And now the number is 469.
. Two times the quotient, i.e. three, is 6. Now choose a name with six which on multiplication will give you 469. Here if you multiply 67 with 7. It gives you 469. So, now the remainder is zero.you have to write one seven beside six and the other beside your first quotient, i.e. 3.
. Hence the quotient is 37....which is your result. Thus the square root of 1369 is 37.
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class I-V Tuition
1
Board
International Baccalaureate, State, ICSE, CBSE
IB Subjects taught
English, Mathematics
CBSE Subjects taught
Mathematics, English
ICSE Subjects taught
Mathematics, English
Taught in School or College
Yes
State Syllabus Subjects taught
Mathematics, English
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 9 Tuition
1
Board
ICSE, CBSE, State
CBSE Subjects taught
English, Mathematics
ICSE Subjects taught
Mathematics, English, English Literature
Taught in School or College
Yes
State Syllabus Subjects taught
English, Mathematics
Answered on 28/11/2019 Learn CBSE/Class 9
Group the numbers 1st, 13in one group, 69 in another
. Now find a number whose square root is equal to or less than 13. The name is 3. So, 3 is both your divisor and quotient.
. Now 3*3=9. So, subtract nine from 13.The remainder in 4. Bring down the numbers from other pair and write it besides 4. Here the number is 69. And now the number is 469.
. Two times the quotient, i.e. three, is 6. Now choose a name with six which on multiplication will give you 469. Here if you multiply 67 with 7. It gives you 469. So, now the remainder is zero.you have to write one seven beside six and the other beside your first quotient, i.e. 3.
. Hence the quotient is 37....which is your result. Thus the square root of 1369 is 37.
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