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Sumita D.

Hauz Khas, Delhi, India - 110016

Sumita D. Class 6 Tuition trainer in Delhi

Sumita D.

Tutor

Hauz Khas, Delhi, India - 110016.

13 Students taught

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Overview

I am taking home tuition for last 8 yrs. I have a capability to work harder and i know how to manage a student, how to deal with them and i have a good teaching skills and knowledge. I have a capability of to be a good teacher.

Languages Spoken

Hindi

English

Education

Delhi University 1998

Master of Arts (M.A.)

Address

Hauz Khas, Delhi, India - 110016

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Class 6 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 6 Tuition

12

Board

CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

No

Reviews

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FAQs

1. Which school boards of Class 8 do you teach for?

CBSE

2. Have you ever taught in any School or College?

No

3. Which classes do you teach?

I teach Class 10 Tuition, Class 11 Tuition, Class 12 Tuition, Class 6 Tuition, Class 7 Tuition, Class 8 Tuition, Class 9 Tuition, Class I-V Tuition, Drawing and Vedic Maths Training Classes.

4. Do you provide a demo class?

Yes, I provide a paid demo class.

5. How many years of experience do you have?

I have been teaching for 12 years.

Answers by Sumita D. (1)

Answered on 19/06/2019 CBSE/Class 10/Mathematics/UNIT II: Algebra/Quadratic Equations/NCERT Solutions/Exercise 4.3

Let the sides of first and second square be X and Y . Area of first square = (X)²And,Area of second square = (Y)²According to question,(X)² + (Y)² = 468 m² ------------(1).Perimeter of first square = 4 × Xand,Perimeter of second square = 4 × YAccording... ...more

Let the sides of first and second square be X and Y .


Area of first square = (X)²

And,

Area of second square = (Y)²

According to question,

(X)² + (Y)² = 468 m² ------------(1).

Perimeter of first square = 4 × X

and,

Perimeter of second square = 4 × Y

According to question,

4X - 4Y = 24 -----------(2)

From equation (2) we get,

4X - 4Y = 24

4(X-Y) = 24

X - Y = 24/4 

X - Y = 6

X = 6+Y ---------(3)

Putting the value of X in equation (1)

(X)² + (Y)² = 468

(6+Y)² + (Y)² = 468

(6)² + (Y)² + 2 × 6 × Y + (Y)² = 468

36 + Y² + 12Y + Y² = 468

2Y² + 12Y - 468 +36 = 0

2Y² + 12Y -432 = 0

2( Y² + 6Y - 216) = 0

Y² + 6Y - 216 = 0

Y² + 18Y - 12Y -216 = 0

Y(Y+18) - 12(Y+18) = 0

(Y+18) (Y-12) = 0

(Y+18) = 0 Or (Y-12) = 0

Y = -18 OR Y = 12

Putting Y = 12 in EQUATION (3)

X = 6+Y = 6+12 = 18

Side of first square = X = 18 m

and,

Side of second square = Y = 12 m.
Answers 14 Comments
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Class 6 Tuition 5.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 6 Tuition

12

Board

CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

No

Class 7 Tuition 5.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 7 Tuition

8

Board

ICSE, CBSE

CBSE Subjects taught

Science, Mathematics

ICSE Subjects taught

Mathematics

Taught in School or College

No

Class 8 Tuition 5.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 8 Tuition

8

Board

ICSE, CBSE

CBSE Subjects taught

Science, Mathematics

ICSE Subjects taught

Mathematics

Taught in School or College

No

Class 9 Tuition 5.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 9 Tuition

10

Board

CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

Yes

Class 10 Tuition 5.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 10 Tuition

10

Board

CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

No

Drawing Classes 5.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Drawing Classes

7

Class 11 Tuition 5.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 11 Tuition

8

Board

CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

No

Class 12 Tuition 5.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 12 Tuition

8

Board

CBSE

CBSE Subjects taught

Mathematics, Accountancy, Economics

Taught in School or College

No

Vedic Maths Training classes 5.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Vedic Maths Training classes

12

Class I-V Tuition 5.0

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class I-V Tuition

8

Board

State, International Baccalaureate, CBSE, ICSE

IB Subjects taught

Mathematics, Science

CBSE Subjects taught

Science, EVS, Mathematics

ICSE Subjects taught

Science, EVS, Mathematics

Taught in School or College

No

State Syllabus Subjects taught

EVS, Science, Mathematics

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No Reviews yet! Be the first one to Review

Answers by Sumita D. (1)

Answered on 19/06/2019 CBSE/Class 10/Mathematics/UNIT II: Algebra/Quadratic Equations/NCERT Solutions/Exercise 4.3

Let the sides of first and second square be X and Y . Area of first square = (X)²And,Area of second square = (Y)²According to question,(X)² + (Y)² = 468 m² ------------(1).Perimeter of first square = 4 × Xand,Perimeter of second square = 4 × YAccording... ...more

Let the sides of first and second square be X and Y .


Area of first square = (X)²

And,

Area of second square = (Y)²

According to question,

(X)² + (Y)² = 468 m² ------------(1).

Perimeter of first square = 4 × X

and,

Perimeter of second square = 4 × Y

According to question,

4X - 4Y = 24 -----------(2)

From equation (2) we get,

4X - 4Y = 24

4(X-Y) = 24

X - Y = 24/4 

X - Y = 6

X = 6+Y ---------(3)

Putting the value of X in equation (1)

(X)² + (Y)² = 468

(6+Y)² + (Y)² = 468

(6)² + (Y)² + 2 × 6 × Y + (Y)² = 468

36 + Y² + 12Y + Y² = 468

2Y² + 12Y - 468 +36 = 0

2Y² + 12Y -432 = 0

2( Y² + 6Y - 216) = 0

Y² + 6Y - 216 = 0

Y² + 18Y - 12Y -216 = 0

Y(Y+18) - 12(Y+18) = 0

(Y+18) (Y-12) = 0

(Y+18) = 0 Or (Y-12) = 0

Y = -18 OR Y = 12

Putting Y = 12 in EQUATION (3)

X = 6+Y = 6+12 = 18

Side of first square = X = 18 m

and,

Side of second square = Y = 12 m.
Answers 14 Comments
Dislike Bookmark

Sumita D. describes herself as Tutor. She conducts classes in Class 10 Tuition, Class 11 Tuition and Class 12 Tuition. Sumita is located in Hauz Khas, Delhi. Sumita takes at students Home. She has 12 years of teaching experience . Sumita has completed Master of Arts (M.A.) from Delhi University in 1998. She is well versed in Hindi and English.

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