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Gurukul Education Center Class 11 Tuition institute in Pune

Gurukul Education Center

Dhanori, Pune, India - 411015

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Overview

Our institute is dedicated towards preparing students for competitive exams.
We can teach subjects with ease and make the subject enjoyable for students.

Address

Off. No no. 203, 2nd floor, L square building, survey no. 283, behind Orchid Hopital, Porwal road

Dhanori, Pune, India - 411015

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Teaches

Class 11 Tuition

Class Location

Online (video chat via skype, google hangout etc)

At the Institute

Board

State, ISC/ICSE, CBSE

ISC/ICSE Subjects taught

Chemistry, Mathematics, Physics

CBSE Subjects taught

Chemistry, Mathematics, Physics

State Syllabus Subjects taught

Physics, Mathematics, Chemistry

Class 12 Tuition

Class Location

Online (video chat via skype, google hangout etc)

At the Institute

Board

State, ISC/ICSE, CBSE

ISC/ICSE Subjects taught

Chemistry, Mathematics, Physics

CBSE Subjects taught

Chemistry, Mathematics, Physics

State Syllabus Subjects taught

Physics, Mathematics, Chemistry

Class 9 Tuition

Class Location

Online (video chat via skype, google hangout etc)

At the Institute

Board

CBSE, ICSE, State

CBSE Subjects taught

Mathematics, Science

ICSE Subjects taught

Mathematics, Physics, Chemistry, Biology

State Syllabus Subjects taught

Mathematics, Science

Class 10 Tuition

Class Location

Online (video chat via skype, google hangout etc)

At the Institute

Board

CBSE, ICSE, State

CBSE Subjects taught

Mathematics, Science

ICSE Subjects taught

Mathematics, Physics, Chemistry, Biology

State Syllabus Subjects taught

Mathematics, Science

Engineering Entrance Coaching classes

Class Location

Online (video chat via skype, google hangout etc)

At the Institute

Engineering Entrance Exams

IIT JEE Coaching Classes

Type of class

Regular Classes

IIT JEE Coaching

IIT JEE Foundation Course, IIT JEE Mains Coaching, IIT JEE Advanced Coaching

IIT-JEE Subjects

Maths, Physics , Chemistry

Courses

Reviews

No Reviews yet! Be the first one to Review

FAQs

1. Which school boards of of Class 11 do you teach for?

State, ISC/ICSE, CBSE

2. Which classes do you teach?

We teach Class 10 Tuition, Class 11 Tuition, Class 12 Tuition, Class 9 Tuition and Engineering Entrance Coaching Classes

3. Do you provide a demo class?

Yes, We provide a free demo class.

4. Where are you located?

We are located in Dhanori, Pune.

Answers by Gurukul Education Center (23)

Answered on 24/12/2017 Learn CBSE/Class 10/Mathematics +1 Tuition/Class IX-X Tuition

Perimeter of a cube with side a = 12 *a = 20cm Therefore , a = 5/3 cm Now , volume of cube = a^3 = (5/3)^3 = 125/27 cm3 If weight of 1cm3 is 8.25g Then weight of 125/27 cm3 will be = 8.25 * 125/27 gm = 38.19gm (ans)
Answers 5 Comments
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Answered on 10/12/2017 Learn CBSE/Class 9/Mathematics +1 Tuition/Class IX-X Tuition

14x6 - 45x3y3-14y6 = 14x6-49x3y3+4x3y3-14y6 = 7x3 (2x3-7y3)+2y3(2x3-7y3) = (7x3+2y3)(2x3-7y3)
Answers 11 Comments
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Answered on 10/12/2017 Learn CBSE/Class 10/Mathematics +1 Tuition/Class IX-X Tuition

Let a = A-2D b = A-D c = A d = A+D e = A+2D a-4b+6c-4d+e = A-2D-4(A-D)+6A-4(A+D)+A+2D = 0
Answers 11 Comments
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Answered on 10/12/2017 Learn CBSE/Class 10/Mathematics +1 Tuition/Class IX-X Tuition

Let the marks scored in Mathematics be x. As given, 9( x+10 ) = x2 x2-9x-90 = 0 x2-15x+6x-90 =0 x(x-15)+6(x-15) = 0 (x+6)(x-15) = 0 Therefore, x = 15 ( x = -6 is ignored)
Answers 6 Comments
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Answered on 10/12/2017 Learn CBSE/Class 10/Mathematics +1 Tuition/Class IX-X Tuition

Let the speed of stream by x kmph. then speed of boat in water upstream = (18-x) kmph Speed of boat in water downstream = (18+x) kmph Now time = Distance / speed Time taken to go upstream - time taken to go downstream = 1 hr 24/(18-x) - 24/(18+x) = 1 24( 18+x-18+x ) = (18^2 - x2) 48x = 324... ...more
Let the speed of stream by x kmph. then speed of boat in water upstream = (18-x) kmph Speed of boat in water downstream = (18+x) kmph Now time = Distance / speed Time taken to go upstream - time taken to go downstream = 1 hr 24/(18-x) - 24/(18+x) = 1 24( 18+x-18+x ) = (18^2 - x2) 48x = 324 - x2 x2+48x-324 = 0 x2+54x-6x-324 = 0 x(x+54)-6(x+54) = 0 (x-6)(x+54) = 0 therefore, speed of stream (x) = 6kmph
Answers 5 Comments
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Teaches

Class 11 Tuition

Class Location

Online (video chat via skype, google hangout etc)

At the Institute

Board

State, ISC/ICSE, CBSE

ISC/ICSE Subjects taught

Chemistry, Mathematics, Physics

CBSE Subjects taught

Chemistry, Mathematics, Physics

State Syllabus Subjects taught

Physics, Mathematics, Chemistry

Class 12 Tuition

Class Location

Online (video chat via skype, google hangout etc)

At the Institute

Board

State, ISC/ICSE, CBSE

ISC/ICSE Subjects taught

Chemistry, Mathematics, Physics

CBSE Subjects taught

Chemistry, Mathematics, Physics

State Syllabus Subjects taught

Physics, Mathematics, Chemistry

Class 9 Tuition

Class Location

Online (video chat via skype, google hangout etc)

At the Institute

Board

CBSE, ICSE, State

CBSE Subjects taught

Mathematics, Science

ICSE Subjects taught

Mathematics, Physics, Chemistry, Biology

State Syllabus Subjects taught

Mathematics, Science

Class 10 Tuition

Class Location

Online (video chat via skype, google hangout etc)

At the Institute

Board

CBSE, ICSE, State

CBSE Subjects taught

Mathematics, Science

ICSE Subjects taught

Mathematics, Physics, Chemistry, Biology

State Syllabus Subjects taught

Mathematics, Science

Engineering Entrance Coaching classes

Class Location

Online (video chat via skype, google hangout etc)

At the Institute

Engineering Entrance Exams

IIT JEE Coaching Classes

Type of class

Regular Classes

IIT JEE Coaching

IIT JEE Foundation Course, IIT JEE Mains Coaching, IIT JEE Advanced Coaching

IIT-JEE Subjects

Maths, Physics , Chemistry

Courses

No Reviews yet! Be the first one to Review

Answers by Gurukul Education Center (23)

Answered on 24/12/2017 Learn CBSE/Class 10/Mathematics +1 Tuition/Class IX-X Tuition

Perimeter of a cube with side a = 12 *a = 20cm Therefore , a = 5/3 cm Now , volume of cube = a^3 = (5/3)^3 = 125/27 cm3 If weight of 1cm3 is 8.25g Then weight of 125/27 cm3 will be = 8.25 * 125/27 gm = 38.19gm (ans)
Answers 5 Comments
Dislike Bookmark

Answered on 10/12/2017 Learn CBSE/Class 9/Mathematics +1 Tuition/Class IX-X Tuition

14x6 - 45x3y3-14y6 = 14x6-49x3y3+4x3y3-14y6 = 7x3 (2x3-7y3)+2y3(2x3-7y3) = (7x3+2y3)(2x3-7y3)
Answers 11 Comments
Dislike Bookmark

Answered on 10/12/2017 Learn CBSE/Class 10/Mathematics +1 Tuition/Class IX-X Tuition

Let a = A-2D b = A-D c = A d = A+D e = A+2D a-4b+6c-4d+e = A-2D-4(A-D)+6A-4(A+D)+A+2D = 0
Answers 11 Comments
Dislike Bookmark

Answered on 10/12/2017 Learn CBSE/Class 10/Mathematics +1 Tuition/Class IX-X Tuition

Let the marks scored in Mathematics be x. As given, 9( x+10 ) = x2 x2-9x-90 = 0 x2-15x+6x-90 =0 x(x-15)+6(x-15) = 0 (x+6)(x-15) = 0 Therefore, x = 15 ( x = -6 is ignored)
Answers 6 Comments
Dislike Bookmark

Answered on 10/12/2017 Learn CBSE/Class 10/Mathematics +1 Tuition/Class IX-X Tuition

Let the speed of stream by x kmph. then speed of boat in water upstream = (18-x) kmph Speed of boat in water downstream = (18+x) kmph Now time = Distance / speed Time taken to go upstream - time taken to go downstream = 1 hr 24/(18-x) - 24/(18+x) = 1 24( 18+x-18+x ) = (18^2 - x2) 48x = 324... ...more
Let the speed of stream by x kmph. then speed of boat in water upstream = (18-x) kmph Speed of boat in water downstream = (18+x) kmph Now time = Distance / speed Time taken to go upstream - time taken to go downstream = 1 hr 24/(18-x) - 24/(18+x) = 1 24( 18+x-18+x ) = (18^2 - x2) 48x = 324 - x2 x2+48x-324 = 0 x2+54x-6x-324 = 0 x(x+54)-6(x+54) = 0 (x-6)(x+54) = 0 therefore, speed of stream (x) = 6kmph
Answers 5 Comments
Dislike Bookmark

Contact

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Gurukul Education Center conducts classes in Class 10 Tuition, Class 11 Tuition and Class 12 Tuition. It is located in Dhanori, Pune. It takes Regular Classes- at the Institute.

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