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Overview

Shambhavi S. conducts classes in C Language, C++ Language and Class 10 Tuition. Shambhavi is located in Samaypur Badli, Delhi. Shambhavi takes at students Home and Regular Classes- at her Home. She is well versed in Marathi, Hindi and English.

Languages Spoken

Marathi

Hindi

English

Address

Samaypur Badli, Delhi, India - 110042

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Teaches

Class 6 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE

CBSE Subjects taught

Computers, Mathematics, English

Taught in School or College

No

Class 7 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE

CBSE Subjects taught

Computers, Mathematics, English

Taught in School or College

No

Class 8 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE

CBSE Subjects taught

Computers, Mathematics, English

Taught in School or College

No

Class 9 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE

CBSE Subjects taught

Information and Comunication Technology, English, Mathematics

Taught in School or College

No

Class 10 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE

CBSE Subjects taught

Information and Comunication Technology, English, Mathematics

Taught in School or College

No

Class I-V Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE

CBSE Subjects taught

Mathematics, English, Computers

Taught in School or College

No

C Language Classes

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

C++ Language Classes

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Proficiency level taught

Basic C++, Advanced C++

Reviews

No Reviews yet!

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 C Language, C++ Language, Class 10 Tuition, Class 6 Tuition, Class 7 Tuition, Class 8 Tuition, Class 9 Tuition, Class I-V Tuition and Online Tuition Classes 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 less than a year.

Contact

Answers by Shambhavi S. (14)

Answered on 08/12/2016 Learn Tuition/Class I-V Tuition +1 Tuition/Class I-V Tuition/Mathematics

How to find a percentage?

The word "percent", means "for each cent", that it, "for each 100", as cent is also called 100. Now, if you want to find how much percent is 50 of 100, this means, for every 100, we have a 50, thus 50"percent". Similarly, how much percent is 20 for 50, we do, (20*100)/50, that amounts to, 40 percent.
Answers 240 Comments
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Answered on 07/12/2016 Learn CBSE/Class 10/Mathematics +2 Tuition/Class IX-X Tuition Expansions

If (square of x) + (inverse square of x) = 102, evaluate x + (inverse of x)?

Squaring (x+1/x) , we get, (x^2+x^-2+2), but we know that x^2+x^-2 is equal to 102, thus (x+1/x)+2=102+2=104.
Answers 1 Comments
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Answered on 07/12/2016 Learn CBSE/Class 10/Mathematics +2 Tuition/Class IX-X Tuition Linear Equations

Solve: x + y = 3.3, 0.6 / (3x - 2y) = -1, 3x - 2y = 0?

The second equation is undefined as denominator is 0, which is not possible. Thus these set of equations are inconsistent.
Answers 1 Comments
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Answered on 07/12/2016 Learn CBSE/Class 10/Mathematics +2 Tuition/Class IX-X Tuition Profit, Loss and Discount

What is the total gain or loss percent if a shopkeeper sells one transistor for Rs 840 at a gain of 20%... ...more
What is the total gain or loss percent if a shopkeeper sells one transistor for Rs 840 at a gain of 20% and another for Rs 960 at a loss of 4%?

C.P=S.P/(1+loss/gain%) Thus , C.P(1+/- loss/gain%)=S.P In first case, we gain 20%, so cp(1+0.2)=S.P, C.P(1.2)=840, solving which we get C.P as 700. Now taking second case of loss of 4%, Cp(1-0.04)=S.P, solving which we get C.P as 1000. Now, total C.P =700+1000=1700, and total S.P as 840+960=1800 Thus,... ...more
C.P=S.P/(1+loss/gain%) Thus , C.P(1+/- loss/gain%)=S.P In first case, we gain 20%, so cp(1+0.2)=S.P, C.P(1.2)=840, solving which we get C.P as 700. Now taking second case of loss of 4%, Cp(1-0.04)=S.P, solving which we get C.P as 1000. Now, total C.P =700+1000=1700, and total S.P as 840+960=1800 Thus, total gain is, (1800-1700)/1700*100=5.8% gain.
Answers 1 Comments
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Answered on 07/12/2016 Learn CBSE/Class 10/Mathematics +2 Tuition/Class IX-X Tuition Factorization

Prove that: (1 / a3) + (1 / b3) + (1 / c3) = (1 / a + 1 / b + 1/c)3 + 3 / abc?

we know that, a3 + b3 + c3 = (a + b + c)3 -- 3(a + b)(b + c)(c + a) a + b + c = 0 implies a + b = -- c, b + c = -- a, c + a = -- b. (1/a)3 + (1/b)3 + (1/c)3 = (1/a + 1/b + 1/c)3 -- 3(1/a + 1/b)(1/b + 1/c)(1/c + 1/a) = (1/a + 1/b + 1/c)3 -- 3(a+b / ab)(b + c / bc)(c + a / ac) = (1/a + 1/b + 1/c)3 --... ...more
we know that, a3 + b3 + c3 = (a + b + c)3 – 3(a + b)(b + c)(c + a) a + b + c = 0 implies a + b = – c, b + c = – a, c + a = – b. (1/a)3 + (1/b)3 + (1/c)3 = (1/a + 1/b + 1/c)3 – 3(1/a + 1/b)(1/b + 1/c)(1/c + 1/a) = (1/a + 1/b + 1/c)3 – 3(a+b / ab)(b + c / bc)(c + a / ac) = (1/a + 1/b + 1/c)3 – 3(– c / ab)(– a / bc)(– b / ac) = (1/a + 1/b + 1/c)3 – 3(abc / a2b2c2) = (1/a + 1/b + 1/c)3 – 3 / abc. Thus, proved.
Answers 1 Comments
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Contact

View All

Teaches

Class 6 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE

CBSE Subjects taught

Computers, Mathematics, English

Taught in School or College

No

Class 7 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE

CBSE Subjects taught

Computers, Mathematics, English

Taught in School or College

No

Class 8 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE

CBSE Subjects taught

Computers, Mathematics, English

Taught in School or College

No

Class 9 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE

CBSE Subjects taught

Information and Comunication Technology, English, Mathematics

Taught in School or College

No

Class 10 Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE

CBSE Subjects taught

Information and Comunication Technology, English, Mathematics

Taught in School or College

No

Class I-V Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Board

CBSE

CBSE Subjects taught

Mathematics, English, Computers

Taught in School or College

No

C Language Classes

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

C++ Language Classes

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Proficiency level taught

Basic C++, Advanced C++

No Reviews yet!

Answers by Shambhavi S. (14)

Answered on 08/12/2016 Learn Tuition/Class I-V Tuition +1 Tuition/Class I-V Tuition/Mathematics

How to find a percentage?

The word "percent", means "for each cent", that it, "for each 100", as cent is also called 100. Now, if you want to find how much percent is 50 of 100, this means, for every 100, we have a 50, thus 50"percent". Similarly, how much percent is 20 for 50, we do, (20*100)/50, that amounts to, 40 percent.
Answers 240 Comments
Dislike Bookmark

Answered on 07/12/2016 Learn CBSE/Class 10/Mathematics +2 Tuition/Class IX-X Tuition Expansions

If (square of x) + (inverse square of x) = 102, evaluate x + (inverse of x)?

Squaring (x+1/x) , we get, (x^2+x^-2+2), but we know that x^2+x^-2 is equal to 102, thus (x+1/x)+2=102+2=104.
Answers 1 Comments
Dislike Bookmark

Answered on 07/12/2016 Learn CBSE/Class 10/Mathematics +2 Tuition/Class IX-X Tuition Linear Equations

Solve: x + y = 3.3, 0.6 / (3x - 2y) = -1, 3x - 2y = 0?

The second equation is undefined as denominator is 0, which is not possible. Thus these set of equations are inconsistent.
Answers 1 Comments
Dislike Bookmark

Answered on 07/12/2016 Learn CBSE/Class 10/Mathematics +2 Tuition/Class IX-X Tuition Profit, Loss and Discount

What is the total gain or loss percent if a shopkeeper sells one transistor for Rs 840 at a gain of 20%... ...more
What is the total gain or loss percent if a shopkeeper sells one transistor for Rs 840 at a gain of 20% and another for Rs 960 at a loss of 4%?

C.P=S.P/(1+loss/gain%) Thus , C.P(1+/- loss/gain%)=S.P In first case, we gain 20%, so cp(1+0.2)=S.P, C.P(1.2)=840, solving which we get C.P as 700. Now taking second case of loss of 4%, Cp(1-0.04)=S.P, solving which we get C.P as 1000. Now, total C.P =700+1000=1700, and total S.P as 840+960=1800 Thus,... ...more
C.P=S.P/(1+loss/gain%) Thus , C.P(1+/- loss/gain%)=S.P In first case, we gain 20%, so cp(1+0.2)=S.P, C.P(1.2)=840, solving which we get C.P as 700. Now taking second case of loss of 4%, Cp(1-0.04)=S.P, solving which we get C.P as 1000. Now, total C.P =700+1000=1700, and total S.P as 840+960=1800 Thus, total gain is, (1800-1700)/1700*100=5.8% gain.
Answers 1 Comments
Dislike Bookmark

Answered on 07/12/2016 Learn CBSE/Class 10/Mathematics +2 Tuition/Class IX-X Tuition Factorization

Prove that: (1 / a3) + (1 / b3) + (1 / c3) = (1 / a + 1 / b + 1/c)3 + 3 / abc?

we know that, a3 + b3 + c3 = (a + b + c)3 -- 3(a + b)(b + c)(c + a) a + b + c = 0 implies a + b = -- c, b + c = -- a, c + a = -- b. (1/a)3 + (1/b)3 + (1/c)3 = (1/a + 1/b + 1/c)3 -- 3(1/a + 1/b)(1/b + 1/c)(1/c + 1/a) = (1/a + 1/b + 1/c)3 -- 3(a+b / ab)(b + c / bc)(c + a / ac) = (1/a + 1/b + 1/c)3 --... ...more
we know that, a3 + b3 + c3 = (a + b + c)3 – 3(a + b)(b + c)(c + a) a + b + c = 0 implies a + b = – c, b + c = – a, c + a = – b. (1/a)3 + (1/b)3 + (1/c)3 = (1/a + 1/b + 1/c)3 – 3(1/a + 1/b)(1/b + 1/c)(1/c + 1/a) = (1/a + 1/b + 1/c)3 – 3(a+b / ab)(b + c / bc)(c + a / ac) = (1/a + 1/b + 1/c)3 – 3(– c / ab)(– a / bc)(– b / ac) = (1/a + 1/b + 1/c)3 – 3(abc / a2b2c2) = (1/a + 1/b + 1/c)3 – 3 / abc. Thus, proved.
Answers 1 Comments
Dislike Bookmark

Contact

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Shambhavi S. conducts classes in C Language, C++ Language and Class 10 Tuition. Shambhavi is located in Samaypur Badli, Delhi. Shambhavi takes at students Home and Regular Classes- at her Home. She is well versed in Marathi, Hindi and English.

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