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Online Tuition Classes > Shambhavi S.

Shambhavi S. Class 6 Tuition trainer in Delhi

Shambhavi S.

locationImg Samaypur Badli, Delhi
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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.

Languages Spoken

Marathi

Hindi

English

Address

Samaypur Badli, Delhi, India - 110042

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Teaches

Class 6 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

CBSE

Subjects taught

Mathematics, English, Computers

Taught in School or College

No

Class 7 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

CBSE

Subjects taught

Computers, Mathematics, English

Taught in School or College

No

Class 8 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

CBSE

Subjects taught

English, Mathematics, Computers

Taught in School or College

No

Class 9 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

CBSE

Subjects taught

Mathematics, English, Information and Comunication Technology

Taught in School or College

No

Class 10 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

CBSE

Subjects taught

Information and Comunication Technology, Mathematics, English

Taught in School or College

No

Class I-V Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

CBSE

Subjects taught

English, Mathematics, Computers

Taught in School or College

No

C Language Classes

Class Location

Online class via Zoom

Student's Home

Tutor's Home

C++ Language Classes

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Proficiency level taught

Advanced C++, Basic C++

Reviews

No Reviews yet!

Answers by Shambhavi S.

Answered on 08/12/2016 +1 Tuition/Class I-V Tuition/CBSE - Class 1/Maths

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Post a Lesson Post a Lesson

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.
Answer Answers 240 Comments Comments
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Answered on 07/12/2016 Learn CBSE - Class 8/Maths/Algebraic Expressions And Identities +2 Tuition/Class IX-X Tuition CBSE - Class 10/Mathematics

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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.
Answer Answers 1 Comments Comments
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Answered on 07/12/2016 Learn CBSE - Class 10/Mathematics +2 Tuition/Class IX-X Tuition Linear Equations

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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.
Answer Answers 1 Comments Comments
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Answered on 07/12/2016 Learn CBSE - Class 8/Maths +2 Tuition/Class IX-X Tuition CBSE - Class 10/Mathematics

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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%...

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.
Answer Answers 1 Comments Comments
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Answered on 07/12/2016 Learn CBSE - Class 9/Mathematics/Polynomials +2 Tuition/Class IX-X Tuition CBSE - Class 10/Mathematics

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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.
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Teaches

Class 6 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

CBSE

Subjects taught

Mathematics, English, Computers

Taught in School or College

No

Class 7 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

CBSE

Subjects taught

Computers, Mathematics, English

Taught in School or College

No

Class 8 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

CBSE

Subjects taught

English, Mathematics, Computers

Taught in School or College

No

Class 9 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

CBSE

Subjects taught

Mathematics, English, Information and Comunication Technology

Taught in School or College

No

Class 10 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

CBSE

Subjects taught

Information and Comunication Technology, Mathematics, English

Taught in School or College

No

Class I-V Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

CBSE

Subjects taught

English, Mathematics, Computers

Taught in School or College

No

C Language Classes

Class Location

Online class via Zoom

Student's Home

Tutor's Home

C++ Language Classes

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Proficiency level taught

Advanced C++, Basic C++

No Reviews yet!

Answers by Shambhavi S.

Answered on 08/12/2016 +1 Tuition/Class I-V Tuition/CBSE - Class 1/Maths

Ask a Question Ask a Question

Post a Lesson Post a Lesson

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.
Answer Answers 240 Comments Comments
Dislike Bookmark

Answered on 07/12/2016 Learn CBSE - Class 8/Maths/Algebraic Expressions And Identities +2 Tuition/Class IX-X Tuition CBSE - Class 10/Mathematics

Ask a Question Ask a Question

Post a Lesson Post a Lesson

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.
Answer Answers 1 Comments Comments
Dislike Bookmark

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

Ask a Question Ask a Question

Post a Lesson Post a Lesson

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.
Answer Answers 1 Comments Comments
Dislike Bookmark

Answered on 07/12/2016 Learn CBSE - Class 8/Maths +2 Tuition/Class IX-X Tuition CBSE - Class 10/Mathematics

Ask a Question Ask a Question

Post a Lesson Post a Lesson

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%...

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.
Answer Answers 1 Comments Comments
Dislike Bookmark

Answered on 07/12/2016 Learn CBSE - Class 9/Mathematics/Polynomials +2 Tuition/Class IX-X Tuition CBSE - Class 10/Mathematics

Ask a Question Ask a Question

Post a Lesson Post a Lesson

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.
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