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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

English, Mathematics, 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, English, Mathematics

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

Mathematics, Computers, English

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

Information and Comunication Technology, Mathematics, English

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, English, Mathematics

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

Mathematics, Computers, English

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

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Answers by Shambhavi S.

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

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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.
Answers 240 Comments
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Answered on 07/12/2016 Learn CBSE - Class 10/Mathematics +2 Expansions

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

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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.
Answers 1 Comments
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Answered on 07/12/2016 Learn CBSE - Class 10/Mathematics +2 Profit, Loss and Discount

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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.
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Answered on 07/12/2016 Learn CBSE - Class 10/Mathematics +2 Factorization

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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

English, Mathematics, 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, English, Mathematics

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

Mathematics, Computers, English

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

Information and Comunication Technology, Mathematics, English

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, English, Mathematics

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

Mathematics, Computers, English

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 CBSE - Class 1/Maths

Ask a Question

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.
Answers 240 Comments
Dislike Bookmark

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

Ask a Question

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.
Answers 1 Comments
Dislike Bookmark

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

Ask a Question

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.
Answers 1 Comments
Dislike Bookmark

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

Ask a Question

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.
Answers 1 Comments
Dislike Bookmark

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

Ask a Question

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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