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Raj Aryan Class 6 Tuition trainer in Mumbai

Raj Aryan

Trainer

Marol Naka, Mumbai, India - 400059.

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Referral Discount: Get ₹ 500 off when you make a payment to start classes. Get started by Booking a Demo.

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Overview

I can teach each and every topics very easily and make it stronger for the students.

Languages Spoken

Hindi

English

Education

Indian Institute of Management, Indore 2016

Integrated Diploma in Management

Address

Marol Naka, Mumbai, India - 400059

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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, ICSE

CBSE Subjects taught

Science, Mathematics, Social Science

ICSE Subjects taught

Physics, Chemistry, Mathematics

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, ICSE

CBSE Subjects taught

Science, Mathematics, Social Science

ICSE Subjects taught

Physics, Chemistry, Mathematics

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, ICSE

CBSE Subjects taught

Science, Mathematics, Social Science

ICSE Subjects taught

Physics, Chemistry, Mathematics

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, ICSE

CBSE Subjects taught

Mathematics, Science

ICSE Subjects taught

Mathematics, Physics, Chemistry, Economic Application

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, ICSE

CBSE Subjects taught

Mathematics, Science

ICSE Subjects taught

Mathematics, Physics, Chemistry, Economic Application

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

Years of Experience in Class I-V Tuition

2

Board

CBSE

CBSE Subjects taught

Science, Mathematics

Taught in School or College

No

Microsoft Excel Training classes

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Microsoft Excel Training classes

2

Teaches following Excel features

Advanced Excel, Basic Excel

Marketing Training classes

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Marketing Training classes

2

Nursery-KG Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Subject

Mathematics

Taught in School or College

No

Reviews

No Reviews yet!

FAQs

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

CBSE, ICSE

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

No

3. Which classes do you teach?

I teach Class 10 Tuition, Class 6 Tuition, Class 7 Tuition, Class 8 Tuition, Class 9 Tuition, Class I-V Tuition, Marketing Training, Microsoft Excel Training and Nursery-KG Tuition 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 Raj Aryan (10)

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

If the zeroes of polynomial x3 – ax2 + bx – c are in AP then show that 2a3 – 9ab + 27c = 0

Let the zeros of given polynomial is m-d, m, and m+d. Here m and d are not zero. substituting the value of zeros in the polynomial, we have=>(m-d)^3-a(m-d)^2+b(m-d)-c=0 =>m^3-d^3-3m^2b+3md^2-a*(m^2+d^2-2*m*d)+b*m-b*d-c=0 =>m^3-d^3-3m^2b+3md^2-a*m^2-ad^2+2*a*m*d+b*m-b*d-c=0----------------(i) Similarily,... ...more
Let the zeros of given polynomial is m-d, m, and m+d. Here m and d are not zero. substituting the value of zeros in the polynomial, we have=>(m-d)^3-a(m-d)^2+b(m-d)-c=0 =>m^3-d^3-3m^2b+3md^2-a*(m^2+d^2-2*m*d)+b*m-b*d-c=0 =>m^3-d^3-3m^2b+3md^2-a*m^2-ad^2+2*a*m*d+b*m-b*d-c=0----------------(i) Similarily, we get a new equation by putting the (m+d), =>m^3+d^3+3m*d^2-a*m^2-a*d*d^2-2a*m*d+b*m+b*d-c=0---------------------(ii) Again, Putting the value of m in polynomial, we have m^3-a*m^2+b*m-c=0-----------------------------------------(iii) Adding equation (i) & (ii), we have=>2*m^3+6*m*d^2-2a*m^2-2a*d^2+2b*m-2*c=0 =>2*(3^3-a*m^2+b*m-c)+6m*d^2-2*a*d^2=0 =>6*m*d^2=2*a*d^2 =>m=a/3------------------------------(iv) Now, Putting the value of m in the polynomial, =>(a/3)^3-a*(a/3)^2+b*a/3-c=0 =>a^3/27-a^3/9+a*b/3-c=0 =>a^3-3*a^3+9*a*b-27*c=0 =>2*a^3-9*a*b+27*c=0 (Proved)
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Answered on 09/12/2017 Learn CBSE/Class 9/Mathematics +1 Tuition/Class IX-X Tuition

Solve : 55x + 52y = 217; 52x + 55y = 217

=>55x+52y=217..........................(i) =>52x+55y=217..........................(ii) Adding equation (i) & (ii), We have 77 (x+y)=434 =>x+y=434/77..................(iii) Subtracting (ii) from (i), We have=>x-y=0-----------------------(iv) From equation (iii) & (iv), we have x=217/77 y=217...
Answers 8 Comments
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Answered on 08/12/2017 Learn CBSE/Class 10/Mathematics +1 Tuition/Class IX-X Tuition

A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same... ...more
A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/hr more. Find the original speed of the train.

Distance=360 KM Let speed be S time (t) =360/s Other Speed=S+5 Time=360/(S+5) As per question, 360/S-360(S+5)=48/60 =>x2+5x -2250 = 0=>(x+50)(x-45)=0 Speed = 45 km/hr as speed cannot be negative
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Answered on 08/12/2017 Learn CBSE/Class 10/Mathematics +1 Tuition/Class IX-X Tuition

If the numbers a, b, c, d and e form an A.P., then find the value of a – 4b + 6c – 4d + e.

Let Consider c=m since a, b, c, d, e are in AP Therefore, b=m-d (Here d is common difference) a=m-2d d=m+d e=m+2d Now, Putting the values of a, b, c, d and e in the final equation, we have =(m-2d)-4(m-d)+6m-4(m+d)+(m+2d) =m-2d-4m+4d+6m-4m-4d+m+2d =0
Answers 11 Comments
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Answered on 08/12/2017 Learn CBSE/Class 9/Mathematics +1 Tuition/Class IX-X Tuition

The mean of 100 observations is 50. If one of the observation which was 50 is replaced by 150 then what... ...more
The mean of 100 observations is 50. If one of the observation which was 50 is replaced by 150 then what will be the resulting mean?

Mean of the observation = 50 Number of observation =100 Total Value =50*100 =5000 Observation value removed = 50 Observation value added =150 Net observation Value =5000-50+150 =5100 New Mean = 5100/100 = 51
Answers 49 Comments
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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, ICSE

CBSE Subjects taught

Science, Mathematics, Social Science

ICSE Subjects taught

Physics, Chemistry, Mathematics

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, ICSE

CBSE Subjects taught

Science, Mathematics, Social Science

ICSE Subjects taught

Physics, Chemistry, Mathematics

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, ICSE

CBSE Subjects taught

Science, Mathematics, Social Science

ICSE Subjects taught

Physics, Chemistry, Mathematics

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, ICSE

CBSE Subjects taught

Mathematics, Science

ICSE Subjects taught

Mathematics, Physics, Chemistry, Economic Application

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, ICSE

CBSE Subjects taught

Mathematics, Science

ICSE Subjects taught

Mathematics, Physics, Chemistry, Economic Application

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

Years of Experience in Class I-V Tuition

2

Board

CBSE

CBSE Subjects taught

Science, Mathematics

Taught in School or College

No

Microsoft Excel Training classes

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Microsoft Excel Training classes

2

Teaches following Excel features

Advanced Excel, Basic Excel

Marketing Training classes

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Marketing Training classes

2

Nursery-KG Tuition

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Subject

Mathematics

Taught in School or College

No

No Reviews yet!

Answers by Raj Aryan (10)

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

If the zeroes of polynomial x3 – ax2 + bx – c are in AP then show that 2a3 – 9ab + 27c = 0

Let the zeros of given polynomial is m-d, m, and m+d. Here m and d are not zero. substituting the value of zeros in the polynomial, we have=>(m-d)^3-a(m-d)^2+b(m-d)-c=0 =>m^3-d^3-3m^2b+3md^2-a*(m^2+d^2-2*m*d)+b*m-b*d-c=0 =>m^3-d^3-3m^2b+3md^2-a*m^2-ad^2+2*a*m*d+b*m-b*d-c=0----------------(i) Similarily,... ...more
Let the zeros of given polynomial is m-d, m, and m+d. Here m and d are not zero. substituting the value of zeros in the polynomial, we have=>(m-d)^3-a(m-d)^2+b(m-d)-c=0 =>m^3-d^3-3m^2b+3md^2-a*(m^2+d^2-2*m*d)+b*m-b*d-c=0 =>m^3-d^3-3m^2b+3md^2-a*m^2-ad^2+2*a*m*d+b*m-b*d-c=0----------------(i) Similarily, we get a new equation by putting the (m+d), =>m^3+d^3+3m*d^2-a*m^2-a*d*d^2-2a*m*d+b*m+b*d-c=0---------------------(ii) Again, Putting the value of m in polynomial, we have m^3-a*m^2+b*m-c=0-----------------------------------------(iii) Adding equation (i) & (ii), we have=>2*m^3+6*m*d^2-2a*m^2-2a*d^2+2b*m-2*c=0 =>2*(3^3-a*m^2+b*m-c)+6m*d^2-2*a*d^2=0 =>6*m*d^2=2*a*d^2 =>m=a/3------------------------------(iv) Now, Putting the value of m in the polynomial, =>(a/3)^3-a*(a/3)^2+b*a/3-c=0 =>a^3/27-a^3/9+a*b/3-c=0 =>a^3-3*a^3+9*a*b-27*c=0 =>2*a^3-9*a*b+27*c=0 (Proved)
Answers 5 Comments
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Answered on 09/12/2017 Learn CBSE/Class 9/Mathematics +1 Tuition/Class IX-X Tuition

Solve : 55x + 52y = 217; 52x + 55y = 217

=>55x+52y=217..........................(i) =>52x+55y=217..........................(ii) Adding equation (i) & (ii), We have 77 (x+y)=434 =>x+y=434/77..................(iii) Subtracting (ii) from (i), We have=>x-y=0-----------------------(iv) From equation (iii) & (iv), we have x=217/77 y=217...
Answers 8 Comments
Dislike Bookmark

Answered on 08/12/2017 Learn CBSE/Class 10/Mathematics +1 Tuition/Class IX-X Tuition

A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same... ...more
A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/hr more. Find the original speed of the train.

Distance=360 KM Let speed be S time (t) =360/s Other Speed=S+5 Time=360/(S+5) As per question, 360/S-360(S+5)=48/60 =>x2+5x -2250 = 0=>(x+50)(x-45)=0 Speed = 45 km/hr as speed cannot be negative
Answers 11 Comments
Dislike Bookmark

Answered on 08/12/2017 Learn CBSE/Class 10/Mathematics +1 Tuition/Class IX-X Tuition

If the numbers a, b, c, d and e form an A.P., then find the value of a – 4b + 6c – 4d + e.

Let Consider c=m since a, b, c, d, e are in AP Therefore, b=m-d (Here d is common difference) a=m-2d d=m+d e=m+2d Now, Putting the values of a, b, c, d and e in the final equation, we have =(m-2d)-4(m-d)+6m-4(m+d)+(m+2d) =m-2d-4m+4d+6m-4m-4d+m+2d =0
Answers 11 Comments
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Answered on 08/12/2017 Learn CBSE/Class 9/Mathematics +1 Tuition/Class IX-X Tuition

The mean of 100 observations is 50. If one of the observation which was 50 is replaced by 150 then what... ...more
The mean of 100 observations is 50. If one of the observation which was 50 is replaced by 150 then what will be the resulting mean?

Mean of the observation = 50 Number of observation =100 Total Value =50*100 =5000 Observation value removed = 50 Observation value added =150 Net observation Value =5000-50+150 =5100 New Mean = 5100/100 = 51
Answers 49 Comments
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Raj Aryan describes himself as Trainer. He conducts classes in Class 10 Tuition, Class 6 Tuition and Class 7 Tuition. Raj is located in Marol Naka, Mumbai. Raj takes at students Home and Online Classes- via online medium. He has 2 years of teaching experience . Raj has completed Integrated Diploma in Management from Indian Institute of Management, Indore in 2016. He is well versed in Hindi and English.

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