https://www.urbanpro.com/bangalore/mona-a/14116745 # Mona A.

## Experienced Maths Teacher

## Overview

## Languages Spoken

## Education

## Address

## Verified Info

## Demo Class

## Class 10 Tuition Overview

## Reviews (1)

## FAQs

## Answers by Mona (9)

## Answers by Mona (9)

Mona A.

Ashok Nagar, Bangalore, India - 560025

Featured

Ashok Nagar, Bangalore, India - 560025.

1

5.0

Details verified of Mona A.✕

Identity

Education

Know how UrbanPro verifies Tutor details

Identity is verified based on matching the details uploaded by the Tutor with government databases.

1. Have taught Mathematics in schools in Delhi and Indore.

2. Bachelor's degree in Mathematical Statistics from Lady Shri Ram college, Delhi

3. Masters degree in Operational Research, Delhi University.

4. PG diploma in Computer Science , NIIT, Delhi

5. Have given personalised coaching for class 12 and GRE as well.

2. Bachelor's degree in Mathematical Statistics from Lady Shri Ram college, Delhi

3. Masters degree in Operational Research, Delhi University.

4. PG diploma in Computer Science , NIIT, Delhi

5. Have given personalised coaching for class 12 and GRE as well.

Hindi Mother Tongue (Native)

English Proficient

Lady Shri Ram College, Delhi University 1989

Bachelor of Arts (honours) Mathematical Statistics

Delhi University 1992

M.A. Operational Research

Ashok Nagar, Bangalore, India - 560025

ID Verified

Education Verified

Phone Verified

Email Verified

Yes, not charged

Report this Profile

✕

x

Is this listing inaccurate or duplicate? Any other problem?

Please tell us about the problem and we will fix it.

Class 10 Tuition

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 10 Tuition

5

Board

CBSE

CBSE Subjects taught

Mathematics

Experience in School or College

St. Xavier's school, Delhi Army Public School, Delhi Indore Public School, Delhi

Taught in School or College

Yes

this is test message this is test message this is test message this is test message this is test message this is test message this is test message

5.0 out of 5.0 1 review

5.0051

S

08 Sep, 2019

Sunayana attended Class 9 Tuition

"Can't get a better teacher than her. She won't help you to just get the answers to the problems but also help in getting the basics right. Her understanding of child psychology is beyond compare. I am recommending her for not just for class 9 but for all classes from 7 - 10. At least have one session with her and you will know what I am talking about. She is awesome! "

Have you attended any class with Mona? Write a Review

1. Which school boards of Class 10 do you teach for?

CBSE

2. Do you have any prior teaching experience?

Yes

3. Which classes do you teach?

I teach Class 10 Tuition, Class 7 Tuition, Class 8 Tuition and Class 9 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 5 years.

Answered on 26 Aug CBSE/Class 9/Mathematics/Surface Area and Volumes/NCERT Solutions/Exercise 13.7

V = (pi*r*r*h)/3 = (3.14*5*5*12)/3 = 3.14*100 =314 cu.cm

Like 0

Answers 3 Comments Answered on 13 Aug CBSE/Class 9/Mathematics/Surface Area and Volumes/NCERT Solutions/Exercise 13.7

V = πr²h/3 1570 = 3.14*r²*15/3 1570 = 15.7r² r² = 100 r = 10 cm

Like 0

Answers 6 Comments Answered on 13 Aug CBSE/Class 9/Mathematics/Surface Area and Volumes/NCERT Solutions/Exercise 13.7

V = πr²h/3 48π = 9πr²/3 = 3πr² 48/3 = 16 = r² r = 4 cm diameter = 2r = 8cm

Like 0

Answers 4 Comments Answered on 12 Aug CBSE/Class 9/Mathematics/Surface Area and Volumes/NCERT Solutions/Exercise 13.7

V = (1/3)*π*r*r*h r =10.5/2= 5.25 V= (1/3)*(22/7)*5.25*5.25*3 = 86.625 m^3 Area of the canvas = curved surface of heap = π*r*l l = √(r*r+h*h) = √(36.5625) =6.046 m (approx) πrl = 3.14*5.25*6.046 = 99.668 m^2 ...more

V = (1/3)*π*r*r*h

r =10.5/2= 5.25

V= (1/3)*(22/7)*5.25*5.25*3

= 86.625 m^3

Area of the canvas = curved surface of heap = π*r*l

l = √(r*r+h*h) = √(36.5625)

=6.046 m (approx)

πrl = 3.14*5.25*6.046

= 99.668 m^2

Like 0

Answers 4 Comments Answered on 12 Aug CBSE/Class 9/Mathematics/Surface Area and Volumes/NCERT Solutions/Exercise 13.7

(i)V = 1/3 * π* r*r*h 9856 = 1/3 *(22/7) *14*14*h h= (9856*3*7)/(22*14*14) =48 cm (ii) slant height l = √(r*r + h*h) = √ (196+2304) = √2500 = 50 cm (iii) curved surface area π*r*l = (22/7)*14*50 =2200 cm^2 ...more

(i)V = 1/3 * π* r*r*h

9856 = 1/3 *(22/7) *14*14*h

h= (9856*3*7)/(22*14*14)

=48 cm

(ii) slant height

l = √(r*r + h*h) = √ (196+2304)

= √2500 = 50 cm

(iii) curved surface area

π*r*l = (22/7)*14*50

=2200 cm^2

Like 0

Answers 5 Comments MonaDirections

x Class 10 Tuition 5.0

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 10 Tuition

5

Board

CBSE

CBSE Subjects taught

Mathematics

Experience in School or College

St. Xavier's school, Delhi Army Public School, Delhi Indore Public School, Delhi

Taught in School or College

Yes

Class 9 Tuition 5.0

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 9 Tuition

5

Board

CBSE

CBSE Subjects taught

Mathematics

Experience in School or College

St Xavier's school , Delhi Army Public School, Delhi Indore Public School, Indore

Taught in School or College

Yes

Class 8 Tuition 5.0

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 8 Tuition

20

Board

ICSE, CBSE

CBSE Subjects taught

Mathematics

ICSE Subjects taught

Mathematics

Experience in School or College

St Xavier's school Delhi Army Public School, Delhi Indore Public School, Indore

Taught in School or College

Yes

Teaching Experience in detail in Class 8 Tuition

Have taught in St Xavier's Delhi and Indore Public School, Indore.

Class 7 Tuition 5.0

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 7 Tuition

20

Board

ICSE, CBSE

CBSE Subjects taught

Mathematics

ICSE Subjects taught

Mathematics

Experience in School or College

St Xavier's school Delhi Army Public School, Delhi Indore Public School, Indore

Taught in School or College

Yes

Teaching Experience in detail in Class 7 Tuition

St Xavier's school Delhi Indore Public School Indore

this is test message this is test message this is test message this is test message this is test message this is test message this is test message

5.0 out of 5.0 1 review

S

08 Sep, 2019

Sunayana attended Class 9 Tuition

"Can't get a better teacher than her. She won't help you to just get the answers to the problems but also help in getting the basics right. Her understanding of child psychology is beyond compare. I am recommending her for not just for class 9 but for all classes from 7 - 10. At least have one session with her and you will know what I am talking about. She is awesome! "

Have you attended any class with Mona? Write a Review

Answered on 26 Aug CBSE/Class 9/Mathematics/Surface Area and Volumes/NCERT Solutions/Exercise 13.7

V = (pi*r*r*h)/3 = (3.14*5*5*12)/3 = 3.14*100 =314 cu.cm

Like 0

Answers 3 Comments Answered on 13 Aug CBSE/Class 9/Mathematics/Surface Area and Volumes/NCERT Solutions/Exercise 13.7

V = πr²h/3 1570 = 3.14*r²*15/3 1570 = 15.7r² r² = 100 r = 10 cm

Like 0

Answers 6 Comments Answered on 13 Aug CBSE/Class 9/Mathematics/Surface Area and Volumes/NCERT Solutions/Exercise 13.7

V = πr²h/3 48π = 9πr²/3 = 3πr² 48/3 = 16 = r² r = 4 cm diameter = 2r = 8cm

Like 0

Answers 4 Comments Answered on 12 Aug CBSE/Class 9/Mathematics/Surface Area and Volumes/NCERT Solutions/Exercise 13.7

V = (1/3)*π*r*r*h r =10.5/2= 5.25 V= (1/3)*(22/7)*5.25*5.25*3 = 86.625 m^3 Area of the canvas = curved surface of heap = π*r*l l = √(r*r+h*h) = √(36.5625) =6.046 m (approx) πrl = 3.14*5.25*6.046 = 99.668 m^2 ...more

V = (1/3)*π*r*r*h

r =10.5/2= 5.25

V= (1/3)*(22/7)*5.25*5.25*3

= 86.625 m^3

Area of the canvas = curved surface of heap = π*r*l

l = √(r*r+h*h) = √(36.5625)

=6.046 m (approx)

πrl = 3.14*5.25*6.046

= 99.668 m^2

Like 0

Answers 4 Comments Answered on 12 Aug CBSE/Class 9/Mathematics/Surface Area and Volumes/NCERT Solutions/Exercise 13.7

(i)V = 1/3 * π* r*r*h 9856 = 1/3 *(22/7) *14*14*h h= (9856*3*7)/(22*14*14) =48 cm (ii) slant height l = √(r*r + h*h) = √ (196+2304) = √2500 = 50 cm (iii) curved surface area π*r*l = (22/7)*14*50 =2200 cm^2 ...more

(i)V = 1/3 * π* r*r*h

9856 = 1/3 *(22/7) *14*14*h

h= (9856*3*7)/(22*14*14)

=48 cm

(ii) slant height

l = √(r*r + h*h) = √ (196+2304)

= √2500 = 50 cm

(iii) curved surface area

π*r*l = (22/7)*14*50

=2200 cm^2

Like 0

Answers 5 Comments Load More

Share this Profile

X

Post your Learning Need

Let us shortlist and give the best tutors and institutes.

or

Send Enquiry to Mona

Let Mona know you are interested in their class

Reply to 's review

Enter your reply*

Your reply has been successfully submitted.