Ashok Nagar, Bangalore, India - 560025.
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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
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Class Location
Online Classes (Video Call via UrbanPro LIVE)
Student's Home
Tutor's Home
Years of Experience in Class 10 Tuition
20
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 Location
Online Classes (Video Call via UrbanPro LIVE)
Student's Home
Tutor's Home
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 Location
Online Classes (Video Call via UrbanPro LIVE)
Student's Home
Tutor's Home
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 Location
Online Classes (Video Call via UrbanPro LIVE)
Student's Home
Tutor's Home
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
4.7 out of 5 7 reviews
Harshitha
Class 10 Tuition
"Mona mam is an amazing teacher who would explain what you expect detailedly . She have cleared all my doubts from the beginning. "
Smitha
"Mona mam's explanation is very fluent and flexible in classes too. Very happy to take classes from her. "
Rajasekhar
"It's very nice and very happy to hire you for the course to my son. He is happy and still continuing the sessions. "
Neha
"Good experience with Ms Mona, My son enjoys the subject now and looks forward to the classes surprisingly. A very hardworking teacher. "
Reply by Mona
Am so happy that he looks forward to studying Maths. Thanks Neha
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 20 years.
Answered on 26/08/2019 Learn CBSE/Class 9/Mathematics/Surface Area and Volumes/NCERT Solutions/Exercise 13.7
Answered on 13/08/2019 Learn CBSE/Class 9/Mathematics/Surface Area and Volumes/NCERT Solutions/Exercise 13.7
Answered on 13/08/2019 Learn CBSE/Class 9/Mathematics/Surface Area and Volumes/NCERT Solutions/Exercise 13.7
Answered on 12/08/2019 Learn 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
Answered on 12/08/2019 Learn 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
Class Location
Online Classes (Video Call via UrbanPro LIVE)
Student's Home
Tutor's Home
Years of Experience in Class 10 Tuition
20
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 Location
Online Classes (Video Call via UrbanPro LIVE)
Student's Home
Tutor's Home
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 Location
Online Classes (Video Call via UrbanPro LIVE)
Student's Home
Tutor's Home
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 Location
Online Classes (Video Call via UrbanPro LIVE)
Student's Home
Tutor's Home
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
Answered on 26/08/2019 Learn CBSE/Class 9/Mathematics/Surface Area and Volumes/NCERT Solutions/Exercise 13.7
Answered on 13/08/2019 Learn CBSE/Class 9/Mathematics/Surface Area and Volumes/NCERT Solutions/Exercise 13.7
Answered on 13/08/2019 Learn CBSE/Class 9/Mathematics/Surface Area and Volumes/NCERT Solutions/Exercise 13.7
Answered on 12/08/2019 Learn 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
Answered on 12/08/2019 Learn 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
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Ashok Nagar, Udaipur 
