Ganapathy, Coimbatore, India - 641006.
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Tamil Mother Tongue (Native)
English Proficient
Hindi Basic
Malayalam Basic
GOBI ARTS COLLEGE 1994
Bachelor of Science (B.Sc.)
Gobi Arts college 1996
Master of Science (M.Sc.)
Ganapathy, Coimbatore, India - 641006
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Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 9 Tuition
15
Board
State, IGCSE, CBSE
CBSE Subjects taught
Mathematics
IGCSE Subjects taught
Mathematics
Taught in School or College
Yes
State Syllabus Subjects taught
Mathematics, Science
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 10 Tuition
15
Board
State, IGCSE, CBSE
CBSE Subjects taught
Mathematics
IGCSE Subjects taught
Mathematics, Physics
Taught in School or College
Yes
State Syllabus Subjects taught
Mathematics
1. Which school boards of Class 10 do you teach for?
State, IGCSE and CBSE
2. Do you have any prior teaching experience?
Yes
3. Which classes do you teach?
I teach Class 10 Tuition and Class 9 Tuition Classes.
4. Do you provide a demo class?
No, I don't provide a demo class.
5. How many years of experience do you have?
I have been teaching for 15 years.
Answered on 20/08/2019 Learn CBSE/Class 10/Mathematics/UNIT II: Algebra/Arithmetic Progression/NCERT Solutions/Exercise 5.2
This A.P. can be written in reverse order as
253, 248, 243, …, 13, 8, 3
For this A.P.,
a = 253
d = 248 − 253 = −5
n = 20
a20 = a + (20 − 1) d
= 253 + (19) (−5)
a20= 158
Therefore, 20th term from the last term is 158.
Answered on 19/08/2019 Learn CBSE/Class 10/Mathematics/UNIT II: Algebra/Pair of linear equations in two variable/NCERT Solutions/Exercise 3.7
Let the capital of the first =x
capital of the second = y
According to first statement,
x+100=2(y-100)
⇒ x-2y=-300------(1)
according to second statement,
y+10=6(x-10)
6x-y=70--------(2)
solving (1) & (2)
(2)×2⇒12x-2y=140
x-2y=-300
(2)-(1)⇒ 11x=440
x=440/11=40
substitute x=40 in (1)
40-2y=-300
-2y=-300-40=-340
y=-340/(-2) =170
First capital=40
second capital=170
Answered on 19/08/2019 Learn CBSE/Class 10/Mathematics/UNIT II: Algebra/Arithmetic Progression/NCERT Solutions/Exercise 5.3
Given a=5 l=45 Sn=400
Sn= n/2(a+l)
400 =(n/2) (5+45)
400=50n/2=25n
n=400/25
=16 = number of terms
To find common difference 'd'
a16=45 ⇒5+15d=45
15d=40
d= 8/3
Answered on 23/07/2019 Learn CBSE/Class 10/Mathematics/UNIT II: Algebra/Arithmetic Progression/NCERT Solutions/Exercise 5.2
Given a11=a+10d=38→(1)
a16=a+15d=73→(2)
(2)-(1)⇒ 5d= 35⇒d= 7
Substitute d=7 in (1) we get
a+10*7 =38
⇒a=38-70=-32
a31=-32+30*7=-32+210
= 178
Answered on 22/07/2019 Learn CBSE/Class 10/Mathematics/UNIT II: Algebra/Arithmetic Progression/NCERT Solutions/Exercise 5.2
The first 3 digit number divisible by 7 is 105
Last 3 digit number divisible by 7 is 994
formula- =a+(n-1)d
Number of terms n = (l-a/d) +1
=(994-105/7) +1
=889/7 +1
=127+1
=128
Hence there are 128 three digit numbers divisible by 7.
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 9 Tuition
15
Board
State, IGCSE, CBSE
CBSE Subjects taught
Mathematics
IGCSE Subjects taught
Mathematics
Taught in School or College
Yes
State Syllabus Subjects taught
Mathematics, Science
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 10 Tuition
15
Board
State, IGCSE, CBSE
CBSE Subjects taught
Mathematics
IGCSE Subjects taught
Mathematics, Physics
Taught in School or College
Yes
State Syllabus Subjects taught
Mathematics
Answered on 20/08/2019 Learn CBSE/Class 10/Mathematics/UNIT II: Algebra/Arithmetic Progression/NCERT Solutions/Exercise 5.2
This A.P. can be written in reverse order as
253, 248, 243, …, 13, 8, 3
For this A.P.,
a = 253
d = 248 − 253 = −5
n = 20
a20 = a + (20 − 1) d
= 253 + (19) (−5)
a20= 158
Therefore, 20th term from the last term is 158.
Answered on 19/08/2019 Learn CBSE/Class 10/Mathematics/UNIT II: Algebra/Pair of linear equations in two variable/NCERT Solutions/Exercise 3.7
Let the capital of the first =x
capital of the second = y
According to first statement,
x+100=2(y-100)
⇒ x-2y=-300------(1)
according to second statement,
y+10=6(x-10)
6x-y=70--------(2)
solving (1) & (2)
(2)×2⇒12x-2y=140
x-2y=-300
(2)-(1)⇒ 11x=440
x=440/11=40
substitute x=40 in (1)
40-2y=-300
-2y=-300-40=-340
y=-340/(-2) =170
First capital=40
second capital=170
Answered on 19/08/2019 Learn CBSE/Class 10/Mathematics/UNIT II: Algebra/Arithmetic Progression/NCERT Solutions/Exercise 5.3
Given a=5 l=45 Sn=400
Sn= n/2(a+l)
400 =(n/2) (5+45)
400=50n/2=25n
n=400/25
=16 = number of terms
To find common difference 'd'
a16=45 ⇒5+15d=45
15d=40
d= 8/3
Answered on 23/07/2019 Learn CBSE/Class 10/Mathematics/UNIT II: Algebra/Arithmetic Progression/NCERT Solutions/Exercise 5.2
Given a11=a+10d=38→(1)
a16=a+15d=73→(2)
(2)-(1)⇒ 5d= 35⇒d= 7
Substitute d=7 in (1) we get
a+10*7 =38
⇒a=38-70=-32
a31=-32+30*7=-32+210
= 178
Answered on 22/07/2019 Learn CBSE/Class 10/Mathematics/UNIT II: Algebra/Arithmetic Progression/NCERT Solutions/Exercise 5.2
The first 3 digit number divisible by 7 is 105
Last 3 digit number divisible by 7 is 994
formula- =a+(n-1)d
Number of terms n = (l-a/d) +1
=(994-105/7) +1
=889/7 +1
=127+1
=128
Hence there are 128 three digit numbers divisible by 7.
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