Let first number be x then second number be 27-x.
Saddu, Raipur, India - 492014.
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Punjabi Mother Tongue (Native)
English Proficient
Hindi Basic
Pt. Ravishankar university 2018
Bachelor of Science (B.Sc.)
D12
Avinash suncity Daldal seoni near urkura road
Saddu, Raipur, India - 492014
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Years of Experience in Class 6 Tuition
1
Board
ICSE, CBSE
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Mathematics, EVS, Punjabi, English, Hindi, Science, Computers, Social Science
ICSE Subjects taught
Computer Science, Physics, Geography, English, Mathematics, History, Punjabi, Chemistry, Biology, EVS
Taught in School or College
No
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1
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ICSE, CBSE
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Mathematics, EVS, Punjabi, English, Hindi, Science, Computers, Social Science
ICSE Subjects taught
Computer Science, Physics, Geography, English, Mathematics, History, Punjabi, Chemistry, Biology, EVS
Taught in School or College
No
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Online (video chat via skype, google hangout etc)
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1
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ICSE, CBSE
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Mathematics, EVS, Punjabi, English, Hindi, Science, Computers, Social Science
ICSE Subjects taught
Computer Science, Physics, Geography, English, Mathematics, History, Punjabi, Chemistry, Biology, EVS
Taught in School or College
No
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Online (video chat via skype, google hangout etc)
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Tutor's Home
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Student
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Reading, Grammar and Vocabulary improvement, Hindi Speaking, Writing
Awards and Recognition
No
Mother Tongue
No
Languages apart from english in which classes are conducted
Yes
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Online (video chat via skype, google hangout etc)
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1
Fees
₹ 135.0 per hour
Board
ICSE, CBSE
CBSE Subjects taught
English, Mathematics, Computers, Hindi, Punjabi, Science, EVS
ICSE Subjects taught
Punjabi, Science, Social Studies, Hindi, English, EVS, Mathematics
Taught in School or College
No
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Online (video chat via skype, google hangout etc)
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1
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EVS, English, Mathematics
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No
Teaching Experience in detail in Nursery-KG Tuition
Good speaking skills with good knowledge of subjects. Teaching students from 1-12.
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Online (video chat via skype, google hangout etc)
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CBSE
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Social science, Information and Comunication Technology, Science, English, Punjabi, Mathematics
Taught in School or College
No
Teaching Experience in detail in Class 10 Tuition
u
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CBSE
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Social science, Information and Comunication Technology, Science, English, Punjabi, Mathematics
Taught in School or College
No
Teaching Experience in detail in Class 9 Tuition
u
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16 yrs to 25 yrs, 10 yrs to 15 yrs, Below 10 yrs
Lived or Worked in English Speaking Country
No
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IELTS
Awards and Recognition
No
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None
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Tutor
Language of instruction offered
English to English, Hindi to English
Curriculum Expertise
CBSE
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Yes
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Group Classes, One on one/ Private Tutions
Teaching done in
Basic Spoken English, Vocabulary
Teaching at
Home
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Years of Experience in Gym
1
1. Which school boards of Class 8 do you teach for?
ICSE, CBSE
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, Gym, Hindi Language, Nursery-KG Tuition and Spoken English 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 1 year.
Answered on 08/08/2019 Learn Tuition
From the question we know (1,-2)i.e x=1 &y=-2
Putting the values of x and y in the equation
2x-3y=8
2*1-3*(-2)=8 (2*1=2 and 3*(-2)=-6)
2-(-6)=8 (-*-=+)
2+6=8
8=8
It proves that both sides are equal.
Therefore ,(1,-2)is a solution of equation 2x-3y=8
Answered on 19/06/2019 Learn CBSE/Class 10/Mathematics/UNIT II: Algebra/Quadratic Equations/NCERT Solutions/Exercise 4.3
Let the average speed of the passenger train be 'x' km/hr and the average speed of the express train be (x + 11) km/hr
Distance between Mysore and Bangalore = 132 km
It is given that the time taken by the express train to cover the distance of 132 km is 1 hour less than the passenger train to cover the same distance.
So, time taken by passenger train = 132/x hr
The time taken by the express train = {(132)/x+11} hr
Now, according to the question
{132/(x+11)} = 132/x + 1
After taking L.C.M. of 132/x +1 and then solving it we get (132+x)/x.
Now,
{132/(x+11)} = (132+x)/x
By cross multiplying, we get
132x = x²+132x+11x+1452
x² + 11x - 1452 = 0
x² + 44x - 33x - 1452 = 0
x(x+44) - 33(x+44) = 0
(x+33) (x+44) = 0
x - 44 or x = 33
As the speed cannot be in negative therefore, x = 33 or the speed of the passenger train = 33 km/hr and the speed is 33 + 11 = 44 km/hr.
Answered on 19/06/2019 Learn CBSE/Class 10/Mathematics/UNIT II: Algebra/Quadratic Equations/NCERT Solutions/Exercise 4.3
The longest side of the triangles i.e. the hypotenuse.
So, by Pythagoras Theorem,
(Diagonal)² = (Smaller Side)² + (Longer Side)²
(x + 60)² = (x)² + (x + 30)²
x² + 120x + 3600 = x² + x² + 60x + 900
x² + 60x - 120x + 900 - 3600 = 0
x² - 60x - 2700 = 0
x² - 90x + 30x - 2700 = 0
x(x - 90) + 30(x - 90) = 0
(x - 90) (x + 30) = 0
x = 90 because x = -30 as length cannot be possible.
So the length of the shorter side is 90 meters and the length of the longer side is 90 + 30 = 120 meters.
Answered on 19/06/2019 Learn CBSE/Class 10/Mathematics/UNIT II: Algebra/Quadratic Equations/NCERT Solutions/Exercise 4.3
Let the time taken by the smaller diameter tap = x
larger = x-10
total time taken = 75 /8
portion filled in one hour by smaller diameter tap = 1/x
and by larger diamter tap = 1/x-10
1/x + 1/x-10 = 8/75
x-10+x/x(x-10) = 8/75
2x+10/x²-10x = 8/75
8(x²-10x) = 75 ×2 (x-5)
8/2 (x²-10x) = 75 (x-5)
4x²-40x = 75x-375
4x² -40x-75x +375 = 0
4x²-115x + 375 = 0
4x²-100x-15x +375 = 0
4x(x-25)-15(x-25)=0
(x-25)(4x-15)
x= 25
x= 15/4
If x= 25
the x-10 = 25-10 = 15
if x = 15/4
x-10 = 15/4 - 10 = 15-40/4 = -25/4
since time cannot be negative therefore x = 25.
Answered on 19/06/2019 Learn CBSE/Class 10/Mathematics/UNIT II: Algebra/Quadratic Equations/NCERT Solutions/Exercise 4.2
Let first number be x then second number be 27-x.
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 6 Tuition
1
Board
ICSE, CBSE
CBSE Subjects taught
Mathematics, EVS, Punjabi, English, Hindi, Science, Computers, Social Science
ICSE Subjects taught
Computer Science, Physics, Geography, English, Mathematics, History, Punjabi, Chemistry, Biology, EVS
Taught in School or College
No
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 7 Tuition
1
Board
ICSE, CBSE
CBSE Subjects taught
Mathematics, EVS, Punjabi, English, Hindi, Science, Computers, Social Science
ICSE Subjects taught
Computer Science, Physics, Geography, English, Mathematics, History, Punjabi, Chemistry, Biology, EVS
Taught in School or College
No
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class 8 Tuition
1
Board
ICSE, CBSE
CBSE Subjects taught
Mathematics, EVS, Punjabi, English, Hindi, Science, Computers, Social Science
ICSE Subjects taught
Computer Science, Physics, Geography, English, Mathematics, History, Punjabi, Chemistry, Biology, EVS
Taught in School or College
No
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Profession
Student
Outcomes taught for
Reading, Grammar and Vocabulary improvement, Hindi Speaking, Writing
Awards and Recognition
No
Mother Tongue
No
Languages apart from english in which classes are conducted
Yes
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Class I-V Tuition
1
Fees
₹ 135.0 per hour
Board
ICSE, CBSE
CBSE Subjects taught
English, Mathematics, Computers, Hindi, Punjabi, Science, EVS
ICSE Subjects taught
Punjabi, Science, Social Studies, Hindi, English, EVS, Mathematics
Taught in School or College
No
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Nursery-KG Tuition
1
Subject
EVS, English, Mathematics
Taught in School or College
No
Teaching Experience in detail in Nursery-KG Tuition
Good speaking skills with good knowledge of subjects. Teaching students from 1-12.
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Board
CBSE
CBSE Subjects taught
Social science, Information and Comunication Technology, Science, English, Punjabi, Mathematics
Taught in School or College
No
Teaching Experience in detail in Class 10 Tuition
u
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Board
CBSE
CBSE Subjects taught
Social science, Information and Comunication Technology, Science, English, Punjabi, Mathematics
Taught in School or College
No
Teaching Experience in detail in Class 9 Tuition
u
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Age groups catered to
16 yrs to 25 yrs, 10 yrs to 15 yrs, Below 10 yrs
Lived or Worked in English Speaking Country
No
Exams Attended
IELTS
Awards and Recognition
No
Certification
None
Profession
Tutor
Language of instruction offered
English to English, Hindi to English
Curriculum Expertise
CBSE
Citizen of English Speaking Country
Yes
Class strength catered to
Group Classes, One on one/ Private Tutions
Teaching done in
Basic Spoken English, Vocabulary
Teaching at
Home
Class Location
Online (video chat via skype, google hangout etc)
Student's Home
Tutor's Home
Years of Experience in Gym
1
Answered on 08/08/2019 Learn Tuition
From the question we know (1,-2)i.e x=1 &y=-2
Putting the values of x and y in the equation
2x-3y=8
2*1-3*(-2)=8 (2*1=2 and 3*(-2)=-6)
2-(-6)=8 (-*-=+)
2+6=8
8=8
It proves that both sides are equal.
Therefore ,(1,-2)is a solution of equation 2x-3y=8
Answered on 19/06/2019 Learn CBSE/Class 10/Mathematics/UNIT II: Algebra/Quadratic Equations/NCERT Solutions/Exercise 4.3
Let the average speed of the passenger train be 'x' km/hr and the average speed of the express train be (x + 11) km/hr
Distance between Mysore and Bangalore = 132 km
It is given that the time taken by the express train to cover the distance of 132 km is 1 hour less than the passenger train to cover the same distance.
So, time taken by passenger train = 132/x hr
The time taken by the express train = {(132)/x+11} hr
Now, according to the question
{132/(x+11)} = 132/x + 1
After taking L.C.M. of 132/x +1 and then solving it we get (132+x)/x.
Now,
{132/(x+11)} = (132+x)/x
By cross multiplying, we get
132x = x²+132x+11x+1452
x² + 11x - 1452 = 0
x² + 44x - 33x - 1452 = 0
x(x+44) - 33(x+44) = 0
(x+33) (x+44) = 0
x - 44 or x = 33
As the speed cannot be in negative therefore, x = 33 or the speed of the passenger train = 33 km/hr and the speed is 33 + 11 = 44 km/hr.
Answered on 19/06/2019 Learn CBSE/Class 10/Mathematics/UNIT II: Algebra/Quadratic Equations/NCERT Solutions/Exercise 4.3
The longest side of the triangles i.e. the hypotenuse.
So, by Pythagoras Theorem,
(Diagonal)² = (Smaller Side)² + (Longer Side)²
(x + 60)² = (x)² + (x + 30)²
x² + 120x + 3600 = x² + x² + 60x + 900
x² + 60x - 120x + 900 - 3600 = 0
x² - 60x - 2700 = 0
x² - 90x + 30x - 2700 = 0
x(x - 90) + 30(x - 90) = 0
(x - 90) (x + 30) = 0
x = 90 because x = -30 as length cannot be possible.
So the length of the shorter side is 90 meters and the length of the longer side is 90 + 30 = 120 meters.
Answered on 19/06/2019 Learn CBSE/Class 10/Mathematics/UNIT II: Algebra/Quadratic Equations/NCERT Solutions/Exercise 4.3
Let the time taken by the smaller diameter tap = x
larger = x-10
total time taken = 75 /8
portion filled in one hour by smaller diameter tap = 1/x
and by larger diamter tap = 1/x-10
1/x + 1/x-10 = 8/75
x-10+x/x(x-10) = 8/75
2x+10/x²-10x = 8/75
8(x²-10x) = 75 ×2 (x-5)
8/2 (x²-10x) = 75 (x-5)
4x²-40x = 75x-375
4x² -40x-75x +375 = 0
4x²-115x + 375 = 0
4x²-100x-15x +375 = 0
4x(x-25)-15(x-25)=0
(x-25)(4x-15)
x= 25
x= 15/4
If x= 25
the x-10 = 25-10 = 15
if x = 15/4
x-10 = 15/4 - 10 = 15-40/4 = -25/4
since time cannot be negative therefore x = 25.
Answered on 19/06/2019 Learn CBSE/Class 10/Mathematics/UNIT II: Algebra/Quadratic Equations/NCERT Solutions/Exercise 4.2
Let first number be x then second number be 27-x.
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