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Deepika Agrawal Class 11 Tuition trainer in Bangalore

Deepika Agrawal

"Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge."

Begur Road, Bangalore, India - 560068.

6 Students

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Overview

With a decade of dedicated teaching experience, I have honed my skills in fostering a dynamic and engaging learning environment. My journey as an educator spans across various levels and subjects, demonstrating a consistent commitment to student success and academic excellence.

Key Highlights:
Diverse Teaching Portfolio: Experience teaching a wide range of subjects, catering to different educational levels from elementary to high school, and specialized subjects.
Innovative Instructional Strategies: Implemented creative teaching methods, including technology integration, project-based learning, and differentiated instruction to accommodate diverse learning styles.
Student-Centered Approach: Focused on building strong relationships with students, understanding their individual needs, and encouraging a growth mindset.
Curriculum Development: Successfully designed and updated curricula to align with educational standards and student needs, ensuring relevance and rigor.
Professional Development: Committed to continuous learning and professional growth, attending workshops, conferences, and obtaining certifications to stay current with educational trends and best practices.
Assessment and Feedback: Developed effective assessment tools and strategies to measure student progress, providing timely and constructive feedback to support their academic journey.
Classroom Management: Maintained a positive and productive classroom atmosphere, utilizing proactive behavior management techniques to minimize disruptions and maximize learning time.
Collaboration and Leadership: Worked collaboratively with colleagues, administrators, and parents to foster a supportive educational community, taking on leadership roles such as mentoring new teachers and leading professional learning communities.

Languages Spoken

English Proficient

Hindi Proficient

Education

DDU 2008

Bachelor of Commerce (B.Com.)

ITM Gorakhpur 2011

Master of Business Administration (M.B.A.)

Address

Begur Road, Bangalore, India - 560068

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Taught Students from these Schools

D

Delhi Public School

Electronic City, Bangalore

N

Narayana E-Techno School

Kammanahalli Sena Vihar, Bangalore

S

Sri Chaitanya Techno School

Arunodaya Nagar, Hayathnagar_Khalsa

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Teaches

Class 11 Tuition

Class Location

Online Classes (Video Call via UrbanPro LIVE)

Student's Home

Tutor's Home

Years of Experience in Class 11 Tuition

10

Board

CBSE, ISC/ICSE

Preferred class strength

One on one/ Private Tutions, Group Classes

ISC/ICSE Subjects taught

Accounts

CBSE Subjects taught

Accountancy

Taught in School or College

Yes

BCom Tuition

Class Location

Online Classes (Video Call via UrbanPro LIVE)

Student's Home

Tutor's Home

Years of Experience in BCom Tuition

5

BCom Subject

Financial Accounting, Cost Accounting

Type of class

Crash Course, Regular Classes

Class strength catered to

One on one/ Private Tutions, Group Classes

Taught in School or College

Yes

Class 12 Tuition

Class Location

Online Classes (Video Call via UrbanPro LIVE)

Student's Home

Tutor's Home

Years of Experience in Class 12 Tuition

10

Board

CBSE, ISC/ICSE

Preferred class strength

One on one/ Private Tutions, Group Classes

ISC/ICSE Subjects taught

Accounts

CBSE Subjects taught

Accountancy

Taught in School or College

Yes

Reviews

No Reviews yet!

FAQs

1. Which school boards of Class 12 do you teach for?

CBSE and ISC/ICSE

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

Yes

3. Which classes do you teach?

I teach BCom Tuition, Class 11 Tuition and Class 12 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 10 years.

Lessons (5)

Introduction of Cost Accounting

QUESTIONS: 1.Define profit centre. Ans: Profit Centre is a part of business accountable for costs and revenues. State the objective of Cost Accounting. Ans: Objectives of Cost Accounting The...

17 May
0 0
0
Accounting for partnership: Basic concept

QUESTIONS: Define Partnership Deed. Answer: A partnership deed is an agreement among the partners which contains ai! the terms of the Partnership. It generally contains the details about all the...

10 May
0 0
1
Indian Economy on the Eve of Independence

Name some notable economists who estimated India’s per capita income during the colonial period. Ans: The following economists estimated India’s per capita income during the colonial...

10 May
0 0
0

Answers by Deepika Agrawal (187)

Answered on 03 Jul Learn CBSE/Class 9/Mathematics/Geometry/Quadrilaterals

Given: In square ABCD, AK = BL = CM = DN.To prove: KLMN is a square. In square ABCD, AB = BC = CD = DA And, AK = BL = CM = DN (All sides of a square are equal.) (Given) So, AB - AK = BC - BL = CD - CM = DA - DN ⇒ KB = CL = DM = AN.......... (1) In △NAKand△KBL∠NAK=∠KBL=900 (Each angle of... ...more

Given: In square ABCD, AK = BL = CM = DN.
To prove: KLMN is a square.

In square ABCD,

AB = BC = CD = DA

And, AK = BL = CM = DN

(All sides of a square are equal.) (Given)

So, AB - AK = BC - BL = CD - CM = DA - DN

⇒ KB = CL = DM = AN.......... (1)

In NAKandKBL
NAK=KBL=900 (Each angle of a square is a right angle.)
AK=BL (Given)
AN=KB [From (1)]
So, by SAS congruence criteria,
\(\triangle NAK ≅ \triangle KBL \)
⇒NK=KL (Cpctc ...(2)
Similarly,
\(\triangle MDN≅ \triangle NAK \)

\(\triangle DNM≅ \triangle CML \)

\(\triangle MCL≅ \triangle LBK \)
\(\rightarrow MN = NK and \angle DNM=\angle KNA \) (Cpctc )… 3)
MN = JM and DNM=CML (Cpctc )… 4)
ML = LK and CML=BLK (Cpctc )… (5)
From (2), (3), (4) and (5), we get
NK = KL = MN = ML…….....(6)
And, DNM=AKN=KLB=LMC
Now,
In NAK
NAK=900
Let AKN=x0

So, DNK=900+x0 (Exterior angles equals sum of interior opposite angles.)
DNM+MNK=900+x0

x0+MNK=900+x0

⇒ MNK=900
Similarly,
NKL=KLM=LMN=900 ...(7)
Using (6) and (7), we get
All sides of quadrilateral KLMN are equal and all angles are \ ( 90^0 \)
So, KLMN is a square.

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Answered on 03 Jul Learn CBSE/Class 9/Mathematics/Geometry/Construction

It is given that AB: BC: CA = 3: 4: 5 3x + 4x + 5x = 12 12x = 12 x = 1 AB = 3 cm, BC = 4 cm and CA = 5 cm Steps of construction: (1) Draw a sufficiently long line segment using a ruler. (2) Locate points A and B on it such that AB = 3 cm. (3) With A as the centre and radius 5 cm, draw an arc. (4) With... ...more

It is given that AB: BC: CA = 3: 4: 5

3x + 4x + 5x = 12

12x = 12

x = 1

AB = 3 cm, BC = 4 cm and CA = 5 cm

Steps of construction:

(1) Draw a sufficiently long line segment using a ruler.

(2) Locate points A and B on it such that AB = 3 cm.

(3) With A as the centre and radius 5 cm, draw an arc.

(4) With B as the centre and radius 4 cm, draw another arc that cuts the previous arc at C.

(5) Join AC and BC.

Then, ABC is the required triangle.

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Answered on 03 Jul Learn CBSE/Class 9/Mathematics/Geometry/Construction

Steps of construction:1. Draw a line segment PQ = 11.6 cm.2. Construct an angle of 45° and bisect it to get ∠QPX.3. Construct an angle of 60° and bisect it to get ∠PQY.4. The ray XP and YQ intersect at A.5. Draw the right bisectors of AP and AQ, cutting PQ at B and C, respectively.6.... ...more

Steps of construction:
1. Draw a line segment PQ = 11.6 cm.
2. Construct an angle of 45° and bisect it to get ∠QPX.
3. Construct an angle of 60° and bisect it to get ∠PQY.
4. The ray XP and YQ intersect at A.
5. Draw the right bisectors of AP and AQ, cutting PQ at B and C, respectively.
6. Join AB and AC.
Thus, △ABC is the required triangle.

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Answered on 01 Jul Learn CBSE/Class 9/Mathematics/Geometry/Quadrilaterals

Quadrilateral PQRS has angle bisectors PT,QA,RA,SC. ΔPQB,ΔQBT,ΔSDC are right angled triangle. Let angle P=2x so, ∠PQB=90−x=∠BQT ∴∠QTB=(90−(90−x))=x ∠CTR=180−x In triangle SDR, ∠RDS=90∘, in parallelogram DCTR ∠DCT & ∠CDR=90∘ ∴∠DRT=x... ...more
Quadrilateral PQRS has angle bisectors PT,QA,RA,SC.
ΔPQB,ΔQBT,ΔSDC are right angled triangle.
Let angle P=2x
so, PQB=90x=BQT
QTB=(90(90x))=x
CTR=180x
In triangle SDR,
RDS=90, in parallelogram DCTR
DCT & CDR=90
DRT=x & DRS=x
DSR=90x
sum of adjacent angles, P+Q=180
Opposite angles P=R,Q=C
 PQRS is parallelogram
1368583_1177888_ans_554d740da33a4cd79ab411e844ef8bcc.png
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Answered on 01 Jul Learn CBSE/Class 9/Mathematics/Geometry/Construction

Given: In ∆ABCAB + BC + CA = 10 cm, ∠B = 60° and ∠C = 45°.Required: To construct ∆ABC.Steps of Construction :1. Draw DE = 10 cm2. At D, construct ∠EDP = 12 of 60°= 30° and at E, construct ∠DEQ =12 of 45° = 22 12∘3. Let DP and EQ meet at A.4. Draw perpendicular bisector... ...more

Given: In ∆ABC
AB + BC + CA = 10 cm, ∠B = 60° and ∠C = 45°.
Required: To construct ∆ABC.

Steps of Construction :
1. Draw DE = 10 cm
2. At D, construct ∠EDP = 12 of 60°= 30° and at E, construct ∠DEQ =12 of 45° = 22 12
3. Let DP and EQ meet at A.
4. Draw perpendicular bisector of AD to meet DE at B.
5. Draw perpendicular bisector of AE to meet DE at C.
6. Join AB and AC. Thus, ABC is the required triangle.

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Teaches

Class 11 Tuition

Class Location

Online Classes (Video Call via UrbanPro LIVE)

Student's Home

Tutor's Home

Years of Experience in Class 11 Tuition

10

Board

CBSE, ISC/ICSE

Preferred class strength

One on one/ Private Tutions, Group Classes

ISC/ICSE Subjects taught

Accounts

CBSE Subjects taught

Accountancy

Taught in School or College

Yes

BCom Tuition

Class Location

Online Classes (Video Call via UrbanPro LIVE)

Student's Home

Tutor's Home

Years of Experience in BCom Tuition

5

BCom Subject

Financial Accounting, Cost Accounting

Type of class

Crash Course, Regular Classes

Class strength catered to

One on one/ Private Tutions, Group Classes

Taught in School or College

Yes

Class 12 Tuition

Class Location

Online Classes (Video Call via UrbanPro LIVE)

Student's Home

Tutor's Home

Years of Experience in Class 12 Tuition

10

Board

CBSE, ISC/ICSE

Preferred class strength

One on one/ Private Tutions, Group Classes

ISC/ICSE Subjects taught

Accounts

CBSE Subjects taught

Accountancy

Taught in School or College

Yes

No Reviews yet!

Answers by Deepika Agrawal (187)

Answered on 03 Jul Learn CBSE/Class 9/Mathematics/Geometry/Quadrilaterals

Given: In square ABCD, AK = BL = CM = DN.To prove: KLMN is a square. In square ABCD, AB = BC = CD = DA And, AK = BL = CM = DN (All sides of a square are equal.) (Given) So, AB - AK = BC - BL = CD - CM = DA - DN ⇒ KB = CL = DM = AN.......... (1) In △NAKand△KBL∠NAK=∠KBL=900 (Each angle of... ...more

Given: In square ABCD, AK = BL = CM = DN.
To prove: KLMN is a square.

In square ABCD,

AB = BC = CD = DA

And, AK = BL = CM = DN

(All sides of a square are equal.) (Given)

So, AB - AK = BC - BL = CD - CM = DA - DN

⇒ KB = CL = DM = AN.......... (1)

In NAKandKBL
NAK=KBL=900 (Each angle of a square is a right angle.)
AK=BL (Given)
AN=KB [From (1)]
So, by SAS congruence criteria,
\(\triangle NAK ≅ \triangle KBL \)
⇒NK=KL (Cpctc ...(2)
Similarly,
\(\triangle MDN≅ \triangle NAK \)

\(\triangle DNM≅ \triangle CML \)

\(\triangle MCL≅ \triangle LBK \)
\(\rightarrow MN = NK and \angle DNM=\angle KNA \) (Cpctc )… 3)
MN = JM and DNM=CML (Cpctc )… 4)
ML = LK and CML=BLK (Cpctc )… (5)
From (2), (3), (4) and (5), we get
NK = KL = MN = ML…….....(6)
And, DNM=AKN=KLB=LMC
Now,
In NAK
NAK=900
Let AKN=x0

So, DNK=900+x0 (Exterior angles equals sum of interior opposite angles.)
DNM+MNK=900+x0

x0+MNK=900+x0

⇒ MNK=900
Similarly,
NKL=KLM=LMN=900 ...(7)
Using (6) and (7), we get
All sides of quadrilateral KLMN are equal and all angles are \ ( 90^0 \)
So, KLMN is a square.

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Answered on 03 Jul Learn CBSE/Class 9/Mathematics/Geometry/Construction

It is given that AB: BC: CA = 3: 4: 5 3x + 4x + 5x = 12 12x = 12 x = 1 AB = 3 cm, BC = 4 cm and CA = 5 cm Steps of construction: (1) Draw a sufficiently long line segment using a ruler. (2) Locate points A and B on it such that AB = 3 cm. (3) With A as the centre and radius 5 cm, draw an arc. (4) With... ...more

It is given that AB: BC: CA = 3: 4: 5

3x + 4x + 5x = 12

12x = 12

x = 1

AB = 3 cm, BC = 4 cm and CA = 5 cm

Steps of construction:

(1) Draw a sufficiently long line segment using a ruler.

(2) Locate points A and B on it such that AB = 3 cm.

(3) With A as the centre and radius 5 cm, draw an arc.

(4) With B as the centre and radius 4 cm, draw another arc that cuts the previous arc at C.

(5) Join AC and BC.

Then, ABC is the required triangle.

Answers 1 Comments
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Answered on 03 Jul Learn CBSE/Class 9/Mathematics/Geometry/Construction

Steps of construction:1. Draw a line segment PQ = 11.6 cm.2. Construct an angle of 45° and bisect it to get ∠QPX.3. Construct an angle of 60° and bisect it to get ∠PQY.4. The ray XP and YQ intersect at A.5. Draw the right bisectors of AP and AQ, cutting PQ at B and C, respectively.6.... ...more

Steps of construction:
1. Draw a line segment PQ = 11.6 cm.
2. Construct an angle of 45° and bisect it to get ∠QPX.
3. Construct an angle of 60° and bisect it to get ∠PQY.
4. The ray XP and YQ intersect at A.
5. Draw the right bisectors of AP and AQ, cutting PQ at B and C, respectively.
6. Join AB and AC.
Thus, △ABC is the required triangle.

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Answered on 01 Jul Learn CBSE/Class 9/Mathematics/Geometry/Quadrilaterals

Quadrilateral PQRS has angle bisectors PT,QA,RA,SC. ΔPQB,ΔQBT,ΔSDC are right angled triangle. Let angle P=2x so, ∠PQB=90−x=∠BQT ∴∠QTB=(90−(90−x))=x ∠CTR=180−x In triangle SDR, ∠RDS=90∘, in parallelogram DCTR ∠DCT & ∠CDR=90∘ ∴∠DRT=x... ...more
Quadrilateral PQRS has angle bisectors PT,QA,RA,SC.
ΔPQB,ΔQBT,ΔSDC are right angled triangle.
Let angle P=2x
so, PQB=90x=BQT
QTB=(90(90x))=x
CTR=180x
In triangle SDR,
RDS=90, in parallelogram DCTR
DCT & CDR=90
DRT=x & DRS=x
DSR=90x
sum of adjacent angles, P+Q=180
Opposite angles P=R,Q=C
 PQRS is parallelogram
1368583_1177888_ans_554d740da33a4cd79ab411e844ef8bcc.png
Answers 1 Comments
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Answered on 01 Jul Learn CBSE/Class 9/Mathematics/Geometry/Construction

Given: In ∆ABCAB + BC + CA = 10 cm, ∠B = 60° and ∠C = 45°.Required: To construct ∆ABC.Steps of Construction :1. Draw DE = 10 cm2. At D, construct ∠EDP = 12 of 60°= 30° and at E, construct ∠DEQ =12 of 45° = 22 12∘3. Let DP and EQ meet at A.4. Draw perpendicular bisector... ...more

Given: In ∆ABC
AB + BC + CA = 10 cm, ∠B = 60° and ∠C = 45°.
Required: To construct ∆ABC.

Steps of Construction :
1. Draw DE = 10 cm
2. At D, construct ∠EDP = 12 of 60°= 30° and at E, construct ∠DEQ =12 of 45° = 22 12
3. Let DP and EQ meet at A.
4. Draw perpendicular bisector of AD to meet DE at B.
5. Draw perpendicular bisector of AE to meet DE at C.
6. Join AB and AC. Thus, ABC is the required triangle.

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Lessons (5)

Introduction of Cost Accounting

QUESTIONS: 1.Define profit centre. Ans: Profit Centre is a part of business accountable for costs and revenues. State the objective of Cost Accounting. Ans: Objectives of Cost Accounting The...

17 May
0 0
0
Accounting for partnership: Basic concept

QUESTIONS: Define Partnership Deed. Answer: A partnership deed is an agreement among the partners which contains ai! the terms of the Partnership. It generally contains the details about all the...

10 May
0 0
1
Indian Economy on the Eve of Independence

Name some notable economists who estimated India’s per capita income during the colonial period. Ans: The following economists estimated India’s per capita income during the colonial...

10 May
0 0
0

Book a Demo

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Deepika Agrawal describes herself as "Balancing minds, one ledger at a time." "Counting on expertise to balance your knowledge.". She conducts classes in BCom Tuition, Class 11 Tuition and Class 12 Tuition. Deepika Agrawal is located in Begur Road, Bangalore. Deepika Agrawal takes Online Classes- via online medium. She has 10 years of teaching experience . Deepika Agrawal has completed Bachelor of Commerce (B.Com.) from DDU in 2008 and Master of Business Administration (M.B.A.) from ITM Gorakhpur in 2011. She is well versed in English and Hindi.

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