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MPT 1386
  • Male, 54 Years
  • Activity Score547

Prabhat R

Stay Focused Stay Sharp
  • I teach at My Home
  • I go to Student's Home
  • Online
  • Qualification:
    M.Sc
  • Experience:
    I impart tuition in Maths, Science and School-level Computer based on MINIMUM STUDIES, MAXIMUM SCORING. All my students are quite satisfied with my way of teaching. It`s absolutely my responsibility to deal with their performance ... More [+]
  • Teaches:
    School level computer, Physics, Mathematics, Computer Science, Chemistry, Bengali, IT & Computer Subjects, Business Mathematics, Polytechnic Entrance, JECA Exam, IIT JEE Mains, CET, BCA Entrance, CS - Foundation, PSAT, MBA Entrance, LSAT, IBSAT
  • Board:
    CBSE/ICSE, All Boards
  • Areas:
  • Pincode:
    700031
Profile Details
Profile Details

Qualification :

M.Sc University of Calcutta 1979

Total Experience :

25 Years

I impart tuition in Maths, Science and School-level Computer based on MINIMUM STUDIES, MAXIMUM SCORING. All my students are quite satisfied with my way of teaching. It`s absolutely my responsibility to deal with their performance and make them excel in academics. I am teaching for last 27+ years in private/group tuition.

Tuitions offered for classes and subjects are as under:

 

VII - X  : Physics, Chemistry, Maths and Computer and IT;

XI - XII : Maths and Computer and IT

Thank you.

Tutoring Option:

I Can Manage Both

Tutoring Approach:

My Tuition is based on 'MINIMUM STUDIES, MAXIMUM SCORING' for which I ensure understanding of the CORE KNOWLEDGE, again for which if I require to teach the subject at a much lower level than his/her present level, I ensure doing that also.

Hourly Fees [INR]:

500.00

Tuition Schedule:

  • Sunday : --
  • Monday : --
  • Tuesday : --
  • Wednesday : --
  • Thursday : --
  • Friday : --
  • Saturday : --
Class 9 - 10 Mathematics, Physics, Chemistry, Computer Science, School level computer, Bengali, CBSE/ICSE, English Medium INR 500.00 /hour
Class 11 - 12 Mathematics, Business Mathematics, Computer Science, IT & Computer Subjects, All Boards, English/Regional Language INR 500.00 /hour
Engineering Entrance & IITJEE Mathematics, IIT JEE Mains, CET, JECA Exam, BCA Entrance, Polytechnic Entrance INR 500.00 /hour
Test Preparation MBA Entrance, IBSAT, LSAT, PSAT INR 600.00 /hour
CA CPT CS - Foundation INR 600.00 /hour
Class 6 - 8 Mathematics, Physics, Chemistry, Computer Science, School level computer, Bengali, CBSE/ICSE INR 500.00 /hour
Profile Details
Services Offered

Prior Online Tutoring Experience?

Yes

Tuition Fees [INR]:

USD10.00

Online Tutoring Experience :

I am giving online tuition only for 6 year and 9 months although my experience as Home tutor and centre tutor is 20+ years.

Online Tools Details :

I do use white boards and other teaching aids and yahoo board along with Skype and Yahoo messenger.

Online Availability :

Monday to Friday: 00:00 to 03:30 and 16:30 to 20:30. Saturdays: 11:00 to 20:30. Sundays: : 00:00 to 07:00 and 08:30 to 20:30. - (All timings are in GMT)

Answer
Answer
  • Answer:

    Question is not clearly framed; however, if we presume Rs. 325 to be the gain amount, then the solution is as under:

    Let the Cost Price(CP) be Rs. 100;

    Therefore, Marked Price(MP) = Rs. 100 + 40% of Rs. 100 = Rs. 100 + Rs. 40 = Rs. 140;

    Therefore, Selling Price(SP) = MP – 10% Discount on MP = Rs. 140 – 10% of Rs. 140 = Rs. 140 – Rs. 14 = Rs. 126; Therefore, his Gain = SP – CP = Rs. 126 – Rs. 100 = Rs. 26; => Gain %-age =(Gain/CP) x 100 = (26/100) x 100 = 26%(of CP);

    Now, Gain Amount, i.e., 26% of CP = Rs. 325 => CP = Rs. 325/26% = Rs. 325/(26/100) = Rs. 325 x (100/26) = Rs. 32500/26 = Rs. 1250. – Ans.

  • Question: Factorise :15pq+15+9q+25p

    Posted in: Mathematics | Date: 27/11/2018

    Answer:

    15pq + 15 + 9q + 25p =(15pq + 25p) + (9q + 15) = 5p(3q + 5) + 3(3q + 5) = (3q + 5)(5p + 3). - Ans.

  • Question: (-15)+(-6)+3.. find the sum to 30terms

    Posted in: Mathematics | Date: 06/12/2018

    Answer:

    It is a case of an A.P. with 1st term a = -15, (2nd term = - 6),

    common difference d = 2nd term - 1st term = (- 6) - (-15) = -6 + 15 = 9

    and number of terms n = 30.

    Therefore, Sum of 1st 30 terms of the series = n/2{2a +(n-1)d} = 30/2{2 x (-15) +(30-1)9} = 15{-30 + 29 x 9} = 15{-30 + 261} = 15 x 231 = 3465(Ans.)

Reviews
Reviews
  • As a teacher, Prabhat sir is excellent. His explanation power is too good. He teaches very well. His advice was very useful for me to build a successful career.
    Reviewed by: Payal
  • Extensive study of every topic with personal attention made every session a wonderful experience.
    Reviewed by: Vaibhav Ghosh

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