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MPT 2096
  • Male, 38 Years
  • Activity Score415

Vishal G

Profile Details
Profile Details

Qualification :


Total Experience :

7 Years

I am having 5+ years of teaching experience. I have taught subjects like Engineering Mechanics, Engineering Graphics, Thermodynamics. Also, I am having good experience in teaching programming language "C".

Tutoring Option:

Home Tuition Only

Tutoring Approach:

My approach will be depend on the student's needs and learning style. However my teaching is based on encouraging students to think for themselves, using real-world examples wherever possible. I usually focus on exam-style or past paper questions in order to check, practice and perfect the student's understanding and technique.

Hourly Fees [INR]:


Tuition Schedule:

  • Sunday : --
  • Monday : --
  • Tuesday : --
  • Wednesday : --
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  • Saturday : --
Class 9 - 10 Mathematics, Physics, Chemistry, Algebra, ICSE, CBSE, Local State Board, IGCSE, English Medium INR 275.00 /hour
Class 11 - 12 Mathematics, Physics, Chemistry, Statistics, CBSE, ICSE, Local State Board, English Medium INR 350.00 /hour
Engineering Subjects Mathematics, Mechanical, Production INR 350.00 /hour
Class 6 - 8 Mathematics, Physics, Chemistry, Algebra INR 250.00 /hour
  • Answer:

    (SI)1 is the simple interest for first scheme whereas (SI)2 is the simple interedt for second scheme. (SI)1:(SI)2= (P1*R*T)/100:(P2*R*T)/100=4*8*T:9*10*T =32/90=16/45 Hence Ratio of Simple interests is 16:45

  • Answer:

    Let amount to be disributed be: Aftabt =2a Bikash =4a and Cathrene=5a Total Amount=2a+4a+5a=11a Amount distributed: Aftab :1b Bikash=3b and catherene=6b hence total amount =10b But 11a= 10b Hence , a=(10/11)b Now, Bikas got 70 less than initia i,.e. 4a-3b=70 4(10/11)b-3b=70 (40b-33b)/11=70 7b=770 hence b=110 hence a=(10/11)*110=100 hence a=100 Hence amount got by each: aftab= 1b=110 Bikash = 3b=3*(110)=330 Catherene=6b=6*(110)=660 hence total amount=110+330+660=1100

  • Answer:

    Use the Lense Formula: (1/v)-(1/u)=(1/f) where u : object distance v : image distance and f: focal length Do rememeber, object distance is always negative as objects are placed on left side. hence , in this example: u=-10 v=20cm

  • Answer:

    CP=[(100*SP)/(100+P%)] CP=[(100*8500)/(100+25)] =[(850000)/125]=Rs.6800 Hence Cost Price of Alloy =Rs.6800 Let Amount of Silver in the alloy=2x g Amount of Copper in the alloy=3x g Amount of lead in the alloy=5x g Amount of zinc in the alloy=6x g Total weight=2x+3x+5x+6x=16x g But given that total weight of the alloy=1 kg =1000 g hence 16x=1000g x= (1000/16)=125/2 Hence Amount of Silver in the alloy=2(125/2) =125 g Amount of Copper in the alloy=3(125/2)=375/2 g Amount of lead in the alloy=5(125/2)=625/2 g Amount of zinc in the alloy=6(125/2)=375g Price per kg of Zinc=Rs.400/kg Cost of zinc in Alloy= (375/1000)400 =Rs.150 Price per kg of lead=2*400=800 per kg Cost of lead in Alloy=[(625/2)/1000]*800=Rs.250 Price per kg of silver =400*125=Rs.50000 per kg Cost of silver in Alloy=50000*(125/1000)=Rs.6250 We have Cost price of alloy= cost of Silver+cost of lead+cost of zinc+ cost of copper 6800=6250+250+150+ cost of Copper cost of copper in the alloy=Rs.150 Hence,for (375/2=187.5g) copper costs Rs.150 hence price of copper per kg=(150/187.5)*1000=Rs.800 per kg. Ratio between price of copper to price of lead=800:800=1:1

  • I took Mathematics and Science classes for more than a year. He is a very good teacher. He explains very well.
    Reviewed by: Omkar Rane
  • As a teacher, Vishal sir is excellent. His explanation power is too good. He teaches very well.
    Reviewed by: Nupur
  • Vishal Sir is one of the best knowledgeable people I have ever met. He is a very good and caring teacher and I learned a lot from him. Apart from his excellent teaching, his tips and advice were really helpful for me to build a good career.
    Reviewed by: Akash
  • He is a very good tutor. My son is taking Math and Science tuition from him. He teaches very well.
    Reviewed by: Siddharth Patil
  • Vishal sir is a very good tutor. He cleared all my doubts and explained every chapter in details before solving the questions. I am taking Mathematics tuition from him and my results have improved.
    Reviewed by: Raul Philip
  • Took Engineering Math tuition for 2 years, elucidates the concepts from the basic level, repeats the explanations till the concepts are clarified.
    Reviewed by: Akshay Bhargava
  • Mr. Vishal is a very good tutor. He taught Mathematics to my child, and he showed improvement in his result.
    Reviewed by: Mrs. Bhave
  • He is a very good teacher. He taught my daughter for 1.5 years and after taking Math tuition from him her results had improved.
    Reviewed by: Mrs. Maheshwari
  • Vishal sir is an excellent Math tutor. He has a very strong sense of responsibility. He also conducts mock tests which helps him to assess his students.
    Reviewed by: Mrs. Dabir
  • My daughter took tuition from Sir for Math and Science in standard 10. He is an excellent teacher and helped her score good marks in her exam
    Reviewed by: Mrs. Agnihotri
  • Mr. Vishal Gavhane is a very good. My son took tuition from him for Math and Science and showed considerable improvement. he is also very sincere about his students.
    Reviewed by: Chandresh Babu
  • Good experience, took tuition for Physics, Chemistry and Math in the 9th and 10th standard, employs loads of examples for explaining the topics.
    Reviewed by: Nivin Vijayan
  • My son took tuition from Sir for 7th and 8th standard in Math and Science. Sir teaches quite well and makes sure to clarify all the fundamental concepts of the topics. My son's marks improved considerably.
    Reviewed by: Anant Jain
  • I took Physics, Chemistry and Math classes from sir for my 11th and 12th standard. He is a great teacher, he explains everything well and employs a very detailed method to help students understand every part of a concept. He also uses examples to explain the concepts.
    Reviewed by: Ritesh Panhalkar

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