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MPT 297086
  • Male, 26 Years
  • Activity Score295

Vinit K

Associated for 2 Years
  • I teach at My Home
  • I go to Student's Home
  • Qualification:
    B.Tech/B.E.
  • Experience:
    I believe that breaking any complex concepts to smaller chunks to the extent that you can start playing around with the info till you enjoy exploring it is what makes the journey easy at the same time thought provoking.
  • Teaches:
    Social Studies, Physics, Mathematics, Chemistry, History, Political Science
  • Board:
    CBSE, CBSE Board
  • Areas:
  • Pincode:
    201307
Profile Details
Profile Details

Qualification :

B.Tech.

Total Experience :

3 Years

I believe that breaking any complex concepts to smaller chunks to the extent that you can start playing around with the info till you enjoy exploring it is what makes the journey easy at the same time thought provoking.

Tutoring Option:

Home Tuition Only

Tutoring Approach:

I make teaching experience interesting by conducting discussions and quizes as to have a better understanding that how much the student has imbibed and what more we need to work on. I work according to a planned schedule so the syllabus is completed on time and we have ample time of revision. This makes student work under no pressure.

Hourly Fees [INR]:

200.00

Class 9 - 10 Physics, History, Chemistry, Social Studies, CBSE INR 200.00 /hour
Class 11 - 12 History, Political Science, CBSE Board INR 200.00 /hour
Class 6 - 8 Mathematics, Physics, Chemistry, Social Studies, CBSE INR 200.00 /hour
Answer
Answer
  • Answer:

    Speed = Distance/Time Let us assume the length of the train be L1 and as it doesn't mention about the speed, so the speed is same for crossing the pole or the 130m long platform. As it crosses the pole it moves the distance of its length. So, Speed(S) =L1/10 Now again when it crosses the platform it will travel (L1+130). So, Speed(S) =(L1+130)/20 Solve these two equations and we'll get the L1's value as 130m.

  • Answer:

    See the problem u may be encountering is its language. The question is saying that the house is sold for Rs 45000. It means is sold @45000 after giving the discount. 

    so, let's assume X be the price at which, if sold at no discount. And if we subtract 10% of X from X then we'll get the amount at which it is sold which is Rs45000.

    X-10%of X= 45000

    X(1-10/100)= 45000

    X=45000*100/90

    X=50000. This is ur answer.

  • Answer:

    Ajay=A

    Bijay=B

    Chaman=C

    A takes 4days for a piece of work i.e. 96 hours. So the amount of work done in 1hour will be 1/96 work.

    96 hour----------1 work

    1hour-------1/96 work

    Now, if A & C work together and complete it in 2hr.

    2hour= 1/96 +1/Y   --------(lets Y be the time taken by C) Equation-1

    Solving this equation we'll get Y=96/191 ----------Equation-2

    Now, 3hour = 1/Y + 1/X ---------(lets X be the time taken by B)-----equation-3

    Putting the value of Y from equation 2 in equation 3, we'll get the value of X as 97/96hour.

    X=97/96 hours

    X=97/96*60 minutes

    X=60.625 minutes.

  • Answer:

    B does the work in 2o days. So, the amount of work he'll do in 1day is:

    20days----------------------1 work

    1day-------------------------1/20 work

    similarly, S does in 1 day-------------------1/30 work

    Ratio of their work= Wb/Ws =(1/20)/(1/30) 

    Wb/Ws=3/2--------------------------------Eq-1

    now if they work together then the time taken by them is: 1/T = 1/20 +1/30 

    T=30*20/(30+20)

    T= 12days 

    For 12 days they get Rs.3840. So,

    for 1 day they'll get = 3840/12= Rs. 320

    now, their work ratio is 3:2 so, they'll get their payment in their work ratio.

    Rb= 320*3/(3+2) = 320*3/5= 192 for 1 day. So, B will get for 12 days= 192*12 = Rs. 2304

    Rs= 320*2/(3+2) = 320*2/5=128 for 1 day. So, S will get for 12days =128*12= 1536

    check: 1536+2304= 3840. 

     

  • Answer:

    Let the distance travelled by Prasun be D with a speed of 50kmph. 

    Speed= distance/time

    50=D/T1 => T1=D/50   equation-1----------(let T1 be the time taken)

    When he returns, he travels at S2 and  So, S2 =D/T2 => T2=D/S2  equation-2--

    Average Speed = total distance travelled/ total time taken

    60kmph=(D+D)/(T1+T2)

    putting from eq-1 and eq-2, and by solving

    60=2D/(D/50+D/S2)

    60= 2*50*S2/(50+S2)

    60*(50+S2)=100S2

    S2=300/4= 75

    So, we'll get S2=75kmph

     

  • Answer:

    pipe B fills in 10hrs so, in 1hr it will fill 1/10 portion of the tank.

    i.e. 10 hours------1 tank

           1 hour -------1/10 tank

    similarly pipe F,

    1 hour--------1/15 tank

    And now M is the outlet pipe which is releasing the water. We can say this pipe is adding negative water.

    So, the time was taken together :

    1/30 = 1/10+ 1/15 -1/M ---------------(suppose M be the time taken by pipe M to empty the full tank)

    solving this,  M= 7.5 hours. So, the pipe M will take 7.5hrs to empty the tank.

     

     

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