MPT - 322951

Saurabh S Male, 30 Years

Associated for 3 Years 2 Months
Class 11 - 12 Tutor

Activity Score - 170

Qualification :
  • M.Tech (NIT Durgapur - 2012)
  • B.Tech/B.E. (National Institute of Technology - 2009)
  • Total Experience:
    1 Year
  • Hourly Fees:
    INR 1000
Tutoring Experience :

I am here to help the students develop core concepts and thinking in maths.Maths is my strongholds and I will help you develop critical thinking and analysis.I am graduate and post graduate from National Institute of Technology,India. I work at one of the maharatna PSU's.     

Tutoring Option :
Home Tuition Only
Tutoring Approach :

The approach will be tailored according to student needs and level of grasping.

Teaches:
Class 9 - 10 Mathematics Computer Science School level computer NTSE CBSE/ICSE INR 500 / Hour
Class 11 - 12 Mathematics Statistics IT & Computer Subjects Algebra CBSE ICSE Local State Board INR 1000 / Hour
College Level Mathematics Chhattisgarh Swami Vivekanand Technical University Bhilai INR 500 / Hour
Test Preparation MAT INR 100 / Hour
Class 6 - 8 Mathematics Computer Science School level computer NTSE CBSE/ICSE INR 500 / Hour
Competitive Exams SSC Exams Insurance Exams Bank PO PSU Exam INR 1000 / Hour
  • Answer:

    Hello Anu, Probability is a type of ratio where we compare how many times an outcome can occur compared to all possible outcomes. As per question, 1)In deck of cards there are 52 cards. There are 13 cards each of spades,hearts, diamond and clubs.Out of which there is one ace card from each spades,hearts, diamond and clubsStandard card decks usually have Hearts and Diamond in Red and Clubs and Spades in Black. 1)There are 4 ace cards out of 52 cards.This is our favorable event. p(ace)=no.of times a favorable event occurs/ Total no.of possible events. P(ace)=4/52=1/13 2)There are 26 red cards and rest are black. As per the definition P(not a red card)=no.of black cards/total no.of cards =26/52=1/2

  • Question: Horse power is unit of

    Posted in: Physics | Date: 17/11/2017

    Answer:

    Power

  • Answer:

    In geometry two triangles, △ABC and △A′B′C′, are similar if and only if corresponding angles have the same measure. It can be shown that two triangles having congruent angles (equiangular triangles) are similar, that is, the corresponding sides can be proved to be proportional. This is known as the AAA similarity theorem. If The triangles have two congruent angles, which in Euclidean geometry implies that all their angles are congruent. That is: If ∠BAC is equal in measure to ∠B′A′C′, and ∠ABC is equal in measure to ∠A′B′C′, then this implies that ∠ACB is equal in measure to ∠A′C′B′ and the triangles are similar. If the triangles are similar all the corresponding sides have lengths in the same ratio: AB / A′B′ = BC / B′C′ = AC / A′C′ . This is equivalent to saying that one triangle (or its mirror image) is an enlargement of the other. Now coming to the question. △COA~△DOB AAA axiom Hence AC/BD=OC/OD------1 Area(△COA)=1/2*base * height=1/2*OC*AC=60 On solving AC=24 cm From 1st equation, 24/BD=5/8 BD=192/5 cm AreaDOB=1/2*DO*BD=1/2*8*192/5=153.6 sq.cm

  • Answer:

    Let the length of straight wire be L units.

    One half is bent in the shape of square i.e. perimeter of square = L/2 units.

    Let each side of square be A units.

    Then as per question,

    4A=L/2

    =>A=L/8

    Area of square=A²=L²/64 sq.units;

    Now,the other half of wire is bent into sector,

    Length of sector,S =L/2

    Let R be radius of the sector

    Converting 120 degrees into radians=(π/360)*120=π/3 radians

    Therefore ,Length of sector,S=R*π/3 units

    =>R=3S/π,

    =>R=3L/2π

    Area of sector=1/2 R²Φ=1/2* length of arc *radius;

    where Φ is in radians

    =>Area of sector=1/2 R²Φ=1/2*(9L²/4π²)*π/3 sq.unit

     

        Area of sector=3L²/8π sq.unit

    Area of square:Area of sector=L²/64:3L²/8π=11/84 sq.units

     

  • Answer:

    Theorem:Angles in the same arc of a circle are equal.

    In the above diagram,

    ∠PTR=∠STQ;

    ∠PRT=∠SQT(ByTheorem)

    ∠TPR=∠TSQ(ByTheorem)

    In ∆STQ According to Pythagoras theorem, (ST)^2 = (SQ)^2-(TQ)^2 Or ST =√36-25 = √11

    By AAA similarity theorem,

    ∆PTR~∆STQ

    Hence,

    TR/TQ =PT/ST=4/√11

    TR=20/√11=6.03

    Option c

     

  • Answer:

    Let (x1,y1) be the one of the vertices of above Isosceles triangle.

    As per question,

    √{(x1-8)^2 + (y1-3)^2}=√{(x1-14)^2 + (y1-3^2}

    =>√{(x1-8)^2={(x1-14)^2

    =>x1=11;

    Putting value of x1=11 in below eqn

    √{(x1-8)^2 + (y1-3)^2}=5

    on solving y1=-1 or 7

    Hence there can be 2 possible coordinates for vetices of this triangle (11,-1) or(11,7)

     As per question,

    Distance between two possible vertices=√{(11-11)^2 + (-1-7)^2 =8

    Option A is the correct answer

  • Answer:

    Let PR be the Pole.

     

    As per the question,

     

    ∟DQR=∟PQD=30°

     

    We have to find PD:DR

     

    In r PQR

    tan 60°= PR/QR=---------eqn1

     

    In r QDR

    tan 30°=DR/QR ------eqn2

     

    From eqn2

     

    QR= DR

     

    From eqn1

    (PD+DR)/QR=,

    =>(PD+DR/) DR= DR,

    =>PD=2DR, or PD/DR=1/2

  • Answer:

    As per question,

    The given cylinder is hollow ,

    Let r and h be radius of base and height of cylinder respectively.

    Surface Area of hollow cylinder

    =2π*r*h

    =>2π*r*h=2500

    finding r from above equation,

    we get r=250/3π;

    Now ,Volume of hollow cylinder =π*r² *h

    =>π*r² *h=(π*r*h) *r

    =>1250*(250/3π)

    on solving we get volume of hollow cylinder=33144 cm3..which when rounded gives 33150.06 cm3

    Option B is correct

  • Answer:

    In tri ABC,

    height of actual tree=AC+BC=a+b

    AB=c=6cm

    tan60=b/c;

    b=6√3 cm

    cos60=c/a;

    a=2c=12cm;

    now,height of tree=12+6√3 cm=22.39cm

    None of the options are correct

  • Answer:

    Consider Image uploaded for the full explanation

  • Answer:

    In triangleTQR,

    tan 60=QR/QT,

    QR/QT=√3,

    QT=10√3

    tan 45=PQ/QT

    =>PQ=QT

    Height of Tower=PQ+QR=PR

    =>30+10√3

    =>47.32

    Option C is correct

     

     

  • Answer:

    In Triangle PSR,

    sinx=PR/PS=PR/L

    =>PR=LSinx;

    In Triangle QTR,

    Siny=QR/RT=(QP+PR)/L

    LSiny=a+LSinx;

    =>L=a/(Siny-Sinx)

     

  • Answer:

    Rate of change of total revenue==Marginal value=d(total revenue)/dx=d(R(x))/dx=6x+36

    when x=5,marginal value=6x5+36=Rs.66

    you can see that

    when x=5,total revenue=R(x)=3x²+36x+5=Rs.260

     

  • Answer:

    Two vectors are perpendicular then their dot product is 0.

    i.e. if a nd b are vectors and they are perpinducular to each other ,Ω=90 then a.b=a . b cos(Ω)=0

    As per question

    ar = iˆ – jˆ + 7kˆ and b r = 5iˆ – jˆ + λkˆ

    ar +br=6iˆ-2jˆ +(7+λ)kˆ-------vector1

    ar-br=-4iˆ+(7-λ)kˆ -----vector2

    Now As per question

    (ar +br) . (ar-br)=0

    =>  λ²=25

    =>λ=±5

     

  • Answer:

    Data (or graph) on sales of the branches missing. The problem cannot be answered

  • Answer:

    Let x ,y,z be the age of husband,wife,child three years ago.

    As per question,

    (x+y+z)/3=27--1

    x+y+z=81;

    ((y-2) +(z-2))/2=20

    y+z=44--2

    putting eqn 2in eqn 1

    we get x=37;

    Hence present age of husband=x+3=37+3=40

    option(B) is the correct answer

     

  • Answer:

    area of court=3.78*5.25=1109;

    let a be the side of square tiles and let there be n square tiles that are required to pave;

    as per question,

    n x a² =1109;

    2 variables and only one equation.

    option D is correct

     

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