LearnPick Navigation
Close
N
  • Male, 38 Years
  • Activity Score245

N.s.balaji

Associated for 1 Year 6 Months
I can handle cbse maths for 6 to 11
  • I teach at My Home
  • Qualification:
    B.Tech/B.E.
  • Experience:
    I have taught unofficially for a span of 1 year for class 8 to 10 mathematics while working. I have handled quantitative aptitude unofficially for a span of 1 year exhaustively.
  • Teaches:
    Chemistry, Mathematics, Physics, English, Algebra, Bank Clerical, IBPS, SBI Exam, TNPSC
  • Board:
    All Boards, CBSE
  • Areas:
  • Pincode:
    600044
Profile Details
Profile Details

Qualification :

B.Tech/B.E. Igniting the minds 2002

Total Experience :

3 Years

I have taught unofficially for a span of 1 year for class 8 to 10 mathematics while working. I have handled quantitative aptitude unofficially for a span of 1 year exhaustively.

Tutoring Option:

Home Tuition Only

Hourly Fees [INR]:

400.00

Class 9 - 10 Mathematics, Physics, Chemistry, Algebra, All Boards, All Medium INR 400.00 /hour
Class 11 - 12 Mathematics, Physics, Chemistry, All Boards, All Medium INR 350.00 /hour
Class 6 - 8 Mathematics, Physics, English, Chemistry, CBSE INR 400.00 /hour
Competitive Exams Bank Clerical, IBPS, SBI Exam, TNPSC INR 800.00 /hour
Educational Resources
Educational Resources

Notes written by me [3]

Speed Analogy In Escalators
  • File(s) contain : 1
Not Downloaded Yet

It Is A Detailed Analogy Of Speed Analogy In Escalators.

Download
Speed Analogy Of Vehicles Moving Towards Each...
  • File(s) contain : 1
Not Downloaded Yet

It is a detailed analysis of speed and time of vehicles moving towards each other.

Download
Relative Speed
  • File(s) contain : 1
1 time downloaded

Relative Speed Concept.

Download
Answer
Answer
  • Answer:

    Number of students good in mathematics is 72% of 25. That's (72/100)*25=18. Hence number of students not good in mathematics is 25-18=7

  • Answer:

    Let variables P,M,C denote the marks obtained in subjects Physics, Mathematics,Chemistry respectively. Then, P+M+C-120=C P+M=120 Average =120/2 =60

  • Answer:

    Number of students learning mathematics and phscics = 30............ (1) Number of students learning mathematics, physics,chemistry =18.......(2) Hence number of students learning mathematics and physics only = (1)-(2)=12....... (3) Similarly, number of students learning chemistry = 28-18=10.......... (4) Hence number of students learning mathematics alone = 100-(10+12+18)=60

  • Answer:

    students studying maths alone= students studying maths -(students studying maths and physics + students studying maths and chemistry -students studying all 3 subjects) M= 120-(40+50-x)................................... (1) Similarly, P=90-(30+40-x).........................................(2) and C=70-(50+30-x)......................................... (3) In the above equations, M,P, C denote students studying maths, physics, chemistry alone respectively and x denotes students studying all subjects. Hence 200= M+P+C+x+20+(40-x)+(50-x)+(30-x) Solving we get x=20

  • Answer:

    teachers teaching only mathematics = teachers teaching maths - teachers teaching physics and mathematics =12-4=8 20-12=8=teachers teaching only physics and teachers teaching physics =12

  • Answer:

    Their next immediate ringing at the same moment will occur in the lcm of their ringing frequency. LCM =120 seconds. Hence theiron next immediate ringing together will occur at 11.27 am

  • Answer:

    Let the unit digit be y and let the tenth's digit be x. As unit digit is 5 more than the tenth's digit, x+5=y.....................(1) Exact value of the number =10x+y............ (2) Upon reversing the digits, Value becomes 10y+x.................. (3) After reversing, the new number is 4 less than twice the original number. Hence, 2 (10x+y)=10y+x+4 2 (11x+5)=11x+54 (substituting the value of your from equation (1)) Solving, we get x=4,y=9 Hence the original number is 49

  • Answer:

    Let the first number be x. As the second number is 3 more than the first, it is (x+3). As sum of their squares is 369, x^2 +(x+3)^2 =369 2x^2 +6x -360=0 Solving by prime factorization, x=12 and x+3=15. Hence sum of the 2 numbers =27

  • Answer:

    Let the number of horses be x. Let the number of ducks be y. As the sum total of their legs =120, 4x+2y = 120..................................... (1) As their total heads is 50, x+y=50................................................ (2) Multiplying equation (2) throughout and by elimination by Equating the coefficients, Equation (1)-Equation (2) gives x=10 Hence number of horses =10

  • Answer:

    Let the 2 numbers be x and y. As product of the numbers is 93, x*y=93........................ (1) Also, their difference is 28 x-y=28........................... (2). Substituting the value of y from equation (1), x-(93/x)=28 x^2-93=28x x^2-93-28x=0 Solving by prime factorization, we get, x=31. So, y=3. Difference of their reciprocally is, (1÷31)-(1÷3)=(-28/93)

  • Answer:

    Let the quantity of first fuel be x. It is akin to the concept of average speed. Here it is average cost. Average cost =22.5= (21x+(24*30))/(x+30) Solving, we get x =30 litres.

  • Answer:

    As there is no net loss or gain in the transaction, average cost price per kg equals average selling price per kg. Let x be the selling price per kg of 2nd variety of sugar. Then, average C.P=AVERAGE S.P=28=((32*60)+(x*60))/(60+48) 3024=1920+48x Solving, we get x=23. Hence the price of the new type of sugar=Rs23/kg

  • Answer:

    As there is 25% gain in the transaction, C.P of the product is given as, (450-x)/x=25/100 where x denotes C.P (Cost Price of the product ). Hence average C.P =360 Let z denote the quantity of first type of fluor and y denote the cost/kg of another type of fluor. Average C.P=360= (620z+130y)/(z+130) As the above equation has 2 unknowns,it has infinite solutions. one of the solution is got by assuming cost per kg of second fluor. Let it be 200. Now 360 (z+130)=620z+(130×200) here we get z=80 Hence by assuming second fluor cost/kg to be 200, we get the quantity of first fluor=80 kg

  • Answer:

    In both the containers, initially 7 litres of milk and 3 litres of water are added.Further, 4 litres of mixture in container 2 is added to container 1 and it gets filled. Hence capacity of container 1 is 14 litres. Hence capacity of container 2 is 140 litres (10 times first). Let x be the quantity of milk added to container 2.subsequently, 134-x becomes the quantity of water added. From the existing 6 litres in container 2, milk=(9/5)and water=(21/5) so, ((9/5)+x)/((21/5)+(134-x))=2/3 solving, we get x=54.2 litres water added =134-54.2=79.8 litres Hence ratio of milk to water added =54.2:79.8 =0.68:1

  • Answer:

    In the total 60 litres, acid and water are in the ratio 2:1.Hence 2x+x=60 hence x=20 so, acid and water are 40 and 20 litres respectively. 40/(20+y)=1/2 where y denotes quantity of water to be added to make the ratio of acid: water 1:2 hence y=60 litres.

  • Answer:

    By selling the mixture at Rs350 per quintal, the merchant gets a profit of 25% Hence, (350-x)/x=25/100 where x is the average cost price per quintal of the mixture . hence average C.P =Rs 280 let the quantity of first quality tea mixed be y quintal. average C.P =280=(290y+(250×150))/(y+150) Solving, we get y=450 quintal Hence 450 quintal of first quality has to be added.

  • Answer:

    80% of 50 litres amountso to 40 litres and hence in a 50 litre quantity, wine=40 litres and water=10 litres. Let the amount of water added to the mixture to make water 50% be x. so, (10+x)÷(50+x)=50÷100 solving, we get x=30 litres. Hence 30 litres of water need to be added.

  • Answer:

    By selling the rice mixture at Rs16/kg,the profit % attained is 15. Hence average C.P /kg of rice mixture is given by, (16-x)/x=15/100 where x denotes average C.P /kg of rice mixture. Solving, C.P=Rs 13.91 Let the quantity of rice costing Rs 12.85 to be mixed with another rice costing Rs14.8 be y. so, 13.91=(12.85y+(14.8*25))/(y+25) Solving, we get y=21 kg Hence 21 kg of rice worth Rs12.85/kg has to be mixed.

  • Answer:

    Speed of the bicycle is 125/5=25m/s speed of the man=(4/5)×Speed of bicycle =20m/s For 20 metres, the man takes 1 second. Therefore for 40 metres, the man will take 2 seconds.

  • Answer:

    Upstream speed= Speed in still water-Speed of current = 5-1= 4 km/hr Downstream speed =Speed in still water+speed of current=5+1=6 km/hr. Let the distance covered be 'd' . As both in upstream and downstream, distance covered is sameternal and the total time taken is 1 hour, (d/6)+(d/4)=1 Solving, we get d= 2.4 km Hence distance =2.4 km

  • Answer:

    Let the speed of boat in still water be x km/hr As distance upstream and downstream are both same, (x+3)*1=(x-3)*1.5 Solving, we get, x=15 Speed of boat in still water = 15 km/hr

  • Answer:

    Speed in the upstream = (d/3)=U............(1) Speed in the downstream=(d/2)=D.........(2) where 'd' denotes the distance covered,'U' denotes upstream speed and 'D' denotes downstream speed. Speed of boat in still water=( D+U)/2=5d/12 Speed of stream = (D-U)/2=d/12 Hence ratio of speed of boat in still water to speed of stream =5:1

  • Answer:

    As shika runs slower than vikash, let us say shika has to start x metres ahead of vikash so as to finish the race in dead heat. Distance covered by shika is such that she completes the race in same 4 minutes. For 4 minutes and 10 seconds (25/6) minutes, shika travels 25m For 4 minutes, she would have travelled (4×2)/(6/25)km=1.92km x=(2-1.92)=0.08km=80m Hence vikash must give shika a start of 80 metres.

  • Answer:

    As both the boats move towards each other, One is moving with the tide and the other,against it. As speed of boat in still water is 20k/hr, t*(20+x+20-x)=50 Where x denotes speed of the stream and t,the time taken by both boats from starting pount to meeting point. t=5/4 hours=75 minutes. Hence they will meet after 75 minutes.

  • Answer:

    Let the time taken in the first half be denoted by t1. In order to compensate the idle time of 2 hours, the train's speed is increased by 40 km/hr .Let the time taken in the second half be t2. t2=t1-2....................... (1) If S1 denotes the speed in first half, (S1*t1)+((S1+40)*t2)=1000 (S1*t1)=500................... (2) ((S1+40)*t2)=500............ (3) Equating (2) and (3), we get, S1*t1=(S1+40)*t2 S1*t1=(S1+40)*(t1-2) (from equation (1)).... (4) Replacing S1 with 500/t1 obtained from equation (2) in equation (4) and solving the quadratic equation, we get, t1=6.385 hours t1-2 = 4.385 hours S1/(S1+40)=(t1-2)/t1=4.385/6.385 =0.6867:1 Hence ratio of speeds is 0.6867:1

  • Answer:

    Length of the train = relative speed of train and person*time taken to pass the person completely. Hence, d= (S-2)*(9/3600)...... (1) d= (S-4)*(10/3600)....... (2) where d and S denotes the length and speed of the train respectively. EquationS (1) and (2) are applicable for persons travelling at 2km/hr and 4km/hr respectively in the direction of train. Equating (1) and (2) we get S=22km/hr Hence length of train (22-2)*(9/3600)km =50metres

  • Answer:

    The given data is sufficient. As Lucy'so mother is 43 years 5 yearso ago,her mother 'so current age is 48years. As Lucy is half her mother's age , she is (1/2)*48=24 years. As Kate is (1/3) of her mother's age , she is (1/3)*48=16. Hence difference in ages of Kate and Lucy is (24-16)=8 years and hence Kate is 8 years younger to lucy

  • Answer:

    1. Let the initial speed and initial time taken by ram be denoted by 'S' m/s and 't' seconds respectively. If he travels at 25% greater speed, he reaches 20 seconds earlier. Hence (1.25S)*(t-20)=(t*S) Cancelling S and solving for 't', we get t=100 seconds. 2.As per 2nd condition, (t/2) +50=the Solving, we get, t=100 seconds. 3.As per condition 3, (S/4)*(t+300)=(t*S) Solving, we get, t= 100 seconds. Hence we observe that in all cases , time taken by Ram is 100 seconds.

  • Answer:

    Let the initial number of apples possessed by John and James be x and y respectively . By giving 7 apples to James, John will have only half the apples as that of James. So, 2*(x-7)=y+7........................... (1) If James gives one apple to John, both will have equal number of apples. So, x+1=y-1............................... (2) Solving (1) and (2), we get x=23,y=25. Therefore John had initially 23 apples and James had initially 25 apples.

  • Answer:

    As heath is neither taller nor shorter than Ricky, both are of equal height. Hence heath is 9 inches taller than stacy (as given that Ricky is 9 inches taller than stacy).

Similar Tutors

MPT 204614

Md. S

Male, 25 Years
MPT 229842

Mohamed S

Male, 26 Years
MPT 203023

Muthu P

Male, 34 Years
MPT 170394

Balu M

Male, 43 Years
MPT 279091

Harshitha R

Female, 22 Years
  • Qualification: B.Tech/B.E.
  • Teaches:Science, Mathematics, English, All Subjects, Physics, NSTSE, NTSE, Che...
  • Areas: Tambaram, Velachery
MPT 274172

Siva

Male, 28 Years
MPT 315959

Daison.s

Male, 23 Years
MPT 305284

Amit K

Male, 21 Years
MPT 299781

Shubham

Male, 19 Years
D

Divya

Female, 26 Years
  • Qualification: B.Tech/B.E.
  • Teaches:Zoology, Computer Science, Biology, English, All Subjects, Business En...
  • Areas: Chromepet

Can't Find The Right Tutor Yet?

Post your requirement in LearnPick

Query submitted.

Thank you!

Drop Us a Query:

Drop Us a Query