B.Tech/B.E. | IIT(ISM)-Dhanbad | 2017 |
1Year
I have been teaching mathematics to the students of class XI - XII of all boards and also to the aspirants of Engineering Entrance since 1 years. My teaching style is unique it is beneficial for boards as well as competitive examinations like IIT JEE Mains, BITSAT, IIT JEE Advanced.
Home Tuition Only
I approach tutoring in such a way that I show the students the basics, and then give them a variety of practice exercises and urge them to figure out the answers themselves. If in doubt, I would keep prompting them to find the right answers thereby helping them to grasp the right concepts.
600.00
Class 9 - 10 | Mathematics, Physics, ICSE, CBSE, Local State Board, IGCSE | INR 500.00 /hour |
Class 11 - 12 | Mathematics, Physics, CBSE, ICSE, Local State Board | INR 600.00 /hour |
Class 6 - 8 | Mathematics, Physics, ICSE, CBSE, Local State Board, IGCSE | INR 400.00 /hour |
Dimension of a=LT^-1
Dimension of b=LT^-2
Dimension of c=LT^-3
magnitude of field intensity vector of each mass at the midpoint, will be equal.
now direction will be towards mass because gravitational forces are always attractive in nature.
hence magnitude of field intensity vector of each mass at the midpoint, will be equal and directed opposite to each other.
so resultant field intensity at midpoint will be ZERO.
First of all initial electric potential=qQ/4пεd.
Now distance between them is 2d.
then final electric potential=qQ/8пεd.
now change in electric potential =|(final electric potential=qQ/8пεd)-(initial electric potential=qQ/4пεd)|
=qQ/8пεd
first of all sum of ages of 15 persons=15*29=435
sum of remaining 13 =435-55-55=325
now the average age of =325/13=25
clear the basic concepts of each topics and focus on its application.
first of all number of boys in school=720*7/12=420 and number of girls =720-420=300
hence to make the strength of girls equal to boys, we need to admit 420-300=120girls
loss=20%
12C=15S
S=0.8C
%loss=(S-C)*100/C=20%
let correct number=x;
now 0.82x=0.28x+540
0.82x-0.28x=540
0.54x=540
x=540/0.54
x=1000
actual answer=0.28*/1000=280
let correct number=x;
now 0.82x=0.28x+540
0.82x-0.28x=540
0.54x=540
x=540/0.54
x=1000
actual answer=0.28*1000=280
sum of ages of 5 members 5 year before=27*5=135
hence some of ages now=135+5*5=160
average now =160/5=32
average of 6 members=36
sum of ages of 6 members=36*6=216
now age of 6th person=216-160=56
number of male=600
literate male=600*20/100=120
total literate=250
literate female=250-120=130
total female=400
%literate female=130*100/400=32.5%
let amount received by tuhin,sushant,rita=3x,5x,7x respectively
7x=5x+4000
2x=4000
x=2000
total amount received by tuhin and sushant=8x=8*2000=16000
2.
sum of zeroes=-b/a=-3/2
product of zeroes=c/a=-2/2=1
4km in easten direction
8300*7-(7500+7900+7800+8200+8500+7500)
58100-47400=10700
general equation of a circle:x^2+y^2+2gx+2fy+c=0: points lie on the circle are (0,.0)(a,0) and (0,b).
by putting (0,0),we get c=0;
by putting (a,0),we get a^2+2ga =0:
by putting (0,b),we get b^2+2fb =0:
hence a =-2g,b =-2f
and equation of circle will be x^2+y^2-ax-by =0
Question should be rechecked.
leaf share:wood share= 3:2 = 12:8
forest share:leaf share= 5:4 = 15:12
forest share:leaf share:wood share= 15:12:8
hence leaf's share = 1400 X 12/35 = 480.
question has no proper details..
incomplete question
AB/4
incomplete question
angle OAP will be right angle.
so AP=√(61X61 - 11X11)=60cm
now perimeter=2*(60+11)=142cm
Anita= 2*bidesh = 6 *simran = 6/4 *Kajal
Kajol can complete in 12 days
so Anita takes 12 X 4/6= 8 days
let the distance be x km.
average speed=total distance/total time
avarage speed = 2x/ (x/24 + x/36) = 144/5 = 28.8 km/hr
let original avg=x;
then=(12*x+69)/13=x-2;
12*x+69=13*x-26
x=95
(20-18)*100/20=10%
1-5/20-5/18
17/36th part will be left.
4km/hr=20/18m/s
8=130/v-(20/18)
v=62.5km/hr
Work done by both per day = 1/20+1/30 = 1/12
for 12 days wages = 3840/-
For one day wages = 3840/12 = 320/
Question has no limiting conditions.
Question is incorrect..
total interest=20988.10-18250.50=2737.6
20988.10*x/100=2737.6
x=15%
1/v - 1/u = 1/f
u= -10cm and f= 20cm
v<0=>1/v= 1/f-1/u
1/v= 1/20 -1/10
1/v = -1/20
v = -20cm
-ve sign shows it is real image
Use Pythagoras theorem for both circles and add the distance
(15/2)^2= 5^2 + x^2
x^2= 31.25
x= 5.59cm
(15/2)^2 = 4^2 + y^2
y= 6.34
x+ y = 11.93cm
Cp= 50 per kg
. Profit he wants is 20 means 120÷100×50=60
Now Discount is 10 %so he will get 90%
means 90/100×60=54
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speed=(180+270)/10.8=450/10.8=41.6667m/s=150km/hr
speed of second train=150-60=90km/hr
let distance is L km.Time to reach=T min.
then L/40=(T+10)/60;
and L/50=(T-5)/60
now (L/40)-(L/50)=((T+10)-(T-5))/60;
L/200=15/60;
L=200/4;
L=50KM;
in quadratic equation ax2+bx+c=0;
the product of zeroes/roots=c/a
and it is only equal to 1/7 in option c.
product of roots=-35/1=-35;
root1*root2=-35;
5*root2=-35;
root2=-35/7=-7;