B.Tech.
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.
Home Tuition Only
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.
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 |
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.
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.
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.
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.
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
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.