0 |
7 Years
I am having 5+ years of teaching experience. I have taught subjects like Engineering Mechanics, Engineering Graphics, Thermodynamics. Also, I am having good experience in teaching programming language "C".
Home Tuition Only
My approach will be depend on the student's needs and learning style. However my teaching is based on encouraging students to think for themselves, using real-world examples wherever possible. I usually focus on exam-style or past paper questions in order to check, practice and perfect the student's understanding and technique.
350.00
Class 9 - 10 | Mathematics, Physics, Chemistry, Algebra, ICSE, CBSE, Local State Board, IGCSE, English Medium | INR 275.00 /hour |
Class 11 - 12 | Mathematics, Physics, Chemistry, Statistics, CBSE, ICSE, Local State Board, English Medium | INR 350.00 /hour |
Engineering Subjects | Mathematics, Mechanical, Production | INR 350.00 /hour |
Class 6 - 8 | Mathematics, Physics, Chemistry, Algebra | INR 250.00 /hour |
(SI)1 is the simple interest for first scheme whereas (SI)2 is the simple interedt for second scheme. (SI)1:(SI)2= (P1*R*T)/100:(P2*R*T)/100=4*8*T:9*10*T =32/90=16/45 Hence Ratio of Simple interests is 16:45
Let amount to be disributed be: Aftabt =2a Bikash =4a and Cathrene=5a Total Amount=2a+4a+5a=11a Amount distributed: Aftab :1b Bikash=3b and catherene=6b hence total amount =10b But 11a= 10b Hence , a=(10/11)b Now, Bikas got 70 less than initia i,.e. 4a-3b=70 4(10/11)b-3b=70 (40b-33b)/11=70 7b=770 hence b=110 hence a=(10/11)*110=100 hence a=100 Hence amount got by each: aftab= 1b=110 Bikash = 3b=3*(110)=330 Catherene=6b=6*(110)=660 hence total amount=110+330+660=1100
Use the Lense Formula: (1/v)-(1/u)=(1/f) where u : object distance v : image distance and f: focal length Do rememeber, object distance is always negative as objects are placed on left side. hence , in this example: u=-10 v=20cm
CP=[(100*SP)/(100+P%)] CP=[(100*8500)/(100+25)] =[(850000)/125]=Rs.6800 Hence Cost Price of Alloy =Rs.6800 Let Amount of Silver in the alloy=2x g Amount of Copper in the alloy=3x g Amount of lead in the alloy=5x g Amount of zinc in the alloy=6x g Total weight=2x+3x+5x+6x=16x g But given that total weight of the alloy=1 kg =1000 g hence 16x=1000g x= (1000/16)=125/2 Hence Amount of Silver in the alloy=2(125/2) =125 g Amount of Copper in the alloy=3(125/2)=375/2 g Amount of lead in the alloy=5(125/2)=625/2 g Amount of zinc in the alloy=6(125/2)=375g Price per kg of Zinc=Rs.400/kg Cost of zinc in Alloy= (375/1000)400 =Rs.150 Price per kg of lead=2*400=800 per kg Cost of lead in Alloy=[(625/2)/1000]*800=Rs.250 Price per kg of silver =400*125=Rs.50000 per kg Cost of silver in Alloy=50000*(125/1000)=Rs.6250 We have Cost price of alloy= cost of Silver+cost of lead+cost of zinc+ cost of copper 6800=6250+250+150+ cost of Copper cost of copper in the alloy=Rs.150 Hence,for (375/2=187.5g) copper costs Rs.150 hence price of copper per kg=(150/187.5)*1000=Rs.800 per kg. Ratio between price of copper to price of lead=800:800=1:1