MY CART (5)

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price:

There are no items in this cart.
Continue Shopping
                   

sir i got 79% marks in hs nd got 74% of marks in pcm....... will im able to sit for iitjee 2011......................... plz help................................ rply fast............

4 years ago

Share

Answers : (2)

                                        

Dear student,


Yeah you will be eligible for IITJEE 2011....




We are all IITians and here to help you in your IIT JEE preparation.

All the best.


 If you like this answer please approve it....


win exciting gifts by answering the questions on Discussion Forum


Sagar Singh


B.Tech IIT Delhi



4 years ago
                                        

The Eligibility Criteria for IIT-JEE is 60% and that for AIEEE is 50%


You cant get admission into BITS though as the eligibilty crtiteria is 80% for BITSAT,so relax and prepare seriously for IIT-JEE and AIEEE


ALL THE BEST


PLZZZZ APPROVE MY ANSWER IF YOU LIKE IT

4 years ago

Post Your Answer

More Questions On IIT JEE Entrance Exam

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!!
Click Here for details
Next number 2, 5, 11, 20, 32, 47?
 
 
Yeah absolutely keep adding the multiples of 3 in your given series. 3+2=5 6+5=11 9+11=20 12+20=23 15+32=47 18+47=65 So the next number will 65.
 
vikas yadav 2 months ago
 
3 rd on1: Solution: i) t₂ - t₁ = 5 - 2 = 3 t₃ - t₂ = 6 t₄ - t₃ = 9 and so on... So let the nth term - (n-1)th term be = x. ii) Sum of all the above = nth term - t₁ = 3 + 6 + 9 + ----- [This...
 
vikas yadav 2 months ago
 
second method: Series 2, 5, 11, 20, 32, 47 Find = next number In this series each digit has the difference of 3,6,9,12,15 etc.. so next series will have the difference of 18. It means next...
 
vikas yadav 2 months ago
Hello I am in grade 11 th . I’m sorry if you find my question a little erratic. I am planning for engineering. What do you deem to be better, doing engineering from IIT or some foreign...
 
 
As you know IIT's are premier institute for engineering, so there is no doubt about that. But both options have their own pros and cons. At IIT's, the competition you will find is...
  img
Harsh Patodia one month ago
 
  img
Harsh Patodia one month ago
Hi, I’m aiming b/w 3500 to 4000 rank in jee advanced. Can someone tell me the required books and material needed for this rank. Also, is HC Verma enough for this rank?
 
 
yap . But H.C.Verma is nice book . See if you can get Haliday,Resnick,Walker. For Maths, books of Asit Das Gupta and R.D.Sharma are best .
 
Piku 9 months ago
 
Firstly ur basics need to be strong. You can practice from H C Verma. If your concepts and basics are clear then u can refer I.E.Irodov.
 
MAlav Shah 9 months ago
 
its better to carry even (pc.agarwal) book with you, for other subjects just give more preference to arihant series or cengage series
 
dheeraj 9 months ago
How should we prepare for bitsat english?
 
 
Synonyms and Antonyms Questions from this topic are commonly asked in all English language exams. For theBITSATEnglish section there are generally 2 questions from this topic. To understand...
  img
Saurabh Kumar one month ago
 
Synonyms and Antonyms Questions from this topic are commonly asked in all English language exams. For theBITSATEnglish section there are generally 2 questions from this topic. To understand...
  img
Saurabh Kumar one month ago
 
there is a book A COMPLETE SUCCESS PACKAGE FOR ARIHANT it includes all the tpoics that are asked in bitsat i highly recommend to buy this book and satrt doing this book bcoz it will tell u...
 
ng29 one month ago
prove that (1²×2²÷1!)+(2²×3²÷3!)+--- = 27e
 
 
I think that is wrong... And plz continue the equation for a little more next time Thanks Mohammed Askiitans Student
 
MOHAMMED ANAS 5 months ago
 
n^4/n!+n^2/n!+2(n^3/n!) =15e+2e+2(5e)=27e
 
vamshiakena 5 months ago
 
n^4/n!+n^2/n!+2(n^3/n!) =15e+2e+2(5e)=27e
 
vamshiakena 5 months ago
DEAR SIR , name the following : – 1 . the physical quantity corresponds to the rate of change of momentum .
 
 
Dear student, The Physical Quantity that corresponds to the rate of change of momentum is Force. The change of momentum is defined as Impulse and the rate of change of Impulse with time is...
  img
Sumit Majumdar 8 months ago
 
It is force. NOW , HOW IS FORCE= RATE OF CHANGE OF MOMENTUM.(answer is below) = newtons second law – force = m( final velocity – initial velocity / time) it can be written as force = m*final...
 
Saloni gordhan Rakholiya 8 months ago
View all Questions »