What is the best way to approach DI questions? As they tend to be calculation extensive, what is the best strategy to optimise the time?

First of all if you pretty sure that a set is time-consuming and calculation intensive, it is always better not to start with that set. The reason being that if you solve that set 15-18 minutes then it will press the panic button. So look out for other sets which you can solve. Once you are done with all other solvable sets, then you can come back to this set. Now have a look at questions where options are not ver close. In these questions you can use approximations. For the questions with close option choices there is no choice but to calculate. Thanks.

Find the digit in 469th place from right of 90^468

90^468 can be written as (9X10)468Now the number will have 468 zeroes in the left because of 10^468.So basically the question is to find the unit digit of 9^468 or 3^936.powers of 3 has cyclicity.3,9,7,1,3,9,7,1,3,9,7,1....After every 4 terms, the series repeat.936 is divisible by 4. This means that the last term will correspond to 4th term in the cycle, i.e. 1So the answer is 1.

the rightmost non-zero digit of the number 20^2520?

The zeros in this number will come only from 10 in the 20. Break the number in the following way-(2X10)2520All the zeroes will come from 10^2520So the unit digit if 2^2520 is the rightmost non zero valu.Now powers of 2 has cyclicity.2,4,8,6,2,4,8,6...You will notice that after every 4 terms, the series repeat. 2520 is divisble by 4. This means the last digit should be the forth term i.e 6.So, 6 is the answer

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find minimal vvalue of the expression a^6+b^6 if a^2+b^2=1?( if ur answer is 1/4 tell me how 3a^2b^2 becomes 3/4)

We have to minimize $$a^{6}+b^{6}$$since $$x^{3}+y^{3}=(x+y)(x^{2}+y^{2}-xy)$$So,$$a^{6}+b^{6}=(a^{2}+b^{2})(a^{4}+b^{4}-a^{2}b^{2})$$$$(a^{2}+b^{2})^{2}=1$$$$(a^{4}+b^{4}-a^{2}b^{2})+3a^{2}b^{2}=1$$which is $$1-3a^{2}b^{2}$$to minimize this expression $$a^{2}b^{2}$$ should be maximum. We know that $$a^{2}+b^{2}=1$$Let a^2=m and b^2=nm+n=1GM<=AM$$\sqrt{mn}$$<=$$\frac{m+n}{2}$$This means, $$mn$$<=$$(\frac{1}{2})^{2}$$this means its maximum value is 1/4.thus minimum value of expression becomes- 1-3/4=1/4.

The sum of all natural numbers less than 101 which are co-prime to 2, 3 and 5??

The sum of all co-primes below a given number = E(n)*N/2. Here E(N) is Euler's Totient Function for that number. So total numbers below 100 which are coprime to 2,3 and 5 is given by100*1*2*4/(2*3*5) = 20*8/6 = 160/6 So sum of all these numbers = 100*160/12 = 1333

Suppose overall cut off is some X, if I qualified in all the sections but my overall score is bellow X then, am I eligible for further stages?

No, you need to clear overall cutoffs as well as the sectional cutoffs to be eligible for further process.

Suppose overall cut off is some X, if I qualified in all the sections but my overall score is bellow X then, am I eligible for further stages?

No, you need to clear overall cutoffs as well as the sectional cutoffs to be eligible for further process.

how do solve parajumbal sentences

We have a blog post on this. Please go through it to understand the tips and tricks in solving parajumbles.Here's the link. https://cracku.in/blog/solve-parajumbles-cat/Hope that helps.