attempting such questions with an on-screen calculator could be a tough task, (in this case) you check for d=5,6,7,8,.... and hope you get answer before 19 otherwise your 1.5 minutes are wasted. Speaking for myself in examination I felt, relying on repetitive calculations increases the panic level. A better approach would be

We know

*ϕ*(n)=192 and e=35, so

(35*d)%192=1 OR 35*d=192k+1 for some k

now look at RHS, last digit of 192k (be it any k) will be {2,4,6,8,0}, add 1 to them and you'll see last digit of RHS will always be {3,5,7,9,1}

as you can see on LHS the last digit is 5, there is no way you can get {3,7,9,1} on the last digit of 35*d, be it any d you pick. Now you can conclude that last digit of 35*d is definitely 5. Therefore you need to check for odd values only. Check for d=5,7,9,11,... time halved

@happy singh: in RSA crypto, they settle for smallest d found. Chalo by chance you got 203, what you should have done is check for any smaller than 203 (which is relatively easy)

subtract a number (say A) from 35*203 such that it is a multiple of both 35 and 192, not necessarily LCM.

As 35*203 > product of (35,192), go for 35*192

35*203 - 35*192 = 35*11

you could have got 11