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1
The baud rate is the rate at which information is transferred in a communication channel. Serial ports use two-level (binary) signaling, so the data rate in bits per second is equal to the symbol rate in bauds. Ref: https://en.wikipedia.org/wiki/Serial_port#Speed. ... bit $=12\text{ bits}$. So, number of characters transmitted per second $=\dfrac{9600}{12}=800$. Correct Answer: $B$
posted Mar 23, 2020 in From GO Admins Arjun 2,815 views
3
Option is D. $foo$ is printing the lowest digit. But the $printf$ inside it is after the recursive call. This forces the output to be in reverse order $2, 0, 4, 8$ The final value $sum$ printed will be $0$ as $C$ uses pass by value and hence the modified value inside $foo$ won't be visible inside $main$.
posted Apr 8, 2019 in Preparation Advice Akash Mishra 1,012 views
8
There is no limit on a large value being considered. f(n)= n^0.0000001 = n^(10^-7) g(n)=lg n(base 2) n= 10^8 f(n)= 1.000001842 , g(n)=26.565 Now, I take n = 10^(10^8) f(n) = n^(10^-7) = (10^(10^8))^(10^-7) = 10^(10^(8-7)) = 10^10 g(n) = 10^8 log10 2 So, f(n) becomes larger. Thus for any x, we can have an n, where nx becomes larger than log n and stays larger from there on wards for any higher n.
posted Jun 1, 2018 in Preparation Advice Shikha Mallick 1,542 views
9
The function terminates for all powers of $2$ (which is infinite), hence (i) is false and (ii) is TRUE. Let $n = 5. $ Now, recursive calls will go like $5 - 14 - 7 - 20 - 10 - 5 -$ And this goes into infinite recursion. And if we multiply $5$ with ... possible, there are infinite recursions possible (even considering this case only). So, (iv) is TRUE and (iii) is false. So, correct answer is (D).
posted May 19, 2018 in Preparation Advice Shikha Mallick 2,386 views
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