# Recent questions tagged test-series

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How to solve this?
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AFAIK "In direct we do specify the address of the operand right? Its just that we specify it in the instruction directly, ie hardcode the address in the instruction. In implied there is no address given since address is deduced from the operation itself." Is this right?
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Consider a system with 2 level paging applicable the page table is divided into 8k pages of size 16kB.The memory is byte addresable if the physical address space is 128MB which is divided into 4KB frames.The page table entry of outer page table is 32 bitand page table entry size ... .What will be the page table size of inner and outer page table. A-32KB,128KB B-32KB,32KB C-128KB,128KB D-128KB,32KB
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int x=0; int A(n) { statement //takes O(1) time if(n==1) return 1; else { $X+=8.A\left ( \frac{n}{2}\right )+n^3$; } return X; } What is the time complexity f the above code?
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what will be differnece if we use c)option and d) option explain ??
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How many rows will be there in the output of the following query?
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A computer has 20 tape drivers, with n processes competing for them. Each process may need five drivers. What is the maximum value of n for the system to be deadlock free?
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$\frac{1}{10^{9}}[30*10^{-3}]+(1-\frac{1}{10^{9}})[300*10^{-9}]$ Answer should be in nano seconds
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Why is the answer D? How to solve it in simple way other than learning Rice Theorem? Does anyone know Rice thm in short?
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Four different pens (1, 2, 3, 4) are to be distributed at random in four pen stands marked as 1, 2, 3, 4. What is the probability that none of the pen occupies the place corresponding to its number ? a. 17/18 b. 3/8 c. 1/2 d. 5/8 How will we use the formula for derangements here? Using the direct formula of !n is giving wrong answer.
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the need of heap allocation at runtime is a) to support dynamic data structure b)to support dynamic scoping c) to support recursion d) all of the above explain?
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How will I find SLR(1) for such a big CFG? I could find error in LL(1) and conflict in LR(0)? But finding SLR(1) is time-consuming if using item generation approach?
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After taking |A|=0, I am getting x3+1=0 and not x3-1=0. Is it x3+1=0 or x3-1=0?
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Find the value of F. Answer seems to be $Q \bigodot R$ Whereas I'm getting PQR + $\bar{P}\bar{Q}\bar{R}$
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The number of min heap trees are possible with 13 elements such that every leaf node must be greater than all non-leaf nodes of the tree are ___ A. 627 B. 747 C. 657 D. 757
f(n) = $\Theta (n^{2})$ g(n) = $\Omega (n)$ h(n)=O(log n) then [ f(n) . g(n) ] + [h(n) . f(n) ] is $\Omega (n)$ $\Theta (n^{2})$ O(log n) None