IBPS IT Officer MCQ – 6 (Data Structure)

The IBPS Specialist Officer (SO) Exam 2016 for recruitment of specialist officer vacancies in public sector banks has been announced. So we D2G Team is here with IBPS SO Material.



41. Which of the following is not a limitation of binary search algorithm ?
(A) binary search algorithm is not efficient when the data elements are more than 1000.
(B) must use a sorted array
(C) requirement of sorted array is expensive when a lot of insertion and deletions are needed
(D) there must be a mechanism to access middle element directly

42. Two dimensional arrays are also called
(A) tables arrays
(B) matrix arrays
(C) both of the above
(D) none of the above

43. The term “push” and “pop” is related to the
(A) Array
(B) Lists
(C) stacks
(D) all of above

44. A data structure where elements can be added or removed at either end but not in the middle is referred as
(A) Linked lists
(B) Stacks
(C) Queues
(D) Deque

45. The following sorting algorithm is of divideand-conquer type
(A) Bubble sort
(B) Insertion sort
(C) Quick sort
(D) None of the above

46. An algorithm that calls itself directly or indirectly is known as
(A) Recursion
(B) Polish notation
(C) Traversal algorithm
(D) None of the above

47. The elements of an array are stored successively in memory cells because
(A) by this way computer can keep track only the address of the first element and the addresses of other elements can be calculated
(B) the architecture of computer memory does not allow arrays to store other than serially
(C) A and B both false
(D) A and B both true

48. The memory address of the first element of an array is called
(A) base address
(B) floor address
(C) foundation address
(D) first address

49. The memory address of fifth element of an array can be calculated by the formula
(A) LOC(Array[5])=Base(Array[5])+(5-lower boun(D), where w is the number of words per memory cell for the array
(B) LOC(Array[5])=Base(Array[4])+(5-Upper boun(D), where w is the number of words per memory cell for the array
(C) LOC(Array[5]=Base(Array)+w(5-lower boun(D), where w is the number of words per memory cell for the array
(D) None of the above

50. The following data structure can’t store the non-homogeneous data elements
(A) Arrays
(B) Records
(C) Pointers
(D) None of the above

Answers:  41. (A) 42. (C) 43. (C) 44. (D) 45. (B) 46. (A) 47. (A) 48. (A) 49. (C) 50. (A)

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