Professional Knowledge Quiz for SBI, IBPS IT Officer - Set 3
In IBPS IT Officer Exam, the Professional Knowledge Section comprises 50 questions and 75 marks. If you want to perform better in upcoming IBPS IT Officer Exam, try to get at least 50 marks in this section.
But How?
Don't worry at all! Just try our various Quizzes on Professional Knowledge. We have already provided lots of Quizzes on Professional Knowledge for IT Officer. You can check them Here.
But How?
Don't worry at all! Just try our various Quizzes on Professional Knowledge. We have already provided lots of Quizzes on Professional Knowledge for IT Officer. You can check them Here.
- A program counter contains a number 825 and address part of the instruction contains the number 24. The effective address in the relative address mode, when an instruction is read from the memory is:
(A) 849
(B) 850
(C) 801
(D) 802
(E) 807
- Software testing techniques are most effective if applied immediately after
(A) Requirement Specification
(B) Design
(C) Coding
(D) Integration
(E) None
- In a unit testing of a module, it is found that for a set of Test data, at the maximum 90% of the code alone were tested with the probability of success 0.9. The reliability of the module is
(A) greater than 0.9
(B) equal to 0.9
(C) at most 0.81
(D) at least 1/0.81
(E) at least 1/0.9
- Which of the following is/are syntactically correct in Java?
(A) int [ ] wer;
(B) int wer [ ];
(C) wer [int]:
(D) A and B
(E) B and C
- Which of the following 4 bit numbers equals in 2’s complement?
(A) 1010
(B) 0101
(C) 1000
(D) All
(E) None of these
- Which of the following is the minimum error code?
(A) Octal Code
(B) Binary
(C) Bray Code
(D) Excess-3-Code
(E) None
- How many characters per sec (7bit+parity) can be transmitted over a 2400 bps line in asynchronous mode?
(A) 300
(B) 240
(C) 320
(D) 430
(E) None
- In a positive edge triggered J.K flipflop, a low J and a low K produces
(A) High State
(B) Low State
(C) Toggle State
(D) No Change
(E) Both B and C
- An algorithm is made up of 2 modules M1 and M2. If order of M1 is f(n) and M2 is g(n) then order of the algorithm is
(A) max(f(n), g(n))
(B) min (f(n), g(n))
(C) (f(n)+g(n))
(D) f(n)×g(n)
(E) None
- The post fix expression of (A+B)*(C*D-E)*f/b is
(A) AB+CD*E-FB/**
(B) AB+CD*E-F**G/
(C) AB+CD*E*F*G/
(D) AB+CDE*-*F*G/
(E) None
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