  # Computer Organization and Architecture Basics MCQs with Answers Part 3

## Computer Organization and Architecture Basics MCQs with Answers Part 3. This mcqs most helpful in SPPU exam 2020 and in other Inportant Exams.

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1. Throughput is also called as………
a. execution time
b. response time
c. bandwidth
d. elapsed time

Ans - C

2. Decreasing response time almost always improves……..
a. execution time
b. response time
c. bandwidth
d. elapsed time

Ans - C

3. ……response time almost always improves throughput
a. Decreasing
b. Increasing
c. Constant
d. both a & b

Ans - A

4. …….response time for some tasks maximizes computer performance
a. Maximizing
b. Minimizing
c. Constant
d. both a & b

Ans - B

5. If computer A runs a program in 10 seconds & computer B runs the same programs in 15 seconds, how much faster is A than B?
a. 1
b. 1.5
c. 10
d. 15

Ans - B

6. If computer A runs a program in 1000 seconds & computer B runs the same programs in 1500 seconds, how much faster is A than B?
a. 50
b. 100
c. 150
d. 200

Ans - C

7. The actual time the CPU spends computing for a specific task is called as…..
a. CPU time
b. user CPU time
c. system CPU time
d. none of the above

Ans - A

8. CPU time spent in a program itself is called as……….
a. CPU time
b. user CPU time
c. system CPU time
d. none of the above

Ans - B

9. If program runs in 10 seconds on a computer, which has 2 GHZ clock , then number of clock cycles required for program A will be ......
a. 1x109
b. 2x109
c. 10x109
d. 20x109

Ans - D

10. Computer will run program in………seconds if it has a clock rate of 2 GHZ and 20x109 no. of clock cycles
a. 6
b. 10
c. 12
d. 15

Ans - B

11. Computer will run program in………seconds if it has a clock rate of 4 GHZ and 16x109 no. of clock cycles
a. 4
b. 6
c. 12
d. 15

Ans - A

12. If program runs in 6 seconds on   a computer , which has 4 GHZ clock , then number of clock cycles required for program A will be ......
a. 1x109
b. 2x109
c. 10x109
d. 24x109

Ans - D

13. If computer A has  24x109 CPU clock cycles and CPI  of 2 , then ........ instructions will  be present in a program.
a. 1x109
b. 12x109
c. 10x109
d. 20x109

ANs - B

14. Computer with program holding 10x109 instructions & CPU clock cycles of  20x109 will have CPI of………….
a. 1
b. 2
c. 10
d. none of the above

Ans - B

15. Computer with program holding 8x109 instructions & CPU clock cycles of  24x109 will have CPI of………….
a. 1
b. 2
c. 3
d. none of the above

Ans - C

16. Computer executing a program holding 10x109 instructions and having CPI  of 1.1 will need ..... CPU clock cycles.
a. 10x109
b. 11x109
c. 12x109
d. 20x109

Ans - B

17. Instruction count in a program depends on………….
a. architecture
b. exact implementation
c. processor structure
d. memory system

Ans - A

18. CPI depends upon……………
a. architecture
b. processor structure
c. memory system
d. both B & C

Ans - D

19. Algorithm of a program affects ....
a. instruction count
b. clock rate
c.CPU time
d. Throughput

Ans - A

20. In computer system compiler affects ....
a. CPI
b. CPU time
c. bandwidth
d. clock rate

Ans - A

21. In computer system instruction set architecture affects ....
a. clock rate
b. CPI
c. throughput
d. CPU time

Ans - B

22. Amdahl’s Law defines the ........ that can be gained by using a particular Feature of a computer.
a. Speedup
b. response time
c. throughput
d. bandwidth

Ans - A

23. Enhancement in the processor  possible using  Amdahl’s Law  is always ......
a. less than 1
b. less than or equal to 1
c. greater than or equal to 1
d. greater than 1

Ans - B

24. With Amdahl’s Law  if an enhancement is only usable for a fraction of a task, we can’t speed up the task by more than the reciprocal of  .....
a. Fraction
b. 1 plus that fraction
c. 1 minus that fraction
d. 2 minus that fraction

Ans - C

25. Faster machines have .........     MIPS
a. lower
b. medium
c. higher
d. both B & C

Ans - C

26. Different  programs on the same computer has .......      MIPS values.
a. same
b. different
c. Both of a and b
d. None of these

Ans - B

27. MIPS is....... dependent.
a. instruction set
b. response time
c. throughput
d. bandwidth

Ans - A

28. MIPS can vary ...... to performance of a computer.
a. in proportion
b. inversely
c. directly
d. Both b and c

Ans - B

29. MFLOPS is....... dependent.
a. program
b. response time
c. throughput
d. bandwidth

Ans - A

30. The system performs 2*107 floating point operations and execuation time of system is 2 seconds. Calculate MFLOPS?
a. 1
b. 10
c. 100
d. 1000

Ans - B

31. What will be a result of performing multiplication operation on nos. (+13)10  & (-6)10
a. 1110110010
b. 1110111010
c. 1010110010
d. 1110110001

Ans - A

32. What will be a result of performing multiplication operation on nos. (-13)10  & (-6)10
a. 1111010
b. 1001110
c. 1010110
d. 1110110

Ans - B

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