McqMate
These multiple-choice questions (MCQs) are designed to enhance your knowledge and understanding in the following areas: Computer Science Engineering (CSE) .
| 51. |
The addressing mode, where you directly specify the operand value is |
| A. | immediate |
| B. | direct |
| C. | definite |
| D. | relative |
| Answer» A. immediate | |
| Explanation: none. | |
| 52. |
addressing mode is most suitable to change the normal sequence of execution of instructions. |
| A. | relative |
| B. | indirect |
| C. | index with offset |
| D. | immediate |
| Answer» A. relative | |
| Explanation: the relative addressing mode is used for this since it directly updates the pc. | |
| 53. |
Which method/s of representation of numbers occupies a large amount of memory than others? |
| A. | sign-magnitude |
| B. | 1’s complement |
| C. | 2’s complement |
| D. | 1’s & 2’s compliment |
| Answer» A. sign-magnitude | |
| Explanation: it takes more memory as one bit used up to store the sign. | |
| 54. |
Which method of representation has two representations for ‘0’? |
| A. | sign-magnitude |
| B. | 1’s complement |
| C. | 2’s complement |
| D. | none of the mentioned |
| Answer» A. sign-magnitude | |
| Explanation: one is positive and one for negative. | |
| 55. |
When we perform subtraction on -7 and 1 the answer in 2’s complement form is |
| A. | 1010 |
| B. | 1110 |
| C. | 0110 |
| D. | 1000 |
| Answer» D. 1000 | |
| Explanation: first the 2’s complement is found and that is added to the number and the overflow is ignored. | |
| 56. |
The processor keeps track of the results of its operations using flags called |
| A. | conditional code flags |
| B. | test output flags |
| C. | type flags |
| D. | none of the mentioned |
| Answer» A. conditional code flags | |
| Explanation: these flags are used to indicate if there is an overflow or carry or zero result occurrence. | |
| 57. |
The register used to store the flags is called as |
| A. | flag register |
| B. | status register |
| C. | test register |
| D. | log register |
| Answer» B. status register | |
| Explanation: the status register stores the condition codes of the system. | |
| 58. |
In some pipelined systems, a different instruction is used to add to numbers which can affect the flags upon |
| A. | and gate |
| B. | nand gate |
| C. | nor gate |
| D. | xor gate |
| Answer» D. xor gate | |
| Explanation: none. | |
| 59. |
The most efficient method followed by computers to multiply two unsigned numbers is |
| A. | booth algorithm |
| B. | bit pair recording of multipliers |
| C. | restoring algorithm |
| D. | non restoring algorithm |
| Answer» B. bit pair recording of multipliers | |
| Explanation: none. | |
| 60. |
For the addition of large integers, most of the systems make use of |
| A. | fast adders |
| B. | full adders |
| C. | carry look-ahead adders |
| D. | none of the mentioned |
| Answer» C. carry look-ahead adders | |
| Explanation: in this method, the carries for each step are generated first. | |
| 61. |
In a normal n-bit adder, to find out if an overflow as occurred we make use of |
| A. | counter |
| B. | flip flop |
| C. | shift register |
| D. | push down stack |
| Answer» C. shift register | |
| Explanation: the shift registers are used to store the multiplied answer. | |
| 62. |
The smallest entity of memory is called |
| A. | cell |
| B. | block |
| C. | instance |
| D. | unit |
| Answer» A. cell | |
| Explanation: each data is made up of a number of units. | |
| 63. |
The collection of the above mentioned entities where data is stored is called |
| A. | block |
| B. | set |
| C. | word |
| D. | byte |
| Answer» C. word | |
| Explanation: each readable part of the data is called blocks. | |
| 64. |
If a system is 64 bit machine, then the length of each word will be |
| A. | 4 bytes |
| B. | 8 bytes |
| C. | 16 bytes |
| D. | 12 bytes |
| Answer» B. 8 bytes | |
| Explanation: a 64 bit system means, that at a time 64 bit instruction can be executed. | |
| 65. |
The type of memory assignment used in Intel processors is |
| A. | little endian |
| B. | big endian |
| C. | medium endian |
| D. | none of the mentioned |
| Answer» A. little endian | |
| Explanation: the method of address allocation to data to be stored is called as memory assignment. | |
| 66. |
When using the Big Endian assignment to store a number, the sign bit of the number is stored in |
| A. | the higher order byte of the word |
| B. | the lower order byte of the word |
| C. | can’t say |
| D. | none of the mentioned |
| Answer» A. the higher order byte of the word | |
| Explanation: none. | |
| 67. |
To get the physical address from the logical address generated by CPU we use |
| A. | mar |
| B. | mmu |
| C. | overlays |
| D. | tlb |
| Answer» B. mmu | |
| Explanation: memory management unit, is used to add the offset to the logical address generated by the cpu to get the physical address. | |
| 68. |
method is used to map logical addresses of variable length onto physical memory. |
| A. | paging |
| B. | overlays |
| C. | segmentation |
| D. | paging with segmentation |
| Answer» C. segmentation | |
| Explanation: segmentation is a process in which memory is divided into groups of variable length called segments. | |
| 69. |
During the transfer of data between the processor and memory we use |
| A. | cache |
| B. | tlb |
| C. | buffers |
| D. | registers |
| Answer» D. registers | |
| Explanation: none. | |
| 70. |
Physical memory is divided into sets of finite size called as |
| A. | frames |
| B. | pages |
| C. | blocks |
| D. | vectors |
| Answer» A. frames | |
| Explanation: none. | |
| 71. |
Add #%01011101,R1 , when this instruction is executed then |
| A. | the binary addition between the operands takes place |
| B. | the numerical value represented by the binary value is added to the value of r1 |
| C. | the addition doesn’t take place, whereas this is similar to a mov instruction |
| D. | none of the mentioned |
| Answer» A. the binary addition between the operands takes place | |
| Explanation: this performs operations in binary mode directly. | |
| 72. |
If we want to perform memory or arithmetic operations on data in Hexa- decimal mode then we use symbol before the operand. |
| A. | ~ |
| B. | ! |
| C. | $ |
| D. | * |
| Answer» C. $ | |
| Explanation: none. | |
| 73. |
When generating physical addresses from a logical address the offset is stored in |
| A. | translation look-aside buffer |
| B. | relocation register |
| C. | page table |
| D. | shift register |
| Answer» B. relocation register | |
| Explanation: in the mmu the relocation register stores the offset address. | |
| 74. |
The technique used to store programs larger than the memory is |
| A. | overlays |
| B. | extension registers |
| C. | buffers |
| D. | both extension registers and buffers |
| Answer» A. overlays | |
| Explanation: in this, only a part of the program getting executed is stored on the memory and later swapped in for the other part. | |
| 75. |
The unit which acts as an intermediate agent between memory and backing store to reduce process time is |
| A. | tlb’s |
| B. | registers |
| C. | page tables |
| D. | cache |
| Answer» D. cache | |
| Explanation: the cache’s help in data transfers by storing most recently used memory pages. | |
| 76. |
Does the Load instruction do the following operation/s? |
| A. | loads the contents of a disc onto a memory location |
| B. | loads the contents of a location onto the accumulators |
| C. | load the contents of the pcb onto the register |
| D. | none of the mentioned |
| Answer» B. loads the contents of a location onto the accumulators | |
| Explanation: the load instruction is basically used to load the contents of a memory location onto a register. | |
| 77. |
Complete the following analogy:- Registers are to RAM’s as Cache’s are to |
| A. | system stacks |
| B. | overlays |
| C. | page table |
| D. | tlb |
| Answer» D. tlb | |
| Explanation: none. | |
| 78. |
The BOOT sector files of the system are stored in |
| A. | harddisk |
| B. | rom |
| C. | ram |
| D. | fast solid state chips in the motherboard |
| Answer» B. rom | |
| Explanation: the files which are required for the starting up of a system are stored on the rom. | |
| 79. |
The transfer of large chunks of data with the involvement of the processor is done by |
| A. | dma controller |
| B. | arbitrator |
| C. | user system programs |
| D. | none of the mentioned |
| Answer» A. dma controller | |
| Explanation: this mode of transfer involves the transfer of a large block of data from the memory. | |
| 80. |
Which of the following techniques used to effectively utilize main memory? |
| A. | address binding |
| B. | dynamic linking |
| C. | dynamic loading |
| D. | both dynamic linking and loading |
| Answer» C. dynamic loading | |
| Explanation: in this method only when the routine is required is loaded and hence saves memory. | |
| 81. |
RTN stands for |
| A. | register transfer notation |
| B. | register transmission notation |
| C. | regular transmission notation |
| D. | regular transfer notation |
| Answer» A. register transfer notation | |
| Explanation: this is the way of writing the assembly language code with the help of register notations. | |
| 82. |
The instruction, Add Loc,R1 in RTN is |
| A. | addsetcc loc+r1 |
| B. | r1=loc+r1 |
| C. | not possible to write in rtn |
| D. | r1<-[loc]+[r1] |
| Answer» D. r1<-[loc]+[r1] | |
| Explanation: none. | |
| 83. |
Can you perform an addition on three operands simultaneously in ALN using Add instruction? |
| A. | yes |
| B. | not possible using add, we’ve to use addsetcc |
| C. | not permitted |
| D. | none of the mentioned |
| Answer» C. not permitted | |
| Explanation: you cannot perform an addition on three operands simultaneously because the third operand is where the result is stored. | |
| 84. |
The instruction, Add R1,R2,R3 in RTN is |
| A. | r3=r1+r2+r3 |
| B. | r3<-[r1]+[r2]+[r3] |
| C. | r3=[r1]+[r2] |
| D. | r3<-[r1]+[r2] |
| Answer» D. r3<-[r1]+[r2] | |
| Explanation: in rtn the first operand is the destination and the second operand is the source. | |
| 85. |
In a system, which has 32 registers the register id is long. |
| A. | 16 bit |
| B. | 8 bits |
| C. | 5 bits |
| D. | 6 bits |
| Answer» C. 5 bits | |
| Explanation: the id is the name tag given to each of the registers and used to identify them. | |
| 86. |
While using the iterative construct (Branching) in execution instruction is used to check the condition. |
| A. | testandset |
| B. | branch |
| C. | testcondn |
| D. | none of the mentioned |
| Answer» B. branch | |
| Explanation: branch instruction is used to check the test condition and to perform the memory jump with the help of offset. | |
| 87. |
The condition flag Z is set to 1 to indicate |
| A. | the operation has resulted in an error |
| B. | the operation requires an interrupt call |
| C. | the result is zero |
| D. | there is no empty register available |
| Answer» C. the result is zero | |
| Explanation: this condition flag is used | |
| 88. |
converts the programs written in assembly language into machine instructions. |
| A. | machine compiler |
| B. | interpreter |
| C. | assembler |
| D. | converter |
| Answer» C. assembler | |
| Explanation: an assembler is a software used to convert the programs into machine instructions. | |
| 89. |
The instructions like MOV or ADD are called as |
| A. | op-code |
| B. | operators |
| C. | commands |
| D. | none of the mentioned |
| Answer» A. op-code | |
| Explanation: this op – codes tell the | |
| 90. |
The assembler directive EQU, when used in the instruction: Sum EQU 200 does |
| A. | finds the first occurrence of sum and assigns value 200 to it |
| B. | replaces every occurrence of sum with 200 |
| C. | re-assigns the address of sum by adding 200 to its original address |
| D. | assigns 200 bytes of memory starting the location of sum |
| Answer» B. replaces every occurrence of sum with 200 | |
| Explanation: this basically is used to replace the variable with a constant value. | |
| 91. |
The directive used to perform initialization before the execution of the code is |
| A. | reserve |
| B. | store |
| C. | dataword |
| D. | equ |
| Answer» C. dataword | |
| Explanation: none. | |
| 92. |
directive is used to specify and assign the memory required for the block of code. |
| A. | allocate |
| B. | assign |
| C. | set |
| D. | reserve |
| Answer» D. reserve | |
| Explanation: this instruction is used to allocate a block of memory and to store the object code of the program there. | |
| 93. |
directive specifies the end of execution of a program. |
| A. | end |
| B. | return |
| C. | stop |
| D. | terminate |
| Answer» B. return | |
| Explanation: this instruction directive is used to terminate the program execution. | |
| 94. |
The last statement of the source program should be |
| A. | stop |
| B. | return |
| C. | op |
| D. | end |
| Answer» D. end | |
| Explanation: this enables the processor to load some other process. | |
| 95. |
The assembler stores all the names and their corresponding values in |
| A. | special purpose register |
| B. | symbol table |
| C. | value map set |
| D. | none of the mentioned |
| Answer» B. symbol table | |
| Explanation: the table where the assembler stores the variable names along with their corresponding memory locations and values. | |
| 96. |
The assembler stores the object code in |
| A. | main memory |
| B. | cache |
| C. | ram |
| D. | magnetic disk |
| Answer» D. magnetic disk | |
| Explanation: after compiling the object code, the assembler stores it in the magnetic disk and waits for further execution. | |
| 97. |
The utility program used to bring the object code into memory for execution is |
| A. | loader |
| B. | fetcher |
| C. | extractor |
| D. | linker |
| Answer» A. loader | |
| Explanation: the program is used to load the program into memory. | |
| 98. |
To overcome the problems of the assembler in dealing with branching code we use |
| A. | interpreter |
| B. | debugger |
| C. | op-assembler |
| D. | two-pass assembler |
| Answer» D. two-pass assembler | |
| Explanation: this creates entries into the symbol table first and then creates the object code. | |
| 99. |
The return address of the Sub-routine is pointed to by |
| A. | ir |
| B. | pc |
| C. | mar |
| D. | special memory registers |
| Answer» B. pc | |
| Explanation: the return address from the subroutine is pointed to by the pc. | |
| 100. |
The location to return to, from the subroutine is stored in |
| A. | tlb |
| B. | pc |
| C. | mar |
| D. | link registers |
| Answer» D. link registers | |
| Explanation: the registers store the return address of the routine and is pointed to by the pc. | |
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