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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

1.

In Reverse Polish notation, expression A*B+C*D is written as __________(a) AB*CD*+(b) A*BCD*+(c) AB*CD+*(d) A*B*CD+The question was asked in unit test.Question is taken from Storage Allocation topic in chapter Storage Management of Compiler

Answer»

Correct OPTION is (a) AB*CD*+

The best I can explain: RPN is AB*CD*+.

2.

(2FAOC) 16 is equivalent to __________(a) (195 084)10(b) (001011111010 0000 1100)2(c) (195 084)10 & (001011111010 0000 1100)2(d) None of the mentionedI had been asked this question during an online interview.My doubt stems from Storage Allocation topic in division Storage Management of Compiler

Answer» RIGHT OPTION is (b) (001011111010 0000 1100)2

Easy explanation: It is EQUIVALENT to (001011111010 0000 1100)2.
3.

What characteristic of RAM memory makes it not suitable for permanent storage?(a) Too slow(b) Unreliable(c) It is volatile(d) Too bulkyThis question was posed to me during an online exam.Origin of the question is Storage Allocation in chapter Storage Management of Compiler

Answer» CORRECT answer is (c) It is VOLATILE

Easy explanation: RAM is volatile.
4.

In computers, subtraction is generally carried out by ______(a) 9’s complement(b) 10’s complement(c) 1’s complement(d) 2’s complementThe question was asked in an interview for job.Question is from Storage Allocation topic in chapter Storage Management of Compiler

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Correct answer is (d) 2’s COMPLEMENT

The best EXPLANATION: SUBTRACTION is DONE by 2’s complement.

5.

SIMD represents an organization that ______________(a) Refers to a computer system capable of processing several programs at the same time(b) Represents organization of single computer containing a control unit, processor unit and a memory unit(c) Includes many processing units under the supervision of a common control unit(d) None of the mentionedThis question was posed to me by my college professor while I was bunking the class.My enquiry is from Storage Allocation topic in section Storage Management of Compiler

Answer»

The correct choice is (c) Includes many PROCESSING UNITS under the SUPERVISION of a common control unit

Easy EXPLANATION: SIMD includes processing units under the super vision of a common control.

6.

The idea of cache memory is based __________(a) On the property of locality of reference(b) On the heuristic 90-10 rule(c) On the fact that references generally tend to cluster(d) All of the mentionedI have been asked this question in an internship interview.My question comes from Storage Allocation topic in section Storage Management of Compiler

Answer» CORRECT choice is (a) On the property of locality of reference

To ELABORATE: CACHE memory is based on the locality of references.
7.

Assembly language ________(a) Uses alphabetic codes in place of binary numbers used in machine language(b) Is the easiest language to write programs(c) Need not be translated into machine language(d) None of the mentionedThis question was addressed to me in quiz.This is a very interesting question from Storage Allocation in portion Storage Management of Compiler

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Correct ANSWER is (a) Uses alphabetic codes in place of binary numbers USED in MACHINE language

To explain I would say: Uses binary numbers in machine language.

8.

Computers use addressing mode techniques for ____________(a) Giving programming versatility to the user by providing facilities as pointers to memory counters for loop control(b) To reduce number of bits in the field of instruction(c) Specifying rules for modifying or interpreting address field of the instruction(d) All of the mentionedThe question was asked in examination.I'd like to ask this question from Storage Allocation topic in portion Storage Management of Compiler

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The correct CHOICE is (d) All of the mentioned

To explain: All of these are ADDRESSING MODE techniques.

9.

The amount of time required to read a block of data from a disk into memory is composed of seek time, rotational latency, and transfer time. Rotational latency refers to ______(a) The time it takes for the platter to make a full rotation(b) The time it takes for the read-write head to move into position over the appropriate track(c) The time it takes for the platter to rotate the correct sector under the head(d) None of the mentionedThis question was addressed to me by my college professor while I was bunking the class.This key question is from Storage Allocation topic in chapter Storage Management of Compiler

Answer»

The CORRECT answer is (a) The time it takes for the platter to MAKE a full rotation

The BEST I can explain: Rotational latency is the time TAKEN to make full rotation.

10.

The average time required to reach a storage location in memory and obtain its contents is called the _____(a) Seek time(b) Turnaround time(c) Access time(d) Transfer timeI have been asked this question in exam.Origin of the question is Storage Allocation topic in section Storage Management of Compiler

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Correct choice is (C) ACCESS time

The EXPLANATION is: Times used to access the CONTENTS.

11.

Floating point representation is used to store ____________(a) Boolean values(b) Whole numbers(c) Real integers(d) IntegersI have been asked this question in a national level competition.This interesting question is from Storage Allocation topic in portion Storage Management of Compiler

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The correct OPTION is (C) REAL integers

Easiest EXPLANATION: They are real Integers.

12.

Suppose that a bus has 16 data lines and requires 4 cycles of 250 nests each to transfer data. The bandwidth of this bus would be 2 Megabytes/sec. If the cycle time of the bus was reduced to 125 nsecs and the number of cycles required for transfer stayed the same what would the bandwidth of the bus?(a) 1 Megabyte/sec(b) 4 Megabytes/sec(c) 8 Megabytes/sec(d) 2 Megabytes/secThis question was addressed to me in an international level competition.My enquiry is from Storage Allocation in portion Storage Management of Compiler

Answer» RIGHT CHOICE is (d) 2 Megabytes/sec

Best EXPLANATION: The BANDWIDTH is 2 mb/s.
13.

The circuit used to store one bit of data is known as_______(a) Register(b) Encoder(c) Decoder(d) Flip FlopThis question was posed to me by my college director while I was bunking the class.Origin of the question is Storage Allocation in chapter Storage Management of Compiler

Answer»

The correct CHOICE is (d) Flip FLOP

To explain I would say: 1 bit CIRCUIT is known as Flip Flop.

14.

Which of the following is not a weighted code?(a) Decimal Number system(b) Excess 3-cod(c) Binary number System(d) None of the mentionedThe question was asked in an internship interview.I want to ask this question from Storage Allocation topic in section Storage Management of Compiler

Answer»

Correct OPTION is (B) Excess 3-cod

The explanation: Excess 3 is not a WEIGHTED CODE.