If an integer occupies 4 bytes and a character occupies 1 byte of memory, each element of the following structure would occupy how many bytes ? struct name { int age; char name[30]; }; A) 30 B) 32 C) 34 D) 36

If an integer occupies 4 bytes and a character occupies 1 byte of memory, each element of the  following structure would occupy how many bytes ?  struct name {  int age;  char name[30];  };  A) 30  B) 32  C) 34  D) 36
by

1 Answer

Answer :

C) 34
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Last Answer : D

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Last Answer : (3) 15 and 14

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Description : Station A uses 32 byte packets to transmit messages to station B using sliding window protocol. The round trip delay between A and B is 40 milli seconds and the bottleneck bandwidth on the path between A and B is 64 kbps. The optimal window size of A is (1) 20 (2) 10 (3) 30 (4) 40

Last Answer : (2) 10

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Last Answer : Answer: A,B,C,D

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Description : A unix file system has 1-KB blocks and 4-byte disk addresses. What is the maximum file size if i-nodes contain 10 direct entries and one single, double and triple indirect entry each? (A) 32 GB (B) 64 GB (C) 16 GB (D) 1 GB

Last Answer : (C) 16 GB

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Last Answer : B) public void myMethod(int a, int

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Description : The correct way to round off a floating number x to an integer value is (A) y = (int)(x+0.5) (B) y = int(x+0.5) (C) y = (int)x+0.5 (D) y = (int)((int)x+0.5)

Last Answer : (A) y = (int)(x+0.5)

A three dimensional array in ‘C’ is declared as int A[x][y][z]. Here, the address of an item at the location A[p][q][r] can be computed as follows (where w is the word length of an integer): (A) &A[0][0][0]+w(y*z*q+z*p+r) (B) &A[0][0][0]+w(y*z*p+z*q+r) (C) &A[0][0][0]+w(x*y*p+z*q+r) (D) &A[0][0][0]+w(x*y*q+z*p+r)

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Last Answer : Answer: B

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Consider a 32 - bit microprocessor, with a 16 - bit external data bus, driven by an 8 MHz input clock. Assume that the microprocessor has a bus cycle whose minimum duration equals four input clock cycles. What is the maximum data transfer rate for this microprocessor? (A) 8x106 bytes/sec (B) 4x106 bytes/sec (C) 16x106 bytes/sec (D) 4x109 bytes/sec

Description : Consider a 32 - bit microprocessor, with a 16 - bit external data bus, driven by an 8 MHz input clock. Assume that the microprocessor has a bus cycle whose minimum duration equals four input clock cycles. What ... (A) 8x106 bytes/sec (B) 4x106 bytes/sec (C) 16x106 bytes/sec (D) 4x109 bytes/sec 

Last Answer : (B) 4x106 bytes/sec

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Last Answer : (B) 80 msec

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50'. What number is missing? -Riddles

Description : 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50'. What number is missing? -Riddles

Last Answer : 22

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A gas occupies a volume of 0.2 L at 176 mm Hg. What volume will the gas occupy at 760 mm Hg?

Description : A gas occupies a volume of 0.2 L at 176 mm Hg. What volume will the gas occupy at 760 mm Hg?

Last Answer : 0.05 L

A gas occupies a volume of 0.2 L at 176 mm Hg. What volume will the gas occupy at 760 mm Hg?

Description : A gas occupies a volume of 0.2 L at 176 mm Hg. What volume will the gas occupy at 760 mm Hg?

Last Answer : 0.05 L

Four channels are multiplexed using TDM. If each channel sends 100 bytes/second and we multiplex 1 byte per channel, then the bit rate for the link is __________. a. 400 bps b. 800 bps c. 1600 bps d. 3200 bps

Description : Four channels are multiplexed using TDM. If each channel sends 100 bytes/second and we multiplex 1 byte per channel, then the bit rate for the link is __________. a. 400 bps b. 800 bps c. 1600 bps d. 3200 bps

Last Answer : d. 3200 bps

A UNIX file system has 1 KB block size and 4-byte disk addresses. What is the maximum file size if the inode contains ten direct block entries, one single indirect block entry, one double indirect block entry and one triple indirect block entry? (A) 30 GB (B) 64 GB (C) 16 GB (D) 1 GB

Description : A UNIX file system has 1 KB block size and 4-byte disk addresses. What is the maximum file size if the inode contains ten direct block entries, one single indirect block entry, one double indirect block entry and one triple indirect block entry? (A) 30 GB (B) 64 GB (C) 16 GB (D) 1 GB

Last Answer : (C) 16 GB

Eight years ago, the ratio of the ages of micky and Donald was 6:5, 12years hence, the ratio of their ages will be 11:10. What is Donald age at present a) 30 years b) 28 years c) 34 years d) 36 years

Description : Eight years ago, the ratio of the ages of micky and Donald was 6:5, 12years hence, the ratio of their ages will be 11:10. What is Donald age at present a) 30 years b) 28 years c) 34 years d) 36 years

Last Answer : B Let 8 years ago the age of Micky and Donald be 6x and 5x resp. then, ((6x+8)+12) / ((5x+8)+12) = 11/10 10(6x+20) = 11(5x+20) 5x = 20 => x = 4 So Donald age is (5x+8) = 28

My brother is 3 years elder to me. My father was 28 years of age when my sister was born while my mother was 26 years of age when I was born. If my sister was 4 years of age when my brother was born, then what was the age my father when my brother was born? a) 35 years b) 34 years c) 33 years d) 32 years e) 30 years

Description : My brother is 3 years elder to me. My father was 28 years of age when my sister was born while my mother was 26 years of age when I was born. If my sister was 4 years of age when my brother was born, ... father when my brother was born? a) 35 years b) 34 years c) 33 years d) 32 years e) 30 years

Last Answer : Let my age = x Then My brother's age = x + 3 My mother's age = x + 26 My sister's age = (x + 3) + 4 = x + 7 My Father's age = (x + 7) + 28 = x + 35 => age my father when my brother was born = x + 35 – (x + 3) = 32. Answer: d)

Five years ago, the age of Arun was 4 times the age of Sarmi after 10 years, Arun will be twice as old as Sarmi. Find the Present ages of Arun and sarmi? a) 30 years, 10.5 years b) 32 years, 11.5 years c) 34 years, 14.5 years d) 35 years, 12.5 years e) None of these

Description : Five years ago, the age of Arun was 4 times the age of Sarmi after 10 years, Arun will be twice as old as Sarmi. Find the Present ages of Arun and sarmi? a) 30 years, 10.5 years b) 32 years, 11.5 years c) 34 years, 14.5 years d) 35 years, 12.5 years e) None of these

Last Answer : Five years ago, x – 5/y – 5 = 4/1 x – 4y = – 15 —-(i) After 10 years, x + 10/y + 10 = 2/1 x – 2y = 10 —–(ii) From eqn (i) and (ii) => x = 35, y = 12.5 Answer: d)

What is the output of the following program ? (Assume that the appropriate pre-processor directives are included and there is no syntax error) main() { char S[ ] = "ABCDEFGH"; printf ("%C",* (& S[3])); printf ("%s", S+4); printf ("%u", S); /* Base address of S is 1000 */ } (A) ABCDEFGH1000 (B) CDEFGH1000 (C) DDEFGHH1000 (D) DEFGH1000

Description : What is the output of the following program ? (Assume that the appropriate pre-processor directives are included and there is no syntax error) main() { char S[ ] = "ABCDEFGH"; printf ("%C",* (& ... Base address of S is 1000 */ } (A) ABCDEFGH1000 (B) CDEFGH1000 (C) DDEFGHH1000 (D) DEFGH1000

Last Answer : (D) DEFGH1000

A dynamic RAM has refresh cycle of 32 times per msec. Each refresh operation requires 100 nsec and a memory cycle requires 250 nsec. What percentage of memory’s total operating time is required for refreshes? (A) 0.64 (B) 0.96 (C) 2.00 (D) 0.32

Description : A dynamic RAM has refresh cycle of 32 times per msec. Each refresh operation requires 100 nsec and a memory cycle requires 250 nsec. What percentage of memory’s total operating time is required for refreshes? (A) 0.64 (B) 0.96 (C) 2.00 (D) 0.32

Last Answer : Answer: D Explanation: in 1ms : refresh = 32 times Memory cycle = 1ms/250ns = 106ns/250ns = 4000 times Therefore, % of refresh time = (32 x 100ns)/(4000 x 250ns) = 3200ns/1000000 x 100% = 0.32%

Assume that variable x resides at memory location 1234, y at 1111 and p at 2222. Int x=1, y=2, *p; p=&x; y=*p; What will be the value of y after execution of above code? A) 2 B) 1 C) 1234 D) 1111

Description : Assume that variable x resides at memory location 1234, y at 1111 and p at 2222. Int x=1, y=2, *p; p=&x; y=*p; What will be the value of y after execution of above code? A) 2 B) 1 C) 1234 D) 1111

Last Answer : C) 1234

What are the types of casting shown by the following examples: (i) char c = (char)120; (ii) int x = ‘t’;

Description : What are the types of casting shown by the following examples: (i) char c = (char)120; (ii) int x = ‘t’;

Last Answer : Ans. (i) Explicit casting (ii) Implicit casting

Classify the following as primitive or non-primitive data types: (i) char (ii) arrays (iii) int (iv) classes

Description : Classify the following as primitive or non-primitive data types: (i) char (ii) arrays (iii) int (iv) classes

Last Answer : Ans. (i) char – Primitive (ii) arrays – Non primitive (iii) int – Primitive (iv) Classes – Non primitive

Classify the following as primitive or non-primitive data types: (i) char (ii) arrays (iii) int (iv) classes

Description : Classify the following as primitive or non-primitive data types: (i) char (ii) arrays (iii) int (iv) classes

Last Answer : Ans. (i) char – Primitive (ii) arrays – Non primitive (iii) int – Primitive (iv) Classes – Non primitive

Give output for following code: class student { int roll no; char name [14]; } s[6]; void main() { cout<<sizeof(s); }

Description : Give output for following code: class student { int roll no; char name [14]; } s[6]; void main() { cout<<sizeof(s); }

Last Answer : Considering roll_no(Single variable) the output is: 96 OR Considering roll, no (Two variables) the output is: 108 OR Considering roll no the output is: error – space between roll and no

Find the root value of 36.1 / 102.4 (1) 61 / 34 (2) 19 / 31 (3) 19 / 32 (4) 19 / 33

Description : Find the root value of 36.1 / 102.4 (1) 61 / 34 (2) 19 / 31 (3) 19 / 32 (4) 19 / 33

Last Answer : (3) 19 / 32

If there are n integers to sort, each integer has d digits and each digit is in the set {1,2, ..., k}, radix sort can sort the numbers in: (A) O(d n k) (B) O(d nk) (C) O((d+n)k) (D) O(d(n+k))

Description : If there are n integers to sort, each integer has d digits and each digit is in the set {1,2, ..., k}, radix sort can sort the numbers in: (A) O(d n k) (B) O(d nk) (C) O((d+n)k) (D) O(d(n+k))

Last Answer : (D) O(d(n+k))

For the implementation of a paging scheme, suppose the average process size be x bytes, the page size be y bytes, and each page entry requires z bytes. The optimum page size that minimizes the total overhead due to the page table and the internal fragmentation loss is given by (A) x/2 (B) xz/2 (C) √2xz (D) (√xz)/2

Description : For the implementation of a paging scheme, suppose the average process size be x bytes, the page size be y bytes, and each page entry requires z bytes. The optimum page size that minimizes the total overhead due to the page ... fragmentation loss is given by (A) x/2 (B) xz/2 (C) √2xz (D) (√xz)/2

Last Answer : (C) √2xz 

Which of the following information about the UNIX file system is not correct? (A) Super block contains the number of i-nodes, the number of disk blocks, and the start of the list of free disk blocks. (B) An i-node contains accounting information as well as enough information to locate all the disk blocks that holds the file’s data. (C) Each i-node is 256-bytes long. (D) All the files and directories are stored in data blocks.

Description : Which of the following information about the UNIX file system is not correct? (A) Super block contains the number of i-nodes, the number of disk blocks, and the start of the list of free disk blocks. (B ... Each i-node is 256-bytes long. (D) All the files and directories are stored in data blocks. 

Last Answer : (C) Each i-node is 256-bytes long.

(2) Exclusion Principle of Pauli Explanation: The Pauli exclusion principle is the quantum mechanical principle that no two identical fermions (particles with half-integer spin) may occupy the same quantum state simultaneously. A more rigorous statement is that the total wave function for two identical fermions is anti-symmetric with respect to exchange of the particles.

Description : (2) Exclusion Principle of Pauli Explanation: The Pauli exclusion principle is the quantum mechanical principle that no two identical fermions (particles with half-integer spin) may occupy the same ... function for two identical fermions is anti-symmetric with respect to exchange of the particles.

Last Answer : The three elements most needed in common fertilizers are – (1) Sulphur, Phosphorous and Sodium (2) Nitrogen, Phosphorous and Sodium (3) Phosphorous, Sodium and Nitrogen (4) Calcium, Phosphorous and Potassium

A memory unit stores 2^16 bytes in 32-bit words. How many address bits are necessary in order to retrieve a single word from this memory unit?

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Last Answer : Each word is 4 bytes, which occupies 4 consecutive addresses at 16 bits each. The first 14 bits of these addresses will be the same, differing only in the two least significant bits. You only need to specify the first 14 bits, because there are only 2^14 words occupying those 2^16 bytes.

Ten signals, each requiring 3000 Hz, are multiplexed on to a single channel using FDM. How much minimum bandwidth is required for the multiplexed channel ? Assume that the guard bands are 300 Hz wide. (A) 30,000 (B) 32,700 (C) 33,000 (D) None of the above

Description : Ten signals, each requiring 3000 Hz, are multiplexed on to a single channel using FDM. How much minimum bandwidth is required for the multiplexed channel ? Assume that the guard bands are 300 Hz wide. (A) 30,000 (B) 32,700 (C) 33,000 (D) None of the above

Last Answer : (B) 32,700

In Frame Relay, the EA field defines thenumber of bytes; it is _____ in the last byte of the address. A) 0 B) 1 C) 2 D) 3

Description : In Frame Relay, the EA field defines thenumber of bytes; it is _____ in the last byte of the address. A) 0 B) 1 C) 2 D) 3

Last Answer : 1

In IPv4, if the fragment offset hasa value of 100, it means that _______. A) thedatagram has notbeen fragmented B) thedatagram is 100 bytes in size C) thefirst byte of the datagram is byte100 D) the first byteof the datagram is byte800

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Last Answer : the first byteof the datagram is byte800

Given the symbols A, B, C, D, E, F, G and H with the probabilities 1/30, 1/30, 1/30, 2/30, 3/30, 5/30, 5/30 and 12/30 respectively. The average Huffman code size in bits per symbol is: (A) 67/30 (B) 70/34 (C) 76/30 (D) 78/30

Description : Given the symbols A, B, C, D, E, F, G and H with the probabilities 1/30, 1/30, 1/30, 2/30, 3/30, 5/30, 5/30 and 12/30 respectively. The average Huffman code size in bits per symbol is: (A) 67/30 (B) 70/34 (C) 76/30 (D) 78/30

Last Answer : (C) 76/30

12 years ago M was half of N in age. If the ratio of their present ages is 4 : 5, what will be the total of their present ages a) 30 b) 35 c) 36 d) 32

Description : 12 years ago M was half of N in age. If the ratio of their present ages is 4 : 5, what will be the total of their present ages a) 30 b) 35 c) 36 d) 32

Last Answer : C Let M's age 12 years ago = x years. Then, N's age 12 years ago = 2x years. (x + 12) / (2x+ l2) = 4/5 => x = 4. So, the total of their present ages =(x + 12 + 2x + 12) = (3x + 24) = 36 years.

A thread is usually defined as a light weight process because an Operating System (OS) maintains smaller data structure for a thread than for a process. In relation to this, which of the following statement is correct? (A) OS maintains only scheduling and accounting information for each thread. (B) OS maintains only CPU registers for each thread. (C) OS does not maintain a separate stack for each thread. (D) OS does not maintain virtual memory state for each thread.

Description : A thread is usually defined as a light weight process because an Operating System (OS) maintains smaller data structure for a thread than for a process. In relation to this, which of the following ... separate stack for each thread. (D) OS does not maintain virtual memory state for each thread.

Last Answer : (B) OS maintains only CPU registers for each thread.

A DMA controller transfers 32-bit words to memory using cycle Stealing. The words are assembled from a device that transmits characters at a rate of 4800 characters per second. The CPU is fetching and executing instructions at an average rate of one million instructions per second. By how much will the CPU be slowed down because of the DMA transfer? (A) 0.06% (B) 0.12% (C) 1.2% (D) 2.5%

Description : A DMA controller transfers 32-bit words to memory using cycle Stealing. The words are assembled from a device that transmits characters at a rate of 4800 characters per second. The CPU is fetching and executing instructions at an average ... of the DMA transfer? (A) 0.06% (B) 0.12% (C) 1.2% (D) 2.5%

Last Answer : Answer: B

What will be the output? numberGames = {} numberGames[(1,2,4)] = 8 numberGames[(4,2,1)] = 10 numberGames[(1,2)] = 12 sum = 0 for k in numberGames: sum += numberGames[k] print len(numberGames) + sum Page No 36 a) 30 b) 24 c) 33 d) 12

Description : What will be the output? numberGames = {} numberGames[(1,2,4)] = 8 numberGames[(4,2,1)] = 10 numberGames[(1,2)] = 12 sum = 0 for k in numberGames: sum += numberGames[k] print len(numberGames) + sum Page No 36 a) 30 b) 24 c) 33 d) 12

Last Answer : a) 30