SAT – Sample Quiz
A score of 80% in the practice exam is a good check for basic knowledge. It is a good practice to give the test at one shot. . . .
Questions and Answers
- 1.
A subnet has been assigned a subnet mask of 255.255.255.192. What is the maximum number of hosts that can belong to this subnet?
- A.
14
- B.
30
- C.
62
- D.
126
Correct Answer
C. 62Explanation
The subnet mask 255.255.255.192 indicates that the first 26 bits of the IP address are used to identify the network, while the remaining 6 bits are used to identify the hosts within that network. With 6 bits, there are 2^6 = 64 possible combinations, but 2 of them are reserved (all 0s and all 1s). Therefore, the maximum number of hosts that can belong to this subnet is 64 - 2 = 62.Rate this question:
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- 2.
Which one of the following is NOT shared by the threads of the same process ?
- A.
Stack
- B.
Address Space
- C.
File Descriptor Table
- D.
Message Queue
Correct Answer
A. StackExplanation
Threads within the same process share the same address space, file descriptor table, and message queue. However, the stack is not shared among threads. Each thread has its own stack, which is used for storing local variables and function call information. This separation ensures that each thread has its own execution context and can independently execute functions without interfering with other threads.Rate this question:
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- 3.
The following C function takes two ASCII strings and determines whether one is an anagram of the other. An anagram of a string s is a string obtained by permuting the letters in s. int anagram (char *a, char *b){int count [128], j;for (j = 0;j < 128; j++) count[j] = 0;j = 0;while (a[J] && b[J]){A;B; }for (j = 0;j < 128; j++) if(count[j]) return 0;return 1;}Choose the correct alternative for statements A and B.
- A.
A : count [a[j]]++and B : count[b[j]]-
- B.
A : count [a[j]]++and B : count[b[j]]++
- C.
A : count [a[j++]]++and B : count[b[j]]-
- D.
A : count [a[j]]++and B : count[b[j++]]-
Correct Answer
A. A : count [a[j]]++and B : count[b[j]]- -
- 4.
Let M = (K, å, d, s, F) be a finite state automaton, whereK = {A, B}, å = {a, b}, s = A, F = {B},d(A, a) = A, d(A, b) = B, d(B,a) = B and d(B, b) = AA grammar to generate the language accepted by M can be specified as G = (V, å, R, S), where V = K È S, and S = AWhich one of the following set of rules will make L(G) = L(M) ?
- A.
{A ® aB, A ® bA, B ® bA, B ® aA, B ® e}
- B.
{A ® aA, A ® bB, B ® aB, B ® bA, B ® e}
- C.
{A ® bB, A ® aB, B ® aA, B ® bA, B ® e}
- D.
{A ® aA, A ® bA, B ® aB, B ® bA, A ® e}
Correct Answer
B. {A ® aA, A ® bB, B ® aB, B ® bA, B ® e}Explanation
The set of rules {A ® aA, A ® bB, B ® aB, B ® bA, B ® e} will make L(G) = L(M). This set of rules specifies that starting from the initial symbol A, we can either replace A with aA or bB. If we replace A with aA, we can continue replacing A with aA or bB, and if we replace A with bB, we can continue replacing B with aB or bA. This set of rules ensures that the grammar generates the same language as the given finite state automaton M.Rate this question:
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- 5.
Which one of the following regular expressions is NOT equivalent to the regular expression (a + b + c)* ?
- A.
(a* + b* + c*)*
- B.
(a*b*c*)*
- C.
((ab)* + c*)*
- D.
(a*b* + c*)*
Correct Answer
C. ((ab)* + c*)*Explanation
The regular expression ((ab)* + c*)* is not equivalent to the regular expression (a + b + c)* because it allows for the sequence "ab" to repeat any number of times, including zero, before being followed by "c". In the original regular expression, the sequence "ab" is not allowed.Rate this question:
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- 6.
Let G be a weighted undirected graph and e be an edge with maximum weight in G. Suppose there is a minimum weight spanning tree in G containing the edge e. Which of the following statements is always TRUE?
- A.
There exists a cutset in G having all edges of maximum weight.
- B.
There exists a cycle in G having all edges of maximum weight.
- C.
Edge e cannot be contained in a cycle.
- D.
All edges in G have the same weight.
Correct Answer
C. Edge e cannot be contained in a cycle.Explanation
In a minimum weight spanning tree, all edges are included to connect all vertices with the minimum total weight. If edge e, which has the maximum weight, were to be contained in a cycle, removing it would still allow the remaining edges to connect all vertices. However, since e has the maximum weight, removing it would result in a spanning tree with a lower total weight, contradicting the assumption that the tree is a minimum weight spanning tree. Therefore, edge e cannot be contained in a cycle.Rate this question:
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- 7.
A software organization has been assessed at SEI CMM Level 4. Which of the following does the organization need to practice beside Process Change Management and Technology Change Management in order to achieve Level 5 ?
- A.
Defect Detection
- B.
Defect Prevention
- C.
Defect Isolation
- D.
Defect Propagation
Correct Answer
B. Defect PreventionExplanation
To achieve SEI CMM Level 5, the organization needs to practice defect prevention in addition to process change management and technology change management. This means that the organization should focus on identifying potential defects in the early stages of the software development process and implementing measures to prevent them from occurring. This could involve implementing quality assurance techniques, conducting thorough code reviews, using automated testing tools, and improving the overall software development process to minimize the occurrence of defects. By emphasizing defect prevention, the organization can strive for continuous improvement and higher levels of quality in their software products.Rate this question:
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- 8.
A CPU has only three instructions II, 12 and 13, which use the following signals in time steps T1-T5:I1: T1: Ain, Bout, Cin T2: PCout, Bin T3: Zout, Ain T4: PCin, Bout T5: EndI2: Tl: Cin, Bout, Din T2: Aout,Bin T3: Zout, Ain T4: Bin, Cout T5: EndI3: Tl: Din, out T2: Din,Bout T3: Zout, Ain T4: Dout,Ain T5: EndWhich of the following logic functions will generate the hardwired control for the signal Ain?
- A.
T1.I1 + T2.I3 + T4 . I3 + T3
- B.
(T1 + T2 + T3) . I3 + T1.I1
- C.
(T1 + T2 ) . I1 + (T2 + T4) . I3 + T3
- D.
(T1 + T2 ) . I2 + (Tl + T3) . I1+ T3
Correct Answer
A. T1.I1 + T2.I3 + T4 . I3 + T3Explanation
The correct answer is T1.I1 + T2.I3 + T4 . I3 + T3. This logic function includes the signals T1, T2, T3, and T4, which are the time steps in which the instruction I1 is executed. It also includes the signals I1 and I3, which are the instructions that use the signal Ain. By combining these signals using logical operators (AND, OR), this function generates the hardwired control for the signal Ain.Rate this question:
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- 9.
Suppose that two parties A and B wish to setup a common secret key (D-H key) between themselves using the Diffle-Hellman key exchange technique. They agree on 7 as the modulus and 3 as the primitive root. Party A chooses 2 and party B chooses 5 as their respective secrets. Their D-H key is
- A.
3
- B.
4
- C.
5
- D.
6
Correct Answer
C. 5Explanation
In the Diffle-Hellman key exchange technique, both parties agree on a modulus and a primitive root. Party A chooses a secret number (2) and party B chooses a secret number (5). Each party then calculates their own public key by raising the primitive root to the power of their secret number modulo the modulus. Party A calculates (3^2) mod 7 = 2, and party B calculates (3^5) mod 7 = 5. The D-H key is the result of raising the other party's public key to the power of their own secret number modulo the modulus. Party A calculates (5^2) mod 7 = 4, and party B calculates (2^5) mod 7 = 6. Therefore, the correct D-H key is 5.Rate this question:
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- 10.
Consider the following message M = 1010001101. The cyclic redundancy check (CRC) for this message using the divisor polynomial x5 + x4 + x2 + 1 is :
- A.
01110
- B.
01011
- C.
10101
- D.
10110
Correct Answer
C. 10101Explanation
The correct answer is 10101 because the given message M is divided by the divisor polynomial x5 + x4 + x2 + 1 using the CRC method. The remainder obtained after the division is 10101, which is the CRC for the message.Rate this question:
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- 11.
Which of the following statements is FALSE regarding a bridge
- A.
Bridge is a layer 2 device
- B.
Bridge reduces collision domain
- C.
Bridge is used to connect two or more LAN segments
- D.
Bridge reduces broadcast domain
Correct Answer
D. Bridge reduces broadcast domainExplanation
A bridge is a layer 2 device that is used to connect two or more LAN segments. It operates at the data link layer of the OSI model and forwards data packets between different LAN segments based on the MAC addresses. One of the main functions of a bridge is to reduce the collision domain by creating separate collision domains for each LAN segment. However, a bridge does not reduce the broadcast domain. Broadcasts are still forwarded to all LAN segments connected to the bridge, allowing devices on each segment to receive the broadcast messages.Rate this question:
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- 12.
What is the availability of a software with the following reliability figures?Mean Time Between Failure (MTBF) = 25 daysMean Time To Repair (MTTR) = 6 hours
- A.
1%
- B.
24%
- C.
99%
- D.
99.009%
Correct Answer
B. 24%Explanation
The availability of a software can be calculated using the formula: Availability = MTBF / (MTBF + MTTR). In this case, the MTBF is given as 25 days and the MTTR is given as 6 hours. To calculate the availability, we need to convert the MTTR from hours to days by dividing it by 24 (since there are 24 hours in a day). So, the MTTR becomes 6/24 = 0.25 days. Plugging these values into the formula, we get: Availability = 25 / (25 + 0.25) = 25 / 25.25 ≈ 0.99 or 99%. Therefore, the correct answer is 99%, not 24%.Rate this question:
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- 13.
In a population of N families, 50% of the families have three children, 30% of the families have two children and the remaining families have one child. What is the probability that a randomly picked child belongs to a family with two children?
- A.
3/23
- B.
6/23
- C.
3/10
- D.
3/5
Correct Answer(s)
B. 6/23
D. 3/5Explanation
The probability that a randomly picked child belongs to a family with two children can be calculated by dividing the number of families with two children by the total number of families. In this case, since 30% of the families have two children, the probability is 30% or 6/20. Therefore, the correct answer is 6/23. The answer 3/5 is incorrect because it represents the probability of a randomly picked child belonging to a family with three children, which is not what the question is asking for.Rate this question:
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- 14.
A student wishes to create symbolic links in a computer system running Unix. Three text files named “file 1”, “file 2” and “file 3” exist in her current working directory, and the student has read and write permissions for all three files. Assume that file 1 contains information about her hobbies, file 2 contains information about her friends and file 3 contains information about her courses. The student executes the following sequence of commands from her current working directoryIn -s file 1 file 2 In -s file 2 file 3Which of the following types of information would be lost from her file system?(I) Hobbies (II) Friends (III) Courses
- A.
(I) and (II) only
- B.
(II) and (III) only
- C.
(II) only
- D.
(I) and (III) only
Correct Answer
B. (II) and (III) onlyExplanation
By executing the command "In -s file 1 file 2", the student creates a symbolic link named "file 2" that points to "file 1". This means that "file 2" now contains the same information as "file 1", which is about her hobbies.
Then, by executing the command "In -s file 2 file 3", the student creates another symbolic link named "file 3" that points to "file 2". This means that "file 3" now contains the same information as "file 2", which is about her hobbies.
Therefore, the original information about her friends (in "file 2") and her courses (in "file 3") would be lost from her file system.
Hence, the correct answer is (II) and (III) only.Rate this question:
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- 15.
A channel has a bit rate of 4 kbps and one-way propagation delay of 20 ms. The channel uses stop and wait protocol. The transmission time of the acknowledgement frame is negligible. To get a channel efficiency of at least 50%, the minimum frame size should be
- A.
80 bytes
- B.
80 bits
- C.
160 bytes
- D.
160 bits
Correct Answer
D. 160 bitsExplanation
To calculate the minimum frame size, we need to consider the channel bit rate and the one-way propagation delay. The channel efficiency is given by the formula: Efficiency = (Frame Size / (Frame Size + 2 * Propagation Delay)) * Bit Rate. We need to find the minimum frame size that gives us an efficiency of at least 50%. By substituting the given values, we can solve for the frame size. In this case, a frame size of 160 bits gives us an efficiency of exactly 50%, making it the minimum frame size required.Rate this question:
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- 16.
Consider an XML file called intro.xml and a document type defintion (DTD) file intro. dtd as follows:intro.xmlWelcome to XML intro.dtd A validating parser will classify intro.xml as
- A.
Well-formed and validated
- B.
Well-formed but not validated
- C.
Validated but not well-formed
- D.
Neither validated not well-formed
Correct Answer
A. Well-formed and validatedExplanation
A validating parser checks if an XML document adheres to the rules specified in the DTD or schema. In this case, the intro.xml file is referencing the intro.dtd file, indicating that it is using a DTD for validation. Since the question does not mention any issues with the XML file, we can assume that it is well-formed. Therefore, a validating parser will classify intro.xml as well-formed and validated, meaning it adheres to the syntax rules and the content rules specified in the DTD.Rate this question:
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- 17.
In an inventory management system implemented at a trading corporation, there are several tables designed to hold all the information. Amongst these, the following two tables hold information on which items are supplied by which suppliers, and which warehouse keeps which items along with the stock-level of these items.Supply = (supplierid, itemcode)Inventory = (itemcode, warehouse, stocklevel)For a specific information required by the management, following SQL query has been writtenSelect distinct STMP supplieridFrom Supply as STMPWhere not unique (Select ITMP . supplieridFrom Inventory, Supply as ITMPWhere STMP . supplierid = ITMP . supplieridAnd ITMP . itemcode = Inventory . itemcodeAnd Inventory . warehouse = ‘Nagpur’);For the warehouse at Nagpur, this query will find all suppliers who
- A.
Do not supply any item
- B.
Supply exactly one item
- C.
Supply one or more items
- D.
Supply two or more items
Correct Answer
D. Supply two or more itemsExplanation
The given SQL query is selecting distinct supplier IDs from the Supply table where the supplier ID is not unique in the result of a subquery. The subquery checks for supplier IDs in the Inventory table that match the supplier ID from the outer query, the item code from the Inventory table, and the warehouse 'Nagpur'.
Since the subquery is checking for supplier IDs that have a match in the Inventory table, the outer query will return supplier IDs that have two or more items supplied to the warehouse at Nagpur. Therefore, the correct answer is "supply two or more items".Rate this question:
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- 18.
In a depth-first traversal of a graph G with n vertices, k edges are marked as tree edges. The number of connected components in G is
- A.
K
- B.
K+1
- C.
N-k-l
- D.
N-k
Correct Answer
C. N-k-lExplanation
In a depth-first traversal of a graph, the number of connected components can be determined by subtracting the number of tree edges (k) from the total number of vertices (n) and the number of back edges (l). Therefore, the correct answer is n - k - l.Rate this question:
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- 19.
Which of the following statements is TRUE about CSMA/CD
- A.
IEEE 802.11 wireless LAN runs CSMA/CD protocol
- B.
Ethernet is not based on CSMA/CD protocol
- C.
CSMA/CD is not suitable for a high propagation delay network like satellite network
- D.
There is no contention in a CSMA/CD network
Correct Answer
C. CSMA/CD is not suitable for a high propagation delay network like satellite networkExplanation
CSMA/CD (Carrier Sense Multiple Access with Collision Detection) is a protocol used in Ethernet networks to control access to the shared medium. It is not suitable for a high propagation delay network like a satellite network because the time it takes for a signal to travel from one end to the other is significantly longer in a satellite network compared to a wired network. This increased delay makes it difficult for CSMA/CD to detect collisions and respond in a timely manner, leading to decreased efficiency and performance.Rate this question:
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Mar 22, 2023Quiz Edited by
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Nov 03, 2009Quiz Created by
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