JT(IT) 2016 PART-B Computer Science
Question 1 |
If a random variable takes a finite set of values it is called:
Continuous variate | |
Normal variate
| |
Discrete variate
| |
Exponential variate |
Question 1 Explanation:
→ A discrete variable is a variable whose value is obtained by counting.
Examples:
number of students present
number of red marbles in a jar
number of heads when flipping three coins
students’ grade level
→ A discrete random variable X has a countable number of possible values.
Example:
Let X represent the sum of two dice.
→ A discrete random variable is one which may take on only a countable number of distinct values such as 0,1,2,3,4,........ Discrete random variables are usually (but not necessarily) counts. If a random variable can take only a finite number of distinct values, then it must be discrete.
Examples of discrete random variables include the number of children in a family, the Friday night attendance at a cinema, the number of patients in a doctor's surgery, the number of defective light bulbs in a box of ten.
Examples:
number of students present
number of red marbles in a jar
number of heads when flipping three coins
students’ grade level
→ A discrete random variable X has a countable number of possible values.
Example:
Let X represent the sum of two dice.
→ A discrete random variable is one which may take on only a countable number of distinct values such as 0,1,2,3,4,........ Discrete random variables are usually (but not necessarily) counts. If a random variable can take only a finite number of distinct values, then it must be discrete.
Examples of discrete random variables include the number of children in a family, the Friday night attendance at a cinema, the number of patients in a doctor's surgery, the number of defective light bulbs in a box of ten.
Question 2 |
In client-server computing vertical scaling means:
Adding or removing client workstations with only a slight performance impact
| |
Migrating servers to a new group of client workstations | |
Migrating client workstations to a larger and faster server machine or multi servers | |
Combining two or more client workstation groups
|
Question 2 Explanation:
→ Vertical scaling refers to adding more resources (CPU/RAM/DISK) to your server (database or application server is still remains one) as on demand.
→ Vertical Scaling is most commonly used in applications and products of middle-range as well as small and middle-sized companies. One of the most common examples of Virtual Scaling is to buy an expensive hardware and use it as a Virtual Machine hypervisor (VMWare ESX).
→ Vertical Scaling usually means upgrade of server hardware. Some of the reasons to scale vertically includes increasing IOPS (Input / Output Operations), amplifying CPU/RAM capacity, as well as disk capacity.
→ However, even after using virtualization, whenever an improved performance is targeted, the risk for downtimes with it is much higher than using horizontal scaling.
→ Vertical Scaling is most commonly used in applications and products of middle-range as well as small and middle-sized companies. One of the most common examples of Virtual Scaling is to buy an expensive hardware and use it as a Virtual Machine hypervisor (VMWare ESX).
→ Vertical Scaling usually means upgrade of server hardware. Some of the reasons to scale vertically includes increasing IOPS (Input / Output Operations), amplifying CPU/RAM capacity, as well as disk capacity.
→ However, even after using virtualization, whenever an improved performance is targeted, the risk for downtimes with it is much higher than using horizontal scaling.
Question 3 |
If the mean of a poisson distribution is m, then standard deviation of the distribution is:
m2 | |
m | |
2*m | |
√m |
Question 3 Explanation:
Mean and Variance of the Poisson distribution.
There is also a formula for the standard deviation, σ, and variance, σ2.
Question 4 |
The 10Base5 cabling scheme of ethernet uses:
Twisted pairs
| |
Fiber optics | |
Thin coax
| |
Thick coax |
Question 4 Explanation:
The original cabling standard for Ethernet that uses coaxial cables. The name derives from the fact that the maximum data transfer speed is 10 Mbps, it uses baseband transmission, and the maximum length of cables is 500 meters.
10Base5 is also called thick Ethernet, ThickWire, and ThickNet.
→ The number 10: At the front of each identifier, 10 denotes the standard data transfer speed over these media - ten megabits per second (10Mbps).
→ The word Base: Short for Baseband, this part of the identifier signifies a type of network that uses only one carrier frequency for signaling and requires all network stations to share its use.
→ The segment type or segment length: This part of the identifier can be a digit or a letter:
- Digit - shorthand for how long (in meters) a cable segment may be before attenuation sets in.
For example, a 10Base5 segment can be no more than 500 meters long.
- Letter - identifies a specific physical type of cable.
For example, the T at the end of 10BaseT stands for twisted-pair.
10Base5 is also called thick Ethernet, ThickWire, and ThickNet.
→ The number 10: At the front of each identifier, 10 denotes the standard data transfer speed over these media - ten megabits per second (10Mbps).
→ The word Base: Short for Baseband, this part of the identifier signifies a type of network that uses only one carrier frequency for signaling and requires all network stations to share its use.
→ The segment type or segment length: This part of the identifier can be a digit or a letter:
- Digit - shorthand for how long (in meters) a cable segment may be before attenuation sets in.
For example, a 10Base5 segment can be no more than 500 meters long.
- Letter - identifies a specific physical type of cable.
For example, the T at the end of 10BaseT stands for twisted-pair.
Question 5 |
The standard deviation of binomial distribution with n observations and probability of success p, probability of failure is q is:
√npq
| |
Pq | |
Np | |
√pq |
Question 5 Explanation:
Mean and Variance of the Binomial Distribution:
The binomial distribution for a random variable X with parameters n and p represents the sum of n independent variables Z which may assume the values 0 or 1. If the probability that each Z variable assumes the value 1 is equal to p, then the mean of each variable is equal to l*p + 0*(l-p) = p, and the variance is equal to p(l-p).
By the addition properties for independent random variables, the mean and variance of the binomial distribution are equal to the sum of the means and variances of the n independent Z variables, so
These definitions are intuitively logical. Imagine, for example 8 flips of a coin. If the coin is fair, then p = 0.5. One would expect the mean number of heads to be half the flips, or np = 8*0.5 = 4.
The variance is equal to np(l-p) = 8*0.5*0.5 = 2.
The binomial distribution for a random variable X with parameters n and p represents the sum of n independent variables Z which may assume the values 0 or 1. If the probability that each Z variable assumes the value 1 is equal to p, then the mean of each variable is equal to l*p + 0*(l-p) = p, and the variance is equal to p(l-p).
By the addition properties for independent random variables, the mean and variance of the binomial distribution are equal to the sum of the means and variances of the n independent Z variables, so
These definitions are intuitively logical. Imagine, for example 8 flips of a coin. If the coin is fair, then p = 0.5. One would expect the mean number of heads to be half the flips, or np = 8*0.5 = 4.
The variance is equal to np(l-p) = 8*0.5*0.5 = 2.
Question 6 |
One of the ad-hoc solutions to count to infinity problem in network routing is:
The split horizon hack
| |
Flow based routing | |
Flooding
| |
Shortest path routing
|
Question 6 Explanation:
There are 2 possible solutions are:
1. Route Poison
2. The split horizon hack
1. Route Poison
2. The split horizon hack
Question 7 |
The normal curve is symmetrical about its:
Standard deviation | |
Mean
| |
Variance | |
Probability
|
Question 7 Explanation:
Symmetrical distribution occurs when the values of variables occur at regular frequencies and the mean, median and mode occur at the same point. In graph form, symmetrical distribution often appears as a bell curve. If a line were drawn dissecting the middle of the graph, it would show two sides that mirror each other.
The probability density of the normal distribution is
Where
• μ is the mean or expectation of the distribution (and also its median and mode).
• σ is the standard deviation, and
• 2 is the variance
The probability density of the normal distribution is
Where
• μ is the mean or expectation of the distribution (and also its median and mode).
• σ is the standard deviation, and
• 2 is the variance
Question 8 |
Given a bit rate of b bits/sec, the time required to send 16 bits is:
16*b sec
| |
16/b sec
| |
16b sec | |
b16 sec
|
Question 8 Explanation:
The speed of the data is expressed in bits per second (bits/s or bps).
The data rate R is a function of the duration of the bit or bit time (TB).
R = 1/TB
The data rate R is a function of the duration of the bit or bit time (TB).
R = 1/TB
Question 9 |
The density of uniform distribution over the interval -⍺ < a < b < ⍺ is given by:
f(x) = λe-λx , x>=0
| |
f(x) = qkp | |
f(x) = 1/(b-a), a | |
f(x) = (⍺/c)x⍺-1
|
Question 9 Explanation:
The probability density function of the continuous uniform distribution is:
The values of f(x) at the two boundaries a and b are usually unimportant because they do not alter the values of the integrals of f(X) dX over any interval, not of X f(X) dX or any higher moment. Sometimes they are chosen to be zero, and sometimes chosen to be 1/(b – a). The latter is appropriate in the context of estimation by the method of maximum likelihood. In the context of Fourier analysis, one may lake the value of f(a) to be 1|2(b – a), since then the inverse transform of may integral transform of this uniform function will yield back the function itself, rather than a function which equal ‘almost everywhere’, i.e except on a set of points with zero measure. Also, it is consistent with the sign function which has no such ambiguity.
In terms of mean and variance σ2, the probability density may be written as:
The values of f(x) at the two boundaries a and b are usually unimportant because they do not alter the values of the integrals of f(X) dX over any interval, not of X f(X) dX or any higher moment. Sometimes they are chosen to be zero, and sometimes chosen to be 1/(b – a). The latter is appropriate in the context of estimation by the method of maximum likelihood. In the context of Fourier analysis, one may lake the value of f(a) to be 1|2(b – a), since then the inverse transform of may integral transform of this uniform function will yield back the function itself, rather than a function which equal ‘almost everywhere’, i.e except on a set of points with zero measure. Also, it is consistent with the sign function which has no such ambiguity.
In terms of mean and variance σ2, the probability density may be written as:
Question 10 |
The built in HTTP request method to request to read a web page is:
HEAD | |
PUT
| |
GET
| |
POST |
Question 10 Explanation:
Request Method:
The request method indicates the method to be performed on the resource identified by the given Request-URI. The method is case-sensitive and should always be mentioned in uppercase. The following table lists all the supported methods in HTTP/1.1.
The request method indicates the method to be performed on the resource identified by the given Request-URI. The method is case-sensitive and should always be mentioned in uppercase. The following table lists all the supported methods in HTTP/1.1.
Question 11 |
Exponential distribution is special case of ____ distribution.
Theta | |
Alpha | |
Beta | |
Gamma |
Question 11 Explanation:
The exponential distribution is not the same as the class of exponential families of distributions, which is a large class of probability distributions that includes the exponential distribution as one of its members, but also includes the normal distribution, binomial distribution, gamma distribution, Poisson, and many others.
Question 12 |
The traditional cryptographic cipher that records the letters but do not disguise them is:
Substitute cipher | |
One-time pads
| |
Secret key algorithms
| |
Transposition cipher |
Question 12 Explanation:
A transposition cipher reorders the letters but does not disguise them. The key is a word or phrase not containing any repeated letters. Its purpose is to number the columns, column 1 being under the letter closest to the start of the alphabet, and so on.
Also this code can be easily broken.
Also this code can be easily broken.
Question 13 |
The algebraic sum of the deviations of all the variables from their mean i.e.,
is:
is:0 | |
1 | |
-1 | |
⍺ |
Question 13 Explanation:
The sum of the deviations from the mean is zero. This will always be the case as it is a property of the sample mean, i.e., the sum of the deviations below the mean will always equal the sum of the deviations above the mean. However, the goal is to capture the magnitude of these deviations in a summary measure.
Question 14 |
The segmentation memory management scheme suffers from:
External fragmentation | |
Internal fragmentation
| |
Starvation
| |
Ageing
|
Question 14 Explanation:
• Segmentation avoids internal fragmentation but still it suffers from external fragmentation.
Paging avoids external fragmentation but still it suffers from internal fragmentation. < br> • Internal fragmentation is the wasted space within each allocated block because of rounding up from the actual requested allocation to the allocation granularity.
• External fragmentation is the various free spaced holes that are generated in either your memory or disk space. External fragmented blocks are available for allocation, but may be too small to be of any use.
• Resource starvation is a problem encountered in concurrent computing where a process is perpetually denied necessary resources to process its work. Starvation may be caused by errors in a scheduling or mutual exclusion algorithm.
• Ageing is a scheduling technique used to avoid starvation.
Paging avoids external fragmentation but still it suffers from internal fragmentation. < br> • Internal fragmentation is the wasted space within each allocated block because of rounding up from the actual requested allocation to the allocation granularity.
• External fragmentation is the various free spaced holes that are generated in either your memory or disk space. External fragmented blocks are available for allocation, but may be too small to be of any use.
• Resource starvation is a problem encountered in concurrent computing where a process is perpetually denied necessary resources to process its work. Starvation may be caused by errors in a scheduling or mutual exclusion algorithm.
• Ageing is a scheduling technique used to avoid starvation.
Question 15 |
The mean, mode and median are connected by the empirical relationship:
Mean-mode = 2(mean-median)
| |
Mean-mode = 3(mean-median) | |
Mean-mode = (mean-mode)/2
| |
Mean-mode = (mean-mode)/3
|
Question 15 Explanation:
A distribution in which the values of mean, median and mode coincide (i.e. mean = median = mode) is known as a symmetrical distribution. Conversely, when values of mean, median and mode are not equal the distribution is known as asymmetrical or skewed distribution. In moderately skewed or asymmetrical distribution a very important relationship exists among these three measures of central tendency. In such distributions the distance between the mean and median is about one-third of the distance between the mean and mode, as will be clear from the diagrams 1 and 2. Karl Pearson expressed this relationship as:
Mode = mean - 3 [mean - median]
Mode = 3 median - 2 mean
and Median = mode + ⅔ [mean-mode]
Mode = mean - 3 [mean - median]
Mode = 3 median - 2 mean
and Median = mode + ⅔ [mean-mode]
Question 16 |
The signed 2’s complement representation of -33 is:
11011111
| |
00100001 | |
01011111
| |
10100001
|
Question 16 Explanation:
Step-1: -33 corresponding positive number 00100001
Step-2: Perform 2’s complement.
Step-2: Perform 2’s complement.
Question 17 |
The root mean square deviation when measured from the mean is:
Greatest
| |
Positive
| |
Least
| |
Negative
|
Question 17 Explanation:
The root mean square deviation when measured from the mean is least value.
Question 18 |
Using signed 2’s complement subtraction the result of 11111010-11110011 is:
10000111 | |
00000111 | |
10001101 | |
00001101 |
Question 18 Explanation:
11111010 → 250(Decimal)
-11110011 → -243(Decimal)
Step1: convert signed number into 2’s complement
11110011
00001100 → (1’s complement)
+1
-------------
00001101 → (2’s complement)
Step 2: Add 11111010 into 2’s complement number
11111010
00001101
--------------
00000111 → solution
-11110011 → -243(Decimal)
Step1: convert signed number into 2’s complement
11110011
00001100 → (1’s complement)
+1
-------------
00001101 → (2’s complement)
Step 2: Add 11111010 into 2’s complement number
11111010
00001101
--------------
00000111 → solution
Question 19 |
The values which divide the frequency into four equal parts are called:
Coefficient of variance
| |
Range | |
Dispersion | |
Quartiles
|
Question 19 Explanation:
The values of a variable that divide a distribution into four equal parts are called quartiles. Since three values are needed to divide a distribution into four parts, there are three quartiles, viz. Q1, Q2 and Q3, known as the first, second and the third quartile respectively. For a discrete distribution, the first quartile (Q1) is defined as that value of the variate such that at least 25% of the observations are less than or equal to it and at least 75% of the observations are greater than or equal to it.
Question 20 |
A digital computer has a memory unit of 64k x 16 and a cache memory of 210 words. The cache uses Direct mapping with a block size of four words. How many bits are there in the tag, index and block Fields of address format?
1, 6, 16
| |
28
| |
6, 8, 2 | |
24
|
Question 20 Explanation:
MM size = 64K × 16 = 216 × 16 i.e., MM has 216 words
Therefore Physical address = PA = 16 bits
No. of blocks in cache = cache-size/block-size = 210/ 22 = 28 = 256
∴ 8 bits for block
As block size = 4 words = 22 words
∴ 2 bits for offset
Now, tag = 16 - 8 - 2 = 6 bits
a) Tag = 6 bits, Index = block = 8 bits, offset = word = 2 bits
Therefore Physical address = PA = 16 bits
No. of blocks in cache = cache-size/block-size = 210/ 22 = 28 = 256
∴ 8 bits for block
As block size = 4 words = 22 words
∴ 2 bits for offset
Now, tag = 16 - 8 - 2 = 6 bits
a) Tag = 6 bits, Index = block = 8 bits, offset = word = 2 bits
Question 21 |
How many 3 digit numbers are there with all different odd digits?
16 | |
48 | |
54 | |
60 |
Question 21 Explanation:
• Three digit odd numbers implies the numbers would only be made of digits 1 , 3 , 5 , 7 , 9 With repetition of digits we would have had 5 * 5 * 5 = 125
• But for every hundreds, maximum of 4 tens is possible (avoiding the duplicate of digit used in hundreds)
• And each of these 4 tens, maximum of 3 units is possible (avoiding the duplicate of digits used in tens)
• 5 * 4 * 3 = 60
• But for every hundreds, maximum of 4 tens is possible (avoiding the duplicate of digit used in hundreds)
• And each of these 4 tens, maximum of 3 units is possible (avoiding the duplicate of digits used in tens)
• 5 * 4 * 3 = 60
Question 22 |
A computer employs RAM chips of 256x8 ROM chip of 1024x8. The computer system needs 2k bytes of RAM, 4k bytes of ROM and 4 interface units each with four registers. A memory mapped I/O configuration is used. How many RAM and ROM chips are used?
4, 16 | |
16, 4 | |
4, 8 | |
8, 4 |
Question 22 Explanation:
→ Size of RAM chip is = 256 x 8
Memory size required = 2k bytes = 2*1024*8 bits
= [Memory size required] / [Size of RAM chip] = 8
→ Size of ROM chip is = 1024*8
Memory size required = 4k bytes = 4*1024*8 bits
= [Memory size required] / [Size of ROM chip] = 4
Memory size required = 2k bytes = 2*1024*8 bits
= [Memory size required] / [Size of RAM chip] = 8
→ Size of ROM chip is = 1024*8
Memory size required = 4k bytes = 4*1024*8 bits
= [Memory size required] / [Size of ROM chip] = 4
Question 23 |
In how many ways can a committee of 4 people be chosen from a group of 12?
495 | |
595
| |
395
| |
295
|
Question 23 Explanation:
As the order of people does not matter, it is C4 12
C(n,r) = C(12,4)
= 12! / [(4!(12−4)!)]
= 495
Hence, a committee of 4 people be selected from a group of 12 people in 495 ways.
C(n,r) = C(12,4)
= 12! / [(4!(12−4)!)]
= 495
Hence, a committee of 4 people be selected from a group of 12 people in 495 ways.
Question 24 |
Which of the following method is used for repetitive in which each action is started in terms of a previous result:
Recursion
| |
Iteration
| |
Looping
| |
Structures
|
Question 24 Explanation:
→ Recursion is the process of repeating items in a self-similar way. In programming languages, if a program allows you to call a function inside the same function, then it is called a recursive call of the function.
Question 25 |
A straight line which cuts a curve on two points at an infinite distance from the origin and yet is not itself wholly at infinity is called:
Spiral
| |
Asymptote
| |
Parallel
| |
Polar
|
Question 25 Explanation:
A straight line which cuts a curve in two points at an infinite distance from the origin, but which is not itself wholly at infinity, is called an asymptote to a curve.
Question 26 |
Which of the following is wrong while inserting a node in the beginning of list?
Obtain a node from available list
| |
Make the next pointer of the current head pointer to new node | |
Make the node pointer of the list pointer to new node
| |
Make the next pointer of the new node pointer to current head of the list
|
Question 26 Explanation:
• Option A: If we want to insert a node but the node is already in available list we can’t insert a node at beginning.
• Option B, C and D: It shows actual procedure to insert a node at beginning. So, Option A is wrong
• Option B, C and D: It shows actual procedure to insert a node at beginning. So, Option A is wrong
Question 27 |
If k parallel lines of a determinant Δ become identical when x=a, then ____ is a factor of Δ.
(x-a)+(k+1)
| |
(x-a)/(k-1)
| |
(x-a)*(k+1)
| |
(x-a)(k-1)
|
Question 27 Explanation:
• If a determinant D vanishes for x = a, then (x - a) is a factor of D, in other words, if two rows (or two columns) become identical for x = a, then (x-a) is a factor of D.
• In general, if k rows (or k columns) become identical (x=a) when a is substituted for x, then (x-a)r-1 is a factor of D.
• In general, if k rows (or k columns) become identical (x=a) when a is substituted for x, then (x-a)r-1 is a factor of D.
Question 28 |
A minimal subgraph G’ of G such that V(G’)=V(G) and G’ is connected is called:
A spanning tree
| |
A connected graph
| |
A directed graph
| |
A biconnected component |
Question 28 Explanation:
Given a connected graph G, a connected subgraph that is both a tree and contains all the vertices of G is called a spanning tree for G.
Question 29 |
Rank of nonsingular square matrix of order r is:
r | |
0 | |
r-1 | |
1 |
Question 29 Explanation:
• A square matrix of order ’r’ is nonsingular if its determinant is non zero and therefore its rank is “r”. It's all rows and columns are linearly independent and it is invertible.
• Rank of singular matrix is less than “r”.
• Rank of singular matrix is less than “r”.
Question 30 |
For the recurrence relation T(n) = 2 + T(n - 1), where T(0)=2, T(n)=?
n2
| |
2n+2 | |
log(n) | |
2n |
Question 30 Explanation:
T(n) = T(n-1)+ 1
T(0) = 1
T(n-1) = T(n-1-1)+1
T(n) = [T(n-2)+1] +1 = T(n-2)+ 2
T(n-2) = T(n-2-1)+1
T(n) = [T(n-3)+1]+1
= T(n-3)+3
= T(n) = T(n-k)+ k
Note: Let k = n
Then T(n) = T(0) + n = 1 + n
∴ O(n)
T(0) = 1
T(n-1) = T(n-1-1)+1
T(n) = [T(n-2)+1] +1 = T(n-2)+ 2
T(n-2) = T(n-2-1)+1
T(n) = [T(n-3)+1]+1
= T(n-3)+3
= T(n) = T(n-k)+ k
Note: Let k = n
Then T(n) = T(0) + n = 1 + n
∴ O(n)
Question 31 |
The newton’s Raphson iterative formula for finding 1/N is:
½(xn + N/xn) | |
xn(1 - Nxn)
| |
½(xn+1/Nxn) | |
1/k((k-1)xn + N/xnk-1)
|
Question 31 Explanation:
• Newton's method is a method for finding successively better approximations to the roots (or zeroes) of a real-valued function. It is one example of a root-finding algorithm.
• The iterations xk+1 = xk − ( f(xk)/ f ′(xk) ) are called Newton’s iterations.
• The iterations xk+1 = xk − ( f(xk)/ f ′(xk) ) are called Newton’s iterations.
Question 32 |
Portability is not a quality factor of:
Software coding
| |
Software design
| |
Software Process | |
Software testing
|
Question 32 Explanation:
McCall’s Quality Factors:
This model classifies all software requirements into 11 software quality factors. The 11 factors are grouped into three categories – product operation, product revision, and product transition factors.
1. Product operation factors − Correctness, Reliability, Efficiency, Integrity, Usability.
2. Product revision factors − Maintainability, Flexibility, Testability.
3. Product transition factors − Portability, Reusability, Interoperability.
This model classifies all software requirements into 11 software quality factors. The 11 factors are grouped into three categories – product operation, product revision, and product transition factors.
1. Product operation factors − Correctness, Reliability, Efficiency, Integrity, Usability.
2. Product revision factors − Maintainability, Flexibility, Testability.
3. Product transition factors − Portability, Reusability, Interoperability.
Question 33 |
The number of strips required in simpson’s 3/8th rule is a multiple of:
1 | |
2 | |
3 | |
6 |
Question 33 Explanation:
Simpson’s 3/8th rule is also known as Simpson's 2nd rule:
Area = 3h/ 8 [( a + 3 b + 3 c + d )]
Simpson's Second Rule:
Multipliers:
Area = 3h/ 8 [( a + 3 b + 3 c + d )]
Simpson's Second Rule:
Multipliers:
Question 34 |
Travelling salesperson problem belongs to which category of problems?
Satisfiable
| |
Non Solvable
| |
Decision | |
Optimization
|
Question 34 Explanation:
It uses dynamic programming. So, Travelling salesperson problem belongs to optimization category.
Question 35 |
A declarative sentence which is either true(1) or false(0) is called:
Lattice
| |
Tautology
| |
Contradiction
| |
Proposition
|
Question 35 Explanation:
• Declarative sentences are propositions.
• Sentences that assert a fact that could either be true or false.
• Sentences that assert a fact that could either be true or false.
Question 36 |
If customer is to be involved in all phases of software development, which software model can be selected:
Prototype model
| |
Waterfall model
| |
RAD model
| |
Component model
|
Question 36 Explanation:
Rapid Application Development focuses on gathering customer requirements through workshops or focus groups, early testing of the prototypes by the customer using iterative concept, reuse of the existing prototypes (components), continuous integration and rapid delivery.
Question 37 |
The points at which the function attains extreme values are called:
Turning points | |
End points
| |
Higher points | |
Extreme points
|
Question 37 Explanation:
The value of the function, the value of y, at either a maximum or a minimum is called an extreme value. The points at which the function attains extreme values are called Turning points.
Question 38 |
In page replacement, ‘adding more frames may cause more page faults’ is referred to as:
Thrashing
| |
Belady’s anomaly
| |
Banker’s anomaly
| |
Ageing |
Question 38 Explanation:
In FIFO page replacement policy sometimes behaves differently. The abnormal behaviour is nothing but Belady’s anomaly. Belady’s anomaly is nothing but adding more frames may cause more page faults.
Question 39 |
If f(x) = ax2 + bx + c the f(x-(b/2a)) is:
An even function for all a except a=0
| |
An even function for all a
| |
Neither even nor odd
| |
An odd function for all a except a=0 |
Question 39 Explanation:
• A function f is even if the graph of f is symmetric with respect to the y-axis. Algebraically, f is even if and only if f(-x) = f(x) for all x in the domain of f.
• A function f is odd if the graph of f is symmetric with respect to the origin. Algebraically, f is odd if and only if f(-x) = -f(x) for all x in the domain of f.
• A function f is odd if the graph of f is symmetric with respect to the origin. Algebraically, f is odd if and only if f(-x) = -f(x) for all x in the domain of f.
Question 40 |
____ is an IEEE 1003.1C standard API for thread creation and synchronization in operating system.
Mac OS X
| |
Solaris
| |
POSIX
| |
Kernel |
Question 40 Explanation:
POSIX Threads, usually referred to as pthreads, is an execution model that exists independently from a language, as well as a parallel execution model. It allows a program to control multiple different flows of work that overlap in time. Each flow of work is referred to as a thread, and creation and control over these flows is achieved by making calls to the POSIX Threads API. POSIX Threads is an API defined by the standard POSIX.1c, Threads extensions (IEEE Std 1003.1c-1995).
Question 41 |
The sum of n terms of 1/(1*2) + 1/(2*3) + 1/(3*4) + ... is
(n+1)/n
| |
n/(n+1)
| |
n/(2n+1)
| |
(2n+1)/n
|
Question 41 Explanation:
Sum upto n terms = 1/(1*2) + 1/(2*3) + 1/(3*4) + ........ + 1/(n*(n+1))
where
1st term = 1/(1*2)
2nd term = 1/(2*3)
3rd term = 1/(3*4)
.
.
.
.
n-th term = 1/(n*(n+1))
n-th term = 1/(n*(n+1))
i.e. the k-th term is of the form 1/(k*(k+1))
which can further be written as k-th term = 1/k - 1/(k+1)
So, sum upto n terms can be calculated as:
(1/1 - 1/1+1) + (1/2 - 1/2+1) + (1/3 - 1/3+1) + ......... + (1/n-1 - /1n) + (1/n - 1/n+1)
= (1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ......... + (1/n-1 - 1/n) + (1/n - 1/n+1)
= 1 - 1/n+1
= ((n+1) - 1)/n+1
= n/n+1
where
1st term = 1/(1*2)
2nd term = 1/(2*3)
3rd term = 1/(3*4)
.
.
.
.
n-th term = 1/(n*(n+1))
n-th term = 1/(n*(n+1))
i.e. the k-th term is of the form 1/(k*(k+1))
which can further be written as k-th term = 1/k - 1/(k+1)
So, sum upto n terms can be calculated as:
(1/1 - 1/1+1) + (1/2 - 1/2+1) + (1/3 - 1/3+1) + ......... + (1/n-1 - /1n) + (1/n - 1/n+1)
= (1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ......... + (1/n-1 - 1/n) + (1/n - 1/n+1)
= 1 - 1/n+1
= ((n+1) - 1)/n+1
= n/n+1
Question 42 |
If a system has multiple instances of resources, to avoid deadlock which of the following algorithms is used?
Deadlock avoidance algorithm
| |
Aging algorithm
| |
Resource allocation graph algorithm
| |
Banker’s algorithm |
Question 42 Explanation:
Banker's Algorithm :
Banker's algo can be applied even when Resource-system have multiple instances of each resource type. It can also be applied when Resource-system have single instance of each resource type But in that case, It would be less efficient than Resource-Allocation Graph Algorithm.
Banker's algo can be applied even when Resource-system have multiple instances of each resource type. It can also be applied when Resource-system have single instance of each resource type But in that case, It would be less efficient than Resource-Allocation Graph Algorithm.
Question 43 |
How many integers are between 1 and 200 which are divisible by any one of the integers 2,3 and 5(Hint: use set operation)?
125 | |
145 | |
146 | |
136 |
Question 43 Explanation:
A) numbers divisible by 2: 200/2 = 100
B) numbers divisible by 3: 200/3 = 66
C) numbers divisible by 5: 200/5 = 40
Counting twice:
AB) numbers divisible by 6: 200/6 = 33
AC) numbers divisible by 10: 200/10 = 20
BC) numbers divisible by 15: 200/15 = 13
Counting 3 times:
ABC) numbers divisible by 30: 200/30 = 6
Total of numbers = A + B + C - AB - AC - BC + ABC = 100 + 66 + 40 - 33 - 20 -13 + 6 = 146
B) numbers divisible by 3: 200/3 = 66
C) numbers divisible by 5: 200/5 = 40
Counting twice:
AB) numbers divisible by 6: 200/6 = 33
AC) numbers divisible by 10: 200/10 = 20
BC) numbers divisible by 15: 200/15 = 13
Counting 3 times:
ABC) numbers divisible by 30: 200/30 = 6
Total of numbers = A + B + C - AB - AC - BC + ABC = 100 + 66 + 40 - 33 - 20 -13 + 6 = 146
Question 44 |
Which of the following is not a CPU scheduling criteria:
Dispatch latency | |
CPU utilization
| |
Throughput | |
Turnaround time
|
Question 44 Explanation:
The term dispatch latency describes the amount of time it takes for a system to respond to a request for a process to begin operation. With a scheduler written specifically to honor application priorities, real-time applications can be developed with a bounded dispatch latency.
Note: The time taken by the dispatcher to stop one process and start another running is known as dispatch latency.
Note: The time taken by the dispatcher to stop one process and start another running is known as dispatch latency.
Question 45 |
Which of the following checks are not included in semantic analysis done by the compiler:
Type checks
| |
Spelling checks
| |
Uniquencess checks | |
Flow of control checks
|
Question 45 Explanation:
The following tasks should be performed in semantic analysis:
1. Scope resolution
2. Type checking
3. Array-bound checking
1. Scope resolution
2. Type checking
3. Array-bound checking
Question 46 |
To add attributes to an existing relation __ commands is used:
Update table
| |
Alter table
| |
Change table
| |
Add table |
Question 46 Explanation:
ALTER TABLE: changing a relation schema.
SQL allows the owner of the database relation to change it by:
1. Adding one or more attributes to the relation.
2. Removing one or more attributes from the relation.
3. Adding one or more constraints to the relation.
4. Removing one or more constraints from the relation.
SQL allows the owner of the database relation to change it by:
1. Adding one or more attributes to the relation.
2. Removing one or more attributes from the relation.
3. Adding one or more constraints to the relation.
4. Removing one or more constraints from the relation.
Question 47 |
In boolean algebra, (x ⋀ y)’ = x’ V y’ and (x V y)’ = x’ ⋀ y’ is known as ___ law.
Demorgan’s law
| |
Absorption
| |
Dominance
| |
Idempotent
|
Question 47 Explanation:
The rules can be expressed in English as:
1. The negation of a disjunction is the conjunction of the negations.
2. The negation of a conjunction is the disjunction of the negations.
1. The negation of a disjunction is the conjunction of the negations.
2. The negation of a conjunction is the disjunction of the negations.
Question 48 |
A relation that is not of connected model but is made visible to a user as a virtual relation is called:
Table | |
Query
| |
View
| |
Joined relations |
Question 48 Explanation:
→ In SQL, a view is a virtual table based on the result-set of an SQL statement.
→ A view contains rows and columns, just like a real table. The fields in a view are fields from one or more real tables in the database.
→ You can add SQL functions, WHERE, and JOIN statements to a view and present the data as if the data were coming from one single table.
CREATE VIEW Syntax:
CREATE VIEW view_name AS
SELECT column1, column2, ...
FROM table_name
WHERE condition;
Note:
A view always shows up-to-date data! The database engine recreates the data, using the view's SQL statement, every time a user queries a view.
→ A view contains rows and columns, just like a real table. The fields in a view are fields from one or more real tables in the database.
→ You can add SQL functions, WHERE, and JOIN statements to a view and present the data as if the data were coming from one single table.
CREATE VIEW Syntax:
CREATE VIEW view_name AS
SELECT column1, column2, ...
FROM table_name
WHERE condition;
Note:
A view always shows up-to-date data! The database engine recreates the data, using the view's SQL statement, every time a user queries a view.
Question 49 |
In algebra of logic, the conjunction of two tautologies is:
Contradiction
| |
Tautology
| |
Negation | |
Disjunction
|
Question 49 Explanation:
Some properties are tautologies:
1. The negation of a contradiction is a tautology.
2. The disjunction of two contingencies can be a tautology.
3. The conjunction of two tautologies is a tautology.
1. The negation of a contradiction is a tautology.
2. The disjunction of two contingencies can be a tautology.
3. The conjunction of two tautologies is a tautology.
Question 50 |
In transaction management od database, ‘After a transaction completes successfully the changes it has made to the database persists even if there are system failures’. This property is referred to as:
Atomicity | |
Consistency | |
Isolation
| |
Durability |
Question 50 Explanation:
• The database should be durable enough to hold all its latest updates even if the system fails or restarts.
• If a transaction updates a chunk of data in a database and commits, then the database will hold the modified data.
• If a transaction commits but the system fails before the data could be written on to the disk, then that data will be updated once the system springs back into action.
• If a transaction updates a chunk of data in a database and commits, then the database will hold the modified data.
• If a transaction commits but the system fails before the data could be written on to the disk, then that data will be updated once the system springs back into action.
Question 51 |
The partial differential equation ∂2z/∂t2 = c2(∂2z/∂x2) represents:
Harmonic function | |
Laplace equation
| |
Wave equation
| |
Homogeneous
|
Question 51 Explanation:
Let us derive the d’Alembert’s formula in an alternate way.
Note that the wave equation can be factored as
Note that the wave equation can be factored as
Question 52 |
For the relation loan(loan_number, branch_name, amount), the query{t|t ∈ loan ∧ t[amount] > 1200} gives:
The loan_number, branch_name and amount for loans over $1200
| |
The loan_number for each loan of an amount greater than $1200
| |
The branch_name for each loan of an amount greater than $1200
| |
The amount for each loan of an amount greater than $1200 |
Question 52 Explanation:
branch (branch-name, branch-city, assets)
customer (customer-name, customer-street, customer-city)
account (account-number, branch-name, balance)
loan (loan-number, branch-name, amount)
depositor (customer-name, account-number)
borrower (customer-name, loan-number)
→ The loan-number, branch-name, and amount for loans of over $1200
{t | t ∈ loan ∧ t[amount] > 1200}
customer (customer-name, customer-street, customer-city)
account (account-number, branch-name, balance)
loan (loan-number, branch-name, amount)
depositor (customer-name, account-number)
borrower (customer-name, loan-number)
→ The loan-number, branch-name, and amount for loans of over $1200
{t | t ∈ loan ∧ t[amount] > 1200}
Question 53 |
The number of two digit numbers divisible by the product of digits is:
8 | |
14 | |
13 | |
5 |
Question 53 Explanation:
11=1*1=1 which is a factor of 11
12=1*2=2 which is a factor of 12
15=1*5=5 which is a factor of 15
24=2*4=8 which is a factor of 24
36=3*6=12 which is a factor of 36
12=1*2=2 which is a factor of 12
15=1*5=5 which is a factor of 15
24=2*4=8 which is a factor of 24
36=3*6=12 which is a factor of 36
Question 54 |
In concurrency control a situation where ‘ A transaction may be waiting for an x-lock on an item, while a sequence of other transactions request and are granted on s-lock on the same item’. May lead to:
Deadlock
| |
Starvation | |
Conflict | |
Lock failure
|
Question 54 Explanation:
In concurrency control a situation where ‘ A transaction may be waiting for an x-lock on an item, while a sequence of other transactions request and are granted on s-lock on the same item’ may lead to starvation.
Reasons of Starvation:
1. If waiting scheme for locked items is unfair. ( priority queue )
2. Victim selection. ( same transaction is selected as a victim repeatedly )
3. Resource leak. (Via denial-of-service attack)
Reasons of Starvation:
1. If waiting scheme for locked items is unfair. ( priority queue )
2. Victim selection. ( same transaction is selected as a victim repeatedly )
3. Resource leak. (Via denial-of-service attack)
Question 55 |
If a relation < from A={1,2,3,4} to B={1,3,5} i.e., (a,b)∈R if a < b, then R-1 is:
{(1,3)(1,5)(2,3)(2,5)(3,5)(4,5)}
| |
{(3,1)(5,1)(3,2)(5,2)(5,3)(5,4)} | |
{(3,3)(3,5)(5,3)(5,5)}
| |
{(3,3)(3,4)(4,5)} |
Question 55 Explanation:
The relation R is {(1,3), (1,5),(2,3),(2,5),(3,5),(4,5) } where (a,b) ∈ R if a
The R-1 is { (3,1),(5,1),(3,2),(5,2),(5,3),(5,4) } where (a,b) ∈ R-1 if a>b.
Question 56 |
Which of the following is not an entity reference in XML?
&qt | |
&
| |
< | |
>
|
Question 56 Explanation:
Predefined Character Entities:
They are introduced to avoid the ambiguity while using some symbols.
For example, an ambiguity is observed when less than ( < ) or greater than ( > ) symbol is used with the angle tag (<>).
Character entities are basically used to delimit tags in XML.
Following is a list of predefined character entities from XML specification. These can be used to express characters without ambiguity.
Ampersand − &
Single quote − '
Greater than − >
Less than − <
Double quote − "
They are introduced to avoid the ambiguity while using some symbols.
For example, an ambiguity is observed when less than ( < ) or greater than ( > ) symbol is used with the angle tag (<>).
Character entities are basically used to delimit tags in XML.
Following is a list of predefined character entities from XML specification. These can be used to express characters without ambiguity.
Ampersand − &
Single quote − '
Greater than − >
Less than − <
Double quote − "
Question 57 |
If a lattice (L,R) has a greatest and least element then it is said to be:
Sub Lattice | |
Complemented Lattice
| |
Bounded Lattice
| |
Distributive Lattice
|
Question 57 Explanation:
→ In a Lattice if Upper Bound and Lower exists then it is called Bounded Lattice.
→ A bounded lattice is an algebraic structure of the form (L, ∨, ∧, 0, 1) such that (L, ∨, ∧) is a lattice, 0 (the lattice's bottom) is the identity element for the join operation ∨, and 1 (the lattice top) is the identity element for the meet operation ∧.
→ Let 'L' be a lattice w.r.t R if there exists an element I∈L such that (aRI)∀x∈L, then I is called Upper Bound of a Lattice L.
Similarly, if there exists an element O∈L such that (ORa)∀a∈L, then O is called Lower Bound of Lattice L.
→ A bounded lattice is an algebraic structure of the form (L, ∨, ∧, 0, 1) such that (L, ∨, ∧) is a lattice, 0 (the lattice's bottom) is the identity element for the join operation ∨, and 1 (the lattice top) is the identity element for the meet operation ∧.
→ Let 'L' be a lattice w.r.t R if there exists an element I∈L such that (aRI)∀x∈L, then I is called Upper Bound of a Lattice L.
Similarly, if there exists an element O∈L such that (ORa)∀a∈L, then O is called Lower Bound of Lattice L.
Question 58 |
____ element is used to create destination anchors in HTML?
< d > | |
< a > | |
< da > | |
< ad > |
Question 58 Explanation:
• HTML offers many of the conventional publishing idioms for rich text and structured documents, but what separates it from most other markup languages is its features for hypertext and interactive documents.
• This section introduces the link (or hyperlink, or Web link), the basic hypertext construct. A link is a connection from one Web resource to another. Although a simple concept, the link has been one of the primary forces driving the success of the Web.
• A link has two ends -- called anchors -- and a direction. The link starts at the "source" anchor and points to the "destination" anchor, which may be any Web resource (e.g., an image, a video clip, a sound bite, a program, an HTML document, an element within an HTML document, etc.)
• This section introduces the link (or hyperlink, or Web link), the basic hypertext construct. A link is a connection from one Web resource to another. Although a simple concept, the link has been one of the primary forces driving the success of the Web.
• A link has two ends -- called anchors -- and a direction. The link starts at the "source" anchor and points to the "destination" anchor, which may be any Web resource (e.g., an image, a video clip, a sound bite, a program, an HTML document, an element within an HTML document, etc.)
Question 59 |
For the context free grammar G:
R → XRX|S, S → aTbTa, T → XTX|Xϵ X → a|b
The strings which are not in L(G) are:
ab,ba,aab
| |
abb,aabab
| |
a,b,aa
| |
a,b,ϵ
|
Question 59 Explanation:
We can’t generate the strings “a” and “b” because the Starting symbol is R→ XRX|S.
Question 60 |
The sum of the series: (1/2) + (1/3) + (1/4) - (1/6) + (1/8) + (1/9) + (1/16) - (1/12) + ... + α is:
(1/2) - log(√2)3
| |
1 + log(√2)3
| |
3/2 | |
1/2 + log(√2)3
|
Question 60 Explanation:
The sum of the series: (1/2) + (1/3) + (1/4) - (1/6) + (1/8) + (1/9) + (1/16) - (1/12) + ... + α is 1 + log(√2)3.
There are 60 questions to complete.