HCU PHD CS MAY 2017
Question 1 
Two motorists set out at the same time from A to B, a distance of 100 miles. They both followed the same route and travelled at different though uniform speeds of integral number of miles /hour. The difference in their speeds was a prime number of miles/hour. After they had been driving for 2 hrs, the distance of the slower car from A was five times that of the faster car from B. How fast did the motorists drive?
30 and 37 miles/hour  
40 and 42 miles/hour  
47 and 49 miles/hour  
None of the above 
Question 1 Explanation:
The two motorists:
Let d=the distance the faster car has traveled. Then it still has 100d miles to go, and the slow car has, therefore, gone 5005d miles. Let A and B be the rates of the fast and slow cars in mph. Then d/A = (5005d)/B = 2
We get B=2505B
Then AB=6f250
=(3A125).
But since AB is prime 3A125=1
hence A=42mph and B=40 mph
Let d=the distance the faster car has traveled. Then it still has 100d miles to go, and the slow car has, therefore, gone 5005d miles. Let A and B be the rates of the fast and slow cars in mph. Then d/A = (5005d)/B = 2
We get B=2505B
Then AB=6f250
=(3A125).
But since AB is prime 3A125=1
hence A=42mph and B=40 mph
Question 2 
A job is done by M men in D days. Then M + N men can do the same job in
(MD)/M+N days  
D/M((M +N) days  
D(D/M)*N days  
D(M/D)*N days 
Question 2 Explanation:
let's assume 1 man will take m time to do the same job, so rate of working for i man is: 1(job)/d*m(days)= 1/d*m
More man means job will done faster so the new rate will be: (m+n)*(rate of 1 man) = m+n/d*m
Now convert this to m+r men by dividing top and bottom will take d*m/m+n
More man means job will done faster so the new rate will be: (m+n)*(rate of 1 man) = m+n/d*m
Now convert this to m+r men by dividing top and bottom will take d*m/m+n
Question 3 
An academic institute starts a class at 10:00 A.M. and ends at 1:47 P.M. It has 4
periods of equal distribution of time. After each period there is a gap of 5 minutes to
start another period. What is the exact duration of the period?
51  
52  
53  
57 
Question 3 Explanation:
let’s count total academic minutes in a day from 10:00 am to 1:47 pm i.e 227 minutes.
Total period is =4
Time gap after each period is = 5 minutes. First class starts at 10:00 am so total gap will be 5*3=15 minutes. Deduct gap from total academic minutes = 227 15=212
Total 4 periods we have given in the question so, 212/4= 53 minutes.
Total period is =4
Time gap after each period is = 5 minutes. First class starts at 10:00 am so total gap will be 5*3=15 minutes. Deduct gap from total academic minutes = 227 15=212
Total 4 periods we have given in the question so, 212/4= 53 minutes.
Question 4 
A contractor estimated that one of his bricklayers would take 9 hrs to build a certain
wall and the other 10 hours. However, he knew from experience that when they worked
together, 10 fewer bricks laid per hour. Since he was in a hurry, he put both men
on the job and found it took exactly 5 hours to build the wall. How many bricks did
it contain?
18  
90  
900  
It can not be determined from the given data 
Question 5 
The number of ways in which 3 men and 2 women can sit in a row so that no two men
are adjacent is
12  
24  
72  
19 
Question 5 Explanation:
The only possibility is
M W M W M
Where M represents men and W represents women.
Now three men can be arranged among themselves in 3! Ways and two women can be arranged in 2! ways.Hence the answer is 3! * 2! =12
M W M W M
Where M represents men and W represents women.
Now three men can be arranged among themselves in 3! Ways and two women can be arranged in 2! ways.Hence the answer is 3! * 2! =12
Question 6 
Given the statement, "I always cut the topright corner of a milk packet for opening it", a milk packet with its topright corner cut is
a necessary condition for the packet to have been opened by me  
a sufficient condition for the packet to have been opened by me  
both Ii necessary and a sufficient condition for the packet to have been opened by
me
 
sometimes a necessary condition and sometimes a sufficient condition for the
packet to have been opened by me

Question 6 Explanation:
The given statement is necessary condition that the packet was opened by me but it is not sufficient condition, because some other person also might open the milk packet in same way. But if it was opened by me then it has to be cut from top right corner ,hence nessecary condition.
Question 7 
A common problem in Computer Science research is to minimise an equation of the Form
where λ is a parameter, E_{t} , E_{m} and E_{d} are the total, model and data errors. Model
error reflects the mismatch between predicted and actual values, while data error is
the error due to noise in the data. For noisy data, the value of λ should be
Close to 0  
Close to 1  
Close to 0.5  
None of the above 
Question 8 
Using general knowledge about Computer Science and by reading the following paragraph carefully, answer the Questions 8  12 below:
Princeton asked me to develop a course in automata theory. Since there were no courses or books on the subject, I asked McCluskey to recommend some materials for a course on automata theory. He gave me a list of six papers and told me that the material would probably give students a good background in automata theory. McCluskey'S list included works by Warren McCulloch and Walter Pitts, Michael Rabin and Dana Scott, John Backus and Peter Naur, Noam Chomsky, Juris Hartmanis and RichardStearns, and, of course, Alan Turing.
In 1943 McCulloch and Pitts, working in neurophysiology, published a paper on a logical calculus for describing events in neuron nets. The paper had a notation for describing how these strings of zeros and ones combine in neurons to produce new strings of zeros and ones. This notation was subsequently developed into the language of regular expressions for describing sets of strings. Rabin and Scott were mathematicians who developed a model of a computer with a finite amount of memory. They called this model the finitestate automaton, and showed that the possible behaviors of finitestate automata were precisely those behaviors that could be described by the regular expressions that grew out of the work of McCulloch and Pitts. ...... The work of Hartmanis and Stearns attracted researchers and focused· attention on the topic of complexity. Among the more significant advances that resulted were the classification of the complexity of most major mathematical theories, the reducibility of many combinatorial problems, the concept of NPcompleteness, and a deeper understanding of concepts such as randomness. Also, through this course I met Jeffrey Ullman and Alfred Aho, with whom I subsequently collaborated for many years. Formal Languages and Automata Theory, which I wrote with Ullman, also evolved from this course.
Who is the ’I’, in the passage?
McCluskey·  
Jeffrey Ullman  
John Hopcroft  
Edsger Dijkstra 
Question 9 
Using general knowledge about Computer Science and by reading the following paragraph carefully, answer the Questions 8  12 below:
Princeton asked me to develop a course in automata theory. Since there were no courses or books on the subject, I asked McCluskey to recommend some materials for a course on automata theory. He gave me a list of six papers and told me that the material would probably give students a good background in automata theory. McCluskey'S list included works by Warren McCulloch and Walter Pitts, Michael Rabin and Dana Scott, John Backus and Peter Naur, Noam Chomsky, Juris Hartmanis and RichardStearns, and, of course, Alan Turing.
In 1943 McCulloch and Pitts, working in neurophysiology, published a paper on a logical calculus for describing events in neuron nets. The paper had a notation for describing how these strings of zeros and ones combine in neurons to produce new strings of zeros and ones. This notation was subsequently developed into the language of regular expressions for describing sets of strings. Rabin and Scott were mathematicians who developed a model of a computer with a finite amount of memory. They called this model the finitestate automaton, and showed that the possible behaviors of finitestate automata were precisely those behaviors that could be described by the regular expressions that grew out of the work of McCulloch and Pitts. ...... The work of Hartmanis and Stearns attracted researchers and focused· attention on the topic of complexity. Among the more significant advances that resulted were the classification of the complexity of most major mathematical theories, the reducibility of many combinatorial problems, the concept of NPcompleteness, and a deeper understanding of concepts such as randomness. Also, through this course I met Jeffrey Ullman and Alfred Aho, with whom I subsequently collaborated for many years. Formal Languages and Automata Theory, which I wrote with Ullman, also evolved from this course.
...... denotes some missing sentences in the passage. What are the most likely missing sentences?
Sentences about Hopcroft's early work in the course  
Sentences about work by Hartmanis and Stearns  
Sentences about work by Backus, Naur and Chomsky  
Sentences about Princeton university 
Question 9 Explanation:
The missing sentences was about Backus naur and Chomsky because the second paragraph contains all the names of the McCluskey'S list except these two names.
Question 10 
Using general knowledge about Computer Science and by reading the following paragraph carefully, answer the Questions 8  12 below:
Princeton asked me to develop a course in automata theory. Since there were no courses or books on the subject, I asked McCluskey to recommend some materials for a course on automata theory. He gave me a list of six papers and told me that the material would probably give students a good background in automata theory. McCluskey'S list included works by Warren McCulloch and Walter Pitts, Michael Rabin and Dana Scott, John Backus and Peter Naur, Noam Chomsky, Juris Hartmanis and RichardStearns, and, of course, Alan Turing.
In 1943 McCulloch and Pitts, working in neurophysiology, published a paper on a logical calculus for describing events in neuron nets. The paper had a notation for describing how these strings of zeros and ones combine in neurons to produce new strings of zeros and ones. This notation was subsequently developed into the language of regular expressions for describing sets of strings. Rabin and Scott were mathematicians who developed a model of a computer with a finite amount of memory. They called this model the finitestate automaton, and showed that the possible behaviors of finitestate automata were precisely those behaviors that could be described by the regular expressions that grew out of the work of McCulloch and Pitts. ...... The work of Hartmanis and Stearns attracted researchers and focused· attention on the topic of complexity. Among the more significant advances that resulted were the classification of the complexity of most major mathematical theories, the reducibility of many combinatorial problems, the concept of NPcompleteness, and a deeper understanding of concepts such as randomness. Also, through this course I met Jeffrey Ullman and Alfred Aho, with whom I subsequently collaborated for many years. Formal Languages and Automata Theory, which I wrote with Ullman, also evolved from this course.
Who invented the finite state automaton?
Hopcroft and Ullman  
Chomsky  
McCulloch and Pitts  
Rabin and Scott 
Question 10 Explanation:
From the line “Rabin and Scott were mathematicians who developed a model
of a computer with a finite amount of memory. They called this model the finitestate
Automaton” it is clear that the answer is Rabin and Scott.
Question 11 
Using general knowledge about Computer Science and by reading the following paragraph carefully, answer the Questions 8  12 below:
Princeton asked me to develop a course in automata theory. Since there were no courses or books on the subject, I asked McCluskey to recommend some materials for a course on automata theory. He gave me a list of six papers and told me that the material would probably give students a good background in automata theory. McCluskey'S list included works by Warren McCulloch and Walter Pitts, Michael Rabin and Dana Scott, John Backus and Peter Naur, Noam Chomsky, Juris Hartmanis and RichardStearns, and, of course, Alan Turing.
In 1943 McCulloch and Pitts, working in neurophysiology, published a paper on a logical calculus for describing events in neuron nets. The paper had a notation for describing how these strings of zeros and ones combine in neurons to produce new strings of zeros and ones. This notation was subsequently developed into the language of regular expressions for describing sets of strings. Rabin and Scott were mathematicians who developed a model of a computer with a finite amount of memory. They called this model the finitestate automaton, and showed that the possible behaviors of finitestate automata were precisely those behaviors that could be described by the regular expressions that grew out of the work of McCulloch and Pitts. ...... The work of Hartmanis and Stearns attracted researchers and focused· attention on the topic of complexity. Among the more significant advances that resulted were the classification of the complexity of most major mathematical theories, the reducibility of many combinatorial problems, the concept of NPcompleteness, and a deeper understanding of concepts such as randomness. Also, through this course I met Jeffrey Ullman and Alfred Aho, with whom I subsequently collaborated for many years. Formal Languages and Automata Theory, which I wrote with Ullman, also evolved from this course.
What was the difficulty in offering the course on Finite Automata Theory at Princeton?
There were no textbooks for the subject  
There were no teachers for the subject  
There were no students for the subject  
None of the above 
Question 11 Explanation:
From the line”Since there were no courses or books on the subject” it is clear that the answer is option A.
Question 12 
Using general knowledge about Computer Science and by reading the following paragraph carefully, answer the Questions 8  12 below:
Princeton asked me to develop a course in automata theory. Since there were no courses or books on the subject, I asked McCluskey to recommend some materials for a course on automata theory. He gave me a list of six papers and told me that the material would probably give students a good background in automata theory. McCluskey'S list included works by Warren McCulloch and Walter Pitts, Michael Rabin and Dana Scott, John Backus and Peter Naur, Noam Chomsky, Juris Hartmanis and RichardStearns, and, of course, Alan Turing.
In 1943 McCulloch and Pitts, working in neurophysiology, published a paper on a logical calculus for describing events in neuron nets. The paper had a notation for describing how these strings of zeros and ones combine in neurons to produce new strings of zeros and ones. This notation was subsequently developed into the language of regular expressions for describing sets of strings. Rabin and Scott were mathematicians who developed a model of a computer with a finite amount of memory. They called this model the finitestate automaton, and showed that the possible behaviors of finitestate automata were precisely those behaviors that could be described by the regular expressions that grew out of the work of McCulloch and Pitts. ...... The work of Hartmanis and Stearns attracted researchers and focused· attention on the topic of complexity. Among the more significant advances that resulted were the classification of the complexity of most major mathematical theories, the reducibility of many combinatorial problems, the concept of NPcompleteness, and a deeper understanding of concepts such as randomness. Also, through this course I met Jeffrey Ullman and Alfred Aho, with whom I subsequently collaborated for many years. Formal Languages and Automata Theory, which I wrote with Ullman, also evolved from this course.
Which of the following did not grow out of the work by Hartmanis and Stearns?
Regular expressions  
Reducibility of combinatorial problems  
Classification of complexity  
NPCompleteness 
Question 12 Explanation:
From the line “The work of Hartmanis and Stearns attracted researchers
and focused· attention on the topic of complexity. Among the more significant advances
that resulted were the classification of the complexity of most major mathematical theories,
the reducibility of many combinatorial problems, the concept of NPcompleteness, and a
deeper understanding of concepts such as randomness” it is clear that regular expression was not grown out of the work by Hartmanis and Stearns.
Question 13 
Which plots show the relationship between two or three variables when comparing data
sets consisting of multiple observations?
Histograms  
Scatter Plots  
Probability Plots  
All the above 
Question 14 
Variability in groups of observations with widely differing means can be compared
using the following measure
Coefficient of variation  
Mean deviation
 
Measure of skewness  
None of the above 
Question 15 
Features / attributes of patterns, which can be measured, are called
Qualitative measure  
Data  
Variables  
All 
Question 16 
Find the sample variance, standard deviation and range of the following data: 572, 572, 573, 568, 569, 575, 565, 570
Variance =10, standard deviation = 3.162, range = 11  
Variance =13, standard deviation = 4.162, range = 10  
Variance =10, standard deviation = 3.162, range = 09  
Variance =10, standard deviation = 3.162, range = 10 
Question 17 
The president and treasurer are to be chosen from a student club consisting of 50
people. How many different choices of officers are possible if there are no restrictions?
2450  
2500  
1225  
1250 
Question 17 Explanation:
We are picking 2 out of a field of 50, and order does matter.So
answer is P(50,2) = 2450
answer is P(50,2) = 2450
Question 18 
In a college football training session, the defensive coordinator needs to have 10 players
standing in a row. Among these 10 players, there are 1 freshman, 2 sophomore, 4 juniors
and 3 seniors. How many different ways can they be arranged in a row if only their
class level will be distinguished?
14600  
12600  
12800  
None of the above

Question 18 Explanation:
Since only their class level will be distinguished. So 2 sophomore will be treated same, 4 juniors will be treated same, and 3 seniors will be treated same.
Hence the no. of ways they can be arranged in a row,
=12600
Hence the no. of ways they can be arranged in a row,
=12600
Question 19 
In how many ways can 5 different trees be planted in a circle?
24  
12  
6  
120 
Question 19 Explanation:
There are a total of 5! (=120) ways to plant five different trees. However, the trees are in a circle, in which each "way" is counted five times, due to rotations. This leaves us 5!/5 = 4! = 24 ways to plant five different trees in a circle.
Question 20 
If at least 85% of students in a class are good in Physics, at least 80% are good in
Computer Science and at least 75% are good in Mathematics, then the percentage of
students who are good in all the three subjects is at least
25  
40  
20  
60 
Question 21 
A family has two children. What is the probability that both the children are girls
given that at least one of them is a girl?
1/4  
1/2  
3/4  
1/3 
Question 21 Explanation:
Given that one of them is atleast girl. So the sample is
{girl, boy}, {girl, girl}
Now the probability that both are girls,
= 1/2
{girl, boy}, {girl, girl}
Now the probability that both are girls,
= 1/2
Question 22 
A person has 2 bank cards, each with a 4, digit number. The 1st number is 4 times the
2nd. The 1st number is the reverse of the 2nd. Which of these is the first number?
8421  
2178  
8712  
None of the above 
Question 22 Explanation:
Lets check option wise
A)8421 can never be the answer because anything multiplied by 4 is even no. and 8421 is odd no. So it cant be the answer.
B)1st no. is 2178 then 2nd no. will be 8712. And it is said that 1st no. is 4 times the 2nd no.
But we can see that 2178 != 4*8712
C)1st no. is 8712 then 2nd no. is 2178.And it is said that 1st no. is 4 times the 2nd no. And yes 8712 = 4*2178.Hence it is the answer.
A)8421 can never be the answer because anything multiplied by 4 is even no. and 8421 is odd no. So it cant be the answer.
B)1st no. is 2178 then 2nd no. will be 8712. And it is said that 1st no. is 4 times the 2nd no.
But we can see that 2178 != 4*8712
C)1st no. is 8712 then 2nd no. is 2178.And it is said that 1st no. is 4 times the 2nd no. And yes 8712 = 4*2178.Hence it is the answer.
Question 23 
A man passed 1/6th of his life in childhood, 1/12th as youth and 1/7th more as a
bachelor. Five years after his marriage, a son was born who died 4 years before his
father at half his fathers final age. What was the final age of the man?
48  
84  
96  
64 
Question 24 
Ram said, "When I am as old as my father is now, I shall be five times as old as my
son is now. By then, my son will be eight years older than I am now. The combined
ages of my father and myself are 100 years.". How old is Ram's son?
13  
15  
17  
None of the above 
Question 24 Explanation:
Let the age of Ram = n
Let the age of Ram’s father = x
Let the age of Ram’s son = y
Now ATQ,
x = 5y (1)
8 + n = y + (x  n) (2)
x + n = 100 (3)
Solving three equations, we get
y = 13
Let the age of Ram’s father = x
Let the age of Ram’s son = y
Now ATQ,
x = 5y (1)
8 + n = y + (x  n) (2)
x + n = 100 (3)
Solving three equations, we get
y = 13
Question 25 
If a flag has three horizontal stripes which can be colored out of five different colors,
how many flags can be constructed such that they are not identical and the adjacent
stripes are not of the same color?
15  
30  
45  
50 
Question 26 
A bag contains 6 white balls, 6 black balls and 8 green balls. What is the probability
of 3 balls which were drawn randomly of same colour?
3/1140  
20/1140  
56/1140  
96/1140 
Question 26 Explanation:
Question 27 
A person walks 8 KM towards east. He took a left turn and walks for 1 KM towards
north. He took right turn and walks towards the east for 7 KM. Now, he turns to right
and walks 9 KM towards south. Now, how much of distance; this person is away from
the starting point?
25 KM  
23 KM  
19 KM  
17 KM 
Question 27 Explanation:
Question 28 
Suppose today is Saturday, after 72 days what will be the day?
Saturday  
Sunday  
Monday  
Tuesday 
Question 28 Explanation:
Since there are 7 days in a week. So we will get remainder as 2 after dividing 72 by 7. So there are 2 odd days .Hence the required day will be 1st odd day will be saturday day and the next odd day will be sunday.Hence the answer is sunday.
Question 29 
A shopkeeper sold two toys at Rs. 120 each. While he made a profit of 20% on one,
he incurred a loss of 20% on other. Then, overall, he
made a profit of Rs. 10  
incurred a loss of Rs. 10  
incurred a loss of Rs. 12  
neither made profit nor incurred loss 
Question 29 Explanation:
CP of one toy in which he made 20% profit is = 120/1.2 = 100
CP of another toy in which he made 20% loss is = 120/0.8 = 150
Hence total cost price = 100 + 150 = 250
And total selling price = 240
So he incurred a loss of = 250  240 = Rs.10
CP of another toy in which he made 20% loss is = 120/0.8 = 150
Hence total cost price = 100 + 150 = 250
And total selling price = 240
So he incurred a loss of = 250  240 = Rs.10
Question 30 
Read the following text carefully, so as to answer the questions 3033
Five Companies A, B, C, D and E saw growth rates ranging from 10% to 50% in the year 2015. The company A with the least revenues of Rs. 600 crores in 2015 saw the maximum growth rate of 50% and the Company D with the highest revenue saw the least growth rate of 10%. Company B's revenues in 2016 was equal to that of Company D in 2015, while Company C's 2016 revenue was equal to that of Company B's in 2015, Company A's 2016 revenue was equal to that of Company E in 2015.
John, an accountant observes that one of the companies has twice the growth rate of another. Mathew, his colleague corrects him and says that this is the case in two different instances.
Company E has a revenue equal to the average seen in Company A and D, and growth rate equal to the average growth rate of A and D. Ram, known for his crypticspeak mentioned that if company C had grown at the rate seen by company A in 2015 would have reached the revenues seen by Company B in 2016.
What is the overall average growth rate seen by all 5 companies put together?
23.5%  
27%  
24.2%  
28.5% 
Question 31 
Read the following text carefully, so as to answer the questions 3033
Five Companies A, B, C, D and E saw growth rates ranging from 10% to 50% in the year 2015. The company A with the least revenues of Rs. 600 crores in 2015 saw the maximum growth rate of 50% and the Company D with the highest revenue saw the least growth rate of 10%. Company B's revenues in 2016 was equal to that of Company D in 2015, while Company C's 2016 revenue was equal to that of Company B's in 2015, Company A's 2016 revenue was equal to that of Company E in 2015.
John, an accountant observes that one of the companies has twice the growth rate of another. Mathew, his colleague corrects him and says that this is the case in two different instances.
Company E has a revenue equal to the average seen in Company A and D, and growth rate equal to the average growth rate of A and D. Ram, known for his crypticspeak mentioned that if company C had grown at the rate seen by company A in 2015 would have reached the revenues seen by Company B in 2016.
Which company had the third highest growth rate7
B  
C  
E  
D 
Question 32 
Read the following text carefully, so as to answer the questions 3033
Five Companies A, B, C, D and E saw growth rates ranging from 10% to 50% in the year 2015. The company A with the least revenues of Rs. 600 crores in 2015 saw the maximum growth rate of 50% and the Company D with the highest revenue saw the least growth rate of 10%. Company B's revenues in 2016 was equal to that of Company D in 2015, while Company C's 2016 revenue was equal to that of Company B's in 2015, Company A's 2016 revenue was equal to that of Company E in 2015.
John, an accountant observes that one of the companies has twice the growth rate of another. Mathew, his colleague corrects him and says that this is the case in two different instances.
Company E has a revenue equal to the average seen in Company A and D, and growth rate equal to the average growth rate of A and D. Ram, known for his crypticspeak mentioned that if company C had grown at the rate seen by company A in 2015 would have reached the revenues seen by Company B in 2016.
Which company had the median revenue in 20167
A  
B  
C  
E 
Question 33 
Read the following text carefully, so as to answer the questions 3033
Five Companies A, B, C, D and E saw growth rates ranging from 10% to 50% in the year 2015. The company A with the least revenues of Rs. 600 crores in 2015 saw the maximum growth rate of 50% and the Company D with the highest revenue saw the least growth rate of 10%. Company B's revenues in 2016 was equal to that of Company D in 2015, while Company C's 2016 revenue was equal to that of Company B's in 2015, Company A's 2016 revenue was equal to that of Company E in 2015.
John, an accountant observes that one of the companies has twice the growth rate of another. Mathew, his colleague corrects him and says that this is the case in two different instances.
Company E has a revenue equal to the average seen in Company A and D, and growth rate equal to the average growth rate of A and D. Ram, known for his crypticspeak mentioned that if company C had grown at the rate seen by company A in 2015 would have reached the revenues seen by Company B in 2016.
In absolute terms, which company added the maximum revenue in 20167
A  
B  
C  
E 
Question 34 
The following chart represents number of students of AMS careers at its
Hyderabad centre who passed the CAT, XAT, CET or none of these exams.
(Assume that there are no students who passed more than one exam).
Answer questions 3437 based upon the information provided in this chart.
What was the percentage of students who cleared CAT exam in 2000?
19.56%  
12.65%  
14.28%  
11.76%  
None of the above 
Question 34 Explanation:
Total no. of students who gave exam in 2000
= 20 + 80 + 140 + 170
= 410
No. of students who cleared CAT exam in 2000 = 20
∴ Required percentage =(20/420)x100=4.8%
Question 35 
The following chart represents number of students of AMS careers at its
Hyderabad centre who passed the CAT, XAT, CET or none of these exams.
(Assume that there are no students who passed more than one exam).
Answer questions 3437 based upon the information provided in this chart.
What was the percentage of students who succeeded in at least one of the three exams in 2000?
82.4%
 
82.8%  
82.35%  
83.3%  
None of the given option is correct

Question 35 Explanation:
Total no. of students who gave exam in 2000 = 410
No. of students who passed atleast one of the three exam = 240
∴ Required percentage =(240/420)x100=58.5%
No. of students who passed atleast one of the three exam = 240
∴ Required percentage =(240/420)x100=58.5%
Question 36 
The following chart represents number of students of AMS careers at its
Hyderabad centre who passed the CAT, XAT, CET or none of these exams.
(Assume that there are no students who passed more than one exam).
Answer questions 3437 based upon the information provided in this chart.
Which year showed the best results in CAT entrance exams for the institute (in terms of the percentage of students who cleared)?
2000  
2001  
2002  
Can't be determined 
Question 36 Explanation:
Total no. of students who gave exam in 2000 = 410
No. of students who cleared CAT exam in 2000 = 20
∴ Required percentage = (20/410)x100=4.8%
Total no. of students who gave exam in 2001
= 30 + 90 + 150 + 180
= 120 + 330
= 450
No. of students who cleared CAT exam in 2001 = 30
∴ Required percentage= (30/450)x100=6.6%
Total no. of students who gave exam in 2002
= 40 + 100 + 160 + 200
= 500
No. of students cleared CAT exam in 2002 = 40
∴ Required percentage= (40/500)x100=8%
Hence the best result was shown in 2002.
No. of students who cleared CAT exam in 2000 = 20
∴ Required percentage = (20/410)x100=4.8%
Total no. of students who gave exam in 2001
= 30 + 90 + 150 + 180
= 120 + 330
= 450
No. of students who cleared CAT exam in 2001 = 30
∴ Required percentage= (30/450)x100=6.6%
Total no. of students who gave exam in 2002
= 40 + 100 + 160 + 200
= 500
No. of students cleared CAT exam in 2002 = 40
∴ Required percentage= (40/500)x100=8%
Hence the best result was shown in 2002.
Question 37 
The following chart represents number of students of AMS careers at its
Hyderabad centre who passed the CAT, XAT, CET or none of these exams.
(Assume that there are no students who passed more than one exam).
Answer questions 3437 based upon the information provided in this chart.
What is the percentage increase in number of students in year 2002 over year 2000?
30%  
17.64%  
117.6%  
85%  
None of the given option is correct 
Question 37 Explanation:
Total no. of students in 2000 = 410
Total no. of students in 2002 = 500
∴ Increase percentage = (500410)/410
=(90/410) x 100
=21.9%
Total no. of students in 2002 = 500
∴ Increase percentage = (500410)/410
=(90/410) x 100
=21.9%
Question 38 
Answer questions 3840 based upon the following information.
A group of 7 people Salman, Shahrukh, Aamir, Ranbir, Imran, Shahid and Akshay are to be arranged in a row of 7 chairs (not necessarily in the same order), such that 2 adjacent chairs are facing opposite directions but not facing each other. Given below are some of the conditions to be followed for the seating arrangement:
a) Akshay sits in a chair whose direction is opposite to that of Imran
b) None of Salman, Shahrukh or Aamir can sit adjacent to each other
c) Ranbir and Shahid are best friends, so they always sit together
d) Imran has 4 people sitting to his right
e) Aamir is sitting 2 positions to the right of Ranbir
Which of the following can never occupy adjacent chairs?
Sharukh & Shahid  
Akshay & Imran  
Aamir & Imran  
Ranbir & Salman 
Question 39 
Answer questions 3840 based upon the following information.
A group of 7 people Salman, Shahrukh, Aamir, Ranbir, Imran, Shahid and Akshay are to be arranged in a row of 7 chairs (not necessarily in the same order), such that 2 adjacent chairs are facing opposite directions but not facing each other. Given below are some of the conditions to be followed for the seating arrangement:
a) Akshay sits in a chair whose direction is opposite to that of Imran
b) None of Salman, Shahrukh or Aamir can sit adjacent to each other
c) Ranbir and Shahid are best friends, so they always sit together
d) Imran has 4 people sitting to his right
e) Aamir is sitting 2 positions to the right of Ranbir
If Ranbir is 3 places to the right of Imran, then who is 2 places to the left of Akshay?
Either Shahrukh or Salman  
Aamir  
Either Aamir or Shahrukh  
None of the above 
Question 40 
Answer questions 3840 based upon the following information.
A group of 7 people Salman, Shahrukh, Aamir, Ranbir, Imran, Shahid and Akshay are to be arranged in a row of 7 chairs (not necessarily in the same order), such that 2 adjacent chairs are facing opposite directions but not facing each other. Given below are some of the conditions to be followed for the seating arrangement:
a) Akshay sits in a chair whose direction is opposite to that of Imran
b) None of Salman, Shahrukh or Aamir can sit adjacent to each other
c) Ranbir and Shahid are best friends, so they always sit together
d) Imran has 4 people sitting to his right
e) Aamir is sitting 2 positions to the right of Ranbir
If Akshay is 3 places to the left of Shahid, then who can occupy the corner positions (in any order)?
Salman and Aamir  
Shahrukh and Salman  
Shahrukh and Aamir  
None of the above 
Question 41 
When an IP datagram is received by a system, the following happens to the currently
running process:
Its state is changed to BLOCKED  
Its state is changed to READY  
Process is killed  
Process is suspended 
Question 42 
When an IP datagram is received by a system, the following happens to the currently
running process:
Its state is changed to BLOCKED  
Its state is changed to READY  
Process is killed  
Process is suspended 
Question 43 
In a uniprocessor system that is running a nonpreemptive kernel, which of the following statements is TRUE:
I. Deadlock never happens
II. Multithreading cannot be implemented on this system
III. Atomic instructions prevent mutual exclusion problems
I and III
 
I, II and III  
III only  
II and III 
Question 43 Explanation:
S1 is false because nonpreemption is one of the necessary condition for deadlock.So if rest of three condition for deadlock is satisfied then there will be a deadlock.
S2 is false because multithreading can be used in both preemptive and onopreemptive kernel.
S3 is true
S2 is false because multithreading can be used in both preemptive and onopreemptive kernel.
S3 is true
Question 44 
Which of the following statements is NOT TRUE for runtime binding?
Process cannot be relocated  
Process cannot be swapped out and into a different memory location  
Process execution is highly efficient  
Two process address spaces may have a conflict 
Question 45 
Which of the following statements is TRUE of Zombie processes?
I. Zombies are an issue primarily in server systems and not in clients
II. Zombies make the system run slower
III. Zombies are eliminated by inheritance by the init process
II and III  
II only  
I and III  
I, II and III 
Question 45 Explanation:
S1 is false
S2 is true
S3 is false because orphans processes are eliminated by inheritance by the init process and not the zombie process.
S2 is true
S3 is false because orphans processes are eliminated by inheritance by the init process and not the zombie process.
Question 46 
Which of the following is NOT possible?
TLB miss with no page fault  
TLB hit with no page fault  
TLB miss with page fault  
TLB hit with page fault 
Question 46 Explanation:
Whenever there is TLB hit ,there cant be page fault,because in TLB only those page table entries are present for which pages are present in main memory .Hence there can never be page fault on TLB hit.So option D is false.
Question 47 
Which of the following  client or server or both  does an active or passive open of
sockets?
Both can do passive open  
Both can do active open  
Clients can do only passive open  
Servers can do only passive open 
Question 47 Explanation:
A passive open is the creation of a listening socket, to accept incoming connections. It uses socket(), bind(), listen(), followed by an accept() loop. An active open is the creation of a connection to a listening port by a client.
Hence servers can only do passive opens.
Hence servers can only do passive opens.
Question 48 
Given the IP address 202.41.85.117/22, how many hosts can this network support?
1022  
1024  
512  
254 
Question 48 Explanation:
In the IP address 202.41.85.117/22 the host id is 3222=10 bits.Hence no. of hosts per network is 2^102 = 1022.We have subtracted two because one IP address is reserved for network id and other ip address is reserved for Directed broadcast address.
Question 49 
A router R1 receives the following advertisements from its neighbors when using RIPv1 that uses the distance vector algorithm as the routing protocol:
R2→ R1:: ((N1, 1), (N2, ))
R3→ R1:: ((N2, 1), (N3, 1))
Which of the following can we infer from these advertisements?
1. R2 and R3 are neighbors of each other
II. Counttoinfinity problem cannot occur in this network
III. Split horizon is enabled in RIPv1 implementation being used by R2 and R3
I, II and III are all TRUE  
I and II are TRUE  
II and III are TRUE  
Only I is TRUE 
Question 50 
A disk is highly errorprone especially in sectors which are heavily used. In such a disk
which of the following mechanisms for maintaining metadata of the files is WORSE?
Regular Linked Allocation Scheme  
Regular Indexed Allocation  
inode form of indexed allocation  
FAT form of linked allocation 
Question 51 
A computer has a 256 KByte, 4way set associative, write back data cache with block
size of 32 Bytes. The processor sends 32 bit addresses to the cache controller. Each
cache tag directory entry contains, in addition to address tag, 2 valid bits, 1 modified
bit and 1 replacement bit. The number of bits in the tag field of an address is
11  
14  
16  
27 
Question 51 Explanation:
32 bits addresses can be divided as,
Question 52 
A processor that has carry, overflow and sign flag bits as part of its program status
word (PSW) performs addition of the following two 2's complement numbers: 01001101
and 11101001. After the execution of this addition operation, the status of the carry,
overflow and sign flags, respectively will be:
1,1, 0  
0, 0, 0  
1,0, 0  
1,0,1 
Question 52 Explanation:
Question 53 
A RAM chip has a capacity of 1024 words of 8 bits each (IK x 8). The number of 2 x 4 decoders with enable line needed to construct a 16K x 16 RAM from lK x 8 RAM is
4  
5  
6  
7 
Question 53 Explanation:
Actually we need decodes for selection of row from address lines not data lines.
So we have to concentrate only at the first term of each of RAM size and chip size which are 16K and 1K.
Now, No. of address lines,
= 16K/1K
= 16
Hence we need a 4×16 decoder for this.
So no. of 2×4 decoders needed to create 4×16 decoder is,
4 line  1 decoder
1 line  ¼ decoder
16 lines  16/4 = 4 decoder
4 line  4/4 = 1 decoder
Hence total no. of 2×4 decoder needed is,
4+1 = 5
So we have to concentrate only at the first term of each of RAM size and chip size which are 16K and 1K.
Now, No. of address lines,
= 16K/1K
= 16
Hence we need a 4×16 decoder for this.
So no. of 2×4 decoders needed to create 4×16 decoder is,
4 line  1 decoder
1 line  ¼ decoder
16 lines  16/4 = 4 decoder
4 line  4/4 = 1 decoder
Hence total no. of 2×4 decoder needed is,
4+1 = 5
Question 54 
Consider the following sequence of microoperations.
MBR ← PC
MAR ← X
PC ← Y
Memory ← MBR
Which one of the following is a possible operation performed by this sequence?
Instruction fetch  
Operand fetch  
Conditional branch  
Initiation of interrupt service 
Question 54 Explanation:
The given sequence is nothing but the initiation of interrupt service.
Question 55 
Let R = {A, B, C, D, E, F} be a relation schema with C → F, E → A, EC → D and
A→ B. Which of the following is a key for R?
CD  
EC  
AE  
AC 
Question 55 Explanation:
ABDF is present at the right side of productions. So check for the CE
CE  CE // CE derives CE
CEF // C derives F
CEFA // E derives A
CEFAB // A derives B
CEFABD // EC derives D
So the key for R is “ EC”
(EC)^{+ }= { A,B,C,D,E,F}
(CD)^{+ }= {C,D,F}
(AE)^{+ } = {A,B,E}
(AC)^{+ } = {A,C,F,B}
CE  CE // CE derives CE
CEF // C derives F
CEFA // E derives A
CEFAB // A derives B
CEFABD // EC derives D
So the key for R is “ EC”
(EC)^{+ }= { A,B,C,D,E,F}
(CD)^{+ }= {C,D,F}
(AE)^{+ } = {A,B,E}
(AC)^{+ } = {A,C,F,B}
Question 56 
Which of the following statements is false?
Any relation with two attributes is in BCNF  
A relation in which every key has one attribute is in 2NF  
A prime attribute can be transitively dependent on a key in 3NF relation  
A prime attribute can be transitively dependent on a key in BCNF relation 
Question 56 Explanation:
Option A is true.
Option B is true because there cant be any partial functional dependencies.
Option C is true because the condition for 3NF is either left side of functional dependencies is super key or right side of functional dependencies is prime attribute.
Option D is false because the condition for BCNF is that left side of functional dependencies must be super key.
Option B is true because there cant be any partial functional dependencies.
Option C is true because the condition for 3NF is either left side of functional dependencies is super key or right side of functional dependencies is prime attribute.
Option D is false because the condition for BCNF is that left side of functional dependencies must be super key.
Question 57 
Relations produced from ER model will always be in
3NF  
BCNF  
4NF  
2NF 
Question 57 Explanation:
3NF is a database schema design approach for relational databases which uses normalizing principles to reduce the duplication of data, avoid data anomalies, ensure referential integrity, and simplify data management. It was defined in 1971 by Edgar F. Codd, an english computer scientist. Entity relation diagram is most widely used in data representation
Question 58 
Consider a schema R = {A, B, C, D} and functional dependencies A → B and C → D.
Then the decomposition R1 = {A, B} and R2 = {C, D} is
Dependency preserving but not lossless join  
Dependency preserving and lossless join  
Lossless join but not dependency preserving  
Neither lossless join nor dependency preserving 
Question 58 Explanation:
The given decomposition does not have any attribute in common, so not lossless decomposition.
The relation R1 covers FD A → B and relation R2 covers FD C → D.Hence the decomposition is functional dependency preserving.
The relation R1 covers FD A → B and relation R2 covers FD C → D.Hence the decomposition is functional dependency preserving.
Question 59 
How many subgraphs with at least one vertex does a labeled complete graph with 3
vertices have?
7  
10  
12  
17 
Question 59 Explanation:
No. of graphs with one verex = 1
No. of graphs with 2 vertces and no edges = 1
No. of graphs with 2 vertices and 1 edges = 1
No. of graphs with 3 vertices and 0 edges = 1
No. of graphs with 3 vertices and 1 edges = C(3,1) ,because since the vertices are labelled so all the edges will be cosidered different.Hence no. of ways of selecting one edges out of 3 edges is =C(3,1) =3
No. of graphs with 3 vertices and 2 edges = C(3,2) = 3
No. of graphs with 3 vertices and 3 edges = C(3,3)=1
Hence total no. of subgraphs possible is = 11.But since none of the option matches so you can remove the complete graph itself.So total no. of subgraphs is 111 = 10
No. of graphs with 2 vertces and no edges = 1
No. of graphs with 2 vertices and 1 edges = 1
No. of graphs with 3 vertices and 0 edges = 1
No. of graphs with 3 vertices and 1 edges = C(3,1) ,because since the vertices are labelled so all the edges will be cosidered different.Hence no. of ways of selecting one edges out of 3 edges is =C(3,1) =3
No. of graphs with 3 vertices and 2 edges = C(3,2) = 3
No. of graphs with 3 vertices and 3 edges = C(3,3)=1
Hence total no. of subgraphs possible is = 11.But since none of the option matches so you can remove the complete graph itself.So total no. of subgraphs is 111 = 10
Question 60 
The number of paths of length 3 between a single pair of two distinct vertices in a
complete graph with 4 vertices is
5  
6  
7  
8 
Question 61 
For a set S, let P(S) denote the power set of S, i.e., the set of all subsets of S. Suppose S1 and S2 are any two sets. Consider the following statements regarding S1 and S2
I. P(S1) U P(S2)=P(S1US2)
II. P(S1) n P(S2)= P(S1 n S2)
III. p(S1)= P(S2) <==> S1=S2
IV. P(0) = 0
Then
Only I and IV are true  
Only II and III are true  
Only I, II and III are true  
Only III and IV are true 
Question 61 Explanation:
I.
Let S1 = {a}
S2 = {b}
LHS:
P(S1) = {Ø, {a}}
P(S2) = {Ø, {b}}
P(S1) U P(S2) = {Ø, {a}, {b}}
RHS:
S1 U S2 = {a,b}
P(S1 U S2) = {Ø, {a}, {b}, {a,b}}
Hence, LHS ≠ RHS
So, False.
II.
Let S1 = {a,b}
S2 = {b,c}
LHS:
P(S1) = {Ø, {a}, {b},{a,b}}
P(S2) = {Ø, {b}, {c}, {b,c}}
P(S2) ⋂ P(S2) = {Ø, {b}}
RHS:
S1 ⋂ S2 = {b}
P(S1 ⋂ S2) = {Ø, {b}}
Hence, LHS = RHS
So, True.
III. Clearly, it is True.
IV. P(0) = {Ø, {0}}
Hence, False.
Let S1 = {a}
S2 = {b}
LHS:
P(S1) = {Ø, {a}}
P(S2) = {Ø, {b}}
P(S1) U P(S2) = {Ø, {a}, {b}}
RHS:
S1 U S2 = {a,b}
P(S1 U S2) = {Ø, {a}, {b}, {a,b}}
Hence, LHS ≠ RHS
So, False.
II.
Let S1 = {a,b}
S2 = {b,c}
LHS:
P(S1) = {Ø, {a}, {b},{a,b}}
P(S2) = {Ø, {b}, {c}, {b,c}}
P(S2) ⋂ P(S2) = {Ø, {b}}
RHS:
S1 ⋂ S2 = {b}
P(S1 ⋂ S2) = {Ø, {b}}
Hence, LHS = RHS
So, True.
III. Clearly, it is True.
IV. P(0) = {Ø, {0}}
Hence, False.
Question 62 
Consider that 135 cities in India are to be connected by highspeed fibre optic links.
Each link will connect a pair of cities so that the entire. network of 135 cities is connected.
The additional requirement is that the network will remain connected if any
single link fails. What is the minimum number of links needed to set up the network?
268  
9045  
270  
135 
Question 62 Explanation:
Total no. of minimum links required to make the network of cities connected is 135 in the cycle form.So even if one singe link get fails then also the network will remain connected.
Question 63 
Consider an unordered doubly linked list with n elements. Given a key, all the elements
less than the key are moved to the left of the key, and all the elements greater. than
the key are moved to the right of the key preserving the same order as in the original
list. What is the time complexity for doing this?
O(n)  
O(n^{2}))  
O(nlogn)  
None of the above 
Question 64 
Consider a weighted undirected graph G with positive edge weights. Let (u, v) be an
edge in the graph. It is known that the shortest path from a vertex s to u has weight
35 and the shortest path from s to v has weight 56. Which of the following is always
true?
Weight of (u, v) ≤ 21  
Weight of (u, v) = 21  
Weight of (u, v) ≥ 21
 
Nothing can be said about the weight of (u,v) 
Question 64 Explanation:
Question 65 
Consider a graph G with six vertices. Which of the following sequences can not represent
the degree of its vertices
3, 3, 3, 2, 2, 1  
3, 3, 3, 3, 2, 2  
3, 3, 3, 2, 2, 2  
3, 2, 2, 2, 2, 1 
Question 65 Explanation:
by applying Havel hakimi theorem we could easily find out the incorrect sequence.
Question 66 
Suppose there is a polynomial time reduction from problem A to problem B. Which
of the following can be inferred from this fact?
If the best algorithm for B takes exponential time, there is no polynomial time
algorithm for A.
 
If the best algorithm for A takes exponential time, there is no polynomial time
algorithm for B.
 
If we have a polynomial time algorithm for A, we must also have a polynomial
time algorithm for B.
 
If we don't know whether there is a polynomial time algorithm for B, there cannot
be a polynomial time algorithm for A.

Question 67 
Time complexity of 0/1 Knapsack Problem when there are n items and total weight
that can be carried in the knapsack is no more than some fixed number M is
O(M x n)  
O(n^{M})
 
O(M^{n})  
O(M +n) 
Question 67 Explanation:
Time complexity of 0/1 knapsack problem is O(M*n).
Question 68 
The source symbols are listed in order of decreasing probability, p = .4, q = .2, r = .2,
s = .1, t = .1. If a binary tree is generated using Huffman code greedy algorithm, with·
assigning 0 to every left edge and 1 to every right edge, then the average code length
is
3.0
 
2.4
 
2.2  
2.8 
Question 68 Explanation:
Question 69 
Let A be n x n real matrix and A^{3} = A, then A^{36} + 2A is
I +A^{2}  
A(A +2I)  
2I +A^{2}  
I +2A^{2} 
Question 69 Explanation:
Question 70 
Which algorithm for string matching preprocesses the pattern to find matches of
prefixes of the pattern with the pattern itself
KnuthMorrisPratt's algorithm  
Boyer Moore's algorithm  
Robin Karp's algorithm  
None of these 
Question 70 Explanation:
KnuthMorrisPratt algorithm uses prefixfunction (┌┐): It preprocesses the pattern to find matches of prefixes of the pattern with the pattern itself. It is defined as the size of the largest prefix of P[0.. j − 1] that is also a suffix of P[1.. j].
Question 71 
The best case performance of heap sort for sorting a given list of numbers into descending
order occurs if the given list is in
ascending order  
descending order  
random order  
all of the above 
Question 71 Explanation:
Whatever the order of the given list is ,the heap sort always takes O(nlogn) times.
Question 72 
Consider a node X in a binary tree. Given that X has two children, let Y be inorder
successor of X. Which of the following is true about Y?
Y has no right child  
Y has no left child  
Y has both children  
None of the above 
Question 72 Explanation:
Since in order successor of the binary search tree is the least element of the right sub tree. So the inorder successor cannot have left child since it itself is the smallest and having left child will violate it to be the minimum.
Question 73 
If a sorted list is provided, which is the best strategy to search for an element in the
list?
linear search  
binary search  
ternary search  
none of the above 
Question 73 Explanation:
Binary search is the best since it has the time complexity of O(logn).
Binary search is better than ternary search because the constant factor of binary search is less than ternary search.
Time Complexity for Binary search = 2clog2n + O(1)
Time Complexity for Ternary search = 4clog3n + O(1)
Therefore, the comparison of Ternary and Binary Searches boils down the comparison of expressions 2Log3n and Log2n . The value of 2Log3n can be written as (2 / Log23) * Log2n . Since the value of (2 / Log23) is more than one, Ternary Search does more comparisons than Binary Search in worst case.
And linear search have time complexity of O(n).
So binary search is best strategy in searching from given sorted list.
Binary search is better than ternary search because the constant factor of binary search is less than ternary search.
Time Complexity for Binary search = 2clog2n + O(1)
Time Complexity for Ternary search = 4clog3n + O(1)
Therefore, the comparison of Ternary and Binary Searches boils down the comparison of expressions 2Log3n and Log2n . The value of 2Log3n can be written as (2 / Log23) * Log2n . Since the value of (2 / Log23) is more than one, Ternary Search does more comparisons than Binary Search in worst case.
And linear search have time complexity of O(n).
So binary search is best strategy in searching from given sorted list.
Question 74 
You are given pointers to the first and the last nodes of a singly linked list, which of
the following operation is dependent on the length of the linked list?
Delete the first element.  
Insert a new element as the first element.  
Delete the last element of the list  
Add a new element at the end of the list. 
Question 74 Explanation:
Deletion of last element is dependent on the lenght of the list because to delete the last element of the list we need pointer to the previous node which is not there .Hence to get the second last node we have to travese the list from starting,hence dependent on the length of the list.
Question 75 
Let G be a Context Free Grammar in Chomsky Normal Form. To derive a string of
terminals of length ℓ using G, the number of productions to be used is
2ℓ1  
2ℓ  
2ℓ+ 1  
not fixed and depends on actual productions 
Question 75 Explanation:
For CNF no. of productions to be used is 2ℓ1 and for GNF it is just ℓ.
Question 76 
L1, L2 and L3 are three languages. If L1 and L2 are regular and if L1 = L2L3, then
L3 has to be regular  
L3 can never be regular  
L3 need not be regular  
L3 can never be a context free 
Question 76 Explanation:
If L1 is regular and L2 are regular then L3 need not be regular. For eg: L3=a^n b^n which is not regular but context free and let L2 = Φ which is regular,then L1=L2.L3 =Φ ,which is regular. Hence if L1 and L2 are regular then L3 need not be regular.
Question 77 
Consider the function f defined below:
struct item
{
int data;
struct item * next;
};
int f(struct item *p)
{
Return ((p == NULL) II (p>next == NULL) 
((p>data <= p>next>data) && f(p>next)));
}
For a given linked list p, the function f returns 1 if and only if
The elements in the list are sorted in nondecreasing order of data value.  
The clements in the list are sorted in nonincreasing order of data value.  
Not all elements in the list have the same value.
 
None of the above. 
Question 77 Explanation:
For a given linked list p, the function f returns 1 if and only if, the elements in the list are sorted in nondecreasing order of data value.
Question 78 
What is the output of following function when called with start pointing to the first node of the following singly linked list?
1→ 2 → 3 → 4 → 5 → 6
void fun(struct node* start){
if(start == NULL)
return;
printf ("%d ", start>data);
if(start>next != NULL)
fun(start>next>next);
printf("%d ", start>data);
135 531  
135  
1 23456
 
1 353 1

Question 78 Explanation:
The output will be the elements in odd positions in the same order given and then in the reverse order. So the output will be
135531.
135531.
Question 79 
Following is pseudo code of a function that takes a queue Q as an argument, and uses a stack S to do processing.
void fun (Queue *Q){
}
Stack S; // creates an empty stack S
// Run while Q is not empty
while (!isEmpty(Q)){
// Dequeue an item from Q and push the dequeued item to S
push(&S, deQueue(Q));
}
// Run while Stack S is not empty
while (!isEmpty(&S)){
// Pop an item from S and enqueue the popped item to Q
enQueue(Q, pop(&S));
}
What does the above function do in general?
Removes the last item from Q.  
Keeps the Q same as it was before the call.  
Makes Q empty  
Reverses the Q. 
Question 79 Explanation:
The given function reverse the given Q.
There are 79 questions to complete.