HCU PHD CS 2018 December
Question 1 
A pilgrim covers half of his journey by bus at 60 km/h, two thirds of the remainder by auto at 20 km/h and the rest by walk at 4 km/h. The average speed of the tourist in km/h during his entire journey is
12  
15  
20  
25  
24 
Question 1 Explanation:
Let ‘x’ be the total distance,
Speed = distance/ time
Total speed = (total distance)/(total time); Time =distance/speed
Time taken for train journey = (x⁄2)/60kmph=x/120
Time taken for bus journey = (x⁄4)/30kmph=x/120
Time taken for cycle journey = (x⁄4)/10kmph=x/40
Total time = x/120+x/120+x/40=5x/120
Total speed = (total distance)/(total time)=x/(5x⁄120)=(120×x)/5x=24kmph
Question 2 
The number of divisors of 2022 is
6  
7  
8  
3 
Question 2 Explanation:
The number of divisors are 8.
1, 2, 3, 6, 337, 674, 1011 and 2022
1, 2, 3, 6, 337, 674, 1011 and 2022
Question 3 
The product of the ages of ravi and ramu is 260. If twice the age Ravi is more than ram's age by 6 years, what is Ravi's age?
16  
14  
12  
13 
Question 3 Explanation:
B ⟶ A
A ⟶ C
BC ⟶ D
AC ⟶ BE
B^{+ }= BACDE
A^{+} = ACBED
So A & B are candidate keys.
There is no partial dependency so in 2NF.
But in the BC⟶D, neither BC is key nor D is prime attribute hence not in 3NF.
A ⟶ C
BC ⟶ D
AC ⟶ BE
B^{+ }= BACDE
A^{+} = ACBED
So A & B are candidate keys.
There is no partial dependency so in 2NF.
But in the BC⟶D, neither BC is key nor D is prime attribute hence not in 3NF.
Question 4 
The value of the expression 12^{112}(mod 17)
1  
4  
13  
16 
Question 4 Explanation:
12^{112}(mod 17)=1.
Question 5 
How many positive integers less than 1000 are coprime with 14?
571  
142  
429  
None of the these 
Question 5 Explanation:
14 can be written as 7×2. So any no. which is coprime to 14 should not be divisible by 7 nor divisible by 2.
Let’s first find no. of numbers divisible by 7 or 2.
Let the no. of numbers divisible by 7 is A, and no. of numbers divisible by 2 is B.
Now no. numbers divisible by 7,
A = 999/7 = 142
No. of numbers divisible by 2 is,
B = 999/2 = 499
No. of numbers divisible by both 7 and 2, i.e., 14
A∩B = 999/14 = 71
∴ No. of numbers divisible by 7 or 2 is
A∪B = A + B  A∩B = 142 + 499  71 = 570
∴ No. of numbers that are coprime to 14 are,
999  570 = 429
Let’s first find no. of numbers divisible by 7 or 2.
Let the no. of numbers divisible by 7 is A, and no. of numbers divisible by 2 is B.
Now no. numbers divisible by 7,
A = 999/7 = 142
No. of numbers divisible by 2 is,
B = 999/2 = 499
No. of numbers divisible by both 7 and 2, i.e., 14
A∩B = 999/14 = 71
∴ No. of numbers divisible by 7 or 2 is
A∪B = A + B  A∩B = 142 + 499  71 = 570
∴ No. of numbers that are coprime to 14 are,
999  570 = 429
Question 6 
There are 25 horses which you need to find out the fastest 3 horses. You can conduct race among at most 5 to find out their relative speed. At no point you can find out the actual speed of the horse in a race. find out how many races are required to get the top 3 horses.
5  
6  
7  
8 
Question 6 Explanation:
Divide all the 25 horses into 5 groups of five each .Then conduct races between the horses of each group and choose the fastest horse of each group.Till now 5 races are conducted. NOw we left with a total of 5 fastest horses and then at last conducted one more race between the remaining horses and chose the three fastest. So this is how three fastest horses are selected with a total 5 races.
Question 7 
A cube of side 1 unit is placed in such a way that the origin coincides with one of its vertices and the three(positive) axes along three of its edges. What are the coordinates of the vertex which is diagonally opposite to the vertex whose coordinates are (1,0,1)?
(0,0,0)  
(1,1,0)  
(0,1,0)
 
(1,1,1) 
Question 7 Explanation:
∴ The coordinates of the vertex which is diagonally opposite to the vertex whose coordinates are (1,0,1) is (0,1,0).
Question 8 
If pqr ≠ 0 and p^{x} =1/q, q^{y }=1/r, r^{z }=1/p, what is the value of the product xyz?
1  
1/pqr  
1  
pqr 
Question 8 Explanation:
pqr≠0
→ p^{x} = 1/q ⟹ 1/p^{x} = 1/q
⟹ q = p^{x}
⟹ log q = log p^{x}
⟹ x log p = log q
⟹ x = log q/ log p
→ q^{y} = 1/r ⟹ 1/q^{y} = 1/r
⟹ q^{y} = r
⟹ y log q = log r
⟹ y = log r/ log q
→ r^{z} = 1/p ⟹ 1/r^{z} = 1/p
⟹ p = r^{z}
⟹ log r^{z} = log p
⟹ z log r = log p
⟹ z = log p/ log r
XYZ = log q/ log p * log r/ log q * log p/ log r =1
→ p^{x} = 1/q ⟹ 1/p^{x} = 1/q
⟹ q = p^{x}
⟹ log q = log p^{x}
⟹ x log p = log q
⟹ x = log q/ log p
→ q^{y} = 1/r ⟹ 1/q^{y} = 1/r
⟹ q^{y} = r
⟹ y log q = log r
⟹ y = log r/ log q
→ r^{z} = 1/p ⟹ 1/r^{z} = 1/p
⟹ p = r^{z}
⟹ log r^{z} = log p
⟹ z log r = log p
⟹ z = log p/ log r
XYZ = log q/ log p * log r/ log q * log p/ log r =1
Question 9 
The minimum number of cards to e dealt from an arbitrarily shuffled deck of 52 cards to guarantee that three cards are with same number is
14  
27  
30  
39  
None of the above 
Question 9 Explanation:
We know that in a deck of 52 cards there are 4 suits.
Now to guarantee that three cards are from same suit,
⇒ Let's first draw 4 cards of different sizes.
⇒ Now again draw 4 cards of different suits. Now we have 8 cards of 2 cards from each suit.
⇒ Now take 1 card from any suit, which will guarantee that 3 cards are from the same suit.
So, in total 9 cards are required.
Now to guarantee that three cards are from same suit,
⇒ Let's first draw 4 cards of different sizes.
⇒ Now again draw 4 cards of different suits. Now we have 8 cards of 2 cards from each suit.
⇒ Now take 1 card from any suit, which will guarantee that 3 cards are from the same suit.
So, in total 9 cards are required.
Question 10 
A container originally contains 10 litres of pure spirit. From this container 1 litre of spirit is replaced with 1 litre of water. Subsequently, 1 litre of the mixture is again replaced with 1 litre of water and this process is replaced is repeated one more time. how much spirit is now left in the container?
7 litre  
7.21 litres  
7.29 litres  
7.31 litres

Question 10 Explanation:
10(729/1000×1=7.29 litres)
Question 11 
Please answer the following questions Q1113 based on the flowchart given below:
Note: Here T0 represents the least significant bit; C,S,T are stored in consecutive 9 bits where each variable can be accessed independently and shift operation is performed on the 9 bits representing C,S,T.
If the values of P and T are 1011 and 1101, respectively, then the contents of S and T at the end of execution are
S=101, T=1111  
S=1000, T=1111  
S=1010, T=1110  
S=1010, T=1100

Question 12 
Please answer the following questions Q1113 based on the flowchart given below:
Note: Here T0 represents the least significant bit; C,S,T are stored in consecutive 9 bits where each variable can be accessed independently and shift operation is performed on the 9 bits representing C,S,T.
What kind of operation the flowchart performs?
unsigned binary Division  
Signed binary Division  
Unsigned binary multiplication  
Signed binary Multiplication 
Question 13 
Please answer the following questions Q1113 based on the flowchart given below:
Note: Here T0 represents the least significant bit; C,S,T are stored in consecutive 9 bits where each variable can be accessed independently and shift operation is performed on the 9 bits representing C,S,T.
Given that T=1010, the number of shift right operations performed is?
1  
2  
3  
4 
Question 14 
What would be the smallest natural number which when divided by any of the numbers 20,44 and 76 leaves a remainder 7 in each case is,
4187  
6047  
7987  
63847 
Question 14 Explanation:
Find the LCM of 20, 42 and 76.
20 = 2x2x5
42 = 2x3x7
76 = 2x2x19
The LCM is 2x2x3x5x7x19 = 7980.
Next add 7 to 7980 to get 7987 as the answer.
20 = 2x2x5
42 = 2x3x7
76 = 2x2x19
The LCM is 2x2x3x5x7x19 = 7980.
Next add 7 to 7980 to get 7987 as the answer.
Question 15 
25 people are in a room. 15 of them play hockey, 17 of them play football, 5 of them play chess, 13 of them play both hockey and football, 2 of them play both chess hockey, 2 of them play both football and chess and none play all three games. then the number of persons who play neither hockey nor football nor chess is:
4  
5  
7  
8 
Question 15 Explanation:
Let A be the people who play hockey.
Let B be the people who play football.
Let C be the people who play chess.
Now, according to question
A = 15
B = 17
C = 5
A ∩ B = 13
A ∩ C = 2
B ∩ C = 2
A ∩ B ∩ C =0
And we have to find,
A’ ∩ B’ ∩ C’ = (A∪B∪C)’ = 25  (A∪B∪C)
Now let’s first find A∪B∪C,
A∪B∪C = A + B + C  (A∩B)  (B∩C)  (C∩A) + (A∩B∩C)
= 15 + 17 + 5  13  2  2 + 0
= 37  17
= 20
∴ The answer will be,
25  20 = 5
Let B be the people who play football.
Let C be the people who play chess.
Now, according to question
A = 15
B = 17
C = 5
A ∩ B = 13
A ∩ C = 2
B ∩ C = 2
A ∩ B ∩ C =0
And we have to find,
A’ ∩ B’ ∩ C’ = (A∪B∪C)’ = 25  (A∪B∪C)
Now let’s first find A∪B∪C,
A∪B∪C = A + B + C  (A∩B)  (B∩C)  (C∩A) + (A∩B∩C)
= 15 + 17 + 5  13  2  2 + 0
= 37  17
= 20
∴ The answer will be,
25  20 = 5
Question 16 
There are 5 bags labeled 1 to 5. All the coins in a given bag have the same weight. Some bags have coins of weight 10 gm, others have coins of weight 11 gm. 1 pick 1,2,4,8,16 coins respectively from bags 1 to 5. Their total weight comes out to 323 gm. Then the labels of the bags having 11 gm coins is
1,2,5  
2,3,4  
1,3,4  
2,3,5 
Question 16 Explanation:
Bags: 1 2 3 4 5
No. of coins picked: 1 2 4 8 16
There are two types of weights of coin, i.e., 10gm and 11gm. And the total weight of picked coins is 323gm.
Let there be x number of 10gm coins and y number of 11gm coins.
So, 10x + 11y = 323  (1)
And we also know that total number of coins picked is
1 + 2 + 4 + 8 + 16 = 31 which is equal to x + y, so,
= 31  (2)
Solving equation (1) and (2), we get
y = 13
Means total there are 13 coins of 11gm.
Now we can choose bag number 1, 3 and 4, we will get a total sum of 13 coins picked from them.
So product of labelled bag number = 1×3×4 = 12
No. of coins picked: 1 2 4 8 16
There are two types of weights of coin, i.e., 10gm and 11gm. And the total weight of picked coins is 323gm.
Let there be x number of 10gm coins and y number of 11gm coins.
So, 10x + 11y = 323  (1)
And we also know that total number of coins picked is
1 + 2 + 4 + 8 + 16 = 31 which is equal to x + y, so,
= 31  (2)
Solving equation (1) and (2), we get
y = 13
Means total there are 13 coins of 11gm.
Now we can choose bag number 1, 3 and 4, we will get a total sum of 13 coins picked from them.
So product of labelled bag number = 1×3×4 = 12
Question 17 
Two parallel chords of length 30cm and 16cm are drawn on the opposite sides of the center of a circle with radius 17cm. The distance between the chords is
23cm  
21cm  
19cm  
none of the above 
Question 17 Explanation:
A = distance of chord of length 30cm from the centre
B = distance of chord of length 16cm from the centre
Now,
Question 18 
In the year 1980, the age (in years) of a person was 1/89th of his year of birth. What is the age (in years) of this person in 2018?
54  
64  
80  
60 
Question 18 Explanation:
Let the year of birth of a person be x, then according to the question,
Question 19 
Read the passage carefully and answer Questions Q1922.
The problem of heuristic programmingof making computers solve really difficult problems are divided into five main areas. Search, pattern recognition, Learning, Planning, and Induction. Whenever appropriate, the discussion is supported by extensive citation of the literature and by descriptions of a few of the most successful heuristic (Problemsolving) programs constructed to date.
The adjective "heuristic," as used here and widely in the literature, means related to improving problemsolving performance; as a noun it is also used in regard to any method or trick used to improve the efficiency of a problemsolving system. A "heuristic program," to be considered successful, must work well on a variety of problems, and may often be executed if it fails on some. We often find it worthwhile to introduce a heuristic method, which happens to cause occasional failures, if there is an overall improvement in performance. But imperfect methods are not necessarily heuristic, nor vice versa. Hence “heuristic” should not be regarded as opposite to “foolproof”; this has caused some confusion in the literature.
What is heuristic programming?
Making computers solve hard problems  
Making computers do patterns recognition  
Making computers solve problems by searching for a solution  
Programming a unique solution to a hard problems 
Question 20 
Read the passage carefully and answer Questions Q1922.
The problem of heuristic programmingof making computers solve really difficult problems are divided into five main areas. Search, pattern recognition, Learning, Planning, and Induction. Whenever appropriate, the discussion is supported by extensive citation of the literature and by descriptions of a few of the most successful heuristic (Problemsolving) programs constructed to date.
The adjective "heuristic," as used here and widely in the literature, means related to improving problemsolving performance; as a noun it is also used in regard to any method or trick used to improve the efficiency of a problemsolving system. A "heuristic program," to be considered successful, must work well on a variety of problems, and may often be executed if it fails on some. We often find it worthwhile to introduce a heuristic method, which happens to cause occasional failures, if there is an overall improvement in performance. But imperfect methods are not necessarily heuristic, nor vice versa. Hence “heuristic” should not be regarded as opposite to “foolproof”; this has caused some confusion in the literature.
Which of the following is NOT a feature of a heuristic program?
Work well on a large number of problems  
It is “foolproof”  
Does not matter if it fails on certain instances of a problem  
Causes overall performance improvement 
Question 21 
Read the passage carefully and answer Questions Q1922.
The problem of heuristic programmingof making computers solve really difficult problems are divided into five main areas. Search, pattern recognition, Learning, Planning, and Induction. Whenever appropriate, the discussion is supported by extensive citation of the literature and by descriptions of a few of the most successful heuristic (Problemsolving) programs constructed to date.
The adjective "heuristic," as used here and widely in the literature, means related to improving problemsolving performance; as a noun it is also used in regard to any method or trick used to improve the efficiency of a problemsolving system. A "heuristic program," to be considered successful, must work well on a variety of problems, and may often be executed if it fails on some. We often find it worthwhile to introduce a heuristic method, which happens to cause occasional failures, if there is an overall improvement in performance. But imperfect methods are not necessarily heuristic, nor vice versa. Hence “heuristic” should not be regarded as opposite to “foolproof”; this has caused some confusion in the literature.
To what does the boldface text, the discussion, in Paragraph I refer?
Extremely hard problems  
Development of “foolproof” solutions  
Improving algorithm performance  
Heuristic programming 
Question 22 
Read the passage carefully and answer Questions Q1922.
The problem of heuristic programmingof making computers solve really difficult problems are divided into five main areas. Search, pattern recognition, Learning, Planning, and Induction. Whenever appropriate, the discussion is supported by extensive citation of the literature and by descriptions of a few of the most successful heuristic (Problemsolving) programs constructed to date.
The adjective "heuristic," as used here and widely in the literature, means related to improving problemsolving performance; as a noun it is also used in regard to any method or trick used to improve the efficiency of a problemsolving system. A "heuristic program," to be considered successful, must work well on a variety of problems, and may often be executed if it fails on some. We often find it worthwhile to introduce a heuristic method, which happens to cause occasional failures, if there is an overall improvement in performance. But imperfect methods are not necessarily heuristic, nor vice versa. Hence “heuristic” should not be regarded as opposite to “foolproof”; this has caused some confusion in the literature.
What, according to the author, are divided into five main areas?
Discussions on using computers to solve problems  
Problems that require the use of heuristics  
Descriptions of the most successful heuristic programs  
Tricks to improve problem solving performance 
Question 23 
Which of the following equations has the greatest number of real solutions?
x^{2 }+ 5x7=x+8  
x^{3}=10x  
7x+5=13x  
e^{x}=x 
Question 24 
In a class, chumley, peter, kennel, Donald, and Sentil are sitting on a bench. Chumley is sitting next to Peter, Kennel is sitting next to Donald. Donald is not sitting with Senthil. If Chumley and Senthil sit on either end of the bench where does Peter sit?
Between Chumley and Donald  
Between Donald and Senthil  
Between chumley and Kennel  
Between Kennel and Senthil 
Question 24 Explanation:
Let
C = Chumley
P = Peter
K = Kennel
D = Donald
S = Sentil
Now, according to question
S K D P C
So clearly Peter is sitting between Donals and Chumley.
C = Chumley
P = Peter
K = Kennel
D = Donald
S = Sentil
Now, according to question
S K D P C
So clearly Peter is sitting between Donals and Chumley.
Question 25 
In an archery match, Peter’s team got more score than David’s team but not as many as Smith’s team. Smith’s team got more scores than Taiwa’s team. Taiwa’s team got less score than David’s team. Which team is in second place in descending order of scores?
Smith’s team  
Taiwa’s team  
David’s team  
Peter’s team 
Question 25 Explanation:
Let,
P = Peter’s team
D = Davis’s team
S = Smith’s team
T = Taiwis team
Now, according to question
P > D
P < S
S > T
T < D
So looking at above four relation we can arrange them is descending order as,
S > P > D > T
So the team in the second place is Peter’s team
P = Peter’s team
D = Davis’s team
S = Smith’s team
T = Taiwis team
Now, according to question
P > D
P < S
S > T
T < D
So looking at above four relation we can arrange them is descending order as,
S > P > D > T
So the team in the second place is Peter’s team
Question 26 
Use the following table to answer Questions 2628
Which year had the maximum pass percentages?
2014  
2015  
2016  
2017 
Question 26 Explanation:
For 2014,
No. of students passed in PT = 0.45 × 2000 = 900
No. of students passed in FT = 0.05 × 200 = 10
∴ Total pass percentage = 910/2200 × 100 = 41.36%
For 2015,
No. of students passed in PT = 0.4 × 1600 = 640
No. of students passed in FT = 400 × 0.4 = 16
∴ Total pass percentage = 800/2000 × 100 = 40%
For 2016,
No. of students passed in PT = 0.4 × 1200 = 480
No. of students passed in FT = 400 × 0.7 = 280
∴ Total pass percentage = 760/1600 × 100 = 47.5%
For 2017,
No. of students passed in PT = 0.5 × 1000 = 500
No. of students passed in FT = 500 × 0.35 = 175
∴ Total pass percentage = 675/1500 × 100 = 45%
∴ 2016 had maximum pass percentage.
No. of students passed in PT = 0.45 × 2000 = 900
No. of students passed in FT = 0.05 × 200 = 10
∴ Total pass percentage = 910/2200 × 100 = 41.36%
For 2015,
No. of students passed in PT = 0.4 × 1600 = 640
No. of students passed in FT = 400 × 0.4 = 16
∴ Total pass percentage = 800/2000 × 100 = 40%
For 2016,
No. of students passed in PT = 0.4 × 1200 = 480
No. of students passed in FT = 400 × 0.7 = 280
∴ Total pass percentage = 760/1600 × 100 = 47.5%
For 2017,
No. of students passed in PT = 0.5 × 1000 = 500
No. of students passed in FT = 500 × 0.35 = 175
∴ Total pass percentage = 675/1500 × 100 = 45%
∴ 2016 had maximum pass percentage.
Question 27 
Use the following table to answer Questions 2628
Which year had the maximum number of students passing?
2014  
2015  
2016  
2017 
Question 27 Explanation:
From the previous solution we can see that maximum no. of students passed in 2014 with a no. 910.
Question 28 
Use the following table to answer Questions 2628
Which year had the minimum number of students failing?
2014  
2015  
2016  
2017 
Question 28 Explanation:
Extending the solution from the previous solution.
No. of students failed in 2014 = 2200910 = 1290
No. of students failed in 2015 = 2000800 = 1200
No. of students failed in 2016 = 1600760 = 840
No. of students passed in 2017 = 1500  675 = 825
Therefore minimum no. of students failed in the year 2017
No. of students failed in 2014 = 2200910 = 1290
No. of students failed in 2015 = 2000800 = 1200
No. of students failed in 2016 = 1600760 = 840
No. of students passed in 2017 = 1500  675 = 825
Therefore minimum no. of students failed in the year 2017
Question 29 
If QKKQUGQL is the code of OMISSION, which word is coded as RYVIWZB?
PATKUBZ  
BZWIVYR  
BZWVIYR  
PTAKBZU 
Question 29 Explanation:
Question 30 
In a class of 90 students, where the number of girls is twice that of boys, Shridhar ranked fourteenth from the top. If there are 10 girls ahead of Shridhar, how many boys are after him in rank?
22  
23  
25  
26 
Question 30 Explanation:
No of boys = x; No of girls = 2x;
x+2x = 90 => 3x = 90
x (Boys)= 30 ; 2x(Girls) = 60
Number of students behind Shridhar = 90 – 14 = 76
No of girls behind Shridhar = 60 – 10 = 50
No of boys behind Shridhar = 76 – 50 = 26
x+2x = 90 => 3x = 90
x (Boys)= 30 ; 2x(Girls) = 60
Number of students behind Shridhar = 90 – 14 = 76
No of girls behind Shridhar = 60 – 10 = 50
No of boys behind Shridhar = 76 – 50 = 26
Question 31 
The average age of a group of 10 students is 14. The average age increases by 1 year when two new students join the group. What is the average age of the two new students who joined the group?
15  
20  
21  
40 
Question 31 Explanation:
Let the total age of 10 students be A.
And let the age of the new students be x1and x2.
Now according to question,
A/10 = 14
A = 140
And, according to question,
And let the age of the new students be x1and x2.
Now according to question,
A/10 = 14
A = 140
And, according to question,
Question 32 
If a constant 1 is added to all the samples of a set of observations, which of the following statements is TRUE?
While all the measures of central tendency of the set remain the same, all the measures of dispersion change.  
While mode of the set change, mean and median of the set remain the same.  
While Mean and median of the set remain the same, range and standard Deviation of the change.  
While Mean and median of the set change, Range and standard Deviation of the set remain the same. 
Question 32 Explanation:
If a constant 1 is added to all the samples of a set of observations then the mean will increase by 1 and the median will also change but the range and standard deviation will remain the same.
Question 33 
A set of alphabet consisting of n (n even) distinct symbols is used to construct a string of length m ≤ n. If m=n/2, then how many strings can be constructed when no repetitions are allowed.
(n)(n1)(n2)...((n/2)+1)  
n!  
n(n+1)
 
n^{2} 
Question 33 Explanation:
Question 34 
When the curves Y=X^{3} and y=X^{5 }are drawn in the XY plane, how many times do they intersect?
1  
2  
3  
5 
Question 35 
A young boy counted in the following way on the fingers of his left hand.he started by calling the thumb 1, the index finger 2, middle finger 3, ring finger 4, little finger 5, then reversed direction, calling the ring finger 6, middle finger 7, index finger 8, thumb 9, then back to the index finger for 10, middle finger for 11, and so on. He counted up to 1954; he ended up on his:
Thumb  
Index Finger  
Middle Finger
 
Ring Finger 
Question 36 
How many real solutions does the equation 2^{x}  2x=0 have?
No solutions
 
Infinite solutions
 
One  
Two 
Question 37 
Let f(x)=x^{2} sin 1/x^{2} for x > 0. Which of the following is a correct statement
f is unbounded  
f is bounded by lim_{z→ ∞} f(x) does not exist  
lim_{z→ ∞} f(x) = 0
 
lim_{z→ ∞ }f(x) = 1 
Question 37 Explanation:
Question 38 
The equation of the circle that passes through the points (1,0) and (0,1) having the smallest radius is
x^{2} + y^{2} =1  
x^{2} + y^{2} xy=0  
x^{2} + y^{2} 2x 2y+1=0
 
All of the above 
Question 38 Explanation:
Circle whose diameter endpoints are (1, 0) and (0, 1) will be of smallest radius.
⇒ (x  1)(x  0) + (y  0)(y  1) = 0
⇒ x^{2 }+ y^{2}  x  y = 0
⇒ (x  1)(x  0) + (y  0)(y  1) = 0
⇒ x^{2 }+ y^{2}  x  y = 0
Question 39 
The minimum value of x^{2} + y^{2} subject to x+y=1 is
½  
1/√2  
¼  
1 
Question 39 Explanation:
=1/2
Question 40 
The postfix form for the prefix expression A/B*C+DE is
ABCDE+*/  
ABC*/DE+  
ABC/DE+*  
None of the above 
Question 40 Explanation:
The given prefix expression is,
 A/B*C+DE
Let’s draw binary tree from given prefix expression,
Now traversing the tree in the form (Left, right, root), to get the postfix expression.
Hence the postfix expression is,
A B C D E + * / 
 A/B*C+DE
Let’s draw binary tree from given prefix expression,
Now traversing the tree in the form (Left, right, root), to get the postfix expression.
Hence the postfix expression is,
A B C D E + * / 
Question 41 
A full binary tree with n leaf nodes contains in total
2^{n} 1 nodes  
2n 1 nodes
 
2n +1 nodes  
n^{2 }1 nodes

Question 41 Explanation:
A full binary tree is a tree in which each node has either 0 children or two children. And total no. of nodes in such a tree with n leaf nodes is 2n1.
Question 42 
For a zombie process, which of the following choices is more appropriate?
Process which has completed its execution by exit() system call but still has an entry in Process table
 
The process in terminated state  
Both (A) and (B) are true  
Only (A) is true

Question 42 Explanation:
A zombie process or defunct process is a process that has completed execution (via the exit system call) but still has an entry in the process table: it is a process in the "Terminated state".
Question 43 
Let the page fault service time be 10ms in a computer with average memory access time being 20ns. If one page fault is generated for every 106 memory access, what is the closest effective access time for the memory?
21ns  
30ns  
23ns  
35ns 
Question 43 Explanation:
P = page fault rate
EA = p × page fault service time + (1 – p) × Memory access time
=1/10^{6}×10×106+(11/106)×20 ≅29.9 ns
EA = p × page fault service time + (1 – p) × Memory access time
=1/10^{6}×10×106+(11/106)×20 ≅29.9 ns
Question 44 
Consider the methods used by processes P1 and P2 for accessing their critical sections whenever needed, as given below. The initial values of shared boolean variables S1 and S2 are randomly assigned.
Which one of the following statements describes the properties achieved?
Mutual exclusion but not progress  
Progress but not mutual exclusion  
Neither mutual exclusion nor progress
 
Both mutual exclusion and progress

Question 44 Explanation:
In this mutual exclusion is satisfied because at any point of time either S1 = S2 or S1 ≠ S2, but not both. But here progress is not satisfied because suppose S1 = 1 and S2 = 0 and P1 is not interested to enter into critical section but P2 wants to enter into critical section, and P2 will not be able to enter, because until P1 will not enter critical section, S1 will not become equal to S2. So if one process do not interested in entering critical section, will not allow other process to enter critical section which is interested. So progress is not satisfied.
Question 45 
A process executes the following code
for(i=0; i
The total number of child processes created is
n  
2^{n} 1  
2^{n}  
2^{n+1} 1 
Question 45 Explanation:
Fork is a system call, implements in kernel.
It is an operation where process creates a copy of itself.
1,3,7,15,31,... ⇒ 2^{n}1
It is an operation where process creates a copy of itself.
1,3,7,15,31,... ⇒ 2^{n}1
Question 46 
A digital signaling system is required to operate at 9600bps. If a signal element encodes a 4bit word, what is the minimum required bandwidth of the channel?
1200 Hz  
1400 Hz
 
1600 Hz  
1900 Hz 
Question 47 
Calculate the polynomial code checksum (CRC) for a frame 1101011011 using the generator
G(x)=x^{4} + x + 1
1111  
1110  
1101  
1100 
Question 47 Explanation:
Question 48 
This elementary problem begins to explore propagation delay and transmission delay, two central concepts in data networking. Consider two hosts, A and b, connected by a single link of rate R bps. Suppose that the two hosts are separated by m meters, and suppose the propagation speed along the link is s meters/sec. Host A is to send a packet of size L bits to Host B. Express the propagation delay, d_{prop}, in terms of m and s in seconds
m/s
 
(ms)  
m*s  
m+s

Question 48 Explanation:
d_{prop}= m/s seconds.
Question 49 
Which of the OSI layers handles each of the following:
(i) Dividing the transmitted bit stream into frames
(ii) Determining which route through the subnet to use
(i) Data link layer. (ii) network layer  
(i) Data link layer. (ii) transport layer  
(i) session layer. (ii) network layer  
(i) physical layer. (ii) MAC layer 
Question 49 Explanation:
Dividing the transmitted bit stream into frames is called framing and framing is one by Data link layer.
Determining which route through the subnet to use is called switching and it is done by network layer
Determining which route through the subnet to use is called switching and it is done by network layer
Question 50 
The hexadecimal equivalence of the hexadecimal IEEE format floating point number 0xC20F0000 is
35.25
 
35.25
 
35.75  
35.75 
Question 50 Explanation:
First, we need to write down the binary representation of 35.75: 100011.11
Next, we need to put normalize it: 1.0001111 * 2^{5}
Now we can get the 3 pieces of the number we need:
S: 1 (negative)
E: 132 (127 + 5) = 1000 0100
M: 000 1111 0000 0000 0000 0000
Now we just put all the bits together: 1100 0010 0000 1111 0000 0000 0000 0000 and convert each group of 4 to a hex digit:
=0xC20F0000
Next, we need to put normalize it: 1.0001111 * 2^{5}
Now we can get the 3 pieces of the number we need:
S: 1 (negative)
E: 132 (127 + 5) = 1000 0100
M: 000 1111 0000 0000 0000 0000
Now we just put all the bits together: 1100 0010 0000 1111 0000 0000 0000 0000 and convert each group of 4 to a hex digit:
=0xC20F0000
Question 51 
In the following code snippet, which variable reference(s) exhibit temporal locality
sum=0;
for(i=0; i
sum +=a[i];
Return sum;
sum, i  
A[i]  
N  
None 
Question 52 
Answer the following two questions Q53 And Q54 Based on the following information
Suppose a processor executes instructions in the following 4 stages (no pipeline), IF&ID, EX, MEM and WB. the IF&ID stage takes 10ns, EX stage 5ns, MEM stage 20ns and WB stage takes 5 ns to finish.
What is the average time per instruction for the unpipelined implementation
20ns  
10ns  
40ns  
5ns 
Question 52 Explanation:
Unpipelined = 10+5+20+5 = 40ns/instruction
Question 53 
Answer the following two questions Q53 And Q54 Based on the following information
Suppose a processor executes instructions in the following 4 stages (no pipeline), IF&ID, EX, MEM and WB. the IF&ID stage takes 10ns, EX stage 5ns, MEM stage 20ns and WB stage takes 5 ns to finish.
What is approximately the average time per instruction for the 4 stage pipelined implementation
20ns  
10ns  
40ns  
5ns 
Question 53 Explanation:
Pipelined = 20ns/instruction (assuming no stalls), stages take 20ns each
Question 54 
What is approximately the average time per instruction for the 4 stage pipelined implementation
20ns  
10ns  
40ns  
5ns 
Question 54 Explanation:
Pipelined = 20ns/instruction (assuming no stalls), stages take 20ns each
Question 55 
A computer has a cache, main memory, and a disk used for virtual memory. If a referenced word is in the cache, 20 ns are required to access it. If it is in main memory but not in the cache, 60 ns are needed to load it into the cache, and then the reference is started again. If the word is not in main memory, 12 ms (12*10^{6} ns) are required to fetch the word from disk, followed by 60ns to copy it to the cache, and then the reference is started again. The cache hit ratio is 0.9 and the main memory hit ratio is 0.6
What is the average time in nanoseconds(ns) required to access a referenced word on this system?
20  
80  
1200089  
480026 
Question 55 Explanation:
Question 56 
The fourbyte sequence 0x86, 0x65, 0x53, 0x82 stored in consecutive memory cells in Little Endian architecture represents which of the following, when represented as a 32bi signed integer in Hexadecimal format
82536586  
7DAC799A  
7DAC9A7A  
None of the above

Question 56 Explanation:
it’s littleendian, so we are going to reverse the order of the bits
our result will be signed
Step 1: Reverse the bytes
Take the bits in blocks of two and work right to left.
0x86 0x65 0x53 0x82 becomes 0x82536586
Step 2: Look at the most significant bit and determine if there will be a negative result
0x82536586 < that's this dude in red here
The most significant bit here contains an 8.
We know that in hex a mostsignificant bit of 7 or more means we are looking at a negative number. (If this were a positive number, ie: the most significant bit is between 0 and 6 inclusive, then skip ahead to Step 4.)
Step 3: Since we are working with a negative number, flip the bits (subtract our hex sequence from FFFFFFFF) and add 1
FFFFFFFF
 82536586
7DAC9A79
Add one to the result:
7DAC9A79
+1
7DAC9A7A
The result is the hex sequence we will use for the next step.
7DAC9A7A
our result will be signed
Step 1: Reverse the bytes
Take the bits in blocks of two and work right to left.
0x86 0x65 0x53 0x82 becomes 0x82536586
Step 2: Look at the most significant bit and determine if there will be a negative result
0x82536586 < that's this dude in red here
The most significant bit here contains an 8.
We know that in hex a mostsignificant bit of 7 or more means we are looking at a negative number. (If this were a positive number, ie: the most significant bit is between 0 and 6 inclusive, then skip ahead to Step 4.)
Step 3: Since we are working with a negative number, flip the bits (subtract our hex sequence from FFFFFFFF) and add 1
FFFFFFFF
 82536586
7DAC9A79
Add one to the result:
7DAC9A79
+1
7DAC9A7A
The result is the hex sequence we will use for the next step.
7DAC9A7A
Question 57 
List which of the following is/are NOT TRUE with respect to a double linked list.
1. Can be traversed in both forward and backward directions
2. Insertion is possible easily
3. No extra space is needed for a previous pointer
Only 1
 
Both 1 and 2
 
Only 3
 
All of 1,2,3

Question 57 Explanation:
Advantages:
1) A DLL can be traversed in both forward and backward direction.
2) The delete operation in DLL is more efficient if pointer to the node to be deleted is given.
3) We can quickly insert a new node before a given node.
In a singly linked list, to delete a node, pointer to the previous node is needed. To get this previous node, sometimes the list is traversed. In DLL, we can get the previous node using previous pointer.
Disadvantages:
1) Every node of DLL Require extra space for a previous pointer. It is possible to implement DLL with single pointer though (See this and this).
2) All operations require an extra pointer previous to be maintained. For example, in insertion, we need to modify previous pointers together with next pointers. For example in following functions for insertions at different positions, we need 1 or 2 extra steps to set previous pointer.
1) A DLL can be traversed in both forward and backward direction.
2) The delete operation in DLL is more efficient if pointer to the node to be deleted is given.
3) We can quickly insert a new node before a given node.
In a singly linked list, to delete a node, pointer to the previous node is needed. To get this previous node, sometimes the list is traversed. In DLL, we can get the previous node using previous pointer.
Disadvantages:
1) Every node of DLL Require extra space for a previous pointer. It is possible to implement DLL with single pointer though (See this and this).
2) All operations require an extra pointer previous to be maintained. For example, in insertion, we need to modify previous pointers together with next pointers. For example in following functions for insertions at different positions, we need 1 or 2 extra steps to set previous pointer.
Question 58 
The relation scheme student performance(name, courseno, rollNo, grade) has the following functional dependencies:
Name, courseNo → grade
rollNo, courseNo → grade
Name → rollNo
rollNo → name
The highest normal form of this relation scheme is
2NF  
BCNF
 
4NF  
3NF 
Question 58 Explanation:
Student Performance (name, courseNo, rollNo, grade)
name, courseNo → grade →(I)
rollNo, courseNo → grade →(II)
name → rollNo →(III)
rollNo → name →(IV)
Candidate keys: name, courseNo (or) rollNo
Its is not BCNF, because the relation III, there is no relationship from super key.
name → rollNo
It is not BCNF, name is not super key.
It belongs to 3NF, because if X→Y, Y is prime then it is in 3NF.
name, courseNo → grade →(I)
rollNo, courseNo → grade →(II)
name → rollNo →(III)
rollNo → name →(IV)
Candidate keys: name, courseNo (or) rollNo
Its is not BCNF, because the relation III, there is no relationship from super key.
name → rollNo
It is not BCNF, name is not super key.
It belongs to 3NF, because if X→Y, Y is prime then it is in 3NF.
Question 59 
From the following instance of a relation scheme R(X,Y,Z), we can conclude that:
X Y Z
1 1 1
1 1 0
2 3 2
2 3 2
X functionally determines Y and Y functionally determines Z  
Y does not functionally determine Z  
X does not functionally determine Y and Y does not functionally determine Z  
None of the above 
Question 59 Explanation:
From the given instanceof relation it can be seen that X functionally determines Y, but we cannot conclude this for the entire relation.
But for the given instance it can be seen that Y does not functionally determines Z, and it can be concluded for the entire relation
But for the given instance it can be seen that Y does not functionally determines Z, and it can be concluded for the entire relation
Question 60 
The statement that is executed automatically by the system as a side effect of the modification of the database is
Backup  
Trigger  
Assertion  
recovery 
Question 60 Explanation:
Statement executed automatically by the system as a side effect of modification to database  Trigger.A database trigger is what is executed automatically in response to certain events on a particular table or view in a database.
Question 61 
Which is the equivalent serial schedule for the following transactions?
T1T2T3
 
T2T1T3  
T1T3T2
 
T3T1T2

Question 61 Explanation:
Question 62 
Given the STUDENTS relation s shown below,
(StudentName, StudentAge) to be the key for this instance, the value X should NOT be equal to
18  
19  
20  
21 
Question 62 Explanation:
There is already one entry with StudentName as “Shankar” and Age as 19. So, X value should not be 19.
Question 63 
The following numbers are entered into an empty binary search tree in the given order: 50, 25, 27, 81, 19, 9, 24, 31, 17, 90. What is the height of the binary search tree.
3  
4  
5  
6 
Question 63 Explanation:
Question 64 
Struct Node
{
Struct Node *left;
Struct node *right;
Void *data;
}
int doSomething(struct Node* root)
{
if(root == NULL) return 0;
if(root > left == NULL && root > right == NULL)
return 1;
return (doSomething (root > left) + doSomething(root > right));
}
What does the function doSomething compute, if we pass the root of a nonempty binary tree.
The number of nodes in the binary tree  
The height of the binary tree  
Number of nonleaf nodes of the binary tree  
Number of leaf nodes of the binary tree 
Question 64 Explanation:
The function doSomething compute no. of leaf nodes of the binary tree because the function outputs 1 and gets added only when both its left child and right child s NULL means the node should be the leaf.
Question 65 
Consider the following pseudo code. What is the total number of multiplications being performed?
p=2
for i =1 to n do
for j=1 to n do
for k=j+1 to n do
p=p*3;
((n1)n(n+1))/6
 
((n1)n(n+1))/12  
(n(n1)(2n+1))/6  
n(n+1)/2 
Question 65 Explanation:
D = 2
for i = 1 to n do
for j = i to n do
for k = j + 1 to n do
D = D * 3;
Also you can try for smaller ‘n’.
for i = 1 to n do
for j = i to n do
for k = j + 1 to n do
D = D * 3;
Also you can try for smaller ‘n’.
Question 66 
Which of the following correctly determine a tight solutions of the recurrence relation T(n)=2T(n/2)+log_{2 }n, with T(1)=1?
Θ(n log n)  
Θ(n^{2})  
Θ(log n)
 
Θ(n) 
Question 66 Explanation:
a=2, b=2 and k=0, p=1
a>b^{k}
O(n^log_{2}^{2}n)
=O(n)
a>b^{k}
O(n^log_{2}^{2}n)
=O(n)
Question 67 
Consider an undirected random graph of 8 vertices. The probability that there is an edge between a pair of vertices is ½. What is the expected number of unordered cycles of length 3?
⅛  
1  
7  
8 
Question 67 Explanation:
So, the total probability that all three edges of the above exists
= 1/2 × 1/2 × 1/2 (as they are independent events)
= 1/8
Total number of ways in which we can select ‘3’ such vertices among ‘8’ vertices = ^{8}C_{3} = 56
Total number of cycles of length ‘3’ out of 8 vertices = 56 × 1/8 = 7
Question 68 
The problem 3SAT and 2SAT are
Both in P  
Both NP complete
 
NPcomplete and in P respectively
 
Undecidable and NPcomplete respectively 
Question 68 Explanation:
The Boolean satisfiability problem (SAT) is a decision problem, whose instance is a Boolean expression written using only AND, OR, NOT, variables, and parentheses. The problem is: given the expression, is there some assignment of TRUE and FALSE values to the variables that will make the entire expression true? A formula of propositional logic is said to be satisfiable if logical values can be assigned to its variables in a way that makes the formula true.
3SAT and 2SAT are special cases of ksatisfiability (kSAT) or simply satisfiability (SAT), when each clause contains exactly k = 3 and k = 2 literals respectively.
2SAT is P while 3SAT is NP Complete.
Question 69 
Given the following system of linear equations with solutions of the form(w,x,y,z), where w,x,yz being real, which of the following statements is NOT TRUE.
w+3x+2y+2z=0
w+4x+y=0
3w+5x+10y+14z=0
2w+5x+5y+6z=0
The system has infinitely many solutions  
The sum of any two solutions is a solution  
(5, 1, 1, 0) is a solution  
The system has no other solution other than a scalar multiple of (5,1,1,0) 
Question 69 Explanation:
The matrix for the given equations is,
Since the rank of the matrix is less than the no. of variables in the equations, so there exists infinite no. of solution.
Since the rank of the matrix is less than the no. of variables in the equations, so there exists infinite no. of solution.
Question 70 
Suppose L_{1 }and L_{2 }are languages on Σ={0,1}. Then the following statement is NOT True.
If L_{1} is regular then L’_{1 } is also regular  
If L_{1} and L_{2} are regular then L_{1} ∩ L_{2} is regular  
If L_{1} and L_{2} are Context Free languages(CFL) then L_{1} ∩ L_{2} is a CFL
 
If L_{1} and L_{2} are Context Free languages(CFL) then L_{1} . L_{2} is a CFL

Question 70 Explanation:
Option A is true because regular language is closed under complementation.
Option B is true since regular language closed under intersection.
Option C is false since context free language is not closed under intersection.
Option D is true since context free language is closed under concatenation.
Option B is true since regular language closed under intersection.
Option C is false since context free language is not closed under intersection.
Option D is true since context free language is closed under concatenation.
Question 71 
A string that is NOT generated by the Regular Expression 1*(01)*0* is
00  
10  
010  
011 
Question 71 Explanation:
All the strings given in options except option D are generated by the given regular expression.Means 011 is not generated by given regular expression.
Question 72 
Please answer the following questions Q??>> based on the flowchart given below:
Given the list A={2,6,13,8,15} of the numbers, the value of d when i=2 and j=3 is
2  
0  
3  
4 
Question 73 
Please answer the following questions Q??>> based on the flowchart given below:
The flowchart determines the
GCD of the numbers in A  
Smallest of the numbers in A  
Smallest difference between any pair of numbers in A  
Largest difference between any pair of numbers in A 
Question 74 
Please answer the following questions Q??>> based on the flowchart given below:
What is the time complexity if j=0 is replaced by j=i+1 in the flowchart?
Θ(n)
Θ(n)  
Θ(n^{2})  
Θ(nlogn)  
None of the above 
Question 75 
What is the output of the following program:
int main()
{
int a[5], i, b=16;
for(i=0; i<5; i++)
a[i]=2*i;
f(a,b);
for(i=0; i<5; i++)
printf(“%d”,a[i]);
printf(“%d”, b);
}
f(int *x, int y)
{
int i;
for(i=0; i<5; i++)
*(x+i)+=2;
y+=2;
}
4,6,8,10,12,16
 
2,4,6,8,10,26  
2,4,6,8,10,18  
2,4,6,8,10,16

Question 75 Explanation:
Now in statement 4, function f will be called.
Now in the function f every array element is increased by 2, now the new elements in the array will be,
But the value of ‘b’ will not get affected, since its address is not passed.
Now the output printed will be,
2, 4, 6, 8, 10, 16
Question 76 
What is the output of the following program:
int main()
{
int n[3][3] ={2, 4, 3, 6, 8, 5,3,5,1};
int i,j;
for(i=0; i<2; i++)
for(j=0; j<2; j++)
printf(“%d %d”, n[i][j], *(*(n+j)+i));
}
2,2,4,4,6,6,8,8
 
2,2,4,4,3,3,6,6  
2,2,4,6,6,4,8,8
 
2,2,6,4,4,6,8,8 
Question 76 Explanation:
Given,
N[3][3] = {2, 4, 3, 6, 8, 5, 3, 5, 1}
Which can also be written as,
Now,
*(*(n+j)+i) = n[j][i]
Now, loop iteration,
→ i=0, j=0, printf will print
n[0][0], n[0][0]
⇒ 2, 2
→ i=0, j=1, printf will print
n[0][1], n[1][0]
⇒ 4, 6
→ i=1, j=0, printf will print
n[1][0], n[0][1]
⇒ 6, 4
→ i=1, j=1, printf will print
n[1][1], n[1][1]
⇒ 8, 8
Therefore the sequence in which the elements will be printed is 2,2,4,6,6,4,8,8
N[3][3] = {2, 4, 3, 6, 8, 5, 3, 5, 1}
Which can also be written as,
Now,
*(*(n+j)+i) = n[j][i]
Now, loop iteration,
→ i=0, j=0, printf will print
n[0][0], n[0][0]
⇒ 2, 2
→ i=0, j=1, printf will print
n[0][1], n[1][0]
⇒ 4, 6
→ i=1, j=0, printf will print
n[1][0], n[0][1]
⇒ 6, 4
→ i=1, j=1, printf will print
n[1][1], n[1][1]
⇒ 8, 8
Therefore the sequence in which the elements will be printed is 2,2,4,6,6,4,8,8
Question 77 
Let n=4224165165. Consider the statements p,q,r and s as below:
P: 3 divides n
Q: 5 divides n
R: 7 divides n
S: 11 divides n
Which of the following logical expressions is TRUE?
P’ ⋀ q ⋀ r ⋀ s  
P ⋀ q’ ⋀ r ⋀ s  
P ⋀ q ⋀ r’ ⋀ s  
P ⋀ q’ ⋀ r ⋀ s’ 
Question 77 Explanation:
First let’s find the truth value of P, Q, R, S.
n = 4224165165
P = 3 divides n = True
Q = 5 divides n = True
R = 7 divides n = False
S = 11 divides n = True
Now, let’s check option wise,
n = 4224165165
P = 3 divides n = True
Q = 5 divides n = True
R = 7 divides n = False
S = 11 divides n = True
Now, let’s check option wise,
Question 78 
Given that 1591=37*43, the value of 2^{1513} mod 1591 is
1  
2  
4  
1024 
Question 78 Explanation:
Question 79 
Which one of the following statements describes the properties achieved?
Mutual exclusion but not progress
 
Progress but not mutual exclusion
 
Neither mutual exclusion nor progress
 
Both mutual exclusion and progress

Question 79 Explanation:
In this mutual exclusion is satisfied because at any point of time either S1 = S2 or S1 ≠ S2, but not both. But here progress is not satisfied because suppose S1 = 1 and S2 = 0 and P1 is not interested to enter into critical section but P2 wants to enter into critical section, and P2 will not be able to enter, because until P1 will not enter critical section, S1 will not become equal to S2. So if one process do not interested in entering critical section, will not allow other process to enter critical section which is interested. So progress is not satisfied.
There are 79 questions to complete.