IV Year â Computer Science & Information Technology - Amazon Web ...
IV Year – Computer Science & Information Technology Answer Keys: 1
A
2
A
3
A
4
A
5
C
6
B
7
D
8
A
9
B
10
B
11
C
12
B
13
A
14
A
15
D
16
C
17
C
18
A
19
B
20
B
21
B
22
C
23
B
24
C
25
C
26
C
27
C
28
A
29
B
30
D
31
C
32
B
33
D
34
D
35
B
36
C
37
A
38
A
39
B
40
C
41
C
42
C
43
B
44
B
45
B
46
B
47
D
48
A
49
D
50
D
Explanations: Section-I: General Ability 2.
By using the formula, n(A B) = n(A - B) + n(A B) + n(B - A) 50 = 20 +15 + n(B - A) n(B - A) = 50 - 35 n(B - A) = 15 Now, n(B) = n(A B) + n(B - A) = 15 +15 = 30
4.
Given R = 5% per annum P = 2, 00, 000 n = from 2000 to 2002 = 3 years. R Population at the end of year 2002 = P 1 + 100
n
= 200000 1 + = 200000 ×
5 100
21 20
×
3
21 20
×
21 20
= 231525 So, the estimated population = 231525.
6.
First number + last number = 8 + 12 = 20. Second number + second last number = 2 + 18 = 20 and so on. So, 11 + 9 = 20. Hence, 11 will be the answer. Email: [email protected], Website: www.engineeringolympiad.in
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IV Year – Computer Science & Information Technology 7.
German that is a Nationality not a country.
8.
Convert speed (km/hr) in to (m/sec). 5 So, 18 5m sec. 18 Dis tan ce 200 Time 40seconds. speed 5
9.
Since her arrival at the native village has been trying to the best of her power spread awareness against dowry system. Replace “is” with “has been” in option B.
10.
Let the dimension of the cuboid be 2x,3x,4x.
Then,
2 2x 3x 3x 4x 4x 2x 208 2 6x 2 12x 2 8x 2 208 2 26x 2 208 52x 2 208 x2 4 x 2cm.
Therefore volumeof cuboid 2(2) 3 2 4(2) 192cm3 . 11.
The structure of the sentence implies that the words which fit into the two blanks are contrasting words. Option (C) is most logical in the given blank. Inspite of the squalid (filthy) surrounding in which she lived, her dress was immaculate (completely tidy). The remaining options do not make sense.
12.
Let „x‟ be the monthly salary of Ram.
100 31 20 % of 100 30 15 %of x 15000 49%of 55%of x 15000 49 55 x 15000 x Rs. 55658 / 100 100
14.
Note: If a trader to sell his goods at cost price but uses false weights, then
Gain% =
error
×100 %
True value - error 200 So, here profit percentage = ×100 % 1000 - 200 200 = ×100 % = 25% 800 Email: [email protected], Website: www.engineeringolympiad.in
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IV Year – Computer Science & Information Technology 15.
The error is in D. The correction is „known to stand up‟.
16.
Let the total number of mangoes be x. Then,
x +1 . 2 x x Remaining Mangoes = x - +1 = -1 . 2 2 Mangoes sold to 1st Customer =
1 x x 1 x 2 Mangoessold to 2nd Customer = - 1 +1 = - +1 = + 3 2 6 3 6 3 x x 2 x x 2 x 5 Remaining Mangoes = -1 - + = - - 1+ = - 2 6 3 2 6 3 3 3 1 x 5 x 2 Mangoes sold to3rd customer = - +1 = + 5 3 3 15 3 x 5 x 2 x x 5 2 4x 7 Remaining Mangoes = - - + = - - + = - . 3 3 15 3 3 15 3 3 15 3 4x 7 ∴ - =3 15 3 4x 16 ⇒ = 15 3 16 15 ⇒ x = × = 20. 3 4 ∴ Number of Mangoes originally the man had = 20. 17.
This question tests your ability to properly draw an inference from the statements that are given to you. Remember that on inference questions, your task is to evaluate each answer choice, looking for the one that has to be true if the stimulus is true. Two of the answer choices are extreme. Extreme answer choices should be avoided in inference questions because they are very unlikely to be correct. Remember that we want the choice that has to be true; extreme choices are unlikely to be the ones that have to be true. Choices (B) should be eliminated for this reason. It does not have to be true that government statisticians are the "only" source. It also does not have to be true that high incomes "never" indicate high net worth. We are only told that there is no correlation, or consistent pattern. That does not mean that there will not be some families with both high income and high net worth. Choices (A) and (D) are not extreme statements, but both do not have to follow from the statements given. Choice (A) could be true, but does not have to be true. A family could also develop high net worth by saving all their income for a short number of years. Choice (D) also could be true, but does not have to be true. Choice (C) is correct because it has to be true given the stimulus. Since we know that there is no correlation between higher income and higher net worth, it is impossible to know what the result will be of in increase in income. Email: [email protected], Website: www.engineeringolympiad.in
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IV Year – Computer Science & Information Technology 18.
So, average of prime numbers between 37 to 50 will be 41 43 47 3 43.66 .... 43.67
20.
Required Percentage =
726 + 846 + 786 + 956 + 870
980 +1050 +1020 +1240 + 940 4184 = ×100 % 5230
×100 %
= 80%
Section-II: Technical 21.
Z = XY + X (X+Y)
XY XY XX XX 0 Y(X X) Y
2k 1 3 3 3 2k 1 3 0 for singular, A 0 3 3 2k 1
22.
k
5 or 2(integer) 2
100 x 102 x 2 100 100 2x 10200 102x 5 500 10200 100x 100 9,700 100x x 97%
23.
5
25.
Value returned by fun () is „no. of invocations‟. (i) Fun(i)
1
2
3
4
5
6
7
8
9
10
1
1
3
5
9
15
25
41
67
109
No. of invocations to execute fun (10) = 109. 26. 1
2
3
5
4
All edges should 90o in one direction, which is not true here Email: [email protected], Website: www.engineeringolympiad.in
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IV Year – Computer Science & Information Technology 27.
A simple graph with n vertices, is necessarily connected if G has more than n -1C 2 edges E
28.
n 1 n 2 E 5 4 10 2
2
Minimized DFA is 0
A
0,1
1 1
BD
CE 0
29.
f(1)=1 f(2)=1 f(3)=2 f(4)=2 f is a function and is onto but not one to one.
30.
Let‟s denote minimum number of nodes in an AVL tree of height h by N(H), then
1, H0 N H 2, H 1 N H 1 N H 2 1, From this, N(6) = 33
H 1
31.
P1 is reducible to P2 implies that P2 is the harder problem and P1 is the easier one. If the easier one is undecidable then the harder one is obviously undecidable.
32.
Since, rank = 1 k 2 0 k 1 2 4
34.
p r p r p r
q r q r q r Therefore, p r q r p r p r q r q r 35.
This will not generate any error in case of call by name because when the body will be substituted, there is nothing to replace as b only variable a is in the body of the procedure and a is global variable hence replaced by a and it will prints 5.
36.
Gantt chart is: P4
P1
0 4 9 Average waiting time = (4+15+9+0)/4 =7
P2
P3 15
24
Email: [email protected], Website: www.engineeringolympiad.in
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IV Year – Computer Science & Information Technology 37.
(i)
When e and f are not to be included 6 P4
(ii) When e and f are to be included 8 P4
38.
6! 360 6 4 !
8! 1680 8 4!
Initially the congestion window size was 1MSS=8B. In slow start the window size increases exponentially until path is congested. So after 1st RTT window size is 8 2=16
2nd 16 2 32 3rd 32 2 64 4th 64 2 128 In additive increase, initially 8 B after 1st RTT window size is 8 2=16 2nd 8 3 24 3rd 8 4 32 4th 8 5 40 39.
MUL A, B MOV C, A
40.
Number of records in data file = 60 Number of records in each block = 3
60 20 . 3 In spare index we need one pointer for each block of data file. Number of pointers in each block = 10. Number of blocks required for data file
41.
n 1 1, n 1 Use p n p k .p n k , n 2 k 1 p 6 42
42. X
1
2
3
4
------
n
P(X)
1 n
1 n
1 n
1 n
------
1 n
1 1 1 1 n 1 Mean 1. 2. 3. .... n. n n n n 2
Email: [email protected], Website: www.engineeringolympiad.in
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IV Year – Computer Science & Information Technology 43.
The default action of YACC is always favor shift operation over reduction in case of shift reduce conflict. The above grammar will generate the shift reduce conflict for the given grammar.
44.
Total number of frames
242 232 10 2
So frame number 32 bit
246 236 10 2 36 36 Size of page table 2 32bit 2 4 Byte 238 Byte Total number of pages in virtual program
Number of pages for page table 45.
238 Byte 228 10 2 Byte
subnet mask 255.255.11100000.0 first subnetwork Address is 132.55.0.0
first host in first subnetwork is 132.55.0.1 second subnetwork address is 132.55.32.0 second subnet and second host is 132.55.32.2 46.
47.
48.
GLB and LUB of any two elements in a lattice/poset must be unique. In (iii), LUB of a & b does not exists and GLB of c & d does not exist.
220 215 25 212 # of Cache block= 5 27 128 2 100 100 4 # of block used by array= 625 2 1250 32 Each block will result in cache miss in X loop. So total misses = 1250 In Y loop, 128 blocks will be available and then all misses Total number of misses = 1250+1122=2372 # of RAM block =
Source a a, b a, b,c a, b,c,d
b 2 2 2 2
c d e 5 3 9 3 4 7 3 4 6
So, sum of shortest distance = 2 + 3 + 4 + 6 = 15 49.
C is a week entity so relation R2 is total so combining C and R2 1 table For each of A, R1 and B one table So total 4 tables required. Email: [email protected], Website: www.engineeringolympiad.in
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IV Year – Computer Science & Information Technology 50.
Identification No.
H LEN 20 0 4 version
5
00
00 56
0003
service 16 5 12 92 type Total length
0000
Time to Line
TTL 59
flags & fragmentation offset
06
upper layer protocol
From the above figure, the time to live (TTL) field is 0X59 in hexadecimal = 5 16 + 9 = 89 in decimal. Therefore, the packet will be dropped after taking 89 hops from the source.
Email: [email protected], Website: www.engineeringolympiad.in
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A
2
A
3
A
4
A
5
C
6
B
7
D
8
A
9
B
10
B
11
C
12
B
13
A
14
A
15
D
16
C
17
C
18
A
19
B
20
B
21
B
22
C
23
B
24
C
25
C
26
C
27
C
28
A
29
B
30
D
31
C
32
B
33
D
34
D
35
B
36
C
37
A
38
A
39
B
40
C
41
C
42
C
43
B
44
B
45
B
46
B
47
D
48
A
49
D
50
D
Explanations: Section-I: General Ability 2.
By using the formula, n(A B) = n(A - B) + n(A B) + n(B - A) 50 = 20 +15 + n(B - A) n(B - A) = 50 - 35 n(B - A) = 15 Now, n(B) = n(A B) + n(B - A) = 15 +15 = 30
4.
Given R = 5% per annum P = 2, 00, 000 n = from 2000 to 2002 = 3 years. R Population at the end of year 2002 = P 1 + 100
n
= 200000 1 + = 200000 ×
5 100
21 20
×
3
21 20
×
21 20
= 231525 So, the estimated population = 231525.
6.
First number + last number = 8 + 12 = 20. Second number + second last number = 2 + 18 = 20 and so on. So, 11 + 9 = 20. Hence, 11 will be the answer. Email: [email protected], Website: www.engineeringolympiad.in
IEO-2017 Partners & Sponsors:
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IV Year – Computer Science & Information Technology 7.
German that is a Nationality not a country.
8.
Convert speed (km/hr) in to (m/sec). 5 So, 18 5m sec. 18 Dis tan ce 200 Time 40seconds. speed 5
9.
Since her arrival at the native village has been trying to the best of her power spread awareness against dowry system. Replace “is” with “has been” in option B.
10.
Let the dimension of the cuboid be 2x,3x,4x.
Then,
2 2x 3x 3x 4x 4x 2x 208 2 6x 2 12x 2 8x 2 208 2 26x 2 208 52x 2 208 x2 4 x 2cm.
Therefore volumeof cuboid 2(2) 3 2 4(2) 192cm3 . 11.
The structure of the sentence implies that the words which fit into the two blanks are contrasting words. Option (C) is most logical in the given blank. Inspite of the squalid (filthy) surrounding in which she lived, her dress was immaculate (completely tidy). The remaining options do not make sense.
12.
Let „x‟ be the monthly salary of Ram.
100 31 20 % of 100 30 15 %of x 15000 49%of 55%of x 15000 49 55 x 15000 x Rs. 55658 / 100 100
14.
Note: If a trader to sell his goods at cost price but uses false weights, then
Gain% =
error
×100 %
True value - error 200 So, here profit percentage = ×100 % 1000 - 200 200 = ×100 % = 25% 800 Email: [email protected], Website: www.engineeringolympiad.in
IEO-2017 Partners & Sponsors:
2 of 8
IV Year – Computer Science & Information Technology 15.
The error is in D. The correction is „known to stand up‟.
16.
Let the total number of mangoes be x. Then,
x +1 . 2 x x Remaining Mangoes = x - +1 = -1 . 2 2 Mangoes sold to 1st Customer =
1 x x 1 x 2 Mangoessold to 2nd Customer = - 1 +1 = - +1 = + 3 2 6 3 6 3 x x 2 x x 2 x 5 Remaining Mangoes = -1 - + = - - 1+ = - 2 6 3 2 6 3 3 3 1 x 5 x 2 Mangoes sold to3rd customer = - +1 = + 5 3 3 15 3 x 5 x 2 x x 5 2 4x 7 Remaining Mangoes = - - + = - - + = - . 3 3 15 3 3 15 3 3 15 3 4x 7 ∴ - =3 15 3 4x 16 ⇒ = 15 3 16 15 ⇒ x = × = 20. 3 4 ∴ Number of Mangoes originally the man had = 20. 17.
This question tests your ability to properly draw an inference from the statements that are given to you. Remember that on inference questions, your task is to evaluate each answer choice, looking for the one that has to be true if the stimulus is true. Two of the answer choices are extreme. Extreme answer choices should be avoided in inference questions because they are very unlikely to be correct. Remember that we want the choice that has to be true; extreme choices are unlikely to be the ones that have to be true. Choices (B) should be eliminated for this reason. It does not have to be true that government statisticians are the "only" source. It also does not have to be true that high incomes "never" indicate high net worth. We are only told that there is no correlation, or consistent pattern. That does not mean that there will not be some families with both high income and high net worth. Choices (A) and (D) are not extreme statements, but both do not have to follow from the statements given. Choice (A) could be true, but does not have to be true. A family could also develop high net worth by saving all their income for a short number of years. Choice (D) also could be true, but does not have to be true. Choice (C) is correct because it has to be true given the stimulus. Since we know that there is no correlation between higher income and higher net worth, it is impossible to know what the result will be of in increase in income. Email: [email protected], Website: www.engineeringolympiad.in
IEO-2017 Partners & Sponsors:
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IV Year – Computer Science & Information Technology 18.
So, average of prime numbers between 37 to 50 will be 41 43 47 3 43.66 .... 43.67
20.
Required Percentage =
726 + 846 + 786 + 956 + 870
980 +1050 +1020 +1240 + 940 4184 = ×100 % 5230
×100 %
= 80%
Section-II: Technical 21.
Z = XY + X (X+Y)
XY XY XX XX 0 Y(X X) Y
2k 1 3 3 3 2k 1 3 0 for singular, A 0 3 3 2k 1
22.
k
5 or 2(integer) 2
100 x 102 x 2 100 100 2x 10200 102x 5 500 10200 100x 100 9,700 100x x 97%
23.
5
25.
Value returned by fun () is „no. of invocations‟. (i) Fun(i)
1
2
3
4
5
6
7
8
9
10
1
1
3
5
9
15
25
41
67
109
No. of invocations to execute fun (10) = 109. 26. 1
2
3
5
4
All edges should 90o in one direction, which is not true here Email: [email protected], Website: www.engineeringolympiad.in
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IV Year – Computer Science & Information Technology 27.
A simple graph with n vertices, is necessarily connected if G has more than n -1C 2 edges E
28.
n 1 n 2 E 5 4 10 2
2
Minimized DFA is 0
A
0,1
1 1
BD
CE 0
29.
f(1)=1 f(2)=1 f(3)=2 f(4)=2 f is a function and is onto but not one to one.
30.
Let‟s denote minimum number of nodes in an AVL tree of height h by N(H), then
1, H0 N H 2, H 1 N H 1 N H 2 1, From this, N(6) = 33
H 1
31.
P1 is reducible to P2 implies that P2 is the harder problem and P1 is the easier one. If the easier one is undecidable then the harder one is obviously undecidable.
32.
Since, rank = 1 k 2 0 k 1 2 4
34.
p r p r p r
q r q r q r Therefore, p r q r p r p r q r q r 35.
This will not generate any error in case of call by name because when the body will be substituted, there is nothing to replace as b only variable a is in the body of the procedure and a is global variable hence replaced by a and it will prints 5.
36.
Gantt chart is: P4
P1
0 4 9 Average waiting time = (4+15+9+0)/4 =7
P2
P3 15
24
Email: [email protected], Website: www.engineeringolympiad.in
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IV Year – Computer Science & Information Technology 37.
(i)
When e and f are not to be included 6 P4
(ii) When e and f are to be included 8 P4
38.
6! 360 6 4 !
8! 1680 8 4!
Initially the congestion window size was 1MSS=8B. In slow start the window size increases exponentially until path is congested. So after 1st RTT window size is 8 2=16
2nd 16 2 32 3rd 32 2 64 4th 64 2 128 In additive increase, initially 8 B after 1st RTT window size is 8 2=16 2nd 8 3 24 3rd 8 4 32 4th 8 5 40 39.
MUL A, B MOV C, A
40.
Number of records in data file = 60 Number of records in each block = 3
60 20 . 3 In spare index we need one pointer for each block of data file. Number of pointers in each block = 10. Number of blocks required for data file
41.
n 1 1, n 1 Use p n p k .p n k , n 2 k 1 p 6 42
42. X
1
2
3
4
------
n
P(X)
1 n
1 n
1 n
1 n
------
1 n
1 1 1 1 n 1 Mean 1. 2. 3. .... n. n n n n 2
Email: [email protected], Website: www.engineeringolympiad.in
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IV Year – Computer Science & Information Technology 43.
The default action of YACC is always favor shift operation over reduction in case of shift reduce conflict. The above grammar will generate the shift reduce conflict for the given grammar.
44.
Total number of frames
242 232 10 2
So frame number 32 bit
246 236 10 2 36 36 Size of page table 2 32bit 2 4 Byte 238 Byte Total number of pages in virtual program
Number of pages for page table 45.
238 Byte 228 10 2 Byte
subnet mask 255.255.11100000.0 first subnetwork Address is 132.55.0.0
first host in first subnetwork is 132.55.0.1 second subnetwork address is 132.55.32.0 second subnet and second host is 132.55.32.2 46.
47.
48.
GLB and LUB of any two elements in a lattice/poset must be unique. In (iii), LUB of a & b does not exists and GLB of c & d does not exist.
220 215 25 212 # of Cache block= 5 27 128 2 100 100 4 # of block used by array= 625 2 1250 32 Each block will result in cache miss in X loop. So total misses = 1250 In Y loop, 128 blocks will be available and then all misses Total number of misses = 1250+1122=2372 # of RAM block =
Source a a, b a, b,c a, b,c,d
b 2 2 2 2
c d e 5 3 9 3 4 7 3 4 6
So, sum of shortest distance = 2 + 3 + 4 + 6 = 15 49.
C is a week entity so relation R2 is total so combining C and R2 1 table For each of A, R1 and B one table So total 4 tables required. Email: [email protected], Website: www.engineeringolympiad.in
IEO-2017 Partners & Sponsors:
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IV Year – Computer Science & Information Technology 50.
Identification No.
H LEN 20 0 4 version
5
00
00 56
0003
service 16 5 12 92 type Total length
0000
Time to Line
TTL 59
flags & fragmentation offset
06
upper layer protocol
From the above figure, the time to live (TTL) field is 0X59 in hexadecimal = 5 16 + 9 = 89 in decimal. Therefore, the packet will be dropped after taking 89 hops from the source.
Email: [email protected], Website: www.engineeringolympiad.in
IEO-2017 Partners & Sponsors:
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