2010
Question 1 
S = 2T  
S = T  1  
S = T  
S = T + 1 
Question 1 Explanation:
i_{d}= no. of vertices of degree ‘d’ in ‘G’
Eg:
No. of vertices with degree ‘2’ = 3
ξ(G')=3×2='6' i.e., sum of degrees
By Handshaking Theorem,
The sum of degrees would be equal to twice the no. of edges
V=2E
It is given that ξ(G)=ξ(S) then
Sum of degrees of vertices in G is equal to sum of degrees of vertices in S
i.e., 2*(no. of edges in G)=2*no. of edges in S no. of edges in G=no. of edges in S
Eg:
ξ(G)=(2×2)+(2×3)=4+6=10
ξ(S)=2×5=10
You can observe that, though no. of vertices are different, but still no. of edges are same.
Question 2 
NewtonRaphson method is used to compute a root of the equation x^{2}  13 = 0 with 3.5 as the initial value. The approximation after one iteration is
3.575  
3.676  
3.667  
3.607

Question 2 Explanation:
Note: Out of syllabus.
Question 3 
What is the possible number of reflexive relations on a set of 5 elements?
2^{10}
 
2^{15}  
2^{20}  
2^{25} 
Question 3 Explanation:
Let set = ‘A’ with ‘n’ elements,
Definition of Reflexive relation:
A relation ‘R’ is reflexive if it contains xRx ∀ x∈A
A relation with all diagonal elements, it can contain any combination of nondiagonal elements.
Eg:
A={1, 2, 3}
So for a relation to be reflexive, it should contain all diagonal elements. In addition to them, we can have possible combination of (n^{2}n)nondiagonal elements (i.e., 2^{n2n})
Ex:
{(1,1)(2,2)(3,3)}  ‘0’ nondiagonal element
{(1,1)(2,2)(3,3)(1,2)}  ‘1’ nondiagonal element
{(1,1)(2,2)(3,3)(1,2)(1,3)} “
___________ “
___________ “
{(1,1)(2,2)(3,3)(1,2)(1,3)(2,1)(2,3)(3,1)(3,2)} (n^{2}n) diagonal elements
____________________
Total: 2^{n2n}
For the given question n = 5.
The number of reflexive relations =2^{(255)}=2^{20}
Definition of Reflexive relation:
A relation ‘R’ is reflexive if it contains xRx ∀ x∈A
A relation with all diagonal elements, it can contain any combination of nondiagonal elements.
Eg:
A={1, 2, 3}
So for a relation to be reflexive, it should contain all diagonal elements. In addition to them, we can have possible combination of (n^{2}n)nondiagonal elements (i.e., 2^{n2n})
Ex:
{(1,1)(2,2)(3,3)}  ‘0’ nondiagonal element
{(1,1)(2,2)(3,3)(1,2)}  ‘1’ nondiagonal element
{(1,1)(2,2)(3,3)(1,2)(1,3)} “
___________ “
___________ “
{(1,1)(2,2)(3,3)(1,2)(1,3)(2,1)(2,3)(3,1)(3,2)} (n^{2}n) diagonal elements
____________________
Total: 2^{n2n}
For the given question n = 5.
The number of reflexive relations =2^{(255)}=2^{20}
Question 4 
Consider the set S = {1, ω, ω^{2}}, where ω and ω^{2} are cube roots of unity. If * denotes the multiplication operation, the structure (S, *) forms
A group
 
A ring  
An integral domain  
A field 
Question 4 Explanation:
A Group is an algebraic structure which satisfies
1) closure
2) Associativity
3) Have Identity element
4) Invertible
Over ‘*’ operation the S = {1, ω, ω^{2}} satisfies the above properties.
The identity element is ‘1’ and inverse of 1 is 1, inverse of ‘w’ is 'w^{2}' and inverse of 'w^{2}' is 'w'.
1) closure
2) Associativity
3) Have Identity element
4) Invertible
Over ‘*’ operation the S = {1, ω, ω^{2}} satisfies the above properties.
The identity element is ‘1’ and inverse of 1 is 1, inverse of ‘w’ is 'w^{2}' and inverse of 'w^{2}' is 'w'.
Question 6 
m_{2}+m_{4}+m_{6}+m_{7}  
m_{0}+m_{1}+m_{3}+m_{5}  
m_{0}+m_{1}+m_{6}+m_{7}  
m_{2}+m_{3}+m_{4}+m_{5} 
Question 6 Explanation:
Convert PQ + QR' + PR' into canonical form
= PQR + PQR' + PQR' + P'QR' + PQR' + PQ'R'
= PQR + PQR' + P'QR' + PQ'R'
=m_{7} + m_{6} + m_{2} + m_{4}
= PQR + PQR' + PQR' + P'QR' + PQR' + PQ'R'
= PQR + PQR' + P'QR' + PQ'R'
=m_{7} + m_{6} + m_{2} + m_{4}
Question 7 
A main memory unit with a capacity of 4 megabytes is built using 1M 1bit DRAM chips. Each DRAM chip has 1K rows of cells with 1K cells in each row. The time taken for a single refresh operation is 100 nanoseconds. The time required to perform one refresh operation on all the cells in the memory unit is
100 nanoseconds  
100*2^{10} nanoseconds  
100*2^{20} nanoseconds  
3200*2^{20} nanoseconds 
Question 7 Explanation:
Each chip capacity = 1M x 1bit
Required capacity = 4MB
Number of chips needed = 4M*8 bits / 1M x 1bit = 32 (1M x 1bit)/(1M x 1bit) = 32
Irrespective of the number of chips, all chips can be refreshed in parallel.
And all the cells in a row are refreshed in parallel too. So, the total time for refresh will be number of rows times the refresh time of one row.
Here we have 1K rows in a chip and refresh time of single row is 100ns.
So total time required =1K×100
=100 ×2^{10} nanoseconds
Required capacity = 4MB
Number of chips needed = 4M*8 bits / 1M x 1bit = 32 (1M x 1bit)/(1M x 1bit) = 32
Irrespective of the number of chips, all chips can be refreshed in parallel.
And all the cells in a row are refreshed in parallel too. So, the total time for refresh will be number of rows times the refresh time of one row.
Here we have 1K rows in a chip and refresh time of single row is 100ns.
So total time required =1K×100
=100 ×2^{10} nanoseconds
Question 8 
P is a 16bit signed integer. The 2's complement representation of P is (F87B)_{16}. The 2's complement representation of 8*P is
(C3D8)_{16}  
(187B)_{16}  
(F878)_{16}  
(987B)_{16} 
Question 8 Explanation:
(F87B)_{16} is 2's complement representation of P.
(F87B)_{16}=(1111 1000 0111 1011)_{2}. (It is a negative number which is in 2's complement form)
P=1111 1000 0111 1011 (2's complement form)
8 * P = 2^{3}* P= 1100 0011 1101 1000. ( NOTE: Left shift k times is equivalent to Multiplication by 2^{k})
Hence, 1100 0011 1101 1000 is 2's complement representation of 8P.
1100 0011 1101 1000 = (C3D8)_{16}.
(F87B)_{16}=(1111 1000 0111 1011)_{2}. (It is a negative number which is in 2's complement form)
P=1111 1000 0111 1011 (2's complement form)
8 * P = 2^{3}* P= 1100 0011 1101 1000. ( NOTE: Left shift k times is equivalent to Multiplication by 2^{k})
Hence, 1100 0011 1101 1000 is 2's complement representation of 8P.
1100 0011 1101 1000 = (C3D8)_{16}.
Question 9 
P⊕Q⊕R
 
P+Q+R  
Question 9 Explanation:
f= P’Q’ R + P’Q R’ + PQ’ R’ + PQR
= (P’Q’ + PQ)R + (P’Q+PQ’)R’
= (P⊕Q)’ R + (P⊕Q)R’
= (P⊕Q⊕R)
= (P’Q’ + PQ)R + (P’Q+PQ’)R’
= (P⊕Q)’ R + (P⊕Q)R’
= (P⊕Q⊕R)
Question 10 
In a binary tree with n nodes, every node has an odd number of descendants. Every node is considered to be its own descendant. What is the number of nodes in the tree that have exactly one child?
0  
1  
(n1)/2  
n1 
Question 10 Explanation:
Given a Binary Tree with n nodes.
Every node has an odd number of descendants.
Also given every node is considered to be its own descendant.
― This is even number of descendants (2), because A is also considered as a descendant.
― This is odd number of descendants (3), A, B, C are descendants here.
Condition satisfied – odd number, but number of nodes in a tree that has exactly one child is 0.
Every node has an odd number of descendants.
Also given every node is considered to be its own descendant.
― This is even number of descendants (2), because A is also considered as a descendant.
― This is odd number of descendants (3), A, B, C are descendants here.
Condition satisfied – odd number, but number of nodes in a tree that has exactly one child is 0.
Question 11 
2 2  
2 1  
0 1  
0 2 
Question 11 Explanation:
Both pointers points to j = 1
now *p = 2
where j is updated with value 2.
printf (i, j) i = 0, j = 2
Question 12 
Two alternative packages A and B are available for processing a database having 10^{k} records. Package A requires 0.0001n^{2} time units and package B requires 10nlog_{10}n time units to process n records.What is the smallest value of k for which package B will be preferred over A?
12  
10  
6  
5 
Question 12 Explanation:
As per given information Package B 10nlog_{10}n is lesser than or equals to Package A 0.0001n^{2} 0 because n^{2} is asymptotically larger than nlogn.
Finally, 10nlog_{10}n ≤ 0.0001n^{2}
Let n = 10^{k} records. Substitute into 10nlog_{10}n ≤ 0.0001n^{2}
10(10^{k})log_{10}10^{k} ≤ 0.0001(10^{k})^{2}
10^{k+1}k ≤ 0.0001 × 10^{2k}
k ≤ 10^{2k−k−1−4}
k ≤ 10^{k−5}
According to the problem value 6 is suitable for K.
Let n = 10^{k} records. Substitute into 10nlog_{10}n ≤ 0.0001n^{2}
10(10^{k})log_{10}10^{k} ≤ 0.0001(10^{k})^{2}
10^{k+1}k ≤ 0.0001 × 10^{2k}
k ≤ 10^{2k−k−1−4}
k ≤ 10^{k−5}
According to the problem value 6 is suitable for K.
Question 13 
Which data structure in a compiler is used for managing information about variables and their attributes?
Abstract syntax tree  
Symbol table  
Semantic stack  
Parse Table

Question 13 Explanation:
Symbol tables are data structures that are used by compilers to hold information about sourceprogram constructs. The information is collected incrementally by the analysis phases of a compiler and used by the synthesis phases to generate the target code. Entries in the symbol table contain information about an identifier such as its character string (or lexeme) , its type, its position in storage, and any other relevant information.
Question 14 
Which languages necessarily need heap allocation in the runtime environment?
Those that support recursion  
Those that use dynamic scoping
 
Those that allow dynamic data structures  
Those that use global variables

Question 14 Explanation:
Dynamic memory is allocated on the heap by the system. So the languages which allow dynamic data structure require heap allocation at runtime.
Question 15 
One of the header fields in an IP datagram is the Time to Live (TTL) field. Which of the following statements best explains the need for this field?
It can be used to prioritize packets
 
It can be used to reduce delays
 
It can be used to optimize throughput  
It can be used to prevent packet looping 
Question 15 Explanation:
Time to Live (TTL) is a limit on the period of time or transmissions in computer and computer network technology that a unit of data (e.g. a packet) can experience before it should be discarded. If the limit is not defined then the packets can go into an indefinite loop. The packet is discarded when the Time to Live field reaches 0 to prevent looping.
Question 16 
Which one of the following is not a client server application?
Internet chat  
Web browsing
 
Email  
ping 
Question 16 Explanation:
Ping is used for knowing status of a host by another host. it is not a client server application.
Question 17 
Let L1 be a recursive language. Let L2 and L3 be languages that are recursively enumerable but not recursive. Which of the following statements is not necessarily true?
L2 – L1 is recursively enumerable
 
L1 – L3 is recursively enumerable
 
L2 ∩ L1 is recursively enumerable
 
L2 ∪ L1 is recursively enumerable 
Question 17 Explanation:
L2 − L1 means L2 ∩ L1^{c} , since L1 is recursive hence L1^{c} must also be recursive, So L2∩L1^{c} is equivalent to (Recursive enum ∩ Recursive) , as every recursive is recursive enum also, so it is equivalent to (Recursive enum ∩ Recursive enum) and recursive enum is closed under intersection, so L2− L1 must be recursive enumerable. Hence A is always true.
L1 − L3 means L1 ∩ L3^{c} , since recursive enumerable is not closed under complement, so L3^{c} may or may not be recursive enumerable, hence we cannot say that L1 − L3 will always be recursive enumerable. So B is not necessarily true always.
L2 ∩ L1 means (Recursive enum ∩ Recursive) , as every recursive is recursive enum also, so it is equivalent to (Recursive enum ∩ Recursive enum) and recursive enum is closed under intersection, so L2∩ L1 must be recursive enumerable. Hence C is always true.
L2 ∪ L1 means (Recursive enum ∪ Recursive) , as every recursive is recursive enum also, so it is equivalent to (Recursive enum ∪ Recursive enum) and recursive enum is closed under union, so L2 ∪ L1 must be recursive enumerable. Hence D is always true.
L1 − L3 means L1 ∩ L3^{c} , since recursive enumerable is not closed under complement, so L3^{c} may or may not be recursive enumerable, hence we cannot say that L1 − L3 will always be recursive enumerable. So B is not necessarily true always.
L2 ∩ L1 means (Recursive enum ∩ Recursive) , as every recursive is recursive enum also, so it is equivalent to (Recursive enum ∩ Recursive enum) and recursive enum is closed under intersection, so L2∩ L1 must be recursive enumerable. Hence C is always true.
L2 ∪ L1 means (Recursive enum ∪ Recursive) , as every recursive is recursive enum also, so it is equivalent to (Recursive enum ∪ Recursive enum) and recursive enum is closed under union, so L2 ∪ L1 must be recursive enumerable. Hence D is always true.
Question 18 
Consider a B^{+}tree in which the maximum number of keys in a node is 5. What is the minimum number of keys in any nonroot node?
1  
2  
3  
4 
Question 18 Explanation:
― In B^{+} tree, the root contains minimum two block pointers and maximum ‘p’ block pointers.
― Here,
p = order
key = order – 1 = p – 1
― In the nonroot node, the minimum no. of keys =⌈p/2⌉1
― So, key = 5
order = 6
― So, minimum no. of keys in root node =⌈6/2⌉1=2
― Here,
p = order
key = order – 1 = p – 1
― In the nonroot node, the minimum no. of keys =⌈p/2⌉1
― So, key = 5
order = 6
― So, minimum no. of keys in root node =⌈6/2⌉1=2
Question 19 
1, 0  
1, 2  
1, 3  
1, 5 
Question 19 Explanation:
Passenger:
― 1, 3 Pids are returned
― 1, 3 Pids are returned
Question 20 
I only
 
II only  
Both I and II  
Neither I nor II 
Question 20 Explanation:
― Twophase locking protocol (2PLP) ensures the conflict serializable schedule but it may not free from deadlock.
― Timestamp ordering protocol ensures conflict serializability and free from deadlock.
― Timestamp ordering protocol ensures conflict serializability and free from deadlock.
Question 22 
P3, Q2, R4, S1  
P2, Q3, R1, S4  
P3, Q2, R1, S4  
P2, Q3, R4, S1 
Question 22 Explanation:
Note: Out of syllabus.
Question 23 
Mutual exclusion but not progress
 
Progress but not mutual exclusion  
Neither mutual exclusion nor progress
 
Both mutual exclusion and progress 
Question 23 Explanation:
In this mutual exclusion is saisfied 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 24 
A system uses FIFO policy for page replacement. It has 4 page frames with no pages loaded to begin with. The system first accesses 100 distinct pages in some order and then accesses the same 100 pages but now in the reverse order. How many page faults will occur?
196  
192  
197  
195 
Question 24 Explanation:
The first 100 accesses causes 100 page faults.
Page 1 …….. Page 100 causes 100 faults.
Now, when they are accesses in reverse order page 100, 99, 98, 97 are already present. So we get 96 page faults. So, total of 196 page faults.
Page 1 …….. Page 100 causes 100 faults.
Now, when they are accesses in reverse order page 100, 99, 98, 97 are already present. So we get 96 page faults. So, total of 196 page faults.
Question 25 
I only
 
I and III only
 
II and III only  
I, II and III 
Question 25 Explanation:
 In SRTF longer bursts will suffer from starvation.
 Preemptive scheduling causes starvation (for example lower priority jobs are waiting).
 Best Response time is given by RR.
 Preemptive scheduling causes starvation (for example lower priority jobs are waiting).
 Best Response time is given by RR.
Question 26 
Consider a company that assembles computers. The probability of a faulty assembly of any computer is p. The company therefore subjects each computer to a testing process. This testing process gives the correct result for any computer with a probability of q. What is the probability of a computer being declared faulty?
pq + (1  p)(1  q)
 
(1  q)p
 
(1  p)q
 
pq 
Question 26 Explanation:
The probability of the computer being declared faulty is,
= Probability of testing process gives the correct result × Probability that computer is faulty + Probability of tetsing process giving incorrect result × Probability that computer is not faulty
= p × q + (1  p) (1  q)
= Probability of testing process gives the correct result × Probability that computer is faulty + Probability of tetsing process giving incorrect result × Probability that computer is not faulty
= p × q + (1  p) (1  q)
Question 27 
What is the probability that divisor of 10^{99} is a multiple of 10^{96}?
1/625  
4/625  
12/625
 
16/625 
Question 27 Explanation:
Probability that divisor of 10^{99} is a multiple of 10^{96}
We can write 10^{99} as 10^{96}×10^{3}
So, (10^{99})/(10^{96}) to be a whole number, [10^{96}×10^{3}/10^{96}]➝ (1)
We can observe that every divisor of 10^{3} is a multiple of 10^{96}
So number of divisor of 10^{3} to be found first
⇒ 10^{3}=(5×2)^{3}=2^{3}×5^{3}
No. of divisors = (3 + 1) (3 + 1) = 16
Total number of divisor of 10^{99} are 10^{99}=2^{99}×5^{99}=100×100=10000
Probability that divisor of 10^{99} is a multiple of 10^{96} is
⇒16/10,000
We can write 10^{99} as 10^{96}×10^{3}
So, (10^{99})/(10^{96}) to be a whole number, [10^{96}×10^{3}/10^{96}]➝ (1)
We can observe that every divisor of 10^{3} is a multiple of 10^{96}
So number of divisor of 10^{3} to be found first
⇒ 10^{3}=(5×2)^{3}=2^{3}×5^{3}
No. of divisors = (3 + 1) (3 + 1) = 16
Total number of divisor of 10^{99} are 10^{99}=2^{99}×5^{99}=100×100=10000
Probability that divisor of 10^{99} is a multiple of 10^{96} is
⇒16/10,000
Question 28 
I and II
 
III and IV  
IV only  
II and IV 
Question 28 Explanation:
Havel Hakimi theorem:
⇾ Arrange the degree of vertices in descending order
eg. d_{1},d_{2}, d_{3}...d_{n}
⇾ Discard d_{1}, subtrack ‘1’ from the next 'd_{1}'degrees
eg:
⇒ 1 1 0 1
⇾ We should not get any negative value if its negative, this is not valid sequence
⇾ Repeat it till we get ‘0’ sequence
I. 7, 6, 5, 4, 4, 3, 2, 1
➡️5, 4, 3, 3, 2, 1, 0
➡️3, 2, 2, 1, 0, 0
➡️1, 1, 0, 0, 0
➡️0, 0, 0, 0
[valid]
II. 6, 6, 6, 6, 3, 3, 2, 2
➡️5, 5, 5, 2, 2, 1, 2
put them in descending order
➡️5, 5, 5, 2, 2, 2, 1
➡️4, 4, 1, 1, 1, 1
➡️3, 0, 0, 0, 1 (descending order)
➡️3, 1, 0, 0, 0
➡️0, 1, 1, 0
[This is not valid]
III. 7, 6, 6, 4, 4, 3, 2, 2
➡️5, 5, 3, 3, 2, 1, 1
➡️4, 2, 2, 1, 0, 1
➡️4, 2, 2, 1, 1, 0 (descending order)
➡️1, 1, 0, 0, 0
➡️0, 0, 0, 0
[valid]
IV. 8, 7, 7, 6, 4, 2, 1, 1
There is a degree ‘8’, but there are only ‘8’ vertices.
A vertex cannot have edge to itself in a simple graph. This is not valid sequence.
⇾ Arrange the degree of vertices in descending order
eg. d_{1},d_{2}, d_{3}...d_{n}
⇾ Discard d_{1}, subtrack ‘1’ from the next 'd_{1}'degrees
eg:
⇒ 1 1 0 1
⇾ We should not get any negative value if its negative, this is not valid sequence
⇾ Repeat it till we get ‘0’ sequence
I. 7, 6, 5, 4, 4, 3, 2, 1
➡️5, 4, 3, 3, 2, 1, 0
➡️3, 2, 2, 1, 0, 0
➡️1, 1, 0, 0, 0
➡️0, 0, 0, 0
[valid]
II. 6, 6, 6, 6, 3, 3, 2, 2
➡️5, 5, 5, 2, 2, 1, 2
put them in descending order
➡️5, 5, 5, 2, 2, 2, 1
➡️4, 4, 1, 1, 1, 1
➡️3, 0, 0, 0, 1 (descending order)
➡️3, 1, 0, 0, 0
➡️0, 1, 1, 0
[This is not valid]
III. 7, 6, 6, 4, 4, 3, 2, 2
➡️5, 5, 3, 3, 2, 1, 1
➡️4, 2, 2, 1, 0, 1
➡️4, 2, 2, 1, 1, 0 (descending order)
➡️1, 1, 0, 0, 0
➡️0, 0, 0, 0
[valid]
IV. 8, 7, 7, 6, 4, 2, 1, 1
There is a degree ‘8’, but there are only ‘8’ vertices.
A vertex cannot have edge to itself in a simple graph. This is not valid sequence.
Question 29 
x=4, y=10  
x=5, y=8  
x=3, y=9  
x=4, y=10 
Question 29 Explanation:
Trace = {Sum of diagonal elements of matrix}
Here given that eigen values are 4, 8
Sum = 8 + 4 = 12
Trace = 2 + y
⇒ 2 + y = 12
y = 10
Determinant = 2y  3x
Product of eigen values = 8 × 4 = 32
2y  3x = 32
(y = 10)
20  3x = 32
12 = 3x
x = 4
∴ x = 4, y = 10
Question 30 
Suppose the predicate F(x, y, t) is used to represent the statement that person x can fool person y at time t. Which one of the statements below expresses best the meaning of the formula ∀x∃y∃t(¬F(x, y, t))?
Everyone can fool some person at some time  
No one can fool everyone all the time  
Everyone cannot fool some person all the time
 
No one can fool some person at some time 
Question 30 Explanation:
F(x,y,t) ⇒ Person 'x' can fool person 'y' at time 't'.
For better understanding propagate negation sign outward by applying Demorgan's law.
∀x∃y∃t(¬F(x, y, t)) ≡ ¬∃x∀y∀t(F(x,y,t))
Now converting ¬∃x∀y∀t(F(x,y,t)) to English is simple.
¬∃x∀y∀t(F(x,y,t)) ⇒ There does not exist a person who can fool everyone all the time.
Which means "No one can fool everyone all the time".
Hence, Option (B) is correct.
For better understanding propagate negation sign outward by applying Demorgan's law.
∀x∃y∃t(¬F(x, y, t)) ≡ ¬∃x∀y∀t(F(x,y,t))
Now converting ¬∃x∀y∀t(F(x,y,t)) to English is simple.
¬∃x∀y∀t(F(x,y,t)) ⇒ There does not exist a person who can fool everyone all the time.
Which means "No one can fool everyone all the time".
Hence, Option (B) is correct.
Question 31 
Question 31 Explanation:
f = ((P’Q’ + Q’R’)’ + ( P’R’ + Q’R’)’ )’
= (P’Q’ + Q’R’)( P’R’ + Q’R’)
= (P’Q’P’R’ + P’Q’Q’R’ + Q’R’P’R’ + Q’R’Q’R’)
= (P’Q’R’ + P’Q’R’ + P’Q’R’ + Q’R’)
= (P’Q’R’ + Q’R’)
= (Q’R’)
= (Q+R)’
= (P’Q’ + Q’R’)( P’R’ + Q’R’)
= (P’Q’P’R’ + P’Q’Q’R’ + Q’R’P’R’ + Q’R’Q’R’)
= (P’Q’R’ + P’Q’R’ + P’Q’R’ + Q’R’)
= (P’Q’R’ + Q’R’)
= (Q’R’)
= (Q+R)’
Question 32 
11, 10, 01, 00
 
10, 11, 01, 00  
10, 00, 01, 11  
11, 10, 00, 01

Question 32 Explanation:
The next four values of Q_{1}Q_{0} are 11, 10, 01, 00.
Question 33 
13  
15  
17  
19 
Question 33 Explanation:
It is given that there is operand forwarding. In the case of operand forwarding the updated value from previous instruction’s PO stage is forwarded to the present instruction’s PO stage. Here there’s RAW dependency between I1I2 for R5 and between I2I3 for R2. These dependencies are resolved by using operand forwarding as shown in the below timeline diagram. The total number of clock cycles needed is 15.
Question 34 
The weight of a sequence a_{0}, a_{1}, ..., a_{n1} of real numbers is defined as a_{0}+a_{1}/2+...+ a_{n1}/2^{n1}. A subsequence of a sequence is obtained by deleting some elements from the sequence, keeping the order of the remaining elements the same. Let X denote the maximum possible weight of a subsequence of a_{0}, a_{1}, ...,a_{n1} and Y the maximum possible weight of a subsequence of a_{1}, a_{2}, ..., a_{n1}. Then X is equal to
max(Y, a_{0}+Y)  
max(Y, a_{0}+Y/2)
 
max(Y, a_{0}+2Y)  
a_{0}+Y/2

Question 34 Explanation:
Using concepts of Dynamic Programming, to find the maximum possible weight of a subsequence of X, we will have two alternatives:
1. Do not include a0 in the subsequence: then the maximum possible weight will be equal to maximum possible weight of a subsequence of {a_{1}, a_{2},….a_{n}} which is represented by Y
2. Include a_{0}: then maximum possible weight will be equal to : a_{0} + (each number picked in Y will get divided by 2) = a_{0} + Y/2.
Key point here is Y will itself pick optimal subsequence to maximize the weight. The value of a_{0} can be anything (negative ori∈R it is possible that Y>a_{0}+Y/2.
Note: Y/2 meaning: Each number picked in Y will get divided by 2
1. Do not include a0 in the subsequence: then the maximum possible weight will be equal to maximum possible weight of a subsequence of {a_{1}, a_{2},….a_{n}} which is represented by Y
2. Include a_{0}: then maximum possible weight will be equal to : a_{0} + (each number picked in Y will get divided by 2) = a_{0} + Y/2.
Key point here is Y will itself pick optimal subsequence to maximize the weight. The value of a_{0} can be anything (negative or
Note: Y/2 meaning: Each number picked in Y will get divided by 2
Question 35 
9  
5  
15  
19 
Question 35 Explanation:
int a[ ] = {12, 7, 13, 4, 11, 6}
if (n <= 0)
return 0;
else if (*a % 2 = = 0)
return *a + f(a+1, n1);
else
return *a – f(a+1, n1);
⇒12+(7(13(4+(11(6)))))
⇒12+(7(13(4+5)))
⇒12+7(4)
⇒12+3
⇒15
if (n <= 0)
return 0;
else if (*a % 2 = = 0)
return *a + f(a+1, n1);
else
return *a – f(a+1, n1);
⇒12+(7(13(4+(11(6)))))
⇒12+(7(13(4+5)))
⇒12+7(4)
⇒12+3
⇒15
Question 36 
q = NULL; p→next = head; head = p;
 
q→next = NULL; head = p; p→next = head;  
head = p; p→next = q; q→next = NULL;  
q→next = NULL; p→next = head; head = p;

Question 36 Explanation:
C function takes a simple linked list as input argument.
The function modifies the list by moving the last element to the front of the list.
Let the list be 1 → 2 → 3 → 4 & the modified list must be 4 → 1 → 2 → 3.
Algorithm:
Traverse the list till last node. Use two pointers. One to store the address of last node & other for the address of second last node.
After completion of loop. Do these.
(i) Make second last as last
(ii) Set next of last as head
(iii) Make last as head
while (p → !=NULL)
{
q = p;
p = p → next;
}
― p is travelling till the end of list and assigning q to whatever p had visited & p takes next new node, so travels through the entire list.
Now the list looks like
According to the Algorithm new lines must be
q → next = NULL; p → next = head; head = p
The function modifies the list by moving the last element to the front of the list.
Let the list be 1 → 2 → 3 → 4 & the modified list must be 4 → 1 → 2 → 3.
Algorithm:
Traverse the list till last node. Use two pointers. One to store the address of last node & other for the address of second last node.
After completion of loop. Do these.
(i) Make second last as last
(ii) Set next of last as head
(iii) Make last as head
while (p → !=NULL)
{
q = p;
p = p → next;
}
― p is travelling till the end of list and assigning q to whatever p had visited & p takes next new node, so travels through the entire list.
Now the list looks like
According to the Algorithm new lines must be
q → next = NULL; p → next = head; head = p
Question 37 
2  
3  
4  
6 
Question 37 Explanation:
Here a, b, and c all have 3 different values so we need atleast 3 registers r1, r2 and r3.
Assume 'a' is mapped to r1, 'b' to r2 and 'c' to r3.
d = a + b, after this line if u notice 'a' is never present on right hand side, so we can map 'd' to r1.
e = c + d, after this line 'd' is never present on rhs, so we can map 'e' to r1.
at this time mapping is
r1  e
r2  b
r3  c
We have 3 registers for e, b and c.
f = c + e
b = c + e
These two are essentially doing same thing, after these two line 'b' and 'f' are same so we can skip computing 'f' or need not give any new register for 'f'. And wherever 'f' is present we can replace it with 'b', because neither of 'f' and 'b' are changing after these two lines, so value of these will be 'c+e' till the end of the program.
At second last line "d = 5 + e"
here 'd' is introduced, we can map it to any of the register r1 or r3, because after this line neither of 'e' or 'c' is required. Value of 'b' is required because we need to return 'd+f', and 'f' is essentially equal to 'b'
finally code becomes
r1 = 1
r2 = 10
r3 = 20
r1 = r1 + r2
r1 = r3 + r1
r2 = r3 + r1
r2 = r3 + r1
r1 = r2 + r2
r3 = 5 + r1
return r3 + r2
Therefore minimum 3 registers needed.
Assume 'a' is mapped to r1, 'b' to r2 and 'c' to r3.
d = a + b, after this line if u notice 'a' is never present on right hand side, so we can map 'd' to r1.
e = c + d, after this line 'd' is never present on rhs, so we can map 'e' to r1.
at this time mapping is
r1  e
r2  b
r3  c
We have 3 registers for e, b and c.
f = c + e
b = c + e
These two are essentially doing same thing, after these two line 'b' and 'f' are same so we can skip computing 'f' or need not give any new register for 'f'. And wherever 'f' is present we can replace it with 'b', because neither of 'f' and 'b' are changing after these two lines, so value of these will be 'c+e' till the end of the program.
At second last line "d = 5 + e"
here 'd' is introduced, we can map it to any of the register r1 or r3, because after this line neither of 'e' or 'c' is required. Value of 'b' is required because we need to return 'd+f', and 'f' is essentially equal to 'b'
finally code becomes
r1 = 1
r2 = 10
r3 = 20
r1 = r1 + r2
r1 = r3 + r1
r2 = r3 + r1
r2 = r3 + r1
r1 = r2 + r2
r3 = 5 + r1
return r3 + r2
Therefore minimum 3 registers needed.
Question 38 
The grammar S → aSabSc is
LL(1) but not LR(1)  
LR(1) but not LR(1)  
Both LL(1) and LR(1)
 
Neither LL(1) nor LR(1) 
Question 38 Explanation:
The LL(1) parsing table for the given grammar is:
As there is no conflict in LL(1) parsing table, hence the given grammar is LL(1) and since every LL(1) is LR(1) also, so the given grammar is LL(1) as well as LR(1).
As there is no conflict in LL(1) parsing table, hence the given grammar is LL(1) and since every LL(1) is LR(1) also, so the given grammar is LL(1) as well as LR(1).
Question 39 
Let L = {w ∈ (0 + 1)*  w has even number of 1s}, i.e. L is the set of all bit strings with even number of 1s. Which one of the regular expression below represents L?
(0 * 10 * 1)*
 
0 * (10 * 10*)*
 
0*(10 * 1*)*0*  
0 * 1(10 * 1)*10* 
Question 39 Explanation:
The best way to find correct answer is option elimination method. We will guess strings which has even number of 1’s and that is not generated by wrong options OR which generate strings which doesn’t have even number of 1’s.
Option A: (reg expr: (0*10*1)* ) doesn’t generate string such as { 110, 1100,....}
Option C: (reg expr: 0*(10*1*)*0* generate string such as {1, 111,....} which have odd number of 1’s.
Option D: (reg expr: 0*1(10*1)*10* doesn’t generate strings such as { 11101, 1111101, ….}.
Option A: (reg expr: (0*10*1)* ) doesn’t generate string such as { 110, 1100,....}
Option C: (reg expr: 0*(10*1*)*0* generate string such as {1, 111,....} which have odd number of 1’s.
Option D: (reg expr: 0*1(10*1)*10* doesn’t generate strings such as { 11101, 1111101, ….}.
Question 40 
Only L2 is context free  
Only L2 and L3 are context free  
Only L1 and L2 are context free  
All are context free 
Question 40 Explanation:
All languages viz L1, L2, L3 and L4 has only one comparison and it can be accepted by PDA (single stack), hence all are Context Free Languages.
Question 41 
Let w be any string of length n is {0, 1}*. Let L be the set of all substrings of w. What is the minimum number of states in a nondeterministic finite automaton that accepts L?
n1  
n  
n+1  
2^{n1} 
Question 41 Explanation:
In order to accept any string of length “n” with alphabet {0,1}, we require an NFA with “n+1” states. For example, consider a strings of length “3” such as “101”, the NFA with 4 states is given below:
Since L is set of all substrings of “w” (Substring of a string is obtained by deleting any prefix or any suffix from string), so if we consider “w” as “101” , then the substrings of w are { ϵ, 0, 1, 10, 01, 101}.
Since the string “101” is also its substring, so we require 4 states (i.e. for n length string, n+1 states are required) and the NFA would be:
Since L is set of all substrings of “w” (Substring of a string is obtained by deleting any prefix or any suffix from string), so if we consider “w” as “101” , then the substrings of w are { ϵ, 0, 1, 10, 01, 101}.
Since the string “101” is also its substring, so we require 4 states (i.e. for n length string, n+1 states are required) and the NFA would be:
Question 42 
T1 → T3 → T2  
T2 → T1 → T3  
T2 → T3 → T1  
T3 → T1 → T2 
Question 42 Explanation:
The given schedule is
― Precedence graph for the above schedule is,
― From the precedence graph the correct serialization is,
― Precedence graph for the above schedule is,
― From the precedence graph the correct serialization is,
Question 43 
100  
200  
300  
2000 
Question 43 Explanation:
Given
R(A, B, C) – 200 tuples
S(B, D, E) – 100 tuples
FD’s:
B → A
A → C
― ‘B’ is primary key for R and foreign key of S from the given FDs.
― Maximum tuples in natural join of R and S is min(200, 100) = 100.
R(A, B, C) – 200 tuples
S(B, D, E) – 100 tuples
FD’s:
B → A
A → C
― ‘B’ is primary key for R and foreign key of S from the given FDs.
― Maximum tuples in natural join of R and S is min(200, 100) = 100.
Question 44 
T1, T2, T3
 
T2, T4  
T3, T4
 
T1, T2, T4 
Question 44 Explanation:
T1 covers S1
T2 covers S3
T4 covers S2, S4.
T2 covers S3
T4 covers S2, S4.
Question 45 
At least twice  
Exactly twice  
Exactly thrice  
Exactly once 
Question 45 Explanation:
S_{0}=1
S_{1}=0
S_{2}=0
P_{0} enters the critical section first,
prints (‘0’)
releases S_{1},S_{2}(i.e., S_{1}=1 & S_{2}=1)
Now P_{1} & P_{2} both can enter critical section releases S_{0} & prints (‘0’)
This process continues, hence the number of zero’s printed ≥2.
S_{1}=0
S_{2}=0
P_{0} enters the critical section first,
prints (‘0’)
releases S_{1},S_{2}(i.e., S_{1}=1 & S_{2}=1)
Now P_{1} & P_{2} both can enter critical section releases S_{0} & prints (‘0’)
This process continues, hence the number of zero’s printed ≥2.
Question 46 
n = 40, k = 26  
n = 21, k = 12  
n = 20, k = 10
 
n = 41, k = 19

Question 46 Explanation:
Consider the case where i = 10 & i = 11, n = 21 & k = 12
P_{10} requests R_{10} & R_{11}
P_{11} requests R_{10} & R_{8}
Hence P_{10} & P_{11} inorder in deadlock.
P_{10} requests R_{10} & R_{11}
P_{11} requests R_{10} & R_{8}
Hence P_{10} & P_{11} inorder in deadlock.
Question 47 
Suppose computers A and B have IP addresses 10.105.1.113 and 10.105.1.91 respectively and they both use the same net mask N. Which of the values of N given below should not be used if A and B should belong to the same network?
255.255.255.0  
255.255.255.128  
255.255.255.192  
255.255.255.224

Question 47 Explanation:
When we perform bitwise AND operation between IP Address and Subnet Mask, it gives Network ID. If for both IP results is same Network ID. It means, both IP are belong to the same network else it's on different network.
When we perform AND operation between IP address 10.105.1.113 and 255.255.255.224 result is 10.105.1.96 and when we perform AND operation between IP address 10.105.1.91 and 255.255.255.224 result is 10.105.1.64.
Therefore, 10.105.1.96 and 10.105.1.64 are different network, so D is correct answer.
When we perform AND operation between IP address 10.105.1.113 and 255.255.255.224 result is 10.105.1.96 and when we perform AND operation between IP address 10.105.1.91 and 255.255.255.224 result is 10.105.1.64.
Therefore, 10.105.1.96 and 10.105.1.64 are different network, so D is correct answer.
Question 48 
2 nanoseconds  
20 nanoseconds  
22 nanoseconds  
88 nanoseconds 
Question 48 Explanation:
Between L1 cache block is of size 4 words, and the bus width between L1 and L2 is also of size 4 words. So only one time the data needs to be transferred from L2 to L1.
And it requires one access of L1 cache and one access of L2 cache. So time taken = 20+2 = 22ns
And it requires one access of L1 cache and one access of L2 cache. So time taken = 20+2 = 22ns
Question 49 
222 nanoseconds
 
888 nanoseconds  
902 nanoseconds
 
968 nanoseconds

Question 49 Explanation:
Between L1 cache block is of size 4 words, and the bus width between L1 and L2 is also of size 4 words. So only one time the data needs to be transferred from L2 to L1.
And it requires one access of L1 cache and one access of L2 cache. So time taken = 20+2 = 22ns
L2 cache block is of size 16 words and nothing is mentioned about the main memory block, so we assume the main memory block is also of size 16 words. But the bus between L2 and main memory is only 4 words..so when the data has to be transferred from main memory to L2 cache it has to send 4 times through the data bus.
When a data request comes to L1, if there is a cache miss in L1 then it will be searched in L2 if there is a hit in L2 then the required data is transferred from L2 to L1 in a single transfer through the bus. If there is a miss in L2 then it will be searched in main memory. Then the 16 words data from main memory will be transferred to L2 in 4 transfers through the data bus. Time taken for this = 4*(200+20) = 4*220 = 880 ns
Then from L2 to L1 only 4 words of data will be transferred through the data bus in a single transfer. We know time taken for this is 22ns.
So total time taken = 880 + 22 = 902 ns
And it requires one access of L1 cache and one access of L2 cache. So time taken = 20+2 = 22ns
L2 cache block is of size 16 words and nothing is mentioned about the main memory block, so we assume the main memory block is also of size 16 words. But the bus between L2 and main memory is only 4 words..so when the data has to be transferred from main memory to L2 cache it has to send 4 times through the data bus.
When a data request comes to L1, if there is a cache miss in L1 then it will be searched in L2 if there is a hit in L2 then the required data is transferred from L2 to L1 in a single transfer through the bus. If there is a miss in L2 then it will be searched in main memory. Then the 16 words data from main memory will be transferred to L2 in 4 transfers through the data bus. Time taken for this = 4*(200+20) = 4*220 = 880 ns
Then from L2 to L1 only 4 words of data will be transferred through the data bus in a single transfer. We know time taken for this is 22ns.
So total time taken = 880 + 22 = 902 ns
Question 50 
7  
8  
9  
10 
Question 50 Explanation:
Here the point to be noted is that vertex 0 is a leaf node. So degree of vertex 0 cannot be more than or equal to 2.
So, the edges of the spanning tree given that vertex 0 is a leaf node in the tree,
So, the minimum possible weight of spanning tree will be
= 1 + 4 + 2 + 3
= 10
So, the edges of the spanning tree given that vertex 0 is a leaf node in the tree,
So, the minimum possible weight of spanning tree will be
= 1 + 4 + 2 + 3
= 10
Question 51 
7  
8  
9  
10 
Question 51 Explanation:
The minimum possible weight of a path p from vertex 1 to vertex 2 such that p contains atmost 3 edges,
= 1 + 4 + 3
= 8
Question 52 
46, 42, 34, 52, 23, 33
 
34, 42, 23, 52, 33, 46
 
46, 34, 42, 23, 52, 33  
42, 46, 33, 23, 34, 52

Question 52 Explanation:
Hash Table consists of 10 slots, uses Open Addressing with hash function and linear probing.
After inserting 6 values, the table looks like
The possible order in which the keys are inserted are:
34, 42, 23, 46 are at their respective slots 4, 2, 3, 6.
52, 33 are at different positions.
― 52 has come after 42, 23, 34 because, it has the collision with 42, because of linear probing, it should have occupy the next empty slot. So 52 is after 42, 23, 34.
― 33 is after 46, because it has the clash with 23. So it got placed in next empty slot 7, which means 2, 3, 4, 5, 6 are filled.
42, 23, 34 may occur in any order but before 52 & 46 may come anywhere but before 33.
So option (C)
After inserting 6 values, the table looks like
The possible order in which the keys are inserted are:
34, 42, 23, 46 are at their respective slots 4, 2, 3, 6.
52, 33 are at different positions.
― 52 has come after 42, 23, 34 because, it has the collision with 42, because of linear probing, it should have occupy the next empty slot. So 52 is after 42, 23, 34.
― 33 is after 46, because it has the clash with 23. So it got placed in next empty slot 7, which means 2, 3, 4, 5, 6 are filled.
42, 23, 34 may occur in any order but before 52 & 46 may come anywhere but before 33.
So option (C)
Question 53 
10  
20  
30  
40 
Question 53 Explanation:
Total 6 insertions
― 33 must be inserted at last (only one possibility)
― 46 can be inserted in any of the 5 places remaining. So 5 ways.
― 52 must be inserted only after inserting 42, 23, 34. So only one choice for 52.
〈42,23,34〉can be sequenced in 3! ways.
Hence, 5×3! = 30 ways
― 33 must be inserted at last (only one possibility)
― 46 can be inserted in any of the 5 places remaining. So 5 ways.
― 52 must be inserted only after inserting 42, 23, 34. So only one choice for 52.
〈42,23,34〉can be sequenced in 3! ways.
Hence, 5×3! = 30 ways
Question 54 
4  
3  
2  
1 
Question 54 Explanation:
Link R1 R2 will not be used because its cost is 6 and link R1R3R2 costs 5, which is lesser than R1R2 link.
Similarly, link R4R6 will not be used, instead this link we can use R4R5R6 link which costs only 5 unit.
Similarly, link R4R6 will not be used, instead this link we can use R4R5R6 link which costs only 5 unit.
Question 55 
0  
1  
2  
4 
Question 55 Explanation:
Now Graph will look like
And only link that will be removed is R5R6 link.
And only link that will be removed is R5R6 link.
Question 56 
masked  
belied
 
betrayed  
suppressed

Question 56 Explanation:
Masked  opposite to given
Belied  Not appropriate to given
Betrayed  Most appropriate to given (Reveal intentionally)
Suppressed  Irrevalent to given
Answer is option C.
Belied  Not appropriate to given
Betrayed  Most appropriate to given (Reveal intentionally)
Suppressed  Irrevalent to given
Answer is option C.
Question 57 
Which of the following options is closest in meaning to the word Circuitous.
cyclic  
indirect
 
confusing  
crooked 
Question 57 Explanation:
Synonyms for Circuitous are indirect, oblique, winding, meandering etc.
Question 58 
uphold  
restrain  
cherish  
conserve 
Question 58 Explanation:
Uphold  cause to remain
Restrain  keep under control
Cherish  befond of
Conserve  protect from harm i.e., keeping safety, loss of destruction
Answer is option d.
Restrain  keep under control
Cherish  befond of
Conserve  protect from harm i.e., keeping safety, loss of destruction
Answer is option d.
Question 59 
25 persons are in a room. 15 of them play hockey, 17 of them play football and 10 of them play both hockey and football. Then the number of persons playing neither hockey nor football is:
2  
17  
13  
3 
Question 59 Explanation:
Total number of persons = a+b+c = 25 → (1)
Number of persons who play hockey = a+b = 15 → (2)
Number of persons who play football = b+c = 17 → (3)
Number of persons who play hockey and football = b = 10 → (4)
From (2) ⇒ a=5
From (3) ⇒ c=7
From (1) ⇒ d = 3
Number of persons who play neither hockey nor football = d = 3
Question 60 
fallow: land  
unaware: sleeper  
wit: jester  
renovated: house 
Question 60 Explanation:
→ A worker who is working is called unemployed.
→ Same as a land which is not used called fallow.
→ Same as a land which is not used called fallow.
Question 61 
If 137+276 = 435 how much is 731+672?
534  
1403
 
1623  
1513 
Question 61 Explanation:
Let, base = x
(137)_{x} + (276)_{x} = (435)_{x}
x^{2} + 3x + 7 + 2x^{2} + 7x + 6 = 4x^{2} + 3x + 5
x^{2}  7x  8 = 0
x = 8 (or) 1
∴ x = 8 (1 cannot be a base)
(731)_{x} + (672)_{x} = (731)_{8} +( 672)_{8}
= 7 × 8^{2} + 3× 8 + 1×1 + 6 × 8^{2} + 7 × 8 + 2 × 1
= 915
From the options, 915 can be written as 1623 in base 8.
(137)_{x} + (276)_{x} = (435)_{x}
x^{2} + 3x + 7 + 2x^{2} + 7x + 6 = 4x^{2} + 3x + 5
x^{2}  7x  8 = 0
x = 8 (or) 1
∴ x = 8 (1 cannot be a base)
(731)_{x} + (672)_{x} = (731)_{8} +( 672)_{8}
= 7 × 8^{2} + 3× 8 + 1×1 + 6 × 8^{2} + 7 × 8 + 2 × 1
= 915
From the options, 915 can be written as 1623 in base 8.
Question 62 
HSIG  
SGHI  
IGSH  
IHSG 
Question 62 Explanation:
Let us solve using option elimination approach .
Option A: HSIG
from (ii) , we can say that Gita and Saira are successive siblings.
Hence, option A is not true.
Option C: IGSH
In any of the above 4 cases, (i) is not true.
Hence, option C is not true.
Option D: IHSG
In any of the above 4 cases, (i) is not true.
Hence, option D is not true.
Option B: SGHI
In last two cases, all the facts are true.
∴ The order is SGHI.
Option A: HSIG
from (ii) , we can say that Gita and Saira are successive siblings.
Hence, option A is not true.
Option C: IGSH
In any of the above 4 cases, (i) is not true.
Hence, option C is not true.
Option D: IHSG
In any of the above 4 cases, (i) is not true.
Hence, option D is not true.
Option B: SGHI
In last two cases, all the facts are true.
∴ The order is SGHI.
Question 63 
Modern warfare has resulted in civil strife.  
Chemical agents are useful in modern warfare.  
Use of chemical agents in warfare would be undesirable.  
People in military establishments like to use chemical agents in war.

Question 63 Explanation:
→ People in military establishments like to use chemical agents in war.
Question 64 
5 skilled workers can build a wall in 20 days: 8 semiskilled workers can build a wall in 25 days; 10 unskilled workers can build a wall in 30 days. If a team has 2 skilled, 6 semiskilled and 5 unskilled workers, how long will it take to build the wall?
20  
10  
16  
15 
Question 64 Explanation:
5 skilled workers can build the wall in 20 days
⇒ 1 skilled worker can build the wall in 100 days
Capacity = 1/100
8 semiskilled workers can build the wall in 25 days
⇒ 1 semiskilled worker can build the wall in 200 days
Capacity = 1/200
10 unskilled workers can build the wall in 30 days
⇒ 1 unskilled worker can build the wall in 300 days
Capacity = 1/300
1 day work (2 skilled + 6 semiskilled + 5 unskilled) = 2(1/100) + 6(1/200) + 5(1/300) = 1/15
∴ They can complete the work in 15 days.
Capacity = 1/100
8 semiskilled workers can build the wall in 25 days
⇒ 1 semiskilled worker can build the wall in 200 days
Capacity = 1/200
10 unskilled workers can build the wall in 30 days
⇒ 1 unskilled worker can build the wall in 300 days
Capacity = 1/300
1 day work (2 skilled + 6 semiskilled + 5 unskilled) = 2(1/100) + 6(1/200) + 5(1/300) = 1/15
∴ They can complete the work in 15 days.
Question 65 
Given digits 2, 2, 3, 3, 3, 4, 4, 4, 4 how many distinct 4 digit numbers greater than 3000 can be formed?
50  
51  
52  
54 
Question 65 Explanation:
Greater than 3000
⇒ First digit: 3 or 4
(i) First digit  3:
We have to choose 3 digits from (2, 2, 3, 3, 4, 4, 4, 4).
Any place can have either 2 or 3 or 4, but (222, 333) is not possible as we have only two 2's and two 3's.
Total = 3 × 3 × 3  2 = 25
(ii) First digit  4:
We have to choose 4 digits from (2, 2, 3, 3, 4, 4, 4, 4).
Any place can have either 2 or 3 or 4, but (222) is not possible we have only two 2's.
Total = 3 × 3 × 3  1 = 26
∴ Total number possible = 25 + 26 = 51
⇒ First digit: 3 or 4
(i) First digit  3:
We have to choose 3 digits from (2, 2, 3, 3, 4, 4, 4, 4).
Any place can have either 2 or 3 or 4, but (222, 333) is not possible as we have only two 2's and two 3's.
Total = 3 × 3 × 3  2 = 25
(ii) First digit  4:
We have to choose 4 digits from (2, 2, 3, 3, 4, 4, 4, 4).
Any place can have either 2 or 3 or 4, but (222) is not possible we have only two 2's.
Total = 3 × 3 × 3  1 = 26
∴ Total number possible = 25 + 26 = 51
There are 65 questions to complete.