Gate 2011
Question 2 
The minimum number of D flipflops needed to design a mod258 counter is
9  
8  
512  
258 
Question 2 Explanation:
Let n is the number of flipflops.
The max Mod values is 2n.
So 2^{n} ≥ 258 ⇒ n = 9
The max Mod values is 2n.
So 2^{n} ≥ 258 ⇒ n = 9
Question 3 
Which one of the following circuits is NOT equivalent to a 2input XNOR (exclusive NOR) gate?
Question 3 Explanation:
All options except option ‘D’ gives EXNOR gates
Question 4 
A thread is usually defined as a "light weight process" because an operating system (OS) maintains smaller data structures for a thread than for a process. In relation to this, which of the following is TRUE?
On perthread basis, the OS maintains only CPU register state  
The OS does not maintain a separate stack for each thread  
On perthread basis, the OS does not maintain virtual memory state  
On per thread basis, the OS maintains only scheduling and accounting information

Question 4 Explanation:
A) False, because on per thread basis OS maintains register, stack and program counter.
B) False, OS do maintain a separate stack for each thread.
C) True
D) False
B) False, OS do maintain a separate stack for each thread.
C) True
D) False
Question 5 
K4 is planar while Q3 is not  
Both K4 and Q3 are planar  
Q3 is planar while K4 is not  
Neither K4 not Q3 is planar 
Question 5 Explanation:
Both the above graphs are planar.
Question 6 
If the difference between the expectation of the square of a random variable (E[X^{2}]) and the square of the expectation of the random variable (E[X])^{2} is denoted by R, then
R = 0  
R < 0  
R ≥ 0  
R > 0 
Question 6 Explanation:
We know that difference of E(X^{2}) and E(X))^{2} is nothing but variance or V(X) which is always greater than or equal to zero.
So the answer will be R≥0.
So the answer will be R≥0.
Question 7 
The lexical analysis for a modern computer language such as Java needs the power of which one of the following machine models in a necessary and sufficient sense?
Finite state automata  
Deterministic pushdown automata  
NonDeterministic pushdown automata  
Turing machine 
Question 7 Explanation:
Lexical Analysis is implemented by finite automata
Question 8 
Let the page fault service time to 10 ms in a computer with average memory access time being 20 ns. If one page fault is generated for every 10^{6} memory accesses, what is the effective access time for the memory?
21 ns  
30 ns  
23 ns  
35 ns 
Question 8 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 9 
Consider a hypothetical processor with an instruction of type LW R1, 20 (R2), which during execution reads a 32bit word from memory and stores it in a 32bit register R1. The effective address of the memory location is obtained by the addition of constant 20 and the contents of register R2. Which of the following best reflects the addressing mode implemented by this instruction for the operand memory?
Immediate Addressing  
Register Addressing  
Register Indirect Scaled Addressing  
Base Indexed Addressing 
Question 9 Explanation:
Here 20 will act as base and content of R2 will be index.
Question 10 
GATE2011  
E2011  
2011  
011 
Question 10 Explanation:
p is the starting address of array.
p[3]  p[1] = 4, and p+4 will be pointing to the fifth position in the array 'c'. So printf starts printing from 2 and prints 2011.
p[3]  p[1] = 4, and p+4 will be pointing to the fifth position in the array 'c'. So printf starts printing from 2 and prints 2011.
Question 11 
A maxheap is a heap where the value of each parent is greater than or equal to the value of its children. Which of the following is a maxheap?
Question 11 Explanation:
Heap is a complete binary tree
Option A: It violates the property of complete binary tree.
Option C: 8 is greater than 5. The root value is always greater than his children. So, the above tree is violating the property of max heap
Option D: 10 is greater than 8. The root value is always greater than his children. So, the above tree is violating the property of max heap
Option A: It violates the property of complete binary tree.
Option C: 8 is greater than 5. The root value is always greater than his children. So, the above tree is violating the property of max heap
Option D: 10 is greater than 8. The root value is always greater than his children. So, the above tree is violating the property of max heap
Question 12 
The algorithm uses dynamic programming paradigm  
The algorithm has a linear complexity and uses branch and bound paradigm  
The algorithm has a nonlinear polynomial complexity and uses branch and bound paradigm  
The algorithm uses divide and conquer paradigm. 
Question 12 Explanation:
Key Note of Dynamic programming:
1. Dynamic programming is when you use past knowledge to make solving a future problem easier.
2. Dynamic programming is a technique used to avoid computing multiple time the same subproblem in a recursive algorithm.
Note: It is neither backtracking nor branch and bound because we are not branching anywhere in the solution space. The algorithm is not divide and conquer as we are not dividing the problem and then merging the solution
2. Dynamic programming is a technique used to avoid computing multiple time the same subproblem in a recursive algorithm.
Note: It is neither backtracking nor branch and bound because we are not branching anywhere in the solution space. The algorithm is not divide and conquer as we are not dividing the problem and then merging the solution
Question 13 
Let P be a regular language and Q be a contextfree language such that Q ⊆ P. (For example, let P be the language represented by the regular expression p*q* and Q be [p^{n}q^{n}  n ∈ N]). Then which of the following is ALWAYS regular?
P ∩ Q  
P – Q  
Σ* – P  
Σ* – Q 
Question 13 Explanation:
Exp: Σ*  P is the complement of P so it is always regular,
since regular languages are closed under complementation
since regular languages are closed under complementation
Question 14 
In a compiler, keywords of a language are recognized during
parsing of the program  
the code generation
 
the lexical analysis of the program  
dataflow analysis 
Question 14 Explanation:
Any identifier is also a token so it is recognized in lexical Analysis
Question 15 
A layer4 firewall (a device that can look at all protocol headers up to the transport layer) CANNOT
block entire HTTP traffic during 9:00PM and 5:00AM  
block all ICMP traffic  
stop incoming traffic from a specific IP address but allow outgoing traffic to the same IP address  
block TCP traffic from a specific user on a multiuser system during 9:00PM and 5:00AM 
Question 15 Explanation:
(A) It is possible to block entire HTTP traffic by blocking port no.80.
(B) Possible because it is network layer protocol.
(C) Possible because SP address is present in Network layer.
(D) Not possible, because to block specific user, we need user id which is present in Application layer.
(B) Possible because it is network layer protocol.
(C) Possible because SP address is present in Network layer.
(D) Not possible, because to block specific user, we need user id which is present in Application layer.
Question 16 
If two fair coins are flipped and at least one of the outcomes is known to be a head, what is the probability that both outcomes are heads?
1/3  
1/4  
1/2  
2/3 
Question 16 Explanation:
Sample space = {HH, HT, TH}
Required probability = 1/3
Question 17 
m1:HTTP m2: SMTP m3: POP  
m1:SMTP m2:FTP m3:HTTP  
m1: SMTP m2: POP m3: HTTP  
m1: POP m2: SMTP m3: IMAP 
Question 17 Explanation:
Sending an email will be done through user agent and message transfer agent by SMTP, downloading an email from mail box is done through POP, checking email in a web browser is done through HTTP
Question 18 
4000  
5000  
4333  
4667 
Question 18 Explanation:
Note: Out of syllabus.
LOC L^{2 } = X
L2 = 2X
Total cost of project
×/1000 ×1000000+5+100000=2×/10000 ×750000+50000 ×5
100×+500000=150×+250000
⟹50×=500000250000
LOC L^{2 } = X
L2 = 2X
Total cost of project
×/1000 ×1000000+5+100000=2×/10000 ×750000+50000 ×5
100×+500000=150×+250000
⟹50×=500000250000
Question 19 
Let the time taken to switch between user and kernel modes of execution be t_{1} while the time taken to switch between two processes be t_{2}. Which of the following is TRUE?
(t_{1}) > (t_{2})  
(t_{1}) = (t_{2})  
(t_{1}) < (t_{2})  
Nothing can be said about the relation between t_{1} and t_{2} 
Question 19 Explanation:
Context switch between the processes involves mode switch also.
Question 20 
A company needs to develop digital signal processing software for one of its newest inventions. The software is expected to have 40000 lines of code. The company needs to determine the effort in personmonths needed to develop this software using the basic COCOMO model. The multiplicative factor for this model is given as 2.8 for the software development on embedded systems, while the exponentiation factor is given as 1.20. What is the estimated effort in personmonths?
234.25  
932.50  
287.80  
122.40 
Question 20 Explanation:
Note: Out of syllabus.
Effort person per month
= α.(KDSI)B
KDSI = Kilo LOC
= 2.8 × (40)1.20
= 2.8 × 83.6511
= 234.22 person per month
Effort person per month
= α.(KDSI)B
KDSI = Kilo LOC
= 2.8 × (40)1.20
= 2.8 × 83.6511
= 234.22 person per month
Question 21 
Which of the following pairs have DIFFERENT expressive power?
Deterministic finite automata (DFA) and Nondeterministic finite automata (NFA)  
Deterministic push down automata (DPDA) and Nondeterministic push down automata (NFDA)  
Deterministic singletape Turning machine and Nondeterministic single tape Turning machine  
Singletape Turning machine and multitape Turning machine 
Question 21 Explanation:
NPDA is more powerful than DPDA.
Hence answer is (B)
Hence answer is (B)
Question 22 
HTML (Hyper Text Markup Language) has language elements which permit certain actions other than describing the structure of the web document. Which one of the following actions is NOT supported by pure HTML (without any server or client side scripting) pages?
Embed web objects from different sites into the same page  
Refresh the page automatically after a specified interval  
Automatically redirect to another page upon download  
Display the client time as part of the page 
Question 22 Explanation:
(D) Impossible without javascript.
Question 23 
Which of the following is NOT desired in a good Software Requirement Specifications (SRS) document?
Functional Requirements
 
NonFunctional Requirements
 
Goals of Implementation
 
Algorithms for Software Implementation 
Question 23 Explanation:
Note: Out of syllabus.
Question 24 
Interrupt from Hard Dist  
Interrupt from Mouse  
Interrupt from Keyboard  
Interrupt from CPU temperature sensor 
Question 24 Explanation:
The interrupts of higher priority are assigned to requests which, if delayed or interrupted, can cause serious consequences. Device with high speed transfer such as magnetic disks or hard disks are given higher priority, and slow devices such as keyboard receive low priority. We also know that delaying a CPU temperature sensor could have serious problems like overheat can damage CPU circuitry.
Therefore, we can conclude, CPU temperature sensor > Hard disk > Mouse > Keyboard.
Therefore, we can conclude, CPU temperature sensor > Hard disk > Mouse > Keyboard.
Question 25 
BankAccount_Num is a candidate key  
Registration_Num can be a primary key
 
UID is a candidate key if all students are from the same country  
If S is a superkey such that S∩UID is NULL then S∪UID is also a superkey 
Question 25 Explanation:
In case two students hold joint account then Bank Account_Num will not uniquely determine other attributes.
Question 26 
f_{1}, f_{2}, f_{4}, f_{1}  
f_{3}, f_{2}, f_{1}, f_{4}  
f_{2}, f_{3}, f_{1}, f_{4}  
f_{2}, f_{3}, f_{4}, f_{1} 
Question 26 Explanation:
Explanation: If they are expecting to find an asymptotic complexity functions means
→ Divide functions into 2 categories Polynomial functions Exponential functions above 4 functions we have only one exponential function is f1 (n) = 2n . So, It’s value is higher than to rest of the functions.
Substitute log on both sides then we get an ascending order is f3, f2, f4.
→ Divide functions into 2 categories Polynomial functions Exponential functions above 4 functions we have only one exponential function is f1 (n) = 2n . So, It’s value is higher than to rest of the functions.
Substitute log on both sides then we get an ascending order is f3, f2, f4.
Question 27 
248000  
44000  
19000  
25000 
Question 27 Explanation:
It is basically matrix chain multiplication problem. We get minimum number of multiplications using ((M1 X (M2 X M3)) X M4).
> Total number of multiplications = 100x20x5 (for M2 x M3) + 10x100x5 + 10x5x80 =19000. P2=20
P3=5
P4=80
Algorithm Recurrence relation:
P0=10
P1=100
P2=20
P3=5
P4=80
Algorithm Recurrence relation:
m[1,2]=P0*P1*P2=10*100*20=20000
m[2,3]=10000
m[3,4]=8000
> Total number of multiplications = 100x20x5 (for M2 x M3) + 10x100x5 + 10x5x80 =19000. P2=20
P3=5
P4=80
Algorithm Recurrence relation:
P0=10
P1=100
P2=20
P3=5
P4=80
Algorithm Recurrence relation:
m[1,2]=P0*P1*P2=10*100*20=20000
m[2,3]=10000
m[3,4]=8000
Question 28 
Ordered indexing will always outperform hashing for both queries  
Hashing will always outperform ordered indexing for both queries  
Hashing will outperform ordered indexing on Q1, but not on Q2  
Hashing will outperform ordered indexing on Q2, but not on Q1. 
Question 28 Explanation:
Hashing works well on the "equal" queries, while ordered indexing works well better on range queries.
For example, consider B^{+} tree, once you have searched a key in B^{+}; you can find range of values via the block pointers pointing to another block of values on the leaf node level.
For example, consider B^{+} tree, once you have searched a key in B^{+}; you can find range of values via the block pointers pointing to another block of values on the leaf node level.
Question 29 
1, 4, 3  
3, 7, 3  
7, 3, 2
 
1, 2, 3 
Question 29 Explanation:
Given matrix is upper triangular matrix and its diagonal elements are its eigen values = 1, 4, 3
Question 31 
k+1  
n+1  
2^{n+1}  
2^{k+1} 
Question 31 Explanation:
Given that n is a constant.
So lets check of n = 2,
L = a_{2k}, k>0
Since k>0 than zero.
So L is the language accepting even no. of a's except 'ε'.
So DFA will be,
So, no. of states required is 2+1 = 3.
So for a^{nk}, (n+1) states will be required.
So lets check of n = 2,
L = a_{2k}, k>0
Since k>0 than zero.
So L is the language accepting even no. of a's except 'ε'.
So DFA will be,
So, no. of states required is 2+1 = 3.
So for a^{nk}, (n+1) states will be required.
Question 32 
4864 bits  
6144 bits  
6656 bits  
5376 bits 
Question 32 Explanation:
No. of cache blocks = cache size/size of a block = 8KB/32B = 256
So we need 8 bits for indexing the 256 blocks in the cache. And since a block is 32 bytes we need 5 word bits to address each byte.
So out of remaining (32  8  5), 19 bits should be tag bits.
So tag entry size = 19 + 1 (valid bit) + 1 (modified bit) = 21 bits
∴ Total size of metadata = 21 × Number blocks = 21 × 256 = 5376 bits
So we need 8 bits for indexing the 256 blocks in the cache. And since a block is 32 bytes we need 5 word bits to address each byte.
So out of remaining (32  8  5), 19 bits should be tag bits.
So tag entry size = 19 + 1 (valid bit) + 1 (modified bit) = 21 bits
∴ Total size of metadata = 21 × Number blocks = 21 × 256 = 5376 bits
Question 33 
An application loads 100 libraries at startup. Loading each library requires exactly one disk access. The seek time of the disk to a random location is given as 10 ms. Rotational speed of disk is 6000 rpm. If all 100 libraries are loaded from random locations on the disk, how long does it take to load all libraries? (The time to transfer data from the disk block once the head has been positioned at the start of the block may be neglected).
0.50 s  
1.50 s  
1.25 s  
1.00 s 
Question 33 Explanation:
Transfer Time = Seek Time + Rotational latency/2 + time to read/write
= 10ms + (60/12000)+0
For 100 libraries = 100(10ms + (60/12000)+0) = 1.5seconds
= 10ms + (60/12000)+0
For 100 libraries = 100(10ms + (60/12000)+0) = 1.5seconds
Question 34 
Question 34 Explanation:
(A) It is wrong.
(B) Accepting string 'b' which is not accepted by given DFA.
(C) Same reason as (B).
(D) Accepting string 'bba' which is not accepted by given DFA.
(B) Accepting string 'b' which is not accepted by given DFA.
(C) Same reason as (B).
(D) Accepting string 'bba' which is not accepted by given DFA.
Question 35 
T1, T2, T3, T6  
T1, T3, T4, T5  
T2, T4, T5, T6  
T2, T3, T4, T5 
Question 35 Explanation:
Note: Out of syllabus.
T1 and T2 checking same condition a = 0 hence, any one of T1 and T2 is redundant.
T3, T4: in both case discriminant (D) = b2 – 4ac = 0. Hence any one of it is
T5 : D > 0
T6 : D < 0
T1 and T2 checking same condition a = 0 hence, any one of T1 and T2 is redundant.
T3, T4: in both case discriminant (D) = b2 – 4ac = 0. Hence any one of it is
T5 : D > 0
T6 : D < 0
Question 37 
Consider two binary operators ‘↑’ and ‘↓’ with the precedence of operator ↓ being lower than that of the operator ↑. Operator ↑ is right associative while operator ↓, is left associative. Which one of the following represents the parse tree for expression (7↓3↑4↑3↓2)?
Question 37 Explanation:
7 ↓ 3 ↑ 4 ↑ 3 ↓ 2
⇒ 7 ↓ (3 ↑ (4 ↑ 3)) ↓ 2
⇒ 7 ↓ (3 ↑ (4 ↑ 3))) ↓ 2 as ↓ is left associative
⇒ 7 ↓ (3 ↑ (4 ↑ 3)) ↓ 2
⇒ 7 ↓ (3 ↑ (4 ↑ 3))) ↓ 2 as ↓ is left associative
Question 38 
Push Down Automate (PDA) can be used to recognize L1 and L2  
L1 is a regular language  
All the three languages are context free  
Turing machines can be used to recognize all the languages 
Question 38 Explanation:
L1: regular language
L2: context free language
L3: context sensitive language
L2: context free language
L3: context sensitive language
Question 39 
3.4  
4.4  
5.1  
6.7

Question 39 Explanation:
No. of clock cycles required by using loadstore approach = 2 + 500 × 7 = 3502 and that of by
using DMA = 20 + 500 × 2 = 1020
Required speed up = 3502/1020=3.4
using DMA = 20 + 500 × 2 = 1020
Required speed up = 3502/1020=3.4
Question 40 
We are given a set of n distinct elements and an unlabeled binary tree with n nodes. In how many ways can we populate the tree with the given set so that it becomes a binary search tree?
0  
1  
n!  
Question 40 Explanation:
Corresponding to each set only 1 binary search tree can be formed because inorder is fixed. only 1 tree possible. If Binary trees would be asked n! possible corresponding to each distinct tree set. Here tree structure is fixed and have only 1 possibility for BST as elements are distinct.
Question 41 
P(x) being true means that x is a prime number  
P(x) being true means that x is a number other than 1
 
P(x) is always true irrespective of the value of x
 
P(x) being true means that x has exactly two factors other than 1 and x 
Question 41 Explanation:
Statement: x is not equal to 1 and if there exists some z for all y such that product of y and z is x, then y is either the no. itselfor 1.
This is the definition of prime nos.
This is the definition of prime nos.
Question 44 
Consider a finite sequence of random values X = [x_{1}, x_{2}, …, x_{n}]. Let μ_{x} be the mean and σ_{x} be the standard deviation of X. Let another finite sequence Y of equal length be derived from this as y_{i} = a * x_{i} + b, where a and b are positive constants. Let μ_{y} be the mean and σ_{y} be the standard deviation of this sequence. Which one of the following statements is INCORRECT?
Index position of mode of X in X is the same as the index position of mode of Y in Y.  
Index position of median of X in X is the same as the index position of median of Y in Y.
 
μ_{y} = aμ_{x} + b  
σ_{y} = aσ_{x} + b 
Question 44 Explanation:
σ_{y} is standard deviation then
(σ_{y})^{2} is variance so,
y_{i} = a * x_{i} + b
(σ_{y})^{2} = a^{2 }(σ_{x})^{2}
⇒ σ_{y} = a σ_{x}
Hence option (D) is incorrect.
(σ_{y})^{2} is variance so,
y_{i} = a * x_{i} + b
(σ_{y})^{2} = a^{2 }(σ_{x})^{2}
⇒ σ_{y} = a σ_{x}
Hence option (D) is incorrect.
Question 45 
A deck of 5 cards (each carrying a distinct number from 1 to 5) is shuffled thoroughly. Two cards are then removed one at a time from the deck. What is the probability that the two cards are selected with the number on the first card being one higher than the number on the second.
1/5  
4/25  
1/4  
2/5 
Question 45 Explanation:
The possible events are
(2,1) (3,2) (4,3) (5,4).
So only 4 possibilities are there and sample space will be,
^{5}C_{1} × ^{4}C_{1} = 20
So probability = 4/20 = 1/5
(2,1) (3,2) (4,3) (5,4).
So only 4 possibilities are there and sample space will be,
^{5}C_{1} × ^{4}C_{1} = 20
So probability = 4/20 = 1/5
Question 46 
5.0 ms  
4.33 ms  
6.33 ms  
7.33 ms 
Question 46 Explanation:
P0 P1 Po P2
0 1 5 13 22
Average waiting time = 4+113=5 ms
0 1 5 13 22
Average waiting time = 4+113=5 ms
Question 47 
2  
9  
5  
3 
Question 47 Explanation:
Load R1, a ; R1 ← M[a]
Load R2, b ; R2 ← M[b]
Sub R1, R2 ; R1 ← R1 – R2
Load R2, c ; R2 ← M[c]
Load R3, d ; R3 ← M[d]
Add R2, R3 ; R2 ← R2 + R3
Load R3, e ; R3 ← M[e]
Sub R3, 3 ; R3 ← R3 – R2
Add R1, R3 ; R1 ← R1 + R3
Total 3 Registers are required minimum
Load R2, b ; R2 ← M[b]
Sub R1, R2 ; R1 ← R1 – R2
Load R2, c ; R2 ← M[c]
Load R3, d ; R3 ← M[d]
Add R2, R3 ; R2 ← R2 + R3
Load R3, e ; R3 ← M[e]
Sub R3, 3 ; R3 ← R3 – R2
Add R1, R3 ; R1 ← R1 + R3
Total 3 Registers are required minimum
Question 52 
1/12(11n^{2}5n)  
n^{2} – n + 1  
6n – 11  
2n + 1 
Question 52 Explanation:
Let take example of 5 vertices,
Cost of MST,
= 3+4+6+8 = 21
Only option (B) satisfies it.
Cost of MST,
= 3+4+6+8 = 21
Only option (B) satisfies it.
Question 53 
11
 
25  
31  
41 
Question 53 Explanation:
Lets first draw graph with 10 vertices,
Now MST of above graph is,
∴ The length of path from v_{5} to v_{6} in the MST is,
8+4+3+6+10 = 31
Now MST of above graph is,
∴ The length of path from v_{5} to v_{6} in the MST is,
8+4+3+6+10 = 31
Question 55 
3  
9  
10  
∞ 
Question 55 Explanation:
N3 has neighbors N2 and N4
N2 has made entry ∞
N4 has the distance of 8 to N1
N3 has the distance of 2 to N4
So 2 + 8 = 10
N2 has made entry ∞
N4 has the distance of 8 to N1
N3 has the distance of 2 to N4
So 2 + 8 = 10
Question 56 
If Log (P) = (1/2) Log (Q) = (1/3) Log (R), then which of the following options is TRUE?
P^{2} = Q^{3}R^{2}  
Q^{2} = PR  
Q^{2} = R^{3}P  
R = P^{2}Q^{2} 
Question 56 Explanation:
logP = 1/2 logQ = 1/3 log (R) = k
∴ P = b^{k}, Q = b^{2k}, R = b^{3k}
Now, Q^{2} = b^{4k} = b^{3k} b^{k} = PR
∴ P = b^{k}, Q = b^{2k}, R = b^{3k}
Now, Q^{2} = b^{4k} = b^{3k} b^{k} = PR
Question 57 
to visit  
having to visit  
visiting  
for a visit 
Question 57 Explanation:
Contemplate is a transitive verb and hence is followed by a gerund Hence the correct usage of contemplate is verb + ing form.
Question 58 
hyperbolic  
restrained  
argumentative  
indifferent 
Question 58 Explanation:
the tone of the sentence clearly indicates a word that is similar to understated is needed for the blank. Alternatively, the word should be antonym of strong (fail to make strong impression). Therefore, the best choice is restrained which means controlled/reserved/timid.
Question 59 
merge  
split  
collect  
separate 
Question 59 Explanation:
Amalgamate means combine or unite to form one organization of structure. So the best option here is spilt. Separate on the other hand, although a close synonym, it is too general to be the best antonym in the given question while Merge is the synonym; Collect is not related.
Question 60 
Incomprehensible  
Indelible
 
Inextricable  
Infallible 
Question 60 Explanation:
Inexplicable means not explicable; that cannot be explained, understood, or accounted for. So the best synonym here is incomprehensible.
Question 62 
The variable cost (V) of manufacturing a product varies according to the equation V = 4q, where q is the quantity produced. The fixed cost (F) of production of same product reduces with q according to the equation F = 100/q. How many units should be produced to minimize the total cost (V+F)?
5  
4  
7  
6 
Question 62 Explanation:
Checking with all options in formula: (4q+100/q) i.e. (V+F). Option A gives the minimum
cost.
Question 63 
A transporter receives the same number of orders each day. Currently, he has some pending orders (backlog) to be shipped. If he uses 7 trucks, then at the end of the 4th day he can clear all the orders. Alternatively, if he uses only 3 trucks, then all the orders are cleared at the end of the 10th day. What is the minimum number of trucks required so that there will be no pending order at the end of the 5th day?
4  
5  
6  
7 
Question 63 Explanation:
Let each truck carry 100 units.
2800 = 4n + e; n = normal
300 = 10n + 2; e = excess/pending
n = 100/3, e = 8000/3
5 days: 500 ×(5⋅100/3)+ 8000/3
Minimum possible = 6
2800 = 4n + e; n = normal
300 = 10n + 2; e = excess/pending
n = 100/3, e = 8000/3
5 days: 500 ×(5⋅100/3)+ 8000/3
Minimum possible = 6
Question 64 
A container originally contains 10 litres of pure spirit. Form 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 repeated one more time. How much spirit is now left in the container?
7.58 litres  
7.84 litres  
7 litres  
7.29 litres 
Question 64 Explanation:
10(729/1000×1=7.29 litres)
Question 65 
how to write a letter of condolence  
what emotional stages are passed through in the healing process  
what the leading causes of death are  
how to give support to a grieving friend 
Question 65 Explanation:
The given passage clearly deals with how to deal with bereavement and grief and so after the tragedy occurs and not about precautions. Therefore, irrespective of the causes of death, a school student, rarely gets into details of causes–which is beyond the scope of the context. Rest all the important in dealing with grief.
There are 65 questions to complete.