## Nielit Scientific Assistance CS 15-10-2017

Question 1 |

A decimal has 25 digits. the number of bits needed for its equivalent binary representation is approximately

50 | |

74 | |

40 | |

None of the above |

Question 1 Explanation:

Consider three digits(1,2,3) of decimal numbers.Maximum number, we can generate by that three digits are 10

Then, Decimal number has 25 digits, so maximum number is 10

Similarly, in the binary representation with “n” bits the maximum number is 2

So we can write 10

After taking log

log

n log

n = 25 log

n = 25 x 3.3 [ log

n = 82.5

Note: Original question paper given option D is 60. But actual answer is 82.5.

^{ 3} -1 which is 999.Then, Decimal number has 25 digits, so maximum number is 10

^{ 25} -1Similarly, in the binary representation with “n” bits the maximum number is 2

^{25} -1So we can write 10

^{25} –1 = 2^{ n} – 1 → 10^{ 25} = 2^{ n}After taking log

_{2} on both sideslog

_{2}2^{n} =log_{ 2} 10^{25}n log

_{2} 2=25 log _{2} 10n = 25 log

_{2} 10n = 25 x 3.3 [ log

_{2} 2=1 & log_{2} 10 =3.322]n = 82.5

Note: Original question paper given option D is 60. But actual answer is 82.5.

Question 2 |

Which of the following is minimum error code?

Octal code | |

Binary Code | |

Gray code | |

Excess-3 Code |

Question 2 Explanation:

→ "Gray code" as an alternative name is "reflected binary code". one of those also lists "minimum error code" and "cyclic permutation code" among the names.

→ Gray codes are widely used to facilitate error correction in digital communications such as digital terrestrial television and some cable TV systems.

→ Gray codes are widely used to facilitate error correction in digital communications such as digital terrestrial television and some cable TV systems.

Question 3 |

Match list I with list II and select the correct answer using the codes given below the lists.

1 2 3 4 | |

3 2 4 1 | |

2 3 1 4 | |

1 4 2 3 |

Question 3 Explanation:

3 address instruction:

Two operand locations and a result location are explicitly contained in the instruction word.

e.g., Y = A − B

2 address instruction:

One of the addresses is used to specify both an operand and the result location.

e.g., Y = Y + X

1 address instruction:

Two addresses are implied in the instruction and accumulator based operations.

e.g., A CC = A CC + X

0 address instructions:

They are applicable to a special memory organization called a stack. It interact with a stack using push and pop operations. All addresses are implied as in register based operations.

T = Tap(T − 1 )

Two operand locations and a result location are explicitly contained in the instruction word.

e.g., Y = A − B

2 address instruction:

One of the addresses is used to specify both an operand and the result location.

e.g., Y = Y + X

1 address instruction:

Two addresses are implied in the instruction and accumulator based operations.

e.g., A CC = A CC + X

0 address instructions:

They are applicable to a special memory organization called a stack. It interact with a stack using push and pop operations. All addresses are implied as in register based operations.

T = Tap(T − 1 )

Question 4 |

A binary search tree contains the values- 1,2,3,4,5,6,7 and 8. The tree is traverses in preorder and the values are printed out. Which of the following sequences is a valid output?

5 3 1 2 4 7 8 6 | |

5 3 1 2 6 4 9 7 | |

5 3 2 4 1 6 7 8 | |

5 3 1 2 4 7 6 8 |

Question 4 Explanation:

Preorder traversal yields (Root, left, right)

Option D:

Let draw binary search tree for the given sequence,

After traversing through this tree we will get same sequence.

Option D:

Let draw binary search tree for the given sequence,

After traversing through this tree we will get same sequence.

Question 5 |

The number of the edges in a regular graph of degree 'd' and 'n' vertices is____

Maximum of n,d | |

n+d | |

nd | |

nd/2 |

Question 5 Explanation:

Sum of degree of vertices = 2*no. of edges

d*n = 2*|E|

∴ |E| = (d*n)/2

d*n = 2*|E|

∴ |E| = (d*n)/2

Question 6 |

Consider the following C code segment:

int IsPrime(n)

{

int i,n;

for(i=2;i<= √n;i++)

if(n%i == 0)

{

printf(“Not Prime\n”);

return 0;

}

return 1;

}

Let T(n) denotes the number of times the for loop is executed by the program on input n.

Which of the following is TRUE?

int IsPrime(n)

{

int i,n;

for(i=2;i<= √n;i++)

if(n%i == 0)

{

printf(“Not Prime\n”);

return 0;

}

return 1;

}

Let T(n) denotes the number of times the for loop is executed by the program on input n.

Which of the following is TRUE?

T(n) = O( √n) and T(n) = Ω ( √n) | |

T(n) = O( √n) and T(n) = Ω (1) | |

T(n) = O(n) and T(n) = Ω ( √n) | |

None of the above |

Question 6 Explanation:

Then the time complexity is Best Case : Ω(1) & Worst Case: O(√n)

Question 7 |

A language L for which there exists a TM,, 'T', that accepts every word in L and either rejects or loops for every word that is not in L, is said to be

recursive | |

recursively enumerable | |

NP-Hard | |

None of the above |

Question 7 Explanation:

A language L for which there exists a TM,, 'T', that accepts every word in L and either rejects or loops for every word that is not in L, is said to be recursively enumerable

Question 8 |

Synthesized attribute can easily be simulated by an

LL grammar | |

ambiguous grammar | |

LR grammar | |

none of the above |

Question 8 Explanation:

● A Synthesized attribute is an attribute of the nonterminal on the left-hand side of a production.

● Synthesized attributes represent information that is being passed up the parse tree.

● LR-attributed grammars allow the attributes to be evaluated on LR parsing. As a result, attribute evaluation in LR-attributed grammars can be incorporated conveniently in bottom-up parsing.

● Synthesized attributes represent information that is being passed up the parse tree.

● LR-attributed grammars allow the attributes to be evaluated on LR parsing. As a result, attribute evaluation in LR-attributed grammars can be incorporated conveniently in bottom-up parsing.

Question 9 |

The logic of pumping lemma us a good example of

The pigeonhole principle | |

The divide and conquer technique | |

recursion | |

Iteration |

Question 9 Explanation:

→ The pigeonhole principle is nothing more than the obvious remark: if you have fewer pigeon holes than pigeons and you put every pigeon in a pigeon hole, then there must result at least one pigeon hole with more than one pigeon. It is surprising how useful this can be as a proof strategy.

→ In the theory of formal languages in computability theory, a pumping lemma or pumping argument states that, for a particular language to be a member of a language class, any sufficiently long string in the language contains a section, or sections, that can be removed, or repeated any number of times, with the resulting string remaining in that language. The proofs of these lemmas typically require counting arguments such as the pigeonhole principle.

→ In the theory of formal languages in computability theory, a pumping lemma or pumping argument states that, for a particular language to be a member of a language class, any sufficiently long string in the language contains a section, or sections, that can be removed, or repeated any number of times, with the resulting string remaining in that language. The proofs of these lemmas typically require counting arguments such as the pigeonhole principle.

Question 10 |

Given relations R(w,x) and S(y,z) the result of
SELECT DISTINCT w,x from R<S

R has no duplicates and S is non empty | |

R and S have no duplicates | |

S has no duplicates and R is non empty | |

R and S has the same number of tuples |

Question 10 Explanation:

r has no duplicate, if r can have duplicates it can be remove in the final state. s in non-empty if s is empty then r*s becomes empty.

Question 11 |

E-R model uses this symbol to represent weak entity set?

Dotted rectangle | |

Diamond | |

Doubly outlined rectangle | |

None of these |

Question 11 Explanation:

→ A weak entity is an entity that cannot be uniquely identified by its attributes alone.

→ It must use a foreign key in conjunction with its attributes to create a primary key.

→ The foreign key is typically a primary key of an entity it is related to.

→ It must use a foreign key in conjunction with its attributes to create a primary key.

→ The foreign key is typically a primary key of an entity it is related to.

Question 12 |

By open domain CASE tools we mean

tools available in open domain | |

Software packages which can be downloaded from the internet | |

Software packages to aid each phase of the systems analysis and design which an be downloaded free of cost from the internet | |

Source codes of CASE tools |

Question 12 Explanation:

● Tools are also in the open domain which can be downloaded and used.

● They do not usually have very good user interfaces.

● They do not usually have very good user interfaces.

Question 13 |

If P is risk probability, L is loss, then is computed as

RE=P/L | |

RE=P+L | |

RE=P*L | |

RE=2*P*L |

Question 13 Explanation:

Risk Exposure(RE) is determined using the following relationship
RE = P*L

where P is the probability of occurrence for a risk, and L is the cost to the project should the risk occur.

where P is the probability of occurrence for a risk, and L is the cost to the project should the risk occur.

Question 14 |

A can is filled with 5 paise coins. Another can is filled with 10 paise coins. Another can is filled with 25 paise coins. All the cans are given wrong labels. If the can labeled 25 paise is not having the 10 paise coins, what will the can, labeled 10 paise have?

25 paise | |

5 paise | |

10 paise | |

cannot be determined |

Question 14 Explanation:

A diagram will make it easy to comprehend.

It is easy to find that the can labeled 25 Paise must have 5 paise coins in it. So, the can labeled 10 Paise, must have 25 Paise coins in it.

It is easy to find that the can labeled 25 Paise must have 5 paise coins in it. So, the can labeled 10 Paise, must have 25 Paise coins in it.

Question 15 |

The possible number of boolean function of 3 variables X,Y and Z such that f(X,Y,Z)=f(X',Y',Z')

8 | |

16 | |

64 | |

32 |

Question 15 Explanation:

→ Like above other input combinations also repeat after these inputs.

→ We have 4 input pairs and produces output either 0 or 1.

→ Total functions possible=2*2*2*2 (or) 2

^{4}=16

Question 16 |

What is the modality of relationship, if there is no explicit need for relationship to occur?

0 | |

2 | |

3 | |

1 |

Question 16 Explanation:

→ The modality of a relationship is 0 if there is no explicit need for the relationship to occur or the relationship is optional.

→ The modality is 1 if an occurrence of the relationship is mandatory

→ The modality is 1 if an occurrence of the relationship is mandatory

Question 17 |

If the original size of data is 40 then after adding error detection redundancy bit the size of data length is

26 | |

36 | |

46 | |

56 |

Question 17 Explanation:

→ Imagine that we want to design a code with m message bits and r check bits that will allow all single errors to be corrected.

→ Each of the 2 m legal messages has n illegal codewords at a distance of 11 from it.

→ These are formed by systematically inverting each of the n bits in the n-bit codeword formed from it. Thus, each of the 2

→ Since the total number of bit patterns is 2

→ Using n=m+r, this requirement becomes

= (m+r+1) ≤ 2

= 40+6+1 ≤ 2

= 47 ≤ 2

r=6

Message size will be 6+40=46

→ Each of the 2 m legal messages has n illegal codewords at a distance of 11 from it.

→ These are formed by systematically inverting each of the n bits in the n-bit codeword formed from it. Thus, each of the 2

^{m} legal messages requires n+1 bit patterns dedicated to it.→ Since the total number of bit patterns is 2

^{n} , We must have (n+1)2^{m} ≤ 2^{ n} .→ Using n=m+r, this requirement becomes

= (m+r+1) ≤ 2

^{r}= 40+6+1 ≤ 2

^{6}= 47 ≤ 2

r=6

Message size will be 6+40=46

Question 18 |

Bit stuffing refers to

Inserting a '0' in user data stream to differentiate it with a flag | |

Inserting a '0' in flag stream to avoid ambiguity | |

Appending a nibble to the flag sequence | |

Appending a nibble to the user data stream |

Question 18 Explanation:

● The term "bit stuffing" broadly refers to a technique whereby extra bits are added to a data stream, which do not themselves carry any information, but either assist in management of the communications or deal with other issues.

● The receiver knows how to detect and remove or disregard the stuffed bits.

● The receiver knows how to detect and remove or disregard the stuffed bits.

Question 19 |

A microprogrammed control unit

Is faster than a hardwired unit | |

Facilitates easy implementation of a new instruction | |

Is useful when small programs are to be run | |

All of the above |

Question 19 Explanation:

→ In hardwired control, we saw how all the control signals required inside the CPU can be generated using a state counter and a PLA circuit.

→ There is an alternative approach by which the control signals required inside the CPU can be generated . This alternative approach is known as microprogrammed control unit.

→ In microprogrammed control unit, the logic of the control unit is specified by a microprogram. A microprogram consists of a sequence of instructions in a microprogramming language. These are instructions that specify microoperations.

→ A microprogrammed control unit is a relatively simple logic circuit that is capable of

(1) sequencing through microinstructions.

(2) generating control signals to execute each microinstruction.

→ The concept of microprogram is similar to computer program. In computer program the complete instructions of the program is stored in main memory and during execution it fetches the instructions from main memory one after another.

→ The sequence of instruction fetch is controlled by program counter (PC)

→ There is an alternative approach by which the control signals required inside the CPU can be generated . This alternative approach is known as microprogrammed control unit.

→ In microprogrammed control unit, the logic of the control unit is specified by a microprogram. A microprogram consists of a sequence of instructions in a microprogramming language. These are instructions that specify microoperations.

→ A microprogrammed control unit is a relatively simple logic circuit that is capable of

(1) sequencing through microinstructions.

(2) generating control signals to execute each microinstruction.

→ The concept of microprogram is similar to computer program. In computer program the complete instructions of the program is stored in main memory and during execution it fetches the instructions from main memory one after another.

→ The sequence of instruction fetch is controlled by program counter (PC)

Question 20 |

If the channel is band limited to 6 kHz and signal to noise ratio is 16, what would be the capacity of channel?

16.15kbps | |

23.24 kbps | |

40.12 kbps | |

24.74 kbps |

Question 20 Explanation:

→ Shannon Capacity (Noisy Channel) In presence of Gaussian band-limited white noise, Shannon-Hartley theorem gives the maximum data rate capacity.

C=B log

where S and N are the signal and noise power, respectively, at the output of the channel.

→ This theorem gives an upper bound of the data rate which can be reliably transmitted over a thermal-noise limited channel. =6 * log

=6 * 4.08746284125034

=24.74 kbps

C=B log

_{2} (1+S/N).where S and N are the signal and noise power, respectively, at the output of the channel.

→ This theorem gives an upper bound of the data rate which can be reliably transmitted over a thermal-noise limited channel. =6 * log

_{2} (1+16)=6 * 4.08746284125034

=24.74 kbps

Question 21 |

At 100% modulation, the power in each sideband is__ of that of arrier.

50% | |

40% | |

60% | |

25% |

Question 21 Explanation:

→ Power in sidebands may be calculated as modulation index.

→ Each sideband is 25%. Total modulation index=1.

→ Each sideband is 25%. Total modulation index=1.

Question 22 |

The capacity relationship is given by

C=Wlog _{2} (1+S/N) | |

C=2Wlog _{ 2} (1+S/N) | |

C=Wlog _{2} (1-S/N) | |

C=Wlog _{10} (1+S/N) |

Question 22 Explanation:

→ Shannon Capacity (Noisy Channel) In presence of Gaussian band-limited white noise, Shannon-Hartley theorem gives the maximum data rate capacity.

C=W log

where S and N are the signal and noise power, respectively, at the output of the channel.

→ This theorem gives an upper bound of the data rate which can be reliably transmitted over a thermal-noise limited channel.

C=W log

_{2} (1+S/N).where S and N are the signal and noise power, respectively, at the output of the channel.

→ This theorem gives an upper bound of the data rate which can be reliably transmitted over a thermal-noise limited channel.

Question 23 |

Which of the following algorithm solve the all pair shortest path problem?

A) Dijkstra's algorithm

B) Floyd's algorithm

C) prim's algorithm

D) Warshall's algorithm

A) Dijkstra's algorithm

B) Floyd's algorithm

C) prim's algorithm

D) Warshall's algorithm

Both A and B | |

Both A and C | |

Both C and D | |

Both B and D |

Question 23 Explanation:

The Floyd Warshall Algorithm is for solving the All Pairs Shortest Path problem. The problem is to find shortest distances between every pair of vertices in edge weighted directed Graph.

Question 24 |

An algorithm is made up of two modules M1 and M2. If order of M1 is f(n) and M2 is g(n) then the order of algorithm is

max(f(n),g(n)) | |

min(f(n),g(n)) | |

f(n)+g(n) | |

f(n) * g(n) |

Question 24 Explanation:

In this problem, we have to check all possibilities.

So, we will prefer max(f(n),g(n)).

So, we will prefer max(f(n),g(n)).

Question 25 |

Choose the correct statements

A total recursive function is also a partial recursive function | |

A partial recursive function is also a total recursive function | |

A partial recursive function is also a primitive recursive function | |

None of the above |

Question 25 Explanation:

→ Primitive recursive functions are a class of functions that are defined using composition and primitive recursion

→ They are a strict subset of those μ-recursive functions (also called partial recursive functions) which are also total functions.

→ Primitive recursive functions form an important building block on the way to a full formalization of computability.

→ A total recursive function is also a partial recursive function

→ They are a strict subset of those μ-recursive functions (also called partial recursive functions) which are also total functions.

→ Primitive recursive functions form an important building block on the way to a full formalization of computability.

→ A total recursive function is also a partial recursive function

Question 26 |

Assume transaction A holds a shared lock R. If transaction B also requests for a shared lock on R. It will

result in deadlock situation | |

immediately be granted | |

immediately be rejected | |

be granted as soon as it is released by A |

Question 26 Explanation:

● Shared locks exist when two transactions are granted read access.

● One transaction gets the shared lock on data and when the second transaction requests the same data it is also given a shared lock.

● Both transactions are in a read-only mode, updating the data is not allowed until the shared lock is released.

● There is no conflict with the shared lock because nothing is being updated.

● Shared locks last as long as they need to last; it depends on the level of the transaction that holds the lock.

● One transaction gets the shared lock on data and when the second transaction requests the same data it is also given a shared lock.

● Both transactions are in a read-only mode, updating the data is not allowed until the shared lock is released.

● There is no conflict with the shared lock because nothing is being updated.

● Shared locks last as long as they need to last; it depends on the level of the transaction that holds the lock.

Question 27 |

Disk requests come to a disk driver for cylinders in the order 10, 22, 20, 2, 40, 6 and 38 at a given time when the given disk drive is reading from cylinder 20. The seek time is 6ms per cylinder.What is the total seek time, if the disk arm scheduling algorithm FCFS is used?

900 ms | |

850 ms | |

360 ms | |

876 ms |

Question 27 Explanation:

Total seek time= 10+12+2+18+38+34+32

= 146*6 ms

= 876 ms

Question 28 |

Table Employee has 10 records. It has a non-NULL SALARY column which is also

UNIQUE. The SQL statement

SELECT COUNT(*) FROM Employee WHERE SALARY > ANY (SELECT SALARY FROM EMPLOYEE);

UNIQUE. The SQL statement

SELECT COUNT(*) FROM Employee WHERE SALARY > ANY (SELECT SALARY FROM EMPLOYEE);

10 | |

9 | |

5 | |

0 |

Question 28 Explanation:

This query counts the number of employees who get more than the minimum salary. From the 10 employees, you need to exclude all those employees who are getting the minimum salary. Since the SALARY column is UNIQUE, only one employee will be getting the minimum salary.

Question 29 |

Which level of RAID refers to disk mirroring with block striping?

RAID level 1 | |

RAID level 2 | |

RAID level 0 | |

RAID level 3 |

Question 29 Explanation:

RAID is a technology that is used to increase the performance and/or reliability of data storage.

Following are the various RAID levels:

RAID 0: Stripping

RAID 1: Mirroring

RAID 2: Hamming code for error detection

RAID 3: Byte-level striping with a dedicated parity disk

RAID 4: Block-level striping with block-level striping with two parity blocks parity

RAID 5: Block-level striping with distributed parity

RAID 6: Block-level striping with double distributed parity.

Following are the various RAID levels:

RAID 0: Stripping

RAID 1: Mirroring

RAID 2: Hamming code for error detection

RAID 3: Byte-level striping with a dedicated parity disk

RAID 4: Block-level striping with block-level striping with two parity blocks parity

RAID 5: Block-level striping with distributed parity

RAID 6: Block-level striping with double distributed parity.

Question 30 |

Which of the following is not a form of main memory?

Instruction cache | |

Instruction register | |

Instruction Opcode | |

Translation lookaside buffer |

Question 30 Explanation:

→ The instruction opcode is the portion of a machine language instruction that specifies the operation to be performed. Beside the opcode itself, most instructions also specify the data they will process, in the form of operands. In addition to opcodes used in the instruction set architectures of various CPUs, which are hardware devices, they can also be used in abstract computing machines as part of their byte code specifications.

Question 31 |

The following program fragment prints

int i=5;

do

{

putchar(i+100);

printf("%d",i--);

}while(i);

int i=5;

do

{

putchar(i+100);

printf("%d",i--);

}while(i);

i5h4g3f2el | |

14h3g2f1e0 | |

An error message | |

None of the above |

Question 31 Explanation:

Step-1: putchar(105) will print the ASCII equivalent of 105 is ' i '.

Step-2: The printf statement prints the current value of i=5 because we are given post decrement.

Step-3: The same process will perform until it fail the condition.

Step-2: The printf statement prints the current value of i=5 because we are given post decrement.

Step-3: The same process will perform until it fail the condition.

Question 32 |

The running time of an algorithm T(n),where 'n' is the input size, is given by

T(n)=8T(n/2)+qn, if n>1

=p, if n=1

Where p,q are constants. The order of this algorithm is

T(n)=8T(n/2)+qn, if n>1

=p, if n=1

Where p,q are constants. The order of this algorithm is

n ^{ 2} | |

n ^{ n} | |

n ^{3} | |

N |

Question 32 Explanation:

The above recurrence is in the form of masters theorem.

Case-1: a>b

Here, a=8,b=2,k=1,p=0

O(n(log

=O(n

Case-1: a>b

^{k}Here, a=8,b=2,k=1,p=0

O(n(log

_{2} 8))=O(n(log_{2} 2^{3} ))=O(n^{3} (log_{ 2} 2)) [ log_{ 2} 2=1]=O(n

^{3} )Question 33 |

The convergence of the bisection method is

Cubic | |

Quadratic | |

Linear | |

none |

Question 33 Explanation:

The convergence of the bisection method is very slow. Although the error, in general, does not decrease monotonically, the average rate of convergence is 1/2 and so, slightly changing the definition of order of convergence, it is possible to say that the method converges linearly with rate 1/2.

Note:For the bisection you simply have that ε

Note:For the bisection you simply have that ε

_{i+1} /ε_{ i} =1/2, so, by definition the order of convergence is 1 (linearly).Question 34 |

When transaction T

_{i} requests a data item currently held by T_{ j} , T_{j} is allowed to wait only if it has a timestamp smaller than that of T_{ j} (that is, T_{ i} older than T_{ j} ). Otherwise, T_{ i} is rolled back(dies). this isWait-die | |

Wait-wound | |

Wound-Wait | |

Wait |

Question 34 Explanation:

__Wait-die scheme :__

It is a non-preemptive technique for deadlock prevention. When transaction T

_{i} requests a data item currently held by T

_{j} , T

_{i} is allowed to wait only if it has a timestamp smaller than that of T

_{ j} (That is T

_{ i} is older than T

_{ j} ), otherwise T

_{ i} is rolled back (dies).

→ In this scheme, if a transaction requests to lock a resource (data item), which is already held with a conflicting lock by another transaction, then one of the two possibilities may occur

1. If TS(T

_{ i} ) < TS(T

_{ j} ) − that is T

_{ i} , which is requesting a conflicting lock, is older than T

_{j}−then T

_{i}is allowed to wait until the data-item is available.

2. If TS(T

_{i} ) > TS(T

_{ j} ) − that is T

_{ i} is younger than T

_{ j} − then T

_{ i} dies. T

_{i} is restarted later with a random delay but with the same timestamp.

→ This scheme allows the older transaction to wait but kills the younger one.

__Example:__

Suppose that transaction T22, T23, T24 have time-stamps 5, 10 and 15 respectively. If T22 requests a data item held by T23 then T22 will wait. If T24 requests a data item held by T23, then T24 will be rolled back.

Question 35 |

What will be the output of following?

main()

{

static int a=3;

printf("%d",a--);

if(a)

main();

}

main()

{

static int a=3;

printf("%d",a--);

if(a)

main();

}

3 | |

3 2 1 | |

3 3 3 | |

Program will fall in continuous loop and print 3 |

Question 35 Explanation:

Step-1: initial a=3 but we are given storage class is static.

Step-2: In printf( ) we are given post decrement. So, it will print 3 then decrements.

Step-3: if(2) then we are calling main() function, a=2 because of static otherwise it will be 3 only.

Step-4: It will print 2 and next iteration will print 1.

Step-2: In printf( ) we are given post decrement. So, it will print 3 then decrements.

Step-3: if(2) then we are calling main() function, a=2 because of static otherwise it will be 3 only.

Step-4: It will print 2 and next iteration will print 1.

Question 36 |

Web links are stored within the page itself and when you wish to 'jump' to the page that is linked, we select the hotspot or anchor. This technique is called

Hypertext | |

hypermedia | |

Both (A) and (B) | |

Anchoring |

Question 36 Explanation:

Web links are stored within the page itself and when you wish to 'jump' to the page that is linked, we select the hotspot or anchor. This technique is called hypertext (or) Hypermedia

Question 37 |

Twelve 1 Ω resistances are used as edges to form a cube. The resistance between two diagonally opposite corners of the cube is

5/6 Ω | |

1/6 Ω | |

6/5 Ω | |

3/2 Ω |

Question 37 Explanation:

Question 38 |

The final value theorem is used to find the

Steady state value of the system output | |

Initial value of the system output | |

Transient behaviour of the system output | |

None of these |

Question 38 Explanation:

The final value theorem is used to find the steady state value of the system output

Question 39 |

For the discrete signal x[n]=a

^{ n} u[n], a>0 the z-transform is(z+a)/z | |

(z-a)/z | |

z/(z-a) | |

z/(z+a) |

Question 39 Explanation:

Let x[n] be causal signal given by x[n] = a

→ The Z-Transform of x[n] is given by

The ROC is defined by | az

Region of Convergence for a discrete time signal x[n] is dened as a continuous region in z plane where the Z-Transform converges. In order to determine RoC, it is convenient to represent the Z-Transform as a is rational X(z)=P(z)/Q(z)

1. The roots of the equation P(z) = 0 correspond to the ’zeros’ of X(z)

2. The roots of the equation Q(z) = 0 correspond to the ’poles’ of X(z)

3. The RoC of the Z-transform depends on the convergence of the polynomials P(z) and Q(z).

^{n} u[n]→ The Z-Transform of x[n] is given by

The ROC is defined by | az

^{ −1}| < 1 or | z| > | a| .__Region of Convergence(RoC)__Region of Convergence for a discrete time signal x[n] is dened as a continuous region in z plane where the Z-Transform converges. In order to determine RoC, it is convenient to represent the Z-Transform as a is rational X(z)=P(z)/Q(z)

1. The roots of the equation P(z) = 0 correspond to the ’zeros’ of X(z)

2. The roots of the equation Q(z) = 0 correspond to the ’poles’ of X(z)

3. The RoC of the Z-transform depends on the convergence of the polynomials P(z) and Q(z).

Question 40 |

For a periodic signal v(t)=30 sin 100t+10cos 300t +6 sin (500t + π /4), the fundamental frequency in rad/s

100 | |

300 | |

500 | |

None of the above |

Question 40 Explanation:

Given, the signal

V (t) = 3 0 sin 100t + 10 cos 300t + 6 sin (500t + π/4)

So, we have

ω

ω

ω

∴ The respective time periods are

T

T

T

So, the fundamental time period of the signal is

LCM (T

as T

∴ The fundamental frequency, ω

V (t) = 3 0 sin 100t + 10 cos 300t + 6 sin (500t + π/4)

So, we have

ω

_{1}= 100 radsω

_{2}= 300 radsω

_{3}= 500 rads∴ The respective time periods are

T

_{1}= 2π/ω_{1}= 2π/100 secT

_{2}= 2π/ω_{2}= 2π/300 secT

_{3}= 2π/ω_{3}= 2π/500 secSo, the fundamental time period of the signal is

LCM (T

_{1}, T_{ 2}, T_{ 3}) =LCM (2π,2π,2π)/HCF (100,300,500)as T

_{0}=2π/100∴ The fundamental frequency, ω

_{0}=2π/T_{ 0}= 100 rad/sQuestion 41 |

If the number of bits per sample in a PCM system is increased from a n to n+1, the improvement in signal to quantization noise ratio will be

3dB | |

6db | |

2n dB | |

n dB |

Question 41 Explanation:

(S/N

= (1.76 + 6 n)

(SQNR)

(SQNR)

(SQNR)

So for every one bit increase in bits per sample will result is 6 dB improvement in signal to quantization ratio.

_{q})_{dB}= (1.76 + 6 n)

_{dB}(SQNR)

_{ 1}= 1 .76 + 6 n(SQNR)

_{ 2}= 1 .76 + 6 n(n + 1 ) = 1 .76 + 6 n + 6(SQNR)

_{2}− (SQNR)_{1}= 1 .76 + 6 n + 6 − 1 .76 − 6 n = 6 dBSo for every one bit increase in bits per sample will result is 6 dB improvement in signal to quantization ratio.

Question 42 |

A carrier A

_{c} Cos(w_{ C} )t is frequency modulated by a signal E_{m} Cos(w_{m} )t. The modulation index is m_{f} . the expression for the resulting FM signal isA _{c}, Cos[w_{ c} t+ m _{f} Sin(w _{m} )t] | |

A _{c} Cos[w_{ c} t+ m_{ f} Cos(w _{m} )t] | |

A _{c} Cos[w _{c} t+ π m_{ f} Sin w _{m} t] | |

A _{c} Cos[w_{ c} t+ 2 π m_{ f} E_{ m} Cos(w_{ m} )t/w_{ m} ] |

Question 42 Explanation:

Question 43 |

A sequential circuit using D flip flop and logic gates is shown in figure, Where X and Y are the inputs and Z is the output. The circuit is

S-R flip flop with inputs X=R and Y=S | |

S-R flip flop with inputs X=S and Y=R | |

J-K flip flop with inputs X=J and Y=K | |

J-K flip flop with X=k and Y=J |

Question 43 Explanation:

Here, PI=Present Input

PS=Present state

NS=Next State

Q

D=X’Z+YZ

PS=Present state

NS=Next State

Q

_{n+1} =D=f(PI,PS)=f(x,y,Q_{ n} )D=X’Z+YZ

Question 44 |

A 4 bit ripple counter and a 4 bit synchronous counter are made using flip flops having a propagation delay of 10ns each. If the worst case delay in the ripple counter and the synchronous counter be R and S respectively, then

R=10 ns, S=40ns | |

R=40ns, S=10ns | |

R=10ns, S=30ns | |

R=30 ns, S=10ns |

Question 44 Explanation:

→ In synchronous counter time delay is constant while in Ripple it is additive.

→ In Ripple counter (or) Asynchronous counter each flip flop waits for its previous flip flops output.

R= bit size*propagation delay

= 4*10ns

= 40ns

→ In Synchronous counter all flip flops are triggered by same clock. It will gives output of all four flip flops at the same time.

S=10ns

→ In Ripple counter (or) Asynchronous counter each flip flop waits for its previous flip flops output.

R= bit size*propagation delay

= 4*10ns

= 40ns

→ In Synchronous counter all flip flops are triggered by same clock. It will gives output of all four flip flops at the same time.

S=10ns

Question 45 |

In MOSFET fabrication, the channel length is delayed during the process of

Isolation oxide growth | |

Channel stop implantation | |

Poly-Silicon gate patterning | |

Lithography step leading to the contact pad |

Question 45 Explanation:

In MOSFET fabrication channel length is defined during Poly-Silicon gate patterning process

Question 46 |

The open loop transfer function of a feedback control system is G(s).H(s)=1/(S+1)

^{ 3} . The gain margin of the system is2 | |

4 | |

8 | |

16 |

Question 46 Explanation:

Given data,

The open loop transfer function of a feedback control system is G(s).H(s)=1/(S+1)

Gain margin=?

→ Here, the formula is -3tan

tan

= -60

W

G(s).H(s)=1/(S+1)

|G(s).H(s)|=X= 1/(1+W

=1⁄8

G(s).H(s)=1/X

= 8

The open loop transfer function of a feedback control system is G(s).H(s)=1/(S+1)

^{3}Gain margin=?

→ Here, the formula is -3tan

^{ -1} (W_{ pc} ) = -180^{ 0}tan

^{ -1} (W_{ pc} ) = -180^{ 0} /3= -60

^{0}W

_{pc} =√3 rad/secG(s).H(s)=1/(S+1)

^{3}|G(s).H(s)|=X= 1/(1+W

_{pc}^{2} )=1⁄8

G(s).H(s)=1/X

= 8

Question 47 |

An ideal op-amp is an ideal

Voltage controlled current source | |

voltage controlled voltage source | |

current controlled current source | |

current controlled voltage source |

Question 47 Explanation:

An ideal op-amp lead us to an op-amp with input resistance is infinite, so no current flows into either input terminal (the “current rule”) and that the differential input offset voltage is zero (the “voltage rule”).

Ideal operational amplification will have

➝ High input impedance ( R

➝ R

➝ Infinite voltage gain

➝ So, it is a voltage controlled voltage source device.

Ideal operational amplification will have

➝ High input impedance ( R

_{in}= ∞ ) and low output impedance ( R_{out}= 0 )➝ R

_{in}= ∞ , so zero input current➝ Infinite voltage gain

➝ So, it is a voltage controlled voltage source device.

Question 48 |

In a ripple counter using edge-triggered JK flip-flops, the pulse input is applied to

Clock input of all flip flops | |

j and K input of one flip flop | |

J and K input of all flip flops | |

Clock input of one flip flops |

Question 48 Explanation:

A ripple counter is an asynchronous counter where only the first flip-flop is clocked by an external clock.

→ All subsequent flip-flops are clocked by the output of the preceding flip-flop. Asynchronous counters are also called ripple-counters because of the way the clock pulse ripples it way through the flip-flops.

So, answer is clock input of one flip-flop.

→ All subsequent flip-flops are clocked by the output of the preceding flip-flop. Asynchronous counters are also called ripple-counters because of the way the clock pulse ripples it way through the flip-flops.

So, answer is clock input of one flip-flop.

Question 49 |

If a processor has 32 bit virtual address, 28 bit physical address, 2 KB pages. How many bits are required for the virtual,physical page number?

17,21 | |

21,17 | |

6,10 | |

None |

Question 49 Explanation:

Number of entries for virtual page number = 2

Number of entries for physical page number = 2

The number bits for virtual page number are 21.

The number bits for physical page number are 21.

^{virtual address}/ Pages =2^{ 32} /2KB= 2^{ 32} /2^{ 11} =2^{ 21}Number of entries for physical page number = 2

^{ physical address }/ Pages =2^{28} /2KB= 2^{ 28} /2^{11} =2^{ 17}The number bits for virtual page number are 21.

The number bits for physical page number are 21.

Question 50 |

What is the maximum value of the function f(x)=2x

^{2} -2x +6 in the interval [0,2]?6 | |

10 | |

12 | |

5,5 |

Question 50 Explanation:

An open interval does not include its endpoints, and is indicated with parentheses. For example, (0,1) means greater than 0 and less than 1.

A closed interval is an interval which includes all its limit points, and is denoted with square brackets. For example, [0,1] means greater than or equal to 0 and less than or equal to 1.

A half-open interval includes only one of its endpoints, and is denoted by mixing the notations for open and closed intervals. (0,1] means greater than 0 and less than or equal to 1, while [0,1) means greater than or equal to 0 and less than 1.

Given function f(x)=2x

According to the given interval , we need to check the function at the values 0,1,2.

f(0)=2x0-2x0+6=6

f(1)=2x1

f(2)=2x2

A closed interval is an interval which includes all its limit points, and is denoted with square brackets. For example, [0,1] means greater than or equal to 0 and less than or equal to 1.

A half-open interval includes only one of its endpoints, and is denoted by mixing the notations for open and closed intervals. (0,1] means greater than 0 and less than or equal to 1, while [0,1) means greater than or equal to 0 and less than 1.

Given function f(x)=2x

^{ 2} -2x +6 and the given interval is [0,2]According to the given interval , we need to check the function at the values 0,1,2.

f(0)=2x0-2x0+6=6

f(1)=2x1

^{2} -2x1+6=2-2+6=6f(2)=2x2

^{ 2} -2x2+6=8-4+6=10Question 51 |

The value of the integral

(x+2)/2 | |

2/(π-2) | |

π-2 | |

π+2 |

Question 51 Explanation:

The integral value of x

=[−( π/2)

=[0+ π+0+C-2-C]

=π-2

^{2} sin(x)dx is [−x^{2} cos(x)+2xsin(x)+2cos(x)+C] with the interval [π/2,0]=[−( π/2)

^{2} cos( π/2 )+2 π/2 sin( π/2 )+2cos( π/2 )+C] - [−( 0)^{2} cos( π/2 )+2(0)sin( π/2 )+2cos( 0 )+C]=[0+ π+0+C-2-C]

=π-2

Question 52 |

Which of the following would not be specified in a communication protocol?

Header contents | |

Trailer contents | |

Error Checking | |

Data content of message |

Question 52 Explanation:

→ A communication protocol is a system of rules that allow two or more entities of a communications system to transmit information via any kind of variation of a physical quantity. The protocol defines the rules, syntax, semantics and synchronization of communication and possible error recovery methods.

→ There are many properties of a transmission that a protocol can define. Common ones include: packet size, transmission speed, error correction types, handshaking and synchronization techniques, address mapping, acknowledgement processes, flow control, packet sequence controls, routing, address formatting

→ There are many properties of a transmission that a protocol can define. Common ones include: packet size, transmission speed, error correction types, handshaking and synchronization techniques, address mapping, acknowledgement processes, flow control, packet sequence controls, routing, address formatting

Question 53 |

Which of the definitions generates the same languages as L, where

L={ x

i. E→ xEy |xy

ii.xy|x

iii. x

L={ x

^{ n} y^{ n} , n>=1 }i. E→ xEy |xy

ii.xy|x

^{+} x yy^{ +}iii. x

^{ +} y^{ +}i only | |

i and ii only | |

ii and iii only | |

ii only |

Question 53 Explanation:

The language L={ x

For n=1, xy

n=2,x

n=3,x

In the above language , there are equal number of x’s followed by y’s

Which is nothing but equal number of x’s followed by equal number of y’s

X

+

means one or more x’s

Y

+

means one or more y’s

The definition-2 won’t give equal number of x’s followed by y’s

It generates any number of x’s followed by any number of y’s.

The expression x

The definition-3 won’t give equal number of x’s followed by y’s

It generates any number of x’s followed by any number of y’s.

^{n} y^{ n} , n>=1 }For n=1, xy

n=2,x

^{2} y^{ 2}n=3,x

^{3} y^{ 3}In the above language , there are equal number of x’s followed by y’s

__Definition-1 :__E→ xEy |xy → xEy → xxEyy → xxxEyyy → xxxxyyyyWhich is nothing but equal number of x’s followed by equal number of y’s

X

+

means one or more x’s

Y

+

means one or more y’s

__Definition -2 :__The expression xy|x^{+} x yy^{ +} gives xy, xxyy , xxxxyy , xxyyyyy and so onThe definition-2 won’t give equal number of x’s followed by y’s

It generates any number of x’s followed by any number of y’s.

__Definition -3 :__The expression x

^{+} y^{ +} gives xy, xxy,xxxy,xyy,xyyy and so on strings.The definition-3 won’t give equal number of x’s followed by y’s

It generates any number of x’s followed by any number of y’s.

Question 54 |

Which of the following is a desirable property of module?

Independency | |

Low Cohesiveness | |

High Coupling | |

Multifunctional |

Question 54 Explanation:

● Highly coupled have program units dependent on each other.

● Loosely coupled are made up of units that are independent or almost independent.

● The desirable property is High cohesion and low coupling

● From the above options , the suitable option is independency

● Loosely coupled are made up of units that are independent or almost independent.

● The desirable property is High cohesion and low coupling

● From the above options , the suitable option is independency

Question 55 |

The solution of the recurrence relation a

_{r} =a_{ r-1}+2a_{ r-2} with a_{ 0} =2,a_{ 1} =7a _{ r} =(3)_{ r} +(1)_{ r} | |

2a _{r} =(2)^{ r} /3 -(1)^{ r} | |

a _{ r} =3^{ r+1} -(-1)^{ r} | |

a _{r} =3(2)_{ r} -(-1)_{ r} |

Question 55 Explanation:

● Given the recurrence relation a

● For r =2, a

● From the above options ,Substitute the r values 0,1,2,3 then option -D gives the solution to recurrence relation.

_{ r} =a_{ r-1} +2a_{ r-2}and a_{0} =2,a_{ 1} =7.● For r =2, a

_{2}= a_{2-1} +2a_{ 2-2} =a_{1} +2a_{ 0} =7+2*2=7+4=11 ● For r=3,a 3= a 3-1 +2a 3-2 =a 2 +2a 1 =11+2*7=11+14=25● From the above options ,Substitute the r values 0,1,2,3 then option -D gives the solution to recurrence relation.

Question 56 |

M is a square matrix of order 'n' and its determinant value is 5. If all the elements of M are multiplied by 2, Its determinant value becomes 40. The value of 'n' is

2 | |

3 | |

5 | |

4 |

Question 56 Explanation:

● From the matrix property :
If each element of a row (or a column) of a determinant is multiplied by a constant k, then its value gets multiplied by k.

● Consider matrix order is 3, there are 3 rows and each row is multiplied by 2 means we need to multiply 8 to the existing determinant.

● The given existing determinant is 5 and each row multiplied by 2 means 8 *5 =40.

● Consider matrix order is 3, there are 3 rows and each row is multiplied by 2 means we need to multiply 8 to the existing determinant.

● The given existing determinant is 5 and each row multiplied by 2 means 8 *5 =40.

Question 57 |

The question is based on the following program fragment.

f(int Y[10], int x)

{

int u,j,k;

i=0;j=9;

do

{

k=(i+j)/2;

if(Y[k]
i=k;

else

j=k;

}while((Y[k]!=x) && (i
if(Y[k] ==x)

printf("x is in the array");

else

printf("x is not in the array");

}

On which of the following contents of 'Y' and 'x' does the program fail?

f(int Y[10], int x)

{

int u,j,k;

i=0;j=9;

do

{

k=(i+j)/2;

if(Y[k]

else

j=k;

}while((Y[k]!=x) && (i

printf("x is in the array");

else

printf("x is not in the array");

}

On which of the following contents of 'Y' and 'x' does the program fail?

Y is [1 2 3 4 5 6 7 8 9 10] and x<10 | |

Y is [1 3 5 7 9 11 13 15 17 19] and x<1 | |

Y is [ 2 2 2 2 2 2 2 2 2 2] and x>2 | |

Y is [ 2 4 6 8 10 12 14 16 18 20] and 2 |

Question 57 Explanation:

This above code is similar to binary search

First iteration:

I=0,j=9

K=(0+9)/2=4

Elements with duplicate values of “2” the condition if(Y[k]
Second iteration :

I=0, j=k=4

K= (0+4)/2=2.

Third iteration:

I=0, j=k=2

K= (0+2)/2=1

Fourth iteration:

I=0, j=k=1

K= (0+1)/2=0

The while condition is Y[k] != x && i < j is true

→ The above program doesn’t work for the cases where element to be searched is the last element of Y[ ] or greater than the last element (or maximum element) in Y[ ].

→ For such cases, program goes in an infinite loop because i is assigned value as k in all iterations, and i never becomes equal to or greater than j.

So, while condition never becomes false.

First iteration:

I=0,j=9

K=(0+9)/2=4

Elements with duplicate values of “2” the condition if(Y[k]

I=0, j=k=4

K= (0+4)/2=2.

Third iteration:

I=0, j=k=2

K= (0+2)/2=1

Fourth iteration:

I=0, j=k=1

K= (0+1)/2=0

The while condition is Y[k] != x && i < j is true

→ The above program doesn’t work for the cases where element to be searched is the last element of Y[ ] or greater than the last element (or maximum element) in Y[ ].

→ For such cases, program goes in an infinite loop because i is assigned value as k in all iterations, and i never becomes equal to or greater than j.

So, while condition never becomes false.

Question 58 |

The address sequence generated by tracing a particular program executing in a pure demand paging system with 100 records per page, with 1 free main memory frame is recorded as follows. 0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240, 0260,
0320, 0370, What is the number of page faults?

15,4 | |

6,4 | |

7,2 | |

4,6 |

Question 58 Explanation:

Memory page numbers will be calculated by using corresponding to given address as follows

The page number is 1 from the record numbers 1-100,

The page number is 2 from the record numbers 101 -200

The page number is 3 from the record numbers 201 -300

The page number is 4 from the record numbers 301 -400

The page number is 5 from the record numbers 401 -500

The page number is 6 from the record numbers 501 -600

The page number is 7 from the record numbers 601 -700

For a given address sequence 0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240,0260, 0320, 0370

The page number numbers sequence is 1,2,5,5,6,6,6,2,3,3,3,4,4

The number of available frame is 1, So we need to use only one frame for executing demand page process.

So total number of page faults are 7.

The page number is 1 from the record numbers 1-100,

The page number is 2 from the record numbers 101 -200

The page number is 3 from the record numbers 201 -300

The page number is 4 from the record numbers 301 -400

The page number is 5 from the record numbers 401 -500

The page number is 6 from the record numbers 501 -600

The page number is 7 from the record numbers 601 -700

For a given address sequence 0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240,0260, 0320, 0370

The page number numbers sequence is 1,2,5,5,6,6,6,2,3,3,3,4,4

The number of available frame is 1, So we need to use only one frame for executing demand page process.

So total number of page faults are 7.

Question 59 |

Consider an ε-tree CFG. If for every pair of productions A → u and A → v

if FIRST(u) ∩ FIRST(v) is empty then the CFG has to be LL(1) | |

If the CFG is LL(1) then FIRST(u) ∩ FIRST(v) has to be empty | |

Both (A) and (B) | |

None of the above |

Question 59 Explanation:

The condition for a grammar to be LL(1)

Theorem: A context free grammar G=(V

1. First(⍺) ∩ First (β) ∩ Follow(A)=Θ.

2. If ⍺ *⇒ ε then First(β) ∩ Follow(A)= Θ.

If grammar is epsilon free then condition 2 is not required.

Now as per condition 1, for every non-terminal A, if we have A → u and A→v

And First(u) First(v) = φ and CFG is epsilon free then it must be LL(1) and if epsilon free CFG is LL(1) then it must satisfy the condition 1.

Theorem: A context free grammar G=(V

_{T} , V_{ N} , S,P) is LL(1) if and if only if for every nonterminal A and every strings of symbols ⍺,β such that ⍺≠β and A → ⍺,β we have1. First(⍺) ∩ First (β) ∩ Follow(A)=Θ.

2. If ⍺ *⇒ ε then First(β) ∩ Follow(A)= Θ.

If grammar is epsilon free then condition 2 is not required.

Now as per condition 1, for every non-terminal A, if we have A → u and A→v

And First(u) First(v) = φ and CFG is epsilon free then it must be LL(1) and if epsilon free CFG is LL(1) then it must satisfy the condition 1.

Question 60 |

A solution for the differential equation x'(t) + 2x(t) = δ (t) with initial condition x(~0)=0

e _{-2t} u(t) | |

e _{2t} u(t) | |

e _{-t} u(t) | |

e _{t} u(t) |

There are 60 questions to complete.

## Scientific Assistance CS 15-10-2017

Question 1 |

The difference between the compound interest and the simple earned at the end of 3 rd year on a sum of money at a rate of 10% per annul is Rs. 77.5. What is the sum?

Rs. 3,500 | |

Rs. 2,500 | |

Rs. 3,000 | |

Rs. 2,000 |

Question 1 Explanation:

Let principal be ‘x’

S.I = P T R / 100

S.I = X * 3 * 10 / 100

C.I = [ P (1 + (R / 100)

C.I = [ X (1+ (10 / 100)

C.I = 331X / 1000

Difference between Compound interest and Simple interest is 77.5

( 331X / 1000 ) - ( 3X / 10) = 77.5

31x / 1000 = 77.5

x = 2500.

S.I = P T R / 100

S.I = X * 3 * 10 / 100

C.I = [ P (1 + (R / 100)

^{T}) - P]C.I = [ X (1+ (10 / 100)

^{3}) - X]C.I = 331X / 1000

Difference between Compound interest and Simple interest is 77.5

( 331X / 1000 ) - ( 3X / 10) = 77.5

31x / 1000 = 77.5

x = 2500.

Question 2 |

Aamir and Birju can cut 5000g of wood in 20 min. Birju and Charles can cut 5000g of wood in 40 min. Charles and Aamir cut 5 kg of wood in 30 min. How much time Charles will take to cut 5 kg wood alone?

120 min | |

48 min | |

240 min | |

120 min |

Question 2 Explanation:

Here in this question 5000g means 5kg only

So from that question Aamir (A) Birju (B) Charles (C)

Let Aamir, Birju and Charles alone can complete the work in A, B and C minutes respectively, then

1 min. work of (A+B)=1/20, (B+C)=1/40 and (C+A)=1/30

A + B + C = 13/240

1 min. work of C = (A+B+C) - (A+B) = 13/240 - 1/20 = 1/240

So C alone can complete the work in 240 min.

So from that question Aamir (A) Birju (B) Charles (C)

Let Aamir, Birju and Charles alone can complete the work in A, B and C minutes respectively, then

1 min. work of (A+B)=1/20, (B+C)=1/40 and (C+A)=1/30

A + B + C = 13/240

1 min. work of C = (A+B+C) - (A+B) = 13/240 - 1/20 = 1/240

So C alone can complete the work in 240 min.

Question 3 |

An alloy contains copper and zinc in the ratio 5 : 3 and another contains copper and tin in the ratio 8 : 5, If equal weights of the two are melted together to form a 3 rd alloy, find the weight of tin per kg in the new alloy.

40/129 | |

5/13 | |

5/26 | |

28/5 |

Question 3 Explanation:

Amount of copper in first alloy is 5x/8 and zinc is 3x/8.

Amount of copper in second alloy is 8x/13 and tin is 5x/13.

So after mixing tin is available in 2 alloy only that means in 3rd alloy 5x/13/2x

Tin present in 3rd alloy is 5/26.

Amount of copper in second alloy is 8x/13 and tin is 5x/13.

So after mixing tin is available in 2 alloy only that means in 3rd alloy 5x/13/2x

Tin present in 3rd alloy is 5/26.

Question 4 |

X is a whole number. If the only common factors of X and X2 are 1 and X, then X is _______

1 | |

a perfect square | |

an odd number | |

a prime number |

Question 4 Explanation:

A number that is divisible only by itself and 1 is called prime number.

Question 5 |

Line AB is 24 metres in length and tangent to the inner one of the two concentric circles at point C, Points A and B lie on the circumference of the outer circle. It is known that the radii of the two circles are integers. The radius of the outer circle is

13m | |

5 m | |

7 m | |

4m |

Question 5 Explanation:

Given data,

AB=24cm

CB=12cm

θC=5cm(Assume)

θB=5

^{2} +12

^{ 2} =169

θB=13cm

Note: They forgot to mention the value 5. If we are considering value=5 then answer should be A.

Question 6 |

Monisha is working with a real estate agent to find a location for the kids’ toy store she plans to open in her town. She is looking for a place that is either in the centre or not too far from the centre of town. It should also be attractive for the right kind of footfall too. Which of the following locations should Monisha’s agent call to her attention?

a storefront in a new high-rise building near the train station in the center of town whose occupants are mainly young, childless professionals who use the train to commute to their offices each day | |

a little shop three blocks away from the town’s main street, located across the street from an elementary school and next door to an ice cream store | |

a stand-alone storefront on a quiet residential street ten blocks away from the town’s center | |

a storefront in a small a strip mall located on the outskirts of town that is also occupied by a pharmacy and a dry cleaner |

Question 6 Explanation:

This option is both near the center of town and in a location where children and their parents are sure to be around. This is the only option that meets both of Monisha’s requirements.

Question 7 |

Reading is a psycholinguistic guessing game. To read critically is a skill as it is a demanding process. One must slow down one’s reading and, with a pencil in hand, perform specific operations on the text. Mark up the text with reactions, conclusions and questions. When one reads, one becomes an active participant.

critical reading is a slow, dull, but essential process. | |

the best critical reading happens at critical times in a person’s life. | |

readers should get in the habit of questioning the truth of what they read. | |

critical reading requires thoughtful and careful attention |

Question 7 Explanation:

Option-A FALSE: The author never says that reading is dull.

Option-B,C FALSE: Not related to given paragraph.

Option-D TRUE: The author stresses the need to read critically by performing thoughtful and careful operations on the text.

Option-B,C FALSE: Not related to given paragraph.

Option-D TRUE: The author stresses the need to read critically by performing thoughtful and careful operations on the text.

Question 8 |

Find the missing term:

60, 40, 55, 45, 50, 50,?

60, 40, 55, 45, 50, 50,?

45 | |

50 | |

55 | |

60 |

Question 8 Explanation:

Question 9 |

Find the missing alphabet:

T, r, O, m, J,?

T, r, O, m, J,?

h | |

i | |

I | |

g |

Question 9 Explanation:

In between r and T there is letter is s

Likewise m and O letter is n

Same h is the next letter to fit in that sequence

Likewise m and O letter is n

Same h is the next letter to fit in that sequence

Question 10 |

Here are some words translated from an artificial language.

Qmelaqali means fruitcake

Qalitiimmeo means cakewalk

Useguamao means buttercup

Which word could mean “cupcake”?

Qmelaqali means fruitcake

Qalitiimmeo means cakewalk

Useguamao means buttercup

Which word could mean “cupcake”?

qalitiiqali | |

amaotiimmeo | |

pakitreft | |

amaoqali |

Question 10 Explanation:

Qmelaqali→ fruitcake

We can divide artificial language into parts.

Qmela→ fruit

qali→ cake

Qalitiimmeo→ cakewalk

We can divide artificial language into parts.

Qali→ cake

tiimmeo→ walk

Useguamao → buttercup

We can divide artificial language into parts.

Usegu→ butter

amao→ cup

cupcake→ ?

amao→ cup

qali→ cake

So, option-D is correct answer.

We can divide artificial language into parts.

Qmela→ fruit

qali→ cake

Qalitiimmeo→ cakewalk

We can divide artificial language into parts.

Qali→ cake

tiimmeo→ walk

Useguamao → buttercup

We can divide artificial language into parts.

Usegu→ butter

amao→ cup

cupcake→ ?

amao→ cup

qali→ cake

So, option-D is correct answer.

Question 11 |

A man walks 5 Km towards south and then turns to the right. After walking 3 Km he turns to the left and walks 5 Km. Now in which direction is he from the starting place?

West | |

South | |

North-East | |

South- West |

Question 11 Explanation:

Question 12 |

If the consonants in the word ‘DROVE’ are first arranged alphabetically and the vowels are put in between two pairs of consonants in the alphabetical order, which of the following will be the fourth from the right end after the rearrangement?

D | |

E | |

R | |

O |

Question 12 Explanation:

In the word DROVE we have two vowels

Those are O and E

After consonants are arranged in alphabetically we got the word like DRV

But we have two place vowels in between two pairs of consonants we got finally DEROV

So right from fourth letter is E

Those are O and E

After consonants are arranged in alphabetically we got the word like DRV

But we have two place vowels in between two pairs of consonants we got finally DEROV

So right from fourth letter is E

Question 13 |

There is a queue in a ticketing office. Amanda is 10 th from the front while Murthy is 25 th from behind and Marta is just in the middle of the two. If there be 50 persons in the queue. What position does Marta occupy from the front?

16 | |

18 | |

15 | |

17 |

Question 13 Explanation:

Question 14 |

There are five janitors. Pali, Qureshi, Rohan, Sant and Timber. They all have a different height, Qureshi is shorter than only Timber and Sant is shorter than Pali and Qureshi. Who among them is the shortest?

Rohan | |

Sant | |

Pali | |

Data inadequate |

Question 14 Explanation:

From that question we can decide Pali Quereshi sant and timber but Rohan is not mentioned there. So Data inadequate is answer.

Question 15 |

A $ B means A is the father of B; A # B means A is the sister of B; A * B means A is the daughter of B and A @ B means A is the brother of B. Which of the following indicates that M is the wife of Q?

Q $ R # T @ M | |

Q $ R @ T # M | |

Q $ R * T # M | |

Q $ R @ T * M |

Question 15 Explanation:

Q $ R means R’s father is Q

R @ T means T’s brother is R

T * M means M’s Daughter T

Finally M’s Daughter is T and T’s Brother is R and R’s father is Q so Q’s wife is M.

R @ T means T’s brother is R

T * M means M’s Daughter T

Finally M’s Daughter is T and T’s Brother is R and R’s father is Q so Q’s wife is M.

Question 16 |

Count the number of squares in the given figure.

32 | |

30 | |

29 | |

28 |

Question 16 Explanation:

For this type of questions there is one formula that is

[ n (n+1)(2n+1) ] / 6

Here n is 4 because it’s 4 x 4

[ 4 (4+1)(8+1) ] / 6

30

[ n (n+1)(2n+1) ] / 6

Here n is 4 because it’s 4 x 4

[ 4 (4+1)(8+1) ] / 6

30

Question 17 |

Who was the Viceroy of India, when Quit India Resolution was passed in 1942?

Lord Linlithgow | |

Lord Wavell | |

Lord Willingdon | |

Lord Mountbatten |

Question 18 |

When was the East India Association set up?

1866 | |

1857 | |

1836 | |

1885 |

Question 19 |

Who was the Spanish navigator who set out to discover India, but instead landed on the soil of America?

Christopher Columbus | |

Vasco Da Gama | |

James Cook | |

None of above |

Question 20 |

Which dynasty was ruling over north India when Alexander the great invaded India?

Gupta Dynasty | |

Maurya Dynasty | |

Sakya Dynasty | |

Nanda Dynasty |

Question 21 |

The roads of cities in the Indus Valley Civilization generally divided the city into

Rectangular Blocks | |

Circular Blocks | |

Triangular Blocks | |

None of above |

Question 22 |

In which year was Pulitzer Prize established?

1917 | |

1918 | |

1922 | |

1928 |

Question 23 |

‘Kanchipuram’ is in which of the following states?

Andhra Pradesh | |

Orissa | |

Kerala | |

Tamil Nadu |

Question 24 |

Which of the following is not a chief organ of the United Nations Organizations?

International Labour Organisation | |

Security Council | |

International Court of Justice | |

General Assembly |

Question 25 |

Which of the following is not a member of G-15?

Indonesia | |

Malaysia | |

Columbia | |

India |

Question 26 |

The group of metals Co, Ni, Fe may best called as

Transition metals | |

Main group metals | |

Alkali metals | |

Rare metals |

Question 27 |

Non stick cooking utensils are coated with

Black paint | |

PVC | |

Teflon | |

Polystyrene |

Question 28 |

The international township built near Pondicherry in India in coloration with UNEESCO is called

Elbaville | |

Auroville | |

Gayaville | |

Broadway |

Question 29 |

Irvin sold a book at a profit of 12%. If Irvin had sold it for Rs. 18 more, then 18% would have been gained. Find the cost price.

Rs. 600 | |

Rs. 300 | |

Rs. 400 | |

Rs. 200 |

Question 29 Explanation:

Explanation: Let CP = 100%

Sold on 12% profit means sold on 112%

now 18% would have been gained means 118% if Rs 18 was more it means 118% - 112% = 18 Rs

so 6%=18Rs

so 100% = (18/6)×100 = 3× 100 = 300 Rs

CP = 300

Sold on 12% profit means sold on 112%

now 18% would have been gained means 118% if Rs 18 was more it means 118% - 112% = 18 Rs

so 6%=18Rs

so 100% = (18/6)×100 = 3× 100 = 300 Rs

CP = 300

Question 30 |

In a group of 7 people, the average age is found to be 17 years. Two more people joined with an average age 19 years. One person left the group whose age was 25 years. What is the new average age of the group?

17.5 Years | |

16.5 Years | |

18 Years | |

16 Years |

Question 30 Explanation:

7 people, the average age is found to be 17 years means 17 * 7 = 119

Two more joined with an average age 19 years means 19 * 2 = 38 So total is 157

But one person left from that group whose age is 25 years 157 - 25 = 132

Remaining average of 8 people is 132 / 8

16.5 Year

Two more joined with an average age 19 years means 19 * 2 = 38 So total is 157

But one person left from that group whose age is 25 years 157 - 25 = 132

Remaining average of 8 people is 132 / 8

16.5 Year

Question 31 |

A 300 meter long metro train crosses a platform in a metro station in 39 seconds while it crosses a lamp post in 18 seconds. What is the length of the platform?

250 meter | |

350 meter | |

520 meter | |

300 meter |

Question 31 Explanation:

Speed = Distance/time = 300/18 = 50/3 m/sec

Let the length of the platform be x meters

Distance = Speed * Time

x+300 = (50/3) * 39

x = 350

Let the length of the platform be x meters

Distance = Speed * Time

x+300 = (50/3) * 39

x = 350

Question 32 |

Assume that a sum of money is divided equally among n girls. Each girl will receive $ 60. If another girl is added to the group and the sum is divided equally among all the girls, each child girl a $ 50 share. What is the sum of money?

$ 3000 | |

$ 300 | |

$ 110 | |

$ 10 |

Question 32 Explanation:

From that question 60 * n = 50 (n + 1)

n = 5

so sum of money is 60 * 5 = 300

n = 5

so sum of money is 60 * 5 = 300

Question 33 |

A tank can be filled by one tap in 10 minutes and by another in 30 minutes. Both the taps are kept open for 5 minutes and then the first one is shut off. In how many minutes more is the tank completely filled?

5 | |

7.5 | |

10 | |

12 |

Question 33 Explanation:

A tank can be filled in 10 mins

another one is in 30 mins and those taps are opened for 5 mins

so in one min one tap is 1/10 another one is 1/30

both can tank open for 5 mins so (1/10 + 1/30 ) * 5

= (2/15) * 5

= 10/15

= 2⁄3

Remaining work is 1- (2⁄3)

= 1⁄3

The left work has to be done by first Tap because first is shut off.

Work will be completed by Second Tap in= (1⁄3) * 30 = 10

another one is in 30 mins and those taps are opened for 5 mins

so in one min one tap is 1/10 another one is 1/30

both can tank open for 5 mins so (1/10 + 1/30 ) * 5

= (2/15) * 5

= 10/15

= 2⁄3

Remaining work is 1- (2⁄3)

= 1⁄3

The left work has to be done by first Tap because first is shut off.

Work will be completed by Second Tap in= (1⁄3) * 30 = 10

Question 34 |

From the four choices provided, choose the analogy that is most similar to the one in the question.

Wealth : Poverty

Wealth : Poverty

part: whole | |

good: excellent | |

prodigal : chary | |

wicked : sinful |

Question 34 Explanation:

Wealth is opposite to Poverty and Prodigal opposite word is chary.

Wealth: an abundance of valuable possessions or money

Poverty: The state of being extremely poor.

Prodigal: spending money or using resources freely and recklessly

chary: Cautiously or suspiciously reluctant to do something.

Wealth: an abundance of valuable possessions or money

Poverty: The state of being extremely poor.

Prodigal: spending money or using resources freely and recklessly

chary: Cautiously or suspiciously reluctant to do something.

Question 35 |

From the four choices provided, choose the analogy that is most similar to the one in the question.

Misfortune : Catastrophe

Misfortune : Catastrophe

miniature : big | |

limited : infinite | |

knowledge : learning | |

generosity : parsimony |

Question 35 Explanation:

Misfortune is related to Catastrophe and knowledge is related to learning.

Misfortune: An unfortunate condition or event.

Catastrophe: An event causing great and usually sudden damage or suffering.

Misfortune: An unfortunate condition or event.

Catastrophe: An event causing great and usually sudden damage or suffering.

Question 36 |

From the four choices provided, choose the analogy that is most similar to the one in the question.

Molecule : Atoms

Molecule : Atoms

family : sisters | |

light : bulb | |

tissue : cells | |

body : limb |

Question 36 Explanation:

Molecule is a group of atoms bonded together.

Tissue is any of the distinct types of material of which animals or plants are made, consisting of specialized cells and their products.

Tissue is any of the distinct types of material of which animals or plants are made, consisting of specialized cells and their products.

Question 37 |

From the four choices provided, choose the analogy that is most similar to the one in the question.

Limp : Walk

Limp : Walk

flap : fly | |

run : race | |

stutter : talk | |

chew : digest |

Question 37 Explanation:

Limp is related to walk and shutter is related to talk.

Limp: Walk with difficulty, typically because of a damaged or stiff leg or foot.

Limp: Walk with difficulty, typically because of a damaged or stiff leg or foot.

Question 38 |

From the four choices provided, choose the analogy that is most similar to the one in the question.

Riddle : Solve

Riddle : Solve

mirage : illusion | |

joke : amuse | |

tangle : unravel | |

target : aim |

Question 38 Explanation:

Riddle is related to Solve and tangle is related to unravel

Riddle: A question or statement intentionally phrased so as to require ingenuity in ascertaining its answer or meaning.

Tangle: Twist together into a confused mass.

unravel: undo (twisted, knitted, or woven threads).

Riddle: A question or statement intentionally phrased so as to require ingenuity in ascertaining its answer or meaning.

Tangle: Twist together into a confused mass.

unravel: undo (twisted, knitted, or woven threads).

Question 39 |

Fact 1 : All chickens are birds.

Fact 2 : Some chickens are hens.

Fact 3 : Female birds lay eggs.

If the first three statements are facts, which of the following statements must also be a fact? i. All birds lay eggs.

ii. Hens are birds.

iii. Some chickens are not hens.

Fact 2 : Some chickens are hens.

Fact 3 : Female birds lay eggs.

If the first three statements are facts, which of the following statements must also be a fact? i. All birds lay eggs.

ii. Hens are birds.

iii. Some chickens are not hens.

ii only | |

ii and iii only | |

I, ii and iii | |

None of the statements is a known fact |

Question 39 Explanation:

According to venn diagram, (i) is FALSE because male birds never lay eggs. (ii) and (iii) are TRUE

Question 40 |

What is the most essential thing for election?

President | |

Voter | |

November | |

Nation |

Question 41 |

What is the most essential thing for ovation?

Outburst | |

bravo | |

applause | |

encore |

Question 42 |

Introducing a man to her husband, a woman said, “His brother’s father is the only son of my grandfather.”How is the woman related to this man?

Mother | |

Aunt | |

Sister | |

Daughter |

Question 42 Explanation:

Only son of her grandfather -- Her father;

Man's brother's father -- man's father.

So, man's father is her father i.e., She is the man's sister.

Man's brother's father -- man's father.

So, man's father is her father i.e., She is the man's sister.

Question 43 |

Pointing to a photograph, a man said, “I have no brother or sister but that man’s father is my father’s son.” Whose photograph was it?

His Own | |

His Son | |

His Father | |

His Grandfather |

Question 43 Explanation:

Since the person who is telling has no brother or sister, so his father son is he himself.

So the man in the photograph is his son.

So the man in the photograph is his son.

Question 44 |

Find the number of quadrilaterals in the given figure.

6 | |

7 | |

9 | |

11 |

Question 44 Explanation:

→ The word "quadrilateral" is derived from the Latin words quadri, a variant of four, and latus, meaning "side".

→ A quadrilateral is a polygon with four edges (or sides) and four vertices or corners.

→ According to figure, we can label first and get total quadrilaterals.

Quadrilaterals are 1234, 1245, 1246, 1248, 3481, 3451, 3461, 4517, 4561, 6178 and 1748.

→ A quadrilateral is a polygon with four edges (or sides) and four vertices or corners.

→ According to figure, we can label first and get total quadrilaterals.

Quadrilaterals are 1234, 1245, 1246, 1248, 3481, 3451, 3461, 4517, 4561, 6178 and 1748.

Question 45 |

Count the number of convex pentagons in the adjoining figure.

16 | |

12 | |

8 | |

4 |

Question 45 Explanation:

A convex polygon is a simple polygon (not self-intersecting) in which no line segment between two points on the boundary ever goes outside the polygon.

The pentagons are 12468, 34682, 56824, 78246, 13467, 35681, 57823, 71245, 24578, 46712, 68134 and 82356.

The pentagons are 12468, 34682, 56824, 78246, 13467, 35681, 57823, 71245, 24578, 46712, 68134 and 82356.

Question 46 |

There is a family of six people whose nicknames are Pat, Qat, Rat, Sat, Tat and Uat. Their professionals are Engineer, Doctor, Teacher, Salesman, Manager and Lawyer. There are two married couples in the family. The Manager is the grandfather of Uat, who is an Engineer Rat, the Salesman, is married to the lady Teacher Qat is the mother of Uat and Tat. The Doctor, Sat is married to the
Manager.

How many male members are there in the family?

How many male members are there in the family?

Two | |

Three | |

Four | |

Data Inadequate |

Question 46 Explanation:

Question 47 |

There is a family of six people whose nicknames are Pat, Qat, Rat, Sat, Tat and Uat. Their professionals are Engineer, Doctor, Teacher, Salesman, Manager and Lawyer. There are two married couples in the family. The Manager is the grandfather of Uat, who is an Engineer Rat, the Salesman, is married to the lady Teacher Qat is the mother of Uat and Tat. The Doctor, Sat is married to the
Manager.

What is the professional of Pat?

What is the professional of Pat?

Lawyer | |

Lawyer or Teacher | |

Manager | |

None of these |

Question 47 Explanation:

Question 48 |

There is a family of six people whose nicknames are Pat, Qat, Rat, Sat, Tat and Uat. Their professionals are Engineer, Doctor, Teacher, Salesman, Manager and Lawyer. There are two married couples in the family. The Manager is the grandfather of Uat, who is an Engineer Rat, the Salesman, is married to the lady Teacher Qat is the mother of Uat and Tat. The Doctor, Sat is married to the
Manager.

Who are the married couples in the family?

Who are the married couples in the family?

Pat-Qat and Sat-Rat | |

Rat-Uat and Sat-Tat | |

Pat-Tat and Sat-Rat | |

Pat-Sat and Rat-Qat |

Question 48 Explanation:

Question 49 |

There is a family of six people whose nicknames are Pat, Qat, Rat, Sat, Tat and Uat. Their professionals are Engineer, Doctor, Teacher, Salesman, Manager and Lawyer. There are two married couples in the family. The Manager is the grandfather of Uat, who is an Engineer Rat, the Salesman, is married to the lady Teacher Qat is the mother of Uat and Tat. The Doctor, Sat is married to the Manager.

How Pat is related to Tat?

How Pat is related to Tat?

Father | |

Grandfather | |

Mother | |

Grandmother |

Question 49 Explanation:

Question 50 |

Shirin went to a bakery and bought items worth Rs. 25, out of which 30 paise went on sales tax on taxable purchases. If the tax rate was 6%, then what was the cost of the tax free items?

Rs. 12 | |

Rs. 19.70 | |

Rs. 19.10 | |

Rs. 18.80 |

Question 50 Explanation:

Total cost of the items he purchased = Rs.25

Given that out of this Rs.25, 30 Paise is given as tax => Total tax incurred = 30 Paise = Rs.30/100

Let the cost of the tax free items = x

Given that tax rate = 6%

( 25 - 30/100 - x ) * (6/100) = 30 / 100

6(25−0.3−x)=30

x=25−0.3−5=19.7

Given that out of this Rs.25, 30 Paise is given as tax => Total tax incurred = 30 Paise = Rs.30/100

Let the cost of the tax free items = x

Given that tax rate = 6%

( 25 - 30/100 - x ) * (6/100) = 30 / 100

6(25−0.3−x)=30

x=25−0.3−5=19.7

Question 51 |

The rate of simple interest on a sum of money is 6 percent per annum for the first 3 years, 8 per cent per annum for the next 5 years and 10 per cent per annum for the period beyond 8 years. If the simple interest accrued by the sum for a total period of 10 years is Rs. 1,560, what is the sum?

Rs. 1,500 | |

Rs. 3,000 | |

Rs. 2,000 | |

Data Inadequate |

Question 51 Explanation:

Sum = (1560 * 100 ) / ( 3 * 6 + 8 * 5 + 2 * 10 ) = 2000

Question 52 |

Adrian starts a start-up with a capital of Rs. 85,000. Brian joins in the start-up with Rs. 42,500 after sometime. For how much period does Brian join, if the profits at the end of the year are divided in the ratio of 3 : 1?

5 months | |

6 months | |

7 months | |

8 months |

Question 52 Explanation:

Let B joins for X months.

Then,

(Profit of A)/(Profit of B) = 3/1

(85000 * 12)/(42500 *X) = 3/1

85000 * 12 = 42500 * 3 * X

X = 8.

B Joined for 8 months

Then,

(Profit of A)/(Profit of B) = 3/1

(85000 * 12)/(42500 *X) = 3/1

85000 * 12 = 42500 * 3 * X

X = 8.

B Joined for 8 months

Question 53 |

A car travels at an average of 50 miles per hour for 2.5 hours and then travels at a speed of 70 miles per hour for 1.5 hours. How far did the car travel in the entire 4 hours?

210 miles | |

230 miles | |

250 miles | |

260 miles |

Question 53 Explanation:

Speed= 50 miles/ hour Time=2.5 hour

speed=70 miles / hour Time=1.5 hour

Speed= Distance/ time

Distance= speed x Time

=50×2.5+70×1.5

=125+105

=230 miles

speed=70 miles / hour Time=1.5 hour

Speed= Distance/ time

Distance= speed x Time

=50×2.5+70×1.5

=125+105

=230 miles

Question 54 |

By selling 45 limes for Rs. 40, a woman loses 20%. How many should she sell for Rs. 24 to gain 20% in the transaction?

16 | |

18 | |

20 | |

22 |

Question 54 Explanation:

SP of 45 limes is Rs.40

loss = 20%

So CP of 45 limes would be 40*100/80= Rs. 50

To gain 20% She should sell these 45 limes for =50*20/100=Rs. 60

Now If she sells for Rs. 60 = 45 limes

So for Rs. 24 he would sell = (24/60)*45 = 18 limes

loss = 20%

So CP of 45 limes would be 40*100/80= Rs. 50

To gain 20% She should sell these 45 limes for =50*20/100=Rs. 60

Now If she sells for Rs. 60 = 45 limes

So for Rs. 24 he would sell = (24/60)*45 = 18 limes

Question 55 |

The prestigious Ramon Magsaysay Award was conferred upon Ms. Kiran Bedi for her excellent contribution to which of the following fields?

Literature | |

Community Welfare | |

Government | |

Journalism |

Question 56 |

Who among the following is not a recipient of ‘Dada Saheb Phalke’ Award?

Ramanand Sagar | |

Raj Kapoor | |

V. Shantaram | |

Ashok Kumar |

Question 57 |

What is part of a database that holds only one type of information?

Report | |

Field | |

Record | |

File |

Question 57 Explanation:

Record holds only specific and one type of information.

Question 58 |

‘.JPG’ extension refers usually to what kind of file?

System file | |

Animation/movie file | |

MS Encarta document | |

Image file |

Question 58 Explanation:

‘.JPG’ extension used for image files.

Question 59 |

Which of the following is not written by Munshi Premchand?

Gaban | |

Godan | |

Guide | |

Manasorovar |

Question 60 |

The famous Dilwara Temples are situated in

Uttar Pradesh | |

Rajasthan | |

Maharashtra | |

Madhya Pradesh |

There are 60 questions to complete.