HCF AND LCM

Facts And Formulae:

1.Highest Common Factor:(H.C.F) or Greatest Common Meaure(G.C.M) :

The H.C.F  of two or more than  two numbers
is the greatest

number that divides each of them exactly.


There are two methods :


1.Factorization method:  Express each one

of the given numbers as the product of prime
factors.

The product  of least powers of common  prime factors gives HCF.



Example :  Find HCF of 26 * 32*5*74 ,   22 *35*52 * 76 ,  2*52 *72


Sol: The prime numbers given common numbers are  2,5,7


Therefore HCF  is  22 * 5 *72 .






2.Division Method : Divide  the larger number by

smaller one. Now divide the divisor by
remainder. Repeat the process

of dividing preceding number last obtained till
zero is obtained as

number. The last divisor  is HCF.


Example: Find HCF of 513, 1134, 1215


Sol:

     1134) 1215(1
          1134
        ----------
            81)1134(14
                81
             -----------
                 324
                 324
              -----------
                  0
             -----------
   HCF of this two numbers is 81.

      81)513(6
         486
          --------
          27)81(3
             81
            -----        
              0                
             ---


HCF  of  81 and 513 is 27.






3.Least common multiple[LCM] : The least number which is

divisible by each one of given
numbers is LCM.


There are two methods for this:



1.Factorization method  :


Resolve each one into product of prime factors.

Then LCM is  product of highest powers

of all factors.



2.Common  division method.


Problems:


1.The  HCF of 2   numbers  is 11 and LCM is  693.If one

of numbers is 77.find other.


Sol:  Other number  = 11 *  693/77=99.






2.Find largest number of 4 digits  divisible by  12,15,18,27


Sol: The largest number is 9999.


LCM of 12,15,18,27  is 540.


on dividing 9999 by  540 we get 279 as remainder.


Therefore

number =9999 – 279  =9720.





3.Find least number which when divided by 20,25,35,40

leaves remainders 14,19,29,34.

Sol:

       20–14=6
      25-19=6
      35-29=6
      40-34=6



Therefore number =LCM of (20,25,35,40) - 6=1394






4.252  can be expressed as prime as :

                     2      252
                    2      126
       3      63
       3      21
                      7

prime factor is 2 *2 * 3 * 3  *7





5.1095/1168  when expressed in simple form is
  
   1095)1168(1
        1095
        ------
          73)1095(15
              73
           ---------
              365
              365
             ---------
               0
            ----------
So, HCF is 73


Therefore

1095/1168 = 1095/73/1168/73= 15/16





6.GCD of 1.08,0.36,0.9 is


Sol:

HCF of 108,36,90
        36)90(2
          72
          ----
    18)36(2
             36
            ----
             0
           ----
HCF is 18.

HCF of 18 and 108 is 18
 18)108(6
    108
   -------
      0
   --------
Therefore  HCF  =0.18






7.Three numbers are in ratio 1:2:3 and HCF is 12.Find numbers.


Sol:

Let the numbers be x.


Three numbers are x,2x,3x


Therefore

HCF is 
                  2x)3x(1
                    2x
       -----
                     x)2x(2
                       2x
                     --------
          0
     ------------- 

HCF is x so, x is 12


Therefore numbers are 12,24,36.





8.The  sum of two numbers is 216 and HCF is 27.


Sol:

 Let numbers are
                  27a  + 27 b =216

                  a + b =216/27=8



Co-primes of 8 are (1,7)  and (3,5)

numbers=(27  * 1 ), (27  *  7)
      =27,89






9.LCM of two numbers is  48..The numbers are in ratio 2:3.

The sum of numbers is



Sol: 

Let the number be x.


Numbers are 2x,3x


LCM of  2x,3x is 6x


Therefore

6x=48

x=8.


Numbers are 16 and 24


Sum=16 +24=40.





10.HCF and LCM of two numbers are  84 and 21.If ratio of

two numbers is 1:4.Then largest of two numbers is 



Sol:

Let the numbers be x,4x


Then x *  4x  =  84  *  21


 x2 =84  * 21 /4


          x  = 21

Largest number is 4 * 21.





11.HCF of two numbers is 23,and other factors of LCM  are

13,14.Largest number  is


Sol: 

23  * 14 is Largest number.





12.The maximum number of students among   them 1001 pens

and 910 pencils can be distributed  in such a way that

each student  gets same number of pens and pencils is ?


Sol:

HCF of 1001 and 910
   910)1001(1
       910
      ------------
             91)910(10
                910
                  --------
                     0
                 ---------
      Therefore  HCF=91





13.The least number which should be added to 2497 so that sum is

divisible by 5,6,4,3  ?



Sol:  LCM   of  5,6,4,3   is 60.



On dividing  2497  by 60   we get 37 as remainder.



Therefore number to added is 60 – 37  =23.



Answer is 23.




14.The least number which is a perfect square and is  divisible by

each of numbers 16,20,24 is ?



Sol:  LCM of 16,20,24   is 240.



2 * 2*2*2*3*5=240



To make it a perfect square multiply by 3 * 5



Therefore  240 * 3 * 5=3600



Answer is 3600.



    

NUMBERS

Introduction:

Natural Numbers:

All positive integers are natural numbers.
Ex 1,2,3,4,8,......

There are infinite natural numbers and number 1 is the least natural number.
Based on divisibility there would be two types of natural numbers. They are

Prime and composite.

Prime Numbers:

A natural number larger than unity is a prime number if it
does not have other divisors except for itself and unity.
Note:-Unity i e,1 is not a prime number.

Properties Of Prime Numbers:

->The lowest prime number is 2.
->2 is also the only even prime number.
->The lowest odd prime number is 3.
->The remainder when a prime number p>=5 s divided by 6 is 1 or 5.However,
if a number on being divided by 6 gives a remainder 1 or 5 need not be
prime.
->The remainder of division of the square of a prime number p>=5 divide by
24 is 1.
->For prime numbers p>3, p²-1 is divided by 24.
->If a and b are any 2 odd primes then a²-b² is composite. Also a²+b²
is composite.
->The remainder of the division of the square of a prime number p>=5
divided by 12 is 1.

Process to Check A Number s Prime or not:

Take the square root of the number.
Round of the square root to the next highest integer call this number as Z.
Check for divisibility of the number N by all prime numbers below Z. If
there is no numbers below the value of Z which divides N then the number
will be prime.

Example 239 is prime or not?
√239 lies between 15 or 16.Hence take the value of Z=16.
Prime numbers less than 16 are 2,3,5,7,11 and 13.
239 is not divisible by any of these. Hence we can conclude that 239
is a prime number.


Composite Numbers:

The numbers which are not prime are known as composite numbers.

Co-Primes:

Two numbers a an b are said to be co-primes,if their H.C.F is 1.
Example (2,3),(4,5),(7,9),(8,11).....
Place value or Local value of a digit in a Number:

place value:

Example 689745132
Place value of 2 is (2*1)=2
Place value of 3 is (3*10)=30 and so on.
Face value:-It is the value of the digit itself at whatever
place it may be.

Example 689745132
Face value of 2 is 2.
Face value of 3 is 3 and so on.
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Tests of Divisibility:

Divisibility by 2:-A number is divisible by 2,if its unit's digit is
any of 0,2,4,6,8.

Example 84932 is divisible by 2,while 65935 is not.
Divisibility by 3:-A number is divisible by 3,if the sum of its digits is
divisible by 3.

Example 1.592482 is divisible by 3,since sum of its digits
5+9+2+4+8+2=30 which is divisible by 3.

Example 2.864329 is not divisible by 3,since sum of its digits
8+6+4+3+2+9=32 which is not divisible by 3.

Divisibility by 4:-A number is divisible by 4,if the number formed by last
two digits is divisible by 4.

Example 1.892648 is divisible by 4,since the number formed by the last
two digits is 48 divisible by 4.

Example 2.But 749282 is not divisible by 4,since the number formed by
the last two digits is 82 is not divisible by 4.

Divisibility by 5:-A number divisible by 5,if its unit's digit is either
0 or 5.

Example 20820,50345
Divisibility by 6:-If the number is divisible by both 2 and 3.
example 35256 is clearly divisible by 2
sum of digits =3+5+2+5+21,which is divisible by 3
Thus the given number is divisible by 6.

Divisibility by 8:-A number is divisible by 8 if the last 3 digits
of the number are divisible by 8.

Divisibility by 11:-If the difference of the sum of the digits in the
odd places and the sum of the digitsin the even places is zero or divisible
by 11.

Example 4832718
(8+7+3+4) - (1+2+8)=11 which is divisible by 11.

Divisibility by 12:-All numbers divisible by 3 and 4 are divisible by 12.

Divisibility by 7,11,13:-The difference of the number of its thousands
and the remainder of its division by 1000 is divisible by 7,11,13.

BASIC FORMULAE:

->(a+b)²=a²+b²+2ab
->(a-b)²=a²+b²-2ab
->(a+b)²-(a-b)²=4ab
->(a+b)²+(a-b)²=2(a²+b²)
->a²-b²=(a+b)(a-b)
->(a-+b+c)²=a²+b²+c²+2(ab+b c+ca)
->a³+b³=(a+b)(a²+b²-ab)
->a³-b³=(a-b)(a²+b²+ab)
->a³+b³+c³-3a b c=(a+b+c)(a²+b²+c²-ab-b c-ca)
->If a+b+c=0 then a³+b³+c³=3a b c

DIVISION ALGORITHM

If we divide a number by another number ,then
Dividend = (Divisor * quotient) + Remainder

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MULTIPLICATION BY SHORT CUT METHODS

1.Multiplication by distributive law:

a)a*(b+c)=a*b+a*c
b)a*(b-c)=a*b-a*c

Example
a)567958*99999=567958*(100000-1)
567958*100000-567958*1
56795800000-567958
56795232042

b)978*184+978*816=978*(184+816)
978*1000=978000

2.Multiplication of a number by 5n:-Put n zeros to the right of the
multiplicand and divide the number so formed by 2n

Example 975436*625=975436*54=9754360000/16=609647500.


PROGRESSION:

A succession of numbers formed and arranged in a definite order according
to certain definite rule is called a progression.

1.Arithmetic Progression:-If each term of a progression differs from its
preceding term by a constant.
This constant difference is called the common difference of the A.P.
The n th term of this A.P is Tn=a(n-1)+d.
The sum of n terms of A.P Sn=n/2[2a+(n-1)d].

xImportant Results:

a.1+2+3+4+5......................=n(n+1)/2.
b.12+22+32+42+52......................=n(n+1)(2n+1)/6.
c.13+23+33+43+53......................=n2(n+1)2/4

2.Geometric Progression:-A progression of numbers in which every
term bears a constant ratio with ts preceding term.
i.e a,a r,a r2,a r3...............
In G.P Tn=a r n-1
Sum of n terms Sn=a(1-r n)/1-r

Problems

1.Simplify
a.8888+888+88+8
b.11992-7823-456

Solution: a.8888
888
88
8
9872
b.11992-7823-456=11992-(7823+456)
=11992-8279=3713

2.What could be the maximum value of Q in the following equation?
5PQ+3R7+2Q8=1114

Solution:    5    P   Q
  3   R    7
  2   Q    8
  11   1   4
2+P+Q+R=11
Maximum value of Q =11-2=9 (P=0,R=0)

3.Simplify: a.5793405*9999 b.839478*625

Solution:
a. 5793405*9999=5793405*(10000-1)
57934050000-5793405=57928256595
b. 839478*625=839478*54=8394780000/16=524673750.

4.Evaluate 313*313+287*287

Solution:
a²+b²=1/2((a+b)²+(a-b)²)
1/2(313+287)² +(313-287)²=1/2(600 ² +26 ² )
½(360000+676)=180338
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5.Which of the following is a prime number?
  a.241    b.337    c.391

Solution:
a.241
16>√241.Hence take the value of Z=16.
Prime numbers less than 16 are 2,3,5,7,11 and 13.
241 is not divisible by any of these. Hence we can
conclude that 241 is a prime number.
b. 337
19>√337.Hence take the value of Z=19.
Prime numbers less than 16 are 2,3,5,7,11,13 and 17.
337 is not divisible by any of these. Hence we can conclude
that 337 is a prime number.
c. 391
20>√391.Hence take the value of Z=20.
Prime numbers less than 16 are 2,3,5,7,11,13,17 and 19.
391 is divisible by 17. Hence we can conclude
that 391 is not a prime number.

6.Find the unit's digit n the product 2467 153 * 34172?

Solution: Unit's digit in the given product=Unit's digit in 7 153 * 172
Now 7 4 gives unit digit 1
7 152 gives unit digit 1
7 153 gives 1*7=7.Also 172 gives 1
Hence unit's digit in the product =7*1=7.

7.Find the total number of prime factors in 411 *7 5 *112 ?

Solution: 411 7 5 112= (2*2) 11 *7 5 *112
= 222 *7 5 *112
Total number of prime factors=22+5+2=29

8.Which of the following numbers s divisible by 3?
a.541326
b.5967013

Solution: a. Sum of digits in 541326=5+4+1+3+2+6=21 divisible by 3.
b. Sum of digits in 5967013=5+9+6+7+0+1+3=31 not divisible by 3.

9.What least value must be assigned to * so that th number 197*5462 is
divisible by 9?

Solution: Let the missing digit be x
Sum of digits = (1+9+7+x+5+4+6+2)=34+x
For 34+x to be divisible by 9 , x must be replaced by 2
The digit in place of x must be 2.

10.What least number must be added to 3000 to obtain a number exactly
divisible by 19?

Solution:On dividing 3000 by 19 we get 17 as remainder
Therefore number to be added = 19-17=2.

11.Find the smallest number of 6 digits which is exactly divisible by 111?

Solution:Smallest number of 6 digits is 100000
On dividing 10000 by 111 we get 100 as remainder
Number to be added =111-100=11.
Hence,required number =10011.

12.On dividing 15968 by a certain number the quotient is 89 and the remainder
is 37.Find the divisor?

Solution:Divisor = (Dividend-Remainder)/Quotient
=(15968-37) / 89
=179.

13.A number when divided by 342 gives a remainder 47.When the same number
is divided by 19 what would be the remainder?

Solution:Number=342 K + 47 = 19 * 18 K + 19 * 2 + 9=19 ( 18K + 2) + 9.
The given number when divided by 19 gives 18 K + 2 as quotient and 9 as
remainder.
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14.A number being successively divided by 3,5,8 leaves remainders 1,4,7
respectively. Find the respective remainders if the order of
divisors are reversed?

Solution:Let the number be x.

3    x    5    y    - 1    8   z    - 4      1    - 7 z=8*1+7=15
y=5z+4 = 5*15+4 = 79
x=3y+1 = 3*79+1=238
Now
  8    238
  5    29   - 6
  3    5    - 4
  1    - 2
Respective remainders are 6,4,2.

15.Find the remainder when 231 is divided by 5?

Solution:210 =1024.unit digit of 210 * 210 * 210 is 4 as
4*4*4 gives unit digit 4
unit digit of 231 is 8.
Now 8 when divided by 5 gives 3 as remainder.
231 when divided by 5 gives 3 as remainder.

16.How many numbers between 11 and 90 are divisible by 7?

Solution:The required numbers are 14,21,28,...........,84
This is an A.P with a=14,d=7.
Let it contain n terms
then T =84=a+(n-1)d
=14+(n-1)7
=7+7n
7n=77 =>n=11.

17.Find the sum of all odd numbers up to 100?

Solution:The given numbers are 1,3,5.........99.
This is an A.P with a=1,d=2.
Let it contain n terms 1+(n-1)2=99
=>n=50
Then required sum =n/2(first term +last term)
=50/2(1+99)=2500.

18.How many terms are there in 2,4,6,8..........,1024?

Solution:Clearly 2,4,6........1024 form a G.P with a=2,r=2
Let the number of terms be n
then 2*2 n-1=1024

TECH

Technical Questions

1. A 2MB PCM(pulse code modulation) has

a) 32 channels
b) 30 voice channels & 1 signaling channel.
c) 31 voice channels & 1 signaling channel.
d) 32 channels out of which 30 voice channels, 1 signaling channel, & 1 Synchronization channel.

Ans: (c)

2. Time taken for 1 satellite hop in voice communication is

a) 1/2 second
b) 1 seconds
c) 4 seconds
d) 2 seconds

Ans: (a)

3. Max number of satellite hops allowed in voice communication is :

a) only one
b) more han one
c) two hops
d) four hops

Ans: (c)

4. What is the max. decimal number that can be accomodated in a byte.

a) 128
b) 256
c) 255
d) 512

Ans: (c)

5. Conditional results after execution of an instruction in a micro processor is stored in

a) register
b) accumulator
c) flag register
d) flag register part of PSW(Program Status Word)

Ans: (d)

6. Frequency at which VOICE is sampled is

a) 4 Khz
b) 8 Khz
c) 16 Khz
d) 64 Khz

Ans: (a)

7. Line of Sight is

a) Straight Line
b) Parabolic
c) Tx & Rx should be visible to each other
d) none

Ans: (c)

8. Purpose of PC(Program Counter) in a MicroProcessor is

a) To store address of TOS(Top Of Stack)
b) To store address of next instruction to be executed.
c) count the number of instructions.
d) to store base address of the stack.

Ans: (b)

9. What action is taken when the processor under execution is interrupted by a non-maskable interrupt?

a) Processor serves the interrupt request after completing the execution of the current instruction.
b) Processor serves the interupt request after completing the current task.
c) Processor serves the interupt request immediately.
d) Processor serving the interrupt request depends upon the priority of the current task under execution.

Ans: (a)

10. The status of the Kernel is

a) task
b) process
c) not defined.
d) none of the above.

Ans: (b)

11. To send a data packet using datagram , connection will be established

a) before data transmission.
b) connection is not established before data transmission.
c) no connection is required.
d) none of the above.

Ans: (c)

12. Word allignment is

a) alligning the address to the next word boundary of the machine.
b) alligning to even boundary.
c) alligning to word boundary.
d) none of the above.

Ans: (a)

13 When a 'C' function call is made, the order in which parameters passed to the function are pushed into the stack is

a) left to right
b) right to left
c) bigger variables are moved first than the smaller variales.
d) smaller variables are moved first than the bigger ones.
e) none of the above.

Ans: (b)

14 What is the type of signalling used between two exchanges?

a) inband
b) common channel signalling
c) any of the above
d) none of the above.

Ans: (a)

15 Buffering is
a) the process of temporarily storing the data to allow for small variation in device speeds
b) a method to reduce cross talks
c) storage of data within transmitting medium until the receiver is ready to receive.
d) a method to reduce routing overhead.

Ans: (a)

16. Memory allocation of variables declared in a program is

a) allocated in RAM.
b) allocated in ROM.
c) allocated on stack.
d) assigned to registers.

Ans: (c)

17. A software that allows a personal computer to pretend as a computer terminal is

a) terminal adapter
b) bulletin board
c) modem
d) terminal emulation

Ans: (d)

18. Find the output of the following program

int *p,*q;
p=(int *)1000;
q=(int *)2000;
printf("%d",(q-p));

Ans: 500

19. Which addressing mode is used in the following statements:

(a) MVI B,55
(b) MOV B,A
(c) MOV M,A

Ans. (a) Immediate addressing mode.
(b) Register Addressing Mode
(c) Direct addressing mode

20. RS-232C standard is used in _____________.

Ans. Serial I/O

21. Memory. Management in Operating Systems is done by

a) Memory Management Unit
b) Memory management software of the Operating System
c) Kernel

Ans: (b)

22. What is done for a Push operation?

Ans: SP is decremented and then the value is stored.

23. Binary equivalent of 52

Ans. 110100


24. Hexadecimal equivalent of 3452

Ans. 72A


25. Explain Just In Time Concept ?

Ans. Elimination of waste by purchasing manufacturing exactly when needed


26. A good way of unit testing s/w program is

Ans. User test


27. OOT uses

Ans. Encapsulated of detect methods


28.EDI useful in

Ans. Electronic Transmission


29. MRPII different from MRP

Ans. Modular version of man redundant initials


30. Hard disk time for R/W head to move to correct sector

Ans. Latency Time

31. The percentage of times a page number bound in associate register is called

Ans. Bit ratio

32. Expand MODEM

Ans. Modulator and Demodulator


33. RDBMS file system can be defined as

Ans. Interrelated


34. Super Key is

Ans. Primary key and Attribute


35. Windows 95 supports44

(a) Multiuser
(b) n tasks
(c) Both
(d) None

Ans. (a)


36.In the command scanf, h is used for

Ans. Short int


37.A process is defined as

Ans. Program in execution


38.A thread is

Ans. Detachable unit of executable code)


39.What is the advantage of Win NT over Win 95

Ans. Robust and secure

40.How is memory management done in Win95

Ans. Through paging and segmentation


41.What is meant by polymorphism

Ans. Redfinition of a base class method in a derived class


42.What is the essential feature of inheritance

Ans. All properties of existing class are derived


43.What does the protocol FTP do

Ans. Transfer a file b/w stations with user authentification


44.In the transport layer ,TCP is what type of protocol

Ans. Connection oriented


45.Why is a gateway used

Ans. To connect incompatible networks


46.How is linked list implemented

Ans. By referential structures


47.What method is used in Win95 in multitasking

Ans. Non preemptive check


48.What is a semaphore

Ans. A method synchronization of multiple processes


49.What is the precedence order from high to low ,of the symbols ( ) ++ /

Ans.( ) , ++, /

50.Preorder of A*(B+C)/D-G

Ans.*+ABC/-DG

51.What is the efficiency of merge sort

Ans. O(n log n)


52.In which layer are routers used

Ans.In network layer


53.Which of the following sorting algorithem has average sorting behavior --
Bubble sort,merge sort,heap sort,exchange sort

Ans. Heap sort


54.In binary search tree which traversal is used for getting ascending order values--Inorder ,post order,preorder

Ans.Inorder


55.What are device drivers used for

Ans.To provide software for enabling the hardware


56.What is fork command in unix

Ans. System call used to create process


57.What is make command in unix

Ans. Used forcreation of more than one file


58.In unix .profile contains

Ans. Start up program


59.In unix 'ls 'stores contents in

Ans.i-node block

60. Which of the following involves context switch,

(a) system call
(b) priviliged instruction
(c) floating poitnt exception
(d) all the above
(e) none of the above

Ans: (a)

61. In OST, terminal emulation is done in

(a) sessions layer
(b) application layer
(c) presentation layer
(d) transport layer

Ans: (b)

62. For 1 MB memory, the number of address lines required,

(a)11
(b)16
(c)22
(d) 24

Ans. (b)

63. Semaphore is used for

(a) synchronization
(b) dead-lock avoidence
(c) box
(d) none

Ans. (a)

64. Which holds true for the following statement

class c: public A, public B

a) 2 member in class A, B should not have same name
b) 2 member in class A, C should not have same name
c) both
d) none

Ans. (a)

65.Preproconia.. does not do which one of the following

(a) macro
(b) conditional compliclation
(c) in type checking
(d) including load file

Ans. (c)

66. Piggy backing is a technique for

a) Flow control
b) Sequence
c) Acknowledgement
d) retransmition

Ans. (c)

67. Which is not a memory management scheme?

a) buddy system
b) swapping
c) monitors
d) paging

Ans : c

68. There was a circuit given using three nand gates with two inputs and one output.
Find the output.

a) OR
b) AND
c) XOR
d) NOT

Ans. (a)

69.Iintegrated check value(ICV) are used as:

Ans. The client computes the ICV and then compares it with the senders value.

70. When applets are downloaded from web sites , a byte verifier performs _________?

Ans. Status check.

71. An IP/IPX packet received by a computer using... having IP/IPX both how the packet is handled.

Ans. Read the, field in the packet header with to send IP or IPX protocol.

72. The UNIX shell ....

a) does not come with the rest of the system
b) forms the interface between the user and the kernal
c) does not give any scope for programming
d) deos not allow calling one program from with in another
e) all of the above

Ans. (b)

73. In UNIX a files i-node ......?

Ans. Is a data structure that defines all specifications of a file like the file size,
number of lines to a file, permissions etc.

74. The very first process created by the kernal that runs till the kernal process is halts is

a) init
b) getty
c) both (a) and (b)
d) none of these

Ans. (a)

75. In the process table entry for the kernel process, the process id value is

(a) 0
(b) 1
(c) 2
(d) 255
(e) it does not have a process table entry

Ans. (a)

76. Which of the following API is used to hide a window

a) ShowWindow
b) EnableWindow
c) MoveWindow
d) SetWindowPlacement
e) None of the above

Ans. (a)

77. Which function is the entry point for a DLL in MS Windows 3.1

a) Main
b) Winmain
c) Dllmain
d) Libmain
e) None

Ans. (b)

78. The standard source for standard input, standard output and standard error is

a) the terminal
b) /dev/null
c) /usr/you/input, /usr/you/output/, /usr/you/error respectively
d) None

Ans. (a)

79. The redirection operators > and >>

a) do the same function
b) differ : > overwrites, while >> appends
c) differ : > is used for input while >> is used for output
d) differ : > write to any file while >> write only to standard output
e) None of these

Ans. (b)

80. The command grep first second third /usr/you/myfile

a) prints lines containing the words first, second or third from the file /usr/you/myfile
b) searches for lines containing the pattern first in the files
second, third, and /usr/you/myfile and prints them
c) searches the files /usr/you/myfiel and third for lines containing the words first or second and prints them
d) replaces the word first with the word second in the files third and /usr/you/myfile
e) None of the above

Ans. (b)

81. You are creating a Index on EMPNO column in the EMPLOYEE table. Which statement will you use?
a) CREATE INdEX emp_empno_idx ON employee, empno;
b) CREATE INdEX emp_empno_idx FOR employee, empno;
c) CREATE INdEX emp_empno_idx ON employee(empno);
d) CREATE emp_empno_idx INdEX ON employee(empno);

Ans. c

82. Which program construct must return a value?
a) Package
b) Function
c) Anonymous block
d) Stored Procedure
e) Application Procedure

Ans. b

83. Which Statement would you use to remove the EMPLOYEE_Id_PK PRIMARY KEY constraint and all depending constraints fromthe EMPLOYEE table?
a) ALTER TABLE employee dROP PRIMARY KEY CASCAdE;
b) ALTER TABLE employee dELETE PRIMARY KEY CASCAdE;
c) MOdIFY TABLE employee dROP CONSTRAINT employee_id_pk CASCAdE;
d) ALTER TABLE employee dROP PRIMARY KEY employee_id_pk CASCAdE;
e) MOdIFY TABLE employee dELETE PRIMARY KEY employee_id_pk CASCAdE;

Ans. a

84. Which three commands cause a transaction to end? (Chosse three)
a) ALTER
b) GRANT
c) DELETE
d) INSERT
e) UPdATE
f) ROLLBACK

Ans. a ,b ,f

85. Under which circumstance should you create an index on a table?
a) The table is small.
b) The table is updated frequently.
c) A columns values are static and contain a narrow range of values
d) Two columns are consistently used in the WHERE clause join condition of SELECT
statements.

Ans.d

86. What was the first name given to Java Programming Language.

a) Oak - Java
b) Small Talk
c) Oak
d) None

Ans.a

87.When a bicycle is in motion,the force of friction exerted by the ground on the two wheels is such that it acts

(a) In the backward direction on the front wheel and in the forward direction on the rear wheel.
(b) In the forward direction on the front wheel and in the backward direction on the rear wheel.
(c) In the backward direction on both the front and rear wheels.
(d) In the backward direction on both the front and rear wheels.

Ans. (d)

88. A certain radioactive element A, has a half life = t seconds.
In (t/2) seconds the fraction of the initial quantity of the element so far decayed is nearly

(a) 29%
(b) 15%
(c) 10%
(d) 45%

Ans. (a)

89. Which of the following plots would be a straight line ?

(a) Logarithm of decay rate against logarithm of time
(b) Logarithm of decay rate against logarithm of number of decaying nuclei
(c) Decay rate against time
(d) Number of decaying nuclei against time

Ans. (b)

90. A radioactive element x has an atomic number of 100.
It decays directly into an element y which decays directly into element z.
In both processes a charged particle is emitted.
Which of the following statements would be true?

(a) y has an atomic number of 102
(b) y has an atomic number of 101
(c) z has an atomic number of 100
(d) z has an atomic number of 101

Ans. (b)

91. If the sum of the roots of the equation ax² + bx + c=0 is equal to the sum of the squares of their reciprocals
then a/c, b/a, c/b are in

(a) AP
(b) GP
(c) HP
(d) None of these

Ans. (c)

92. A man speaks the truth 3 out of 4 times.
He throws a die and reports it to be a 6.
What is the probability of it being a 6?

(a) 3/8
(b) 5/8
(c) 3/4
(d) None of the above

Ans. (a)

93. If cos²A + cos²B + cos²C = 1 then ABC is a

(a) Right angle triangle
(b) Equilateral triangle
(c) All the angles are acute
(d) None of these

Ans. (a)

94. Image of point (3,8) in the line x + 3y = 7 is

(a) (-1,-4)
(b) (-1,4)
(c) (2,-4)
(d) (-2,-4)

Ans. (a)

95. The mass number of a nucleus is

(a) Always less than its atomic number
(b) Always more than its atomic number
(c) Sometimes more than and sometimes equal to its atomic number
(d) None of the above

Ans. (c)

96. The maximum KE of the photoelectron emitted from a surface is dependent on

(a) The intensity of incident radiation
(b) The potential of the collector electrode
(c) The frequency of incident radiation
(d) The angle of incidence of radiation of the surface

Ans. (c)

97. Which of the following is not an essential condition for interference

(a) The two interfering waves must be propagated in almost the same direction or
the two interfering waves must intersect at a very small angle
(b) The waves must have the same time period and wavelength
(c) Amplitude of the two waves should be the same
(d) The interfering beams of light must originate from the same source

Ans. (c)

98. When X-Ray photons collide with electrons

(a) They slow down
(b) Their mass increases
(c) Their wave length increases
(d) Their energy decreases

Ans. (c)

99. An electron emits energy

(a) Because its in orbit
(b) When it jumps from one energy level to another
(c) Electrons are attracted towards the nucleus
(d) The electrostatic force is insufficient to hold the electrons in orbits

Ans. (b)

100. How many bonds are present in CO₂ molecule?

(a) 1
(b) 2
(c) 0
(d) 4

Ans. (d)

csc

P@Q' means 'P is not greater than Q'.

'P#Q' means 'P is neither greater than nor smaller than Q'.

'P$Q' means 'P is not smaller than Q'.

'PQ' means 'P is neither smaller than nor equal to Q'.

'P%Q' means 'P is neither greater than nor equal to Q'.

In each of the following questions assuming the given statements to be true, find out which of the two conclusions I and II given below them is/are definitely true.

Give answer 1): if only conclusion I is true.

Give answer 2): if only conclusion II is true.

Give answer 3): if either conclusion I or conclusion II is true.

Give answer 4): if neither conclusion I nor conclusion II is true.

Give answer 5): if both conclusion I and II are true.

6. Statements: F$W,W#T,TK

Conclusions: I. FK

II. W$K

7. Statements: R@M,M%D,D$H

Conclusions: I. R@H

II. DR

8. Statements: J$L,L#B,B@E

Conclusions: I. E$L

II. E%L

9. Statements: A$V,V#R,R@U

Conclusions: I. UR

II. U#R

10. Statements: F%G,G@H,HJ

Conclusions: I. F@H

II. G@J

Directions (Q.11-15): In each question below is given a statement followed by two assumptions numbered I and II. An assumption is something supposed or taken for granted. You have to consider the statement and the following assumptions and decide which of the assumptions is implicit in the statement.

Give answer 1): if only assumption I is implicit.

Give answer 2): if only assumption II is implicit.

Give answer 3): if either assumption I or assumption II is implicit.

Give answer 4): if neither assumption I nor assumption II is implicit.

Give answer 5): if both the assumptions I and II are implicit.

11. Statement: “Enroll with us before 30th November to get the advantage of our 20% discount offer”. -An advertisement by a coaching class

Assumptions:

I. Discount offer is bound to attract good students as well.

II. Even those students who cannot afford to pay the fees of coaching classes may join this class.

12. Statement: “Join our Yoga institute to keep yourself completely fit.”----An advertisement

Assumptions:

I. People may prefer exercise to medication.

II. There is an awareness to a great extent about Yoga exercises among people.

13. Statement: If you want to get a good job you must have at least the basic knowledge of computers.

Assumption:

I. All good jobs involve use of computers.

II. Computer knowledge has been made an essential criterion by most of the companies nowadays.

14. Statement: As a measure to avoid occurrence of the epidemics due to monsoon the civic authorities have organized free vaccination camps all over the city.

Assumptions:

I. There may be a good response to the camps organized by civic authorities.

II. people are generally aware about the need for vaccination.

15. Statements: In view of the large number of cases of suicides committed by the farmers in State X the State Government has decided to waive off the agriculture loans granted to the farmers.

Assumptions:

I. This may stop further cases of suicides committed by the farmers in State X.

II. This move of the Government may be welcomed by the public at large.

Answers

1. (4)
2. (2)
3. (2)
4. (5)
5. (5)
6. (5)
7. (2)
8. (4)
9. (5)
10. (4)
11. (2)
12. (5)
13. (3)
14. (5)
15. (5)

APTITUDE

Important Facts:

Cost Price: The price at which an article is purchased,
is called its cost price,abbreviated as C.P.

Selling Price: The price at which an article is sold,
is called its selling price,abbreviated as S.P.

Profit or Gain: If S.P. Is greater than C.P. The seller
is said to have a profit or gain.

Loss:if S.P. Is less than C.P., the seller is said to
have incurred a loss.

Formulae

1.Gain=(S.P-C.P)
2.Loss=(C.P-S.P)
3.Loss or Gain is always reckoned on C.P.
4.Gain%=(gain*100)/C.P
5.Loss%=(loss*100)/C.P
6.S.P=[(100+gain%)/100]*C.P
7.S.P=[(100-loss%)/100]*C.P
8.C.P=(100*S.P)/(100+gain%)
9.C.P=(100*S.P)/(100-loss%)
10.If an article is sold at a gain of say,35%,then S.P=135% of C.P.
11.If an article is sold at a loss of say,35%,then S.P=65% of C.P.
12.When a person sells two similar items, one at a gain of say,
x%,and the other at a loss of x%,then the seller always incurs a
loss given by Loss%=[common loss and gain %/10]2=(x/10)2
13.If a trader professes to sell his goods at cost price,but uses
false weight,then Gain%=[(error/(true value-error))*100]%
14.Net selling price=Marked price-Discount

Top

Simple Problems

1.A man buys an article for Rs.27.50 and sells it for Rs.28.60
Find the gain percent.

Sol: C.P=Rs 27.50 S.P=Rs 28.60
then Gain=S.P-C.P=28.60-27.50=Rs 1.10
Gain%=(gain*100)/C.P%
=(1.10*100)/27.50%=4%

2.If a radio is purchased for Rs 490 and sold for Rs 465.50
Find the loss%?

Sol: C.P=Rs 490 S.P=Rs 465.50
Loss=C.P-S.P=490-465.50=Rs 24.50
Loss%=(loss*100)/C.P%
=(24.50*100)/490%=5%

3.Find S.P when C.P=Rs 56.25 and Gain=20%

Sol: S.P=[(100+gain%)/100]*C.P
S.P=[(100+20)/100]56.25=Rs 67.50

4.Find S.P when C.P=Rs 80.40,loss=5%

Sol: S.P=[(100-loss%)/100]*C.P
S.P=[(100-5)/100]*80.40=Rs 68.34

5.Find C.P when S.P=Rs 40.60,gain=16%?

Sol: C.P=(100*S.P)/(100+gain%)
C.P=(100*40.60)/(100+16)=Rs 35

6.Find C.P when S.P=Rs 51.70 ,loss=12%?

Sol: C.P=(100*S.P)/(100-loss%)
C.P=(100*51.70)/(100-12)=Rs 58.75

7.A person incurs 5% loss by selling a watch for Rs 1140 . At
what price should the watch be sold to earn 5% profit?

Sol: Let the new S.P be Rs x then,
(100-loss%):(1st S.P)=(100+gain%):(2nd S.P)
(100-5)/1140=(100+5)/x
x=(105*1140)/95=Rs 1260

8.If the cost price is 96% of the selling price,then what is
the profit percent?

Sol: let S.P=Rs 100 then C.P=Rs 96
profit=S.P-C.P=100-96=Rs 4
profit%=(profit*C.P)/100%
=(4*96)/100=4.17%

9.A discount dealer professes to sell his goods at cost price
but uses a weight of 960 gms for a Kg weight .Find his gain %?

Sol: Gain%=[(error*100)/(true value-error)]%
=[(40*100)/1000-40)]%=25/6%

10.A man sold two flats for Rs 675,958 each .On one he gains
16% while on the other he losses 16%.How much does he gain or
lose in the whole transaction?

Sol: loss%=[common loss or gain%/10]2=(16/10)2=2.56%

11.A man sold two cows at Rs 1995 each. On one he lost 10% and
on the other he gained 10%.what his gain or loss percent?

Sol: If loss% and gain% is equal to 10
then there is no loss or no gain.

12.The price of an article is reduced by 25% in order to
restore the must be increased by ?

Sol: [x/(100-x)]*100 =[25/(100-25)]*100
=(25/75)*100=100/3%

13.Two discounts of 40% and 20% equal to a single discount of?

Sol: {[(100-40)/100]*[(100-20)/100]}%=(60*80)/(100*100)%
=48%
single discount is equal to (100-48)%=52%
Top
Difficult Problems

1.The cost of an article including the sales tax is Rs 616.The
rate of sales tax is 10%,if the shopkeeper has made a profit
of 12%,then the cost price of the article is?

Sol: 110% of S.P=616
S.P=(616*100)/110=Rs 560
C.P=(100*S.P)/(100+gain%)
C.P. =(100*560)/(100+12)=Rs 500

2.Sam purchased 20 dozens of toys at the rate of 375 Rs per dozen.
He sold each one of then at the rate of Rs 33.What was his
percentage profit?

Sol: C.P of one toy=Rs 375/12=Rs 31.25
S.P of one toy=Rs 33
profit=S.P-C.P=33-31.25=Rs 1.75
profit %=(profit/C.P)*100
=(1.75/31.25)*100
profit% =5.6%

3.Two third of consignment was sold at a profit of 5% and the
remainder at a loss of 2%.If the total was Rs 400,the value of the
consignment was?

Sol: let the total value be Rs x
value of 2/3=2x/3, value of 1/3=x/3
total S.P value be Rs x
total S.P=[(105% of 2x/3)+(98% of x/3)]
=(105*2x)/(100*3)+(98/100)+x/3
=308x/300
(308x/300)-x=400
8x/300=400
x=(300*400)/8=Rs 15000

4.Kunal bought a suitcase with 15% discount on the labelled price.
He sold the suitcase for Rs 2880 with 20% profit on the labelled
price .At what price did he buy the suitcase?

Sol: let the labelled price be Rs x
then 120% of x=2880
x=(2880*100)/120=Rs 2400
C.P=85% of the 2400
(85*2400)/100=Rs 2040

5.A tradesman gives 4% discount on the marked price and gives
article free for buying every 15 articles and thus gains 35%. The
marked price is above the cost price by

Sol: let the C.P of each article be Rs 100
then C.P of 16 articles=Rs (100*16)=Rs 1600
S.P of 15 articles =1600*(135/100)=Rs 2160
S.P of each article =2160/15=Rs 144
If S.P is Rs 96, marked price =Rs 100
If S.P is Rs 144,marked price =(100/96)*144=Rs 15000
therefore marked price=50% above C.P

6.By selling 33m of cloth ,one gains the selling price of 11m.Find
the gain percent?

Sol: gain=S.P of 33m-C.P of 33m
=11m of S.P
S.P of 22m=C.P of 33m
let C.P of each meter be Rs 1,then C.P of 22m=Rs 22
S.P of 22m=Rs 33
gain=S.P-C.P=33-22=Rs 11
gain%=(gain/C.P)*100
=(11/22)*100=50%
Top

7.The price of a jewel, passing through three hands, rises on the
whole by 65%.if the first and second sellers earned 20% and 25%
profit respectively,find the percentage profit earned by the
third seller?

Sol: let the original price of the jewel be Rs P and
let the profit earned by the third seller be x%
then (100+x)% of 125% of P=165% of P
[(100+x)/100]*(125/100)*(120/100)*P=(165/100)*P
100+x=(165*100*100)/(125*120)
100+x=110
x=10%

8.When a producer allows 36% commission on the retail price of his
product ,he earns a profit of 8.8%.what would be his profit
percent if the commission is reduced by 24%

Sol: let retail price =Rs 100
commission=Rs 36
S.P=retail price-commission=100-36=Rs 64
But profit=8.8%
C.P=(100*C.P)/(gain+100)=(100*64)/(100+8.8)=Rs 1000/17
new commission=Rs 12
new S.P=100-12=Rs 88
gain=88-(1000/17)=Rs 496/17
gain%=gain*100/C.P
=(496*17*100)(17*1000)
gain%=49.6%

9.Vikas bought paper sheets for Rs 7200 and spent Rs 200 on
transport. Paying Rs 600,he had 330 boxes made,which he sold
at Rs 28 each. His profit percentage is

Sol: total investments=7200+200+600=Rs 8000
total receipt=330*28=Rs 9240
gain=S.P-C.P
=total receipt-total investments
gain=9240-8000=Rs 1240
gain% =gain*100/C.P=1240*100/8000=15.5%

10.A person earns 15% on investment but loses 10% on another
investment .If the ratio of the two investments be 3:5 ,what is the
gain or loss on the two investments taken together?

Sol: let the investments be 3x and 5x
then total investment=8x
total receipt=115% of 3x+90% of 5x
=115*3x/100+90*5x/100=7.95x
loss=C.P-S.P=8x-7.95x=0.05x
loss%=.05x*100/8x=0.625%

11.The profit earned by selling an article for Rs 900 is double the
loss incurred when the same article is sold for Rs 490 .At what
price should the article be sold to make 25% profit?

Sol: let C.P be Rs x
900-x=2(x-450)
3x=1800
x=Rs 600
C.P=Rs 600 , gain required=25%
S.P=(100+gain%)*C.P/100
S.P=(100+25)*600/100=Rs 750

12.If an article is sold at 5% gain instead of 5% loss,the seller
gets Rs 6.72 more. The C.P of the article is?

Sol: let C.P be Rs x
105% of x-95% of x=6.72
(105/100)*x-(95/100)*x=6.72
x/10=6.72
x=Rs 67.21.