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Technical Questions
1. A 2MB PCM(pulse code modulation) has
a) 32 channels
b) 30 voice channels & 1 signalling channel.
c) 31 voice channels & 1 signalling channel.
d) 32 channels out of which 30 voice channels, 1 signalling channel, & 1
Synchronizatio 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)
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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 nonmaskable
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)
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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
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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:
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(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 opertion?
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
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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 supports
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(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
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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)
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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.inode block
60. Which of the following involves context switch,
(a) system call
(b) priviliged instruction
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(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
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(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
_________?
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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
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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
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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.
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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 ?
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(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
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(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
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(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)

3. DECIMAL FRACTIONS

3. DECIMAL FRACTIONS
IMPORTANT FACTS AND FORMULAE
I. Decimal Fractions : Fractions in which denominators are powers of 10 are known as decimal
fractions.
Thus ,1/10=1 tenth=.1;1/100=1 hundredth =.01;
99/100=99 hundreths=.99;7/1000=7 thousandths=.007,etc
II. Conversion of a Decimal Into Vulgar Fraction : Put 1 in the denominator under the decimal
point and annex with it as many zeros as is the number of digits after the decimal point. Now,
remove the decimal point and reduce the fraction to its lowest terms.
Thus, 0.25=25/100=1/4;2.008=2008/1000=251/125.
III. 1. Annexing zeros to the extreme right of a decimal fraction does not change its value
Thus, 0.8 = 0.80 = 0.800, etc.
2. If numerator and denominator of a fraction contain the same number of decimal
places, then we remove the decimal sign.
Thus, 1.84/2.99 = 184/299 = 8/13; 0.365/0.584 = 365/584=5
IV. Operations on Decimal Fractions :
1. Addition and Subtraction of Decimal Fractions : The given numbers are so
placed under each other that the decimal points lie in one column. The numbers
so arranged can now be added or subtracted in the usual way.
2. Multiplication of a Decimal Fraction By a Power of 10 : Shift the decimal
point to the right by as many places as is the power of 10.
Thus, 5.9632 x 100 = 596,32; 0.073 x 10000 = 0.0730 x 10000 = 730.
3.Multiplication of Decimal Fractions : Multiply the given numbers considering
them without the decimal point. Now, in the product, the decimal point is marked
off to obtain as many places of decimal as is the sum of the number of decimal
places in the given numbers.
Suppose we have to find the product (.2 x .02 x .002). Now, 2x2x2 = 8. Sum of
decimal places = (1 + 2 + 3) = 6. .2 x .02 x .002 = .000008.
4.Dividing a Decimal Fraction By a Counting Number : Divide the given
number without considering the decimal point, by the given counting number.
Now, in the quotient, put the decimal point to give as many places of decimal as
there are in the dividend.
Suppose we have to find the quotient (0.0204 + 17). Now, 204 ^ 17 = 12. Dividend contains
4 places of decimal. So, 0.0204 + 17 = 0.0012.
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5. Dividing a Decimal Fraction By a Decimal Fraction : Multiply both the dividend and the
divisor by a suitable power of 10 to make divisor a whole number. Now, proceed as above.
Thus, 0.00066/0.11 = (0.00066*100)/(0.11*100) = (0.066/11) = 0.006V
V. Comparison of Fractions : Suppose some fractions are to be arranged in ascending or
descending order of magnitude. Then, convert each one of the given fractions in the decimal form,
and arrange them accordingly.
Suppose, we have to arrange the fractions 3/5, 6/7 and 7/9 in descending order.
now, 3/5=0.6,6/7 = 0.857,7/9 = 0.777....
since 0.857>0.777...>0.6, so 6/7>7/9>3/5
VI. Recurring Decimal : If in a decimal fraction, a figure or a set of figures is repeated
continuously, then such a number is called a recurring decimal.
In a recurring decimal, if a single figure is repeated, then it is expressed by putting a dot on it.
If a set of figures is repeated, it is expressed by putting a bar on the set
______
Thus 1/3 = 0.3333….= 0.3; 22 /7 = 3.142857142857.....= 3.142857
Pure Recurring Decimal: A decimal fraction in which all the figures after the decimal point
are repeated, is called a pure recurring decimal.
Converting a Pure Recurring Decimal Into Vulgar Fraction : Write the repeated figures
only once in the numerator and take as many nines in the denominator as is the number of
repeating figures.
thus ,0.5 = 5/9; 0.53 = 53/59 ;0.067 = 67/999;etc...
Mixed Recurring Decimal: A decimal fraction in which some figures do not repeat and some
of them are repeated, is called a mixed recurring decimal.
e.g., 0.17333.= 0.173.
Converting a Mixed Recurring Decimal Into Vulgar Fraction : In the numerator, take the
difference between the number formed by all the digits after decimal point (taking repeated
digits only once) and that formed by the digits which are not repeated, In the denominator,
take the number formed by as many nines as there are repeating digits followed by as many
zeros as is the number of non-repeating digits.
Thus 0.16 = (16-1) / 90 = 15/19 = 1/6;
____
0.2273 = (2273 – 22)/9900 = 2251/9900
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VII. Some Basic Formulae :
1. (a + b)(a- b) = (a2 - b2).
2. (a + b)2 = (a2 + b2 + 2ab).
3. (a - b)2 = (a2 + b2 - 2ab).
4. (a + b+c)2 = a2 + b2 + c2+2(ab+bc+ca)
5. (a3 + b3) = (a + b) (a2 - ab + b2)
6. (a3 - b3) = (a - b) (a2 + ab + b2).
7. (a3 + b3 + c3 - 3abc) = (a + b + c) (a2 + b2 + c2-ab-bc-ca)
8. When a + b + c = 0, then a3 + b3+ c3 = 3abc
SOLVED EXAMPLES
Ex. 1. Convert the following into vulgar fraction:
(i) 0.75 (ii) 3.004 (iii) 0.0056
Sol. (i). 0.75 = 75/100 = 3/4 (ii) 3.004 = 3004/1000 = 751/250 (iii) 0.0056 = 56/10000 = 7/1250
Ex. 2. Arrange the fractions 5/8, 7/12, 13/16, 16/29 and 3/4 in ascending order of magnitude.
Sol. Converting each of the given fractions into decimal form, we get :
5/8 = 0.624, 7/12 = 0.8125, 16/29 = 0.5517, and 3/4 = 0.75
Now, 0.5517<0.5833<0.625<0.75<0.8125
 16/29 < 7/12 < 5/8 < 3/4 < 13/16
Ex. 3. arrange the fractions 3/5, 4/7, 8/9, and 9/11 in their descending order.
Sol. Clearly, 3/5 = 0.6, 4/7 = 0.571, 8/9 = 0.88, 9/111 = 0.818.
Now, 0.88 > 0.818 > 0.6 > 0.571
 8/9 > 9/11 > 3/4 > 13/ 16
Ex. 4. Evaluate : (i) 6202.5 + 620.25 + 62.025 + 6.2025 + 0.62025
(ii) 5.064 + 3.98 + 0.7036 + 7.6 + 0.3 + 2
Sol. (i) 6202.5 (ii) 5.064
620.25 3.98
62.025 0.7036
6.2025 7.6
+ __ 0.62025 0.3
6891.59775 _2.0___
19.6476
Ex. 5. Evaluate : (i) 31.004 – 17.2368 (ii) 13 – 5.1967
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Sol. (i) 31.0040 (ii) 31.0000
– 17.2386 – _5.1967
13.7654 7.8033
Ex. 6. What value will replace the question mark in the following equations ?
(i) 5172.49 + 378.352 + ? = 9318.678
(ii) ? – 7328.96 + 5169.38
Sol. (i) Let 5172.49 + 378.352 + x = 9318.678
Then , x = 9318.678 – (5172.49 + 378.352) = 9318.678 – 5550.842 = 3767.836
(ii) Let x – 7328.96 = 5169.38. Then, x = 5169.38 + 7328.96 = 12498.34.
Ex. 7. Find the products: (i) 6.3204 * 100 (ii) 0.069 * 10000
Sol. (i) 6.3204 * 1000 = 632.04 (ii) 0.069 * 10000 = 0.0690 * 10000 = 690
Ex. 8. Find the product:
(i) 2.61 * 1.3 (ii) 2.1693 * 1.4 (iii) 0.4 * 0.04 * 0.004 * 40
Sol. (i) 261 8 13 = 3393. Sum of decimal places of given numbers = (2+1) = 3.
2.61 * 1.3 = 3.393.
(ii) 21693 * 14 = 303702. Sum of decimal places = (4+1) = 5
2.1693 * 1.4 = 3.03702.
(iii) 4 * 4 * 4 * 40 = 2560. Sum of decimal places = (1 + 2+ 3) = 6
0.4 * 0.04 * 0.004 * 40 = 0.002560.
Ex. 9. Given that 268 * 74 = 19832, find the values of 2.68 * 0.74.
Sol. Sum of decimal places = (2 + 2) = 4
2.68 * 0.74 = 1.9832.
Ex. 10. Find the quotient:
(i) 0.63 / 9 (ii) 0.0204 / 17 (iii) 3.1603 / 13
Sol. (i) 63 / 9 = 7. Dividend contains 2 places decimal.
0.63 / 9 = 0.7.
(ii) 204 / 17 = 12. Dividend contains 4 places of decimal.
0.2040 / 17 = 0.0012.
(iii) 31603 / 13 = 2431. Dividend contains 4 places of decimal.
3.1603 / 13 = 0.2431.
Ex. 11. Evaluate :
(i) 35 + 0.07 (ii) 2.5 + 0.0005
(iii) 136.09 + 43.9
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Sol. (i) 35/0.07 = ( 35*100) / (0.07*100) = (3500 / 7) = 500
(ii) 25/0.0005 = (25*10000) / (0.0005*10000) = 25000 / 5 = 5000
(iii) 136.09/43.9 = (136.09*10) / (43.9*10) = 1360.9 / 439 = 3.1
Ex. 12. What value will come in place of question mark in the following equation?
(i) 0.006 +? = 0.6 (ii) ? + 0.025 = 80
Sol. (i) Let 0.006 / x = 0.6, Then, x = (0.006 / 0.6) = (0.006*10) / (0.6*10) = 0.06/6 = 0.01
(ii) Let x / 0.025 = 80, Then, x = 80 * 0.025 = 2
Ex. 13. If (1 / 3.718) = 0.2689, Then find the value of (1 / 0.0003718).
Sol. (1 / 0.0003718 ) = ( 10000 / 3.718 ) = 10000 * (1 / 3.718) = 10000 * 0.2689 = 2689.
___ ______
Ex. 14. Express as vulgar fractions : (i) 0.37 (ii) 0.053 (iii) 3.142857
__ ___
Sol. (i) 0.37 = 37 / 99 . (ii) 0.053 = 53 / 999
______ ______
(iii) 3.142857 = 3 + 0.142857 = 3 + (142857 / 999999) = 3 (142857/999999)
_ __ _
Ex. 15. Express as vulgar fractions : (i) 0.17 (ii) 0.1254 (iii) 2.536
_
Sol. (i) 0.17 = (17 – 1)/90 = 16 / 90 = 8/ 45
__
(ii) 0.1254 = (1254 – 12 )/ 9900 = 1242 / 9900 = 69 / 550
(iii) 2.536 = 2 + 0.536 = 2 + (536 – 53)/900 = 2 + (483/900) = 2 + (161/300) = 2 (161/300)
Ex. 16. Simplify: 0.05 * 0.05 * 0.05 + 0.04 * 0.04 * 0.04
0.05 * 0.05 – 0.05 * 0.04 + 0.04 * 0.04
Sol. Given expression = (a3 + b3) / (a2 – ab + b2), where a = 0.05 , b = 0.04
= (a +b ) = (0.05 +0.04 ) =0.09 TheOnlineGK

HCF & LCM

2. H.C.F. AND L.C.M. OF NUMBERS
IMPORTANT FACTS AND FORMULA
I. Factors and Multiples : If a number a divides another number b exactly, we say that a
is a factor of b. In this case, b is called a multiple of a.
II. Highest Common Factor (H.C.F.) or Greatest Common Measure (G.C.M.) or
Greatest Common Divisor (G.C.D.): 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 of finding the H.C.F. of a given set of numbers :
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 H.C.F.
2. Division Method: Suppose we have to find the H.C.F. of two given numbers.
Divide the larger number by the smaller one. Now, divide the divisor by the
remainder. Repeat the process of dividing the preceding number by the remainder
last obtained till zero is obtained as remainder. The last divisor is the required
H.C.F.
Finding the H.C.F. of more than two numbers : Suppose we have to find the
H.C.F. of three numbers. Then, H.C.F. of [(H.C.F. of any two) and (the third
number)] gives the H.C.F. of three given numbers.
Similarly, the H.C.F. of more than three numbers may be obtained.
III. Least Common Multiple (L.C.M.) : The least number which is exactly divisible by
each one of the given numbers is called their L.C.M.
1. Factorization Method of Finding L.C.M.: Resolve each one of the given
numbers into a product of prime factors. Then, L.C.M. is the product of highest
powers of all the factors,
2. Common Division Method {Short-cut Method) of Finding L.C.M.:
Arrange the given numbers in a row in any order. Divide by a number which
divides exactly at least two of the given numbers and carry forward the numbers
which are not divisible. Repeat the above process till no two of the numbers are
divisible by the same number except 1. The product of the divisors and the
undivided numbers is the required L.C.M. of the given numbers,
IV. Product of two numbers =Product of their H.C.F. and L.C.M.
V. Co-primes: Two numbers are said to be co-primes if their H.C.F. is 1.
VI. H.C.F. and L.C.M. of Fractions:
1.H C F= H.C.F. of Numerators 2.L C M = L.C.M of Numerators__
L.C.M. of Denominators H.C.F. of Denominators
VII. H.C.F. and L.C.M. of Decimal Fractions: In given numbers, make the same
number of decimal places by annexing zeros in some numbers, if necessary. Considering
these numbers without decimal point, find H.C.F. or L.C.M. as the case may be. Now, in
the result, mark off as many decimal places as are there in each of the given numbers.
VIII. Comparison of Fractions: Find the L.C.M. of the denominators of the given
fractions. Convert each of the fractions into an equivalent fraction with L.C.M. as the
denominator, by multiplying both the numerator and denominator by the same number.
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The resultant fraction with the greatest numerator is the greatest.
SOLVED EXAMPLES
Ex. 1. Find the H.C.F. of 23 X 32 X 5 X 74, 22 X 35 X 52 X 73,23 X53 X72
Sol. The prime numbers common to given numbers are 2,5 and 7.
H.C.F. = 22 x 5 x72 = 980.
Ex. 2. Find the H.C.F. of 108, 288 and 360.
Sol. 108 = 22 x 33, 288 = 25 x 32 and 360 = 23 x 5 x 32.
H.C.F. = 22 x 32 = 36.
Ex. 3. Find the H.C.F. of 513, 1134 and 1215.
Sol.
1134 ) 1215 ( 1
1134
81 ) 1134 ( 14
81
324
324
x
H.C.F. of 1134 and 1215 is 81.
So, Required H.C.F. = H.C.F. of 513 and 81.
______
81 ) 513 ( 6
__486____
27) 81 ( 3
81
0
H.C.F. of given numbers = 27.
Ex. 4. Reduce 391 to lowest terms .
667
to lowest terms.
Sol. H.C.F. of 391 and 667 is 23.
On dividing the numerator and denominator by 23, we get :
391 = 391  23 = 17
667 667 23 29
Ex.5.Find the L.C.M. of 22 x 33 x 5 x 72 , 23 x 32 x 52 x 74 , 2 x 3 x 53 x 7 x 11.
Sol. L.C.M. = Product of highest powers of 2, 3, 5, 7 and 11 = 23 x 33 x 53 x 74 x 11
Ex.6. Find the L.C.M. of 72, 108 and 2100. TheOnlineGK
Sol. 72 = 23 x 32, 108 = 33 x 22, 2100 = 22 x 52 x 3 x 7.
L.C.M. = 23 x 33 x 52 x 7 = 37800.
Ex.7.Find the L.C.M. of 16, 24, 36 and 54.
Sol.
2 16 - 24 - 36 - 54
2 8 - 12 - 18 - 27
2 4 - 6 - 9 - 27
3 2 - 3 - 9 - 27
3 2 - 1 - 3 - 9
2 - 1 - 1 - 3
 L.C.M. = 2 x 2 x 2 x 3 x 3 x 2 x 3 = 432.
Ex. 8. Find the H.C.F. and L.C.M. of 2 , 8 , 16 and 10.
3 9 81 27
Sol. H.C.F. of given fractions = H.C.F. of 2,8,16,10 = 2_
L.C.M. of 3,9,81,27 81
L.C.M of given fractions = L.C.M. of 2,8,16,10 = 80_
H.C.F. of 3,9,81,27 3
Ex. 9. Find the H.C.F. and L.C.M. of 0.63, 1.05 and 2.1.
Sol. Making the same number of decimal places, the given numbers are 0.63, 1.05 and
2.10.
Without decimal places, these numbers are 63, 105 and 210.
Now, H.C.F. of 63, 105 and 210 is 21.
H.C.F. of 0.63, 1.05 and 2.1 is 0.21.
L.C.M. of 63, 105 and 210 is 630.
L.C.M. of 0.63, 1.05 and 2.1 is 6.30.
Ex. 10. Two numbers are in the ratio of 15:11. If their H.C.F. is 13, find the
numbers.
Sol. Let the required numbers be 15.x and llx.
Then, their H.C.F. is x. So, x = 13.
The numbers are (15 x 13 and 11 x 13) i.e., 195 and 143.
Ex. 11. TheH.C.F. of two numbers is 11 and their L.C.M. is 693. If one of the
numbers is 77,find the other.
Sol. Other number = 11 X 693 = 99
TheOnlineGK
77
Ex. 12. Find the greatest possible length which can be used to measure exactly the
lengths 4 m 95 cm, 9 m and 16 m 65 cm.
Sol. Required length = H.C.F. of 495 cm, 900 cm and 1665 cm.
495 = 32 x 5 x 11, 900 = 22 x 32 x 52, 1665 = 32 x 5 x 37.
H.C.F. = 32 x 5 = 45.
Hence, required length = 45 cm.
Ex. 13. Find the greatest number which on dividing 1657 and 2037 leaves
remainders 6 and 5 respectively.
Sol. Required number = H.C.F. of (1657 - 6) and (2037 - 5) = H.C.F. of 1651 and 2032
_______
1651 ) 2032 ( 1 1651
1651_______
381 ) 1651 ( 4
1524_________
127 ) 381 ( 3
381
0
Required number = 127.
Ex. 14. Find the largest number which divides 62, 132 and 237 to leave the same
remainder in each case.
Sol . Required number = H.C.F. of (132 - 62), (237 - 132) and (237 - 62)
= H.C.F. of 70, 105 and 175 = 35.
Ex.15.Find the least number exactly divisible by 12,15,20,27.
Sol.
3 12 - 15 - 20 - 27
4 4 - 5 - 20 - 9
5 1 - 5 - 5 - 9
1 - 1 - 1 - 9
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Ex.16.Find the least number which when divided by 6,7,8,9, and 12 leave the same
remainder 1 each case
Sol. Required number = (L.C.M OF 6,7,8,9,12) + 1
3 6 - 7 - 8 - 9 - 12
4 2 - 7 - 8 - 3 - 4
5 1 - 7 - 4 - 3 - 2
1 - 7 - 2 - 3 - 1
L.C.M = 3 X 2 X 2 X 7 X 2 X 3 = 504.
Hence required number = (504 +1) = 505.
Ex.17. Find the largest number of four digits exactly divisible by 12,15,18 and 27.
Sol. The Largest number of four digits is 9999.
Required number must be divisible by L.C.M. of 12,15,18,27 i.e. 540.
On dividing 9999 by 540,we get 279 as remainder .
Required number = (9999-279) = 9720.
Ex.18.Find the smallest number of five digits exactly divisible by 16,24,36 and 54.
Sol. Smallest number of five digits is 10000.
Required number must be divisible by L.C.M. of 16,24,36,54 i.e 432,
On dividing 10000 by 432,we get 64 as remainder.
Required number = 10000 +( 432 – 64 ) = 10368.
Ex.19.Find the least number which when divided by 20,25,35 and 40 leaves
remainders 14,19,29 and 34 respectively.
Sol. Here,(20-14) = 6,(25 – 19)=6,(35-29)=6 and (40-34)=6.
Required number = (L.C.M. of 20,25,35,40) – 6 =1394.
Ex.20.Find the least number which when divided by 5,6,7, and 8 leaves a remainder
3, but when divided by 9 leaves no remainder .
Sol. L.C.M. of 5,6,7,8 = 840.
 Required number is of the form 840k + 3
Least value of k for which (840k + 3) is divisible by 9 is k = 2.
Required number = (840 X 2 + 3)=1683
Ex.21.The traffic lights at three different road crossings change after every 48 sec.,
72 sec and 108 sec.respectively .If they all change simultaneously at 8:20:00
hours,then at what time they again change simultaneously .
Sol. Interval of change = (L.C.M of 48,72,108)sec.=432sec.
So, the lights will agin change simultaneously after every 432 seconds i.e,7
min.12sec
Hence , next simultaneous change will take place at 8:27:12 hrs. TheOnlineGK
Ex.22.Arrange the fractions 17 , 31, 43, 59 in the ascending order.
18 36 45 60
Sol.L.C.M. of 18,36,45 and 60 = 180.
Now, 17 = 17 X 10 = 170 ; 31 = 31 X 5 = 155 ;
18 18 X 10 180 36 36 X 5 180
43 = 43 X 4 = 172 ; 59 = 59 X 3 = 177 ;
45 45 X 4 180 60 60 X 3 180
Since, 155<170<172<177, so, 155 < 170 < 172 < 177
180 180 180 180
Hence, 31 < 17 < 43 < 59
36 18 45 60
TheOnlineGK

1-NUMBERS

1. NUMBERS
IMPORTANT FACTS AND FORMULA

1.Natural Numbers : Counting numbers 1, 2, 3, 4, 5,..... are called natural
numbers.
2.Whole Numbers : All counting numbers together with zero form the set of whole
numbers. Thus,
(i) 0 is the only whole number which is not a natural number.
(ii) Every natural number is a whole number.
3.Integers : All natural numbers, 0 and negatives of counting numbers i.e.,
{…, - 3 , - 2 , - 1 , 0, 1, 2, 3,…..} together form the set of integers.
(i) Positive Integers : {1, 2, 3, 4, …..} is the set of all positive integers.
(ii) Negative Integers : {- 1, - 2, - 3,…..} is the set of all negative integers.
(iii) Non-Positive and Non-Negative Integers : 0 is neither positive nor
negative. So, {0, 1, 2, 3,….} represents the set of non-negative integers, while
{0, - 1 , - 2 , - 3 ,…..} represents the set of non-positive integers.
4. Even Numbers : A number divisible by 2 is called an even number, e.g., 2, 4, 6, 8,
10, etc.
5. Odd Numbers : A number not divisible by 2 is called an odd number. e.g., 1, 3, 5, 7,
9, 11, etc.
6. Prime Numbers : A number greater than 1 is called a prime number, if it has exactly
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two factors, namely 1 and the number itself.
Prime numbers upto 100 are : 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43,
47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
Prime numbers Greater than 100 : Letp be a given number greater than 100. To find out
whether it is prime or not, we use the following method :
Find a whole number nearly greater than the square root of p. Let k > *jp. Test whether p
is divisible by any prime number less than k. If yes, then p is not prime. Otherwise, p is
prime.
e.g,,We have to find whether 191 is a prime number or not. Now, 14 > V191.
Prime numbers less than 14 are 2, 3, 5, 7, 11, 13.
191 is not divisible by any of them. So, 191 is a prime number.
7.Composite Numbers : Numbers greater than 1 which are not prime, are known as
composite numbers, e.g., 4, 6, 8, 9, 10, 12.
Note : (i) 1 is neither prime nor composite.
(ii) 2 is the only even number which is prime.
(iii) There are 25 prime numbers between 1 and 100.
8. Co-primes : Two numbers a and b are said to be co-primes, if their H.C.F. is 1. e.g.,
(2, 3), (4, 5), (7, 9), (8, 11), etc. are co-primes,
V.TESTS OF DIVISIBILITY
1. Divisibility By 2 : A number is divisible by 2, if its unit's digit is any of 0, 2, 4, 6, 8.
Ex. 84932 is divisible by 2, while 65935 is not.
2. Divisibility By 3 : A number is divisible by 3, if the sum of its digits is divisible by 3.
Ex.592482 is divisible by 3, since sum of its digits = (5 + 9 + 2 + 4 + 8 + 2) = 30, which
is divisible by 3.
But, 864329 is not divisible by 3, since sum of its digits =(8 + 6 + 4 + 3 + 2 + 9) = 32,
which is not divisible by 3.
3. Divisibility By 4 : A number is divisible by 4, if the number formed by the last two
digits is divisible by 4.
Ex. 892648 is divisible by 4, since the number formed by the last two digits is
48, which is divisible by 4.
But, 749282 is not divisible by 4, since the number formed by the last tv/o digits is 82,
which is not divisible by 4.
4. Divisibility By 5 : A number is divisible by 5, if its unit's digit is either 0 or 5. Thus,
20820 and 50345 are divisible by 5, while 30934 and 40946 are not.
5. Divisibility By 6 : A number is divisible by 6, if it is divisible by both 2 and 3. Ex.
The number 35256 is clearly divisible by 2.
Sum of its digits = (3 + 5 + 2 + 5 + 6) = 21, which is divisible by 3. Thus, 35256 is
divisible by 2 as well as 3. Hence, 35256 is divisible by 6.
6. Divisibility By 8 : A number is divisible by 8, if the number formed by the last
three digits of the given number is divisible by 8.
Ex. 953360 is divisible by 8, since the number formed by last three digits is 360, which is
divisible by 8.
But, 529418 is not divisible by 8, since the number formed by last three digits is 418,
which is not divisible by 8.
7. Divisibility By 9 : A number is divisible by 9, if the sum of its digits is divisible
by 9.TheOnlineGK
Ex. 60732 is divisible by 9, since sum of digits * (6 + 0 + 7 + 3 + 2) = 18, which is
divisible by 9.
But, 68956 is not divisible by 9, since sum of digits = (6 + 8 + 9 + 5 + 6) = 34, which is
not divisible by 9.
8. Divisibility By 10 : A number is divisible by 10, if it ends with 0.
Ex. 96410, 10480 are divisible by 10, while 96375 is not.
9. Divisibility By 11 : A number is divisible by 11, if the difference of the sum of its
digits at odd places and the sum of its digits at even places, is either 0 or a number
divisible by 11.
Ex. The number 4832718 is divisible by 11, since :
(sum of digits at odd places) - (sum of digits at even places)
- (8 + 7 + 3 + 4) - (1 + 2 + 8) = 11, which is divisible by 11.
10. Divisibility By 12 ; A number is divisible by 12, if it is divisible by both 4 and
3.
Ex. Consider the number 34632.
(i) The number formed by last two digits is 32, which is divisible by 4,
(ii) Sum of digits = (3 + 4 + 6 + 3 + 2) = 18, which is divisible by 3. Thus, 34632 is
divisible by 4 as well as 3. Hence, 34632 is divisible by 12.
11. Divisibility By 14 : A number is divisible by 14, if it is divisible by 2 as well as 7.
12. Divisibility By 15 : A number is divisible by 15, if it is divisible by both 3 and 5.
13. Divisibility By 16 : A number is divisible by 16, if the number formed by the last4
digits is divisible by 16.
Ex.7957536 is divisible by 16, since the number formed by the last four digits is 7536,
which is divisible by 16.
14. Divisibility By 24 : A given number is divisible by 24, if it is divisible by both3 and
8.
15. Divisibility By 40 : A given number is divisible by 40, if it is divisible by both
5 and 8.
16. Divisibility By 80 : A given number is divisible by 80, if it is divisible by both 5 and
16.
Note : If a number is divisible by p as well as q, where p and q are co-primes, then the
given number is divisible by pq.
If p arid q are not co-primes, then the given number need not be divisible by pq,
even when it is divisible by both p and q.
Ex. 36 is divisible by both 4 and 6, but it is not divisible by (4x6) = 24, since
4 and 6 are not co-primes.
VI MULTIPLICATION BY SHORT CUT METHODS
1. Multiplication By Distributive Law :
(i) a x (b + c) = a x b + a x c (ii) ax(b-c) = a x b-a x c.
Ex. (i) 567958 x 99999 = 567958 x (100000 - 1)
= 567958 x 100000 - 567958 x 1 = (56795800000 - 567958) = 56795232042. (ii) 978 x
TheOnlineGK
184 + 978 x 816 = 978 x (184 + 816) = 978 x 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
Ex. 975436 x 625 = 975436 x 54= 9754360000 = 609647600
16
VII. BASIC FORMULAE
1. (a + b)2 = a2 + b2 + 2ab 2. (a - b)2 = a2 + b2 - 2ab
3. (a + b)2 - (a - b)2 = 4ab 4. (a + b)2 + (a - b)2 = 2 (a2 + b2)
5. (a2 - b2) = (a + b) (a - b)
6. (a + b + c)2 = a2 + b2 + c2 + 2 (ab + bc + ca)
7. (a3 + b3) = (a +b) (a2 - ab + b2) 8. (a3 - b3) = (a - b) (a2 + ab + b2)
9. (a3 + b3 + c3 -3abc) = (a + b + c) (a2 + b2 + c2 - ab - bc - ca)
10. If a + b + c = 0, then a3 + b3 + c3 = 3abc.
VIII. DIVISION ALGORITHM OR EUCLIDEAN ALGORITHM
If we divide a given number by another number, then :
Dividend = (Divisor x Quotient) + Remainder
IX. {i) (xn - an ) is divisible by (x - a) for all values of n.
(ii) (xn - an) is divisible by (x + a) for all even values of n.
(iii) (xn + an) is divisible by (x + a) for all odd values of n.
X. PROGRESSION
A succession of numbers formed and arranged in a definite order according to certain
definite rule, is called a progression.
1. Arithmetic Progression (A.P.) : If each term of a progression differs from its
preceding term by a constant, then such a progression is called an arithmetical
progression. This constant difference is called the common difference of the A.P.
An A.P. with first term a and common difference d is given by a, (a + d), (a + 2d),(a +
3d),.....
The nth term of this A.P. is given by Tn =a (n - 1) d.
The sum of n terms of this A.P.
Sn = n/2 [2a + (n - 1) d] = n/2 (first term + last term).
SOME IMPORTANT RESULTS :
(i) (1 + 2 + 3 +…. + n) =n(n+1)/2
(ii) (l2 + 22 + 32 + ... + n2) = n (n+1)(2n+1)/6
(iii) (13 + 23 + 33 + ... + n3) =n2(n+1)2
2. Geometrical Progression (G.P.) : A progression of numbers in which every term
bears a constant ratio with its preceding term, is called a geometrical progression.
The constant ratio is called the common ratio of the G.P. A G.P. with first term a and
common ratio r is :
a, ar, ar2,
In this G.P. Tn = arn-1
TheOnlineGK
sum of the n terms, Sn= a(1-rn)
(1-r)
SOLVED EXAMPLES
Ex. 1. Simplify : (i) 8888 + 888 + 88 + 8
(ii) 11992 - 7823 - 456
Sol. i ) 8888 ii) 11992 - 7823 - 456 = 11992 - (7823 + 456)
888 = 11992 - 8279 = 3713-
88 7823 11992
+ 8 + 456 - 8279
9872 8279 3713
Ex. 2, What value will replace the question mark in each of the following equations
?
(i) ? - 1936248 = 1635773 (ii) 8597 - ? = 7429 - 4358
Sol. (i) Let x - 1936248=1635773.Then, x = 1635773 + 1936248=3572021.
(ii) Let 8597 - x = 7429 - 4358.
Then, x = (8597 + 4358) - 7429 = 12955 - 7429 = 5526.
Ex. 3. What could be the maximum value of Q in the following equation? 5P9
+ 3R7 + 2Q8 = 1114
Sol. We may analyse the given equation as shown : 1 2
Clearly, 2 + P + R + Q = ll. 5 P 9
So, the maximum value of Q can be 3 R 7
(11 - 2) i.e., 9 (when P = 0, R = 0); 2 Q 8
11 1 4
Ex. 4. Simplify : (i) 5793405 x 9999 (ii) 839478 x 625
Sol.
i)5793405x9999=5793405(10000-1)=57934050000-5793405=57928256595.b
ii) 839478 x 625 = 839478 x 54 = 8394780000 = 524673750.
16
Ex. 5. Evaluate : (i) 986 x 237 + 986 x 863 (ii) 983 x 207 - 983 x 107
Sol.
(i) 986 x 137 + 986 x 863 = 986 x (137 + 863) = 986 x 1000 = 986000.
(ii) 983 x 207 - 983 x 107 = 983 x (207 - 107) = 983 x 100 = 98300.
Ex. 6. Simplify : (i) 1605 x 1605 ii) 1398 x 1398
Sol.TheOnlineGK
i) 1605 x 1605 = (1605)2 = (1600 + 5)2 = (1600)2 + (5)2 + 2 x 1600 x 5
= 2560000 + 25 + 16000 = 2576025.
(ii) 1398 x 1398 - (1398)2 = (1400 - 2)2= (1400)2 + (2)2 - 2 x 1400 x 2
=1960000 + 4 - 5600 = 1954404.
Ex. 7. Evaluate : (313 x 313 + 287 x 287).
Sol.
(a2 + b2) = 1/2 [(a + b)2 + (a- b)2]
(313)2 + (287)2 = 1/2 [(313 + 287)2 + (313 - 287)2] = ½[(600)2 + (26)2]
= 1/2 (360000 + 676) = 180338.
Ex. 8. Which of the following are prime numbers ?
(i) 241 (ii) 337 (Hi) 391 (iv) 571
Sol.
(i) Clearly, 16 > Ö241. Prime numbers less than 16 are 2, 3, 5, 7, 11, 13.
241 is not divisible by any one of them.
241 is a prime number.
(ii) Clearly, 19>Ö337. Prime numbers less than 19 are 2, 3, 5, 7, 11,13,17.
337 is not divisible by any one of them.
337 is a prime number.
(iii) Clearly, 20 > Ö39l". Prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17, 19.
We find that 391 is divisible by 17.
391 is not prime.
(iv) Clearly, 24 > Ö57T. Prime numbers less than 24 are 2, 3, 5, 7, 11, 13, 17, 19, 23.
571 is not divisible by any one of them.
571 is a prime number.
Ex. 9. Find the unit's digit in the product (2467)163 x (341)72.
Sol. Clearly, unit's digit in the given product = unit's digit in 7153 x 172.
Now, 74 gives unit digit 1.
7152 gives unit digit 1,
 7153 gives unit digit (l x 7) = 7. Also, 172 gives unit digit 1.
Hence, unit's digit in the product = (7 x 1) = 7.
Ex. 10. Find the unit's digit in (264)102 + (264)103
Sol. Required unit's digit = unit's digit in (4)102 + (4)103.
Now, 42 gives unit digit 6.
(4)102 gives unjt digit 6.
(4)103 gives unit digit of the product (6 x 4) i.e., 4.
Hence, unit's digit in (264)m + (264)103 = unit's digit in (6 + 4) = 0.
Ex. 11. Find the total number of prime factors in the expression (4)11 x (7)5 x (11)2.
TheOnlineGK
Sol. (4)11x (7)5 x (11)2 = (2 x 2)11 x (7)5 x (11)2 = 211 x 211 x75x 112 = 222 x 75 x112
Total number of prime factors = (22 + 5 + 2) = 29.
Ex.12. Simplify : (i) 896 x 896 - 204 x 204
(ii) 387 x 387 + 114 x 114 + 2 x 387 x 114
(iii) 81 X 81 + 68 X 68-2 x 81 X 68.
Sol.
(i) Given exp = (896)2 - (204)2 = (896 + 204) (896 - 204) = 1100 x 692 = 761200.
(ii) Given exp = (387)2+ (114)2+ (2 x 387x 114)
= a2 + b2 + 2ab, where a = 387,b=114
= (a+b)2 = (387 + 114 )2 = (501)2 = 251001.
(iii) Given exp = (81)2 + (68)2 – 2x 81 x 68 = a2 + b2 – 2ab,Where a =81,b=68
= (a-b)2 = (81 –68)2 = (13)2 = 169.
Ex.13. Which of the following numbers is divisible by 3 ?
(i) 541326 (ii) 5967013
Sol.
(i) Sum of digits in 541326 = (5 + 4 + 1 + 3 + 2 + 6) = 21, which is divisible by 3.
Hence, 541326 is divisible by 3.
(ii) Sum of digits in 5967013 =(5+9 + 6 + 7 + 0+1 +3) = 31, which is not divisible by 3.
Hence, 5967013 is not divisible by 3.
Ex.14.What least value must be assigned to * so that the number 197*5462 is r 9 ?
Sol.
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 .
Hence, the digit in place of * must be 2.
Ex. 15. Which of the following numbers is divisible by 4 ?
(i) 67920594 (ii) 618703572
Sol.
(i) The number formed by the last two digits in the given number is 94, which is not
divisible by 4.
Hence, 67920594 is not divisible by 4.
(ii) The number formed by the last two digits in the given number is 72, which is
divisible by 4.
Hence, 618703572 is divisible by 4. TheOnlineGK
Ex. 16. Which digits should come in place of * and $ if the number 62684*$ is
divisible by both 8 and 5 ?
Sol.
Since the given number is divisible by 5, so 0 or 5 must come in place of $. But, a
number ending with 5 is never divisible by 8. So, 0 will replace $.
Now, the number formed by the last three digits is 4*0, which becomes divisible by 8, if
* is replaced by 4.
Hence, digits in place of * and $ are 4 and 0 respectively.
Ex. 17. Show that 4832718 is divisible by 11.
Sol. (Sum of digits at odd places) - (Sum of digits at even places)
= (8 + 7 + 3 + 4) - (1 + 2 + 8) = 11, which is divisible by 11.
Hence, 4832718 is divisible by 11.
Ex. 18. Is 52563744 divisible by 24 ?
Sol. 24 = 3 x 8, where 3 and 8 are co-primes.
The sum of the digits in the given number is 36, which is divisible by 3. So, the
given number is divisible by 3.
The number formed by the last 3 digits of the given number is 744, which is
divisible by 8. So, the given number is divisible by 8.
Thus, the given number is divisible by both 3 and 8, where 3 and 8 are co-primes.
So, it is divisible by 3 x 8, i.e., 24.
Ex. 19. What least number must be added to 3000 to obtain a number exactly
divisible by 19 ?
Sol. On dividing 3000 by 19, we get 17 as remainder.
Number to be added = (19 - 17) = 2.
Ex. 20. What least number must be subtracted from 2000 to get a number exactly
divisible by 17 ?
Sol. On dividing 2000 by 17, we get 11 as remainder.
Required number to be subtracted = 11.
Ex. 21. Find the number which is nearest to 3105 and is exactly divisible by 21.
Sol. On dividing 3105 by 21, we get 18 as remainder.
Number to be added to 3105 = (21 - 18) - 3.
Hence, required number = 3105 + 3 = 3108. TheOnlineGK
Ex. 22. Find the smallest number of 6 digits which is exactly divisible by 111.
Sol. Smallest number of 6 digits is 100000.
On dividing 100000 by 111, we get 100 as remainder.
Number to be added = (111 - 100) - 11.
Hence, required number = 100011.-
Ex. 23. On dividing 15968 by a certain number, the quotient is 89 and the remainder
is 37. Find the divisor.
Dividend - Remainder 15968-37
Sol. Divisor = -------------------------- = ------------- = 179.
.Quotient 89
Ex. 24. A number when divided by 342 gives a remainder 47. When the same
number ift divided by 19, what would be the remainder ?
Sol. On dividing the given number by 342, let k be the quotient and 47 as remainder.
Then, number – 342k + 47 = (19 x 18k + 19 x 2 + 9) = 19 (18k + 2) + 9.
The given number when divided by 19, gives (18k + 2) as quotient and 9 as
remainder.
Ex. 25. A number being successively divided by 3, 5 and 8 leaves remainders 1, 4
and 7 respectively. Find the respective remainders if the order of divisors be
reversed,
Sol.
3 X
5 y - 1
8 z - 4
1 - 7
z = (8 x 1 + 7) = 15; y = {5z + 4) = (5 x 15 + 4) = 79; x = (3y + 1) = (3 x 79 + 1) = 238.
Now,
8 238
5 29 - 6
3 5 - 4
1 - 9,
Respective remainders are 6, 4, 2.
Ex. 26. Find the remainder when 231 is divided by 5.
Sol. 210 = 1024. Unit digit of 210 x 210 x 210 is 4 [as 4 x 4 x 4 gives unit digit 4].
Unit digit of 231 is 8.
TheOnlineGK
Now, 8 when divided by 5, gives 3 as remainder.
Hence, 231 when divided by 5, gives 3 as remainder.
Ex. 27. How many numbers between 11 and 90 are divisible by 7 ?
Sol. The required numbers are 14, 21, 28, 35, .... 77, 84.
This is an A.P. with a = 14 and d = (21 - 14) = 7.
Let it contain n terms.
Then, Tn = 84 => a + (n - 1) d = 84
=> 14 + (n - 1) x 7 = 84 or n = 11.
Required number of terms = 11.
Ex. 28. Find the sum of all odd numbers upto 100.
Sol. The given numbers are 1, 3, 5, 7, ..., 99.
This is an A.P. with a = 1 and d = 2.
Let it contain n terms. Then,
1 + (n - 1) x 2 = 99 or n = 50.
Required sum = n (first term + last term)
2
= 50 (1 + 99) = 2500.
2
Ex. 29. Find the sum of all 2 digit numbers divisible by 3.
Sol. All 2 digit numbers divisible by 3 are :
12, 51, 18, 21, ..., 99.
This is an A.P. with a = 12 and d = 3.
Let it contain n terms. Then,
12 + (n - 1) x 3 = 99 or n = 30.
Required sum = 30 x (12+99) = 1665.
2
Ex.30.How many terms are there in 2,4,8,16……1024?
Sol.Clearly 2,4,8,16……..1024 form a GP. With a=2 and r = 4/2 =2.
Let the number of terms be n . Then
2 x 2n-1 =1024 or 2n-1 =512 = 29.
n-1=9 or n=10.
Ex. 31. 2 + 22 + 23 + ... + 28 = ?
Sol. Given series is a G.P. with a = 2, r = 2 and n = 8.
sum = a(rn-1) = 2 x (28 –1) = (2 x 255) =510
(r-1) (2-1) TheOnlineGK