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LIVE NECO 2017/2018 QUESTIONS AND ANSWERS

Discussion in 'Waec, Weac, Neco, Waec GCE And Neco GCE' started by Yoi.shina, Apr 3, 2017.

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    Steps To Re-Print JAMB 2017 Exam Slip
    It is very easy to do that. Just follow the simple instructions below:
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    NECO 2017 SPECIMEN.


    speciment A-frog or toad
    speciment B-thoracic vertabra
    speciment C-lumbar vertabra
    speciment D-flamboyant flower of forest
    speciment E-millipedi
    speciment F-agama lizard
    specimet G-food of domastic fowl
    speciment H-food of domastic duck
    speciment I-tomato fruit
    speciment J-coconut fruit (with it busk)
    speciment K-dry bean pod or dry pride barbados pod or dry flamboyant pod
    speciment L-combretum fruit
    speciment M-adult burterfly
    speciment N-prawn
    speciment O-earthworm
    speciment P-tapeworm

    More are coming.....
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    PHYSICS PRACTICAL ANSWERS WILL BE AVAILABLE SOON....

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    PHYS

    VERIFIED PHY ANSWERS


    3ai)
    tabulate
    s/n1,2,3,4,5,6
    R(pie)6pie,5pie,4pie,3pie,2pie,1pie
    V(v) 2.20 2.15 2.05 1.85 1.70 1.45
    P(v) 0.76,0.81,0.91,1.11,1.26,1.51
    I(A) 0.37,0.43,0.51,0.62,0.85,1.45
    3aii)
    E=2.96v

    3aviii)
    -I ensured tight connections
    -I avoided error due to parallax when reading the voltmetre
    -I ensured the key is opened after each reading to prevent the cell from running down
    3bi)
    The potential difference between two points in an electric circuit is the work done when a coulomb of charge passes between the points in an electric circuits
    3bii)
    E=I(R+r)
    where E=e.m.f of the cell
    Therefore E=IR+Ir=V+Ir
    Where :
    V=IR=P.d lost across the external resistance of the cell
    Ir=Pd lost across the internal resistance of the cell
    E=e.m.f=workdone across external resistance of the cell
    Since V=IR is the workdone across the external resistance R only can be other than V


    2bi)
    he characteristics of imaged formed are :
    i)It is virtual
    ii) It is enlarged or magnified i.e. twice or two times bigger as the object(m=2)

    2bII) The concave mirror ,mounted in its holder ,is moved to and fro in front of the until a sharp image of the cross wire of the ray box is formed on the screen adjacent to the object .The distance between the mirror and the screen was measured as 30.1cm.Since the radius of the curvature r,=2fo ,then half this distance is the focal length of the mirror fo .Thus focal length was determined to be 15.05cm approximately 15cm
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    IMG-20170605-WA0007.jpg


    2ai)
    fo=15cm
    2av)
    a=60.00cm
    b=20.00cm
    Hence L=a/b=60.00/20.00
    L=3
    2avi)
    TABULATE
    S/N:1,2,3,4,5
    b(cm):20.00,25.00,35.00,40.00
    a(cm):60.00,37.50,30.00,26.25,24.00
    L=a/b:3.00,1.50,1.00,0.75,0.60
    2avii)
    Slope=Change in L/Change in a
    =(3-0.25)/(60-18.6)
    =2.75/41.4
    =0.006642
    2aviii)
    S^-1=1/S
    =(1/0.0066425cm)
    S=15.05
    S=15cm
    2aix)
    PRECAUTIONS
    -I ensured that all apparatus are in straight line
    -I avoided error due to parallax when reading the metre rule
    -I avoided zero error on the metre rule
    2bi)
    u=10cm
    f=15cm
    Using 1/v+1/u=1/f
    1/f-1/u=1/v
    1/v=1/15-1/10
    1/v=(2-3)/30
    1/v=-1/30
    v=-30cm
    The characteristics of image formed are:
    -It is virtual
    -It is enlarged and magnified ie twice or two times as big as the object m=2
    2bii)
    The concave mirror mounted in its holder is moved to and fro in front of the screen until a sharp image of the cross wire of the ray box is formed on the screen adjacent of the object.The distance between the mirror and the screen was measured as 30cm since the radius of curvature r=2fo then half is distance


    3ix)
    slope=DI/DP=1.2-0.4/0.7-0.27=0.8/0.43=1.86Av^1
    slope=1.86pie
    3x)
    K=5^-1
    k=1/slope = 1/1.86 = 0.538pie^-1

    3xi)
    i)i ensured tight connections
    ii)i ensured clean terminals
    3b)
    Ohm's lawstates that the currentthrough a conductorbetween two points is directly proportionalto the voltageacross the two points.
    3bii)
    I=Total EmF/R+r
    I=3E/R+3r


    1)
    s/n 1,2,3,4,5
    d(cm) 90.0,80.0,70.0,60.0,50.0
    t(s) 31.80,30.70,30.00,29.00,28.00
    T=t/20(s) 1.590,1.540,1.500,1.450,1.400
    d^-1(cm^-) 0.011,0.013,0.014,0.017,0.020
    slope DT/d^-1
    =1.590-1.400/0.020-0.011=0.19/0.009
    =21.1cms
    precautions
    i)i ensured that the ruler is moving in an oscillatory manner and not rotational
    ii)i ensured that the experiment was carried out in air and tight region for accuracy
    1bi)
    d=70.0cm
    1bii)
    the metre rule will be in equilibrium when the distance at each ends are equal to each other

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    VERY WRONG
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    Data processing

    (3a)
    software that handles the storage, retrieval, and updating of data in a computer system.

    3aii)
    MySQL
    Microsoft Access


    3aiii)
    A file is a collection of records which have common properties. Each file has its own file reference which is unique. The file reference indicates the subject or contexts of the records.
    While
    A record can be an image, text based or in electronic or physical format.

    3bi) Normalization is a process of organizing the data in database to avoid data redundancy, insertion anomaly , update anomaly & deletion anomaly

    3Bii)
    *Insertion Anomalies:Insertion anomalies occur when we
    try to insert data into a flawed table.
    *Deletion Anomalies:- Deletion anomalies occur when we
    delete data from a flawed schema.
    *Update Anomalies:- Update anomalies occur when we
    change data in a flawed schema.


    3c.
    Pick tool
    Shape tool
    Zoom tool
    Freehand tool


    4b) log files - A log file is a recording of everything that goes in and out of a particular server

    Firewall- A firewall is a system designed to prevent unauthorized access to or from a private network.

    Encryption - Encryption is the conversion of electronic data into another form, called ciphertext, which cannot be easily understood by anyone except authorized parties.

    Backup- backup refers to the copying and archiving of computer data so it may be used to restore the original after a data loss event.

    4ci)
    Primary key uniquely identify a record in the table. While,
    Foreign key is a field in the table that is primary key in another table.
    Primary Key can't accept null values. While,
    Foreign key can accept multiple null value.

    4cii) Vision problems
    Headache
    Obesity


    5a)
    i)Transmission is the act of transferring something from one spot to another, like a radio or TV broadcast, or a disease going from one person to another.



    5ai)
    Information Transmission can be defined as the process of sending, propagating and receiving of information from one person to another in the same remote or different location
    5aii)
    i)Tasks can be completed faster because computers work at amazing speed.
    ii)Computers can process large amounts of data and generate error-free results, provided that the data is entered correctly.
    5bi)
    hierarchical database modelis a data modelin which the data is organized into a tree-like structure. The data is stored asrecordswhich are connected to one another throughlinks. A record is a collection of fields, with each field containing only one value. Theentity typeof a record defines which fields the record contains.
    ii)network modelis a database modelconceived as a flexible way of representing objects and their relationships. Its distinguishing feature is that the schema, viewed as a graph in which object types are nodes and relationship types are arcs, is not restricted to being a hierarchy or lattice
    5bii)
    i)Data Dictionary Management is where the DBMS stores definitions of the data elements and their relationships (metadata). The DBMS uses this function to look up the required data component structures and relationships.
    ii)Data Storage Management
    This particular function is used for the storage of data and any related data entry forms or screen definitions, report definitions, data validation rules, procedural code, and structures that can handle video and picture formats.
    iii)Data Transformation and Presentation This function exists to transform any data entered into required data structures. By using the data transformation and presentation function the DBMS can determine the difference between logical and physical data formats.
    iv)Database Communication Interfaces
    This refers to how a DBMS can accept different end user requests through different network environments. An example of this can be easily related to the internet.
    ================================
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    Data processing

    (3a)
    software that handles the storage, retrieval, and updating of data in a computer system.

    3aii)
    MySQL
    Microsoft Access


    3aiii)
    A file is a collection of records which have common properties. Each file has its own file reference which is unique. The file reference indicates the subject or contexts of the records.
    While
    A record can be an image, text based or in electronic or physical format.

    3bi) Normalization is a process of organizing the data in database to avoid data redundancy, insertion anomaly , update anomaly & deletion anomaly

    3Bii)
    *Insertion Anomalies:Insertion anomalies occur when we
    try to insert data into a flawed table.
    *Deletion Anomalies:- Deletion anomalies occur when we
    delete data from a flawed schema.
    *Update Anomalies:- Update anomalies occur when we
    change data in a flawed schema.


    3c.
    Pick tool
    Shape tool
    Zoom tool
    Freehand tool


    4b) log files - A log file is a recording of everything that goes in and out of a particular server

    Firewall- A firewall is a system designed to prevent unauthorized access to or from a private network.

    Encryption - Encryption is the conversion of electronic data into another form, called ciphertext, which cannot be easily understood by anyone except authorized parties.

    Backup- backup refers to the copying and archiving of computer data so it may be used to restore the original after a data loss event.

    4ci)
    Primary key uniquely identify a record in the table. While,
    Foreign key is a field in the table that is primary key in another table.
    Primary Key can't accept null values. While,
    Foreign key can accept multiple null value.

    4cii) Vision problems
    Headache
    Obesity


    5a)
    i)Transmission is the act of transferring something from one spot to another, like a radio or TV broadcast, or a disease going from one person to another.



    5ai)
    Information Transmission can be defined as the process of sending, propagating and receiving of information from one person to another in the same remote or different location
    5aii)
    i)Tasks can be completed faster because computers work at amazing speed.
    ii)Computers can process large amounts of data and generate error-free results, provided that the data is entered correctly.
    5bi)
    hierarchical database modelis a data modelin which the data is organized into a tree-like structure. The data is stored asrecordswhich are connected to one another throughlinks. A record is a collection of fields, with each field containing only one value. Theentity typeof a record defines which fields the record contains.
    ii)network modelis a database modelconceived as a flexible way of representing objects and their relationships. Its distinguishing feature is that the schema, viewed as a graph in which object types are nodes and relationship types are arcs, is not restricted to being a hierarchy or lattice
    5bii)
    i)Data Dictionary Management is where the DBMS stores definitions of the data elements and their relationships (metadata). The DBMS uses this function to look up the required data component structures and relationships.
    ii)Data Storage Management
    This particular function is used for the storage of data and any related data entry forms or screen definitions, report definitions, data validation rules, procedural code, and structures that can handle video and picture formats.
    iii)Data Transformation and Presentation This function exists to transform any data entered into required data structures. By using the data transformation and presentation function the DBMS can determine the difference between logical and physical data formats.
    iv)Database Communication Interfaces
    This refers to how a DBMS can accept different end user requests through different network environments. An example of this can be easily related to the internet.
    ================================
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    ANIMAL -PRACT - Answers
    1a )
    Specimen A is a udder
    Specimen B is a fresh milk
    Specimen C is a fowl egg

    1b)
    SPECIMEN A= Udder . .. .
    The udder is a large gland which is surrounded with skin it is attached to the body by ligament. .. .. .
    WHILE
    SPECIMEN B = fresh milk
    Is a milk from animay ( sheep . cattle ) that has not been pasteurized . protein can be obtained from it.

    1c )
    Pregnancy

    1d)
    - It provides nutrients to the newborn baby
    - It provides antibodies which prevent newborn babies from infection

    1e )
    Cheese ,evaporated milk , yoghourt and ice cream
    Animal Husbandry

    1f )
    i ) By placing in the ray of light
    ii) After 14days of incubation
    iii ) 18 days

    1g)
    Total eggs = 24000
    total removal= 315 +410 +140 = 865
    % hatchability = ( 24000 - 865 ) / 24000 * 100 %
    = 96 .4%
    ===============

    2a )
    SPECIMEN D= Animal bone
    SPECIMEN E= Animal blood
    SPECIMEN F = Fresh fish

    2b)
    SPECIMEN D= it should be burnt to ashes
    SPECIMEN E= Salt should be added to prevent it
    SPECIMEN F = it should be smoked

    2c )
    SPECIMEN D= Calcium
    SPECIMEN E= Mineral
    SPECIMEN F = protein

    2d)
    SPECIMEN E
    - after collection from the animals it will be boiled and salt is added as additive
    It will be implanted into solid pellets

    SPECIMEN F
    - The fish is smoked on fire and dried to a reasonable extent .They are later crushed and mixed into required form

    2f )
    - Limestone
    ===============

    3a )
    G - tapeworm
    H - roundworm
    I - animal tail

    3b)
    Cattle
    Pig

    3c )
    G- stomach
    H -alimentary canal

    3d)
    ( tabulate )
    G- it is segment
    - flat body
    H -not segmented
    Round body

    3e )
    - Through regular environment sanitation
    - through deworming

    3f )
    - it brings about discomfort to farm animals
    - it causes general body weakness

    3g)
    Cattle, sheep and goat

    3h )
    - It reduces the harbour of vectors
    - it reduces cannibalism
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    STAY TUNE FOR UR CHEM
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    Chemistry-Pract-Answers
    1a)
    tabulate
    Volumeofpipette/baseused=25.0cm3
    Indicator used-Phenolphthalein
    Burette
    Readings

    Rough- First,Second,Third
    FinalBurette Reading (cm3) |16.00|
    |31.70| 15.65| 31.25|
    InitialBurette Reading (cm3)- |0.00|
    |16.00| |0.00| |15.65|
    Volumeof acid used (cm3) |16.00|
    |15.70| |15.65| |15.60|
    Average titre value
    =15.70+15.60+15.65cm^3/3=15.65cm3
    1bi)
    Concentration of B in moldm^-3
    H2C2O4(aq)+2XOH(aq)→ X2C2O4(aq)+2H2O(i) CAVA=nA CBVB=nB
    0.08×15.65=1
    CB×25/2
    CB=2×0.08×15.65/25 =0.100 moldm^-3
    1bii)
    Molar mass of XOH
    Concentration of B in gdm^-3
    =1.0×1000/250=4gdm^-3
    Therefore,molar mass of XOH
    =4gdm^-3
    0.100moldm^-3
    =40gmol^-1
    1biii)
    Relative atomic mass of X
    XOH=40
    X +16+1=40
    X +17=40
    X = 40-17
    X=23
    ================================
    2ai)
    Tabulate
    -Test-
    Xn+10cm3of H2O and stirred
    -Observation-
    Xn dissolves to form a colourless solution
    -Inference-
    Xn is a soluble salt
    2aii)
    -Test-
    First portion of Xn+ NaOH(aq) in drops,then
    in excess
    -Observation-
    White precipitate which dissolves in excess to give a colourless solution
    -Inference-
    Zn2+,Pb2+,orAl3+ present
    2aii)
    -Test-
    Second portion of Xn+NH3(aq) in drops,
    then in excess
    -Observation-
    White precipitate in soluble in excess
    -Inference-
    Pb2+orAl3+ present
    2aiii)
    -Test-
    Third portion of Xn+dil.HCl
    -Observation-
    White crystalline/chalky precipitate
    -Inference-
    Pb2+confirmed
    2bi)
    -Test-
    Yn + litmus
    -Observation-
    Red litmus paper turned blue
    -Inference-
    Yn is alkaline
    2bii)
    -Test-
    Portion of Yn+conc. HNO3
    -Observation-
    Yellow colouration
    -Inferrence-
    Protein present
    ================================
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    1ai)
    A=Toad
    B=Thoracic Vertebra
    C=Lumbar vertebra
    D= Pride of Barbados
    E=Millipede
    F=Agama Lizard
    1aii)
    DRAW literal view of Thoracic Vertebra (check image in pdf)
    1aiii)
    B===>located in chest, attached to the ribs
    C===>located in loin
    1aiv)
    Tabulate
    SPECIMEN B
    i)They are 12 in number
    ii) they are large
    iii) They are rigid
    iv)less vulnerable to degenerative conditions
    v) less massive

    2ai)
    G- foot of domestic fowl
    H-foot of domestic duck
    I-tomatoe fruit
    J-coconut fruit
    K-dry bean pod
    L-combretion fruit
    2aii)
    i-they both have claws
    ii-they both have three toes in front and one at the back
    iii-they are both used for walking
    2aiii)
    Spc- H is webbed while sec-G is not
    2iv)
    i-for scratching
    ii-for walking/running i.e locomotion
    2v)
    i-prominent hind limb for running away from treats
    ii-prominent claws for scratching the ground in search of food

    2bi)
    I- simple sacculent fruit - berry
    J-simple sacculent fruit - drupe
    K-simple dry dehicent fruit- legume
    L-simple dry dehiscent fruit.


    2bii) Tabulate I and J

    spc-I
    i-multi-seeded
    ii-endocarp is juicy
    iii-mesocarp is juicy
    iv-several carpel are founded
    v-several ovaries

    spc-J
    i-one seeded
    ii-endocarp is hard
    iii-
    iv-monocarpel
    v-superior ovary.


    3i)
    M-Adult butterfly
    N-prawn
    O-earthworm
    P-tapeworm
    3ii)
    M-the class to which it belongs is INSECTA

    3iii)
    OBSERVATION Feature of M:

    i. A pair of clubbed antenae
    Ii. A head,thorax and abdomen
    Iii.wings.
    3iv)
    OBSERVABLE feature of N:
    i.antenea
    Ii.pincer
    iii.abdomen
    Iv.swimmerets
    V.walking legs
    3v)
    M- pollination of flowers
    N-serve as food to man.


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    1)
    tabulate
    Number-
    5932
    *6141
    num
    3679
    *3113
    den
    Log
    3.7732
    3.7882
    7.5612
    3.5657
    3.4931
    7.0588
    anti log
    7.5612
    7.0588/0.5024/3=0.1675
    anti log
    =1471*10
    ================================

    5)
    tabulate
    Mark- 21-25,26-30,31-35,36-40,41-45,46-50
    F- 6,8,12,4,6,4=40
    X- 23,28,33,38,43,48
    FX- 138,224,396,152,258,192=1360
    Class Boundaries- 20.5-20.5, 25.5-30.5,35.5-40.50,40.5-45.5
    5i)
    MEAN
    (X)=EFX/EF=1360/40=34
    MODE
    L1+[Fm-Fa/2fm-fa-fb]c
    =30.5+[12-8/2(12)-8-4]5
    =30.5+[4/12]5
    =30.5+1.67
    =32.17

    11a)
    Tabulate
    X- |-1| |0| |1| |2| |3| |4|
    6 |6| |6| |6| |6| |6| |6|
    +X |-1| |0| |1| |2| |3| |4|
    -X^2 |-1| |0| |-1| |-4| |-9| |16|
    Y- |4| |6| |6| |4| |0| |-6|
    11b)
    CLICK HERE FOR THE GRAPH
    11ci)
    gradient at x/x
    dy/dx =B-A/B-C
    =7-4/2-1=3/1=3
    11cii)
    roots of equation from the graph
    x=-2.2 and x=3
    11ciii)
    minimum value of y
    =6.4.


    4i)
    difference in latitude=36degree + 36 degree =72degree
    distance travelled=@/360*2pir
    =72/360*2*22/77*64
    =20275200/2520
    =8045.71
    =8050km(35.f)

    4ii)
    given that speed =800km/hr
    distance travelled =804571km
    time=distance travelled/speed
    =8045.71/800
    =10.057
    =10hrs(to the nearest hour)

    2)
    a = 3, b = 20, C = -7
    Sum is
    (Α/1 + 1/Β + Β/1 + 1/ Α) Note Α means Alpha
    Α^2 Β + Β^2 + Α + Β / Α Β
    = -7/3 (-20/3) + (-20/3) = 140/9 - 20/3 / -7/3 (Note / means
    Divided)
    = 80/9 x -3/7 = -80/21
    Product:
    -7/3 + 2 + 1/-7/3 = -7/3 + 2/1 - 3/7
    = -49+42-9/21
    = -16/21
    x^2 - (sum)x + product= 0
    x^2 - 80/21x + (-16/21) = 0
    Final Answer : 21x^2 - 80x - 16 = 0


    NO (3a)
    I=#30,000
    R=3%
    T= 4YRS

    P=100 X 1/RT =100 X 30,000/3 X 4
    =25,000#

    (3B)
    #7000 T0 FRANCS

    8FRANS= #1
    ?FRANCS =7000

    CROSS MULTIPLY
    =#56,000 FRANCS

    SPENDING 49,400 FRANCS

    56,000-49,400 = 6600
    CONVERTING 6,600 FRANCS TO # AT 10 FRANCS TO 1#
    10 FRANCS—– 1#
    6600 FRANCS—?#
    =6,600/10 X1


    6i)
    7+8+x*x=47
    15+2x=47
    2x=47-15
    2x/2 =32/2
    x=16
    6ii)
    Tram only
    7+8+3+y =30
    18+y =30
    y=30-18=12
    Tram only =12
    6iii)
    at least two
    7+16+3+8
    =34//
    6iv)
    16+7+12+16+8+3+9=95
    62+9=95
    9=95-62
    9=33// or car only =33


    7a)
    a+2d=11 —-eq (1)
    -(a+8d)=29 —-eq (2)
    -6d/-6d = -18/-6
    d=3
    sub for d=3 in —-eq (1)
    a+2(3)=11
    a+6=11
    a=11-6=5
    7b)
    L = ar^n-1
    729/3 = 3/3 x 3^n-1
    243 = 3^n-1
    3^5 = 3^-1
    5 = n-1
    N = 5+1
    = 6
    Sn = a(r^-1)/r-1 = 3(3^6 – 1)/3-1
    Sn = 3( 729 -1)/2
    S6 = 3 x 728/2
    S6 = 1092
    7c)
    3x^3/3 + x | 2,1
    (2^3 + 2) – (1^3+1)
    (8+2) – (1+1)
    = 8

    8ai)
    y=(4x+9)^3
    let u =4x+9, y=U^3
    du/dx=4, dy/du=3U^2
    dy/dx=dy/du * du/dx = 4*3u^2=12u^2
    recall= U=4x+9
    dy/dx=12(4x+9)^2
    8aii)
    y=(3x-2)^3 (x^2 + 4)^2
    u=(3x-2)^3 V=(x^2+4)^2
    du/dx=3.3(3x-2)^3-1 , dv/dx=2x.2(x^2+4)^2-1
    =4x(x^2+4)
    in other words
    dy/dx=Vdu/dx + udv/dx
    =(x^2+4)^2 9 (3x-2)^2+(3x-2)^3 4x(3x-2)
    =(x^2+4) (3x-2)^2 (9x^2+36+12x^2-8x)
    =(x^2+4) (3x-2)^2 (21x^2-8x+36)
    8b)
    y-y1=m(x-x1)
    y-6=2(x-2)
    y-6=2x-4
    y=2x-4+6=2x+2

    12a)
    from economics
    3x+2y-12=90
    3x+2y=90+12
    3x+2y=102 -----(1)
    sum of angle in a circle=360degree
    5x+y-9+3x+2y-12+3y
    +4x+21=360
    5x+3x+4x+y+2y+3y-9-12+21=360
    12x+6y=360 -----(2)
    solving equation 1and2
    simulteanously
    13x+2y=102
    12x+6y=360
    using elimination method
    3x+2y=102*4
    12x+8y=408
    12x+6y=360
    2y/2=48/2
    y=24
    substitute y=24 into
    equation (1) for x
    3x + 2y =102
    3x+2*24=102
    3x+48=102
    3x=102-48
    3x/3=54/3
    x=18
    x+y=18+24
    x+y=42
    12b)
    candidate that registerd for literature
    =4x+21/360 * 1200
    =4*18+21/360 * 1200
    =72+21/360 * 1200
    93/360*1200 =111600/360
    =310
    candidate that registered for government
    =3y/360 * 1200
    =3*24/360 * 1200
    =72/360 * 1200 =86400/360
    =240
    310 - 240 = 70
    the candidate that registered for litereature are 70, more than candidate that registered for government

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