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    OBJ – MATHS
    1 ABBCDDBBAA
    11ADAACCDCCC
    21DADBABCDAD
    31BADADBCACB
    41ADDDBDADCA

    MATHS THEORY.

    9a)
    Draw the diagram
    Angles PTR and PSR are similar
    |PT|/|PS| = |TQ|/|SR|
    In angle PTR
    |TQ|²=|PT|²+|PQ|²-2|PT||PQ|cos30degrees
    =4²+6²-2×4×6×cos30
    =16+36-48×0.8660
    =52-41.568
    =10.432
    |TQ|=√10.432 =3.22cm
    4/10 = 3.22/|SR|
    4|SR| = 10×3.22
    |SR| = 32.2/4
    |SR| = 8.05cn
    Approximately 8cm(to the nearest whole number)

    9b)
    Atqrs = AΔPSR – AΔPTR
    AΔPTR = 1/2×4×6×sin30
    =2×6×0.5
    =6cm²
    AanglePTQ/AanglePSR = |PT|²/|PS|²
    6/AanglePSR = 4²/10²
    6/AanglePSR = 16/100
    16×AanglePSR = 6×100
    AanglePSR = 600/16 = 37.5cm2
    ATQRS = 37.5 – 6
    =31.5cm2
    =32cm2

    5a)
    m+n+s+p+q/5=12
    m+n+s+p+q=60……(1)
    Now;
    (m+4)+(n-3)+(5+6)+p-2)+(q+8)/5
    =(m+n+s+p+q)+(4-3+3+6-2+8)/5
    =60+13/5
    =73/5
    =14.6

    5b)
    75% of 500 = 375 people
    Number of people above 65 years = 500-375
    =125
    25% of 500 = 125
    Number of people below 15 years = 125
    Number between 15 years and 65 years
    =500-(125+125)
    =500-250
    =250 people

    8a)
    Cost price for Lami= #300.00
    Profit made by lami = x%
    Ie selling price for lami=(100+x/100)×#300
    =#3(100+x)
    =#(300+3x)
    Bola’s cost price = #3(100+x)
    Profit made by bola =x%
    Selling price for bola =(100+x/100)×#3(100+x)
    =#3/100(100+x)²
    James cost price =#3/100(100+x)²=300+(6x+3/4)
    expanding;
    3/100(10000+200+x²) = 300+3/4+6x
    3(10000+200x+x²)=30000+75+600x
    30000+600x+3x²=30000+75+600x
    3x²=75
    X² = 75/3
    X² = 25
    X = square root 25
    X = 5

    8b)
    3x-2<10+x<2+5x
    3x-2<10+x & 10+x<2+5x
    3x-x<10+2 & 10-2<5x-x
    2x<12 8<4x
    X<12/2 4x>8
    X<6 x>8/4
    X>2
    Also; 3x-2<2+5x
    -4<2x
    2x > -4
    X > -2
    Therefore; Range is -2

    10a)
    Using Pythagoras theorem from SPQ
    |SQ|^2 = 12^2 + 5^2
    = 144+25
    =169
    SQ= sqroot of 169
    = 13cm
    Sin tita= 5/13 = 0.3846
    Tita= Sin^-1(0.3846)
    = 22.6degrees
    From PRQ
    Sin tita= |PR|/12
    Sin 22.6 = PR/12
    Sin 22.6= PR/12
    PR= 12xsin 22.6
    PR= 12×0.3843
    PR= 4.61cm

    10bii)
    Let the height at which m touches the wall= y
    Cos x^degrees= 8/10= 0.8
    x^degrees= Cos^-1(0.8)
    = 36.87degrees
    Sin x^degrees = y/12
    Sin 36.87= y/12
    y= 12xsin36.87
    y= 12×0.60000
    y= 7.2m

    7a)
    X1-X/Y1-Y = X2-X1/Y2-Y1
    2-X/5-Y = -4-2/-7-5
    2-X/5-Y= -6/-12
    -12(2-X)=-6(5-Y)
    -24+12X=-30+6Y
    6Y-12X=30+24
    6Y-12X=-6
    6y-12x+6=0
    y-2x+1=0

    7bi)
    DRAW THE DIAGRAM
    7bii)
    I)
    p^2=q+r^2-2qrcosP
    p^2=8^2+5^2-2*8*5*cos90
    p^2=64+25-0
    p^2=89
    p=sqroot(89)
    p=9.4339km
    therefore |QR|=9.43km(3 sf)
    II)
    q/sinQ=p/sinP
    8/sinQ=9.4339/sin90
    sinQ=(8*sin90/9.4339
    sinq=(8*1)/9.4339 =0.8480
    Q=sin^1(0.8480)=57.99 degrees
    but Q=30+ A
    A=Q-30
    =57.99-30
    A=27.99 degrees
    The bearing of R from Q
    =180-A
    180-27.99
    =155.01
    =>152 degrees

    6a)
    Draw the Venn diagram
    Let the number of cars with faults in brakes only be x

    6b)
    Number that passed = 60% × 240 = 144
    Number that failed =
    240 – 144 = 96
    Therefore; 28+2x+x+14+6+6-x+8 = 96
    2x + 62 = 96
    2x = 96 – 62
    2x = 34
    X = 34/2
    X = 17
    6bi)
    faulty brakes cars = 8+6+x+6-x
    = 8+6+6
    =20

    6bii)
    only one fault = 28+x+2x
    =28+3x
    =28+3(19)
    =28+51
    = 79

    2)
    Given that y = 2pxˆ² – p² x – 14
    AT (3, 10)
    10 = 2p(3)² – p² (3) – 14
    10 = 18p – 3p² – 14
    3p² – 18p + 24 = 0
    p² – 6p + 8 = 0
    using factor method,
    p² – 2p -4p + 8 = 0
    p(p-2) – 4(p-2) = 0
    (p-4)(p-2) = 0
    p-4 = 0 or p-4 = 0
    p= 4 or p =2

    2b)
    The lines must be solved simultenously
    3y – 2x = 21 ——- (1)
    4y + 5x = 5 ——-(2)
    using elimination method,
    (4) 3y – 2x = 21
    (30 4y + 5x = 5
    12y – 8y = 84 ——— (3)
    12y + 15x = 15 ——-(4)
    equ (4) minus equ(3)
    23x = -69
    x = -69/23
    x = -3
    Put this into equation (1)
    3y -2(-3) = 21
    3y = 6 = 21
    3y = 21 -6
    3y = 15
    y =15/3
    y = 5
    coordinates of Q is (-3, 5)

    3a)
    The diagonal = 10.2m and 9.3cm
    Using Pythagoras theory
    Ac² = 10.2² + 9-3²
    Ac² = 104.04 + 86.49
    Ac² = 190.53
    Ac² = √190.53
    Ac² = 13.80

    3b)
    DRAW THE DIAGRAM
    Using Pythagoras theory
    5² = 3² + x²
    x² = 5² – 3²
    X²= 25 – 9
    X² = √16
    X= 4cm
    CosX = adjacent/hyp
    = 4/5
    Tan X = opp/adj. = 3/4
    5cos x – 4tan x
    5(4/5)- 4(3/4)
    20/5 – 12/4
    4-3= 1

    4ai)
    sum of angle in a D =180degree
    xdegree + 90degree + 180degree – (3x+15)=180degree
    xdegree + 90degree + 180degree – 3x+15=180degree
    -2x=180degree – 255
    +2x/2=+75/2
    x=37.5

    4aii)
    <RsQ =180 – (3x+15)
    <RsQ =180-(3*37.5+15)
    =180-(112.5 + 15)
    =180 – 127.5
    <RsQ= 52.5degree

    4b)
    2N4seven =15Nnine
    2*7^2+N*7^1+4*7degree =1*9^2 + 5*9^1+N*9degree
    9*49+N*7+4*1=1*81+5*9+N*1
    98+7N+4=81+45+N
    7N+102=126+N
    7N-N=126-102
    6N/6 =24/6
    N=4

    1)
    On February 28th 2012, value = (100-30/100) * #900,00.00
    = 70/100 * #900,00
    = #630,000.00
    On february 28th 2013, value = (100-22/1000 * #630,00
    = 78/100 8 #630,000
    = #491,400
    On february 28th 2014, value = 78/100 8 #491,400
    =383,292
    On february 28th 2015, value = 78/100 * #383,292
    = #298,967.76

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    Master Solution April 18, 2018 Categories: WAEC 4768


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