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Author Topic: 2018/2019 WAEC WASSCE MATHS Questions And Answer Expo/Runs  (Read 7758 times)

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Offline JEFFREY96

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2018/2019 WAEC WASSCE MATHS Questions And Answer Expo/Runs
« on: April 18, 2018, 02:59:06 am »
Real 2018/2019 WAEC WASSCE MATHS Real Questions And Answers  Expo Runs Runz

===================
Wednesday, 18th April, 2018

Mathematics 2 (Essay) – 09:30 a.m. – 12:00pm

Mathematics 1 (Objective) – 3:00 p.m – 4:30 p.m..
====================

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    Offline JEFFREY96

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    *Maths*


    OBJ - MATHS
    1 ABBCDDBBAA
    11ADAACCDCCC
    21DADBABCDAD
    31BADADBCACB
    41ADDDBDADCA
    (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
    = #299,000


    ======================================


    (2a)
    Given y-2px^2-p^2x-14
    At (3,10)
    10=2p(3)^2-p^2(3)-14
    3p^2-18p+24=0
    p^2-6p+8==0
    Using factor method
    P^2-2p-4p+8=0
    P(p-2)-4(p-2)=0
    (p-4)(p-2)=0
    p=4 or p=2

    (2b)
    The lines must be solved simultaneously
    3y-2x=21--(i)
    4y+5x=5--(ii)
    Using elimination method
    =>12y-8x=84--(iii)
    =>12y+15x=15--(iv)
    eq(iv) minus (eqiii)
    23x=-69
    x=-69/23
    x=-3
    Put x into eq(i)
    3y-2(-3)=21
    3y+6=21
    3y=21-6
    3y=15
    y=15/3
    Coordiantes of Q is (-3,5)

    ================================



    (3a)
    Using pythagoras theorem
    L^2=5.1^2 + 4.65^2
    L^2=26.01 + 21.6225
    L^2=47.6325
    L=sqroot(47.6325)

    L=6.9cm(1 d.p)
    Perimeter of rhombus=4
    =4*6.9
    =27.6cm

    (3b)
    Sin x=3/5
    DRAW THE TRIANGLE
    Using pythagoras tripple the third side=4
    therefore cosx=4/5
    tanx=3/4
    therefore 5cosx-4tanx
    =5(4/5)-4(3/4)
    =4-3
    =1

    ===========================================


    (4ai)
    Draw the diagram
    ai X + 90 = 3x + 15
    90 = 3x - X + 15
    90 = 2x + 15
    2x + 15 = 90
    2x = 90 - 15
    X = 75/2
    X = 37.5•

    (4aii)
    (4b)
    2N4seven = 15Nnine
    Converting both to base 10
    2×7+N×7¹+4×7 = 1×9²+5×9¹+N×9
    98 + 7N + 4 = 81 + 45 + N
    7N + 102 = 126 + N
    7N - Ń = 126 - 102
    6N = 24
    Ń = 24/6
    Ń = 4

    =======================================

    (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
    (b)
    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

    (6)
    Total number of cars on road worthiness = 240
    60% passed ie 60/100×240/1 = 144cars.
    Number that failed = 240-144 = 96cars

    (6a)
    Draw the Venn diagram
    C = clutch
    B = brakes
    S = steering

    (6b)
    From the diagram above
    E = 28+12+8+6+x+6+2x
    96=60+3x
    96-60=3x
    36/3 = 3x/3
    Therefore X = 12

    (i) The no of cars that had faulty brakes
    =12+8+6+x (Since X = 12)
    =12+8+6+12 = 38

    (ii) Only one fault = (no of clutch only) + (no of brakes only) + (no of steering only)
    =28+x+12x = 28+12+24
    =64cars

    ==============================================



    (7a)
    (y-y1)/(x-x1)=(y2-y1)/(x2-x1)
    (y-5)/(x-2)=(-7-5)/(-4-2)
    (y-5)/(x-2)=-12/-6
    (y-5)/(x-2)=2
    Cross multiply
    y-5=2(x-2)
    y-5=2x-4
    2x-y-4+5=0
    2x-y+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

    ==========================================




    (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
    ======================================


    (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

    (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= 12x0.3843
    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= 12x0.60000
    y= 7.2m
       
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