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CONFIRMED WAEC 2017 MATH ANSWER



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  • MATHS OBJ:<br /> 1-10: CBBCCACBBC
    11-20: CACBAAAADB
    21-30: BBADBCCACB
    31-40: BBCACBDADA
    41-50: DDACCCDDCB


    Where ever u c
    ^ it means raise to power
    / division
    * multiplication
    X normal X

    =====================
    SECTION A ANS ALL
    =====================
    1a)
    (y-1)log4^10= ylog16^10
    log4^10 (y-1)= log16^y10
    4^(y-1)=16y
    4^y-1=4^2y
    y-1=2y
    -1=2y=y
    -1=y
    y= -y

    1b)
    let the actual time for 5km/hr be t
    for 4km/hr=30mint + t
    4km/hr=0.5 + t
    distance = 4(0.5+t)
    =2*4t
    for 5km/hr, time= t
    distance =5t
    1+4t=5t
    t=2hrs
    actual distance = 5*2=10km
    ==================

    2a)
    2/3(3x-5)-3/5(2x-3)=3/1
    L C M =15
    10(3x-5)-9(2x-3)=45
    30x-50-18x+27=45
    30x-18x=45+50-27
    12x-23=45
    12x=45+23
    12x=68
    x=68/12
    x=34/6
    x=17/3

    2b)
    U'aS=180-(n+88)
    =180-n-88=92-n
    also, u'TQ=18m
    80degree + 92-n+180-m=180degree
    80+92+180-n-m=180degree
    352-n-m=180degree
    -n-m=180-352
    -n-m=-172
    +(n+m)= +172
    m+n=172dgree
    ==================

    3a)
    Tan 23.6° = h/50
    Cross multiply
    Tan 23.6° x h/50
    h = 50 tan 23.6°
    = 21.844m‎

    3b)
    Area of A = 1/2bh
    45 = 1/2 x 10 x h
    45 = 5h
    h = 9cm
    Area of < QTUS = 1/2 ( QT + US)h
    = 1/2 ( 6 + 16)9
    = 99cm^2
    ==================

    4a)
    T6=37
    T6=a+(6-1)d
    T6=a+5d
    a+5d=37 -----(eq1)
    s6=147
    sn=n/2(2a+(n-1)d)
    147=3(2a+5d)
    49=2a+5d
    2a+5d=49 ----(eq2)
    a+5d=37 ---(eq1)
    2a+5d=49 ---(eq2)
    a=12

    4b)
    S15=15/2(2(12)+14d)
    S15=15/2(24+14d)
    from(1)
    a+5d=37
    12+5d=37
    5d=37-12
    5d=25
    d=5
    S15 = 15/2(24+14(15)
    S15= 15/2(24+70)
    S15=15/2*94
    S15=15*42
    S15=630
    ==================

    5a)
    draw
    U=20
    B= y-45
    S= y-34
    B=bag
    S=shoe
    let n(B)=y
    n(S)=y+11
    for bag only y-45
    for shoe only y-11-45=y-34

    5b)
    y-45+45+y-34=120
    2y-34=120
    2y=154
    y=154/2
    y=77
    number of customers who bought shoe = y+11
    77+11=88

    5c)
    n(bag)=77customers
    probability =77/120
    =0.642

    =====================
    SECTION B ANS 5 QUESTIONS ONLY
    =====================
    10a)
    Sin x = 5/13
    Using pythagoras rule
    M^2 = 13^2 - 5^2 (^ means Raise to power)
    M^2 = 169 - 25
    M ^2 = 144
    M = √144
    M = 12
    Hence:<br /> Cos x - 2sin x / 2tan x
    12/13 - 2(5/13) / 2(5/12)
    ‎= 12/13 - 10/23 / 5/6
    FIND LCM
    = 12 - 10/13 / 5/6
    = 12/65

    10bi)
    Considering < LMB
    /MB/^2. = 12^2 - 9.6^2
    /MB/^2 = 51.84
    /MB/ = √51.84
    /MB/ = 7.2m
    From < AML
    /LA/^2 = 2.8^2 + 9.6^2
    /LA/ ^2 = 100
    /LA/ = √100
    /LA/ = 10m

    10bii)
    Let the angle be. θ
    From Tanθ = 9.6/2.8
    Tan θ = 3.4288
    θ = Tan^-1 ( 3.4288)
    = 73.74°
    =============

    13ai)
    given
    x(*)y=x+y/2
    i)3(*)2/5=3+2/5/2
    =(15+2/5)*1/2
    =17/5*1/2
    =17/10= 1,7/10

    13aii)
    8(*)y=8^1/4
    =8+y/2 =33/4
    32+4y=66
    4y=66-32
    4y=34
    y=34/4
    y=17/2
    y=8^1/2

    13b)
    given DABC
    AB=(^-4/6) and AC =(3/^-8)
    so AP =1/2(^-4/6)
    AP=(^-2/3)
    hence
    CP = CA + AP
    CP= -(3/^8)+(^-2/3)
    CP = (^-5/11)
    =================

    8a)
    In Table Form / Tabular form
    X = 1,2,3,4,5
    F = m+2, m-1, 2m-3, m+5, 3m-4 = 8m - 1‎
    Fx = m+2, 2m-2, 6m-9, 4m+20, 15m-20 = 28m - 9‎
    But x̄ ( this symbol (x̄) means X bar)
    = 75/23‎
    ΣFx / Σf = 75/23 = 28m - 9/8m-1
    75/23 = 28m - 9/8m - 1
    Cross multiply
    75(8m-1) = 23(28m-9)
    600m - 75 = 644m - 207
    -75 + 207 = 644m - 600m
    132 = 44m
    M = 3

    8bi)
    In tabular form
    X = 1,2,3,4,5
    F = 5,2,3,8,5
    Cum Freq= 5,7,10,18,23
    Q1 = (N+1/4) = (23+1/4)
    = 6
    Q3 = (3N + 1/4) = (3*23+1/4)
    = 18
    Inter quarter range = Q3 - Q1
    =. 18-6
    = 12

    8bii)
    Pr. (at least 4 mark)
    = 8+3+2+5/23
    = 18/23‎

    Posted by OGA EASYNET on 20.04 11:51 Reply




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