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    MATHS-OBJ!!
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    Completed
    MATHEMATICS-THEORY-ANSWERS!
    (1a)
    T1 = 7 – 2x
    T2 = 9
    T3 = 5x + 17
    r = T1/T2 = T3/12
    9/(7- 2x) = (5x + 17)/9
    (5x + 17)(7 – 2x) = 81
    35x – 10x² + 119 – 34x = 81
    x – 10x² + 119 = 81
    10x² – x + 81 – 119 =0
    10x² – x – 38 = 0
    10x² – 20x + 19x – 38 =0
    10x(x – 2) + 19 (x – 2) =0
    (10x + 19) (x – 2) = 0
    x = – 19/10 or x =2
    .: x = 2

    (1b)
    p/q = 3/4
    . : 4p = 3q
    .: p = 3q/4
    3p + 2q = 68
    3(3q/4) + 2q = 68
    9q/4 + 2q = 68
    9q + 8q = 272
    17q = 272
    q = 16
    .: p = 3×16/4
    p = 12
    .: The smaller number = 12
    =============================

    (2a)
    y= (Pr/m – P²r) -³/²
    Multiply both index by -⅔
    y-⅔= (Pr/m – P²r)-³/²*-⅔
    y-⅔/1 = Pr/m – P²r/1
    my-⅔ = Pr – mP²r
    my-⅔/P-mP² = r (P-mP²)/P-mP²
    m/(P-mP²)y⅔ = r

    (2b)
    y= -8, m=1, P=3

    r = m/(P-mP²)y⅔
    r = 1/(3-1*3²)* -8⅔
    r = 1/(3-9)*(³√-8)²
    r = 1/-6*(-2)² = 1/-6*4 = 1/-24
    :. r = -1/24
    =============================

    (3)
    Perimeter of minor segment= AB + AB
    AB = 2AM = 2rSin36°
    = 2*24.5*Sin36°
    = 28.8m

    AB = 72°/360° * 2*22/7* 24.5
    = 72/360* 154 = 30.8m
    :. Perimeter= 30.8+28.8
    = 57.6m
    =============================

    (4a)
    2(2y+10) = 5x-35 (angle at centre = twice at circle)
    4y+20= 5x-35
    5x = 4y+55
    x = 4/5y +11……….(1)

    (2y+10)+(2x+40) = 180 (supplementary angle)
    2x+2y+50 = 180
    x+y = 65……..(2)

    Put (1) into (2)
    4/5y +11+y =65
    9/5y = 54
    y = 5/9*54 = 30
    From (2) x=65-y = 65-30= 35

    (4b)
    = 360 – (175-35+40+70+40)
    = 360-290
    = 70°
    =============================

    (5a)
    Given m= tan30° = 1/√3
    n = tan45° = 1
    m -n/mn= 1/√3 -1/(1/√3)(1)
    = √3(1/√3 -1)
    = 1 – √3

    (5b)
    15-x+x+10-x =20
    25-x = 20
    x= 25-20 = 5
    PnB(both) = 5/20 = 1/4
    =============================

    (6a)
    On x-axis: 2cm to 2units
    On y-axis: 2cm to 10units

    (6b)
    (x+2)(x-4) = 0
    x²-4x+2x-8 = 0
    x²-2x-8= 0
    mx² + nx + r = y
    :. m=1, n= -2, r = -8

    (6c)
    Gradient= y2 -y1/x2 -x1
    = -27 -5/-5 -3
    = -32/-8 = 4

    (6d)
    (x+2)(x-4) > 0
    x+2<0 or x-4<0
    x< -2 or x<4
    And x-4>0
    x+2>0 or x-4>0
    x> -2 or x>4
    :. x> -2
    =============================

    (7a)
    Given; cost price = ₦250.00 per book
    Total cost price = 250*180 books
    = ₦45,000

    Selling Price = (₦300*y books) + 95% of ₦250*(180-y)
    = ₦300 + ₦237.5(180-y)
    = ₦(62.5y + 42750)

    Now, Profit = S.P – C.P
    ie 62.5y + 42750 – 45000 = 7125
    62.5y = 7125 +2250
    62.5y = 9375
    y = 9375/62.5
    y = 150

    (7b)
    (i) Profit = S – C
    = 33x – x²/20 – 24x -103
    P = -x²/20 + 9x -103

    (ii) When x = 20
    P= -20²/20 + 9(20) – 103
    = -20+180-103= 57

    Cost price, C = 24x +103
    = 24(20)+103
    = 583
    %Profit= P/C *100%
    = 57/583 * 100%
    = 9.777%
    =============================

    (10a)
    [TABULATE]

    Age X; 3|| 4|| 5|| 6|| 7|| 8|| 9|| 10||

    Number of children F; 2 || 6 ||5 || x=4 || 6 || 9 || 8 || 5||

    Fx; 6 || 24 || 25 || 6x=24 || 42 || 72 || 72 || 50||

    x – x̄; -4 || – 3 || – 2 || – 1 || 0 || 1 || 2 || 3||

    (x – x̄)²; 16|| 9 || 4 || 1 || 0 || 1 || 4 || 9 ||

    f(x – x̄)²; 32 || 54 || 20 || 4 || 0 || 9 || 32 || 45||

    Σf = x + 41
    Σfx = 6x + 291
    Σf(x – x̄)² = 196

    Mean x̄ = Σf/Σf
    x̄ = 7 = (6x + 291)/(x +41)
    7x + 287 = 6x + 291
    7x – 6x = 291 – 287
    x = 4

    (10b)
    S.D = √[(Σf(x – x̄)²]/Σf
    S.D = √(196/x+41)
    S.D = √(196/45)
    S.D = 2.1
    =============================

    (11a)
    Given: y = x +7 – – – – – – (1)
    42 + 38 + 57 + x + (x+y) + (2x – 15) + (3x – y) = 360° (external angle of a polygon)
    122 + 7x = 360°
    7x = 360 – 122 = 238
    x = 238/7
    x = 34
    y = x +7
    y = 34 + 7
    y = 41

    (11b)
    = 180° – 34 – 34
    = 112°

    = 360° – 112 – 146
    = 102°

    But 102 + 2 2 =============================

    (12a)
    Pr (loose) = 1/4
    Pr (Win) = 1- 1/4 = 3/4
    (i) Pr (LWL) = 1/4*3/4*1/4 = 3/64
    (ii) Pr (WWW) = 3/4*3/4*3/4 = 27/64
    (iii) Pr (Only two) = Pr (WWL) or Pr(LWW) or Pr(WLW)
    = (3/3*3/4*1/4) + (1/4*3/4*3/4) + (3/4*1/4*3/4)
    = 9/64 + 9/64 + 9/64 = 27/64

    (12b)
    Volume= 1/3πr²h
    1200 = 1/3*22/7*r²*24
    25200/528 = 528r²/528
    r² = 525/11
    Volume= 1/3πr²h
    Volume= 1/3*22/7*525/11*84
    Volume = 4200cm³
    =============================

    (13a)
    Total surface area, S = 2πrh + 2πr²
    = 2πr(h + r)
    209 = 2 x (22/7) x (7/2)(h+7/2)
    209 = 22(h + 22/7)
    209/22 = h + 7/2
    19/2 = h + 7/2
    h = 19/2 – 7/2
    h = 12/3
    h = 6cm

    (13bi)
    Draw the diagram

    (13bii)
    h/(Fx̄ + 19) = Tan38°
    h/Fx̄ = Tan43°
    (Fx̄ + 19) = Fx̄ Tan43°
    (Fx̄ + 19) = Fx̄ (Tan43°/Tan38°)
    Fx̄ + 19 = 1.19356Fx̄
    1.19356Fx̄ – Fx̄ = 19
    0.19356 Fx̄ = 19
    Fx̄ = 19/0.19356

    But h = Fx̄ Tan43
    h = (19/0.19356) x 0.9325
    Height, h = 91.5meters
    =============================

    (1)

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    (2)

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    (3)

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    (6)

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    (7)

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    (10)

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    (11)

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    (12)

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    (13)

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    Completed!.



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    Master Solution June 2, 2022 Categories: WAEC 210


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