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    Maths-Obj!
    1CBBADCCBEC
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    ====================================

    Maths-THEORY
    All are 101% Authentic.
    ====================================

    (1a)
    Log¹⁰6+ Log¹⁰45 – Log¹⁰27
    Log¹⁰(6*45/27) = Log(270/27)
    = Log¹⁰10 = 1

    (1b)
    8^x = 32
    2^3x = 2⁵
    :. 3x = 5
    x=5/3

    (1c) 81⅙/27-⅑ = 3⁴*⅙/3³*-⅑ = 3⅔/3-⅓
    :. 3⅔-(-⅓) = 3⅔+⅓ = 3³/³
    = 3¹
    = 3
    ====================================

    (2a)
    (i) Gradient (m) = y¹ – y²/x¹- x²
    m = -1-0/2-3 = +1/+1=1

    (ii) y – y¹= m(x-x¹)
    y-0 = 1(x-3)
    y = x-3
    y-x+3=0

    (2b)
    (i) Area of ∆ABC = ½absinϴ
    = ½*6*8*sin60
    = 3*8*sin60
    = 20.7846
    = 20.78(approx)

    (ii) Area of parallelogram = absinϴ
    = 6*8*sin60
    = 48*sin60
    = 48*0.8660
    = 41.6
    ====================================

    (3a)
    |PQ| = 12km, |QP| = 12km
    Speed from P to Q = 6km/h
    Speed from Q to P = (6+x)km/h
    Total time taken 3hrs20mins

    Speed = distance/time

    From P to Q = 6/1*12/t
    6t = 12
    t = 12/6 = 2hrs
    :. Time left = 3hrs 20mins – 2hrs
    = 1hr 20mins

    From Q to P , Speed = distance/time

    6+x = 12/ 1²⁰/⁶⁰
    6+x = 12/⁴/³
    6+x = 12*3/4
    6+x = 9
    x = 9-6 = 3

    (3b)
    x² – (sum)x + product = 0

    Sum = ⅔ + ¾= 8+9/12 = 17/12

    Products = ⅔*¾ = ½

    x² – 17x/12 + ½ = 0
    12x² – 17x + 6 = 0
    ====================================

    (4a)
    Y = x²/ 1+x²
    U=x² , V= 1+x²
    du/dx = 2x , dv/dx = 2x

    dy/dx = (vdu/dx – udv/dx)/v²
    = (1+x²)2x – x² * 2x/(1+x²)²
    = 2x+2x³-2x³/(1+x²)²

    dy/dx = 2x/(1+x²)²

    (4b)
    ⅔(3x +2) = ¾(2x -3

    6x +4/3 = 6x -9/4
    4(6x +4)= 3(6x-9)
    24x +16 = 18x-27
    24x-18x = -27-16
    6x/6 = -43/6

    x = -7⅙
    ====================================

    (5)
    TABULATE

    Mass (kg): 31-40| 41-50| 51-60| 61-70| 71-80| 81-90

    F: 3| 10| 15| 12| 6| 4

    x: 35.5| 45.5| 55.5| 65.5| 75.5| 85.5

    Fx: 106.5| 455| 832.5| 786| 453| 342

    Class boundaries: 30.5-40.5| 40.5-50.5| 59.5-60.5| 60.5-70.5| 70.5-80.5| 80.5-90.5

    (i) Mean (x-bar) = ∑fx/∑f = 2975/50 = 59.5
    :. Mean = 60kg

    (ii) Mode = L¹ + ( fm-fa/2fm-fa-fb) c
    = 50.5. + (15-10/2*15-10-12)10
    = 50.5 +(5/8)10
    = 50.5+6.25
    :. Mode = 56.75
    Mode = 57kg(approx)
    ====================================

    (6a)
    T² = ar = 6 …… (T¹)
    T⁴ = ar³ = 54….. (T²)

    Common ratio = T²/T¹
    ar³/at = 54/6
    r²= 9
    r = ± √9 = ±3
    r =3

    Subtract r=3 in equation T¹
    ar= 6
    3a=6
    a = 6/3 = 2
    :. a = 2 , r =3

    (i) 1st term is 2
    (ii) 5th term T⁵=ar⁴
    T⁵ = 2*3⁴
    = 2*81
    = 162

    (6b)
    (i)
    Let pencil be x
    Let pens be y
    Let Ruler be z

    U= 160
    n(x) = 75
    n(y) = 87
    n(z) = 93
    n(xny) =25
    n(xnz) = 30
    n(ynz) = 47
    pd n(xnynz) = x

    n(xnynz¹) = 25-x
    n(xnzny¹) = 30-x
    n(ynznx¹) = 47-x
    n(xnynz¹) = 75-(25-x+x+30-x)
    = 75 -(55-x)
    = 75-55+x
    = 20+x

    n(ynx¹nz¹) = 87-(25-x+x+47-x)
    = 87-(72-x)
    = 15+x

    n(znx¹ny¹) = 93-(30-x+x+47-x)
    = 93-77+x
    = 16+x

    :. 20+x+25-x+x+30-x+15+x+47-x+16+x=160
    = 153+x =160
    x = 160-153
    x = 7

    (ii)
    n(xny¹nz¹) = 20+7
    = 27
    :. 27 pupils has pencils only
    ====================================

    (7a)
    ∆XAB = ∆ABC (corresponding angle)

    :. ∆BAD + ADC + ∆ACD = 180( sum of angles at triangle)

    ∆CAD + 83+47= 180
    ∆CAD = 180-83-47
    CAD = 50

    :. ∆ADY = CAD ( parallel to each other)
    x = 50°

    (7b)
    Using cosine rule
    c² = a²+b²- 2abcosC
    x² = 6²+8²-2(6)(8)cos120
    = 36+64- 96cos120
    = 36+64+48
    x² = 148
    x = √148
    x = 12km

    Using sine rule
    a/sinA = b/sinB

    12/sin120 = 6/sinϴ
    12sinϴ = 6sin120
    Sinϴ = 6sin120/12
    Sinϴ = 0.4330
    ϴ = sin-¹ 0.4330
    ϴ = 25.66
    = 26°(approx)

    :. The bearing of the boat from its starting point is
    360 -(26+80)
    360 – 106
    = 254°
    ====================================

    (8a)
    Y = 2x²+ 3

    |¹ 2x³/3 + 3x|
    |² |

    = 2(1)/3 + 3(1) – 2(2)²/3 + 3(2)
    =2/3 +3/1 – 16/3 +6
    = 2+9/3 – 16+18/3
    = 11/3 – 34/3 = -23/3
    = -7.67

    (8bi)
    3p +5q = 9500….. (i) *5
    5p +10q = 17500….(ii) *3

    15p +25q = 47500…..(iii)
    15p + 30q = 52500…..(iv)

    Subtracting (iii) from (iv)
    -5q = -5000
    q = -5000/-5
    q = 1000

    Substituting q =1000 into equation (i)
    3p + 5(1000) = 9500
    3p + 5000 = 9500
    3p = 9500-5000
    3p = 4500
    p = 4500/3
    p = 1500

    :. q = ₦1000.00
    p = ₦1500.00

    (8bii)
    8 men and 12 women

    8p +12p
    8(1500) + 12(1000)
    = 12000 + 12000
    = ₦24000.00

    :. ₦24000.00 will be contributed by 8 men and 12 women
    ====================================

    (9a)
    Distance PQ = ϴ/360 * 2πRcosα

    Where ϴ = 11+11 = 22° and α = 12°

    PQ = 22/360*2*3.142*6400*cos12°
    PQ = 884*787.2*0.9781/360

    PQ = 2403.9

    Distance QS= ϴ/360 *2πR
    Where ϴ = 44-12 = 32°

    = 32/360 *2*3.142*6400
    = 1286963.2/360 = 3574.9

    Total distance= 2403.9+3574.9 = 5978.8
    = 5980km( 3 s.f)

    (9b)
    Average speed = Total distance/Total time = 5978.8/8 = 747.35
    = 747km/hr

    (9c)
    No time difference between Q and S because they are on the same longitude
    ====================================

    (12ai)
    TABULATE

    Class interval: 1-20| 21-40| 41-60| 61-80| 81-100|

    Frequency (f): 7| 13| 5| 4| 1

    Class mark(x): 10.5| 30.5| 50.5| 70.5| 90.5|

    d=x-A: -20| 0| 20| 40| 60|

    Fd: -140| 0| 100| 160| 60|

    d²: 400| 0| 400| 1600| 3600|

    fd²: 2800| 0| 2000| 6400| 3600|

    (12aii)
    Mean= A+∑fd/∑f = 30.5+180/30
    = 30.5+6
    = 36.5
    Standard deviation= √ ∑fd²/∑f – (∑fd/∑f)²
    = √14800/30 – (180/30)²
    = √493.33 – 36
    = √ 457.33
    = 21.4

    (12b)
    P(above 60 marks) = (4+1/30) = 5/30
    = ⅙
    ====================================

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    Master Solution August 9, 2021 Categories: NECO 262


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