Verified Neco GCE 2018 Mathematics Obj & Theory Answers

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Verified Neco GCE 2018 Mathematics Obj & Theory Answers
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    NECO GCE MATHEMATICS



    NECO VERIFIED MATHS OBJ
    1 CCBEAEBECE
    11 AADCEECABB
    21 CBBAACEBAD
    31 CAABEEBCAD
    41 ECCECEDDAC
    51 ACABCCCCCB




    PLEASE NOTE THAT
    / means DIVISION
    * means Multiplication
    tita You Should Know that.
    rais means Raise to power or
    ^ means Raise to power


    NECO GCE MATHS THEORY ANSWERS BY NAIJACLASS.com



    *NECO GCE MATHS*

    (1)
    TABULATE:
    No| Log
    3081| 3.4887
    0.775 | 1.8893
    0.456 | 1. 6589
    | UNDER Log
    | 3.4887
    | 1. 5482
    | 3.9405
    | 0.9851
    Square root of 4/3081/0.775.0456
    Antilog = 9663
    = 9.663
    = 2.9





    (2a)
    1101₂ = 2x + 1
    1 x 2² + 0 x 2¹+ 1 x 2º = 2x + 1
    8 + 4 + 1 = 2x +1
    13 – 2x
    X = 12/2 = 6

    (2b)
    Sin x = 12/13 = 0.9231
    Z = sin-¹ 0.9231 = 67.4º
    Therefore 3 sin x + ½ cos x
    = 3 sin 67.4 + ½ cos 67.4
    = 3 x 0.9232 + ½ x 0.3843
    = 7.7696 + 0.1921





    (3a)
    PQ x 90º [angle in a semicircle]
    QPO = 90º - 62 [ angle in a triangle]
    = 28º
    Therefore POZ = 28º [alternative angles]
    (ii)
    PXZ = ½ x 28º = 14º [angle at center is twice angle at cirumfeence]

    (3b)
    6/5 + 3/x+3 – 9/5(x+3)
    = 6 (x+3) + 3 (5) – 9/5(x+3)
    = 6x + 18 + 15 – 9/5 (x+3)
    = 6x + 24/5(x+3)
    = 6(x+4)/5(x+3)





    (4a)
    1/3(y-1)+2>1/2(2y-1)+1
    2y-2+12>6y-3+1
    2y-6y>-2-10
    -4y>-12
    y<3

    (4bi)
    M=y2-y1/X2-X1
    =-1-2/2-3=-3/-1=3

    (4bii)
    2X+1=X+3
    2X-X=3-1
    X=2
    ==========================


    (6)
    126y = 86
    1 x y² 2 x y¹ + 6 x yº = 86
    y² + 2y + 6 = 86
    y² + 2y + 6 – 86 = 0
    y² + 2y – 80 = 0
    y + 10y – 8y – 80 = 0
    y (y+10)-8y (y+10) = 0
    y – 8 = 0 or y + 10 = 0
    the positive value of y

    (6b) Area of triangle = ½ x b x h
    Let the Acheal Area = x
    Area = ½ x b x h
    Base = x – 9x/100 = 93/100
    Height = 9x/100 + x = 107x/100
    Area = ½ x 93x/100 x 107x/100
    =963x/2000
    % error = actual Area – wrong/actual area x 100
    = x – 963x/2000 /x 100
    = 20000x – 963x/20000 x 100
    = 19037/20000 x 100
    = 95.185%

    (6c)
    p/100 + 2p + 7 = 11.02 x 100
    p + 200p + 700 = 1102
    201p = 1102 – 700
    201p = 402
    P = 402/201
    P = N2
    P = 200K
    =============================


    (8)
    S²₁ (3x – 1) (x+2) dx
    By expansion’
    S²₁ 3x² + 6x – x – 2 dx
    S²₁ 3x² + 5x – 2 dx
    By integrating using d formula
    Xn+1/n+1
    Therefore 3x ²+¹/2+1 + 5x¹+¹/1+1
    = 2xº+¹/0+1 + c
    Therefore 3x³/3 + 5x²/2 + 2x/1 + c
    Therefore x³ + 5x²/2 + 2x
    But x = 2 at higher and are at lower by substituting 2 in x than of value of 1
    (2)³ + 5/2(2)² + 2(2) – ((1)³ + 3/2(1) + 2 (1)
    8 + 5/2 x 4 + 4 – (1 + 5/2 + 2)
    8 + 10 + 4 – (3 + 5/2)
    22 – 11/2 = = 22-11/2
    = 44- 11/2
    = 33/2

    (8bi)
    T = 2π square l/g
    Dividing both side by 2π
    T/2π = square root of L/g
    By squaring both side
    (T/2 π)² = (L/g)²
    T²/4 π² = l/g
    Cross multiplication
    gT²/T² = 4π²L/T²
    g = 4π²L/T²

    (8bii)
    T = (0.4)1/2 = square root of 0.4
    L = 0.04 T1 = 3.14
    G = 4π²L/T2
    = 4 x (3.14)² x 0.04/(0.4)²
    = 4 x 9.8596 x 0.04/0.4
    = 1.578/0.4
    g = 3.94
    ===========================

    (9a)
    2p-q=10.......(1)
    3p+q^2=22........(2)
    Eq(1) x3 and eq(2) x2
    6p-3q=30
    6p+2q=44
    Subtracting eq(1) from eq(2)
    -3q-2q= -14
    2q^2+3q= 14
    2q^2+3q-14=0
    2q^2+7q-4q-14=0
    q(2q+7)-2(2q+7)=0
    (q-2)(2q+7)=0
    q-2=0 or 2q+7=0
    q=2 or q= -7/2
    Substitute q into eq(1)
    2p-q=10
    2p-2=10
    2p=10+2
    2p=12
    p=12/2=6
    When q= -7/2
    2p-q=10
    2p-(-7/2)=10
    2p+7/2=10
    4p+7=10
    4p=10-7
    4p=3
    p=3/4

    (9b)
    z^2 -25/z^2-9z+20
    If Z is undefined
    Z^2-9z+20=0
    Using factorization method
    -4z and -5z
    Z^2 -4z-5z+20=0
    Z(z-4) - 5(z-4)=0
    (z-4)(z-5)=0
    Z-4=0 or Z-5=0
    Z=4 or Z=5
    Z=4 or 5
    Z is undefined when it is equal to 4 or 5






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