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NECO GCE – MATHEMATICS

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NECO GCE 2019 – MATHS ANSWERS

MATHS OBJ
01-10: BCBEDDCCDC
11-20: DEBCBBDDEC
21-30: BCBCCBAEED
31-40: EBABEDCDBB
41-50: DDCECDCBAB

51-60: DDBCEACBBE

THEORY
++++++++++++++++++++++++
(1)
√((0.0024)×35000)/0.0105
No | Log
0.024 |_3.3802
35000 |4.5441

  • |=1.9243
    0.0105|_2.0212 –
    |3.9031 ÷2
    |1.9516
    √((0.0024)×35000)/0.0105
    Antilog of .9516 =89.45
    ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
    (2a)
    Given that the roots of the equation are
    X = -2/3 and X = -3/2
    3x = -2 and 2x = -3
    3x + 2 = 0 and 2x + 3 =0
    (3x+2)(2x+3) = 0
    6x² + 9x + 4x + 6 = 0
    6x² + 13x + 6 = 0
    (2b)
    R = [3 4 0 ]
    [2 0 3 ]
    [1 2 2]
    (2ci)
    2/3R
    =2/3[3 4 0]
    [2 0 3]
    [1 2 2]
    = [⅔(3) ⅔(4) ⅔(0)]
    [⅔(2) ⅔(0) ⅔(3)]
    [⅔(1) ⅔(4) ⅔(2)]
    = [2 8/3 0 ]
    [4/3 0 2 ]
    [2/3 4/3 4/3]
    (2cii)
    |R|
    = |3 4 0|
    |2 0 3|
    |1 2 2|
    =3|0 3| -4|2 3| +0|2 0|
    |2 2| |1 2| |1 2|
    =3(2×0-3×2)-4(2×2-3×1)
    +0(2×2-0×1)
    =3(0 – 6)-4(4 – 3) +0(4 – 0)
    =3(-6) -4(1) +0(4)
    = -18 – 4 + 0 = -22
    (2ciii)
    The transpose of R
    = [3 2 1]
    [4 0 2]
    [0 3 2]
    ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
    (3a)
    Given : R(3,5) and S(-2, -6)
    equation to line through them is :
    y-5/x-3 = -6-5/-2-3
    y-5/x-3 = -11/-5
    y-5/x-3 = 11/5
    5(y-5)= 11(x-3)
    5y-25 = 11x-33
    5y-11x = 25-33
    5y-11x = -8 or 11x-5y = 8
    (3b)
    RS= √(X1X2)^2 (y1y2)^2
    √(-2-3)^2 + (-6-5)^2
    √(-5)^2 + (-11)^2
    √25 + 121
    √146
    = 12.08
    ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
    (4a)
    Given: curve; y x² – 3x
    gradient ; dy/dx = 2x – 3
    At (2-2), gradient = 2(2) – 3
    4-3 = 1
    (4b)
    Given; y= 1+x²/1-x²
    dy/dx = (1-x²)(2x) – (1+x²)(-2x)/(1-x²)²
    = (1-x²)(2x) – (1+x²)(2x)/(1-x²)²
    = 2x(1-x² + 1+x²)/(1-x²)²
    = 2x(2)/(1-x²)²
    = 4x/(1-x²)²
    ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
    (5)
    No of blue balls = 6
    No of red balls = 10
    (i)
    Prob (2 balls of some colour)
    = BB or RR
    Total no of balls = 6+10 = 16
    BB or RR
    (6/16 × 5/15) + (10/16 × 9/15)
    =30/240 + 90/240
    =120/240 = 1/2
    (ii)
    Prob (2 balls of different colours)
    = BR or RB
    = (6/16 × 10/15) or (10/16 × 6/15)
    = 60/240 + 60/240 = 120/240
    =1/2
    ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
    (6a)
    27^(2x+1) × 3^-x = 81^(x-2)/9^(x+2)
    = 3^3(2x+1) × 3^-x = 3^4(x-2)/3^2(x+2)
    =3^6x+3-x = 3^4x-8-2x-4
    5x+3 = 2x – 12
    5x – 2x = -12-3
    3x = -15
    3x/3 = -15/3
    X = -5
    (6b)
    X/x+101 = 11/1000
    Since all the members are in binary, convert all to denary (base 10)
    Xbase2 = Xbase10
    101base2 = (1×2^2)+(1×2^0) = 4+1 = 5base10
    11base2 = (1×2¹)+(1×2raise to power 0) = 2+1 = 3base10
    1000base2 = 1×2^3 = 8base10
    X/X+101 = 11/1000 –> X/X+5 = 3/8
    3(x+5) = 8(x)
    3x+15 = 8x
    15 = 8x – 3x
    15 = 5x
    15/5 = 5/5
    X = 3
    Convert X=3 to base 10 to base 2
    2|3
    2|1R1
    |0R1
    .:. x = 11
    (6c)
    Given that log5 base 10 = 0.699
    and log3 base 10 = 0.477
    10
    Log75 base 10 = log(3×5×5) base 10
    =log(3×5²) base 10
    =log3 base 10 + 2log5 base 10
    =0.477 + 2(0.699)
    = 0.477 + 1.398
    = 1.875
    log75 base 10 = 1.875
    ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
    (7a)
    Given that Y=2×2+7x-6
    To find the gradient of the curve at the point x=3
    dy/dx=4x+7 (at x=3)
    dy/dx=4(3)+7
    dy/dx=19
    (7bi)
    Ade bought 7kg of maize + 4kg of meat=N4240
    Kemi bought 3kg of maize + 5kg meat =N4610
    Let x = 1kg of maize and y=1kg of meat
    Therefore 7x+4y=4240—–(eq1)
    3x+5y=4610———(eq2)
    Substituting simultaneously
    Multiply eq1 by 3 and eq2 by 7
    21x+12y=12720—eq3
    21x+35y=32270—eq4
    Substract eq3 from eq4
    23y=19550
    23y/23=19550/23=850
    y=850
    Substitute y=850 in1
    7x+4y=4240
    7x+4(850)=4240
    7x=4240-3400
    x=840/7
    =120
    Hence the total cost price per Kg of maize is N120.00 while the total cost per kg of meat is N850.00
    (7bii)
    Total cost of 10kg of maize and 5kg of meat
    =10x+5y
    =10(120)+5(850)
    1200+4250
    =N545
    ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
    (9)
    Draw the diagram
    Q(34.5°N, 22.3°W)
    P(34.5°N, 38.7°E)
    R(35.4°S, 38.7°E)
    Distance between P and Q along the parallel of latitude
    D = Φ/360 × 2πr
    Where r = RcosΦ
    Φ = 38.7° + 22.3° = 61°
    Q = 34.6°
    R = 6400km
    D = 61/360 × 22/7 ×6400cos34.6
    D =1717760(0.8231)/252
    D = 5,610.66768
    = 5,610km

(9b)
|PR|=(34.6+35.4)/360 * 2 22/76400
=70/360 * 44/7 *6400
=19712000/2520
=7820km(3 sf)
Therefore the shortest distance between P and R = 7820km(3 sf)

(9c)
Circumference of the circle of latitude through R = 2πr where r = RcosΦ
= 2πRcosΦ
= 2 × 22/7 × 6400cos35.4
=281600/7(0.8151)
= 229,532.16/7
= 32,790.3
= 32,800km (3 s.f)
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

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