NECO GCE 2014/2015 Mathematics Objective/Theory Questions And Answers

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NECO GCE 2014/2015 Mathematics Objective/Theory Questions And Answers
  • MATHS OBJ 'TYPE A'
    1-10: CDDCCEDBAB
    11-20: CBABCDADAB
    21-30: CCBCDBDCDA
    31-40: BBDBCCEDBA
    41-50: ECBADBCEAB
    51-60: CDDBDCABDA

    MATHS OBJ TYPE D:
    1 BDEBBCECEE
    11 DCBBCBBDCE
    21 DCBCDADABC
    31 BDCADAEBDC
    41 CBCDCDACDC
    51 ECBBAAAAAC

    (1a)
    R=14cm, tita=74 degree
    lenght of chord= 2r sin tita/2
    =2*14*sin(74/2)
    =2*14*sin 37 degree
    =2*14*sin37 degree
    =2*14*0.6018
    =16.85cm

    (1b)
    R=14cm, L=16.85cm
    let x reps the distance
    R^2=L^2 +X^2
    14^2= (16.85)^2 + x^2
    x^2= 87.92
    x= sqr root (87.92)
    =9.38

    (1c)
    Area of triangle POQ
    =1/2 r^2 sin tita
    =1/2* (14)^2 * sin 74 degree
    =94.21 cm^2

    (2a)
    sum of inferior angle
    =(n-2)180 degree
    =(6-2)180 degree
    =4*180 degree= 720 degree
    720 degree =140 +5x
    5x=720-140
    5x=580
    x=580/5
    =116 degree

    (2b)
    sum is 720 degree
    k+k+15+k+30+35+k+45+k+55=720 degree
    6k+180= 720
    6k=720-180
    6k=540
    k=540/6
    k=90 degree


    (3)
    f5=f4+24-----eqn1
    f4=f3+8---eqn 2
    f5=ar^4----eqn 3
    f4=ar^3------eqn 4
    f3=ar^2------eqn 5
    ar^4=ar^3+24
    ar^4-ar^3=24
    ar^3=ar^2+8
    ar^3-ar^2=8
    ar^3-ar^2=8
    ar^2(r^2-(r))=24
    ar^2(r-1)=8
    (r^2-r)/r-1=24/8
    r(r-1)/r-1=3
    r=3

    (i)common ratio(r)=3
    (ii) first term(a) is thus;
    ar^2(r-1)=8
    a(3)^2(3-1)=8
    a*18=8
    a=8/18=4/9
    first term(a)=4/9

    (iii)
    sn= a(r^n-1)/(r-1)
    s5=(4/9(3^5-(1))/(3-1)
    s5=4/9(243-1)/2
    s5=(4/9*(242))/2
    s5=4*242/(9*2)
    s5=(4*242)/9*2
    s5=968/18
    =53.7
    (4a)
    tan tita=71.5/26
    tan tita= 2.75
    tita= tan^-1(2.75)
    tita= 70 degree

    (4b)
    adjacent = p^2 -q^2, hyp= (p+q)^2
    (p+q)^2= p^2 -q^2+ opp^2
    p^2 + 2pq +q^2= p^2- q^2 +opp^2
    opp^2=2pq- q^2- q^2
    =opp^2=2pq-2q^2
    sin tita= (2pq-(2q^2))/(p+q)^2

    (5a)
    draw the table
    (5b)
    perfect cub=0
    Pr(perfect cube)=0

    (5c)
    divisible by 5=(5,5,5,5,10,10,10)
    divisible by 3=(3,3,6,6,6,6,6,9,9,9,9,12)
    probability=19/36


    (6a)
    Tabulate:
    no:
    3.267,0.483,0.385,
    log:
    0.5141,bar 1.3161,bar 1.4145
    total bar 2.7306
    0.5141
    bar 2.7306
    1.3835
    3.267
    0.43*0.385
    antilog 24.18

    (6b)
    z=(3x-2)/(2x+3)
    z(2x+3)=3x-2
    2zx+3z=3x-2
    2zx-3x=--3z-2
    x(2z-3)=-3z-2
    x(2z-3)=-3z-2
    x=(-3-2)/2z-3
    x=-(3z+2)/2z-3
    x=(-2-3z)/(2z-3)


    (7a)
    A:T:U= 1/3: 7/12: 1/12
    let A reps Ade
    T reps Tayo
    U reps uche
    Total ratio= 1/3 +7/12+ 1/12
    =(4+7+1)/12=12/12=1
    Ade=7000 + U
    U=(1/12)/1*(T)
    T=12U
    A=(1/3)/12U
    A=4U
    4U=7000 + U
    3U=7000
    U=7000/3 =
    N2333.3
    T=12*(7000)/3
    T=4*7000
    T=N28,000
    Total profit = N28,000

    (7bi)
    Given (A/P)^1/n -1 =r/100
    (A/p)^1/n=r/100 + 1
    (A/p)^1/n= (r+100)/100
    A/p=((r+100)/100)^n
    A=p((r+100)/100)^n)

    (7bii)
    given p=7808,n=3, r=5
    A=7808((25+100)/100)^3
    A=7808(125/100)^3
    A=7800*(5/4)^3
    A=7808*(125/64)
    A=15250



    (8ai)
    area of one triangle = 1/2 r^2 sin tita
    =1/2 *(10)^2* sin 45 degree
    =100/2*(sqr root 2/2)
    =50 sqr root 2/ 2 = (25 sqr root 2) m^2
    area of the octagon =8*25 (sqr root 2) =(200 sqr root 2) m^2

    (8aii)
    perimeter of the octagon
    =8* 27.86
    =222.88 cm^2

    (8b)
    ((3x+2)/4)-((2x+5)/3))=1/6
    (3x(3x+2)-4x(2x+5))/12=1/6
    (9x+6-8x-20)/12=1/6
    9x-14)/12=1/6
    x-14=2,x=2+14, x=16

    (9ai)
    Lenght(L)= (p+4)m
    breadth(B)=(p-1)m
    diagonal(D)=(p+5)m
    (p+5)^2=(p-1)^2+(p+4)^2
    p^2 +10p +25= p^2 -2p +1+ p^2 +8p +16 2p^2 +6p +17 - pΛ2 -10p -25 =0
    p^2-4p-8=0.
    p=-(b (+-) b^2 -4ac)/2a
    p=+4 (+-) sqr root ((-4^2)-4*1*-8)/2*1
    p=(4(+-) sqr root (16+32))/2
    p=(4 (+-) sqr root (48))/2 p=( 4(+-)6.93)/2, p=(4-6.93)/2
    p=10.93/2, p=-2.93/2
    p=5.47, p=-1.47m
    the value of p is 5.47m

    (9aii)
    area of the lawn
    =l*b
    =(p+4)*(p-1)
    =(5.47+4)*(5.47-1)
    =9,47*4.47 =42.33m^2

    (9b)
    Base of a pyramid is 4.5m by 2.5m
    Height = 4m
    volume = 1/3*area of base*height
    = 1/3*11.25*4
    volume = 15m^3.

    (9c)
    Using SOHCAHTOA
    Tan tita = 5/12
    x^2=12^2-5^2
    x^2=144-25
    x^2=119 x=root 119
    x= 10.9 = 11 apprx.
    Cos tita = 11/12.

    (11a)
    given cos B= ((sqr root 2) sin B + x^2)/4x
    cos 45 degree= ((sqr root 2) sin 45 degree + x^2)
    (1/ sqr root 2)*4x= (sqr root 2)*(1/ sqr root 2) + x^2
    ((4/sqr root 2)*((sqr root 2)/(sqr root 2))=1+x^2
    (4( sqr root x))/2=1+x ^2 x^2- (2 sqr root 2)x+1=0
    x=((2 sqr root 2)(+-) sqr root (8-4))/2
    x=((2 sqr root 2)(+-)2)/2
    x=((2 sqr root 2)(+)2)/2 or ((2 sqr root 2)(-)2)/2
    x= (sqr root 2)+1 or (sqr root 2)-1

    (11b)
    distance AH=12*3= 36km
    distance AB=18*3=54km
    ab is the distance apart
    AB^2=36^2 +54^2-2(36)(54)cos 66 degree
    AB^2=1296+2916-3888*0.4067 AB^2=4212-1581.25
    AB^2=2630.75
    AB=sqr root (2630.75)=
    51.3km

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