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  • Ondo State Joint Exam 2017: Mathematics Questions and Answers
  • MATHEMATICS ANSWERS
    OBJ:
    1-10: DBABCBAEDD
    11-20: BEBACABCCB
    21-30: DACAEDBDEA
    31-40: BEEDDDEBBB
    41-50: CAADCDDDBC
    ESSAY:
    Answer All The (5) Question in This Section.
    (1B)
    X={2,4,6,8}
    Y={2,3,7,9}
    (XUY) = {2,3,4,6,7,8,9}
    Y={2,3,7,9}
    Z={4,5,6,7,8}(YUZ) = {2,3,4,5,6,7,8,9}
    {XUY} n {YUZ}={2,3,4,6,7,8,9}ANS
    …………………………….
    (2A)
    A=35
    X=7
    Y=4
    A proportion X 1/Y^2
    A = XK/Y^2
    When K is Constant
    35 = 7K/42
    35= 7K/16
    35*16=7K
    560 = 7K
    K = 560/7
    K=80
    .:.
    A = XK/Y^2
    A=5 * 80/8^2
    A=400/64
    A=6.25
    (2B)
    If S = ut 1/At^2 Make T the subject of formula
    of the equation.
    S = ut 1/At^2
    SAt^2 = ut 1
    SAt^2 – ut =1
    SA – u (t^2-t) =1
    t^2 – t = 1/SA – u
    t = 1/SA-u
    (3) Using Ruler And a Pair of Compasses
    Only,construct triangle ABC with /AB/=10cm, /BC/
    =12cm and =120 degree, construct the locus l of
    ponts equidistant from A and B.
    DRAW IT LIKE THE ABOVE IMAGE ON YOUR
    ANSWER SHEET.
    (4A) Using Completing The Square Method Solve
    The equation. X^2 – X – 12 = 0
    X^2-X-12=0
    X^2-X/2 – (1/2)^2 = 12/2 1/4
    (X^2 – 1/2)^2 = 24 1/4 = 25/4
    (X-1/2)^2 = √25/4
    X-1/2 = ± √25/4
    X – 1/2 = ± 5/2
    X = 1/2 5/2OR1/2 – 5/2
    X = 1 5/2OR1-5/2
    X = 6/2 OR -4/2X=3OR-2ANS
    SECTION B: Answer (5) Question in This Section.
    9A)AP = Tn A (n – 1)d
    5th term = T5 = a 4d
    14th term = T14 = a 13d
    T5 = a 4d = 28……………….(equa 1)
    T14 = a 13d = 82 …………..(equa 2)
    Solving the simultaneous equationsSubtract
    equation (1) from equation (2)
    a 4d =28………(1)
    a 13d = 82 …..(2)
    9d = 54
    :. d = 54/9
    d=6
    Substitute (d) in equation (1)
    a 4 (6) = 28
    a 24 = 28
    a = 28 – 24
    a=4
    To find the 21st term
    T21 = a 20d
    =4 20 (6)
    = 4 120
    = 124
    12A)
    Let the distance be a,c
    In ∆ABC ^A=90-205=115°(A FROM glasses
    emoticon
    also ABC=180-205=25°
    :.B=60-25=35
    :.C=180-(35 115)=180-150=30°
    USING SINE RULE
    (i)a/sinA =b/sineB
    a=B sine a/sine B= 15 sin 115°/sin30°=15×0.9404/
    0.9880=14.181/0.9880
    =14.20a=14.20kmii)similarlySin b=C sin B/sin C
    =15 sin
    35°/sin30°=15×0.4281/0.9880
    =6.4215/0.9880
    =6.499km
    :.distance of B from A =14.20km
    :.bearing of B from A= 6.499 km
    For Ondo State joint examination questions and answers for Mathematics 2017 is available.

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