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UNILAG POST UTME 2011 UPDATES

Discussion in 'Universities in Nigeria' started by PROFICEMUGâ„¢, Jun 30, 2011.

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    (17) y is inversely proportional to the square of x. When x = 3, then y = 4. Find the constant of proportionality.
    (a) 48 (b) 4/9 (c) 2.25 (d) 36

    ans= (d) 36
    solu[hr][/hr]
    y ~ 1/x^2,
    x = 3.
    y = 4.
    x^2 y = k
    9 x 4 = 36
    k = 36
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    (18) The solution to the inequality 5 - 2x > 11 - 4x is
    (a) x > 3 (b) x > 3 (c) x > 1 (d) x < 1

    ans= (a) & (b) are correct.
    Solu[hr][/hr]
    5 - 11 > -4x + 2x
    -6 > -2x
    x > 3
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    (19) if |x 2| = |x 3|           
          |1 5| = |2 3|
    find the value of x.
    (a) 2 (b) -2 (c) 0 (d) -1

    ans= (b) -2
    solu[hr][/hr]
    5x - 2 = 3x - 6
    5x - 3x = -6 + 2
    2x = -4
    x = -2
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    (20) The determinant of
    |1 2 0|
    |3 2 1|
    |4 2 1| is
    (a) 0 (b) 2 (c) -2 (d) 1

    ans= (b) 2
    solu[hr][/hr]
    |+ - +|
    |- + -|
    |+ - +|

    1|2 1| -2|3 1| + 0|3 2|
    |2 1|    |4 1|    |4 2|

    1(2 - 2) - 2(3 - 4)
    0 + 2 = 2
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    (21) <AOB = 130 degrees. Find <ACB.
    (a) 115 (b) 135 (c) 70 (d)65

    ans: 115
    solu[hr][/hr]
    first draw a line from A to B and name the angle D
    AOB = 2ADB
    130/2 = ADB
    65 = ADB
    ADB + ACB = 180
    65 + X = 180.
    X = 180 - 65.
    X = 115.
    • Guru Member

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    (22) fig.2 shows a circle of radius 4cm. The area of the shaded segment is
    (a) 4 pi cm^2
    (b) 4 pi - 8cm^2
    (c) 84cm^2
    (d) 2 pi - 4cm^2

    ans= (b)
    solu[hr][/hr]
    area of segment =
    area of sector - area of triangle
    = tita/360 x pi x r^2 - 1/2 b h
    = 90/360 x pi x 4^2 - 1/2x4x4
    = 1/4 x pi x 4^2 - 8
    = 4 x pi - 8cm^2
    = 4pi - 8cm^2
    • Guru Member

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    (23) fig.3 shows a pyramid on top of a cuboid. The height of the cuboid is Hcm, the height of the pyramid is bcm, & the square base of both shapes has scm. Find the volume of the shape.
    (a) s^2(H +h)cm^3
    (b) s^2(H + h)cm^3
    (c) (1/3)s^2(H + h)cm^3
    (d) (1/2)s^2(2H + h)cm^3

    ans= (a) & (b) is correct
    solu[hr][/hr]
    vol of shape=
    = vol of pyramid + vol of cuboid
    = 1/3 x area of base x h + L x b x h
    = 1/3 x s^2 x h + L x b x h

    s^2 = L x b
    they share same base area.
    = 1/3 x (s^2) x h + (s^2) x H
    = s^2 x [h/3 + H]cm^3
    = 1/3 x (s^2) x [h + H]cm^3
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    (24) if y = sin(x^2 + 7), then dy/dx is
    (a) 2x cos (x^2 + 7)
    (b) (2x + 7) cos (x^2 + 7)
    (c) -2 cos (x^2 + 7)
    (d) 2x cos x

    ans= (a) 2x cos (x^2 + 7)
    solu[hr][/hr]
    u= x^2 + 7.
    du/dx = 2x

    y= sin u.
    dy/du = cos u.

    dy/dx = dy/du x du/dx
    = cosu x 2x
    = 2x cos (x ^2 +7)
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    (25) the line y = kx - 3 is perpendicular to the line 2y + 3x = 7. The value of k is
    (a) -1/3 (b) -2/3 (c) 3/2
    (d) 2/3

    ans= (d) 2/3
    solu[hr][/hr]
    y = mx + c
    y = kx - 3 .......(1)
    m1 = k

    2y + 3x = 7
    2y = 7 - 3x
    y = (-3/2)x + 7/2.......(2)
    m2 = -3/2

    gradient of 2 perpendicular line
    m1 x m2 = -1
    k x -3/2 = -1
    -3k = -2
    k = 2/3
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    Network iz pretty bad from my end.
    So thats 16 - 25
    continue later.
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    (26) the midpoint of the line segment joining (-1, 3) and (5, 7) is
    (a) (3, 5)
    (b) (3, 2)
    (c) (2, 5)
    (d) (1, 6)

    ans= (c) (2, 5)
    solu[hr][/hr]
    (-1, 3)= (x1, y1)
    (5, 7)= (x2, y2)
    midpoint=
    ((x1 + x2)/2), ((y1 + y2)/2)
    ((-1 + 5)/2), ((3 + 7)/2
    ((2)), ((5))
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    (27) the solution of the inequality x^2 + 3x - 10 < 0 is
    (a) -2 < x < 5
    (b) x < -5 or x > 2
    (c) 2 < x < 5
    (d) -5 < x < 5

    ans= not any in the option
    solu[hr][/hr]
    x^2 + 3x - 10 < 0
    x^2 + 5x - 2x - 10 < 0
    x(x + 5) - 2(x + 5) < 0
    (x + 5)(x - 2) < 0
    (x < -5) (x < 2)
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    (27) the solution of the inequality x^2 + 3x - 10 < 0 is
    (a) -2 < x < 5
    (b) x < -5 or x > 2
    (c) 2 < x < 5
    (d) -5 < x < 5

    ans= not any in the option
    solu[hr][/hr]
    x^2 + 3x - 10 < 0
    x^2 + 5x - 2x - 10 < 0
    x(x + 5) - 2(x + 5) < 0
    (x + 5)(x - 2) < 0
    (x < -5) (x < 2)
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    (28) A binary operation is defined by a * b = a + b -3.
    The identity is
    (a) 3 (b) -3 (c) 1 (d) 0

    ans= (a) 3
    solu[hr][/hr]
    change b to e.
    a * b = a + b -3
    a * e = a + e -3
    note that for identity
    a * e = a
    so a * e = a + e -3
    a = a + e - 3
    e = a - a + 3
    e = 3
    • Guru Member

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    (28) A binary operation is defined by a * b = a + b -3.
    The identity is
    (a) 3 (b) -3 (c) 1 (d) 0

    ans= (a) 3
    solu[hr][/hr]
    change b to e.
    a * b = a + b -3
    a * e = a + e -3
    note that for identity
    a * e = a
    so a * e = a + e -3
    a = a + e - 3
    e = a - a + 3
    e = 3
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    (29) A binary operation is defined by a * y = xy - x + y.
    The value of (3 * 4) * 5 is
    (a) 81 (b) 61 (c) 57 (d) 73

    ans= (c) 57
    solu[hr][/hr]
    a * y = xy - x + y
    (3 * 4)
    3 * 4 = (3 x 4) - 3 + 4
    (3 * 4) = 12 + 1 = 13
    (3 * 4) = 13

    (3 * 4) * 5
    a * y = xy - x + y
    13 * 5 = (13 x 5) - 13 + 5
    65 - 13 + 5
    = 57
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    (30) Find the difference between the mean & the median of the numbers 1, 2, 3, 4, 5, 7, 8, 9 & 10
    (a) 0 (b) 1/9 (c) 5 (d) 4/9

    ans= (d) 4/9
    solu[hr][/hr]
    mean = (49/9)
    median = 5
    mean - median
    (49/9) - 5
    = (49 - 45)/9
    = 4/9
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    (31) there are eight men & nine women on a committee. In how many ways can a subcommittee of two men & three women be chosen?
    (a) 2,352 (b) 112 (c) 6,188 (d) 28,224

    ans= (a) 2,352
    solu[hr][/hr]
    men = 8 select= 2 men
    women = 9 select= 3 women
    nCr = n!/(n-r)r!
    = 2352
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    (32) change 671 base nine to base 8.
    (a) 550 base eight
    (b) 540 base eight
    (c) 671 base eight
    (d) 1046 base eight.

    Ans= (d)
    solu[hr][/hr]
    convert to 10.
    6 x 9^2 + 7 x 9^1 + 1
    = 486 + 63 + 1
    = 550 base 10.
    Now to base 8.
    We divide.
    8|550|R
    8|  68|6
    8|    8|4
    8|    1|0
      |    0|1
    = 1046 base 8
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    (33) write 14/(√5 -√3) in the form a√5 + b√3. Where a & b are rational.
    (a) 7 (b) 7√5 + 5√3
    (c) 7√5 - 7√3
    (d) 7√3 + 7√5

    ans= (c) is close to the answer.
    Sol[hr][/hr]
    (14/(√5 - √3)) x (√5 + √3)/(√5 + √3)

    = 14√5 + 14√3/2
    = 7√5 + 7√3
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    (34) In the relation
    log base x (y) = z, write x in terms of y & z.
    (a) x = z^y (b) x = z^y
    (c) x = z^(1/y) (d) x = z^(1/y)

    ans= nil
    sol[hr][/hr]
    log base x (y) = z
    x^z = y
    x^(z x 1/z) = y^(1/z)
    x = y^(1/z)
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    (35) solve the equation
    √(x + 7) = x - 5
    (a) 9 (b) 5, -7 (c) 2 (d) 2, 9

    ans= (d)

    x + 7 = (x - 5)^2
    x + 7 = x^2 - 10x + 25
    x^2 - x - 10x + 25 - 7 = 0
    x^2 - 11x + 18 = 0
    x^2 - 9x - 2x + 18 = 0
    x(x - 9) - 2(x - 9)
    (x - 9)(x - 2)
    x = 9
    x = 2
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    (35) solve the equation
    √(x + 7) = x - 5
    (a) 9 (b) 5, -7 (c) 2 (d) 2, 9

    ans= (d)

    x + 7 = (x - 5)^2
    x + 7 = x^2 - 10x + 25
    x^2 - x - 10x + 25 - 7 = 0
    x^2 - 11x + 18 = 0
    x^2 - 9x - 2x + 18 = 0
    x(x - 9) - 2(x - 9)
    (x - 9)(x - 2)
    x = 9
    x = 2
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    (35) solve the equation
    √(x + 7) = x - 5
    (a) 9 (b) 5, -7 (c) 2 (d) 2, 9

    ans= (d)

    x + 7 = (x - 5)^2
    x + 7 = x^2 - 10x + 25
    x^2 - x - 10x + 25 - 7 = 0
    x^2 - 11x + 18 = 0
    x^2 - 9x - 2x + 18 = 0
    x(x - 9) - 2(x - 9)
    (x - 9)(x - 2)
    x = 9
    x = 2
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    (36) let y = (4x + 3)/(2x - 5) write x as a function of y.
    (a) x = (2y - 3)/(5y + 3)
    (b) x = (5y - 3)/(2y + 4)
    (c) x = (5y + 3)/(2y - 4)
    (d) x = (5y - 3)(2y - 4)

    ans= (c)
    sol[hr][/hr]
    y(2x - 5) = 4x + 3
    2xy - 5y = 4x + 3
    2xy - 4x = 3 + 5y
    x(2x - 4) = 5y + 3
    x = (5y + 3)/(2y - 4)
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    (37) the second & fifth terms of a geometric progression are 6 & -48, respectively. The first term is
    (a) -3 (b) 3 (c) 12 (d) -12

    ans= (a)
    sol[hr][/hr]
    u2 = ar = 6.....(1)
    u5 = ar^4 = -48....(2)
    divide (2) by (1)
    a(r^4)/ar = -48/6
    r^3 = -8
    r = 3√(-8)
    r = -2.....(3)
    put (3) into (1)
    ar = 6
    a(-2) = 6
    a = 6/(-2)
    a = - 3
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    (38) find the positive solution of the equation
    log (x + 1) + log (x + 4) = 1
    (a) 6 (b) 0 (c) 2 (d) 1

    ans= (d)
    sol[hr][/hr]
    we arent given the bases so definitely we knw we are working in base 10.
    Also note that
    log base10 (10) = 1

    log base10 (x + 1) + log base10 (x + 4) = log base10 (10)

    log base10 (x + 1)(x + 4) = log base10 (10)

    cancel out the log base10
    (x + 1)(x + 4) = 10

    at this point, to save time try out the options.
    E.g (d) 1.
    (1 + 1)(1 + 4) = 10.
    2 x 5 = 10
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    (39) N72000 is invested at 8% simple interest. After how many years has it reached N87840?
    (a) 2whole 3/4years
    (b) 2 years
    (c) 3 years
    (d) 2whole 1/2years

    ans= (a)
    sol[hr][/hr]
    S.I = P x R x T
    s.i = N87840 - N72000
    s.i = N15840

    15840 = 72000 x 8% x T
    T = 15840/5760
    T = 2.75years.
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    • GL Legend
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    ►►♫O.MICHAEL.O♫◄◄ GL Legend

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