# UNILAG POST UTME 2011 UPDATES

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

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### PROFICEMUGâ„¢Guru Member

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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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### PROFICEMUGâ„¢Guru Member

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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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### PROFICEMUGâ„¢Guru Member

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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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### PROFICEMUGâ„¢Guru Member

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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
65 + X = 180.
X = 180 - 65.
X = 115.
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### PROFICEMUGâ„¢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
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### PROFICEMUGâ„¢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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### PROFICEMUGâ„¢Guru Member

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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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### PROFICEMUGâ„¢Guru Member

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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

m1 x m2 = -1
k x -3/2 = -1
-3k = -2
k = 2/3
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### PROFICEMUGâ„¢Guru Member

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Network iz pretty bad from my end.
So thats 16 - 25
continue later.
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### PROFICEMUGâ„¢Guru Member

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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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### PROFICEMUGâ„¢Guru Member

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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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### PROFICEMUGâ„¢Guru Member

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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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### PROFICEMUGâ„¢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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### PROFICEMUGâ„¢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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### PROFICEMUGâ„¢Guru Member

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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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### PROFICEMUGâ„¢Guru Member

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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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### PROFICEMUGâ„¢Guru Member

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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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### PROFICEMUGâ„¢Guru Member

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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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### PROFICEMUGâ„¢Guru Member

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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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### PROFICEMUGâ„¢Guru Member

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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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### PROFICEMUGâ„¢Guru Member

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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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### PROFICEMUGâ„¢Guru Member

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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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### PROFICEMUGâ„¢Guru Member

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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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### PROFICEMUGâ„¢Guru Member

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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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### PROFICEMUGâ„¢Guru Member

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Good night. :)
• GL Legend
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### ►►♫O.MICHAEL.O♫◄◄GL Legend

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