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Question 1 Mathematics 2010 | JAMB/UTME (2010)
A
1/3
B
C
D
1/6
Explanation.
$$\left(\frac{81}{16}\right)^{\frac{-1}{4}}\times 2^{-1}=\frac{1}{\left(\frac{81}{16}\right)^{\frac{1}{4}}}\times \frac{1}{2}\\ =\left(\frac{16}{81}\right)^{\frac{1}{4}}\times \frac{1}{2}\\ =\left(\frac{2}{3}\right)^{4\times \frac{1}{4}}\\ =\frac{2}{3}\times \frac{1}{2}\\ =\frac{1}{3}$$

Question 2 Mathematics 2010 | JAMB/UTME (2010)
A
(3 √ 15 +11) / 2
B
3 √ 15 -11
C
(3 √ 15 -11) / 2
D
3 √ 15 +11
Explanation.
 23+5 5-3

=2+5
 x 5+3 5+3
5-3

 = 23(5+3)+ 5(5+3) ____(5)2-(3)2

 = 215+2(3) +5+15 _____5 - 3

 = 215+6+5+15 ______2

 = 11+315 ___2

Question 3 Mathematics 2010 | JAMB/UTME (2010)
A
7.14 x 10-3
B
7.14 x 10-2
C
7.14 x 10-1
D
7.14 x 10-4
Explanation.
0.21 x 0.34 = 0.0714
= 7.14 x 10-2

Question 4 Mathematics 2010 | JAMB/UTME (2010)
A
10
B
C
15
D
20
Explanation.
Let C n G = X
n C only = 40 - x
n G only = 30 - x
40 - X + 30 - X = 50
70 - X = 50
X = 70 - 50
X = 20

Question 5 Mathematics 2010 | JAMB/UTME (2010)
A
 x(x-5) 2(x+2)

B
 x(x-5) 2(x-2)

C
 x(x+5) 2(x+2)

D
 x2+5 2x+4

Explanation.
 x2 + 3x2 2x - 8

 = x(x2+3x10) 2(x2-4)

 = x(x+5)(x-2) 2(x2-2 2)

 = x(x+5)(x-2) 2(x-2)(x+2)

 = x(x+5) 2(x+2)

Question 6 Mathematics 2010 | JAMB/UTME (2010)
A
12
B
C
D
Explanation.
y x
y = Kx
K = y/x = 3/16
= 3/4
y = 3/4x
= 3/464 when x = 64
= 3/4 x 8/1
= 6

Question 7 Mathematics 2010 | JAMB/UTME (2010)
A
36
B
25
C
11
D
55
Explanation.
x * y = x + y2
2 * 3 = 2 + 32
= 2 + 9
= 11
(2 * 3) * 5 = 11 + 52
= 11 + 25
= 36

Question 8 Mathematics 2010 | JAMB/UTME (2010)
A
q = 18
B
p <1
C
p = 18
D
q < 1
Explanation.
Using substitution If 18(p+18) = (18 + p)q
let q = 18
imply 18(p+18) = (18+p)18
i.e 18p + 18 x 18 = 18 x 18 + 18p
Since the left hand side = the right hand side
imply that q = 18

Question 9 Mathematics 2010 | JAMB/UTME (2010)
A
B
C
D
Explanation.
$$\left[\begin{array}{cc} x & 3\\ 2 & 7\end{array}\right]$$= 15
x x 7 - 2 x 3 = 15
7x - 6 = 15
7x = 15+6
7x = 21
x = 21/7
x = 3

Question 10 Mathematics 2010 | JAMB/UTME (2010)
A
65o
B
50o
C
75o
D
55o
Explanation.
RTS = 90o ( in semi circle)
UTS = 25o ( in alternate segment)
X + 90 + 25 = 180(sum of s on a str line)
X + 115 = 180
X = 180 - 115
X = 65o

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