2006 - JAMB Mathematics Past Questions and Answers - page 3
21
Evaluate \(\frac{2}{6-5\sqrt{3}}\)
A
\(-\left(\frac{12}{39}-\frac{10\sqrt{3}}{39}\right)\)
B
\(\frac{12}{39}-\frac{10\sqrt{3}}{39}\)
C
\(-\left(\frac{12}{39}+\frac{10\sqrt{3}}{39}\right)\)
D
\(\frac{12}{39}+\frac{10\sqrt{3}}{39}\)
correct option: c
\(\frac{2}{6-5\sqrt{3}} = \frac{2}{6-5\sqrt{3}} \times \frac{6+5\sqrt{3}}{6+5\sqrt{3}}\\
=\frac{2(6+5\sqrt{3})}{(6-5\sqrt{3})(6+5\sqrt{3})}\\
=\frac{12+10-\sqrt{3}}{36-25(3)}\\
=\frac{12+10-\sqrt{3}}{36-75}\\
=\frac{12+10-\sqrt{3}}{-39}\\
=-\left(\frac{12}{39}-\frac{10\sqrt{3}}{39}\right)\)
Users' Answers & Comments22
Compute 1100112 + 111112
A
10010102
B
10100102
C
10001102
D
10001002
23
Simplify (25)-1/2 x (27)1/3 + (121)-1/2 x (625)-1/4
A
34/55
B
9/11
C
14/5
D
3/275
correct option: a
25)-1/2 x (27)1/3 + (121)-1/2 x (625)-1/4
52x-1/2 x 33x1/3 + 112x-1/2 x 54x-1/4
5-1 x 31 x 11-1 x 5-1
1/5 x 3/1 + 1/11 x 1/5
3/5 + 1/5 = 33+1/55
=34/55
Users' Answers & Comments52x-1/2 x 33x1/3 + 112x-1/2 x 54x-1/4
5-1 x 31 x 11-1 x 5-1
1/5 x 3/1 + 1/11 x 1/5
3/5 + 1/5 = 33+1/55
=34/55
24
Convert 22324 to base six
A
4506
B
2546
C
5536
D
5406
correct option: a
1st convert to base 10
22324 = 2x43 + 2x42 + 3x41 + 2x40
= 2x64 + 2x16 + 3x4 + 2x1
= 128 + 32 + 12 + 2
= 174 convert to base 6
6/174
6/29 R 0
6/4 R 5
6/0 R 5
= 4506
Users' Answers & Comments22324 = 2x43 + 2x42 + 3x41 + 2x40
= 2x64 + 2x16 + 3x4 + 2x1
= 128 + 32 + 12 + 2
= 174 convert to base 6
6/174
6/29 R 0
6/4 R 5
6/0 R 5
= 4506
25
A
89 cm2
B
70 cm2
C
90 cm2
D
80 cm2
correct option: c
Area of rectangle PQRS
= 10 x 7 = 70 cm2
Area of semi circle
QR = Dπ = 7 x 22/7 = 22cm2
∴Area of the figure = 70 + 22
= 92cm2
= 90cm2
Users' Answers & Comments= 10 x 7 = 70 cm2
Area of semi circle
QR = Dπ = 7 x 22/7 = 22cm2
∴Area of the figure = 70 + 22
= 92cm2
= 90cm2
26
If tan θ = 5/4, find sin2θ - cos2θ.
A
5/4
B
41/9
C
9/41
D
1
correct option: c
Tan θ = 5/4
x2 = 52 + 42
= 25 + 16
= 41
x = √41
sin2θ - cos2θ = (5/√41)2 - (4/√41)2
= 25/41 - 16/41
= 9/41
Users' Answers & Commentsx2 = 52 + 42
= 25 + 16
= 41
x = √41
sin2θ - cos2θ = (5/√41)2 - (4/√41)2
= 25/41 - 16/41
= 9/41
27
PQ and RS are two parallel lines. If the coordinates of P, Q, R, S are (1,q), (3,2), (3,4), (5,2q) respectively, find the value of q
A
3
B
4
C
1
D
2
correct option: d
Gradient PQ, P(1,q) and Q(3,2)
\(=\frac{(2-q)}{(3-1)} = \frac{(2-q)}{2}\)
Gradient of RS : R(3,4) and S(5,2q)
\(= \frac{(2q-4)}{(5-3)}= \frac{(2q-4)}{2} = \frac{2(q-2)}{2}\)
= q-2
Since PQ and RS are parallel,
their generation are equal
\(∴ \frac{(2-q)}{2} = q-2\)
2-q = 2(q-2)
2-q = 2q-4
2+4 = 2q+q
6 = 3q
q = 2
Users' Answers & Comments\(=\frac{(2-q)}{(3-1)} = \frac{(2-q)}{2}\)
Gradient of RS : R(3,4) and S(5,2q)
\(= \frac{(2q-4)}{(5-3)}= \frac{(2q-4)}{2} = \frac{2(q-2)}{2}\)
= q-2
Since PQ and RS are parallel,
their generation are equal
\(∴ \frac{(2-q)}{2} = q-2\)
2-q = 2(q-2)
2-q = 2q-4
2+4 = 2q+q
6 = 3q
q = 2
28
In triangle XYZ, ∠XYZ = 15o, ∠XZY = 45o and lXYl = 7 cm. Find lYZl.
A
14√2 cm
B
\(7\left(\frac{\sqrt{6}}{2}\right)\)
C
7√2 cm
D
7 cm
correct option: b
∠yxz = 180 - (45 + 15)
= 180 - 60
= 120o
Using sine rule
\(\frac{x}{sinx}=\frac{7}{sinz}\\ \frac{x}{sin 120}=\frac{7}{sin 45}\\ x=\frac{7 sin 120}{sin45}\\ x=\frac{7sin(180-120)}{sin 45}\\ x=\frac{7 sin 60}{sin 45}=\left(7\left(\frac{\sqrt{3}}{2}\right)\div \frac{1}{\sqrt{2}}\right)\\ x =\left(7\left(\frac{\sqrt{3}}{2}\right)\div \frac{1}{\sqrt{2}}\right)\\ x = 7\left(\frac{\sqrt{6}}{2}\right)\)
Users' Answers & Comments= 180 - 60
= 120o
Using sine rule
\(\frac{x}{sinx}=\frac{7}{sinz}\\ \frac{x}{sin 120}=\frac{7}{sin 45}\\ x=\frac{7 sin 120}{sin45}\\ x=\frac{7sin(180-120)}{sin 45}\\ x=\frac{7 sin 60}{sin 45}=\left(7\left(\frac{\sqrt{3}}{2}\right)\div \frac{1}{\sqrt{2}}\right)\\ x =\left(7\left(\frac{\sqrt{3}}{2}\right)\div \frac{1}{\sqrt{2}}\right)\\ x = 7\left(\frac{\sqrt{6}}{2}\right)\)
29
A
55o
B
50o
C
45o
D
40o
correct option: c
y = 55 (alternate angles)
x = x (same reason)
260 + y + x = 360 (∠s at a point)
260 + 55 + x = 360
315 + x = 360
x = 360 - 315
x = 45o
Users' Answers & Commentsx = x (same reason)
260 + y + x = 360 (∠s at a point)
260 + 55 + x = 360
315 + x = 360
x = 360 - 315
x = 45o
30
A
55o
B
45o
C
35o
D
25o
correct option: a
∠PSQ = 90o (∠ in semi circle)
∠PSQ + ∠QSR = 145o
90 + ∠QSR = 145
∠QSR = 145 - 90
∠QSR = 55o
BUt xo = ∠QSR (∠s in the same segment)
∴xo = 55o
Users' Answers & Comments∠PSQ + ∠QSR = 145o
90 + ∠QSR = 145
∠QSR = 145 - 90
∠QSR = 55o
BUt xo = ∠QSR (∠s in the same segment)
∴xo = 55o