2006 - JAMB Mathematics Past Questions and Answers - page 3

(\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))

110011₂ + 11111₂ = 1010010₂

25)^-1/₂ x (27)^1/₃ + (121)^-1/₂ x (625)^-1/₄

5^2x-1/₂ x 3^3x1/₃ + 11^2x-1/₂ x 5^4x-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

1^st 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

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

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

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

∠yxz = 180 - (45 + 15)

= 180 - 60

= 120^o

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))

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 = 45^o

∠PSQ = 90^o (∠ in semi circle)

∠PSQ + ∠QSR = 145^o

90 + ∠QSR = 145

∠QSR = 145 - 90

∠QSR = 55^o

BUt x^o = ∠QSR (∠s in the same segment)

∴x^o = 55^o