1990 - JAMB Mathematics Past Questions and Answers - page 3

Let his assistant work for x days

∴ his master worked (x + 10) day. Amount received by master = 40(x + 10),

amount got by his assistance = 10x

Total amount collected = N2000.00

∴ 40(x + 10) + 10x = 2000

= 40x + 400 + 10x

= 2000

50x + 400 = 2000

50x = 2000 - 400

50x = 1600

x = (\frac{1600}{50})

x = 32 days

(\frac{x}{x + y}) + (\frac{y}{x - y}) - (\frac{x^2}{x^2 - y^2})

(\frac{x}{x + y}) + (\frac{y}{x - y}) - (\frac{x^2}{(x + y)(x - y})

= (\frac{x(x - y) + y(x + y) - x^2}{(x + y)(x - y})

= (\frac{x^2 + xy + xy + y^2 - x^2}{(x + y)(x - y})

= (\frac{y^2}{(x + y)(x - y)})

= (\frac{y^2}{(x^2 - y^2)})

Given that x² + y² + z² = 194, calculate z if x = 7 and (\sqrt{y}) = 3

x = 7

∴ x² = 49

(\sqrt{y}) = 3

∴ y² = 81 = x² + y² + z² = 194

49 + 81 + z² = 194

130 + z² = 194

z² = 194 - 130

= 64

z = (\sqrt{64})

= 8

3 + 6 + 12 + .....18thy term

1st term = 3, common ratio (\frac{6}{3}) = 2

n = 18, sum og GP is given by Sn = a(\frac{(r^n - 1)}{r - 1})

s¹⁸ = 3(\frac{(2^{18} - 1)}{2 - 1})

= 3(2¹⁷ - 1)

x + x + 1

(\frac{dy}{dx}) = 2x + 1

At the turning point, (\frac{dy}{dx}) = 0

2x + 1 = 0

x = -(\frac{1}{2})

(\frac{d^2y}{dx^2}) = 2 > 0(min Pt)

= (\frac{1}{4}) + (\frac{1}{2})

= 1

Length of Arc AB = (\frac{\theta}{360}) 2(\pi)r

= (\frac{48}{360}) x 2(\frac{22}{7}) x (\frac{21}{2})

= (\frac{4 \times 22 \times \times 3}{30}) (\frac{88}{10}) = 8.8cm

Perimeter = 8.8 + 2r

= 8.8 + 2(10.5)

= 8.8 + 21

= 29.8cm

For a regular polygon of n sides

n = (\frac{360}{\text{Exterior angle}})

Exterior < = 180^o - 140^o

= 40^o

n = (\frac{360}{40})

= 9 sides

Cos (\theta) = (\frac{12}{13})

x² + 12² = 13²

x² = 169- 144 = 25

x = 25

= 5

Hence, tan(\theta) = (\frac{5}{12}) and cos(\theta) = (\frac{12}{13})

If cos²(\theta) = 1 + (\frac{1}{tan^2\theta})

= 1 + (\frac{1}{\frac{(5)^2}{12}})

= 1 + (\frac{1}{\frac{25}{144}})

= 1 + (\frac{144}{25})

= (\frac{25 + 144}{25})

= (\frac{169}{25})