JAMB Mathematics Past Questions & Answers - Page 204

Total students = 120

grade = 18

Z^o = \(\frac{18}{120}\) x \(\frac{360}{1}\)

= 54^o

Length of an area = \(\frac{\theta}{360}\) x 2\(\pi\)r

Longitude difference = 147 + 153 = 30N^oN

distance between xy = \(\frac{\theta}{360}\) x 2\(\pi\)R cos60^o

= \(\frac{300}{360}\) x 4 x 10⁴ x \(\frac{1}{2}\)

= 1.667 x 104km(1667 km)

P(R¹) = \(\frac{6}{10}\)

= \(\frac{2}{3}\)

P(R¹ n R¹¹) = P(both red)

\(\frac{3}{5}\) x \(\frac{5}{9}\)

= \(\frac{1}{3}\)

Area of trapezium = 14cm²

Area of trapezium = \(\frac{1}{2}\)(a + b)h

14 = \(\frac{1}{2}\)(4 + 3)h

14 = \(\frac{7}{2}\)h

h = \(\frac{14 \times 2}{7}\)

= 4

Area of SQR = \(\frac{1}{2}\)(3 x 4)

= \(\frac{12}{2}\)

= 6.0

\(\frac{1\sqrt{3}}{(\frac{1}{2})^2}\)

= \(\frac{4}{\sqrt{3}}\)

= \(\frac{\sqrt{3}}{\sqrt{3}}\)

= \(\frac{4\sqrt{3}}{\sqrt{3}}\)

m = 64kg, V = \(\frac{4\pi r^3}{3}\)

= \(\frac{4\pi(4)^3}{3}\)

= \(\frac{256\pi}{3}\) x 10^-6m³

density(P) = \(\frac{\text{Mass}}{\text{Volume}}\)

= \(\frac{64}{\frac{256\pi}{3 \times 10^{-6}}}\)

= \(\frac{64 \times 3 \times 10^{-6}}{256}\)

= \(\frac{3}{4 \times 10^{-6}}\)

m = PV = \(\frac{3}{4 \pi \times 10^{-6}}\) x \(\frac{4}{3}\) \(\pi\)[3² - 2²] x 10^-6

\(\frac{3}{4 \times 10^{-6}}\) x \(\frac{4}{3}\) x 5 x 10^-6

= 5kg