1999 - WAEC Mathematics Past Questions and Answers - page 3

< ABC = 40° (alternate angles)

\(\therefore\) < ACB = \(\frac{180° - 40°}{2}\)

= 70°

\(\therefore\) Bearing of A from C = 360° - 70° 

= 290°

PR = 4 + 4 = 8 cm

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

= \(\frac{1}{2} (4 + 8) \times 5\)

= 30 cm\(^2\)

Volume = \(\frac{1}{3} b^2 h\)

\(\therefore b = \sqrt{\frac{3v}{h}}\)

\(b = \sqrt{\frac{3(80)}{15}} \)

\(b = \sqrt{16} = 4.0 cm\)

24\(_x\) = 31\(_y\)

\(2 \times x^1 + 4 \times x^0 = 3 \times y^1 + 1 \times y^0\)

\(2x + 4 = 3y + 1 \implies 2x = 3y + 1 - 4\)

\(2x = 3y - 3 = 3(y - 1)\)

Travelled x km/h for 1 hour \(\therefore\) traveled x km in the first hour.

Traveled y km/h for 2 hours \(\therefore\) traveled 2y km in the next 2 hours.

Average speed = \(\frac{x + 2y}{1 + 2}\)

= \(\frac{x + 2y}{3} kmh^{-1}\)

Area = \(\frac{1}{2} \times base \times height\)

\(height : base = 1 : 3\)

\(\implies base = 3 \times height\)

Let height = h;

Area = \(\frac{1}{2} \times 3h \times h = 216\)

\(3h^2 = 216 \times 2 = 432\)

\(h^2 = \frac{432}{3} = 144\)

\(h = \sqrt{144} = 12.0 cm\)

\(\therefore base = 3 \times 12 = 36 cm\)

Volume of water after 9 seconds = \(5\pi \times 9 = 45\pi cm^3\)

Volume of cylinder = \(\pi r^2 h\)

\(\therefore \pi r^2 h = 45\pi\)

\(\pi \times 3^2 \times h = 45\pi\)

\(\implies 9h = 45 \)

\(h = 5 cm\)

(where h = height of the water after 9 secs)