1999 - WAEC Mathematics Past Questions and Answers - page 3
The diagram shows the position of three ships A, B and C at sea. B is due north of C such that |AB| = |BC| and the bearing of B from A = 040°. What is the bearing of A from C
< ABC = 40° (alternate angles)
\(\therefore\) < ACB = \(\frac{180° - 40°}{2}\)
= 70°
\(\therefore\) Bearing of A from C = 360° - 70°
= 290°
When we rationalize gives \(\frac{16\sqrt{3}}{3}\)
In the diagram, PQST and QRST are parallelograms. Calculate the area of the trapezium PRST.
PR = 4 + 4 = 8 cm
Area of trapezium = \(\frac{1}{2} (a + b) h\)
= \(\frac{1}{2} (4 + 8) \times 5\)
= 30 cm\(^2\)
The height of a pyramid on square base is 15cm. If the volume is 80cm\(^3\), find the length of the side of the base
Volume = \(\frac{1}{3} b^2 h\)
\(\therefore b = \sqrt{\frac{3v}{h}}\)
\(b = \sqrt{\frac{3(80)}{15}} \)
\(b = \sqrt{16} = 4.0 cm\)
Two numbers 24\(_{x}\) and 31\(_y\) are equal in value when converted to base ten. Find the equation connecting x and y
24\(_x\) = 31\(_y\)
\(2 \times x^1 + 4 \times x^0 = 3 \times y^1 + 1 \times y^0\)
\(2x + 4 = 3y + 1 \implies 2x = 3y + 1 - 4\)
\(2x = 3y - 3 = 3(y - 1)\)
A car travel at x km per hour for 1 hour and at y km per hour for 2 hours. Find its average speed
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 = \(\frac{x + 2y}{1 + 2}\)
= \(\frac{x + 2y}{3} kmh^{-1}\)
The height and base of a triangle are in ratio 1:3 respectively. If the area of the triangle is 216 cm\(^2\), find the length of the base.
Area = \(\frac{1}{2} \times base \times height\)
\(height : base = 1 : 3\)
\(\implies base = 3 \times height\)
Let height = h;
Area = \(\frac{1}{2} \times 3h \times h = 216\)
\(3h^2 = 216 \times 2 = 432\)
\(h^2 = \frac{432}{3} = 144\)
\(h = \sqrt{144} = 12.0 cm\)
\(\therefore base = 3 \times 12 = 36 cm\)
Water flows from a tap into cylindrical container at the rate 5πcm\(^3\) per second. If the radius of the container is 3cm, calculate the level of water in the container at the end of 9 seconds.
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)