1990 - WAEC Mathematics Past Questions and Answers - page 3
Solve the equation 7y\(^2\) = 3y
7y\(^2\) = 3y
7y\(^2\) - 3y = 0
y(7y - 3) = 0
y = 0 or y = \(\frac{3}{7}\)
Find the value of m which makes x\(^2\) + 8 + m a perfect square
x\(^2\) + 8 + m
To make it a perfect square, we add \((\frac{b}{2})^2\)
= \((\frac{8}{2})^2\)
= 16
Solve the equation 2a\(^2\) - 3a - 27 = 0
2a\(^2\) - 3a - 27 = 0
2a\(^2\) - 9a + 6a - 27 = 0
a(2a - 9) + 3(2a - 9) = 0
(a + 3)(2a - 9) = 0
a = -3 or a = \(\frac{9}{2}\)
calculate the surface area of a sphere of radius 7cm [Take π = 22/7]
Surface area of a sphere = \(4\pi r^2\)
= \(4 \times \frac{22}{7} \times 7 \times 7\)
= \(616 cm^2\)
If the radius of the parallel of latitude 30°N is equal to the radius of the parallel of latitude θ°S, what is the value of θ?
A cylindrical container closed at both ends, has a radius of 7cm and height 5cm [Take π = 22/7]
Find the total surface area of the container
TSA of a cylinder = \(2\pi r^2 + 2\pi rh\)
= \(2\pi r (r + h)\)
= \(2 \times \frac{22}{7} \times 7 (7 + 5)\)
= \(44 \times 12\)
= \(528 cm^2\)
A cylindrical container closed at both ends, has a radius of 7cm and height 5cm [Take π = 22/7]. What is the volume of the container?
Volume = \(\pi r^2 h\)
= \(\frac{22}{7} \times 7^2 \times 5\)
= \(770 cm^3\)
In the diagram above, |PQ| = |PR| = |RS| and ∠RPS = 32°. Find the value of ∠QPR
From the figure, < PSR = 32° (base angles of an isos. triangle)
\(\therefore\) < PRS = 180° - (32° + 32°) = 116° (sum of angles in a triangle)
< QRP = 180° - 116° = 64° (angle on a straight line)
< PQR = 64° (base angles of an isos. triangle)
< QPR = 180° - (64° + 64°) = 52°
In the diagram above, WXYZ is a rhombus and ∠WYX = 20°. What is the value of ∠XZY
Diagonals bisect at 90°; < YXZ = 90° - 20° = 70°
But ZY = XY (sides of a rhombus)
\(\therefore\) < XYZ = 70° (base angle of an isos. triangle)