1990 - WAEC Mathematics Past Questions and Answers - page 3

7y\(^2\) = 3y

7y\(^2\) - 3y = 0

y(7y - 3) = 0

y = 0 or y = \(\frac{3}{7}\)

x\(^2\) + 8 + m

To make it a perfect square, we add \((\frac{b}{2})^2\)

= \((\frac{8}{2})^2\)

= 16

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 = \(\frac{9}{2}\)

πr² = 360^o

Surface area of a sphere = \(4\pi r^2\)

= \(4 \times \frac{22}{7} \times 7 \times 7\)

= \(616 cm^2\)

\(\theta_{1} = \theta_{2} = 30°\)

TSA of a cylinder = \(2\pi r^2 + 2\pi rh\)

= \(2\pi r (r + h)\)

= \(2 \times \frac{22}{7} \times 7 (7 + 5)\)

= \(44 \times 12\)

= \(528 cm^2\)

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

= \(\frac{22}{7} \times 7^2 \times 5\)

= \(770 cm^3\)

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°

Diagonals bisect at 90°;  < YXZ = 90° - 20° = 70°

But ZY = XY (sides of a rhombus)

\(\therefore\) < XYZ = 70° (base angle of an isos. triangle)