1993 - WAEC Mathematics Past Questions and Answers - page 2

\(\frac{5}{6r} - \frac{3}{4r}\)

= \(\frac{10 - 9}{12r}\)

= \(\frac{1}{12r}\)

2x\(^2\) - 21x + 45

= 2x\(^2\) - 15x - 6x + 45

= x(2x - 15) - 3(2x - 15)

= (x - 3)(2x - 15)

y = 3x  ... (i);

4y - 5x =14  ... (ii)

Put 3x for y in (ii).

4(3x) - 5x = 14

12x - 5x = 14

7x = 14 \(\implies\) x = 2.

y = 3x = 3(2) = 6

(x, y) = (2, 6)

Area of a sector = θ/360 x πr\(^2\)

= 120/360 x 22/7 x 81/1

= 84.86 cm\(^2\)

\(\approxeq\) 85cm\(^2\) (to the nearest cm)

πrl + πr² = πr(l + r)
²²/₇ x 5(13 + 5)
282 6/7cm²

Volume of a cone = \(\frac{1}{3} \pi r^2 h\)

r = 4\(\frac{1}{2}\) cm; h = 14 cm

Volume of cone = \(\frac{1}{3} \times \frac{22}{7} \times \frac{9}{2} \times \frac{9}{2} \times 14\)

= 297 cm\(^3\)

When half- filled, the volume of the water = \(\frac{297}{2} = 148.5 cm^3\)

Area of rhombus = \(\frac{pq}{2}\)

where p and q are the two diagonals of the rhombus.

\(\therefore 60 = \frac{12 \times q}{2}\)

6q = 60 \(\implies\) q = 10 cm

Perimeter of a sector = \(2r + \frac{\theta}{360} \times 2\pi r\)

= \(2(10.5) + \frac{120}{360} \times 2 \times \frac{22}{7} \times 10.5\)

= \(21 + 22\)

= 43 cm

< STQ = < SQR = 68° (alternate segment)

\(\therefore\) < STQ = 68° 

< TQS = \(\frac{180° - 68°}{2}\)

= \(\frac{112}{2} = 56°\)

\(\therefore\) < PQT = 180° - (68° + 56°)

= 180° - 124°

= 56°

Sum of interior angles in a polygon = \((2n - 4) \times 90°\)

\(\therefore (2n - 4) \times 90° = 30 \times 90°\)

\(\implies 2n - 4 = 30 \)

\(2n = 34 \implies n = 17\)

The polygon has 17 sides.