WAEC Mathematics Past Questions & Answers - Page 111

Let 0 = angle of the minor sector

angle of the major sector = 360 - $\theta$(angle at a point)

2 $\pi r$ = 54 + 126(i.e circumference of minor and major arc)

2$\pi r = 180^o$

r = $\frac{180}{2\pi}$ = $\frac{90}{\pi}$

Lenght of ninor arc

= $\frac{\theta}{360} \times 2 \pi r$

54 = $\frac{\theta}{360} \times 3 \pi r$

$\theta = \frac{360 \times 54}{2 \pi r}$

but r = $\frac{90}{\pi}$ substituting $\frac{90}{\pi}$ for r

$\theta = \frac{360 \times 54 \times \pi}{2 \times \pi \times 90}$

$\theta = 2 \times 54 = 108^o$

angle of the major sector = 360 - 108^o

= 252^o

Curved surface area = 2$\pi h$

704 = 2 x $\frac{22}{7} \times 8 \times h$

704 = $\frac{352}{7}$h

352h = 704 x 7

h = $\frac{704 \times 7}{352}$

= $\frac{4928}{352}$

h = 14cm

$\frac{-m}{2} - \frac{5}{4} \geq \frac{5m}{12} - \frac{7}{6}$

= $\frac{-2m - 5}{4} \geq \frac{5m - 14}{12}$

12(-2m - 5) $\geq$ 4(5m - 14)

-24m - 60 $\geq$ 20m - 56

-24m - 20m $\geq$ -56 + 60

44m $\geq$ 4

m $\leq \frac{4}{-44}$

m $\leq \frac{-1}{11}$

12 = $\frac{n}{3} - 2n = 1$, multiply through by 3

36 + n - 6n = 3

-5n = 3 - 36

-5n = -33

-5n + 33 = 0

5n - 33 = 0

x + y = 2y - x + 1 = 5

x + y = 2y - x + 1

x + x + y - 2y = 1

2x - y = 1....(i)

2y - x + 1 = 5

-x + 2y = 5 + 1

-x = 2y = 4

x - 2y = -4 .....(ii)

solve simultaneously (i) x 2x - y = 1

(ii) x x - 2y = -4

2x - y = 1

=2x - 4y = -8

3y = 9

y = $\frac{9}{3}$

y = 3

substitute y = 3 into equation (i)

2x - y = 1

2x - 3 = 1

2x = 1 + 3

2x = 4

x = $\frac{4}{2}$

= 2