2002 - WAEC Mathematics Past Questions and Answers - page 1

Given, Length of minor arc = 5.5m

Angle subtended by minor arc = 360° - 210° = 150°

$\therefore 5.5=\frac{150}{360}\times 2\times \frac{22}{7}\times r$

$\frac{55r}{21}=5.5$

$r=\frac{5.5\times 21}{55}$

r = 2.1m

Length of major arc = $\frac{210}{360}\times 2\times \frac{22}{7}\times 2.1$

= $7.7m\approxeq 8m$ (to the nearest metre)

$B{D}^2=4^2+4^2$

$BD=\sqrt{16+16}=\sqrt{32}$

$BD=4\sqrt{2}cm$

$(2\sqrt{3}{)}^2=(2\sqrt{2}{)}^2+h^2$

$h^2=12-8=4$

$h=\sqrt{4}=2cm$

Volume of pyramid = $\frac{a^{2}h}{3}$

= $\frac{4^2\times 2}{3}$

= $\frac{32}{3}=10\frac{2}{3}c{m}^3$

Curved surface area or a cone $=\pi rl$
from the information $l^2=4^2+3^2=16+9l=\sqrt{25}=5;\therefore C.S.A=\frac{22}{7}\times 3\times 5Since\frac{22}{7}=\pi \therefore C.S.A=15\pi$

< UQR = 29° (alternate angles)

< PQR = < PQU + < UQR

= 36° + 29°

= 65°

$5\frac{1}{4}÷(1\frac{2}{3}-\frac{1}{2})\frac{21}{4}÷\left(1\frac{4-3}{6}\right)\frac{21}{4}÷\left(1\frac{1}{6}\right)\frac{21}{4}\times \frac{6}{7}=4\frac{1}{2}$

(0.5x + 2.6 = 5x + 0.35<br /> 0.5x - 5x = 0.35-2.6<br /> -4.5x = -2.25<br /> x = \frac{-2.25}{-4.5}<br /> 0.5)

Sum of exterior angle of any polygon is 360^o
(2x+5)^o + 2x^o + (x-20)^o + x^o + (3x+10)^o + (x + 15)^o = 360^o; 10x = 350
x = 35

(M5_{ten} = 1001011_{two}<br /> =1 \times 2^6 + 0\times 2^5 + 0\times 2^4 + 1\times 2^3 + 0\times 2^2 + 1\times 2^1 <br /> =64+8+2+1=75_{ten}<br /> ∴ m = 7 )