WAEC Mathematics Past Questions & Answers - Page 60

From the diagram,

tan 30$^o$ = $\frac{h}{27}$

h = 27 tan 30$^o$

= 27 x $\frac{1}{\sqrt{3}}$

= $\frac{27}{\sqrt{3}}$ x $\frac{\sqrt{3}}{\sqrt{3}}$

= $\frac{27 \sqrt{3}}{3}$

= 9$\sqrt{3m}$

0.612 = $\frac{0.612}{1}$ x $\frac{1000}{1000}$

= $\frac{612}{1000}$

= $\frac{153}{250}$

$\frac{x}{y}$ = $\frac{1}{4} \div \frac{3}{8}$ = $\frac{1}{4} \times \frac{8}{3}$ = $\frac{2}{3}$

$\frac{y}{z}$ = $\frac{1}{3} \div \frac{4}{9}$ = $\frac{1}{3} \times \frac{9}{4}$ = $\frac{3}{4}$

But,

x = $\frac{2}{5}T_1$, y = $\frac{3}{5}T_1$

y = $\frac{3}{7}T_2$, z = $\frac{4}{7}T_2$

Using y = y

$\frac{3}{5}T_1$ = x = $\frac{3}{7}T_2$

 $\frac{T_1}{T_2}$ = $\frac{3}{7}$ x $\frac{5}{3}$ = $\frac{15}{21}$

$T_1 = 15$ and $T_2 = 21$

Thus , x = $\frac{2}{5}$ x 15 = 6

y = $\frac{3}{5}$ x 15 = 9

y = $\frac{3}{7}$ x 21 = 9

z = $\frac{4}{7}$ x 21 = 12

Hence; x : z = 6 : 12

= 1 : 2

$Q_1 = \frac{1}{4}$ (N + 1)th

$\frac{1}{4} \times 12^{th}$ no.

= 3rd no ($\cong$ 4)

$Q_3 = \frac{3}{4}$  (N + 1)th

= $\frac{3}{4}$ x 12th no.

= 9th no. ($\cong$ 12)

Hence, interquartile range

= $Q_3 - Q_1$

= 12 - 4

= 8

3x $\equiv$ 4(mod 5)

In modulo 5, multiples of 5 that give solution to the given equation are 5, 20, 35, 50,... but 5 will yield the leaast value of x.

Thus; 3x = 4 + 5 = 9

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

x = 3