WAEC Past Questions and Answers - Page 2671

Note that \(\frac{10\sqrt{3}}{\sqrt{5}} = \frac{10\sqrt{3}}{\sqrt{5}} \times - \frac{\sqrt{5}}{\sqrt{5}}\)

= \(\frac{10\sqrt{15}}{\sqrt{5}} = 2\sqrt{15}\)

hence, (\(\frac{10\sqrt{3}}{\sqrt{5}} - \sqrt{15}\))² = (\(2\sqrt{15} - \sqrt{15}\))²

= (\(2\sqrt{15} - \sqrt{15}\))(\(2\sqrt{15} - \sqrt{15}\))

= 4\(\sqrt{15 \times 15} - 2\sqrt{15 \times 15} - 2\sqrt{15 x 15} + \sqrt{15 \times 15}\)

= 4 x 15 - 2 x 15 - 2 x 15 + 15

= 60 - 30 - 30 + 15

= 15

(\(\frac{3}{4} - \frac{2}{3}\)) x 1\(\frac{1}{5}\)

= (\(\frac{9 - 8}{12} \times \frac{6}{5}\))

= \(\frac{1}{12} \times \frac{6}{5}\)

= \(\frac{1}{10}\)

23_x + 101_x = 130_x

2 x X¹ + 3 x X^o + 1 x X² + 0 x X¹ + 1 x X^o

= 1 x X^o = 1 x X² + 3 x X¹ + 0 x X^o

= X² + 3x + 0

2x + 3 = x² + 0 + 1 + x² + 3x

2x - 3x + x² - x² = -3 - 1

- x = -4

x = 4

In \(\Delta\) XYZ, 2m + 2n + 68^o = 180^o

2(m + n) + 68^o = 180^o...(1)

in \(\Delta\) XOZ, m + n + q = 180^o ...(2)

(m + n) = 180^o - q...(3)

substituting 180^o - q for (m + n) in (1) gives

2(180^o - q) + 68^o = 180^o

360^o - 2q = 180^o - 68^o

360^o - 2q = 112^o

360^o - 112^o = 2q

248^o = 2q

q = \(\frac{248^o}{2}\)

= 124^o

hence, < XOZ = 124^o

In the diagram

|AM| = |MB| - \(\frac{|AB|}{2}\)

= \(\frac{16\sqrt{3}}{2}\)cm

= 8\(\sqrt{3}\)cm

in \(\Delta\) AMO, r² = |AM|Z + |MO|²

r² = (8\(\sqrt{3}\))²
+ 10²

= 64 x 3 + 100

= 192 + 100

= 292

r = \(\sqrt{292}\)

17.088cm

17cm