2010 - WAEC Mathematics Past Questions and Answers - page 4

x2 + kx + \(\frac{16}{9}\); Perfect square

But, b2 - 4ac = 0, for a perfect square

where a - 1; b = k; c = \(\frac{16}{9}\)

k2 - 4(1) x \(\frac{16}{9}\) = 0

k2 - \(\frac{64}{9}\) = 0

k2 = \(\frac{64}{9}\)

k = \(\sqrt{\frac{64}{9}}\)

k = \(\frac{8}{3}\)

x kmh^-1 = y ms^-1

(\frac{x km}{1 hr}) = y ms^-1

(x \times \frac{1km}{1hr}) = y ms^-1

(x \times \frac{1000m}{60 \times 60s}) = y ms^-1

(x \times \frac{1000}{3600} \frac{m}{s}) = y ms^-1

(x \times \frac{5}{18} ms^{-1})

(x \times \frac{5}{18} ms^{-1}) = y ms^-1

y = (\frac{5}{18})x

mean \(\leq\) 5; \(\frac{2 + 5 + 2x + 7}{4}\) \(\leq\) 5

= \(\frac{14 + 2x}{4} \leq 5\)

= 14 + 2x \(\leq\) 5 x 4

14 + 2x \(\leq\) 20 ; 2x \(\leq\) 20 - 14

2x \(\leq\) 20 - 14

2x \(\leq\) 6

x \(\leq\) \(\frac{6}{2}\)

x \(\leq\) 3

Pr. (winning 100m race) = (\frac{1}{8})

Pr. (losing 100m race) = 1 - (\frac{1}{8}) = (\frac{7}{8})

Pr. (winning high jump) = (\frac{1}{4})

Pr. (losing high jump ) = 1 - (\frac{1}{4}) = (\frac{3}{4})

Pr. (winning only one) = Pr. (Winning 100m race and losing high jump) or Pr.(Losing 100m race and winning high jump)

= ((\frac{1}{8} \times \frac{3}{4})) + ((\frac{7}{8} \times \frac{1}{4}))

= (\frac{3}{32} + \frac{7}{32})

= (\frac{10}{32})

= (\frac{5}{16})

Joining SQ. In (\bigtriangleup) SPQ,

(22^o + a) + 55^o + (41^o + b) = 180^o

121^o + a + b = 180^o

a + b = 180 - 121

a + b = 59^o.....(1)

In (\bigtriangleup) SRPQ; R + a + b = 180^o

R + 59^o = 180^o

(in (1), a + b = 59^o)

R = 180 - 59

R = 121^o

Pass mark = 50%

No. of students that passed = f₅₀ + f₆₅ + f₈₀

= 45 + 25 + 15

= 85

Percentage of students with marks ranging from 35 to 50 = (\frac{f_{35} + f{40} + f{50}}{\sum f})

= (\frac{35 + 40 + 45}{20 + 35 + 40 + 45 + 25 + 15}) x 100%

= (\frac{120}{180}) x 100%

= 66(\frac{2}{3})%

4m - 15^o = m + 75^o

(Vertically opposite angles are equal)

4m - m = 75 + 15

3m = 90

m = (\frac{90}{3})

m = 30^o

In (\bigtriangleup) PQR,

Q = S = 75^o (Corresponding angle)

R = Q = 75^o (Base angles of an isosceles (\bigtriangleup))

But, y + 75^o = 180^o (Sum of angles in a straight line)

y = 180 - 75

y = 105^o

(\bigtriangleup) is similar to (\bigtriangleup) HIJ

< HKL = HJI = x^o

Hence, (\frac{LH}{JH} = \frac{KH}{JH} \frac{KL}{IJ})

(\frac{LH}{JH} = \frac{KL}{JI})