2016 - WAEC Mathematics Past Questions and Answers - page 2

2x + y = 7...(1)

3x - 2y = 3...(2)

From (1), y = 7 - 2x for y in (2)

3x - 2(7 - 2x) = 3

3x - 14 + 4x = 3

7x + 3 + 14 = 17

x = (\frac{17}{7})

Hence, 7 x (\frac{17}{7})

= 17 - 10

= 7

(\frac{2}{1 - x} - \frac{1}{x}) = (\frac{2x - 1(1 - x)}{x(1 - x)})

= (\frac{2x - 1(1 + x)}{x(1 - x)})

= (\frac{3x - 1}{x(1 - x)})

P = S + (\frac{sm^2}{nr})

P = S(1 + (\frac{m^2}{nr}))

P = S(1 + (\frac{nr + m^2}{nr}))

nrp = S(nr + m²)

S = (\frac{nrp}{nr + m^2})

(2x + 3y)² - (x - 4y)²

= (2x + 3y)(2x + 3y) - (x - 4y)(x - 4y)

= 4x² + 12xy + 9y² - (x² - 8xy+ 16y²)

= 4x² + 12xy + 19y² - x² + 8xy - 16y²

= 3x² + 20xy - 7y²

= 3x² + 21xy - xy - 7y²

= 3x(x + 7y) - y(x + 7y)

= (3x - y)(x + 7y)

Curved surface area of cylinder = 2(\pi)rh

110 = 2 x (\frac{22}{7}) x r x 5

r = (\frac{110 \times 7}{44 \times 5})

= 3.5cm

Volume of pyramid = (\frac{1}{3})lbh

90 = (\frac{1}{3} \times x \times 6 \times 15)

x = (\frac{90 \times 33}{6 \times 15})

= 3

Let: (x1, y1) = (1, 2)

(x2, y2) = (5, 8)

The gradient m of \(\bar{PQ}\) is given by

m = \(\frac{y_2 y_1}{x_2 - x_1}\)

= \(\frac{8 - 2}{5 - 1}\)

= \(\frac{6}{4}\)

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

|PQ| = (\sqrt{(x_2 - X- 1) + (y_2 - y_1)^2})

= (\sqrt{(5 - 1)^2 + (8 - 2)^2})

= (\sqrt{4^2 + 6^2})

= (\sqrt{16 + 36})

= (\sqrt{52})

= 2(\sqrt{13})

sin 60^o = x + 0.5 0^o(given)

0.8660 = x + 0.5

0.8660 - 0.5 = x

x = 0.3660

cos(\theta) = x(given)

cos(\theta) = 0.3660

Hence, (\theta) = cos^-1(0.3660)

= 68.53^o

= 69^o (nearest degree)

Number of students in the club is

10 + 24 + 8 + 5 + 3 = 50