2015 - WAEC Mathematics Past Questions and Answers - page 1

N1.00 = $0.075

N X = $15,000

X = \(\frac{1.00 \times 15000}{0.075}\)

= N200,000.00

(\frac{27^x \times 3^{1 - x}}{9^{2x}} = 1)

(\frac{3^{3x} \times 3^{1 - x}}{3^{2(2 - x)}} = 3^0)

(3^{3x} \times 3^{1 - x} \div 3^{4x} = 3^0)

(3^{(3x + 1 - x - 4x)} = 3^0)

(3^{(1 - 2x)} = 3^0)

since the bases are equal,

1 - 2x = 0

• 2x = -1
  
  

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

If log_x 64 = 3, then 64 = x³

4³ = x³

Since the indices are equal,

x = 4

Hence, x log₂8 = log₂8^x

= log₂8⁴

= log₂(2³)₄

= log₂2¹²

= 1 log₂ = 1(1)

= 12

If 2ⁿ = y,

then, 2(^{(2 + \frac{n}{3})}) = 2² x 2(^\frac{n}{3})

= 4 x (2ⁿ)(^{\frac{1}{3}})

But y = 2ⁿ, hence

2(^{(2 + \frac{n}{3})}) = 4 x y(^{\frac{1}{3}})

= 4y(^\frac{1}{3})

6ax - 12by - 9ay + 8bx

= 6ax - 9ay + 8bx - 12by

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

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

Let x = (\frac{3}{4}) or x = -4

i.e. 4x = 3 or x = -4

(4x - 3)(x + 4) = 0

therefore, 4x² + 13x - 12 = 0

If m = 4, n = 9, r = 16,

then (\frac{m}{n}) - 1(\frac{7}{9}) + (\frac{n}{r})

= (\frac{4}{9}) - (\frac{16}{9}) + (\frac{9}{16})

= (\frac{64 - 258 + 81}{144})

= (\frac{-111}{144})

= - 1(\frac{37}{48})