2016 - WAEC Mathematics Past Questions and Answers - page 1

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

((\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})

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

d (\alpha) t²

d = t² k

where k is a constant. d = 45cm, when t = 3s; thus 45 = 3² x t

k = (\frac{45}{9}) = 5

thus equation connecting d and t is d = 5t²

when t = 6s, d = 5 x 6²

= 5 x 36

= 180cm

From the venn diagram, Nigeria footballers from a subset of good footballers.

On a map, 1cm represents 5km. Then it follows that 1cm² represents 25km². Acm² represents 100km². By apparent cross-multiplication, 1cm² x 100km² = Acm²x 25km²

therefore A = (\frac{100}{25}) = 4cm²

(\frac{3^{n - 1} \times 27^{n + 1}}{81^{n}})

= (\frac{3^{n - 1} \times 3^{3(n + 1)}}{3^{4n}})

= 3(^{n - 1 + 3n + 3 - 4n})

= 3(^{4n - 4n - 1 + 3})

= 3²

= 9

A = P + 1

I = A - P

= 10,400 - P

Now using I = (\frac{P \times T \times R}{100})

i.e. 10,400 - P = (\frac{P \times 5 \times 6}{100})

= 100(10,400 - P) = 30P

10(10,400 - P) = 3P

104,000 - 10P = 3P

104,000 - 10P = 3P

104,000 = 3P + 10P

= 104,000 = 13P

P = (\frac{104,000}{100})

P = D8,000

Let x = (\frac{4}{3}), x = -(\frac{3}{7})

Then 3x = 4, 7x = -3

3x - 4 = 0, 7x + 3 = 0

(3x - 4)(7x + 3) = 0

21x² + 9x - 28x - 12 = 0

21x² - 19x - 12 = 0

(\frac{y^2 - 9y + 18}{y^2 + 4y - 21})

Factorize the denominator;

Y² + 7y - 3y - 21

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

= (y - 3)(y + 7)

Hence the expression (\frac{y^2 - 9y + 18}{y^2 + 4y - 21}) is undefined

when y² + 4y - 21 = 0

ie. y = 3 or -7