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