2007 - WAEC Mathematics Past Questions and Answers - page 3

p \(\alpha \frac{I}{Q}\); p \(\frac{k}{q}\) (where k is constant)

q = \(\frac{k}{p}\)

q \(\alpha \frac{1}{p}\)

3 \(\sqrt{45} - 12\sqrt{5} + 16\sqrt{20}\)

= 3 x \(\sqrt{9 \times 5} - 12 \times \sqrt{5} + 16 \times \sqrt{4 \times 5}\)

= 3 x 3 x \(\sqrt{5} - 12 \times \sqrt{5} + 16 \times 2 \times \sqrt{5}\)

= 9\(\sqrt{5} - 2 \sqrt{5} + 32 \sqrt{5}\)

= 9\(\sqrt{5} + 32\sqrt{5} - 12\sqrt{5}\)

= 29\(\sqrt{5}\)

p = {-3.2.....4.9}; Since x is an integer

least value of x is -3

\(\frac{3x^3}{(3x)^3}\) = \(\frac{3 \times x^3}{3^3 \times x^3}\)

= \(\frac{3 \times x^3}{3 \times 3 \times 3 \times x^3}\)

= \(\frac{1}{3^2}\)

= \(\frac{1}{9}\)

Let Abu's share = 2x;

Kayode's = x

Uche's share = \(\frac{x}{2}\)

Total: 2x + x + \(\frac{x}{2}\) = N140,000

\(\frac{7x}{2}\) = N140,000

x = \(\frac{280,000}{7}\) = N40,000

Kayode's share = N40,000

I am x years old; My brother = 3 + x; My brother

last year = (3 + x) - 1 = x + 3 - 1

= (x + 2) years

\(\frac{2x - 1}{x + 3}\) not defined

x + 3 = 0

x = -3

x = 3, y = 2, z = 4

3x² - 2y + z

= 3(3)² - 2(2) + 4

3(9) - 4 + 4 = 27

x + y = 2........... (i)

3x - 2y = 1......(ii)

From (i): x = 2 - y ..........(iii)

subt. (2 - y) for x in (ii)

3(2 - y) 2y = 1

6 - 3y - 2y = 1

-5y = 1 - 6 = -5

y = \(\frac{-5}{-5}\) = 1

subst. (i) for y in (iii)

x = 2 - 1 = 1

x = 1, y = 1

8^x-1 = \(\frac{1}{4}\)

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

3x - 3 = -2

3x = -2 + 3

3x = 1

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