# 2017 - WAEC Mathematics Past Questions and Answers - page 1

Express 0.0000407, correct to 2 significant figures

If x varies inversely as y and y varies directly as z, what is the relationship between x and z?

**correct option:**b

\(x \propto \frac{1}{y}\), y \(\propto\) z

x = \(\frac{k}{y}\)

y = mz

Since y = mz,

x = \(\frac{k}{mz}\), where k and m are constants. Hence,

x \(\propto\) \(\frac{1}{z}\)

Evaluate \(\frac{3\frac{1}{4} \times 1\frac{3}{5}}{11\frac{1}{3} - 5 \frac{1}{3}}\)

**correct option:**b

\(\frac{3\frac{1}{4} \times 1\frac{3}{5}}{11\frac{1}{3} - 5 \frac{1}{3}}\) = \(\frac{\frac{26}{5}}{\frac{18}{3}}\) = \(\frac{26}{5} \div \frac{18}{3}\)

= \(\frac{13}{15}\)

The ages of Tunde and Ola are in the ratio 1:2. If the ratio of Ola's age to Musa's age is 4:5, what is the ratio of Tunde's age to Musa's age?

**correct option:**c

Tunde: Ola \(\to\) 1 : 2 ; Ola; Musa \(\to\) 4 : 5

\(\frac{1}{2}\) x \(\frac{4}{5}\)

= \(\frac{2}{5}\)

If M = {x : 3 \(\leq\) x < 8} and N = {x : 8 < x \(\leq\) 12}, which of the following is true?

i. 8 \(\in\) M \(\cap\) N

ii. 8 \(\in\) M \(\cup\) N

iii. M \(\cap\) N = \(\varnothing\)

Given that a = log 7 and b = \(\log\) 2, express log 35 in terms of a and b.

**correct option:**c

\(\frac{\log 7 \times \log 10}{\log 2}\)

log 7 x log 10 \(\div\) log 2

a + 1 - b

a - b + 1

If x = \(\frac{2}{3}\) and y = - 6, evaluate xy - \(\frac{y}{x}\)

**correct option:**b

x = \(\frac{2}{3}\) and y = - 6

xy - \(\frac{y}{x}\)

\(\frac{2}{3} - (6)^2 - (-6) \div \frac{2}{3}\)

= -4 - (6) x \(\frac{3}{2}\)

= -4 - (-6) x \(\frac{3}{2}\)

= -4 - (-9)

= -4 + 9

= 5

Solve the equation: \(\frac{1}{5x} + \frac{1}{x}\)= 3

**correct option:**b

\(\frac{1}{5x} + \frac{1}{x}\)= 3

\(\frac{1 + 5}{5x}\) = 3

6 = 15x

x = \(\frac{6}{15}\)

= \(\frac{2}{5}\)

A sum of N18,100 was shared among 5 boys and 4 girls with each boy taking N20.00 more than each girl. Find a boy's share.

**correct option:**c

Let a girl's share = x + 20

4x + 5(x + 20) = 18,100

4x + 5x + 100 = 18,100

9x + 100 = 18,100

9x = 18,000

x = \(\frac{18,000}{9}\)

x = 2,000

\(\therefore\) Each boy gets N(2,000 + 20)

= N2,020.

One factor of \(7x^2 + 33x - 10\) is

**correct option:**c

\(7x^2 + 33x - 10\)

\(7x^2 + 35x - 2x - 10\)

7x (x + 5) - 2 (x + 5)

(7x - 2) (x + 5)