# Mathematics 2017 Past Questions | WAEC

Study the following Mathematics past questions and answers for JAMB, WAEC NECO and Post-JAMB. Get prepared with official past questions and answers for upcoming examinations.

Express 0.0000407, correct to 2 significant figures

**A.**0.0**B.**0.00004**C.**0.000041**D.**0.0000407

**Correct Option: Answer is C**

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

**A.**x \(\alpha\) z**B.**x \(\alpha\) \(\frac{1}{z}\)**C.**a \(\alpha\) z\(^2\)**D.**x \(\alpha\) \(\frac{1}{z^2}\)

**Correct Option: Answer is 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 \(\propto\) \(\frac{1}{z}\)

**3. **Evaluate \(\frac{3\frac{1}{4} \times 1\frac{3}{5}}{11\frac{1}{3} – 5 \frac{1}{3}}\)

**A.**\(\frac{14}{15}\)**B.**\(\frac{13}{15}\)**C.**\(\frac{4}{5}\)**D.**\(\frac{11}{15}\)

**Correct Option: Answer is B**

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

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

**4. **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?

**A.**1 : 4**B.**1 : 5**C.**2 : 5**D.**5 : 2

**Correct Option: Answer is C**

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

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

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

**5. **If M = {x : 3 ≤$\le $ x < 8} and N = {x : 8 < x ≤$\le $ 12}, which of the following is true?

i. 8 ∈ M ∩ N

ii. 8 ∈ M $\cup $ N

iii. M ∩ N = ∅

**A.**iii only**B.**i and ii**C.**ii and iii only**D.**i, ii and iii

**Correct Option: Answer is A**

M = {3, 4, 5, 6, 7,}, N = {9, 10, 11, 12}

**6. **Given that a = log 7 and b = $\mathrm{log}$ 2, express log 35 in terms of a and b.

**A.**a + b + 1**B.**ab – 1**C.**a – b + 1**D.**b – a + 1

**Correct Option: Answer is C**

##### Explanation

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

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

a + 1 – b

a – b + 1

**7. **If x = 2/3 and y = – 6, evaluate xy – y/x

**A.**0**B.**5**C.**8**D.**9

**Correct Option: Answer is 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

**8. **Solve the equation: **1/5x + 1/x= 3**

**A.**1/5**B.**2/5**C.**3/5**D.**4/5

**Correct Option: Answer is B**

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

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

6 = 15x

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

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

**9. **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.

**A.**N1,820.00**B.**2,000.00**C.**N2,020.00**D.**N2,040.00

**Correct Option: Answer is 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 = 18,000/9

x = 2,000

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

= N2,020.

**10. **One factor of **7x²+3x−10** is

**A.**7x + 5**B.**x – 2**C.**7x – 2**D.**x – 5

**Correct Option: Answer is C**