# 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.

**21. **Given that cos 30\(^o\) = sin 60\(^o\) = \(\frac{3}{2}\) and sin 30\(^o\) = cos 60\(^o\) = \(\frac{1}{2}\), evaluate \(\frac{tan 60^o – q}{1 – tan 30^o}\)

**A.**\(\sqrt{3 – 2}\)**B.**2 – \(\sqrt{3}\)**C.**\(\sqrt{3}\)**D.**-2

**Correct Option: Answer is C**

**22. **The average age of a group of 25 girls is 10year. If one girl, aged 12 years and 4 months joins the group, find the new average age of the group

**A.**10.1 years**B.**9.3 years**C.**8.7 years**D.**8 . 3 years

**Correct Option: Answer is A**

x = 10 ; 10 = x/25

x = 250

x = 250+12.4/26

x = 10.09

x = 10.1 years

**23. **In what number base was the addition 1 + nn = 100, where n > 0, done?

**A.**n – 1**A.**n + 2**B.**n + 1**C.**n

**Correct Option: Answer is C**

24. Simplify; \(\sqrt{2}(\sqrt{6} + 2\sqrt{2}) – 2\sqrt{3}\)

**A.**4**B.**\(\sqrt{3} + 4\)**C.**4 \(\sqrt{2}\)**D.**4\(\sqrt{3} + 4\)

**Correct Option: Answer is A**

25. Three exterior angles of a polygon are 30º, 40º and 60º. If the remaining exterior angles are 46º each, name the polygon.

**A.**decagon**B.**nonagon**C.**octagon**D.**hexagon

**Correct Option: Answer is C**

Sum of all exterior angles is 360º

360º (30º – 40º)

360 – (130º)

230º

remaining is

46º = 230/4 = 5

5 + 3 = 8 sides; Octagon

**26. **Simplify the expression \(\frac{a^2 b^4 – b^2 a^4}{ab(a + b)}\)

**A.**\(a^2 – b^2\)**B.**\(b^2 – a^2\)**C.**\(a^2b – ab^2\)**D.**\(ab^2 – a^2b\)

**Correct Option: Answer is D**

\(\frac{a^2 b^4 – b^2 a^4}{ab(a + b)}\) = \(\frac{a^2 b^24(b^2 – a^2}{ab(a + b)}\)

= \(\frac{ab [(b – a) (b + a)]}{a + b}\)

= ab(b – a)

= \(ab^2 – a^2b\)

**27. **Find the 6th term of the sequence \(\frac{2}{3} \frac{7}{15} \frac{4}{15}\),…

**A.**-\(\frac{1}{3}\)**B.**-\(\frac{1}{5}\)**C.**-\(\frac{1}{15}\)**D.**\(\frac{1}{9}\)

**Correct Option: Answer is A**

a = \(\frac{2}{3}\), d = \(\frac{7}{15}\) – \(\frac{2}{3}\)

= 7 – 10

= \(\frac{-3}{15}\)

d = – \(\frac{-1}{5}\)

T6 = a + 5d

= \(\frac{2}{3}\) + 5(\(\frac{-1}{5}\)

= \(\frac{2}{3}\) – 1

= \(\frac{2 – 3}{3}\)

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

**28. **The roots of a quadratic equation are \(\frac{-1}{2}\) and \(\frac{2}{3}\). Find the equation.

**A.**\(6x^2 – x + 2 = 0\)**B.**\(6x^2 – x – 2 = 0\)**C.**\(6x^2 + x – 2 = 0\)**D.**\(6x^2 + x + 2 = 0\)

**Correct Option: Answer is B**

(x + \(\frac{1}{2}\)) (n – \(\frac{2}{3}\))

\(x^2 – \frac{2}{3^x} + \frac{x}{2} – \frac{1}{3}\)

\(6x^2 – 4n + 3n – 2 = 0\)

\(6x^2 – x – 2 = 0\)

**29. **Make x the subject of the relation d = \(\sqrt{\frac{6}{x} – \frac{y}{2}}\)

**A.**x = \(\frac{6 + 12}{d^2 + y}\)**B.**x = \(\frac{12}{d^2 – y}\)**C.**x = \(\frac{12}{y} – 2d^2\)**D.**x = \(\frac{12}{2d^2 – y}\)

**Correct Option: Answer is A**

d = \(\sqrt{\frac{6}{x} – \frac{y}{2}}\)

\(d^2 = \frac{6}{x} – \frac{y}{2}\)

\(2xd^2 = 12 – xy\)

\(2xd^2 + xy = 12\)

x = \(\frac{6 + 12}{d^2 + y}\)

**30. **Consider the statements: p it is hot, q: it is raining

Which of the following symbols correctly represents the statement “It is raining if and only if it it is cold”?

**A.**p ⟺$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}$ ∼q**B.**p ⟺$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}$ q**C.**∼p ⟺$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}$ ∼q**D.**q ⟺$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}$ ∼p