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 ⟺ ∼q
- B. p ⟺ q
- C. ∼p ⟺ ∼q
- D. q ⟺ ∼p