Year : 
2017
Title : 
Mathematics (Core)
Exam : 
WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

21 - 30 of 49 Questions

# Question Ans
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

Detailed Solution

Tan 60 = 3; Tan 30 = 1
\(\frac{\tan 60^o - 1}{1 - tan 30^o}\) = \(\frac{\sqrt{3 - 1}}{1 - \frac{1}{\sqrt{3}}}\) = \(\frac{\sqrt{3 - 1}}{\frac{3 - 1}{\sqrt{3}}}\)
= \(\frac{\sqrt{3 - 1}}{1} \times \frac{\sqrt{3 - 1}}{\sqrt{3}}\)
= \(\frac{\sqrt{3 - 1}}{1} \times \frac{\sqrt{3}}{\sqrt{3}}\)
= \(\sqrt{3}\)

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. 3 years

23.

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

A. n - 1

B. n + 2

C. n + 1

D. n

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\(\sqrt{3} + 4\)

Detailed Solution

\(\sqrt{2}(\sqrt{6} + 2\sqrt{2}) - 2\sqrt{3}\)
\(\sqrt{12}\) + 2 x 2 - 2\(\sqrt{3}\)
2 \(\sqrt{3}\) - 2 \(\sqrt{3}\) + 4
= 4
25.

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

A. decagon

B. nonagon

C. octagon

D. hexagon

Detailed Solution

Sum of all exterior angles is 360\(^o\)
360\(^o\) (30\(^o\) - 40\(^o\))
360 - (130\(^o\))
230\(^o\)
remaining is 46\(^o\) = \(\frac{230}{46}\) = 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. \(ab^2 - a^2b\)

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

Detailed Solution

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. \(6x^2 + x + 2 = 0\)

Detailed Solution

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

30.

Consider the statements: p it is hot, q: it is rainingWhich of the following symbols correctly represents the statement "It is raining if and only if it it is cold"?

A. p \(\iff\) \(\sim\)q

B. p \(\iff\) q

C. \(\sim\)p \(\iff\) \(\sim\)q

D. q \(\iff\) \(\sim\)p

D

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

Detailed Solution

Tan 60 = 3; Tan 30 = 1
\(\frac{\tan 60^o - 1}{1 - tan 30^o}\) = \(\frac{\sqrt{3 - 1}}{1 - \frac{1}{\sqrt{3}}}\) = \(\frac{\sqrt{3 - 1}}{\frac{3 - 1}{\sqrt{3}}}\)
= \(\frac{\sqrt{3 - 1}}{1} \times \frac{\sqrt{3 - 1}}{\sqrt{3}}\)
= \(\frac{\sqrt{3 - 1}}{1} \times \frac{\sqrt{3}}{\sqrt{3}}\)
= \(\sqrt{3}\)

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. 3 years

23.

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

A. n - 1

B. n + 2

C. n + 1

D. n

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\(\sqrt{3} + 4\)

Detailed Solution

\(\sqrt{2}(\sqrt{6} + 2\sqrt{2}) - 2\sqrt{3}\)
\(\sqrt{12}\) + 2 x 2 - 2\(\sqrt{3}\)
2 \(\sqrt{3}\) - 2 \(\sqrt{3}\) + 4
= 4
25.

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

A. decagon

B. nonagon

C. octagon

D. hexagon

Detailed Solution

Sum of all exterior angles is 360\(^o\)
360\(^o\) (30\(^o\) - 40\(^o\))
360 - (130\(^o\))
230\(^o\)
remaining is 46\(^o\) = \(\frac{230}{46}\) = 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. \(ab^2 - a^2b\)

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

Detailed Solution

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. \(6x^2 + x + 2 = 0\)

Detailed Solution

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

30.

Consider the statements: p it is hot, q: it is rainingWhich of the following symbols correctly represents the statement "It is raining if and only if it it is cold"?

A. p \(\iff\) \(\sim\)q

B. p \(\iff\) q

C. \(\sim\)p \(\iff\) \(\sim\)q

D. q \(\iff\) \(\sim\)p

D