WAEC Further Mathematics Past Questions & Answers - Page 67

331.

Simplify \(2\log_{3} 8 - 3\log_{3} 2\)

A.

\(-\log_{3} 4\)

B.

\(-\log_{3} 2\)

C.

\(3\log_{3} 2\)

D.

\(3\log_{3} 4\)

Correct answer is C

\(2\log_{3} 8 - 3\log_{3} 2 = \log_{3} 8^{2} - \log_{3} 2^{3}\)

= \(\log_{3}(\frac{64}{8}) \)

= \(\log_{3} 8 = \log_{3} 2^{3}\)

= \(3 \log_{3} 2\)

332.

Calculate, correct to the nearest degree, the angle between the vectors \(\begin{pmatrix} 13 \\ 1 \end{pmatrix}\) and \(\begin{pmatrix} 1 \\ 4 \end{pmatrix}\)

A.

58°

B.

72°

C.

74°

D.

87°

Correct answer is B

\(a . b = |a||b| \cos \theta\)

\(\begin{pmatrix} 13 \\ 1 \end{pmatrix}. \begin{pmatrix} 1 \\ 4 \end{pmatrix} = 13 \times 1 + 1 \times 4 = 13 + 4 = 17\)

\(17 = (\sqrt{13^{2} + 1^{2}})(\sqrt{1^{2} + 4^{2}}) \cos \theta\)

\(17 = (\sqrt{170})(\sqrt{17}) \cos \theta\)

\(\cos \theta = \frac{17}{17\sqrt{10}} = \frac{\sqrt{10}}{10} = 0.3162\)

\(\theta = \cos^{-1} 0.3162 = 72°\)

333.

Two fair dices, each numbered 1, 2, ..., 6, are tossed together. Find the probability that they both show even numbers.

A.

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

B.

\(\frac{1}{4}\)

C.

\(\frac{1}{6}\)

D.

\(\frac{1}{12}\)

Correct answer is B

P(even in 1 dice) = \(\frac{3}{6} = \frac{1}{2}\)

P(even in 2 fair die) = \(\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\)

334.

The position vectors of A and B are (2i + j) and (-i + 4j) respectively; find |AB|.

A.

\(3\sqrt{2}\)

B.

\(\sqrt{34}\)

C.

\(\sqrt{34}\)

D.

\(9\sqrt{2}\)

Correct answer is A

No explanation has been provided for this answer.

335.

Three men, P, Q and R aim at a target, the probabilities that P, Q and R hit the target are \(\frac{1}{2}\), \(\frac{1}{3}\) and \(\frac{3}{4}\) respectively. Find the probability that exactly 2 of them hit the target.

A.

\(1\)

B.

\(\frac{1}{2}\)

C.

\(\frac{5}{12}\)

D.

\(\frac{1}{12}\)

Correct answer is C

\(p(P) = \frac{1}{2}, p(P') = \frac{1}{2}\)

\(p(Q) = \frac{1}{3}, p(Q') = \frac{2}{3}\)

\(p(R) = \frac{3}{4}, p(R') = \frac{1}{4}\)

p(exactly two hit the target) = p(P and Q and R') + p(P and R and Q') + p(Q and R and P')

= \((\frac{1}{2} \times \frac{1}{3} \times \frac{1}{4}) + (\frac{1}{2} \times \frac{3}{4} \times \frac{2}{3}) + (\frac{1}{3} \times \frac{3}{4} \times \frac{1}{2})\)

= \(\frac{1}{24} + \frac{6}{24} + \frac{3}{24}\)

= \(\frac{5}{12}\)