WAEC Further Mathematics Past Questions & Answers - Page 98

486.

P and Q are the points (3, 1) and (7, 4) respectively. Find the unit vector along PQ.

A.

\(\begin{pmatrix} 4 \\ 3 \end{pmatrix}\)

B.

\(\begin{pmatrix} 0.6 \\ 0.8 \end{pmatrix}\)

C.

\(\begin{pmatrix} 0.8 \\ 0.6 \end{pmatrix}\)

D.

\(\begin{pmatrix} -0.8 \\ 0.6 \end{pmatrix}\)

Correct answer is C

\(PQ = \begin{pmatrix} 7 - 3 \\ 4 - 1 \end{pmatrix}\)

\(= \begin{pmatrix} 4 \\ 3 \end{pmatrix}\)

\(\hat{n} = \frac{\overrightarrow{PQ}}{|PQ|} \)

\(|PQ| = \sqrt{4^{2} + 3^{2}} = \sqrt{25} = 5\)

\(\hat{n} = \frac{1}{5}\begin{pmatrix} 4 \\ 3 \end{pmatrix} = \begin{pmatrix} 0.8 \\ 0.6 \end{pmatrix}\) 

487.

Two out of ten tickets on sale for a raffle draw are winning tickets. If a guest bought two tickets, what is the probability that both tickets are winning tickets?

A.

\(\frac{1}{80}\)

B.

\(\frac{1}{45}\)

C.

\(\frac{1}{20}\)

D.

\(\frac{1}{10}\)

Correct answer is B

P(winning) = \(\frac{2}{10}\)

P(both tickets winning) = \(\frac{2}{10} \times \frac{1}{9} = \frac{1}{45}\)

488.

Given that \(P = \begin{pmatrix} 3 & 4 \\ 2 & x \end{pmatrix}; Q = \begin{pmatrix} 1 & 3 \\ -2 & 4 \end{pmatrix}; R = \begin{pmatrix} -5 & 25 \\ -8 & 26 \end{pmatrix}\)  and PQ = R, find the value of x.

A.

-5

B.

-2

C.

2

D.

5

Correct answer is D

\(P = \begin{pmatrix} 3 & 4 \\ 2 & x \end{pmatrix}; Q = \begin{pmatrix} 1 & 3 \\ -2 & 4 \end{pmatrix}; R = \begin{pmatrix} -5 & 25 \\ -8 & 26 \end{pmatrix}\) 

PQ = \(\begin{pmatrix} 3 & 4 \\ 2 & x \end{pmatrix} \begin{pmatrix} 1 & 3 \\ -2 & 4 \end{pmatrix} = \begin{pmatrix} -5 & 25 \\ 2 - 2x & 6 + 4x \end{pmatrix} = R\)

\(\implies 2 - 2x = -8; -2x = -8 - 2 = -10\)

\(6 + 4x = 26 \implies 4x = 26 - 6 = 20\)

\(\implies x = 5\)

489.

Find the upper quartile of the following scores: 41, 29, 17, 2, 12, 33, 45, 18, 43 and 5.

A.

45

B.

41

C.

33

D.

21

Correct answer is B

Arranging the scores in ascending order, we have: 2, 5, 12, 17, 21, 29, 33, 41, 43, 45.

The upper quartile = 41.

490.

If \(2\sin^{2}\theta = 1 + \cos \theta, 0° \leq \theta \leq 90°\), find \(\theta\)

A.

30°

B.

45°

C.

60°

D.

90°

Correct answer is C

\(2\sin^{2}\theta = 1 + \cos \theta \implies 2(1 - \cos^{2}\theta) = 1 + \cos \theta\)

\(2 - 2\cos^{2}\theta = 1 + \cos \theta\)

\(2 - 2\cos^{2}\theta - 1 - \cos \theta = 0\)

\(2\cos^{2}\theta + \cos \theta - 1 = 0\)

\(2\cos^{2}\theta + 2\cos\theta - \cos \theta - 1 = 0 \implies 2\cos \theta(\cos \theta + 1) - 1(\cos \theta + 1) = 0\)

\((2\cos \theta - 1)(\cos \theta + 1) = 0 \implies \cos \theta = \frac{1}{2} \)

\(\theta = \cos^{-1} \frac{1}{2} = 60°\)