WAEC Further Mathematics Past Questions & Answers - Page 42

206.

A lift moving upwards with a uniform acceleration of 5\(ms^{-2}\) carries a body of mass p kg. If the reaction on the floor is 480 N, find the value of p. [Take g = \(10 ms^{-2}\)].

A.

32

B.

36

C.

48

D.

64

Correct answer is A

Reacting force - Weight = Net force

R = W + ma = mg + ma

480 = 10p + 5p

480 = 15p

p = 32 kg

207.

If \(a = \begin{pmatrix} 3 \\ 2 \end{pmatrix}\) and \(b = \begin{pmatrix} -3 \\ 5 \end{pmatrix}\), find a vector c such that \(4a + 3c = b\).

A.

\(\begin{pmatrix} 3 \\ -1 \end{pmatrix}\)

B.

\(\begin{pmatrix} -5 \\ -1 \end{pmatrix}\)

C.

\(\begin{pmatrix} -5 \\ 1 \end{pmatrix}\)

D.

\(\begin{pmatrix} -5 \\ -9 \end{pmatrix}\)

Correct answer is B

\(4 \begin{pmatrix} 3 \\ 2 \end{pmatrix} + 3 \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -3 \\ 5 \end{pmatrix}\)

\(\begin{pmatrix} 12 \\ 8 \end{pmatrix} + \begin{pmatrix} 3x \\ 3y \end{pmatrix} = \begin{pmatrix} -3 \\ 5 \end{pmatrix}\)

\(\begin{pmatrix} 3x \\ 3y \end{pmatrix} = \begin{pmatrix} -3 - 12 \\ 5 - 8 \end{pmatrix}\)

\(\begin{pmatrix} 3x \\ 3y \end{pmatrix} = \begin{pmatrix} -15 \\ -3 \end{pmatrix}\)

\(\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -5 \\ -1 \end{pmatrix}\)

208.

The mean and median of integers x, y, z and t are 5 and z respectively. If x < y < z < t and y = 4, find (x + t).

A.

12

B.

11

C.

10

D.

8

Correct answer is A

\(Mean : \frac{x + 4 + z + t}{4} = 5 \implies x + 4 + z + t = 20\)

\(\implies x + z + t = 16 ... (1)\)

\(Median : \frac{4 + z}{2} = z \implies 4 + z = 2z\)

\(4 = z\)

From 1, 

\(\implies x + 4 + t = 16 \)

\(x + t = 12\)

209.

Find, in surd form, the value of \(\cos 165\).

A.

\(\frac{1}{4}(\sqrt{6} + \sqrt{2})\)

B.

\(\frac{1}{4}(\sqrt{6} - \sqrt{2})\)

C.

\(-\frac{1}{4}(\sqrt{6} - \sqrt{2})\)

D.

\(-\frac{1}{4}(\sqrt{6} + \sqrt{2})\)

Correct answer is D

\(\cos 165 = -\cos (180 - 165) = -\cos 15\)

\(\cos 15 = \cos (45 - 30)\)

\(\cos (x - y) = \cos x \cos y + \sin x \sin y\)

\(\cos (45 - 30) = \cos 45 \cos 30 + \sin 45 \sin 30\)

= \((\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2}) + (\frac{\sqrt{2}}{2})(\frac{1}{2})\)

= \(\frac{1}{4}(\sqrt{6} + \sqrt{2})\)

\(\therefore \cos 165 = -\frac{1}{4}(\sqrt{6} + \sqrt{2})\)

210.

The parallelogram PQRS has vertices P(-2, 3), Q(1, 4), R(2, 6) and S(-1,5). Find the coordinates of the point of intersection of the diagonals.

A.

\((-1, 5)\)

B.

\((-\frac{1}{2}, 3\frac{1}{2})\)

C.

\((0, 4\frac{1}{2})\)

D.

\((1\frac{1}{2}, 5)\)

Correct answer is C

No explanation has been provided for this answer.