Correct Answer: A. 18

Correct Answer: D. 4

Explanation

(a)(i) Position vector of A, \(\overrightarrow{OA} = i + 5j\)

Position vector of B, \(\overrightarrow{OB} = 3i + 9j\)

Position vector of C, \(\overrightarrow{OC} = -i + j\)

The points A, B and C are collinear if \(|\overrightarrow{AB}| = k|\overrightarrow{BC}|\), where k is a constant.

If k > 0, then \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\) are in the same direction.

If k < 0, then they are in opposite directions, in this case they are A, C, B.

\(\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = - \overrightarrow{OA} + \overrightarrow{OB}\)

= \(- i - 5j + (3i + 9j) = 2i + 4j = 2(i + 2j)\)

\(\overrightarrow{BC} = \overrightarrow{BO} + \overrightarrow{OC} = - \overrightarrow{OB} + \overrightarrow{OC}\)

= \(- 3i - 9j + (-i + j) = - 4i - 8j = -4(i + 2j)\)

\(\overrightarrow{AB} = -\frac{1}{2} \overrightarrow{BC}\)

\(\therefore\) A, B and C are collinear with \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\) in opposite directions.

(ii) \(|\overrightarrow{AB}| : |\overrightarrow{BC}| = 1 : 2\)

(b)

We take moments about X.

Clockwise moment = \(500 \times 8 + 100 \times 12 = 5200N\)

Anticlockwise moment = \(T_{2} \times 20\)

From the principles of moment,

\(20T_{2} = 5200N \implies T_{2} = 260N\)

For vertical equilibrium,

\(T_{1} + T_{2} = 600N\)

\(\therefore T_{1} = 600 - T_{2} = 600 - 260 =340N\)

The tensions in the strings are 340N and 260N respectively.

Correct Answer: D. 88

Explanation

First, we need to find the total number of points for his fourth exam. That would be 80?4 or 320. The reason we do this is because we can calculate his point total with the new average and determine his final score by taking the difference. So 82?5=410. We subtract 410 and 320 to get 90 as our answer.

Correct Answer: D. 3

Explanation

\((\sqrt{x} + 1) * (\sqrt{x} - 1) = 4 \implies \frac{\sqrt{x} + 1}{\sqrt{x} - 1} + \frac{\sqrt{x} - 1}{\sqrt{x} + 1} = 4\)

\(\frac{(\sqrt{x} + 1)(\sqrt{x} + 1) + (\sqrt{x} - 1)(\sqrt{x} - 1)}{(\sqrt{x} - 1)(\sqrt{x} + 1)}\)

= \(\frac{x + 2\sqrt{x} + 1 + x - 2\sqrt{x} + 1}{x - 1} \implies \frac{2x + 2}{x - 1} = 4\)

\(2x + 2 = 4x - 4 \therefore 4x - 2x = 2x = 2 + 4= 6\)

\(x = 3\)

Correct Answer: B. 4.0

Correct Answer: D. 377.0cm²

Correct Answer: D. \(\frac{pq}{t}\)

Explanation

q oranges = t naira

1 naira = \(\frac{q}{t}\)

p naira = \(p(\frac{q}{t})\)

= \(\frac{pq}{t}\) oranges

Explanation

(a) \(1001_{2} = 1 \times 2^{3} + 0 \times 2^{2} + 0 \times 2^{1} + 1 \times 2^{0}\)

= \(8 + 0 + 0 + 1\)

= 9

\(\sqrt{9} = 3_{10}\)

\(\therefore \sqrt{1001_{2}} = 11_{2}\)

(b) In \(\Delta QOP, OQ = OP\)

\(\therefore QK = KP\)

\(\Delta QOK = \Delta POK\) (right- angled triangle)

In \(\Delta POK\),

\(OP^{2} = PK^{2} + OK^{2}\)

\(x^{2} = y^{2} + (\frac{z}{2})^{2}\)

\(\frac{z^{2}}{4} = x^{2} - y^{2}\)

\(z^{2} = 4x^{2} - 4y^{2}\)

\(z = \sqrt{4(x^{2} - y^{2})}\)

\(z = 2\sqrt{x^{2} - y^{2}}\)

(c)(i) \(< PSQ = \frac{1}{2}(< POQ) = \frac{1}{2}(160°)\)

= 80°

In \(\Delta PSQ\),

\(< PQS + < QPS + < PSQ = 180°\)

\(40° + < QPS + 80° = 180°\)

\(< QPS = 180° - 120° = 60°\)

(ii) \(< PSR = < PSQ + < QSR\)

= \(80° + 45°\)

= \(125°\)

\(< PQS = < PQO + < RQS \)

= \(40° + < RQS\)

\(< PSR + < PQS = 180°\)

\(\therefore 125° + 40° + < RQS = 180°\)

\(< RQS = 180° - 165°\)

= \(15°\)

Explanation

Let one of the sides be x metres.

The other is (x + 1.5) metres.

(a) \(Area = Length \times breadth\)

\(x(x + 1.5) = 13.5\)

\(x^{2} + 1.5x = 13.5 \implies x^{2} + 1.5x - 13.5 = 0\)

\(x^{2} + 4.5x - 3x - 13.5 = 0 \implies x(x + 4.5) - 3(x + 4.5) = 0\)

\((x - 3)(x + 4.5) = 0 \implies x = 3 ; x = -4.5\)

Since x cannot be negative, therefore x = 3m.

The longer side = (3 + 1.5) = 4.5m

(b) \(N250 = 1m^{2}\)

\(N2000 = \frac{2000}{250} = 8m^{2}\)

\(\therefore \text{The area of the floor that will be covered by the carpet} = 8m^{2}\)