WAEC SSCE - General Mathematics - 2013

Akọwa Nkọwa

We know that sin x = \(\frac{5}{13}\) and 0^o \(\leq\) x \(\leq\) 90^o. First, we can find the value of cos x using the identity: sin² x + cos² x = 1 sin² x + cos² x = 1 \(\frac{25}{169}\) + cos² x = 1 cos² x = \(\frac{144}{169}\) cos x = \(\pm\)\(\frac{12}{13}\) Since 0^o \(\leq\) x \(\leq\) 90^o, we know that cos x is positive. Therefore, cos x = \(\frac{12}{13}\). Next, we can find the value of tan x using the identity: tan x = \(\frac{sin x}{cos x}\) tan x = \(\frac{sin x}{cos x}\) = \(\frac{\frac{5}{13}}{\frac{12}{13}}\) = \(\frac{5}{12}\) Finally, we can find the value of (cos x - tan x) as: cos x - tan x = \(\frac{12}{13}\) - \(\frac{5}{12}\) = \(\frac{79}{156}\) Therefore, the answer is (cos x - tan x) = \(\frac{79}{156}\).

Akọwa Nkọwa

Akọwa Nkọwa

To add fractions, we need to have a common denominator. In this case, the common denominator is \((x+3)(x-2)\). Therefore, we need to convert each fraction to have this denominator. \[\frac{2}{x+3} - \frac{1}{x-2} = \frac{2(x-2)}{(x+3)(x-2)} - \frac{(x+3)}{(x+3)(x-2)}\] Simplifying the above expression, we have: \[\frac{2(x-2) - (x+3)}{(x+3)(x-2)} = \frac{2x - 4 - x - 3}{(x+3)(x-2)} = \frac{x-7}{(x+3)(x-2)}\] Therefore, \(\frac{2}{x + 3} - \frac{1}{x - 2} = \boxed{\frac{x-7}{(x+3)(x-2)}}\). The correct option is (a).

Akọwa Nkọwa

Akọwa Nkọwa

The probability of an event happening can never be greater than 1. Therefore, 1.01 cannot be a probability of Mary scoring 85% in a mathematics test.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

We can start by substituting the given values: \begin{align*} \frac{x^{\frac{1}{2}} - y^{\frac{1}{3}}}{y - x^{\frac{2}{3}}} &= \frac{64^{\frac{1}{2}} - 27^{\frac{1}{3}}}{27 - 64^{\frac{2}{3}}} \\ &= \frac{8 - 3}{27 - 16} \\ &= \frac{5}{11} \end{align*} Therefore, the answer is $\frac{5}{11}$.

Akọwa Nkọwa

To multiply two numbers in scientific notation, we simply multiply their coefficients and add their exponents. (2.7 x 10^-4) x (6.3 x 10⁶) = (2.7 x 6.3) x 10^-4+6 = 17.01 x 10² We can express 17.01 x 10² in standard form by moving the decimal point to the left two places, which gives us: 1.701 x 10³ Therefore, the correct answer is (c) 1.701 x 10³.

Akọwa Nkọwa

Akọwa Nkọwa

In a regular polygon, each exterior angle is equal to 360 degrees divided by the number of sides. Let the number of sides be represented by n. Therefore, the measure of each exterior angle is 360/n. Since the interior angle and exterior angle are supplementary, we can write: Interior angle + Exterior angle = 180 degrees Let x be the measure of each interior angle. We are given that each interior angle is 5 times each exterior angle, so: x = 5(360/n - x) Simplifying and solving for x, we get: x = 150 degrees The sum of the interior angles of an n-sided polygon is (n - 2) × 180 degrees. Since each interior angle in our polygon is 150 degrees, we can write: n × 150 = (n - 2) × 180 Simplifying and solving for n, we get: n = 12 Therefore, the polygon has 12 sides. So, the correct answer is 12.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Given the equations: \begin{align*} \frac{1}{2}x + 2y &= 3 \\ \frac{3}{2}x - 2y &= 1 \\ \end{align*} We can solve for x and y using simultaneous equations: First, multiply the first equation by 2: \begin{align*} x + 4y &= 6 \\ \frac{3}{2}x - 2y &= 1 \\ \end{align*} Then, multiply the second equation by 2: \begin{align*} x + 4y &= 6 \\ 3x - 4y &= 2 \\ \end{align*} Add the equations together: \begin{align*} 4x &= 8 \\ x &= 2 \\ \end{align*} Substitute x = 2 back into the first equation: \begin{align*} \frac{1}{2}(2) + 2y &= 3 \\ 1 + 2y &= 3 \\ 2y &= 2 \\ y &= 1 \\ \end{align*} Finally, substitute x = 2 and y = 1 into (x + y): \begin{align*} x + y &= 2 + 1 \\ &= 3 \\ \end{align*} Therefore, (x + y) = 3, and the answer is.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

To solve this problem, we need to find out in what base the addition statement is true. Let's assume the base is "b". Then, in the units column, we have: - 5 + 4 = 10, so we write down 0 and carry-over 1. - In the "b" column, we have: 6 + 2 + 2 + 1 = 11, so we write down 1 and carry-over 1. - In the "b^2" column, we have: 4 + 2 + 5 + 1 = 12, so we write down 2 and carry-over 1. - In the "b^3" column, we have: 4 + 2 + 2 = 8. Putting these digits together, we get the number 8201 in base "b". Now, we need to check if this number is equal to 1050 in base 10. 8201 in base "b" means: 8 x b^3 + 2 x b^2 + 0 x b + 1 x 1 = 1050 Rearranging the terms, we get: 8 b^3 + 2 b^2 + 1 = 1050 Subtracting 1 from both sides: 8 b^3 + 2 b^2 = 1049 Since b is a positive integer, we can see that b must be greater than 5. Trying b = 6, we get: 8 x 6^3 + 2 x 6^2 = 1048 This is not equal to 1049, so we need to try a larger base. Trying b = 7, we get: 8 x 7^3 + 2 x 7^2 = 1049 This is equal to 1049, so the base is 7. Therefore, the answer is seven.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

To solve this problem, we can use the Pythagorean theorem to find the length of the chord. We know that the radius of the circle is 5cm and the distance from the centre of the circle to the chord is 2cm. We can draw a perpendicular line from the centre of the circle to the chord, which will divide the chord into two equal parts. This perpendicular line will also bisect the chord and form a right triangle with the radius of the circle and half of the chord. The hypotenuse of this triangle is the radius of the circle, which is 5cm, and one leg is half of the chord, which we can call x. The other leg is the distance from the centre of the circle to the chord, which is 2cm. Using the Pythagorean theorem, we can solve for x: 5^2 = x^2 + 2^2 25 = x^2 + 4 x^2 = 21 x = √21 Therefore, the length of the chord is twice the value of x, which is 2√21cm. Hence, the correct answer is option A.

Akọwa Nkọwa

Akọwa Nkọwa

Four oranges sell for Nx, so one orange costs \(\frac{N}{4}x\). Three mangoes sell for Ny, so one mango costs \(\frac{N}{3}y\). Olu bought 24 oranges and 12 mangoes, so he paid: \[24 \cdot \frac{N}{4}x + 12 \cdot \frac{N}{3}y = 6Nx + 4Ny\] Therefore, Olu paid N(6x + 4y) in terms of x and y. The answer is option (B).

Akọwa Nkọwa

Akọwa Nkọwa

Let the base area of the cube be x, then the length of one side of the cube is \(\sqrt[3]{64}\) = 4 cm. Since the base area of the cube and cuboid are equal, the base of the cuboid must also have an area of x. The volume of the cuboid is given as 80cm\(^3\) which can be expressed as: 80 = x × h, where h is the height of the cuboid We know that the length of the base of the cuboid is equal to the length of the side of the cube. Therefore, the dimensions of the cuboid are 4 cm by 4 cm by h cm. Using the formula for the volume of a cuboid, we get: Volume of cuboid = length × width × height = 4 × 4 × h = 16h Substituting 80 for the volume of the cuboid, we get: 16h = 80 Solving for h, we get: h = 5cm Therefore, the height of the cuboid is 5cm. Hence, the correct answer is 5cm.

Akọwa Nkọwa

We are given that y varies directly with the square root of (x+1), which can be written as y=k√(x+1), where k is the constant of variation. To find k, we use the given values of x and y: y=k√(x+1) 6=k√(3+1) 6=k√4 6=2k k=3 So the equation for y in terms of x is y=3√(x+1). To find x when y=9, we substitute these values into the equation and solve for x: 9=3√(x+1) 3=√(x+1) 9=x+1 x=8 Therefore, x = 8 when y = 9.

Akọwa Nkọwa

Let's call the number we are looking for "x". According to the problem, when we subtract x from 2, the result is equal to 4 less than one-fifth of the number. In mathematical terms, this can be written as: 2 - x = (1/5)x - 4 To solve for x, we can start by simplifying the equation by adding x to both sides: 2 = (6/5)x - 4 Next, we can add 4 to both sides: 6 = (6/5)x Finally, we can multiply both sides by 5/6 to isolate x: x = 5 Therefore, the answer is 5.

Akọwa Nkọwa

We want to solve for q, so we need to isolate it on one side of the equation. First, we can isolate the cube root of p by cubing both sides of the equation: p^(1/3) = (3√q)/r (p^(1/3))³ = (3√q)³/r³ p = (27q)/r³ Next, we can isolate q by multiplying both sides by r³/27: (p/27)r³ = q Therefore, the solution is q = (p/27)r³, which is equivalent to q = pr^(1/3).

Akọwa Nkọwa

Akọwa Nkọwa

In a histogram, the mode is the value with the highest frequency, or the tallest bar. Looking at the given histogram, it appears that the tallest bar is the one corresponding to the interval between 52 and 54, which means that the mode lies somewhere in that interval. Since the interval width is 2, we can estimate the mode to be around the middle of the interval, which is (52 + 54)/2 = 53. Therefore, the estimated mode of the distribution is 53.5. So, the answer is (c) 53.5.

Akọwa Nkọwa

The formula for the volume of a pyramid is given by V = 1/3 * B * h, where B is the area of the base and h is the height of the pyramid. In this case, the base of the pyramid is a rectangle with dimensions 12m by 8m, so its area is A = 12m * 8m = 96m². The height of the pyramid is given as 14m. Substituting these values into the formula, we have V = 1/3 * 96m² * 14m = 448m³. Therefore, the volume of the pyramid is 448m³.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

The sales boy gave a change of N68 instead of N72, which means he gave N4 less than he should have. To find the percentage error, we use the formula: Percentage Error = (Error / True Value) × 100% In this case, the True Value is N72 and the Error is N4. Substituting these values into the formula, we get: Percentage Error = (4 / 72) × 100% Percentage Error = 0.055555... × 100% Percentage Error = 5.5555...% Rounding off to the nearest whole number, we get: Percentage Error ≈ 6% Therefore, the sales boy's percentage error is approximately 6%. The closest option to this answer is 5\(\frac{5}{9}\)%, so the answer would be: - 5\(\frac{5}{9}\)%

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

We can start by simplifying the left-hand side of the equation using the laws of square roots: \begin{align*} \sqrt{50} - K\sqrt{8} &= \sqrt{25\cdot 2} - K\sqrt{4\cdot 2} \\ &= 5\sqrt{2} - 2K\sqrt{2} \\ &= \sqrt{2}(5 - 2K). \end{align*} Now, we can rewrite the given equation as: \begin{align*} \sqrt{2}(5 - 2K) &= \frac{2}{\sqrt{2}} \\ 5 - 2K &= \frac{2}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} \\ 5 - 2K &= \frac{2}{2} \\ 5 - 2K &= 1 \\ -2K &= -4 \\ K &= 2. \end{align*} Therefore, the value of K that satisfies the equation is 2.

Akọwa Nkọwa

We are given the equation y = x2 + 2x + k and we know that it passes through the point (2,0). We can substitute x = 2 and y = 0 into the equation to find k. Substituting, we get: 0 = 22 + 2(2) + k 0 = 4 + 4 + k 0 = 8 + k k = -8 Therefore, the value of k is -8, which is the fourth option.

Akọwa Nkọwa

We can begin by factoring out (x + 2) from both terms on the left side of the equation: (x + 2)(x² - 3x + 2) + 2(x + 2)(x - 1) = (x + 2) M (x + 2)[(x² - 3x + 2) + 2(x - 1)] = (x + 2) M Simplifying the expression in the brackets, we get: (x + 2)(x² - x) = (x + 2) M Now, we can cancel out (x + 2) from both sides of the equation: x² - x = M Therefore, the value of M is simply x² - x.

Akọwa Nkọwa

The intersection of two sets x and y1 is represented by the overlapping region of the two circles. From the diagram, we can see that the number of elements in the intersection is 2. Therefore, n(x \(\cap\) y1) = 2.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

The set p represents all values of y such that 2y is greater than or equal to 6. Simplifying the inequality 2y \(\geq\) 6 gives y \(\geq\) 3. Therefore, p can be written as p = {y: y \(\geq\) 3}. Similarly, the set Q represents all values of y such that y - 3 is greater than or equal to 4. Simplifying the inequality y - 3 \(\geq\) 4 gives y \(\geq\) 7. Therefore, Q can be written as Q = {y: y \(\geq\) 7}. The intersection of p and Q, denoted by p\(\cap\)Q, is the set of all values of y that are in both p and Q. Since p contains all values of y greater than or equal to 3 and Q contains all values of y greater than or equal to 7, the intersection of p and Q is {y: y \(\geq\) 7}. Therefore, p\(\cap\)Q = {7, 8, 9, 10, ...}. The correct option is (c) {3, 4, 5, 6, 7}.

Akọwa Nkọwa

To find the median class in a histogram, we need to identify the class interval that contains the median value. The median is the middle value in a dataset, so we need to find the midpoint of the data. To find the median class: 1. Add up the frequencies in the histogram starting from the left-hand side until the total is greater than or equal to the total number of data points divided by 2. 2. The median class is the class interval that contains the midpoint of the data. In this histogram, the total number of data points is 80. The midpoint is (80 + 1)/2 = 40.5. Starting from the left-hand side, we can add up the frequencies: 8 + 19 + 24 = 51. The median falls within the interval 40.5 - 50.5, which is the third interval. Therefore, the median class is 40.5 - 50.5.

Akọwa Nkọwa

Akọwa Nkọwa

(a) Perimeter of the remaining board.

Reading the diagram. The board is a rectangle \(22\text{ cm}\) wide and \(12\text{ cm}\) tall. A circular segment (curved arch) is cut from the bottom edge, leaving flat pieces of \(5\text{ cm}\) on the left and \(3\text{ cm}\) on the right. Hence the chord (span of the cut) is

\[\text{chord} = 22 - 5 - 3 = 14\text{ cm}.\]

The radius is \(1\tfrac{1}{2}\) times the chord:

\[r = \tfrac{3}{2}\times 14 = 21\text{ cm}.\]

Central angle of the segment. Half the chord is \(7\text{ cm}\). If the chord subtends \(2\theta\) at the centre,

\[\sin\theta = \frac{7}{21} = \frac{1}{3} \;\Rightarrow\; \theta = 19.47^\circ,\qquad 2\theta = 38.94^\circ.\]

Arc length of the cut. With \(\pi = \tfrac{22}{7}\):

\[\text{arc} = \frac{2\theta}{360}\times 2\pi r = \frac{38.94}{360}\times 2\times\frac{22}{7}\times 21 = \frac{38.94}{360}\times 132 = 14.28\text{ cm}.\]

Perimeter of the remaining portion = top + two sides + two flat bottoms + arc:

\[P = 22 + 12 + 12 + 5 + 3 + 14.28 = 68.28\text{ cm}.\]

Perimeter \(\approx \mathbf{68.28\text{ cm}}\) (2 d.p.).

(b) Surd evaluation.

\[\frac{3}{\sqrt{3}}\left(\frac{2}{\sqrt{3}} - \frac{\sqrt{12}}{6}\right).\]

First, \(\dfrac{3}{\sqrt{3}} = \dfrac{3}{\sqrt3}\times\dfrac{\sqrt3}{\sqrt3} = \dfrac{3\sqrt3}{3} = \sqrt{3}.\) Inside the bracket, \(\dfrac{2}{\sqrt3} = \dfrac{2\sqrt3}{3}\) and \(\sqrt{12} = 2\sqrt3\), so \(\dfrac{\sqrt{12}}{6} = \dfrac{2\sqrt3}{6} = \dfrac{\sqrt3}{3}.\) Therefore

\[\frac{2}{\sqrt3} - \frac{\sqrt{12}}{6} = \frac{2\sqrt3}{3} - \frac{\sqrt3}{3} = \frac{\sqrt3}{3}.\]

Multiplying:

\[\sqrt{3}\times\frac{\sqrt3}{3} = \frac{3}{3} = \mathbf{1}.\]

Akọwa Nkọwa

(a) Father : son \(= 10 : 3\), and the son is \(15\). So father's age \(= \dfrac{10}{3}\times 15 = 50\) years.

Let the ratio be \(2 : 1\) in \(x\) years: \[\frac{50 + x}{15 + x} = \frac{2}{1} \Rightarrow 50 + x = 2(15 + x) = 30 + 2x.\] \[50 - 30 = 2x - x \Rightarrow x = 20.\] The ratio will be \(2 : 1\) in \(\mathbf{20\ \text{years}}.\)

(b) Mean of \(x, y, z\) is \(6\): \(x + y + z = 18.\)

Mean of \(x, y, z, l, u, v, w\) is \(9\): \(x + y + z + l + u + v + w = 63.\)

So \(l + u + v + w = 63 - 18 = 45.\)

Mean of \(l, u, v, w = \dfrac{45}{4} = \mathbf{11.25}.\)

Akọwa Nkọwa

Akọwa Nkọwa

(a)(i) Sum equals three times the difference: \[p + q = 3(p - q) \Rightarrow p + q = 3p - 3q \Rightarrow 4q = 2p \Rightarrow \mathbf{p = 2q}.\]

(a)(ii) Substitute \(p = 2q\): \[\frac{p^2 + q^2}{pq} = \frac{(2q)^2 + q^2}{(2q)q} = \frac{4q^2 + q^2}{2q^2} = \frac{5q^2}{2q^2} = \frac{5}{2} = \mathbf{2.5}.\]

(b) Selling price of \(100\) articles at \(25\) for \(\text{N}66\): \(\dfrac{100}{25}\times 66 = 4\times 66 = \text{N}264.\)

This gives a \(32\%\) gain, so cost price \(= \dfrac{264}{1.32} = \text{N}200.\)

New selling price at \(20\) for \(\text{N}50\): \(\dfrac{100}{20}\times 50 = 5\times 50 = \text{N}250.\)

New gain \(= 250 - 200 = \text{N}50\), so \[\text{gain}\% = \frac{50}{200}\times 100 = \mathbf{25\%\ \text{gain}}.\]

Akọwa Nkọwa

(a) \(7x + 4 < \tfrac{1}{2}(4x + 3)\).

Multiply the bracket: \(7x + 4 < 2x + \tfrac{3}{2}.\)

\(7x - 2x < \tfrac{3}{2} - 4 \Rightarrow 5x < -\tfrac{5}{2} \Rightarrow x < -\tfrac{1}{2}.\)

Solution: \(\mathbf{x < -0.5}.\)

(b) For every \(\text{N}2.00\) Salem gets, Sunday gets \(50\) kobo \((\text{N}0.50)\): so Salem : Sunday \(= 2 : 0.5 = 4 : 1.\)

For every \(\text{N}4.00\) Sunday gets, Shaka gets \(\text{N}2.00\): so Sunday : Shaka \(= 4 : 2 = 2 : 1.\)

Make Sunday common: Salem : Sunday \(= 8 : 2\) and Sunday : Shaka \(= 2 : 1\), giving \[\text{Salem : Sunday : Shaka} = 8 : 2 : 1.\] Total parts \(= 11\). One part \(= \dfrac{\text{N}1100}{11} = \text{N}100.\)

Shaka's share \(= 1 \times \text{N}100 = \mathbf{\text{N}100.00}.\)

Akọwa Nkọwa

Akọwa Nkọwa

(a) Numerator: \(\tfrac{3}{4}\left(3\tfrac{3}{8} + 1\tfrac{5}{8}\right) = \tfrac{3}{4}\left(\tfrac{27}{8} + \tfrac{13}{8}\right) = \tfrac{3}{4}\times\tfrac{40}{8} = \tfrac{3}{4}\times 5 = \tfrac{15}{4}.\)

Denominator: \(2\tfrac{1}{8} - 1\tfrac{1}{2} = \tfrac{17}{8} - \tfrac{12}{8} = \tfrac{5}{8}.\)

Therefore \(\dfrac{15/4}{5/8} = \dfrac{15}{4}\times\dfrac{8}{5} = \dfrac{120}{20} = \mathbf{6}.\)

(b) Write \(1.125 = \dfrac{9}{8} = \dfrac{3^2}{2^3}.\) Then \[\log_{10}1.125 = 2\log_{10}3 - 3\log_{10}2 = 2(0.4771) - 3(0.3010) = 0.9542 - 0.9030 = 0.0512.\] Correct to 2 significant figures, \(\log_{10}1.125 \approx \mathbf{0.051}.\)

Akọwa Nkọwa

(a) Diagram. A vertical wall meets the horizontal ground at a right angle. The ladder \(LM\) leans with its top \(L\) touching the wall \(5\) m above the ground and its foot \(M\) on the ground, making \(37^\circ\) with the ground. The height (5 m), the ground distance, and the ladder form a right-angled triangle.

(b) Length of the ladder. The 5 m height is opposite the \(37^\circ\) angle:

\[\sin 37^\circ = \frac{5}{|LM|} \;\Rightarrow\; |LM| = \frac{5}{\sin 37^\circ} = \frac{5}{0.6018} = 8.31\text{ m (3 s.f.)}\]

(c) New angle after moving the foot 2 m towards the wall. Original horizontal distance of the foot from the wall:

\[|LM|\cos 37^\circ = 8.308 \times 0.7986 = 6.635\text{ m}\]

Pushing the foot 2 m nearer gives a new base distance \(6.635 - 2 = 4.635\) m, with the ladder length unchanged (8.308 m). If \(\alpha\) is the new angle with the ground:

\[\cos\alpha = \frac{4.635}{8.308} = 0.5579 \;\Rightarrow\; \alpha = 56^\circ \text{ (nearest degree)}\]

Akọwa Nkọwa

First find the radius from the area: \[\pi r^2 = 154 \Rightarrow \frac{22}{7}r^2 = 154 \Rightarrow r^2 = \frac{154\times 7}{22} = 49 \Rightarrow r = 7\,\text{cm}.\]

Let each of the two equal sectors have angle \(a\). The third sector is three times the two put together: \[\text{third} = 3(2a) = 6a.\] Total angle: \(2a + 6a = 8a = 360^\circ \Rightarrow a = 45^\circ.\)

So the third sector has angle \(6a = 270^\circ.\)

Arc length of third sector: \[\frac{270}{360}\times 2\pi r = \frac{3}{4}\times 2\times\frac{22}{7}\times 7 = \frac{3}{4}\times 44 = 33\,\text{cm}.\]

Perimeter of a sector = arc + two radii: \[33 + 2(7) = 33 + 14 = \mathbf{47\,\text{cm}}.\]

Akọwa Nkọwa

Akọwa Nkọwa

(a) Take the pole vertical with the lamp \(L\) at its top \((4.2\,\text{m})\). The boy \(BC\) is \(1.2\,\text{m}\) tall standing \(6\,\text{m}\) from the pole's foot; his shadow \(CS\) stretches away from the pole on the ground. The light ray from \(L\) passes over the boy's head to the shadow tip \(S\).

(b)(i) Let the shadow length be \(s\). By similar triangles (light from \(L\) grazing the boy's head): \[\frac{\text{lamp height}}{\text{pole-to-shadow-tip}} = \frac{\text{boy height}}{\text{boy-to-shadow-tip}} \Rightarrow \frac{4.2}{6 + s} = \frac{1.2}{s}.\] \[4.2 s = 1.2(6 + s) \Rightarrow 4.2s = 7.2 + 1.2s \Rightarrow 3s = 7.2 \Rightarrow s = 2.4.\] Length of the boy's shadow \(= \mathbf{2.4\,\text{m}}.\)

(b)(ii) Angle of elevation of the lamp from the boy (measured from the top of his head): the horizontal distance is \(6\,\text{m}\) and the lamp is \(4.2 - 1.2 = 3.0\,\text{m}\) above him. \[\tan\theta = \frac{3.0}{6} = 0.5 \Rightarrow \theta = 26.57^\circ \approx \mathbf{27^\circ}.\]

Akọwa Nkọwa

(a) Draw a circle for the earth with the equator horizontal and the North pole on top. Mark \(T\) on the equator (longitude \(15^\circ E\)). The plane flies due North along the \(15^\circ E\) meridian to \(P\), then East along the parallel of latitude through \(P\) to \(Q\) on longitude \(65^\circ E\).

(b)(i) Latitude of P: Distance North \(= 950\times 4 = 3800\,\text{km}\), an arc along a meridian: \[3800 = \frac{\theta}{360}\times 2\pi R = \frac{\theta}{360}\times 2\times\frac{22}{7}\times 6400.\] \[2\pi R = 40228.6\,\text{km} \Rightarrow \theta = \frac{3800\times 360}{40228.6} = 34.0^\circ.\] Latitude of \(P \approx \mathbf{34^\circ N}.\)

(b)(ii) Distance P to Q: Along the parallel of latitude \(34^\circ N\), the difference in longitude is \(65^\circ - 15^\circ = 50^\circ\): \[PQ = \frac{50}{360}\times 2\pi R\cos34^\circ = \frac{50}{360}\times 40228.6\times 0.8290.\] \[PQ = 5587.3\times 0.8290 = 4632\,\text{km}.\] Correct to four significant figures, \(|PQ| \approx \mathbf{4632\,\text{km}}.\)

Akọwa Nkọwa

(a) Geometric progression

For three consecutive GP terms, the square of the middle term equals the product of the outer terms:

\[6^2 = (3 - x)(7 - 5x)\]\[36 = 21 - 15x - 7x + 5x^2\]\[36 = 5x^2 - 22x + 21\]\[5x^2 - 22x - 15 = 0\]

(i) Values of x

Using the quadratic formula with \(a = 5,\; b = -22,\; c = -15\):

\[x = \frac{22 \pm \sqrt{(-22)^2 - 4(5)(-15)}}{2(5)} = \frac{22 \pm \sqrt{484 + 300}}{10} = \frac{22 \pm \sqrt{784}}{10} = \frac{22 \pm 28}{10}\]\[x = 5 \quad\text{or}\quad x = -\frac{6}{10} = -\frac{3}{5}\]

(ii) Constant ratio

The ratio is \(r = \dfrac{6}{3 - x}\).

• If \(x = 5\): \(r = \dfrac{6}{3-5} = \dfrac{6}{-2} = -3\) (negative, rejected since \(r > 0\)).
• If \(x = -\dfrac{3}{5}\): \(r = \dfrac{6}{3 + \frac{3}{5}} = \dfrac{6}{\frac{18}{5}} = 6\times\dfrac{5}{18} = \dfrac{5}{3}\) (positive, accepted).

So the admissible value is \(x = -\dfrac{3}{5}\) with constant ratio \(r = \dfrac{5}{3}\). (Check: terms are \(3.6,\; 6,\; 10\), and \(\tfrac{6}{3.6} = \tfrac{10}{6} = \tfrac{5}{3}\).)

(b) Finding \(\angle ADC\)

In the quadrilateral, \(|AB| = 3\text{ cm}\), \(|BC| = 4\text{ cm}\), \(|CD| = 6\text{ cm}\), \(|DA| = 7\text{ cm}\), and the angle at B is \(90^\circ\). Draw the diagonal \(AC\).

In right triangle \(ABC\):

\[|AC|^2 = |AB|^2 + |BC|^2 = 3^2 + 4^2 = 9 + 16 = 25 \;\Rightarrow\; |AC| = 5\text{ cm}\]

In triangle \(ACD\), apply the cosine rule with \(\angle ADC\) opposite \(AC\):

\[|AC|^2 = |DA|^2 + |DC|^2 - 2\,|DA|\,|DC|\cos(\angle ADC)\]\[25 = 7^2 + 6^2 - 2(7)(6)\cos(\angle ADC)\]\[25 = 49 + 36 - 84\cos(\angle ADC)\]\[84\cos(\angle ADC) = 85 - 25 = 60\]\[\cos(\angle ADC) = \frac{60}{84} = 0.7143\]\[\angle ADC = \cos^{-1}(0.7143) = 44.4^\circ \approx 44^\circ\]

Therefore \(\angle ADC \approx 44^\circ\).