WAEC SSCE - General Mathematics - 2018

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To find the value of p, we need to determine the slope of the line passing through the two given points. The slope of a line passing through two points (x1, y1) and (x2, y2) is given by the formula: slope = (y2 - y1) / (x2 - x1) Using the coordinates of the given points, we get: slope = (-2 - 2) / (-8 - 4) = -4 / (-12) = 1/3 Since the equation of the line is given as 3y = px + q, we can rewrite this equation in slope-intercept form, y = (p/3)x + (q/3), by dividing both sides by 3. The slope of the line in slope-intercept form is then (p/3). Since we know the slope of the line passing through the two given points is 1/3, we can set these two expressions equal to each other and solve for p: (p/3) = 1/3 Multiplying both sides by 3, we get: p = 1 Therefore, the value of p is 1.

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The volume of a cylinder is given by the formula V = πr²h, where V is the volume, r is the radius and h is the height of the cylinder. In this question, we are given that the height of the cylindrical tank is 10m and its volume is 385 m². Therefore, we can find the radius of the cylinder as follows: 385 = πr² × 10 r² = 38.5/π r = √(38.5/π) = 3.5m (approximately) Finally, we can find the diameter of the cylinder by doubling the radius: Diameter = 2r = 2 × 3.5 = 7m Therefore, the diameter of the tank is 7m.

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We are given that there are 250 boys and 150 girls in a school. We are also given that 60% of the boys play football, which means that 0.6 x 250 = 150 boys play football. Similarly, 40% of the girls play football, which means that 0.4 x 150 = 60 girls play football. Therefore, the total number of students who play football is 150 + 60 = 210. To find the percentage of the school that plays football, we need to divide the number of students who play football by the total number of students in the school and multiply by 100. The total number of students in the school is 250 + 150 = 400. So, the percentage of the school that plays football is (210/400) x 100 = 52.5%. Therefore, the correct option is 52.5% (option D).

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We are given that the curve passes through the points \((-2, 0)\) and \((3, 0)\). Since the curve is a function of \(x\), we can assume that the equation of the curve is of the form \(y = f(x)\). If the curve passes through the point \((-2, 0)\), then we can substitute \(x = -2\) and \(y = 0\) into the equation of the curve to get: $$ 0 = f(-2) $$ Similarly, if the curve passes through the point \((3, 0)\), then we can substitute \(x = 3\) and \(y = 0\) into the equation of the curve to get: $$ 0 = f(3) $$ Therefore, the curve must have at least two roots, namely \(x = -2\) and \(x = 3\). This suggests that the curve is a quadratic function. We can use the information about the roots of the curve to write the equation of the curve in factored form: $$ y = A(x + 2)(x - 3) $$ where \(A\) is a constant. Since the coefficient of \(x^2\) in the equation is 1, we can simplify this equation to: $$ y = Ax^2 + Bx + C $$ where \(A = 1\), and \(B\) and \(C\) are constants. To find the values of \(B\) and \(C\), we can substitute the points \((-2, 0)\) and \((3, 0)\) into the equation: $$ 0 = A(-2)^2 + B(-2) + C \quad \text{and} \quad 0 = A(3)^2 + B(3) + C $$ Simplifying these equations, we get: $$ 4A - 2B + C = 0 \quad \text{and} \quad 9A + 3B + C = 0 $$ We can use these two equations to solve for \(B\) and \(C\). Adding the two equations, we get: $$ 13A + B = 0 $$ Substituting this expression for \(B\) into one of the previous equations, we get: $$ 4A - 2(-13A) + C = 0 $$ Simplifying, we get: $$ 20A + C = 0 $$ Solving these two equations simultaneously, we get: $$ A = \frac{1}{13}, \quad B = -\frac{4}{13}, \quad C = 0 $$ Therefore, the equation of the curve is: $$ y = \frac{1}{13}(x + 2)(x - 3) = \frac{1}{13}(x^2 - x - 6) $$ So the correct answer is \(y = x^2 - x - 6\).

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To find the distance between two points M(-3, 8) and N(5, -7), we use the distance formula: |MN| = √[(x₂ - x₁)² + (y₂ - y₁)²] where (x₁, y₁) and (x₂, y₂) are the coordinates of M and N, respectively. Substituting the values, we get: |MN| = √[(5 - (-3))² + (-7 - 8)²] |MN| = √[(5 + 3)² + (-15)²] |MN| = √[8² + 15²] |MN| = √(64 + 225) |MN| = √289 |MN| = 17 Therefore, the distance |MN| between the two points is 17 units. So, the correct option is "17 units".

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Since O is the midpoint of WZ and the diameter of the semicircle, it follows that O lies on the circumference of the semicircle. Let angle WOZ be m. Then angle WXY is also m, since |WX| = |XY|. Similarly, angle YXZ is also m. Since the sum of angles in a triangle is 180 degrees, we have: angle WXY + angle YXZ + angle WYZ = 180 degrees Substituting m for angle WXY and angle YXZ, we get: 2m + angle WYZ = 180 degrees Since WXYZ is a trapezium, we have: angle WYZ + angle WXY = 180 degrees Substituting m for angle WXY, we get: angle WYZ + m = 180 degrees Combining this equation with the previous one, we get: 2m + angle WYZ = angle WYZ + m + 180 degrees Simplifying, we get: m = 60 degrees Therefore, the value of m is 60 degrees.

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To simplify \(\frac{2-18m^2}{1+3m}\), we need to factor the numerator and denominator. First, we can factor out a 2 from the numerator: \[\frac{2-18m^2}{1+3m} = \frac{2(1-9m^2)}{1+3m}\] Next, we can factor the numerator further using the difference of squares formula: \[\frac{2(1-9m^2)}{1+3m} = \frac{2(1-3m)(1+3m)}{1+3m}\] Finally, we can cancel out the common factor of \((1+3m)\) in the numerator and denominator: \[\frac{2(1-3m)(1+3m)}{1+3m} = 2(1-3m)\] Therefore, the simplified form of \(\frac{2-18m^2}{1+3m}\) is \(2(1-3m)\).

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To simplify this expression, we need to follow the order of operations (PEMDAS): 1. First, we perform the multiplication of the mixed numbers: 2\(\frac{1}{4} \times 3\frac{1}{2} = \frac{9}{4} \times \frac{7}{2} = \frac{63}{8}\) 2. Then, we perform the division of the mixed numbers: 4 \(\frac{3}{8} = \frac{35}{8}\) \(\frac{63}{8} \div \frac{35}{8} = \frac{63}{8} \times \frac{8}{35} = \frac{9}{5} =\) 1\(\frac{4}{5}\) Therefore, the answer is option (D), 1\(\frac{4}{5}\).

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To solve for x, we can use the following steps: 1. Rewrite the equation using exponential form: 10^(log(6x - 4) - log(2)) = 10^1 2. Use the fact that log(a) - log(b) = log(a/b) to simplify the equation: 10^(log((6x - 4) / 2)) = 10^1 3. Use the definition of logarithms to simplify the equation: (6x - 4) / 2 = 10 4. Solve for x by multiplying both sides of the equation by 2 and then subtracting 4 from both sides: 6x - 4 = 20, 6x = 24, and x = 4 So, the solution to the equation is x = 4.

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To find the value of t, we need to use the information given in the diagram. In a triangle, the sum of all angles is 180 degrees. So, if we know two angles, we can find the third one by subtracting their sum from 180. In this case, we know the angles labeled x and y. So, to find t, we can add x and y and then subtract the sum from 180. That is, t = 180 - (x + y). So, the value of t is 126 degrees.

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To find the value of x, we need to use the given information that the mean of 1, 3, 5, 7, and x is 4. The mean of a set of numbers is found by adding up all the numbers in the set and then dividing by the total number of numbers. So, we can write an equation: (1 + 3 + 5 + 7 + x) / 5 = 4 To solve for x, we can first simplify the equation by multiplying both sides by 5: 1 + 3 + 5 + 7 + x = 20 Then, we can solve for x by subtracting the sum of the known numbers from both sides: x = 20 - (1 + 3 + 5 + 7) x = 20 - 16 x = 4 Therefore, the value of x is 4.

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The area of the rectangle WXYZ is 8cm * 6cm = 48 cm\(^2\). The shaded part is made up of four congruent triangles, each of which has an area of 1/2 * base * height = 1/2 * 4cm * 3cm = 6 cm\(^2\). So the total area of the shaded part is 4 * 6 cm\(^2\) = 24 cm\(^2\). The area of the part of the rectangle that is not shaded is 48 cm\(^2\) - 24 cm\(^2\) = 24 cm\(^2\).

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In this question, we are asked to find the value of x for which \(32_{four} = 22_x\). We know that \(32_{four}\) means 3 fours plus 2 ones, which is equal to 14 in the decimal system. So, we have: $$14_{10} = 2\times x^1 + 2\times x^0 = 2x+2$$ Now we can solve for x: $$14_{10} = 2x+2$$ $$12_{10} = 2x$$ $$x = 6_{10}$$ Therefore, \(32_{four} = 22_6\) and the value of x is 6.

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To find the median of a set of numbers, we first need to arrange them in order from smallest to largest or largest to smallest. In this case, the numbers are: 0, 0, 1, 1, 1, 2, 2, 3, 4 Next, we determine the middle value(s) of the set. If the set has an odd number of values, there will be one middle value. If the set has an even number of values, there will be two middle values, and the median is the average of these two values. In this case, the set has an even number of values (10), so there are two middle values: 1 and 1. To find the median, we take the average of these two values: (1 + 1)/2 = 1 Therefore, the median of the set {2, 1, 0, 3, 1, 1, 4, 0, 1, 2} is 1. So the answer is 1.0.

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The given equation relates the temperature in degrees Celsius (C) to the temperature in degrees Fahrenheit (F). To find C when F = 98.6, we can simply substitute F = 98.6 into the equation and solve for C. So, we have: F = \(\frac{9}{5}\)C + 32 98.6 = \(\frac{9}{5}\)C + 32 (substituting F = 98.6) Subtracting 32 from both sides, we get: 66.6 = \(\frac{9}{5}\)C Multiplying both sides by 5/9, we get: C ≈ 37 Therefore, C is approximately equal to 37 degrees Celsius (not exactly any of the options provided).

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To find x : z, we need to have a common ratio between x, y, and z. We can use the given ratios to find a common ratio involving all three. Since x : y = 1/4 : 3/8, we can simplify this ratio by multiplying both terms by 8 to get: x : y = 2 : 3 Similarly, since y : z = 1/3 : 4/9, we can simplify this ratio by multiplying both terms by 3 to get: y : z = 1 : 4/3 Now we have a common ratio of y between the two ratios. We can use this common ratio to find x : z by multiplying the two simplified ratios: x : y = 2 : 3 y : z = 1 : 4/3 Multiplying these ratios gives: x : z = (2/3) * (1/(4/3)) = 2/4 = 1/2 Therefore, x : z = 1 : 2. In summary, x : z = 1 : 2.

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To calculate the probability that a team selected at random scored either 4 or 7 goals, we need to add the frequency of teams that scored 4 goals to the frequency of teams that scored 7 goals and divide by the total number of teams (25). From the table, we can see that 6 teams scored 4 goals and 3 teams scored 7 goals. Therefore, the total number of teams that scored either 4 or 7 goals is: 6 + 3 = 9 The total number of teams in the competition is 25. Therefore, the probability that a team selected at random scored either 4 or 7 goals is: 9/25 This can be simplified as follows: 9/25 = 0.36 Therefore, the probability that a team selected at random scored either 4 or 7 goals is 0.36 or 36%. Hence, the correct option is: - \(\frac{9}{25}\)

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We can solve this problem using trigonometry, specifically the tangent function. Let's draw a diagram to visualize the situation. We have a right triangle with the height of the tree as one of the legs, the distance from the tree to the point on the ground as the other leg, and the angle of elevation (30 degrees) as the angle opposite the height of the tree.  We can use the tangent function to find the height of the tree: $$\tan(30^\circ) = \frac{\text{height of tree}}{27\text{ m}}$$ We know that the tangent of 30 degrees is equal to 1/\(\sqrt{3}\) (or approximately 0.577), so we can substitute that in and solve for the height of the tree: $$\frac{1}{\sqrt{3}} = \frac{\text{height of tree}}{27\text{ m}}$$ Multiplying both sides by 27 m gives: $$\text{height of tree} = \frac{27\text{ m}}{\sqrt{3}} = 9\sqrt{3}\text{ m}$$ Therefore, the height of the tree is 9\(\sqrt{3}\) meters. Answer option (D) is correct.

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The formula for the surface area of a sphere is \(4\pi r^2\), where \(r\) is the radius of the sphere. Given that the surface area of the sphere is \(\frac{792}{7} cm^2\) and \(\pi = \frac{22}{7}\), we can set up an equation: $$ 4\left(\frac{22}{7}\right)r^2 = \frac{792}{7} $$ Simplifying this equation by canceling out the \(7\)s, we get: $$ 4\left(\frac{22}{1}\right)r^2 = 792 $$ Multiplying both sides by \(\frac{1}{4}\) to isolate \(r^2\), we get: $$ \left(\frac{22}{1}\right)r^2 = 198 $$ Dividing both sides by \(\frac{22}{1}\), we get: $$ r^2 = 9 $$ Taking the square root of both sides, we get: $$ r = 3 $$ Therefore, the radius of the sphere is 3 cm. The formula for the volume of a sphere is \(\frac{4}{3}\pi r^3\). Substituting the value of \(r\) into this formula, we get: $$ \frac{4}{3}\left(\frac{22}{7}\right)(3)^3 \approx 113 $$ Therefore, the volume of the sphere is approximately 113\(cm^3\). So the correct answer is 113\(cm^3\).

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To find the value of (x + y), we need to solve for x and y first. We can do this by using two equations: y + 2x = 4 and y - 3x = -1. First, we can isolate y in one of the equations by adding or subtracting the other equation. For example, if we add the two equations, we get: 2y = 3 Then, we can solve for y by dividing both sides by 2: y = 3/2 Next, we can substitute the value of y back into one of the original equations to solve for x. For example, using the equation y + 2x = 4: 3/2 + 2x = 4 Subtracting 3/2 from both sides: 2x = 7/2 And finally, dividing both sides by 2: x = 7/4 Now that we have found the values of x and y, we can add them to find (x + y): x + y = 7/4 + 3/2 Combining the fractional terms: x + y = 7/4 + 6/4 And finally, adding the whole numbers: x + y = 13/4 = 3.25 So, (x + y) is approximately equal to 3.

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The graph of y = x\(^2\) and y = x intersect at two points where the two curves have the same y-value. To find these points, we need to find the x-values where x\(^2\) = x. We can start by subtracting x from both sides: x\(^2\) - x = 0 Next, we can factor out x: x (x - 1) = 0 This equation tells us that either x = 0 or x - 1 = 0. So, the two points of intersection are (0, 0) and (1, 1).

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Let Manya's present age be \(m\). Then Musa \(= m + 3\).

Seven years ago Musa was twice Manya:

(1) Manya is \(10\) years and Musa is \(13\) years now.

(2) The present sum of ages is \(23\), and it rises by 2 each year. Let it take \(t\) years:

The sum of their ages will be 45 in 11 years.

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In triangle PQR the three sides are marked as follows: \(|PQ| = 2y + x\), \(|QR| = 4y\) and \(|PR| = 6y - 2x + 1\). Since PQR is isosceles the two slant sides are equal, \(|QR| = |PR|\).

(a) Values of x and y

Perimeter equation:

\[ (2y + x) + 4y + (6y - 2x + 1) = 28 \]

\[ 12y - x + 1 = 28 \quad\Rightarrow\quad 12y - x = 27 \quad\text{...(1)} \]

Isosceles equation \((|QR| = |PR|)\):

\[ 4y = 6y - 2x + 1 \quad\Rightarrow\quad 2x - 2y = 1 \quad\text{...(2)} \]

From (1), \(x = 12y - 27\). Substituting into (2):

\[ 2(12y - 27) - 2y = 1 \Rightarrow 24y - 54 - 2y = 1 \Rightarrow 22y = 55 \]

\[ y = \frac{55}{22} = \frac{5}{2} = 2\tfrac{1}{2} \]

\[ x = 12\left(\tfrac{5}{2}\right) - 27 = 30 - 27 = 3 \]

Hence x = 3 and y = 2\(\tfrac{1}{2}\) (i.e. 2.5).

(b) Lengths of the sides

• \(|PQ| = 2y + x = 2\left(\tfrac{5}{2}\right) + 3 = 5 + 3 = \textbf{8 cm}\)
• \(|QR| = 4y = 4\left(\tfrac{5}{2}\right) = \textbf{10 cm}\)
• \(|PR| = 6y - 2x + 1 = 6\left(\tfrac{5}{2}\right) - 2(3) + 1 = 15 - 6 + 1 = \textbf{10 cm}\)

Check: \(8 + 10 + 10 = 28\) cm, which agrees with the given perimeter, and \(|QR| = |PR| = 10\) cm confirms the triangle is isosceles.

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(a)(i) Diagram: a cone standing on its vertex (point downwards), base radius 21 cm at the top, water filling it to height \(h\).

(a)(ii) Height. \(4.158\) litres \(= 4158 \text{ cm}^3\).

(b) When the surface diameter drops to 28 cm, the surface radius is 14 cm. By similar cones the remaining water height is

Volume remaining:

Volume drawn out \(= 4158 - 1232 = 2926 \text{ cm}^3\).

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Purchase price \(= \text{N}900{,}000.00\) on 1 March 2011. The value on 28 February 2015 is after 4 full years.

Year 1 (to Feb 2012): depreciation 30%, so the value is multiplied by \(0.70\).

Years 2, 3, 4: depreciation 22% each year, so each year the value is multiplied by \(0.78\).

Now \((0.78)^3 = 0.474552\), and \(900000 \times 0.70 = 630000\):

Correct to the nearest hundred naira, the value of the car on 28 February 2015 is N299,000.00.

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(a) \(\sin X = \tfrac{3}{5} \Rightarrow \cos X = \tfrac{4}{5}\) (a 3-4-5 triangle).

(b) Volume of the pyramid \(= \tfrac{1}{3} \times (42 \times 11) \times 12 = \tfrac{1}{3} \times 462 \times 12 = 1848 \text{ cm}^3\).

The cone has the same height (12 cm) and the same volume (1848):

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From the diagram, in triangle PSR the point T lies on PS and Q lies on PR, with TQ parallel to SR (shown by the arrows). Given \(|PT| = 4\text{ cm}\), \(|TS| = 6\text{ cm}\), \(|PQ| = 6\text{ cm}\) and \(\angle SPR = 30^\circ\).

Preliminary - use the parallel lines. Because \(TQ \parallel SR\), triangles PTQ and PSR are similar (equiangular). The full side PS is

\[|PS| = |PT| + |TS| = 4 + 6 = 10\text{ cm}.\]

The ratio of similarity is \(\dfrac{PT}{PS} = \dfrac{4}{10} = 0.4\), so

\[\frac{PQ}{PR} = 0.4 \;\Rightarrow\; |PR| = \frac{PQ}{0.4} = \frac{6}{0.4} = 15\text{ cm}.\]

(a) Finding |SR|. In triangle PSR use the Cosine Rule with \(|PS| = 10\), \(|PR| = 15\), \(\angle P = 30^\circ\):

\[|SR|^2 = |PS|^2 + |PR|^2 - 2|PS||PR|\cos 30^\circ.\]

\[|SR|^2 = 10^2 + 15^2 - 2(10)(15)\cos 30^\circ = 100 + 225 - 300(0.8660).\]

\[|SR|^2 = 325 - 259.8 = 65.2.\]

\[|SR| = \sqrt{65.2} = 8.07\ldots \approx 8\text{ cm}.\]

(b) Area of TQRS. TQRS is the trapezium left when the small triangle PTQ is removed from triangle PSR:

\[\text{Area of } TQRS = \text{Area of } \triangle PSR - \text{Area of } \triangle PTQ.\]

Using \(\text{Area} = \tfrac{1}{2}ab\sin C\) with the common angle \(30^\circ\):

\[\text{Area of } \triangle PSR = \tfrac{1}{2}(10)(15)\sin 30^\circ = \tfrac{1}{2}(150)(0.5) = 37.5\text{ cm}^2.\]

\[\text{Area of } \triangle PTQ = \tfrac{1}{2}(4)(6)\sin 30^\circ = \tfrac{1}{2}(24)(0.5) = 6\text{ cm}^2.\]

\[\text{Area of } TQRS = 37.5 - 6 = 31.5 \approx 32\text{ cm}^2.\]

\[\boxed{|SR| \approx 8\text{ cm},\qquad \text{Area of } TQRS \approx 32\text{ cm}^2.}\]

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(a) Class boundaries

Price (₦ million)	Class boundaries (₦ million)	Number of vehicles
1.0 – 1.9	0.95 – 1.95	23
2.0 – 2.9	1.95 – 2.95	48
3.0 – 3.9	2.95 – 3.95	107
4.0 – 4.9	3.95 – 4.95	90
5.0 – 5.9	4.95 – 5.95	32

(b) Histogram

The bars have equal width of ₦1 million and are drawn contiguously, using the class boundaries on the horizontal axis.

The tallest bar is for the class ₦3.0 million to ₦3.9 million. Drawing the two modal lines across this bar and reading their point of intersection on the price axis gives approximately

\[2.95+0.75=3.70.\]

Therefore, the most preferred selling price is

\[\boxed{₦3,700,000.00}\]

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(a)

In right-angled triangle \(SPQ\),

\[SQ^2=PS^2+PQ^2=5^2+12^2=169.\]

Hence, \(SQ=13\text{ cm}\).

\(PR\) is the perpendicular height from \(P\) to the hypotenuse \(SQ\). Equating the two expressions for the area of \(\triangle SPQ\):

\[\frac12(5)(12)=\frac12(13)(PR).\]

\[PR=\frac{5\times12}{13}=4.615\ldots\text{ cm}.\]

\[\boxed{PR=4.62\text{ cm}}\qquad\text{(correct to 3 significant figures).}\]

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(b)(i) The required diagram is:

(b)(ii)

For ladder \(L\), let \(\theta\) be its angle with the horizontal ground. Then

\[\cos\theta=\frac{8}{10}=0.8.\]

Therefore,

\[\sin\theta=\sqrt{1-0.8^2}=\sqrt{0.36}=0.6.\]

Since ladder \(M\) makes the same angle \(\theta\) with the ground, if \(h\) is the height at which it touches the wall,

\[\sin\theta=\frac{h}{12}.\]

\[h=12(0.6)=7.2\text{ m}.\]

\[\boxed{\text{Ladder }M\text{ touches the wall }7.2\text{ m above the ground.}}\]

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(a) Construction (ruler and compasses only).

1. Draw \(|PQ| = 6.8\) cm. At Q construct \(\angle PQR = 120^{\circ}\) and draw the ray along which R lies.
2. Through P draw a line parallel to QR (since \(QR \parallel PS\)) and mark \(|PS| = 10.6\) cm. The diagonal \(|PR| = 9.3\) cm (arc from P) fixes R on its ray. Join R to S to complete trapezium PQRS.
3. \(l_1\): construct the perpendicular bisector of \(PR\) (points equidistant from P and R).
4. \(l_2\): construct the perpendicular bisector of \(QR\) (points equidistant from Q and R). Let \(l_1\) and \(l_2\) meet at Y.

(b) Expected measurements from an accurate drawing:

• \(|QR| \approx 3.8\) cm
• \(\angle PSR \approx 60^{\circ}\)
• \(|QY| \approx 5.4\) cm

(Accept small tolerances, since these are read off the construction.)

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(a)

Given

\[p\begin{pmatrix}2\\3\end{pmatrix}+q\begin{pmatrix}5\\-2\end{pmatrix}=\begin{pmatrix}-4\\13\end{pmatrix}.\]

Equating corresponding components gives

\[2p+5q=-4 \qquad (1)\]

\[3p-2q=13 \qquad (2)\]

Multiply (1) by 2 and (2) by 5:

\[4p+10q=-8\]

\[15p-10q=65\]

Adding,

\[19p=57\]

\[p=3.\]

Substituting into (1),

\[2(3)+5q=-4\]

\[5q=-10\]

\[q=-2.\]

Therefore, \(\boxed{p=3,\ q=-2}\).

(b)(i) and (ii)

Using the scale \(2\text{ cm}\) to \(2\) units on both axes, the quadrilateral and its image are plotted below. The vertices are joined in the order \(W\to X\to Y\to Z\to W\), and similarly for the image.

A rotation of \(90^\circ\) anticlockwise about the origin maps \((x,y)\) onto \((-y,x)\). Hence:

\[\begin{aligned}W(2,3)&\longmapsto W_1(-3,2),\\X(4,-1)&\longmapsto X_1(1,4),\\Y(-3,-4)&\longmapsto Y_1(4,-3),\\Z(-3,2)&\longmapsto Z_1(-2,-3).\end{aligned}\]

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(a) The mean of five numbers is their sum divided by 5.

Since the mean of \(m, n, s, p, q\) is 12, the sum is \(5 \times 12 = 60\).

Adding the constants changes the sum by \(+4 - 3 + 6 - 2 + 8 = +13\).

New sum \(= 60 + 13 = 73\), so new mean \(= \dfrac{73}{5} = 14.6\).

(b) The 25th percentile (15 years) and the 75th percentile (65 years) enclose the middle 50% of the data.

Number of people between 15 and 65 years \(= 50\% \times 500 = 250\).

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(a) The diagonals of a rhombus bisect each other at right angles. Half-diagonals are

A side of the rhombus is the hypotenuse of a right triangle with these legs:

Perimeter \(= 4s = 4 \times 6.9016 = 27.6\ \text{cm (to 1 d.p.).}\)

(b) Given \(\sin x = \tfrac{3}{5}\) with \(0^\circ < x < 90^\circ\), this is a 3-4-5 right triangle, so

Then

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(a) Diagram: from X, line XY runs at bearing \(078^{\circ}\) (285 m); from Y the man walks due south to Z; XZ \(= 307\) m closes the triangle.

Place X at the origin:

Z has the same easting as Y (a due-south move): \(Z = (278.8,\ z)\).

(b)(ii) Distance YZ \(= 59.3 - (-128.5) = 187.8 \approx 188\) m.

(b)(i) Bearing of X from Z. \(\vec{ZX} = (-278.8,\ 128.5)\), pointing north-west.

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From the diagram the track is a rectangle capped by two semicircular ends. The two straight (parallel) sides are each 60 m long, and each semicircular end has diameter 120 m (the marked vertical distance between the two straights).

(a) Distance for one lap, then two laps.

The two semicircular ends together form one complete circle of diameter \(d = 120\) m. So one lap consists of the two straights plus one full circle:

\[ \text{One lap} = 2(60) + \pi d = 120 + \frac{22}{7}\times 120. \]

\[ \frac{22}{7}\times 120 = \frac{2640}{7} = 377.14\text{ m}. \]

\[ \text{One lap} = 120 + \frac{2640}{7} = \frac{840 + 2640}{7} = \frac{3480}{7} = 497.14\text{ m}. \]

Running round the track two times:

\[ D = 2 \times \frac{3480}{7} = \frac{6960}{7} = 994.29\text{ m} \;(\text{to 2 d.p.}). \]

Distance covered \(\approx 994.29\) m.

(b) Speed in km/h.

Time \(= 200\) s. First find the speed in metres per second:

\[ \text{Speed} = \frac{D}{t} = \frac{6960/7}{200} = \frac{6960}{1400} = 4.971\text{ m/s}. \]

Convert to km/h (multiply by \(\tfrac{3600}{1000} = 3.6\)):

\[ \text{Speed} = 4.971 \times 3.6 = 17.90\text{ km/h}. \]

Alternatively: \(D = \dfrac{6960}{7000}\) km \(= 0.99429\) km, \(t = \dfrac{200}{3600}\) h \(= \dfrac{1}{18}\) h, so speed \(= 0.99429 \times 18 = 17.90\) km/h.

Speed \(\approx 17.9\) km/h.

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(a) Finding \(x\).

Reading the diagram. \(Q\), \(R\) and \(S\) lie on a circle with centre \(O\). The straight line \(Q\,O\,S\) passes through the centre, so \(QS\) is a diameter, and it is produced beyond \(S\) to the external point \(T\). The angle at \(Q\) is \(x\), the right-angle mark is at \(R\), and the exterior angle at \(S\) (between \(SR\) and \(ST\)) is \((3x + 15)^\circ\).

Angle in a semicircle. Since \(QS\) is a diameter, the angle it subtends at \(R\) is a right angle:

Exterior-angle theorem in \(\triangle QRS\). The line \(Q\,S\,T\) is straight, so \((3x+15)^\circ\) is the exterior angle at \(S\); it equals the sum of the two remote interior angles \(\angle Q\) and \(\angle R\):

(The follow-up part (ii) is not legible in the source text, so only \(x\) is evaluated here.)

(b) Number-base equation: \(2N4_{\text{seven}} = 15N_{\text{nine}}\).

Expand each numeral in powers of its base, treating \(N\) as an unknown digit.

Left side (base 7):

Right side (base 9):

Equate:

Check: \(N = 4\) is a valid digit in both base 7 and base 9. Left \(= 102 + 28 = 130\); right \(= 126 + 4 = 130\). Both equal \(130_{\text{ten}}\). Correct.

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(a) Determination of \(x\) and \(y\)

Mark, \(m\)	Frequency, \(f\)	\(fm\)
10	1	10
20	1	20
30	\(x\)	\(30x\)
40	5	200
50	\(y\)	\(50y\)
60	1	60
70	4	280
80	3	240
90	1	90
Total	\(16+x+y\)	\(900+30x+50y\)

Since there are 30 students,

\[16+x+y=30 \quad\Rightarrow\quad x+y=14\tag{1}\]

Also,

\[52=\frac{900+30x+50y}{30}\]

\[1560=900+30x+50y\]

\[3x+5y=66.\tag{2}\]

From (1), \(x=14-y\). Hence,

\[3(14-y)+5y=66\]

\[42+2y=66\Rightarrow y=12.\]

Therefore, \(x=14-12=2\).

\(\boxed{x=2,\ y=12}\)

(b) Grouped frequency distribution

Class interval	Class boundaries	Frequency
1 - 10	0.5 - 10.5	1
11 - 20	10.5 - 20.5	1
21 - 30	20.5 - 30.5	2
31 - 40	30.5 - 40.5	5
41 - 50	40.5 - 50.5	12
51 - 60	50.5 - 60.5	1
61 - 70	60.5 - 70.5	4
71 - 80	70.5 - 80.5	3
81 - 90	80.5 - 90.5	1

(c) Histogram

(d) Mode

The modal class is \(41-50\). Reading the peak position from the histogram by the usual intersecting-lines method gives a mode of approximately

\[\boxed{44\text{ marks}}\]

Antwoorddetails

(a) For \(y=2x^{2}+x-10\), the completed table is:

\(x\)	\(-5\)	\(-4\)	\(-3\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)
\(y\)	\(35\)	\(18\)	\(5\)	\(-4\)	\(-9\)	\(-10\)	\(-7\)	\(0\)	\(11\)	\(26\)

(b) The graph of \(y=2x^{2}+x-10\) is shown below. The straight line \(y=2x\), used in part (c)(ii), is also shown.

(c)(i) \[2x^{2}+x=10\iff 2x^{2}+x-10=0.\] The solutions are the \(x\)-coordinates where the parabola cuts the \(x\)-axis. Hence, \[\boxed{x=-2.5\text{ or }x=2}.\]

(c)(ii) The solutions of \(2x^{2}+x-10=2x\) are the \(x\)-coordinates of the points where the parabola intersects the line \(y=2x\). From the graph, \[\boxed{x=-2\text{ or }x=2.5}.\]

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(a) and (b) The required coordinate graph, drawn to equal scales on both axes, is shown below. The vertices are joined in the order in which they are named.

The coordinates plotted are:

For a rotation of \(90^\circ\) anticlockwise about the origin, \((x,y)\mapsto(-y,x)\). Thus, for example, \(P(3,2)\mapsto P_1(-2,3)\).

Reflection in \(y=x\) maps \((x,y)\mapsto(y,x)\). Thus \(P_1(-2,3)\mapsto P_2(3,-2)\).

(c) Combining the two transformations gives

\[(x,y)\mapsto(-y,x)\mapsto(x,-y).\]

Hence \(T\) is a reflection in the x-axis.

(d) The line \(P_1Q_1\) has gradient

\[m=\frac{-1-3}{-5-(-2)}=\frac{-4}{-3}=\frac{4}{3}.\]

Its equation is

\[y-3=\frac{4}{3}(x+2).\]

At the x-axis, \(y=0\), so

\[-3=\frac{4}{3}(x+2),\qquad x=-\frac{17}{4}.\]

Therefore \(W\left(-\frac{17}{4},0\right)\). Since \(P_1S_1\) and \(R_1W\) are both horizontal, they are parallel. Thus \(P_1S_1R_1W\) is a trapezium.

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Price before the first sale. After a 40% cut the price is \(0.6\) of the original; after a further 30% cut it is \(0.7 \times 0.6 = 0.42\) of the original.

Article costing GH\unicode{x20B5}180.00. Final price \(= 0.42 \times 180 = \text{GH}\unicode{x20B5}75.60\).

Total reduction \(= 180 - 75.60 = \text{GH}\unicode{x20B5}104.40\).

Percentage reduction \(= \dfrac{104.40}{180} \times 100\% = 58\%\).

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(a) Cumulative frequency table

Marks (%)	Class boundaries	Frequency	Cumulative frequency
10–19	9.5–19.5	4	4
20–29	19.5–29.5	7	11
30–39	29.5–39.5	12	23
40–49	39.5–49.5	18	41
50–59	49.5–59.5	20	61
60–69	59.5–69.5	14	75
70–79	69.5–79.5	9	84
80–89	79.5–89.5	4	88
90–99	89.5–99.5	2	90

Total frequency, \(N=90\).

(b) Cumulative frequency curve

The less-than ogive is plotted using the upper class boundaries and their cumulative frequencies.

(c)(i) Median

\[\frac{N}{2}=\frac{90}{2}=45.\]

From the curve, the mark corresponding to cumulative frequency 45 is approximately \(51.5\%\).

\[\boxed{\text{Median}=51.5\%}\]

(c)(ii) Probability of obtaining distinction

At \(75\%\), the cumulative frequency read from the curve is approximately \(80\). Hence, the number scoring at least \(75\%\) is

\[90-80=10.\]

Therefore,

\[P(\text{distinction})=\frac{10}{90}=\frac{1}{9}\approx0.111.\]

Antwoorddetails

(a) Since the graph passes through \((3,10)\), substitute \(x=3\) and \(y=10\) into \(y=2px^2-p^2x-14\):

\[10=2p(3)^2-p^2(3)-14.\]

\[10=18p-3p^2-14\]

\[3p^2-18p+24=0\]

\[p^2-6p+8=0\]

\[(p-2)(p-4)=0.\]

Therefore,

\[\boxed{p=2\text{ or }p=4}.\]

For these two values, the corresponding curves are \(y=4x^2-4x-14\) and \(y=8x^2-16x-14\). Both pass through \((3,10)\), as shown below.

(b) The equations of the two lines are

\[3y-2x=21 \quad \text{and} \quad 4y+5x=5.\]

Multiply the first equation by 5 and the second equation by 2:

\[15y-10x=105\]

\[8y+10x=10.\]

Adding the equations gives

\[23y=115\]

\[y=5.\]

Substitute \(y=5\) into \(3y-2x=21\):

\[3(5)-2x=21\]

\[15-2x=21\]

\[-2x=6\]

\[x=-3.\]

Hence, the coordinates of \(Q\) are

\[\boxed{Q=(-3,5)}.\]

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(a)(i) Modulo 5 tables on {2, 3, 4}

Addition \(\oplus\) (add, then take the remainder on division by 5):

Multiplication \(\otimes\) (multiply, then take the remainder on division by 5):

(For example \(3\otimes 4 = 12 = 2\times 5 + 2 \equiv 2\), and \(4\oplus 3 = 7 = 5 + 2 \equiv 2\).)

(a)(ii)(a) Solve \(4\otimes e \oplus 2 \equiv 1 \ (\text{mod }5)\)

\(4\otimes e \equiv 1 \ominus 2 \equiv -1 \equiv 4 \ (\text{mod }5)\). So \(4e \equiv 4 \ (\text{mod }5)\). Multiplying both sides by the inverse of 4 (which is 4, since \(4\times 4 = 16 \equiv 1\)) gives \(e \equiv 16 \equiv 1 \ (\text{mod }5)\). Hence \(e = 1\).

(a)(ii)(b) Find n if \(4 \oplus n\otimes 2 \equiv 1 \ (\text{mod }5)\)

Multiplication first: \(n\otimes 2 = 2n\). Then \(4 \oplus 2n \equiv 1\), so \(2n \equiv 1 - 4 \equiv -3 \equiv 2 \ (\text{mod }5)\). Multiplying by the inverse of 2 (which is 3, since \(2\times 3 = 6 \equiv 1\)) gives \(n \equiv 6 \equiv 1 \ (\text{mod }5)\). Hence \(n = 1\).

(b) Validity of the statements (given \(p \Rightarrow q\))

Here p: Landi has cholera, q: Landi is in the hospital, and we are told \(p \Rightarrow q\) is true.

• (i) "If in hospital, then has cholera" is \(q \Rightarrow p\), the converse. It does not follow from \(p\Rightarrow q\), so it is not valid.
• (ii) "If not in hospital, then no cholera" is \(\lnot q \Rightarrow \lnot p\), the contrapositive of \(p\Rightarrow q\). It is logically equivalent, so it is valid.
• (iii) "If no cholera, then not in hospital" is \(\lnot p \Rightarrow \lnot q\), the inverse. It does not follow, so it is not valid.

Antwoorddetails

(a) Venn diagram

Let the three sets be:

• C = cars with faulty clutch
• B = cars with faulty brakes
• S = cars with faulty steering

(b) Calculation

Number that passed:

\[60\%\text{ of }240=\frac{60}{100}\times240=144\]

Therefore, number that failed:

\[240-144=96\]

The given numbers for the two-set intersections include the cars with all three faults. Hence:

\[C\cap S\text{ only}=14-8=6\]

\[C\cap B\text{ only}=20-8=12\]

Let the number with brakes only be \(x\). Then the number with steering only is \(2x\).

From the Venn diagram:

\[28+6+8+12+6+x+2x=96\]

\[60+3x=96\]

\[3x=36\]

\[x=12\]

Thus, brakes only \(=12\), while steering only \(=2(12)=24\).

(i) Cars with faulty brakes

\[12+12+6+8=38\]

\[\boxed{38\text{ cars}}\]

(ii) Cars with only one fault

\[28+12+24=64\]

\[\boxed{64\text{ cars}}\]

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1. Given

\[\log_a(y+2)=1+\log_a x\]

Since \(1=\log_a a\),

\[\log_a(y+2)=\log_a a+\log_a x=\log_a(ax).\]

Therefore,

\[y+2=ax\]

\[\boxed{x=\frac{y+2}{a}}\]

2(a)(i) Pie chart

Total timber production:

\[600+900+1800+1500+2400=7200\text{ tonnes}.\]

Community	Production (tonnes)	Sector angle
Bibiani	600	\(\frac{600}{7200}\times360^\circ=30^\circ\)
Amenfi	900	\(\frac{900}{7200}\times360^\circ=45^\circ\)
Oda	1800	\(\frac{1800}{7200}\times360^\circ=90^\circ\)
Wiawso	1500	\(\frac{1500}{7200}\times360^\circ=75^\circ\)
Sankore	2400	\(\frac{2400}{7200}\times360^\circ=120^\circ\)

2(a)(ii)

\[\frac{900}{7200}\times100\%=\boxed{12.5\%}.\]

2(a)(iii)

Difference in production:

\[1800-600=1200\text{ tonnes}.\]

Hence, the extra revenue received by Oda is

\[1200\times\$560.00=\boxed{\$672,000.00}.\]

Antwoorddetails

(a) Construction

1. Draw the line segment \(SR=12\text{ cm}\).
2. At \(S\), construct a perpendicular to \(SR\). Mark \(P\) on it so that \(PS=8\text{ cm}\).
3. With centre \(P\) and radius \(6\text{ cm}\), draw an arc. With centre \(R\) and radius \(11\text{ cm}\), draw another arc cutting the first at \(Q\).
4. Join \(PQ\) and \(QR\) to complete quadrilateral \(PQRS\).
5. Construct the perpendicular from \(Q\) to \(SR\), meeting \(SR\) at \(K\).

(b) Measurements from the construction

(i) \(\lvert RK\rvert \approx 9.3\text{ cm}\).

(ii) \(\angle QRS \approx 55^\circ\).

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1. Bar chart

2. Mean score

Score, \(x\)	1	2	3	4	5	6	Total
Frequency, \(f\)	2	5	13	11	9	10	\(50\)
\(fx\)	2	10	39	44	45	60	\(200\)

\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{200}{50}=4\]

Therefore, the mean score is \(4\).

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(a)(i) With profit \(y = 20000\):

(a)(ii) Quadratic formula:

Taking the positive root: \(x = \dfrac{761.0}{2} \approx 380\) (in thousands), i.e. about \(380{,}000\) textbooks.

(b)(i) Diagram: vertical tree \(TQ\) with P on the ground so that \(\angle TPQ = 28^{\circ}\); R is the midpoint of \(TQ\).

(b)(ii) Let \(PQ = d\). Then \(TQ = d\tan 28^{\circ}\) and \(RQ = \tfrac{1}{2}TQ = \tfrac{1}{2}d\tan 28^{\circ}\).

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(a) Solving the inequalities simultaneously

First inequality:

\[ 5 - x > 1 \;\Rightarrow\; -x > 1 - 5 \;\Rightarrow\; -x > -4 \;\Rightarrow\; x < 4 \]

Second inequality:

\[ 9 + x \geq 8 \;\Rightarrow\; x \geq 8 - 9 \;\Rightarrow\; x \geq -1 \]

Combining both conditions:

\[ -1 \leq x < 4 \]

(b) Finding \(\angle QRS\)

From the diagram, \(O\) is the centre and \(PS\) passes through \(O\), so \(PS\) is a diameter. Also \(|PQ| = |QR|\) (equal chords, shown by the tick marks) and \(\angle PSR = 56^{\circ}\). The points \(P, Q, R\) lie on the arc on the same side of the diameter.

Step 1: Use equal chords. Equal chords subtend equal arcs, so \(\text{arc } PQ = \text{arc } QR\).

Step 2: Use the inscribed angle at S. \(\angle PSR = 56^{\circ}\) is the angle subtended at the circumference by the arc \(PQR\) (arc \(PR\) through \(Q\)). Hence

\[ \text{arc } PQR = 2 \times 56^{\circ} = 112^{\circ} \]

and since arc \(PQ = \) arc \(QR\), each equals \(56^{\circ}\).

Step 3: Find the remaining arc. Because \(PS\) is a diameter, the semicircle \(P\)-\(Q\)-\(R\)-\(S\) totals \(180^{\circ}\):

\[ \text{arc } RS = 180^{\circ} - \text{arc } PQ - \text{arc } QR = 180^{\circ} - 56^{\circ} - 56^{\circ} = 68^{\circ} \]

Step 4: Use the cyclic quadrilateral PQRS. Opposite angles of a cyclic quadrilateral are supplementary, so \(\angle QPS + \angle QRS = 180^{\circ}\).

\(\angle QPS\) subtends arc \(QRS = \text{arc } QR + \text{arc } RS = 56^{\circ} + 68^{\circ} = 124^{\circ}\), so

\[ \angle QPS = \tfrac{1}{2}\times 124^{\circ} = 62^{\circ} \]

Therefore

\[ \angle QRS = 180^{\circ} - 62^{\circ} = 118^{\circ} \]

\(\angle QRS = 118^{\circ}\).

Antwoorddetails

(a) Lamin's cost is \(N300\). Selling to Bola at \(x\%\) profit means Bola pays \(300\left(1 + \frac{x}{100}\right)\).

Bola resells at another \(x\%\) profit, so James pays \(300\left(1 + \frac{x}{100}\right)^2\).

James pays \(N\left(6x + \frac{3}{4}\right)\) more than Lamin's \(N300\):

So \(x = 5\) (taking the positive value).

(b) Split \(3x - 2 < 10 + x < 2 + 5x\) into two inequalities.

Left: \(3x - 2 < 10 + x \Rightarrow 2x < 12 \Rightarrow x < 6\).

Right: \(10 + x < 2 + 5x \Rightarrow 8 < 4x \Rightarrow x > 2\).

Therefore \(2 < x < 6\).

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Let the first term be \(a\) and the common difference \(d\).

Sum of 10 terms: \(S_{10} = \tfrac{10}{2}(2a + 9d) = 130 \Rightarrow 2a + 9d = 26\).

Fifth term is three times the first: \(a + 4d = 3a \Rightarrow 4d = 2a \Rightarrow a = 2d\).

Substitute: \(2(2d) + 9d = 26 \Rightarrow 13d = 26 \Rightarrow d = 2\), so \(a = 4\).

Common difference \(= 2\); first term \(= 4\).

Number of terms when the last term is 28:

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(a) Completed table of values

\(x\)	\(0^\circ\)	\(30^\circ\)	\(60^\circ\)	\(90^\circ\)	\(120^\circ\)	\(150^\circ\)	\(180^\circ\)	\(210^\circ\)	\(240^\circ\)	\(270^\circ\)	\(300^\circ\)
\(y=2\cos x-\sin x\)	2.00	1.23	0.13	−1.00	−1.87	−2.23	−2.00	−1.23	−0.13	1.00	1.87

(b) Graph of \(y=2\cos x-\sin x\)

Using 2 cm to \(30^\circ\) on the \(x\)-axis and 2 cm to 1 unit on the \(y\)-axis, plot the points and join them with a smooth curve.

(c)

(i) From the intersections of the curve with \(y=1\),

\[x\approx 37^\circ\quad\text{or}\quad270^\circ.\]

(ii) \[2\cos x-\sin x=0\implies 2\cos x=\sin x\implies \tan x=2.\]

Thus, reading the \(x\)-intercepts of the graph,

\[x\approx 63^\circ\quad\text{or}\quad243^\circ.\]