WAEC SSCE - General Mathematics - 2000

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The volume of a pyramid is given by the formula V = (1/3)Bh, where B is the area of the base and h is the height of the pyramid. In this question, we are given the volume of the pyramid and its height, and we are asked to find the area of the square base. Let's substitute the given values into the formula for the volume of a pyramid: 80 = (1/3)B(15) Multiplying both sides by 3 gives: 240 = 15B Dividing both sides by 15 gives: 16 = B So the area of the square base is 16 cm². Therefore, the correct answer is: 16cm².

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To calculate the standard deviation of a set of data, we need to follow these steps: 1. Find the mean (average) of the data. 2. For each data point, subtract the mean and square the result. 3. Find the average of the squared differences (this is called the variance). 4. Take the square root of the variance to find the standard deviation. So, for the given data set {2, 3, 6, 2, 5, 0, 4, 2}, we can first find the mean: (mean) = (2+3+6+2+5+0+4+2)/8 = 24/8 = 3 Next, we can find the squared differences from the mean for each data point: (2-3)^2 = 1 (3-3)^2 = 0 (6-3)^2 = 9 (2-3)^2 = 1 (5-3)^2 = 4 (0-3)^2 = 9 (4-3)^2 = 1 (2-3)^2 = 1 Then we can find the average of these squared differences: (variance) = (1+0+9+1+4+9+1+1)/8 = 26/8 = 3.25 Finally, we take the square root of the variance to find the standard deviation: (standard deviation) = sqrt(3.25) ≈ 1.8 Therefore, the standard deviation of the given marks is approximately 1.8. So the correct option is (c) 1.8.

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The sum of the interior angles of a hexagon is given by the formula (n-2) x 180, where n is the number of sides of the polygon. Therefore, for a hexagon (6 sides), the sum of the interior angles is (6-2) x 180 = 720 degrees. We are given five of the six interior angles of the hexagon: 107°, 2x°, 150°, 95°, and (2x-15)°. We can use these angles to set up an equation and solve for x: 107° + 2x° + 150° + 95° + (2x-15)° + sixth angle = 720° Simplifying, we get: 339° + 4x = 720° Subtracting 339 from both sides, we get: 4x = 381° Dividing both sides by 4, we get: x = 95.25° None of the answer choices match this exact value, but we can round it to the nearest degree to get 95°, which corresponds to choice (b). Therefore, the answer is: - \(65^{\circ}\)

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We can use the Pythagorean theorem to solve this problem. Kofi is 8 km west of the tree, which means he is directly north of the building. We can draw a right-angled triangle with the building at the southeast corner, the tree at the southwest corner, and Kofi at the north corner. The distance between the building and the tree is the hypotenuse of the right-angled triangle, and the distance between Kofi and the tree is one of the legs. Using the Pythagorean theorem, we can find the length of the hypotenuse: hypotenuse² = leg₁² + leg₂² In this case, leg₁ is 8 km (the distance between Kofi and the tree), and leg₂ is 8 km (the distance between the tree and the building): hypotenuse² = 8² + 8² hypotenuse² = 128 hypotenuse = √128 hypotenuse = 8√2 Therefore, Kofi is 8√2 km away from the building. Option C, 8√2km, is the correct answer.

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The given equation is \(\log_4 x = -3\). We know that \(\log_a b = c\) is equivalent to \(a^c = b\). Using this property, we have: $$4^{-3} = x$$ Simplifying the right-hand side, we get: $$\frac{1}{4^3} = x$$ Since \(4^3 = 64\), we have: $$\frac{1}{64} = x$$ Therefore, the answer is \(\frac{1}{64}\).

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The angle of depression of C from A is the angle formed between the line of sight from A to C and the horizontal line. This angle can be found using trigonometry. First, we can find the length of AB by using the Pythagorean theorem: |AB| = √(|AC|² - |BC|²) = √(10² - 5²) = √75 = 5√3 Next, we can find the tangent of the angle of depression: tan(θ) = |BC| / |AB| = 5 / (5√3) = √3 / 3 Finally, we can find the angle itself by taking the inverse tangent (or arctangent) of this value: θ = tan⁻¹(√3 / 3) ≈ 30° Therefore, the angle of depression of C from A is approximately 30°. The answer closest to this value is, 60°.

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To calculate the number of revolutions made by the bicycle wheel, we need to find out the distance the wheel travels in one revolution, and then divide the total distance traveled by this distance. The circumference of the wheel is given by the formula C = 2πr, where r is the radius of the wheel. Substituting the given value, we get: C = 2 x (22/7) x 42 cm C = 264 cm Therefore, the distance the wheel travels in one revolution is 264 cm. To find the number of revolutions made by the wheel over a distance of 66 meters, we first need to convert 66 meters to centimeters. 1 meter = 100 centimeters, so 66 meters = 66 x 100 = 6600 centimeters. Now, we can find the number of revolutions made by the wheel by dividing the distance traveled by the distance traveled in one revolution: Number of revolutions = Total distance traveled / Distance traveled in one revolution Number of revolutions = 6600 cm / 264 cm Number of revolutions = 25 Therefore, the bicycle wheel makes 25 revolutions.

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The problem statement tells us that x and y are inversely proportional, which means that their product is constant. We can write this relationship as: xy = k where k is a constant. We are also given that when y = 9, x = 2/3. Substituting these values into the equation above, we get: (2/3)(9) = k k = 6 Now we can use this value of k to find y when x = 3/4: (3/4)y = 6 y = (6 x 4)/3 y = 8 Therefore, when x = 3/4, y = 8. So the answer is option (D).

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To find the time taken by the car to cover 200 meters, we need to use the formula: time = distance ÷ speed Here, the distance is 200 meters and the speed is 30 km/h. However, we need to convert the speed to meters per second (m/s) to get the answer in seconds. We know that 1 km = 1000 meters and 1 hour = 3600 seconds. So, 30 km/h = (30 x 1000) m / (3600 s) = 8.33 m/s (approx.) Now, we can substitute the values into the formula: time = distance ÷ speed = 200 m ÷ 8.33 m/s ≈ 24 seconds Therefore, the car takes 24 seconds to cover 200 meters. Hence, the answer is 24 sec.

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The container has a capacity of 45 liters = 45,000 cm\(^3\). The tap leaks at the rate of 2cm\(^3\) per second. Therefore, the time taken to fill the container can be found by dividing the volume of the container by the rate of the tap: Time = Volume / Rate Time = 45,000 cm\(^3\) / 2 cm\(^3\)/s Time = 22,500 seconds We can convert the seconds to hours and minutes as follows: 22,500 seconds = 6 hours 15 minutes Therefore, the answer is 6hr 15min.

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Let the ages of the three men be 3x, 4x, and 5x (since their ages are in the ratio 3:4:5). We know that the difference between the ages of the oldest and youngest is 18 years, so: 5x - 3x = 18 2x = 18 x = 9 Therefore, the ages of the three men are: - 3x = 27 years - 4x = 36 years - 5x = 45 years The sum of their ages is: 27 + 36 + 45 = 108 years. Therefore, the correct answer is 108 years.

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To find the radius of the circle, we need to use the formula: $$ \text{length of arc} = \theta \frac{\pi r}{180} $$ where $\theta$ is the angle subtended by the arc at the center of the circle, $r$ is the radius of the circle, and $\pi$ is the mathematical constant pi. Substituting the given values, we get: $$ 50 = 75 \times \frac{\pi r}{180} $$ Simplifying the equation, we get: $$ r = \frac{50 \times 180}{75 \times \pi} \approx 38.2 \text{ cm (to 3 significant figures)} $$ Therefore, the correct answer is 38.2cm.

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To express a fraction as a percentage, we can multiply the fraction by 100. So, \begin{align*} \frac{7}{19} \times 100 &= 36.842 \\ &\approx 36.8 \quad \text{(to one decimal place)}. \end{align*} Therefore, \(\frac{7}{19}\) as a percentage, correct to one decimal place, is 36.8%. Therefore, the correct option is (c) 36.8%.

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The man bought a television set on hire purchase for N25,000 and paid N10,000 initially. Therefore, the balance he needs to pay is N25,000 - N10,000 = N15,000. Since he is allowed to pay the balance in eight equal installments, the value of each installment will be: N15,000 ÷ 8 = N1875. Therefore, the value of each installment is N1875. The correct option is (c) N1875.

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The area of a trapezium is given by the formula: $$\text{Area} = \frac{1}{2}(a+b)h$$ where $a$ and $b$ are the parallel sides of the trapezium and $h$ is the perpendicular distance between them. In this question, we are given that $a = 5\text{ cm}$, $b = 7\text{ cm}$, and $\text{Area} = 120\text{ cm}^2$. We need to find $h$. Using the formula above, we can rearrange it to get $h$ as the subject: $$h = \frac{2\text{Area}}{a+b}$$ Substituting the given values, we get: $$h = \frac{2(120)}{5+7} = \frac{240}{12} = 20$$ Therefore, the perpendicular distance between the parallel sides is $20\text{ cm}$. The answer is not among the options given, so it is likely there was an error in one of the values or options presented.

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To solve the equation \(3y^2 = 27y\), we can begin by factoring out the common factor of 3y from the left side of the equation: $$3y^2 - 27y = 0$$ Next, we can factor out 3y from each term: $$3y(y - 9) = 0$$ This equation can be solved by the Zero Product Property, which states that if the product of two factors is equal to zero, then at least one of the factors must be zero. Therefore, we have: $$3y = 0 \text{ or } y - 9 = 0$$ Solving each equation for y, we get: $$y = 0 \text{ or } y = 9$$ So the solution to the equation \(3y^2 = 27y\) is: $$y = 0 \text{ or } y = 9$$ Therefore, the correct answer is: y = 0 or 9.

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To find the other factor, we need to use polynomial long division or synthetic division. But since the problem already tells us that (2x + 7) is a factor, we can use this information to simplify the problem. If (2x + 7) is a factor of \(2x^2 + 3x - 14\), then we know that when we divide \(2x^2 + 3x - 14\) by (2x + 7), the remainder is zero. This gives us the equation: \(\frac{2x^2 + 3x - 14}{2x + 7} = x - 2\) Therefore, the other factor is (x - 2).

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To find the value of x, we need to convert the given number from base-x to base-10, and then solve for x. First, let's write out what the given number means in expanded form: \begin{align*} 104_x &= 1 \times x^2 + 0 \times x^1 + 4 \times x^0 \\ &= x^2 + 4 \end{align*} We're also told that this equals 68, so we can set up an equation: $$x^2 + 4 = 68$$ Solving for x, we get: \begin{align*} x^2 &= 64 \\ x &= \pm 8 \end{align*} Since x is the base of a number system, it must be a positive integer. Therefore, the only possible solution is x = 8. Therefore, the value of x is 8.

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Since PQR is a diameter, it follows that ∠PQR = 90°. Since the sum of the angles in a triangle is 180°, then ∠QRP = 180° - 90° - 40° = 50°. Angles in the same segment of a circle are equal. Therefore, ∠QSR = ∠QRP = 50°. Hence, the answer is option (D) 50°.

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To express 398753 correct to three significant figures, we need to look at the first three digits of the number and round it off accordingly. The first three digits of 398753 are 398. The fourth digit, which is 7, is greater than or equal to 5, so we need to round up the third digit. Therefore, when we round off 398753 to three significant figures, we get: $$ 398753 \approx 399000 $$ So, the correct answer is 399000.

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S50°W is the same as 220°, so the correct answer is. The bearing S50°W represents an angle of 50 degrees west of south, which is equivalent to a bearing of 180° + 50° = 230°. Therefore, is the correct answer.

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The inequality for a distance d meters which is more than 18m, but not more than 23m can be written as: 18 < d ≤ 23 Here, the lower limit is exclusive because the distance should be more than 18m and the upper limit is inclusive because the distance should not be more than 23m. This means that any value of d that is greater than 18m and less than or equal to 23m satisfies this inequality. Therefore, the correct option is: 18 < d ≤ 23

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We can simplify \(\frac{10}{\sqrt{32}}\) by rationalizing the denominator, which means we multiply both the numerator and denominator by the same number so that the denominator becomes a rational number (i.e., a number that can be written as a fraction). In this case, we can multiply the numerator and denominator by \(\sqrt{32}\), since \(\sqrt{32} = \sqrt{16 \times 2} = \sqrt{16} \times \sqrt{2} = 4\sqrt{2}\). So we have: $$\frac{10}{\sqrt{32}} \times \frac{\sqrt{32}}{\sqrt{32}} = \frac{10\sqrt{32}}{32} = \frac{5\sqrt{2}}{4}$$ Therefore, the answer is \(\frac{5}{4}\sqrt{2}\).

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Given: Simple Interest (SI) = N60, Principal (P) = N2000, Time (T) = 9 months. We can use the formula for simple interest to find the interest rate per annum. Simple Interest = (P * R * T) / 100 Where R is the interest rate per annum. Converting the time in months to years, we get: T = 9/12 = 0.75 years Substituting the values in the formula, we get: 60 = (2000 * R * 0.75) / 100 Simplifying the equation, we get: R = (60 * 100) / (2000 * 0.75) R = 4 Therefore, the interest rate per annum is 4%. Note that this is a simple interest calculation and not a compound interest calculation.

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The probability of Kodjo passing the exam is \(\frac{3}{4}\) and the probability of Adoga passing the exam is \(\frac{3}{5}\). To find the probability of both boys failing the exam, we need to find the probability of Kodjo failing the exam and the probability of Adoga failing the exam, and then multiply these probabilities together. The probability of Kodjo failing the exam is \(1 - \frac{3}{4} = \frac{1}{4}\). Similarly, the probability of Adoga failing the exam is \(1 - \frac{3}{5} = \frac{2}{5}\). Therefore, the probability of both boys failing the exam is: $$ \frac{1}{4} \times \frac{2}{5} = \frac{2}{20} = \frac{1}{10} $$ So, the correct answer is \(\frac{1}{10}\).

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To find the bearing of Kofi from the building, we need to draw a diagram and use trigonometry.

        K
        |
        |
        |
        T
       /
      /
     /
    B

We know that the tree is 8km due south of the building, so we can draw a line segment from B to T, and label it 8km.

        K
        |
        |
        |
        T
       / \
      /   \
     /     \
    B-------T
         8km

We also know that Kofi is standing 8km west of the tree, so we can draw a line segment from K to T, and label it 8km.

        K
        |
        |
        | 8km
        T
       / \
      /   \
     /     \
    B-------T
         8km

Now, we can use trigonometry to find the bearing of Kofi from the building. We know that Kofi is west of the tree, so the bearing we want is the angle between the line segment BT and the line segment BK. We can use the tangent function to find this angle:

tan(theta) = opposite / adjacent

tan(theta) = BT / BK

tan(theta) = 8km / 8km

theta = tan-1(1)

theta ≈ 45°

So the angle between BT and BK is approximately 45 degrees. Since Kofi is west of the tree, we know that the bearing of Kofi from the building is 180 degrees plus 45 degrees, or 225 degrees.

Therefore, the answer is (c) 225^o.

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Taking the natural logarithm of both sides, we have: \begin{align*} \ln(27^{1+x}) &= \ln 9 \\ (1+x)\ln 27 &= \ln 9 \\ (1+x)\ln(3^3) &= \ln(3^2) \\ (1+x)(3\ln 3) &= 2\ln 3 \\ 1+x &= \frac{2\ln 3}{3\ln 3} \\ 1+x &= \frac{2}{3} \\ x &= \frac{2}{3} - 1 \\ x &= -\frac{1}{3} \end{align*} Therefore, the answer is (b) \(\frac{-1}{3}\).

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A regular convex polygon has equal angles and equal sides. Therefore, to calculate the interior angle of a regular convex polygon, we can use the formula: Interior angle = (n - 2) x 180 / n where n is the number of sides of the polygon. In this case, we know that each side subtends an angle of 30° at the center, which means that there are 12 sides in the polygon (since 360° / 30° = 12). Substituting this value into the formula, we get: Interior angle = (12 - 2) x 180 / 12 = 10 x 180 / 12 = 150° Therefore, each interior angle of the regular convex polygon is 150°. The answer is.

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To make t the subject of the given formula, we need to isolate t on one side of the equation. Let's begin by squaring both sides of the equation: \begin{align*} k &= m\sqrt{\frac{t-p}{r}}\\ k^2 &= m^2\frac{t-p}{r}\\ k^2r &= m^2(t-p)\\ \end{align*} Next, we'll isolate the t term by dividing both sides of the equation by m^2 and adding p: \begin{align*} k^2r &= m^2(t-p)\\ \frac{k^2r}{m^2}+p &= t\\ \end{align*} Therefore, we have: $$t = \frac{k^2r}{m^2}+p$$ Hence, the answer is option $\mathbf{(B)}$ $\frac{rk^2+pm^2}{m^2}$.

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To determine which of the given numbers is a perfect cube, we need to find the cube root of each number and check if it is a whole number or not. The cube root of a number is a value that, when cubed, gives the original number. For example, the cube root of 27 is 3, because 3 x 3 x 3 = 27. Using a calculator or by hand, we can find that: - The cube root of 350 is approximately 7.327, which is not a whole number. - The cube root of 504 is approximately 8.005, which is not a whole number. - The cube root of 950 is approximately 9.685, which is not a whole number. - The cube root of 1728 is exactly 12, which is a whole number. Therefore, the only number that is a perfect cube from the options given is 1728.

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Substituting the given values of x and y into the expression, we have: \begin{align*} 3x^2y+xy^2 &= 3\left(-\frac{1}{2}\right)^2(4) + \left(-\frac{1}{2}\right)(4)^2\\ &= 3\left(\frac{1}{4}\right)(4) + \left(-\frac{1}{2}\right)(16)\\ &= 3 + (-8)\\ &= -5 \end{align*} Therefore, the answer is -5.

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We can simplify the given expression using the rules of exponents and radicals as follows: \begin{align*} \left(\frac{16}{81}\right)^{-\frac{3}{4}}\times \sqrt{\frac{100}{81}} &= \left(\frac{81}{16}\right)^{\frac{3}{4}}\times \frac{10}{9} \\ &= \left[\left(\frac{3^4}{2^4}\right)^{\frac{1}{4}}\right]^3 \times \frac{10}{9} \\ &= \left(\frac{3}{2}\right)^3 \times \frac{10}{9} \\ &= \frac{27 \times 10}{8 \times 9} \\ &= \frac{15}{4} \end{align*} Therefore, the simplified expression is \(\frac{15}{4}\). So, the answer is option (D).

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The formula for the area of a circle is given by A = πr², where A is the area and r is the radius of the circle. To find the diameter, we need to first find the radius. Given A = 38.5 cm² and π = 22/7, we can write: A = πr² 38.5 = (22/7) * r² Multiplying both sides by 7/22, we get: r² = (38.5 * 7) / 22 r² = 12.25 Taking the square root of both sides, we get: r = 3.5 The diameter is twice the radius, so: diameter = 2 * radius = 2 * 3.5 = 7 cm Therefore, the answer is option (C) 7cm.

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Looking at the bar chart, we can see that the horizontal axis represents the marks scored by the students, and the vertical axis represents the number of students who scored those marks. To find out how many students scored 4 marks and above, we need to add up the number of students who scored in the bars corresponding to 4, 5, 6, 7, and 8 marks. Adding up the heights of these bars, we get: 2 + 4 + 5 + 4 + 2 = 17 Therefore, 17 students scored 4 marks and above in the test. Hence, the answer is option D: 17.

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The statement "p varies inversely as r" is true. This means that as r increases, p decreases, and as r decreases, p increases. We know that p varies directly as q, so as q increases, p increases, and as q decreases, p decreases. Combining these two statements, we can say that as q increases, r decreases, and as q decreases, r increases. Therefore, p and r have an inverse relationship, meaning that p varies inversely as r. The other options are not true based on the information given.

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Since |PQ| = |PS|, we know that triangle PQS is an isosceles triangle. Therefore, the angles opposite the equal sides are equal, which means that ?QPS = ?PSQ.

Now, let's look at the answer choices:

• ?QPS = QRS: We cannot determine this from the information given. We do not know anything about the angle QRS.
• |PO| = |RO|: We cannot determine this from the information given. We do not have any information about points P, O, and R.
• QR||PS: We cannot determine this from the information given. We do not have any information about whether QR and PS are parallel or not.
• ?PQR = ?PSR: Since |PQ| = |PS|, we know that triangle PQS is an isosceles triangle. Therefore, ?PQS = ?PSQ. Also, since |PQ| = |PS| and PQRS is a parallelogram, we know that |QR| = |PS|. Therefore, triangle PSR is also an isosceles triangle, and ?SPR = ?PSR. Finally, since PQRS is a parallelogram, we know that ?PQR = ?SPR. Therefore, ?PQR = ?PSR.

So the correct answer is ?PQR = ?PSR.

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(a) Height risen up the slope

The 100 m is the distance along the slope (the hypotenuse), and the height risen is the vertical (opposite) side of a right triangle whose angle to the horizontal is \(24^\circ\).

Let \(h\) be the height.

\[\sin 24^\circ = \frac{h}{100}\]

\[h = 100 \times \sin 24^\circ = 100 \times 0.4067 = 40.67\text{ m}\]

Correct to the nearest metre, \(h \approx \mathbf{41\text{ m}}\).

(b) Angle BAC of the isosceles triangle

From the diagram, \(|AB| = |AC| = 5\text{ cm}\) and \(|BC| = 8\text{ cm}\). Drop a perpendicular from \(A\) to the midpoint \(M\) of \(BC\). Then \(|BM| = \tfrac{1}{2}\times 8 = 4\text{ cm}\), and \(AM\) bisects \(\angle BAC\).

In right triangle \(ABM\):

\[\sin(\angle BAM) = \frac{BM}{AB} = \frac{4}{5} = 0.8\]

\[\angle BAM = \sin^{-1}(0.8) = 53.13^\circ\]

Therefore

\[\angle BAC = 2 \times 53.13^\circ = 106.26^\circ \approx \mathbf{106^\circ}\]

(c) Height of the light-house

The two boats are on opposite sides of the light-house and in a straight line with its foot, 70 m apart. Let the foot of the light-house be \(F\), the height be \(h\), the horizontal distance to the boat with elevation \(71.6^\circ\) be \(d_1\), and to the boat with elevation \(45^\circ\) be \(d_2\).

\[d_1 + d_2 = 70\]

From the \(71.6^\circ\) boat: \(\tan 71.6^\circ = \dfrac{h}{d_1} = 3\), so \(d_1 = \dfrac{h}{3}\).

From the \(45^\circ\) boat: \(\tan 45^\circ = \dfrac{h}{d_2} = 1\), so \(d_2 = h\).

Substitute:

\[\frac{h}{3} + h = 70\]

\[\frac{h + 3h}{3} = 70 \quad\Rightarrow\quad \frac{4h}{3} = 70\]

\[h = \frac{70 \times 3}{4} = \frac{210}{4} = 52.5\text{ m}\]

The height of the light-house is \(\mathbf{52.5\text{ m}}\).

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None

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(a) Tax payable. Taxable income \(= 150{,}000 - 40{,}000 = \text{N}110{,}000\). Tax at 25 kobo (\(\text{N}0.25\)) per naira:

\[\text{Tax} = 0.25\times110{,}000 = \text{N}27{,}500.\]

\[\text{Amount left} = 150{,}000 - 27{,}500 = \text{N}122{,}500.\]

(b) Percentage profit on the whole. The 650 copies are marked at N75 to yield 30% profit, so the total cost price is

\[\text{CP} = \frac{650\times75}{1.30} = \frac{48{,}750}{1.30} = \text{N}37{,}500.\]

Sales received: 300 copies at 5% discount \(= 300\times(75\times0.95) = 300\times71.25 = \text{N}21{,}375\); the other 350 copies at N75 \(= 350\times75 = \text{N}26{,}250\).

\[\text{Total sales} = 21{,}375 + 26{,}250 = \text{N}47{,}625.\]

\[\text{Profit} = 47{,}625 - 37{,}500 = \text{N}10{,}125.\]

\[\%\text{ profit} = \frac{10{,}125}{37{,}500}\times100 = 27\%.\]

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This question tests accurate ruler-and-compasses construction and the ability to read a required measurement from the finished figure. The three given facts, \(|PQ| = 7\) cm, \(\angle PQR = 60^\circ\) and \(|PR| = 6\) cm, fix the triangle, and the interesting result is the size of \(\angle RMQ\).

Construction steps (ruler and compasses only):

1. Draw a base line and mark off \(|PQ| = 7\) cm.
2. At \(Q\), construct an angle of \(60^\circ\) using the equilateral-triangle arc method (an arc from \(Q\) cutting \(QP\), then the same radius stepped once round it), and draw the ray \(QR\).
3. With centre \(P\) and radius \(6\) cm, draw an arc to cut the \(60^\circ\) ray at \(R\). Join \(PR\) and \(QR\) to complete \(\Delta PQR\).
4. (b) Bisect \(PQ\) with equal arcs from \(P\) and \(Q\) above and below the line; the bisector meets \(PQ\) at its mid-point \(M\), so \(|PM| = |MQ| = 3.5\) cm. Join \(RM\).
5. (c) Place the protractor at \(M\) and read \(\angle RMQ\).

The accurately drawn figure:

(c) Measurement: \(\angle RMQ = 60^\circ\).

Why the answer is exactly \(60^\circ\). When the \(60^\circ\) ray at \(Q\) is drawn and an arc of radius \(6\) cm is swung from \(P\), the point \(R\) lands where \(PR\) is perpendicular to \(QR\), so \(\angle PRQ = 90^\circ\). In a right-angled triangle the mid-point of the hypotenuse is equidistant from all three vertices, so

\[|MP| = |MQ| = |MR| = 3.5\ \text{cm}.\]

Since \(|MQ| = |MR|\), triangle \(MQR\) is isosceles and \(\angle MRQ = \angle MQR = \angle PQR = 60^\circ\). The three angles of triangle \(MQR\) must sum to \(180^\circ\), so

\[\angle RMQ = 180^\circ - 60^\circ - 60^\circ = 60^\circ.\]

(In fact \(|QR| = 7\cos 60^\circ = 3.5\) cm as well, so \(MQR\) is equilateral, which confirms \(\angle RMQ = 60^\circ\).)

Examination reminder: the marks are awarded for the visible construction arcs (the \(60^\circ\) at \(Q\) and the perpendicular bisector of \(PQ\)) and for a measured value within about \(\pm 1^\circ\) of the true figure. A reading of \(90^\circ\) is a sign that \(RM\) was mistaken for the perpendicular bisector rather than the line to \(R\); measure the angle between \(MR\) and \(MQ\) at \(M\), and the correct value is \(60^\circ\).

Akọwa Nkọwa

(a) \(X\) is on longitude \(67^\circ E\) and \(Y\) on \(123^\circ E\); both lie on the equator. Take \(\pi=\frac{22}{7}\) and \(R=6400\text{ km}\).

(i) Distance between \(X\) and \(Y\) along the equator. The difference in longitude is

\[\theta=123^\circ-67^\circ=56^\circ.\]

Distance along the equator (a great circle):

\[d=\frac{\theta}{360^\circ}\times 2\pi R=\frac{56}{360}\times 2\times\frac{22}{7}\times 6400.\]\[d=\frac{56}{360}\times\frac{2\times22\times6400}{7}=\frac{56}{360}\times 40228.57\]\[d\approx 6257.78\text{ km}\approx\mathbf{6258\text{ km}}.\]

(ii) Distance of \(X\) from the North pole. The North pole is \(90^\circ\) of latitude from the equator, measured along a meridian (a great circle):

\[d=\frac{90}{360}\times 2\pi R=\frac{1}{4}\times 2\times\frac{22}{7}\times 6400\]\[d=\frac{1}{4}\times 40228.57\approx\mathbf{10057\text{ km}}.\]

(b) In circle centre \(O\), \(N\) is the mid-point of chord \(PQ\) with \(|PQ|=8\text{ cm}\), \(|ON|=3\text{ cm}\) and \(\angle ONR=20^\circ\). Find \(\angle ORN\).

Since \(N\) is the mid-point of the chord, \(ON\perp PQ\) and

\[|PN|=\tfrac{1}{2}|PQ|=4\text{ cm}.\]

The radius is \(|OP|\); by Pythagoras in right triangle \(ONP\):

\[|OP|=\sqrt{|ON|^2+|PN|^2}=\sqrt{3^2+4^2}=\sqrt{25}=5\text{ cm}.\]

\(R\) lies on the circle, so \(|OR|=5\text{ cm}\) too. Now apply the sine rule in triangle \(ONR\), where \(|ON|=3\), \(|OR|=5\) and \(\angle ONR=20^\circ\):

\[\frac{\sin\angle ORN}{|ON|}=\frac{\sin\angle ONR}{|OR|}\]\[\sin\angle ORN=\frac{|ON|\sin 20^\circ}{|OR|}=\frac{3\times0.3420}{5}=0.2052\]\[\angle ORN=\sin^{-1}(0.2052)\approx 11.8^\circ\approx\mathbf{12^\circ}.\]

Akọwa Nkọwa

	Limes	Apples	Total
Good	10	8	18
Bad	6	6	12
Total	16	14	30

There are 30 fruits. Picking is without replacement, so the second denominator is 29.

(i) Both good limes (10 good limes):

\[ \frac{10}{30}\times\frac{9}{29}=\frac{90}{870}=\frac{3}{29}\approx 0.103 \]

(ii) Both bad fruits (12 bad fruits):

\[ \frac{12}{30}\times\frac{11}{29}=\frac{132}{870}=\frac{22}{145}\approx 0.152 \]

(iii) One good apple and one bad lime (either order; 8 good apples, 6 bad limes):

\[ \frac{8}{30}\times\frac{6}{29}+\frac{6}{30}\times\frac{8}{29}=\frac{48+48}{870}=\frac{96}{870}=\frac{16}{145}\approx 0.110 \]

(b) Solve \(\log_3(4x+1)-\log_3(3x-5)=2\). Combine the logs:

\[ \log_3\!\left(\frac{4x+1}{3x-5}\right)=2 \;\Rightarrow\; \frac{4x+1}{3x-5}=3^2=9 \] \[ 4x+1=9(3x-5)=27x-45 \;\Rightarrow\; 46=23x \;\Rightarrow\; x=2 \]

Check: \(3x-5=1>0\) and \(4x+1=9>0\), so \(x=2\) is valid.

Akọwa Nkọwa

Akọwa Nkọwa

(a) Find x and y.

The six numbers in ascending order are \(1,\;2,\;x,\;11,\;y,\;14\).

Use the mean. The mean of the six values is 8, so their sum is \(6\times 8=48\):

\[1+2+x+11+y+14=48\]

\[28+x+y=48\Rightarrow x+y=20\quad(1)\]

Use the median. For six values the median is the average of the 3rd and 4th values, which are \(x\) and \(11\):

\[\frac{x+11}{2}=9\Rightarrow x+11=18\Rightarrow x=7\]

Substitute into (1):

\[7+y=20\Rightarrow y=13\]

Check ascending order: \(1,2,7,11,13,14\) is valid. So \(x=7,\;y=13\).

(b) Area of quadrilateral MPQN.

Since \(MN\parallel PQ\), triangles \(LMN\) and \(LPQ\) are similar (equal angles at \(L\), and corresponding angles equal).

From the diagram, \(|LM|=3\text{ cm}\) and \(|MP|=1\text{ cm}\), so:

\[|LP|=|LM|+|MP|=3+1=4\text{ cm}\]

The ratio of corresponding sides is:

\[\frac{|LM|}{|LP|}=\frac{3}{4}\]

The ratio of areas of similar triangles is the square of the ratio of sides:

\[\frac{\text{Area }\Delta LMN}{\text{Area }\Delta LPQ}=\left(\frac{3}{4}\right)^{2}=\frac{9}{16}\]

Given \(\text{Area }\Delta LMN=18\text{ cm}^{2}\):

\[18=\frac{9}{16}\times\text{Area }\Delta LPQ\]

\[\text{Area }\Delta LPQ=18\times\frac{16}{9}=32\text{ cm}^{2}\]

The quadrilateral \(MPQN\) is the region between the two parallel lines:

\[\text{Area }MPQN=\text{Area }\Delta LPQ-\text{Area }\Delta LMN=32-18=14\text{ cm}^{2}\]

Akọwa Nkọwa

(a) Substitute \(a=1,\ b=2,\ y=3\) into \(y = ax^2 - bx - 12\):

\[3 = x^2 - 2x - 12 \Rightarrow x^2 - 2x - 15 = 0.\]

Factorise: \((x - 5)(x + 3) = 0\), giving

\[x = 5\quad\text{or}\quad x = -3.\]

(b) Square both sides of \(\sqrt{x^2 + 1} = \tfrac{5}{4}\):

\[x^2 + 1 = \frac{25}{16} \Rightarrow x^2 = \frac{25}{16} - 1 = \frac{9}{16}.\]

\[x = \frac{3}{4}\quad(\text{positive value}).\]

Akọwa Nkọwa

None

Akọwa Nkọwa

(a)

\[\frac{4.56\times3.6}{0.12} = \frac{16.416}{0.12} = 136.8 = 1.368\times10^{2}.\]

(b) Use the difference of two squares \(a^2 - b^2 = (a-b)(a+b)\):

\[(73.8)^2 - (26.2)^2 = (73.8 - 26.2)(73.8 + 26.2) = 47.6\times100 = 4760.\]

(c) Convert the mixed number to an improper fraction:

\[1\tfrac{19}{81} = \frac{81 + 19}{81} = \frac{100}{81},\qquad \sqrt{\frac{100}{81}} = \frac{10}{9}.\]

Akọwa Nkọwa

(a)(i) Volume of soil dug from the well.

The well is a cylinder of radius \(r=1\text{ m}\) and depth \(h=8\text{ m}\).

\[V=\pi r^2 h=\frac{22}{7}\times 1^2\times 8=\frac{176}{7}=25.14\text{ m}^3\]

The volume of soil dug out is \(25.14\text{ m}^3\) (to 2 d.p.).

(a)(ii) Thickness of the new layer of soil.

The soil is spread over a rectangular floor \(4\text{ m}\times 12\text{ m}\).

\[\text{Floor area}=4\times 12=48\text{ m}^2\]

Let the thickness be \(t\). Volume of the layer = floor area \(\times\) thickness, and this equals the volume of soil:

\[48\times t=\frac{176}{7}\]\[t=\frac{176}{7\times 48}=\frac{176}{336}=0.5238\text{ m}\]

Converting to centimetres: \(0.5238\times 100=52.38\text{ cm}\).

Correct to the nearest cm, the thickness is \(52\text{ cm}\).

(b) Quadrilateral \(ABCD\).

From the diagram, \(\angle DAB=90^{\circ}\), \(|AB|=3.3\text{ cm}\), \(|AD|=5.6\text{ cm}\), \(|BC|=3.9\text{ cm}\) and \(|DC|=5.2\text{ cm}\).

(i) Length of \(BD\).

Triangle \(ABD\) is right-angled at \(A\), so by Pythagoras:

\[BD^2=AB^2+AD^2=3.3^2+5.6^2=10.89+31.36=42.25\]\[BD=\sqrt{42.25}=6.5\text{ cm}\]

(ii) Show that \(\angle BCD=90^{\circ}\).

In triangle \(BCD\), \(|BC|=3.9\), \(|DC|=5.2\) and \(|BD|=6.5\). Test the converse of Pythagoras:

\[BC^2+DC^2=3.9^2+5.2^2=15.21+27.04=42.25\]\[BD^2=6.5^2=42.25\]

Since \(BC^2+DC^2=BD^2\), the converse of Pythagoras theorem holds, so the angle opposite \(BD\) is a right angle. Therefore \(\angle BCD=90^{\circ}\).