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When a chord is drawn in a circle, it divides the circle into two equal parts. The line connecting the midpoint of the chord and the center of the circle is perpendicular to the chord. Therefore, to find the distance of the chord from the center of the circle, we need to draw a perpendicular bisector to the chord and measure the distance from the center of the circle to the perpendicular bisector. Let O be the center of the circle, AB be the chord of length 6cm and M be the midpoint of AB. Then OM is the perpendicular bisector of AB, and AM = MB = 3cm. Using the Pythagorean theorem in triangle OAM, we have: $$OA^2 = OM^2 + AM^2 = 5^2 - 3^2 = 16$$ Taking the square root of both sides, we have OA = 4cm. Therefore, the distance of the chord from the center of the circle is 4cm. So the correct option is (d) 4.0cm.

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We can simplify this expression by using the rule that states \(\frac{a^m}{a^n}=a^{m-n}\) for any non-zero real number a and integers m and n. Applying this rule, we get: \[\frac{3x^3}{(3x)^3} = \frac{3x^3}{27x^3} = \frac{1}{9}\] Therefore, the answer is \(\frac{1}{9}\).

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To simplify this expression, we need to perform division and then express the answer in standard form. (0.3 x 10⁵) \(\div\) (0.4 x 10⁷) = (0.3/0.4) x 10⁵−7 = 0.75 x 10^-2 = 7.5 x 10^-3 Therefore, the answer is 7.5 x 10^-3.

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The area of the square field is given as 110.25m². We know that the area of a square is given by the formula A = s², where A is the area and s is the side length. Therefore, we can find the side length of the square as follows: s² = A s² = 110.25 s = √110.25 = 10.5m The perimeter of the square is given by the formula P = 4s, where P is the perimeter and s is the side length. Therefore, we can find the perimeter of the square as follows: P = 4s = 4(10.5m) = 42m To find the cost of fencing it round at N75.00 per metre, we need to multiply the perimeter by N75.00: Cost = P × N75.00 = 42m × N75.00 = N3,150.00 Therefore, the cost of fencing the square field at N75.00 per metre is N3,150.00. So the correct option is (B) N3,150.00.

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Let Uche's share be x. Since Kayode has twice as much as Uche, Kayode's share will be 2x. Also, since Abu has twice as much as Kayode, Abu's share will be 2(2x) = 4x. The sum of their shares is N140,000, so we have: x + 2x + 4x = N140,000 7x = N140,000 x = N20,000 Therefore, Kayode's share is 2x = 2(N20,000) = N40,000. Hence, the answer is N40,000.

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The circumference of a circle is given by the formula 2πr, where r is the radius of the circle and π is a constant value approximately equal to 22/7 or 3.14. Using the given formula, the circumference of the circle with radius 16cm is 2 x (22/7) x 16cm = 100.57cm (to 2 decimal places). Similarly, the circumference of the circle with radius 23cm is 2 x (22/7) x 23cm = 144.57cm (to 2 decimal places). Therefore, the difference between the circumferences of the two circles is 144.57cm - 100.57cm = 44.00cm (to 2 decimal places). Hence, the correct option is (C) 44.00cm.

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To solve the inequality 4x - 3(2x - 1) > 1, we can simplify it as follows: 4x - 6x + 3 > 1 -2x + 3 > 1 -2x > -2 x < 1 Therefore, the range of values of x that satisfy the inequality is x < 1. So, the correct option is "x < 1".

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We can use the formula for the volume of a cylinder to solve this problem, which is: V = πr²h where V is the volume of the cylinder, r is the radius of the base, and h is the height of the cylinder. We are given that the volume of the cylinder is 1200cm³, so we can substitute this value into the formula: 1200 = πr²h We are also given that the area of the base is 150cm², and the formula for the area of a circle is: A = πr² where A is the area of the circle and r is the radius. Since the base of the cylinder is a circle, we can use this formula to find the radius: 150 = πr² r² = 150/π r ≈ 6.12 cm (rounded to two decimal places) Now we can substitute this value for r in the first formula and solve for h: 1200 = π(6.12)²h h ≈ 8 cm (rounded to two decimal places) Therefore, the height of the cylinder is approximately 8cm. So the correct option is (b) 8.00cm.

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To simplify the expression, we need to first expand the denominator of the fraction, which is \((3x)^3\). This can be written as \(3^3 x^3\) or \(27x^3\). Substituting this in the original fraction gives: $$\frac{3x^3}{(3x)^3} = \frac{3x^3}{27x^3} = \frac{1}{9}$$ Therefore, the simplified form of the expression is \(\frac{1}{9}\). Hence, the correct option is \(\frac{1}{9}\).

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If you are currently x years old, and your brother is 3 years older than you, then your brother's current age is x + 3. To find out how old your brother was last year, you need to subtract 1 from his current age: (x + 3) - 1 = x + 2 Therefore, your brother was x + 2 years old last year. So, the correct option is (b) (x + 2) years.

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We know that 8^x-1 = \(\frac{1}{4}\). We can write 8^x-1 as (2³)^x-1 = 2^3x-3. Similarly, we can write 1/4 as 2^-2. Therefore, we can write the given equation as 2^3x-3 = 2^-2. Since the bases are equal, we can equate the exponents to get: 3x - 3 = -2 Adding 3 to both sides, we get: 3x = 1 Dividing both sides by 3, we get: x = \(\frac{1}{3}\) Therefore, the value of x is \(\frac{1}{3}\).

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Let's call the number of normal working hours the messenger did as "x" and the number of overtime hours as "y". We can create two equations based on the given information: Equation 1: The messenger worked a total of 10 hours x + y = 10 Equation 2: The total amount paid is N31.00 2.5x + 4y = 31 We can use the first equation to solve for x in terms of y: x = 10 - y Now we can substitute this expression for x into the second equation: 2.5(10 - y) + 4y = 31 Simplifying and solving for y, we get: 25 - 2.5y + 4y = 31 1.5y = 6 y = 4 Therefore, the messenger worked 4 hours of overtime. Answer: (c) 4.

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To find the mean age of the twenty children, we need to calculate the sum of the ages and divide by the total number of children. Sum of the ages = (1 x 2) + (2 x 3) + (3 x 5) + (4 x 6) + (5 x 4) = 2 + 6 + 15 + 24 + 20 = 67 Total number of children = 2 + 3 + 5 + 6 + 4 = 20 Mean age = sum of the ages/total number of children = 67/20 = 3.35 years Therefore, the mean age of the twenty children is 3.35 years. The correct option is (b) 3.35 years.

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We can use the formula for simple interest: Simple Interest = (P * r * t) / 100 where P is the principal, r is the rate of interest per annum, and t is the time period in years. We are given that P = N520.00 and the simple interest, SI = N39.00. We are also given that the time period is 3 years. Substituting these values into the formula, we get: N39.00 = (N520.00 * r * 3) / 100 Simplifying this equation, we get: r = (N39.00 * 100) / (N520.00 * 3) = 2.5% Therefore, the rate of interest per annum is 2.5%. So, the correct option is: - 2\(\frac{1}{2}\)%

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The sum of the interior angles of a polygon can be found by using the formula (n-2) x 180, where n is the number of sides of the polygon. For a regular polygon, all interior angles are equal. Let's call the measure of each interior angle x. Since the polygon is regular, we can use the fact that the sum of the measures of the interior angles of a polygon with n sides is (n-2) x 180, and that each interior angle of this polygon has measure x. Thus, we can set up the equation: n x x = (n-2) x 180 Simplifying this equation, we get: nx = 180n - 360 nx - 180n = -360 n(x - 180) = -360 n = -360 / (x - 180) We know that each interior angle of the polygon has a measure of 162 degrees, so x = 162. Substituting this value into the equation we just derived, we get: n = -360 / (162 - 180) n = -360 / (-18) n = 20 Therefore, the polygon has 20 sides. Answer: 20.

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To solve this system of simultaneous equations, we need to find the values of x and y that satisfy both equations at the same time. One way to do this is to use the method of substitution. From the first equation, we have x + y = 2, which we can rearrange as x = 2 - y. We can then substitute this expression for x into the second equation, giving 3(2-y) - 2y = 1. Simplifying the left-hand side, we get 6 - 5y = 1, and solving for y, we find y = 1. We can then substitute this value for y into either equation to find x. Using the first equation, we get x + 1 = 2, so x = 1. Therefore, the solution to the simultaneous equations is x = 1 and y = 1.

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We can solve the given equation by taking logarithms to the base 2 on both sides. This gives: \begin{align*} 8^{x+1} &= \frac{1}{4}\\ \Rightarrow \quad 2^{3(x+1)} &= 2^{-2}\\ \Rightarrow \quad 3(x+1) &= -2\\ \Rightarrow \quad 3x+3 &= -2\\ \Rightarrow \quad 3x &= -5\\ \Rightarrow \quad x &= -\frac{5}{3} \end{align*} Therefore, the value of $x$ is $-\frac{5}{3}$. So, the correct option is \(-\frac{5}{3}\).

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To make w the subject of the relation \(\frac{a + bc}{wd + f} = g\), we can follow these steps: 1. Multiply both sides by \((wd + f)\) 2. Divide both sides by g 3. Divide both sides by (a + bc) This gives: \begin{align*} \frac{a + bc}{wd + f} &= g \\ (a + bc) &= g(wd + f) \\ a + bc &= gwd + gf \\ gwd &= a + bc - gf \\ w &= \frac{a + bc - gf}{gd} \end{align*} Therefore, the solution is: \begin{equation*} w = \frac{a + bc - gf}{gd} \end{equation*} Hence, the option that matches this answer is: - \(\frac{a + bc - fg}{dg}\)

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To solve this question, we need to convert the number 111 from binary to decimal, then square the result and convert the answer back to binary. Starting with the binary number 111, we can convert it to decimal using the place value system. The rightmost digit represents 1, the second digit from the right represents 2, and the leftmost digit represents 4. Adding these values together, we get: 1 + 2 + 4 = 7 So 111 in binary is equal to 7 in decimal. To square 7, we simply multiply it by itself: 7 x 7 = 49 So the decimal equivalent of (111_two)² is 49. To convert this back to binary, we use the same place value system but in reverse. Starting with the largest power of 2 that is less than or equal to 49, we subtract that value and place a 1 in the corresponding digit. We then repeat this process with the remainder until we reach 0. 49 is greater than or equal to 32, so we subtract 32 and place a 1 in the 6th digit from the right. The remainder is 17. 17 is greater than or equal to 16, so we subtract 16 and place a 1 in the 5th digit from the right. The remainder is 1. 1 is less than 2, so we place a 1 in the 1st digit from the right. The remainder is 0, so we have our final answer: 49 in binary is equal to 110001_two Therefore, the correct answer is (b) 110001_two.

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The expression \(\frac{2x-1}{x+3}\) is not defined when the denominator, \(x+3\), is equal to zero. Therefore, we can solve the equation \(x+3=0\) to find the value of x that makes the expression undefined. \(x+3=0\) \(x=-3\) So, the value of x that makes the expression undefined is -3. Therefore, the correct answer is (-3).

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First, we need to simplify each of the surds in the expression. \[\begin{aligned} 3\sqrt{45} &= 3\sqrt{9\times 5} = 3\times 3\sqrt{5} = 9\sqrt{5} \\ 16\sqrt{20} &= 16\sqrt{4\times 5} = 16\times 2\sqrt{5} = 32\sqrt{5} \end{aligned}\] Now we can substitute these simplified surds back into the original expression and simplify it further: \[\begin{aligned} 3\sqrt{45} - 12\sqrt{5} + 16\sqrt{20} &= 9\sqrt{5} - 12\sqrt{5} + 32\sqrt{5} \\ &= (9-12+32)\sqrt{5} \\ &= 29\sqrt{5} \end{aligned}\] Therefore, the simplified expression is 29\(\sqrt{5}\). The correct option is (a).

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To divide the sum of the given numbers by their mean, we need to first find their sum and mean. Sum = 8 + 6 + 7 + 2 + 0 + 4 + 7 + 2 + 3 = 39 Mean = (8 + 6 + 7 + 2 + 0 + 4 + 7 + 2 + 3) / 9 = 4.33 (rounded to two decimal places) Now, to divide the sum by the mean, we simply divide the sum by the mean: 39 / 4.33 = 9 (rounded to the nearest whole number) Therefore, the answer is 9.

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To expand the expression (3a - xy)(3a + xy), we can use the formula for multiplying two binomials: (a + b)(a - b) = a² - b² If we let a = 3a and b = xy, then we have: (3a - xy)(3a + xy) = (3a)² - (xy)² Expanding the terms on the right side, we get: (3a)² - (xy)² = 9a² - x²y² Therefore, the answer is 9a² - x²y².

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The given set P = {-3.2 ≤ x < 5} contains all the integers that are greater than or equal to -3.2 but less than 5. However, since x has to be an integer, the least possible value of x is -3 (since -3 is the greatest integer less than or equal to -3.2) and the greatest possible value of x is 4 (since 4 is the greatest integer less than 5). Therefore, the statement "least value of x is -3" is true for the set P, and the statement "greatest value of x is 5" is false.

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Substituting x = 3, y = 2 and z = 4 into the expression \(3x^{2}-2y+z\), we get: $$3(3)^2 - 2(2) + 4 = 27 - 4 + 4 = 27$$ Therefore, the value of \(3x^{2}-2y+z\) when x = 3, y = 2 and z = 4 is 27. The correct option is (b) 27.

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To convert a number from base 5 to base 3, we need to first convert it to base 10, and then convert it to base 3. 42₅ can be written as: $42_{5} = 4\cdot5^{1} + 2\cdot5^{0} = 20 + 2 = 22_{10}$ Now we convert 22₁₀ to base 3. We can do this by repeatedly dividing 22 by 3 and taking the remainders, until the quotient is zero. The remainders, read from bottom to top, give us the base 3 numeral. \begin{array}{c|c} \text{Dividend} & \text{Remainder}\\ \hline 22 & 1 \\ 7 & 2 \\ 2 & 2 \\ 0 & \\ \end{array} So, 22₁₀ is equal to 212₃. Therefore, the correct option is (3) 211₃.

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The volume of a cylinder is given by the formula V = πr²h, where r is the radius of the base and h is the height of the cylinder. In this case, the base radius is 10cm and the height is 14cm. Substituting these values into the formula, we have: V = π(10cm)²(14cm) = π(100cm²)(14cm) = 1400πcm³ Using the value of π ≈ 3.14, we can approximate the answer to: V ≈ 1400(3.14) ≈ 4400cm³ Therefore, the volume of the cylinder is approximately 4400cm³. The correct option is (D) 4400cm³.

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The sector of a circle can be folded to form a cone if and only if the arc length of the sector is equal to the circumference of the base of the cone. Given that the radius of the sector is 14 cm and the angle is 60^o, the arc length of the sector is \(\frac{60}{360}\times2\pi(14)=\frac{14\pi}{3}\) cm. Let's denote the radius of the base of the cone by r. Then, the circumference of the base of the cone is \(2\pi r\) cm. Since the arc length of the sector is equal to the circumference of the base of the cone, we have: \[\frac{14\pi}{3}=2\pi r\] Dividing both sides by 2π gives: \[r=\frac{14}{3}\div2=\frac{7}{3}=2\frac{1}{3}\,\text{cm}\] Therefore, the radius of the base of the cone is \(2\frac{1}{3}\,\text{cm}\). Answer:.

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To evaluate (111_two)², we need to convert the binary number 111_two to its decimal equivalent, then square the result, and finally convert the answer back to binary. 111_two = 1 × 2² + 1 × 2¹ + 1 × 2⁰ = 7_ten (111_two)² = (7_ten)² = 49_ten To convert 49_ten back to binary, we repeatedly divide by 2 and write down the remainders in reverse order: 49 ÷ 2 = 24 remainder 1 24 ÷ 2 = 12 remainder 0 12 ÷ 2 = 6 remainder 0 6 ÷ 2 = 3 remainder 0 3 ÷ 2 = 1 remainder 1 1 ÷ 2 = 0 remainder 1 Reading the remainders in reverse order gives us: 49_ten = 110001_two Therefore, the answer is (B) 110001_two.

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The probability of an event occurring is the measure of the likelihood that the event will happen, given as a number between 0 and 1, inclusive. The probability of an event not occurring is the measure of the likelihood that the event will not happen. Since the sum of the probabilities of an event occurring and not occurring is 1, we can find the probability of an event not occurring by subtracting the probability of the event occurring from 1. Therefore, the probability of an event not occurring is equal to 1 minus the probability of the event occurring. Mathematically, if the probability of an event occurring is x, then the probability of the event not occurring is 1 - x. So, option A, 1 - x, is the correct answer.

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Let's start by using variables to represent the amount of money each person gets. Let Uche's share be x. Then, we know that Kayode's share is twice that amount, so his share is 2x. And we also know that Abu's share is twice Kayode's share, so his share is 2(2x) = 4x. The sum of their shares is given as N140,000, so we can set up an equation: x + 2x + 4x = 140,000 Simplifying the left side of the equation, we get: 7x = 140,000 Dividing both sides by 7, we get: x = 20,000 So Uche's share is N20,000. We can then find Kayode's share, which is twice that amount: 2x = 2(20,000) = 40,000 Therefore, Kayode's share is N40,000. So the answer is, N40,000.

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The expression \(\frac{2x-1}{x+3}\) is undefined if the denominator is equal to zero, since division by zero is not allowed in mathematics. Therefore, we need to solve the equation \(x+3=0\) to find the value of x for which the expression is undefined. Solving for x, we have: \begin{align*} x+3 &= 0 \\ x &= -3 \end{align*} Therefore, the expression is not defined for x = -3. Thus, the correct answer is (-3).

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Given that p \(\alpha \frac{I}{Q}\), where \(\alpha\) means 'is proportional to'. To determine the relationship between q and p, we need to manipulate the equation so that q is isolated on one side. We can write p = k\(\frac{I}{Q}\), where k is the constant of proportionality. Multiplying both sides by Q, we get pQ = kI. Dividing both sides by p, we get Q = \(\frac{k}{p}\)I. Since k is a constant of proportionality, we can write it as k = cp for some other constant c. Therefore, Q = \(\frac{c}{p}\)I. This means that q is inversely proportional to p. So, the correct option is q \(\alpha \frac{1}{p}\).

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Substituting x = 3, y = 2, and z = 4 into the expression 3x² - 2y + z gives: 3x² - 2y + z = 3(3)² - 2(2) + 4 = 27 - 4 + 4 = 27 Therefore, the value of 3x² - 2y + z is 27.

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First, we need to simplify each term under the square root sign. We can write: \begin{align*} 3\sqrt{45} &= 3\sqrt{9\times 5} = 3\times 3\sqrt{5} = 9\sqrt{5}\\ 12\sqrt{5} &= 2\times 2\times 3\sqrt{5} = 4\sqrt{5}\times 3\\ 16\sqrt{20} &= 16\sqrt{4\times 5} = 16\times 2\sqrt{5} = 32\sqrt{5} \end{align*} Now we can substitute these simplified expressions back into the original expression and simplify: \begin{align*} 3\sqrt{45}-12\sqrt{5}+16\sqrt{20} &= 9\sqrt{5}-4\sqrt{5}\times 3+32\sqrt{5}\\ &= (9-12+32)\sqrt{5}\\ &= 29\sqrt{5} \end{align*} Therefore, the answer is \(29\sqrt{5}\).

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To solve the simultaneous equations x + y = 2 and 3x - 2y = 1, we can use the method of substitution or elimination. Method of Substitution: We can solve for one variable in terms of the other from the first equation and substitute it into the second equation, then solve for the other variable. Let's solve for y in terms of x from the first equation: x + y = 2 y = 2 - x Now substitute y = 2 - x into the second equation: 3x - 2y = 1 3x - 2(2 - x) = 1 3x - 4 + 2x = 1 5x = 5 x = 1 Now substitute x = 1 into either equation to find y: x + y = 2 1 + y = 2 y = 1 Therefore, the solution is x = 1 and y = 1. So, the correct answer is: x = 1 and y = 1.

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Perimeter of the tunnel cross-section.

The cross-section is made up of the straight base \(AB=100\text{ m}\) together with the arc that domes over the top. The arc has radius \(r=56\text{ m}\), and the diagram shows the arc is the major arc (it rises above and beyond a semicircle).

Step 1: Angle subtended by the chord at the centre.

Let \(O\) be the centre and let the perpendicular from \(O\) bisect the chord \(AB\). Half the chord is \(50\text{ m}\). If \(\theta\) is half the central angle of the minor arc:

\[\sin\theta=\frac{50}{56}=0.8929\Rightarrow\theta=63.26^{\circ}\]

So the minor-arc central angle is:

\[2\theta=126.52^{\circ}\]

Step 2: Reflex angle for the major (tunnel) arc.

\[\text{Reflex angle}=360^{\circ}-126.52^{\circ}=233.48^{\circ}\]

Step 3: Length of the major arc.

\[\text{Arc}=\frac{233.48}{360}\times 2\pi r=\frac{233.48}{360}\times 2\times\frac{22}{7}\times 56\]

\[=\frac{233.48}{360}\times 352=0.6486\times 352=228.3\text{ m}\]

Step 4: Perimeter.

\[\text{Perimeter}=AB+\text{arc}=100+228.3=328.3\text{ m}\]

Correct to the nearest metre, the perimeter of the cross-section is \(328\text{ m}\).

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(a) Factorise numerator and denominator: \[x^2 - 8x + 16 = (x-4)^2,\qquad x^2 - 7x + 12 = (x-3)(x-4).\] \[\frac{(x-4)^2}{(x-3)(x-4)} = \frac{x-4}{x-3}\quad (x\neq 4).\]

(b) If \(\tfrac{1}{2}, \tfrac{1}{x}, \tfrac{1}{3}\) are successive terms of an A.P., the common difference is constant: \[\frac{1}{x} - \frac{1}{2} = \frac{1}{3} - \frac{1}{x}.\] The left side is \(\dfrac{2 - x}{2x}\) and the right side is \(\dfrac{x - 3}{3x}\), so \[\frac{2 - x}{2x} = \frac{x - 3}{3x}.\] Multiplying both sides by \(3x\) and dividing by \((x-3)\): \[\frac{2 - x}{x - 3} = \frac{2x}{3x} = \frac{2}{3}.\] Hence \(\dfrac{2 - x}{x - 3} = \dfrac{2}{3}\), as required.

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(a) Arithmetic progression.

Let the first term be \(a\) and the common difference be \(d\). The \(n\)th term is \(T_n = a + (n-1)d\).

Given \(T_3 = -9\) and \(T_8 = 26\):

(i) Common difference. Subtract (1) from (2):

(ii) First term. Substitute \(d = 7\) into (1):

Common difference \(= 7\); first term \(= -23\).

(b) Area of the trapezium PQZY.

From the diagram, \(PQ \parallel YZ\), with \(|XP| = 2\text{ cm}\), \(|PY| = 3\text{ cm}\), \(|PQ| = 6\text{ cm}\) and area of \(\triangle XPQ = 24\text{ cm}^2\). P lies on XY and Q lies on XZ.

Since \(PQ \parallel YZ\), triangle \(XPQ\) is similar to triangle \(XYZ\) (equal corresponding angles). The ratio of corresponding sides along XY is:

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

Therefore:

The trapezium PQZY is what remains when \(\triangle XPQ\) is removed from \(\triangle XYZ\):

Area of trapezium PQZY \(= 126\text{ cm}^2\).

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(a) Numerator: \(3\tfrac{1}{12} + \tfrac{7}{8} = \tfrac{37}{12} + \tfrac{7}{8} = \tfrac{74}{24} + \tfrac{21}{24} = \tfrac{95}{24}.\)

Denominator: \(2\tfrac{1}{4} - \tfrac{1}{6} = \tfrac{9}{4} - \tfrac{1}{6} = \tfrac{27}{12} - \tfrac{2}{12} = \tfrac{25}{12}.\)

\[\frac{95/24}{25/12} = \frac{95}{24}\times\frac{12}{25} = \frac{95}{2\times 25} = \frac{95}{50} = \frac{19}{10} = \mathbf{1\tfrac{9}{10}}.\]

(b)(i) \(p = \dfrac{m}{2} - \dfrac{n^2}{5m}.\) Multiply through by \(5m\): \[5mp = \frac{5m^2}{2} - n^2 \Rightarrow n^2 = \frac{5m^2}{2} - 5mp \Rightarrow \mathbf{n = \sqrt{\frac{5m^2}{2} - 5mp}}.\]

(b)(ii) With \(p = 14,\ m = -8\): \[n^2 = \frac{5(-8)^2}{2} - 5(-8)(14) = \frac{320}{2} + 560 = 160 + 560 = 720.\] \[n = \sqrt{720} = \mathbf{26.8}\ (\text{to 3 s.f.}).\]

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(a) From the diagram, AOB is a straight line, so the angles on the upper side of the line sum to \(180^\circ\), and the angles on the lower side also sum to \(180^\circ\).

Upper side (angles \(AOC\) and \(COB\)):

Lower side (angles \(AOD\) and \(DOB\)):

From (1), \(y = 45 - x\). Substitute into (2):

Then \(y = 45 - 30 = 15\).

\(x = 30,\quad y = 15\)

Check: \(3(x+y)=135^\circ\), \(135+45=180^\circ\) (upper). \(5x+y=165^\circ\), \(165+15=180^\circ\) (lower). Correct.

(b) Let the base of the pole be \(B\), and let the two points be \(P\) and \(Q\) on opposite sides. The pole \(BT = 33\ \text{m}\) is vertical.

For the point with elevation \(53^\circ\):

For the point with elevation \(67^\circ\):

Since the points are on opposite sides of the pole, the distance between them is:

Distance \(\approx 38.9\ \text{m}\) (to 3 significant figures).

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(a) Using four-figure logarithm tables: \[\log(0.004592) = \bar{3}.6621 = -3 + 0.6621 = -2.3379.\] \[(0.004592)^{1/3}:\quad \frac{1}{3}(-2.3379) = -0.7793 = \bar{1}.2207.\] Antilog of \(0.2207\) is \(1.662\), so the value is \(1.662\times 10^{-1} \approx \mathbf{0.1662}.\)

(b) \(\log_{10} y + 3\log_{10} x = 2 \Rightarrow \log_{10}(yx^3) = 2 \Rightarrow yx^3 = 10^2 = 100.\) \[\mathbf{y = \frac{100}{x^3}}.\]

(c) \(3x - 2y = 21\ \ (1)\), \(4x + 5y = 5\ \ (2).\)

\((1)\times 5:\ 15x - 10y = 105.\quad (2)\times 2:\ 8x + 10y = 10.\) Adding: \(23x = 115 \Rightarrow x = 5.\)

From \((1)\): \(3(5) - 2y = 21 \Rightarrow -2y = 6 \Rightarrow y = -3.\)

Solution: \(\mathbf{x = 5,\ y = -3}.\)

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(a) \[\frac{3.69\times 10^{5}}{1.64\times 10^{-3}} = \frac{3.69}{1.64}\times 10^{5-(-3)} = 2.25\times 10^{8}.\] Answer in standard form: \(\mathbf{2.25\times 10^{8}}.\)

(b) Simple interest for one year. Bank A: \(\dfrac{20000\times y\times 1}{100} = 200y.\) Bank B: \(\dfrac{25000\times 1.5y\times 1}{100} = 375y.\)

Total interest: \[200y + 375y = 4600 \Rightarrow 575y = 4600 \Rightarrow y = \frac{4600}{575} = \mathbf{8}.\]

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Take the class mid-values \(x\) and build a working table (\(N=1+5+10+9+5=30\)):

(a) Mean.

\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{420}{30}=14.\]

(b) Standard deviation.

\[s=\sqrt{\frac{\sum fx^{2}}{N}-\bar{x}^{2}}=\sqrt{\frac{6710}{30}-14^{2}}=\sqrt{223.67-196}=\sqrt{27.67}\approx 5.26.\]

The mean number of absentees is 14 and the standard deviation is 5.26 (to 2 d.p.).

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Out of \(24\) apples, \(6\) are bad, so \(18\) are good. Because the drawing is with replacement, each draw is independent with \[P(\text{good}) = \frac{18}{24} = \frac{3}{4},\qquad P(\text{bad}) = \frac{6}{24} = \frac{1}{4}.\]

(a) First two good, third bad: \[\frac{3}{4}\times\frac{3}{4}\times\frac{1}{4} = \mathbf{\frac{9}{64}}.\]

(b) All three bad: \[\frac{1}{4}\times\frac{1}{4}\times\frac{1}{4} = \mathbf{\frac{1}{64}}.\]

(c) All three good: \[\frac{3}{4}\times\frac{3}{4}\times\frac{3}{4} = \mathbf{\frac{27}{64}}.\]

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\(Y\) is on bearing \(135^\circ\), \(Z\) on bearing \(225^\circ\) from \(X\). The angle between them at \(X\) is \[\angle YXZ = 225^\circ - 135^\circ = 90^\circ,\] so triangle \(YXZ\) is right-angled at \(X\).

(a) By Pythagoras: \[|YZ| = \sqrt{60^2 + 80^2} = \sqrt{3600 + 6400} = \sqrt{10000} = \mathbf{100\,\text{km}}.\]

(b) Using coordinates: \(Y = (60\sin135^\circ, 60\cos135^\circ) = (42.43, -42.43)\) and \(Z = (80\sin225^\circ, 80\cos225^\circ) = (-56.57, -56.57).\) \[\vec{YZ} = (-98.99,\ -14.14),\] which points South-West. The bearing satisfies \(\sin B = \dfrac{-98.99}{100},\ \cos B = \dfrac{-14.14}{100}\) (third quadrant), so \[B = 180^\circ + \tan^{-1}\!\frac{98.99}{14.14} = 180^\circ + 81.9^\circ \approx \mathbf{262^\circ}.\] Bearing of \(Z\) from \(Y\) is about \(262^\circ.\)

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(a) Height of the closed cylinder

For a cylinder closed at both ends, the total surface area is

With \(r = 3.5\text{ cm}\), \(TSA = 209\text{ cm}^2\) and \(\pi = \tfrac{22}{7}\):

Now \(2\times\dfrac{22}{7}\times 3.5 = 22\), so

The height of the cylinder is \(6\text{ cm}\).

(b) Circle with tangent ABC at B

From the diagram, \(EB\) is a diameter (it passes through the centre O). The given angles are the inscribed angle \(\angle BDF = 66^\circ\) at D and the tangent-chord angle \(\angle DBC = 57^\circ\) at B. G is the point where chord \(FD\) crosses the diameter \(EB\).

Finding the arcs

The inscribed angle \(\angle BDF\) stands on arc \(BF\) (not containing D), so

The tangent-chord angle \(\angle DBC\) equals half the intercepted arc \(BD\):

Since \(EB\) is a diameter, each semicircle is \(180^\circ\):

(Check: \(114+66+48+132 = 360^\circ\). Correct.)

(i) \(\angle EBF\)

\(\angle EBF\) is the inscribed angle at B standing on arc \(EF\):

(ii) \(\angle BGF\)

G is the intersection of chords \(EB\) and \(FD\). The angle between two chords equals half the sum of the two arcs it intercepts (arc \(BF\) and the vertically opposite arc \(ED\)):

Therefore \(\angle EBF = 24^\circ\) and \(\angle BGF = 99^\circ\).

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Construction of Triangle PQR and the Required Loci

(a)(i) Construct triangle PQR

1. Using a ruler, draw a line segment PQ = 10 cm.
2. At point Q, construct a perpendicular to PQ using a ruler and compasses. This gives a line making an angle of 90° with PQ.
3. Using the compasses set to 7 cm, measure from Q along the perpendicular line and mark the point R.
4. Join P to R. The required triangle PQR is formed.

Therefore, PQ = 10 cm, QR = 7 cm, and ∠PQR = 90°.

(a)(ii) Construct locus l<sub>1</sub>

The locus of points equidistant from Q and R is the perpendicular bisector of QR.

1. With centres Q and R, draw arcs of equal radius greater than half of QR.
2. Let the arcs intersect at two points.
3. Join the two intersection points with a straight line.

This line is labelled l<sub>1</sub>.

(a)(iii) Construct locus l<sub>2</sub>

The locus of points equidistant from P and Q is the perpendicular bisector of PQ.

1. With centres P and Q, draw arcs of equal radius greater than half of PQ.
2. Join the two points where the arcs intersect.

This line is labelled l<sub>2</sub>.

(b) Locate point O

Point O, which is equidistant from P, Q, and R, is the point of intersection of l<sub>1</sub> and l<sub>2</sub>.

Since triangle PQR is right-angled at Q, the point O is also the midpoint of the hypotenuse PR. It is the centre of the circle passing through P, Q, and R.