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**Question 3**
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Find the \(n^{th}\) term of the sequence 2 x 3, 4 x 6, 8 x 9, 16 x 12...

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**Question 4**
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Find the mean deviation of 20, 30, 25, 40, 35, 50, 45, 40, 20 and 45

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**Question 5**
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In the diagram, WXYZ is a rectangle with dimension 8cm by 6cm. P, Q, R and S are the midpoints of the sides of the rectangle as shown. Using this information, what type of quadrilateral is the shaded region?

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**Question 6**
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The graph of y = x\(^2\) and y = x intersect at which of these points?

**Answer Details**

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).

**Question 7**
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Expression 0.612 in the form \(\frac{x}{y}\), where x and y are integers and y \(\neq\) 0

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**Question 8**
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A curve is such that when y = 0, x = -2 or x = 3. Find the equation of the curve.

**Answer Details**

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 \(y = x^2 - x - 6\).

**Question 9**
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There are 250 boys and 150 girls in a school, if 60% of the boys and 40% of the girls play football, what percentage of the school play football?

**Answer Details**

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).

**Question 10**
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Find the value of x for which \(32_{four} = 22_x\)

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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.

**Question 11**
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Given that Y is 20cm on a bearing of 300\(^o\) from x, how far south of y is x?

**Question 14**
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M and N are two subsets of the universal set (U). If n(U) = 48, n(M) = 20, n(N) = 30 and n(MUN) = 40, find n(M \(\cap\) N)

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**Question 15**
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Find the median of 2, 1, 0, 3, 1, 1, 4, 0, 1 and 2

**Answer Details**

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.

**Question 16**
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In the diagram, WXYZ is a rectangle with diamension 8cm by 6cm. P, Q, R and S are the midpoints of the rectangle as shown. Using this information calculate the area of the part of the rectangle that is not shaded

**Answer Details**

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\(^2\).

**Question 17**
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If tan x = \(\frac{4}{3}\), 0\(^o\) < x < 90\(^o\), find the value of sin x - cos x

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**Question 19**
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The mean of 1, 3, 5, 7 and x is 4. Find the value of x

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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.

**Question 20**
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The diagonals of a rhombus WXYZ intersect at M. If |MW| = 5cm and |MX| = 12cm, calculate its perimeter

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**Question 21**
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The angle of elevation of the top of a tree from a point 27m away and on the same horizontal ground as the foot of the tree is 30\(^o\). Find the height of the tree.

**Answer Details**

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. ![tree diagram](https://i.imgur.com/vQEZw5R.png) 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.

**Question 22**
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Find the inter-quartile range of 1, 3, 4, 5, 8, 9, 10, 11, 12, 14, 16

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**Question 23**
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Find the value of x for which \(\frac{x - 5}{x(x - 1)}\) is defined

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**Question 24**
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The angles of a polygon are x, 2x, 2x, (x + \(30^o\)), (x + \(20^o\)) and (x - \(10^o\)). Find the value of x

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**Question 25**
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Find the value of t in the diagram

**Answer Details**

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.

**Question 27**
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The table shows the distribution of goals scored by 25 teams in a football competition. Calculate the probability that a team selected at randon scored either 4 or 7 goals.

**Answer Details**

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: - \(\frac{9}{25}\)

**Question 28**
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In the diagram, PS and RS are tangents to the circle centre O,

**Question 29**
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The volume of a cylindrical tank, 10m high is 385 m\(^2\). Find the diameter of the tank. [Take \(\pi = \frac{22}{7}\)]

**Answer Details**

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.

**Question 30**
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If y + 2x = 4 and y - 3x = -1, find the value of (x + y)

**Answer Details**

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.

**Question 31**
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Factorise completely the expression

\((x + 2)^2\) - \((2x + 1)^2\)

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**Question 32**
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Given that y varies inversely as the square of x. If x = 3 when y = 100, find the equation connecting x and y.

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**Question 33**
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Simplify; \(\frac{2 - 18m^2}{1 + 3m}\)

**Answer Details**

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 \(2(1-3m)\).

**Question 34**
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The table shows the distribution of goals scored by 25 teams in a football competition. Calculate the probability that a team selected at random scored at most 3 goals.

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**Question 35**
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Simplify; 2\(\frac{1}{4} \times 3\frac{1}{2} \div 4 \frac{3}{8}\)

**Answer Details**

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\(\frac{4}{5}\).

**Question 36**
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If \(\log_{10}\)(6x - 4) - \(\log_{10}\)2 = 1, solve for x.

**Answer Details**

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.

**Question 37**
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If F = \(\frac{9}{5}\)C + 32, find C when F = 98.6

**Answer Details**

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).

**Question 38**
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The equation of the line through the points (4,2) and (-8, -2) is 3y = px + q, where p and q are constants. Find the value of p.

**Answer Details**

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.

**Question 39**
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The total surface area of a hemispher is 75\(\pi cm^2\). Find the radius.

**Question 40**
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In the diagram, which of the following ratios is equal to \(\frac{|PN|}{|PQ|}\)?

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**Question 41**
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Evaluate: \((64^{\frac{1}{2}} + 125^{\frac{1}{3}})^2\)

**Question 42**
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If P and Q are two statements, under what condition would p|q be false?

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**Question 44**
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If x : y = \(\frac{1}{4} : \frac{3}{8}\) and y : z = \(\frac{1}{3} : \frac{4}{9}\), find x : z

**Answer Details**

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.

**Question 45**
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If M and N are the points (-3, 8) and (5, -7) respectively, find |MN|

**Answer Details**

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".

**Question 46**
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In the diagram, PQ is a straight line, (m + n) = 110\(^o\) and (n + r) = 130\(^o\) and (m + r) = 120\(^o\). Find the ratio of m : n : r

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**Question 47**
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Donations during the launching of a church project were sent in sealed envolopes. The table shows the distribution of the amount of money in the envelope. How much was the donation?

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**Question 48**
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The diagram shows a trapezium inscribed in a semi-circle. If O is the mid-point of WZ and |WX| = |XY| = |YZ|, calculate the value of m

**Answer Details**

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.

**Question 49**
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A piece of thread of length 21.4cm is used to form a sector of a circle of radius 4.2cm on a piece of cloth. Calculate, correct to the nearest degree, the angle of the sector. [Take \(\pi = \frac{22}{7}\)]

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**Question 50**
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The surface area of a sphere is \(\frac{792}{7} cm^2\). Find, correct to the nearest whole number, its volume. [Take \(\pi = \frac{22}{7}\)]

**Answer Details**

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\(cm^3\).

**Question 51**
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A container, in the form of a cone resting on its vertex, is full when 4.158 litres of water is poured into it.

(a) If the radius of its base is 21 cm,

(i) represent the information in a diagram;

(ii) calculate the height of the container.

(b) A certain amount of water is drawn out of the container such that the surface diameter of the water drops to 28 cm. Calculate the volume of the water drawn out. (Take \(\pi\) = \(\frac{22}{7}\))

**Question 52**
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(a) In a right-angled triangle, sin X = \(\frac{3}{5}\). Evaluate, leaving the answer as a fraction, 5 (cosX)\(^2\) – 3.

(b) The base of a pyramid, 12 cm high, is a rectangle with dimensions 42 cm by 11 cm. if the pyramid is filled with water and emptied into a conical container of equal height and volume, calculate, leaving the answer in surd form (radicals), the base radius of the container. [Take π=\(\frac{22}{7}\)]

**Question 53**
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a. A textbook company discovered that the profit made from selling its books is given by y = \(\frac{x^2}{8}\) + 5x, where x is the number of textbooks sold (in thousands) and y is the corresponding profit (in Ghana Cedis). If the company made a profit of GH₵ 20,000.00

i. form a quadratic equation in x;

ii. (using the quadratic formula, find, correct to the nearest whole number, the number of textbooks sold to make the profit.

b. The angle of elevation of the top T of a tree from a point P on the same ground level as the foot Q of a tree is 28\(^o\). A bird perched at a point R, halfway up the tree.

i. Represent the information in a diagram.

ii. Calculate, correct to the nearest degree, the angle of elevation of R from P.

**Answer Details**

None

**Question 54**
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(a) Using a scale of 2cm to 2units on both axes, draw on a sheet of graph paper two perpendicular axes 0x and 0y for - 10 \(\leq\) x \(\leq\) 10 and -10 \(\leq\) y \(\leq\)10

(b) Given the points P(3, 2). Q(-1. 5). R(0. 8) and S(3, 7). draw on the same graph, indicating clearly the vertices and their coordinates, the:

(i) quadrilateral PQRS;

(ii) image P\(_1\)Q\(_1\)R\(_1\)S\(_1\) of PQRS under an anticlockwise rotation of 90\(^o\) about the origin where P \(\to\) P\(_1\), Q \(\to\) Q\(_{1}\), R \(\to\) R\(_{1}\) and S \(\to\) S\(_{1}\)

(iii) image P\(_2\)Q\(_2\)R\(_2\)S\(_2\) of P\(_1\)Q\(_1\)R\(_1\)S\(_1\) under a reflection in the line y - x = 0 where P\(_1\) \(\to\) P\(_2\), Q\(_1\) \(\to\) Q\(_{2}\), R\(_1\) \(\to\) R\(_{2}\) and S\(_1\) \(\to\) S\(_{2}\)

(c) Describe precisely the single transformation T for which T : PQRS \(\to\) P\(_2\)Q\(_2\)R\(_2\)S\(_2\)

(d) The side P\(_1\)Q\(_1\) of the quadrilateral P\(_1\)Q\(_1\)R\(_1\)S\(_1\) cuts the x-axis at the point W. What type of quadrilateral is P\(_1\)S\(_1\)R\(_1\)W?

None

**Answer Details**

None

**Question 55**
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A used car was purchased at N900,000.00. Its value depreciated by 30% in the first year. In each subsequent year, the depreciation was 22% of its value at the beginning of the year. If the car was bought on the 1st of March, 2011, calculate, correct to the nearest hundred naira, the value of the car on the 28th of February, 2015.

**Question 56**
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The table shows the distribution of marks scored by students in a test.

Mark (%) |
10 - 19 | 20 - 29 | 30 - 39 | 40 - 49 | 50 - 59 | 60 - 69 | 70 - 79 | 80 - 89 | 90 - 99 |

Frequency | 4 | 7 | 12 | 18 | 20 | 14 | 9 | 4 | 2 |

(a) Construct a cumulative frequency table for the distribution.

(b) Draw a cumulative frequency curve for the distribution.

(c) Use the curve to estimate the:

(i) median;

(ii) probability that a student selected at random obtained distinction, if the lowest mark for distinction is 75%.

None

**Answer Details**

None

**Question 57**
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1. A donkey is tied with a rope to a post which is 15 *m* from a fence. If the length of the rope between the donkey and the post is 17*m*, calculate the length of the fence within the reach of the donkey.

2. The base of a right pyramid with vertex, V, is a square, PQRS, of side 15 *cm*. If the slant height is 32 *cm* long:

3. represent the information in a diagram;

4. calculate its:

5. height, correct to **one** decimal place;

6. volume, correct to the **nearest \(cm^3\)**

**Question 58**
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The table shows the distribution of sources obtained when a fair diwe was rolled 50 times.

Score | 1 | 2 | 3 | 4 | 5 | 6 |

Frequency | 2 | 5 | 13 | 11 | 9 | 10 |

1. Draw a bar chart for the distribution

2. Calculate the mean score of the distribution

**Answer Details**

None

**Question 59**
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(a) Mr John paid N4,800.00 in N1.00 ordinary shares of a company which sold at N2.50 per share. If dividend was declared at 25k per share, how much dividend did he get?

(b) Using the method of completing the square, solve \(\frac{1 - x}{x} + \frac{x}{1 - x} = \frac{5}{2}\)

**Answer Details**

None

**Question 60**
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(a) If the mean of m, n, s, p and q is 12, calculate the mean of (m + 4), (n - 3), (s + 6), (p - 2) and (q + 8).

(b) In a community of 500 people, the 75th percentile age is 65 years while the 25th percentile age is 15 years. How many of the people are between 15 and 65 years?

**Answer Details**

None

**Question 61**
**Report**

a. Find the range of values of x which satisfy the following inequalities simultaneously: 5 - x > 1 and 9 + x \(\geq\) 8