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**Question 1**
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Is the diagram, MN, PQ and RS are three intersecting straight lines. Which of the following statements is/are true? i. t = y ii. x + y + z + m = 180^{o} ii. x + m + n = 180^{o} iv. x + n = m + z

**Question 2**
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The diagram is a circle centre O. If < SPR = 2m and < SQR = n, express m in terms of n

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**Question 3**
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The graph is that of y = 2x^{2} - 5x - 3. For what value of x will y be negative? For what value of x will y be negative?

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To find when y is negative, we need to find the values of x for which 2x^2 - 5x - 3 is negative. One way to do this is to use the quadratic formula: x = (-b ± sqrt(b^2 - 4ac)) / 2a where a = 2, b = -5, and c = -3. Plugging these values in, we get: x = (-(-5) ± sqrt((-5)^2 - 4(2)(-3))) / 2(2) x = (5 ± sqrt(49)) / 4 x = (5 ± 7) / 4 So the solutions are x = -3/2 and x = 2. We can check that when x < -3/2 or x > 2, y is positive, and when -3/2 < x < 2, y is negative. Therefore, the answer is option (C) -\(\frac{1}{2} < x < 3\).

**Question 4**
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If N2,500.00 amounted to N3,50.00 in 4 years at simple interest, find the rate at which the interest was charged

**Answer Details**

Simple interest is calculated by multiplying the principal amount, the interest rate, and the time in years. The formula for simple interest is given as: Simple Interest = (Principal x Rate x Time) / 100 In this problem, we know that the principal amount (P) is N2,500.00, the amount (A) after 4 years is N3,500.00. We need to find the rate (R) at which the interest was charged. Using the formula for simple interest, we can write: A = P + I where I is the interest amount. We can rearrange the formula to solve for I: I = A - P = N3,500.00 - N2,500.00 = N1,000.00 Now, we can substitute the values we know into the formula for simple interest and solve for R: I = (P x R x T) / 100 N1,000.00 = (N2,500.00 x R x 4) / 100 Simplifying, we get: R = (N1,000.00 x 100) / (N2,500.00 x 4) = 10% Therefore, the rate at which the interest was charged is 10%.

**Question 5**
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Convert 35_{10} to number in base 2

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To convert a number from base 10 to base 2 (binary), we need to divide the decimal number by 2 repeatedly until the quotient is zero. Then, we write the remainders in reverse order. Let's convert 35_{10} to binary using this method: 35 / 2 = 17 remainder 1 17 / 2 = 8 remainder 1 8 / 2 = 4 remainder 0 4 / 2 = 2 remainder 0 2 / 2 = 1 remainder 0 1 / 2 = 0 remainder 1 The remainders, in reverse order, are 100011, which is the binary equivalent of 35_{10}. Therefore, the correct answer is option (C) 100011. Each digit in a binary number represents a power of 2, starting from the rightmost digit, which represents 2^0 = 1. The next digit represents 2^1 = 2, the next represents 2^2 = 4, and so on. To convert a binary number to decimal, we multiply each digit by its corresponding power of 2 and add up the results.

**Question 6**
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If x + y = 2y - x + 1 = 5, find the value of x

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We are given two equations, x + y = 5 and 2y - x + 1 = 5, and we need to find the value of x. From the first equation, we can rearrange it to get x = 5 - y. Substituting this value of x into the second equation, we get: 2y - (5 - y) + 1 = 5 Simplifying the equation, we get: 3y - 4 = 5 Adding 4 to both sides, we get: 3y = 9 Dividing both sides by 3, we get: y = 3 Substituting this value of y into x + y = 5, we get: x + 3 = 5 Subtracting 3 from both sides, we get: x = 2 Therefore, the value of x is 2.

**Question 7**
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Mr. Manu travelled from Accra to Pamfokromb a distance of 720km in 8 hours. What will be his speed in m/s?

**Answer Details**

To find the speed in meters per second (m/s), we need to convert the distance and time to meters and seconds respectively. Distance is given as 720km. 1km is equal to 1000m, so 720km = 720 x 1000m = 720,000m. Time taken is given as 8 hours. 1 hour is equal to 3600 seconds, so 8 hours = 8 x 3600 seconds = 28,800 seconds. Speed = Distance ÷ Time = 720,000m ÷ 28,800s = 25m/s. Therefore, Mr. Manu's speed is 25m/s. (25m/s) is the correct answer.

**Question 8**
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The position of three ships P,Q and R at sea are illustrated in the diagram. The arrows indicated the North direction. The bearing of Q from P is 050^{o} and < PQR = 72^{o}. Calculate the bearing of R and Q

**Question 9**
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The sum of the interior angles of regular polygon is 1800^{o}. How many sides has the polygon?

**Answer Details**

The sum of the interior angles of a polygon can be found by the formula: sum of interior angles = (n-2) x 180 degrees where n is the number of sides in the polygon. In this problem, we are given that the sum of the interior angles of the polygon is 1800 degrees. So we can set up an equation: 1800 = (n-2) x 180 Simplifying the equation: 10 = n - 2 n = 12 Therefore, the polygon has 12 sides.

**Question 10**
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A bag contains 4 red and 6 black balls of the same size. If the balls are shuffled briskly and two balls are drawn one after the other without replacement, find the probability of picking balls of different colours

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**Question 11**
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A side and a diagonal of a rhombus are 10cm and 12cm respectively, Find its area

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**Question 12**
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Make p the subject of the relation: q = \(\frac{3p}{r} + \frac{s}{2}\)

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**Question 13**
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The diagram is a circle with centre P. PRST are points on the circle. Find the value of < PRS

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**Question 14**
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If x and y are variables and k is a constant, which of the following describes an inverse relationship between x and y?

**Answer Details**

An inverse relationship between two variables means that as one variable increases, the other variable decreases. Mathematically, an inverse relationship can be represented by an equation where one variable is multiplied or divided by a constant. Out of the given equations, the equation that represents an inverse relationship between x and y is y = \(\frac{k}{x}\). To see why this is the case, let's consider what happens to y as x increases. Suppose x doubles in value. Then, according to the equation y = \(\frac{k}{x}\), y will be halved in value. Similarly, if x triples in value, y will be divided by 3. This means that as x increases, y decreases, which is the characteristic of an inverse relationship. On the other hand, in the equation y = kx, as x increases, y also increases. This is not an inverse relationship but a direct relationship. The same applies to y = k\(\sqrt{x}\) and y = x + k. Therefore, the equation that describes an inverse relationship between x and y is y = \(\frac{k}{x}\).

**Question 15**
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Express 3 - [\(\frac{x - y}{y}\)] as a single fraction

**Answer Details**

To express 3 - [\(\frac{x - y}{y}\)] as a single fraction, we first need to simplify the expression inside the square brackets: 3 - [\(\frac{x - y}{y}\)] = 3 - (\(\frac{x}{y}\) - \(\frac{y}{y}\)) = 3 - \(\frac{x}{y}\) + 1 = 4 - \(\frac{x}{y}\) Therefore, 3 - [\(\frac{x - y}{y}\)] can be expressed as a single fraction: = 4 - \(\frac{x}{y}\) = \(\frac{4y - x}{y}\) Hence, the answer is \(\frac{4y - x}{y}\).

**Question 16**
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In the diagram, O is a circle centre of the circle PQRS and < PSR = 86^{o}. If < PQR = x^{o}, find x

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**Question 17**
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The bar chart shows the frequency distribution of marks scored by students in a class test. Calculate the mean of the distribution.

**Answer Details**

To calculate the mean of a frequency distribution, we need to find the sum of all the values multiplied by their respective frequencies and divide by the total number of values. Looking at the bar chart, we can see that there are 5 values: 1, 2, 3, 4, and 5. The corresponding frequencies are 1, 2, 5, 6, and 1, respectively. To calculate the mean, we first need to calculate the sum of all the values multiplied by their respective frequencies: (1 x 1) + (2 x 2) + (3 x 5) + (4 x 6) + (5 x 1) = 1 + 4 + 15 + 24 + 5 = 49 Next, we need to calculate the total number of values, which is the sum of all the frequencies: 1 + 2 + 5 + 6 + 1 = 15 Finally, we divide the sum of the values multiplied by their frequencies by the total number of values: 49 / 15 = 3.27 (rounded to two decimal places) Therefore, the mean of the distribution is approximately 3.27. The correct answer is not provided in the options, but the closest one is option (C) 2.4.

**Question 18**
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A kite flies on a taut string of length 50m inclined at tan angle 54^{o} to the horizontal ground. The height of the kite above the ground is

**Question 19**
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Factorise completely: 32x^{2}y - 48x^{3}y^{3}

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We can begin by factoring out the greatest common factor of the two terms, which is 16x^{2}y. This gives: 32x^{2}y - 48x^{3}y^{3} = 16x^{2}y(2 - 3xy^{2}) So the correct option is 16x^{2}y(2 - 3xy^{2}).

**Question 20**
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In the diagram, |QR| = 10m, |SR| = 8m

< QPS = 30^{o}, < QRP = 90^{o} and |PS| = x, Find x

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**Question 21**
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Alfred spent \(\frac{1}{4}\) of his money on food, \(\frac{1}{3}\) on clothing and save the rest. If he saved N72,20.00, how much did he spend on food?

**Answer Details**

Let A be the total amount of money that Alfred has. From the given information, he spent \(\frac{1}{4}\) of A on food and \(\frac{1}{3}\) of A on clothing, and he saved the rest. Therefore, we have: Amount spent on food = \(\frac{1}{4}\)A Amount spent on clothing = \(\frac{1}{3}\)A Amount saved = A - (\(\frac{1}{4}\)A + \(\frac{1}{3}\)A) Amount saved = A - \(\frac{7}{12}\)A Amount saved = \(\frac{5}{12}\)A We know that Alfred saved N72,20.00, so we can write: \(\frac{5}{12}\)A = N72,20.00 Multiplying both sides by \(\frac{12}{5}\), we get: A = N172,80.00 Therefore, the amount spent on food is: \(\frac{1}{4}\)A = \(\frac{1}{4}\) x N172,80.00 = N43,20.00 Hence, Alfred spent N43,20.00 on food. Answer is correct.

**Question 22**
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The bar chart shows the frequency distribution of marks scored by students in a class test. What is the median of the distribution?

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**Question 23**
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Solve the inequality: \(\frac{-m}{2} - \frac{5}{4} \geq \frac{5m}{12} - \frac{7}{6}\)

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**Question 25**
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Given that the mean of the scores 15, 21, 17, 26, 18 and 29 is 21, calculate the standard deviation of the scores

**Answer Details**

To calculate the standard deviation of the scores, we first need to find the variance, which is the average of the squared deviations from the mean. 1. Find the mean of the scores: Mean = (15 + 21 + 17 + 26 + 18 + 29)/6 = 126/6 = 21 2. Calculate the deviations from the mean for each score: 15 - 21 = -6 21 - 21 = 0 17 - 21 = -4 26 - 21 = 5 18 - 21 = -3 29 - 21 = 8 3. Square each deviation: (-6)^2 = 36 0^2 = 0 (-4)^2 = 16 5^2 = 25 (-3)^2 = 9 8^2 = 64 4. Find the average of the squared deviations: (36 + 0 + 16 + 25 + 9 + 64)/6 = 150/6 = 25 5. Take the square root of the variance to get the standard deviation: Standard deviation = sqrt(25) = 5 Therefore, the standard deviation of the scores is 5. The answer is 5.

**Question 26**
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In the diagram, MQ//RS, < TUV = 70^{o} and < RLV = 30^{o}. Find the value of x

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**Question 27**
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The n^{th} term of a sequence is T_{n} = 5 + (n - 1)^{2}. Evaluate T_{4} - T_{6}

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To evaluate T_{4} - T_{6}, we first need to find the values of T_{4} and T_{6}. T_{4} = 5 + (4 - 1)^{2} = 5 + 9 = 14 T_{6} = 5 + (6 - 1)^{2} = 5 + 25 = 30 Therefore, T_{4} - T_{6} = 14 - 30 = -16 Hence, the answer is -16.

**Question 28**
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Solve (\(\frac{27}{125}\))^{-\(\frac{1}{3}\)} x (\(\frac{4}{9}\))^{\(\frac{1}{2}\)}

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**Question 29**
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Simplify; \(\frac{3\sqrt{5} \times 4\sqrt{6}}{2 \sqrt{2} \times 3\sqrt{2}}\)

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**Question 30**
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Find the coefficient of m in the expression of (\(\frac{m}{2} - 1 \frac{1}{2}\)) (m + \(\frac{2}{3}\))

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**Question 31**
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The curved surface area of a cylindrical tin is 704cm^{2}. If the radius of its base is 8cm, find the height. [Take \(\pi = \frac{22}{7}\)]`

**Question 32**
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**Answer Details**

The equation that describes an inverse relationship between x and y is y = k/x, which is the second option. An inverse relationship between two variables means that as one variable increases, the other variable decreases in a proportional manner. In other words, if we double the value of x, the value of y would be halved, and vice versa. In the equation y = k/x, as x increases, y decreases, and as x decreases, y increases. The constant k represents the scale of the relationship between x and y. The other options do not describe an inverse relationship between x and y. is a direct proportion where y increases as x increases. is a power relationship where y increases as the square root of x increases. is a linear relationship where y increases at a constant rate as x increases.

**Question 33**
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Express 302.10495 correct to five significant figures

**Question 34**
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The lengths of the minor and major arcs 54cm and 126cm respectively. Calculate the angle of the major sector

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**Question 35**
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Solve for x in the equation; \(\frac{1}{x} + \frac{2}{3x} = \frac{1}{3}\)

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To solve the equation \(\frac{1}{x} + \frac{2}{3x} = \frac{1}{3}\), we can use the following steps: Step 1: Find the common denominator. The denominators in this equation are x and 3x. The common denominator is 3x. Step 2: Multiply both sides of the equation by the common denominator. Multiplying both sides by 3x gives: 3 + 2 = x Step 3: Simplify and solve for x. Simplifying the left side gives: 5 = x Therefore, the solution to the equation is x = 5. x = 5, is the correct answer. Options 2, 3, and 4 are incorrect because they do not equal 5.

**Question 36**
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In the diagram, MN//PO, < PMN = 112^{o}, < PNO = 129^{oo} and < MPN = y^{o}. Find the value of y

**Question 37**
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In 1995, the enrollment of two schools X and Y were 1,050 and 1,190 respectively. Find the ration of the enrollments of X and Y

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To find the ratio of enrollments of schools X and Y, we need to divide the enrollment of school X by the enrollment of school Y. Ratio of enrollments of X and Y = Enrollment of X / Enrollment of Y Enrollment of X = 1,050 Enrollment of Y = 1,190 Ratio of enrollments of X and Y = 1,050 / 1,190 Simplifying this fraction by dividing both numerator and denominator by 10 gives: Ratio of enrollments of X and Y = 105 / 119 Further simplifying by dividing both numerator and denominator by 7 gives: Ratio of enrollments of X and Y = 15 / 17 Therefore, the ratio of the enrollments of X and Y is 15:17. Answer option (B) is correct.

**Question 38**
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The graph is that of y = 2x^{2} - 5x - 3. For what value of x will y be negative? What is the gradient of y = 2x^{2} - 5x - 3 at the point x = 4?

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**Question 39**
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If p = {prime factors of 210} and Q = {prime less than 10}, find p \(\cap\) Q

**Answer Details**

The prime factors of 210 are 2, 3, 5, and 7. The prime numbers less than 10 are 2, 3, 5, and 7. The intersection (p ∩ Q) is the set of elements that are in both p and Q. Therefore, p ∩ Q is {2, 3, 5, 7}. So, the correct option is: - {2,3,5,7}

**Question 40**
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The diagram is a polygon. Find the largest of its interior angles