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**Question 1**
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To arrive on schedule, a train is to cover a distance of 60km at 72km/hr. If it starts 10 minutes late, at what speed must it move to arrive on schedule?

**Answer Details**

To arrive on schedule, the train needs to cover a distance of 60 km at a speed of 72 km/hr. However, the train starts 10 minutes late, which means it has lost 10 minutes of travel time. We need to find the speed the train needs to travel to make up for the lost time and arrive on schedule. We can use the formula: distance = speed x time Let's first calculate the time taken to cover the distance of 60 km at the speed of 72 km/hr. time = distance / speed time = 60 km / 72 km/hr time = 0.8333 hours or 50 minutes (rounded to the nearest minute) Since the train is starting 10 minutes late, it has only 40 minutes to cover the distance of 60 km to arrive on schedule. We can use the same formula to find the speed the train needs to travel to cover the distance in 40 minutes. speed = distance / time speed = 60 km / (40/60) hours speed = 90 km/hr Therefore, the correct answer is 90 km/hr.

**Question 2**
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If (-3, -4) is a point on the line y = mx + 2 find the value of m.

**Answer Details**

The equation of a line in slope-intercept form is given by y = mx + b, where m is the slope of the line and b is the y-intercept. In this question, we are given that the point (-3, -4) lies on the line y = mx + 2. This means that when x = -3, y = -4. We can substitute these values into the equation to get: -4 = m(-3) + 2 Simplifying this expression, we get: -4 = -3m + 2 Subtracting 2 from both sides, we get: -6 = -3m Dividing both sides by -3, we get: m = 2 Therefore, the value of m is 2.

**Question 4**
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The number of goals scored by a school team in 10 netball matches are as follows: 3, 5, 7, 7, 8, 8, 8, 11, 11, 12. Find the probability that in a match, the school team will score at most 8 goals.

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**Question 5**
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Which of the following is not a rational number?

**Answer Details**

A rational number is any number that can be expressed as a ratio of two integers (where the denominator is not zero). - -5 can be expressed as -5/1, which is a ratio of two integers, so it is rational. - \(\sqrt{4}\) is equal to 2, which can be expressed as the ratio 2/1, so it is rational. - \(3\frac{3}{4}\) is equal to 15/4, which is a ratio of two integers, so it is rational. - \(\sqrt{90}\) cannot be expressed as a ratio of two integers. It is irrational. Therefore, the number that is not rational is \(\sqrt{90}\).

**Question 7**
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The lengths of the parallel sides of a trapezium are 9 cm and 12 cm. lf the area of the trapezium is 105 cm^{2}, find the perpendicular distance between the parallel sides.

**Answer Details**

The formula for the area of a trapezium is given as: Area = 1/2 × (sum of the parallel sides) × (perpendicular distance between them) In this case, we have the lengths of the parallel sides as 9 cm and 12 cm, and the area as 105 cm^{2}. Substituting these values in the formula, we get: 105 = 1/2 × (9 + 12) × (perpendicular distance) 105 = 1/2 × 21 × (perpendicular distance) 105 = 10.5 × (perpendicular distance) Dividing both sides by 10.5, we get: Perpendicular distance = 105/10.5 Perpendicular distance = 10 Therefore, the perpendicular distance between the parallel sides is 10 cm. Hence, the correct option is (c) 10cm.

**Question 9**
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In the diagram, |QR| = 10cm, PR⊥QS, angle PSR = 30° and angle PQR = 45°. Calculate in meters |QS|

**Question 11**
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The height of a right circular cone is 4cm. The radius of its base is 3cm. Find the curved surface area

**Answer Details**

The curved surface area of a cone is given by the formula: \[\text{Curved surface area} = \pi rl,\] where $r$ is the radius of the base, $l$ is the slant height, and $\pi$ is the constant pi (approximately 3.14). In this case, the height of the cone is 4 cm and the radius of the base is 3 cm. To find the slant height, we can use the Pythagorean theorem: \[l^2 = r^2 + h^2.\] Plugging in the values we get, \[l^2 = 3^2 + 4^2 = 9 + 16 = 25.\] So, $l = 5$ cm. Now, we can use the formula for curved surface area, substituting $r=3$ cm and $l=5$ cm: \[\text{Curved surface area} = \pi \cdot 3 \cdot 5 = 15\pi \text{ cm}^2.\] Therefore, the answer is (B) $15\pi cm^2$.

**Question 12**
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In the diagram O is the center of the circle, ∠SOR = 64° and ∠PSO = 36°. Calculate ∠PQR

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**Question 13**
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The diagram is the graph of \(y = 6 + x - x^2\). The graph intercepts the x- axis at P and R and the y- axis at Q.

What is the value of y at Q?

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**Question 14**
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If q oranges are sold for t Naira, how many oranges can be bought for p naira?

**Answer Details**

The correct answer is \(\frac{qp}{t}\). Since q oranges are sold for t Naira, we can say that the cost of one orange is \(\frac{t}{q}\) Naira. To find how many oranges can be bought for p Naira, we need to divide p by the cost of one orange: Number of oranges = \(\frac{p}{\frac{t}{q}}\) Simplifying, we get: Number of oranges = \(\frac{pq}{t}\)

**Question 15**
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In the diagram O is the centre of the circle. Which of the following is/are not true? I. a = b II. b + c = 180^{o} III. a + b = c

**Question 16**
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The diagram is the graph of \(y = 6 + x - x^2\). The graph intercepts the x- axis at P and R and the y- axis at Q.

When \(y = 3\frac{1}{3}\), what is the positive value of x?

**Answer Details**

The given function is \(y = 6 + x - x^2\). We are required to find the value of x when y is equal to \(3\frac{1}{3}\), which is also equal to \(\frac{10}{3}\). Substituting \(y = \frac{10}{3}\) in the given equation, we get \[\frac{10}{3} = 6 + x - x^2\] Rearranging the terms, we get \[x^2 - x + \frac{8}{3} = 0\] Solving the quadratic equation, we get two solutions for x, which are \(\frac{4}{3}\) and \(1\frac{1}{2}\). However, we are asked to find the positive value of x, which is \(\boxed{2\frac{1}{5}}\).

**Question 17**
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In the diagram, SQ is a tangent to the circle at P, XP||YQ, ∠XPY = 56 o and ∠PXY = 80 o.Find angle PQY

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**Question 18**
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Simplify \(3\sqrt{12} + 10\sqrt{3} - \frac{6}{\sqrt{3}}\)

**Answer Details**

We can start by simplifying each term separately before adding them up: \(3\sqrt{12} = 3\sqrt{4\cdot3} = 3\cdot2\sqrt{3} = 6\sqrt{3}\) \(10\sqrt{3}\) is already in its simplest form. \(\frac{6}{\sqrt{3}} = \frac{6}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}\) Now we can substitute these values back into the original expression: \(3\sqrt{12} + 10\sqrt{3} - \frac{6}{\sqrt{3}} = 6\sqrt{3} + 10\sqrt{3} - 2\sqrt{3} = 14\sqrt{3}\) Therefore, the simplified form of the expression is \(14\sqrt{3}\), which is.

**Question 19**
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The ratio of the number of men to the number of women in a 20 member committee is 3:1. How many women must be added to the 20-member committee so as to make the ratio of men to women 3:2?

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**Question 20**
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A number is selected at random from the set Y = {18, 19, 20, . . . 28, 29}. Find the probability that the number is prime.

**Answer Details**

To find the probability of selecting a prime number from the set Y = {18, 19, 20, ..., 28, 29}, we need to first identify the prime numbers in this set. The prime numbers in this set are 19, 23, and 29. The total number of elements in the set Y is 12. Therefore, the probability of selecting a prime number from the set Y is 3/12 or 1/4. Therefore, the correct option is (a) \(\frac{1}{4}\).

**Question 21**
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In the diagram, \(\bar{PS}\hspace{1mm} and \hspace{1mm}\bar{QT}\) are two altitudes of ?PQR. Which of the following is equal to ∠RQT?

**Answer Details**

In a triangle, an altitude is a line segment drawn from a vertex perpendicular to the opposite side. In the given diagram, \(\bar{PS}\) and \(\bar{QT}\) are two altitudes of triangle PQR. By definition, the altitude \(\bar{QT}\) is perpendicular to the side \(\bar{PR}\). Therefore, ∠QTR is a right angle. Also, we know that the sum of the angles in a triangle is 180 degrees. So, we can find the measure of ∠RQT as follows: ∠RQT = 180° - ∠QTR - ∠QRT We don't know the value of either ∠QTR or ∠QRT, but we can find one of them using the fact that \(\bar{PS}\) is an altitude. Since \(\bar{PS}\) is perpendicular to \(\bar{QR}\), ∠QPS is a right angle. Therefore, ∠QTR = ∠QTP + ∠RTP Since ∠QTP and ∠RTP are both complementary to ∠QPR, which is a known angle in the diagram, we can find their values: ∠QTP = ∠RTP = 90° - ∠QPR Now we can substitute these values into our equation for ∠RQT: ∠RQT = 180° - (90° - ∠QPR) - (90° - ∠QPR) Simplifying this expression gives: ∠RQT = 2∠QPR - 90° Therefore, the answer is option (B) ?SRP, since ∠QPR and ∠SRP are corresponding angles and thus equal to each other.

**Question 22**
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Points X, Y and Z are located in the same horizontal plane such that Y is 12 km north of X and Z is on a bearing of 270° from X. If |XZ| = 6km, calculate, correct to one decimal place, lYZl

**Question 23**
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Find the nth term of the sequence 4, 10, 16 ,...

**Answer Details**

The sequence 4, 10, 16,... is an arithmetic sequence because there is a common difference between every pair of consecutive terms. To find the common difference, we can subtract any term from the following term. For example: - The common difference between the first and second terms is 10 - 4 = 6. - The common difference between the second and third terms is 16 - 10 = 6. Since the common difference is 6, we can find the nth term of the sequence using the formula: nth term = first term + (n - 1) * common difference The first term is 4 and the common difference is 6, so we get: nth term = 4 + (n - 1) * 6 Simplifying this expression, we get: nth term = 6n - 2 Therefore, the correct answer is (A) 2(3n-1).

**Question 24**
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In the diagram, LMT is a straight line. lf O is the centre of circle LMN, OMN = 20°, LTN = 32° and |NM| = |MT|, find LNM.

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**Question 25**
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Which of the following is represented by the above sketch?

**Answer Details**

The given sketch is a quadratic function graph. From the graph, we can see that the parabola intersects the x-axis at two points which are (-3, 0) and (2, 0). Therefore, the roots or zeros of the quadratic function are -3 and 2. By comparing the options given, we can see that only option B, y = x^{2} - x - 6, has roots of -3 and 2. Thus, option B is the correct answer.

**Question 26**
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The length, in cm, of the sides of a right angled triangle are x, (x+2) and (x+1) where x > 0. Find , in cm, the length of its hypotenuse

**Answer Details**

In a right-angled triangle, the hypotenuse is the longest side, and it is opposite to the right angle. Using the Pythagorean theorem, we know that the sum of the squares of the two shorter sides is equal to the square of the hypotenuse. So, we have: \begin{align*} x^2 + (x+2)^2 &= (x+1)^2 \\ x^2 + x^2 + 4x + 4 &= x^2 + 2x + 1 \\ x^2 + 2x^2 + 4x + 4 &= x^2 + 2x + 1 \\ 3x^2 + 4x + 4 &= x^2 + 2x + 1 \\ 2x^2 + 2x + 3 &= 0 \end{align*} We can solve this quadratic equation using the quadratic formula: \begin{align*} x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ &= \frac{-2 \pm \sqrt{2^2 - 4(2)(3)}}{2(2)} \\ &= \frac{-2 \pm \sqrt{-8}}{4} \\ &= \frac{-1}{2} \pm \frac{\sqrt{2}}{2}i \end{align*} Since x must be greater than 0, the only valid solution is: \begin{align*} x &= \frac{-1}{2} + \frac{\sqrt{2}}{2}i \end{align*} However, we are asked to find the length of the hypotenuse, which is given by: \begin{align*} \sqrt{x^2 + (x+2)^2} &= \sqrt{\left(\frac{-1}{2} + \frac{\sqrt{2}}{2}i\right)^2 + \left(\frac{1}{2} + \frac{\sqrt{2}}{2}i\right)^2} \\ &= \sqrt{\frac{1}{4} - \frac{1}{2}\sqrt{2}i - \frac{1}{4} - \frac{1}{2}\sqrt{2}i + \frac{1}{4} + \frac{\sqrt{2}}{2}i + \frac{1}{4} - \frac{\sqrt{2}}{2}i} \\ &= \sqrt{\frac{1}{2}} \\ &= \frac{\sqrt{2}}{2} \approx 0.707 \end{align*} Therefore, the length of the hypotenuse is approximately 0.707 cm, which is closest to 5 cm.

**Question 27**
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