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**Question 1**
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In the diagram, /PQ/ = /QR/ and /PR/ = /RS/ = /SP/, calculate the side of < QRS

**Question 2**
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In the figure /PX/ = /XQ/, PQ//YZ and XV//QR. What is the ratio of the area of XYZQ to te area of \(\bigtriangleup\)YZR?

**Question 3**
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P naira invested for 4 years invested for 4 years at r% simple interest per annum yields 0.36 p naira interest. Find the value of r

**Answer Details**

We can use the formula for simple interest, which is: I = P * r * t Where I is the interest earned, P is the principal (initial amount invested), r is the annual interest rate, and t is the time in years. In this problem, we know that the interest earned is 0.36P, the principal is P, and the time is 4 years. Substituting these values into the formula, we get: 0.36P = P * r * 4 Simplifying the equation, we can divide both sides by P * 4: 0.36 = r * 4 r = 0.36 / 4 r = 0.09 or 9% Therefore, the answer is option (C) 9.

**Question 4**
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In the diagram, O is the centre of the circle and PQRS is a cyclic quadrilateral. Find the value of x.

**Question 5**
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Given the sets A = [2, 4, 6, 8] and B = [2, 3, 5, 9]. If a number is selected at random from set B, what is the probability that the number is prime?

**Question 6**
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If p = \(\frac{3}{5} \sqrt{\frac{q}{r}}\), express q in terms of p and r

**Answer Details**

Given that p = \(\frac{3}{5} \sqrt{\frac{q}{r}}\), we need to express q in terms of p and r. Let's first isolate the term containing q on one side of the equation: \begin{align*} p &= \frac{3}{5} \sqrt{\frac{q}{r}} \\ \frac{5}{3}p &= \sqrt{\frac{q}{r}} \\ \left(\frac{5}{3}p\right)^2 &= \frac{q}{r} \\ \frac{25}{9} p^2r &= q \\ \end{align*} Therefore, q is equal to \(\frac{25}{9} p^2r\). So the correct option is: \(\frac{25}{9} p^2r\)

**Question 7**
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If p = \(\frac{1}{2}\) and \(\frac{1}{p - 1} = \frac{2}{p + x}\), find the value of x

**Answer Details**

Given that p = \(\frac{1}{2}\) and \(\frac{1}{p - 1} = \frac{2}{p + x}\), we want to find the value of x. We can start by substituting the value of p in the second equation: \begin{aligned} \frac{1}{p - 1} &= \frac{2}{p + x} \\ \\ \frac{1}{\frac{1}{2} - 1} &= \frac{2}{\frac{1}{2} + x} && \text{(Substitute } p = \frac{1}{2} \text{)}\\ \\ \frac{1}{-\frac{1}{2}} &= \frac{2}{\frac{1}{2} + x} && \text{(Simplify)}\\ \\ -2 &= \frac{2}{\frac{1}{2} + x} && \text{(Simplify)}\\ \\ -2\left(\frac{1}{2} + x\right) &= 2 && \text{(Cross-multiply)}\\ \\ -1 - 2x &= 1 && \text{(Simplify)}\\ \\ -2x &= 2 \\ \\ x &= -1 \end{aligned} Therefore, the value of x is -1.5, which corresponds to option B: -1\(\frac{1}{2}\).

**Question 8**
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If the perimeter of \(\bigtriangleup\)PQR in thr diagram is 24cm, what is the area of \(\bigtriangleup\)PRS?

**Question 10**
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Given that x = 2 and y = -\(\frac{1}{4}\), evaluate \(\frac{x^2y - 2xy}{5}\)

**Question 11**
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If cos (x + 25)^{o} = sin 45^{o}, find the value of x

**Answer Details**

We know that cos (90^{o} - A) = sin A. So, cos (x + 25^{o}) = cos (90^{o} - 45^{o}) = cos 45^{o}. Since cos A = cos B implies A = ±B + 360^{o}n, we have: x + 25^{o} = ±45^{o} + 360^{o}n x = -25^{o} ± 45^{o} + 360^{o}n x = 20^{o} + 360^{o}n or x = 70^{o} + 360^{o}n Since we are given the range of x, we take x = 20^{o}. Therefore, the value of x is 20^{o}.

**Question 12**
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If tan x = 1, evaluate sin x + cos x, leaving your answer in the surd form

**Answer Details**

If \(\tan x = 1\), then we can find the values of sin x and cos x using the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\). Since \(\tan x = \frac{\sin x}{\cos x} = 1\), we can say that \(\sin x = \cos x\). So we have, \[\sin^2 x + \cos^2 x = 2\cos^2 x = 1 \implies \cos x = \pm\frac{1}{\sqrt{2}}\] Using \(\sin x = \cos x\), we can say that \(\sin x = \pm\frac{1}{\sqrt{2}}\) Now, \[\sin x + \cos x = \pm\frac{1}{\sqrt{2}} \pm \frac{1}{\sqrt{2}} = \pm\sqrt{2}\] Since \(\tan x = 1\), x must be in the first quadrant, where both \(\sin x\) and \(\cos x\) are positive. Therefore, we have \[\sin x + \cos x = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}\] Hence, the answer is \(\sqrt{2}\). Therefore, the option that represents this answer is: $\sqrt{2}$.

**Question 13**
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Simplify \(\frac{\frac{1}{x} + \frac{1}{y}}{x + y}\)

**Answer Details**

To simplify the expression, we can start by getting a common denominator for the two fractions in the numerator. The common denominator is xy. So, we can write: \begin{align*} \frac{\frac{1}{x} + \frac{1}{y}}{x + y} &= \frac{\frac{y}{xy} + \frac{x}{xy}}{x + y} \\ &= \frac{\frac{x + y}{xy}}{x + y} \\ &= \frac{1}{xy} \end{align*} Therefore, the simplified expression is \(\frac{1}{xy}\).

**Question 15**
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Simplify \(\sqrt{50} + \frac{10}{\sqrt{2}}\)

**Question 16**
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In the figure shown, PQs is a straight line. What is the value of < PRQ?

**Question 17**
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What is the length of an edge of a cube whose total surface area is X cm^{2} and whose total surface area is \(\frac{X}{2}\)cm^{3}?

**Question 19**
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XY is a chord of circle centre O and radius 7cm. The chord XY which is 8cm long subtends an angle of 120^{o} at the centre of the circle. Calculate the perimeter of the minor segment. [Take \(\pi = \frac{22}{7}\)]

**Question 20**
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In the diagram, < QPR = 90^{o}. If q^{2} = 25 - r^{2}. Find the value of p

**Question 22**
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Solve the inequality 1 - 2x < - \(\frac{1}{3}\)

**Answer Details**

The given inequality is: 1 - 2x < -\(\frac{1}{3}\) To solve this inequality, we need to isolate the variable, which in this case is x. First, we can simplify the inequality by adding \(\frac{1}{3}\) to both sides: 1 - 2x + \(\frac{1}{3}\) < 0 Next, we can combine like terms: \(\frac{4}{3}\) - 2x < 0 Now, we can isolate x by subtracting \(\frac{4}{3}\) from both sides: -2x < -\(\frac{4}{3}\) Finally, we can solve for x by dividing both sides by -2, remembering to flip the inequality because we are dividing by a negative number: x > \(\frac{2}{3}\) Therefore, the solution to the inequality is x > \(\frac{2}{3}\).

**Question 23**
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If x \(\alpha\) (45 + \(\frac{1}{2}y\)), which of the following is true>?

**Question 24**
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