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**Question 2**
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If the probability that an event will occur is p and the probability that it will not occur is q, which of the following is true?

**Answer Details**

The sum of the probabilities of an event occurring and not occurring is always equal to 1. So, if p is the probability that an event will occur, then the probability that it will not occur is 1-p (since the total probability is 1). Therefore, we have: p + (1-p) = 1 which simplifies to: p - p + 1 = 1-p + p 1 = 1 So, the correct option is: p + q = 1.

**Question 3**
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A chord of length 30cm is 8cm away from the center of the circle. What is the radius of the circle. What is the perimeter of the sector? Take \(\pi =\frac{22}{7}\)

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**Question 4**
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The bearing S40°E is the same as

**Answer Details**

The bearing S40°E is equivalent to an angle that is 40 degrees east of the southern direction. The direction opposite to south is north, so if we draw a line going north and another line going 40 degrees to the east of south, the angle formed between these two lines is the direction that corresponds to the bearing S40°E. This angle is equal to 180° - 40° = 140°. Therefore, the correct answer is 140^{o}.

**Question 5**
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If 2x : (x +1) = 3:2, what is the value of x?

**Answer Details**

To solve the equation 2x : (x + 1) = 3 : 2, we need to cross-multiply the ratios. That means we multiply the numerator of the left-hand side ratio by the denominator of the right-hand side ratio, and we multiply the denominator of the left-hand side ratio by the numerator of the right-hand side ratio. So we have: 2x * 2 = 3 * (x + 1) Simplifying the right-hand side: 4x = 3x + 3 Subtracting 3x from both sides: x = 3 Therefore, the value of x is 3. To check, we can substitute x = 3 back into the original equation and see if both sides are equal: 2(3) : (3 + 1) = 6 : 4 = 3 : 2 So the equation is true when x = 3, and that is our final answer.

**Question 6**
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Evaluate \(log_{10}5 + log_{10}20\)

**Answer Details**

To evaluate \(log_{10}5 + log_{10}20\), we can use the logarithmic rule that states: $log_{a}x + log_{a}y = log_{a}(xy)$ Applying this rule, we get: $log_{10}5 + log_{10}20 = log_{10}(5 \times 20)$ Simplifying the right-hand side: $log_{10}(5 \times 20) = log_{10}100$ Finally, we know that \(log_{10}100 = 2\), so: \(log_{10}5 + log_{10}20 = log_{10}100 = 2\) Therefore, the answer is 2.

**Question 7**
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In the diagram, MN||PQ, |LM| = 3cm and |LP| = 4cm. If the area of ?LMN is 18cm^{2}, find the area of the quadrilateral MPQN

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**Question 8**
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Express 0.0462 in standard form

**Question 9**
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Evaluate \((20_{three})^2 - (11_{three})^2\) in base three

**Question 10**
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The variance of a given distribution is 25. What is the standard deviation?

**Answer Details**

The standard deviation is the square root of the variance. Therefore, to find the standard deviation when given the variance, we simply take the square root of the variance. In this case, the variance is given as 25. Taking the square root of 25 gives us: sqrt(25) = 5 So the standard deviation of the distribution is 5. To summarize, the formula for finding the standard deviation from the variance is to take the square root of the variance. In this case, the variance is 25, so the standard deviation is 5.

**Question 11**
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Evaluate \(5\frac{2}{5}\times \left(\frac{2}{3}\right)^2\div\left(1\frac{1}{2}\right)^{-1}\)

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**Question 12**
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If \(y \propto \frac{1}{\sqrt{x}}\) and x = 16 when y = 2, find x when y = 24

**Question 13**
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A solid cylinder of radius 7cm is 10 cm long. Find its total surface area.

**Answer Details**

A solid cylinder has two circular bases and a curved lateral surface. To find its total surface area, we need to find the area of both circular bases and the area of the lateral surface, and then add them together. The area of one circular base is \(\pi r^2\), where r is the radius of the cylinder. Since the radius is given as 7 cm, the area of one circular base is: \begin{align*} \text{Area of one circular base} &= \pi \times (7 \text{ cm})^2 \\ &= 49 \pi \text{ cm}^2 \\ \end{align*} Since there are two circular bases, the total area of both circular bases is: \begin{align*} \text{Total area of both circular bases} &= 2 \times \text{Area of one circular base} \\ &= 2 \times 49 \pi \text{ cm}^2 \\ &= 98 \pi \text{ cm}^2 \\ \end{align*} The area of the lateral surface is the curved surface area of the cylinder, which is given by \(2\pi rh\), where r is the radius of the cylinder and h is the height (or length) of the cylinder. Since the radius is given as 7 cm and the length is given as 10 cm, the area of the lateral surface is: \begin{align*} \text{Area of lateral surface} &= 2\pi rh \\ &= 2\pi \times (7 \text{ cm}) \times (10 \text{ cm}) \\ &= 140 \pi \text{ cm}^2 \\ \end{align*} Therefore, the total surface area of the cylinder is the sum of the area of both circular bases and the area of the lateral surface: \begin{align*} \text{Total surface area} &= \text{Area of both circular bases} + \text{Area of lateral surface} \\ &= 98 \pi \text{ cm}^2 + 140 \pi \text{ cm}^2 \\ &= 238 \pi \text{ cm}^2 \\ \end{align*} Therefore, the total surface area of the cylinder is 238\(\pi\) cm\(^2\).

**Question 14**
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Factorize \(6x^2 + 7x - 20\)

**Answer Details**

To factorize \(6x^2 + 7x - 20\), we need to find two numbers that multiply to -120 (the product of the leading coefficient, 6, and the constant, -20) and add up to the coefficient of the middle term, 7. These numbers are 15 and -8. Therefore, we can write: \begin{align*} 6x^2 + 7x - 20 &= 6x^2 + 15x - 8x - 20 \\ &= 3x(2x+5) - 4(2x+5) \\ &= (3x-4)(2x+5) \end{align*} So the factorization of \(6x^2 + 7x - 20\) is \((3x-4)(2x+5)\).

**Question 15**
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Simplify \(\frac{2x-1}{3}-\frac{x+3}{2}\)

**Answer Details**

To subtract two fractions, they need to have a common denominator. The common denominator of 3 and 2 is 6, so we can write: \begin{align*} \frac{2x-1}{3}-\frac{x+3}{2} &= \frac{2(2x-1)}{2 \cdot 3}-\frac{3(x+3)}{3 \cdot 2} \\ &= \frac{4x-2}{6}-\frac{3x+9}{6} \\ &= \frac{4x-3x-2-9}{6} \\ &= \frac{x-11}{6} \end{align*} Therefore, the answer is \(\frac{x-11}{6}\).

**Question 16**
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In a bag of oranges, the ratio of the good ones to the bad ones is 5:4. If the number of bad oranges in the bag is 36, how many oranges are there in the altogether?

**Answer Details**

The ratio of good to bad oranges is 5:4, which means that for every 5 good oranges, there are 4 bad ones. If we know there are 36 bad oranges, we can use this ratio to find the number of good oranges as follows: 5/4 = x/36 where x is the number of good oranges. We can solve for x by cross-multiplying: 4x = 5*36 x = 45 So there are 45 good oranges in the bag. To find the total number of oranges, we add the number of good and bad ones: 45 + 36 = 81 Therefore, there are 81 oranges in the bag altogether.

**Question 17**
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Given that P = {b, d, e, f} and Q = {a, c, f, g} are subsets of the universal set U = {a,b, c, d, e, f, g}. Find P' ∩ Q

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**Question 18**
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Given that sin \(P = \frac{5}{13}\), where p is acute, find the value of cos p - tan p

**Answer Details**

We know that \(\sin p = \dfrac{5}{13}\) and $p$ is acute. Using Pythagorean identity, we can find $\cos p$: \begin{align*} \cos^2 p &= 1 - \sin^2 p \\ \cos^2 p &= 1 - \left(\dfrac{5}{13}\right)^2 \\ \cos^2 p &= \dfrac{144}{169} \\ \cos p &= \dfrac{12}{13} \end{align*} Using the definition of tangent, we can find $\tan p$: \begin{align*} \tan p &= \dfrac{\sin p}{\cos p} \\ \tan p &= \dfrac{\frac{5}{13}}{\frac{12}{13}} \\ \tan p &= \dfrac{5}{12} \end{align*} Therefore, \begin{align*} \cos p - \tan p &= \dfrac{12}{13} - \dfrac{5}{12} \\ &= \dfrac{144}{156} - \dfrac{65}{156} \\ &= \dfrac{79}{156} \end{align*} Hence, the answer is \(\frac{79}{156}\).

**Question 19**
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Which of the following correctly expresses 48 as a product of prime factors?

**Answer Details**

To express 48 as a product of prime factors, we need to keep dividing it by prime numbers until we get to 1. Let's start with 2, which is the smallest prime number that can divide 48: 48 ÷ 2 = 24 We can continue dividing by 2 until we can't divide by 2 anymore: 24 ÷ 2 = 12 12 ÷ 2 = 6 6 ÷ 2 = 3 Now, we can see that we have reached a prime number, 3, so we stop dividing. Therefore: 48 = 2 x 2 x 2 x 2 x 3 So the correct option is 2 x 2 x 2 x 2 x 3 or.

**Question 20**
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A boy estimated his transport fare for a journey as N190 instead of N200. Find the percentage error in his estimate

**Answer Details**

The error in the boy's estimate is the difference between what he estimated and the actual cost of the journey, which is N200 - N190 = N10. To find the percentage error, we divide the error by the actual cost and then multiply by 100%. Percentage error = (error / actual cost) x 100% = (10 / 200) x 100% = 5% Therefore, the percentage error in the boy's estimate is 5%. Option (d) is the correct answer.

**Question 23**
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The nth term of a sequence is \(2^{2n-1}\). Which term of the sequence is \(2^9?\)

**Answer Details**

To find which term of the sequence is \(2^9\), we need to solve for n in the equation \(2^{2n-1}=2^9\). First, we can simplify the equation by dividing both sides by \(2^9\), giving us \(2^{2n-1-9}=2^{2n-10}=1\). Next, we can solve for n by taking the logarithm of both sides of the equation. Since any logarithm base can be used, we can use the natural logarithm, denoted as ln: \begin{align*} 2n-10 &= \ln 1 \\ 2n-10 &= 0 \\ 2n &= 10 \\ n &= 5 \end{align*} Therefore, the fifth term of the sequence is \(2^9\).

**Question 24**
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In the diagram, PQR is a triangle.|PQ| = |PR|, |PS| = |SQ| and PRS = 50°. What is the size of ∠PSQ?

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**Question 25**
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The diagram shows a triangular prism of length 7cm. The right - angled triangle PQR is a cross section of the prism |PR| = 5cm and |RQ| = 3cm. What is the area of the cross-section?

**Answer Details**

The area of a triangle is given by the formula: $$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $$ In triangle PQR, the base is |RQ| = 3cm and the height is |PS|, where S is the foot of the perpendicular from point P to line QR. Since the cross section is a right-angled triangle, we can use Pythagoras theorem to find |PS| as follows: $$ \begin{align*} |PS|^2 &= |PR|^2 - |RS|^2 \\ &= 5^2 - 3^2 \\ &= 16 \\ \end{align*} $$ Therefore, |PS| = 4cm. Substituting the values for base and height into the area formula, we get: $$ \text{Area} = \frac{1}{2} \times 3\text{cm} \times 4\text{cm} = 6\text{cm}^2 $$ Hence, the area of the cross-section is 6 cm^{2}.

**Question 26**
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A box contains 5 red, 3 green and 4 blue balls. A boy is allowed to take away two balls from the box.* *What is the probability that one is green and the other is blue?

**Question 27**
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If \(\frac{y-3}{2}<\frac{2y-1}{3}\), which of the following is true?

**Answer Details**

To solve this inequality, we can start by simplifying both sides. First, we can multiply both sides by 6 to eliminate the denominators: 3(y - 3) < 4(2y - 1) Expanding the right side gives: 3(y - 3) < 8y - 4 Simplifying and collecting like terms, 3y - 9 < 8y - 4 Subtracting 3y from both sides, -9 < 5y - 4 Adding 4 to both sides, -5 < 5y Dividing both sides by 5, -1 < y So the inequality is true for all values of y greater than -1. Therefore, the correct answer is: y > -7.

**Question 28**
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Evaluate \(\frac{x^2 + x - 2}{2x^2 + x -3}\) when x = -1

**Answer Details**

Substituting x = -1, we get: \[\frac{(-1)^2 + (-1) - 2}{2(-1)^2 + (-1) - 3}\] \[=\frac{1 - 1 - 2}{2 - 1 - 3}\] \[=\frac{-2}{-2}\] \[=1\] Therefore, the value of the expression is 1 when x = -1. Hence the correct option is (d) 1.

**Question 29**
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A box contains 5 red, 3 green and 4 blue balls. A boy is allowed to take away two balls from the box. What is the probability that the two balls are red?

**Answer Details**

There are 12 balls in total in the box, out of which 5 are red. When the boy takes out the first ball, there are 11 balls left, out of which 4 are red. Therefore, the probability of the first ball being red is 5/12, and the probability of the second ball being red is 4/11. To find the probability of both events happening together, we multiply the probabilities: \(\frac{5}{12}\times\frac{4}{11}=\frac{5}{33}\) Therefore, the probability that the two balls taken out are both red is 5/33. Answer: \(\frac{5}{33}\)

**Question 30**
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Express 25° 45' in decimal (Hint: 1° = 60')

**Answer Details**

To convert 25° 45' to decimal, we first note that there are 60 minutes in a degree. So, to convert minutes to degrees, we divide by 60. Thus, 25° 45' = 25 + 45/60 degrees Simplifying, 25° 45' = 25 + 0.75 degrees Combining the whole and fractional parts, we get: 25° 45' = 25.75 degrees Therefore, the answer is: 25.75°.

**Question 31**
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A man is four times as old as his son. The difference between their ages is 36 years Find the sum of their ages

**Answer Details**

Let the age of the son be x. Then, the age of the man will be 4x (as he is four times as old as his son). The difference in their ages is 36 years, so we have: 4x - x = 36 Simplifying this, we get: 3x = 36 x = 12 So, the age of the son is 12 and the age of the man is 4 times that, which is 48. Therefore, the sum of their ages is: 12 + 48 = 60 years. Hence, the correct option is (c) 60 years.

**Question 32**
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If \(2x^2 + kx - 14 = (x+2)(2x-7)\), find the value of K

**Answer Details**

To find the value of k, we need to expand the right-hand side of the equation, which is equal to the left-hand side: \begin{align*} (x+2)(2x-7) &= 2x^2 - 3x - 14 \\ \end{align*} Now we can compare the coefficients of the terms on both sides of the equation: \begin{align*} \text{Coefficient of } x^2: \quad &2 = 2 \\ \text{Coefficient of } x: \quad &-3 = k \\ \text{Coefficient of the constant term:} \quad &-14 = -14 \\ \end{align*} Therefore, we have found that the value of k is -3.

**Question 33**
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Subtract (-y + 3x + 5z) from (4y - x - 2z)

**Answer Details**

To subtract (-y + 3x + 5z) from (4y - x - 2z), we can just subtract the corresponding coefficients of each variable. So, - the coefficient of y is 4 - (-1) = 5 - the coefficient of x is -1 - 3 = -4 - the coefficient of z is -2 - 5 = -7 Therefore, the result is: 5y - 4x - 7z. So the correct option is (a) 5y - 4x - 7z.

**Question 34**
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PQRS is a trapezium in which |PS| = 9cm, |QR| = 15cm, |PQ| = \(2\sqrt{3}, \angle PQR = 90^o and \angle QRS = 30^o\). Calculate the area of the trapezium

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**Question 35**
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