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**Question 1**
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The following numbers represent at a set of scores for a class of 32 students, where the maximum score possible was 12, 6, 5, 9, 4, 4, 8, 7, 5, 6, 3, 2, 5, 4, 6, 9, 10, 4, 3, 2, 3, 4, 6, 8, 7, 4, 2, 1, 8, 7, 7, 6, 11. What is the percentage of the class, correct to the nearest whole number, scored above 6?

**Answer Details**

To find the percentage of the class that scored above 6, we need to first count the number of students who scored above 6, and then divide that number by the total number of students and multiply by 100 to get the percentage. We can start by counting the number of students who scored above 6. From the given set of scores, we can see that the following students scored above 6: 9, 9, 8, 7, 8, 7, 7, 11. Counting these students, we get a total of 8 students who scored above 6. To find the percentage of the class that scored above 6, we divide the number of students who scored above 6 (8) by the total number of students (32), and then multiply by 100: 8/32 * 100 = 25% So, the percentage of the class that scored above 6 is 25%, which is closest to 34% when rounded to the nearest whole number. Therefore, the answer is (a) 34%.

**Question 2**
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The diagram above is a rectangle. If the perimeter is 36m, find the area of the rectangle

**Question 3**
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A chord of circle of radius 26cm is 10cm from the center of the circle. calculate the length of the chord

**Question 4**
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Find the value of X if \(cos x = \frac{5}{8}for 0^o\le X\le 180^o\)

**Answer Details**

To find the value of X, we can use the inverse cosine function (also known as the arccosine function) on both sides of the equation: \begin{align*} \cos x &= \frac{5}{8} \\ \Rightarrow \quad x &= \cos^{-1} \left( \frac{5}{8} \right) \end{align*} Using a calculator, we can find that: \begin{align*} x &\approx 51.32^\circ \end{align*} Therefore, the correct answer is 51.3^{o}.

**Question 5**
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The radius of a geographical globe is 60cm. Find the length of the parallel of latitude 60^{o}N

**Question 6**
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*The pie chart above show the distribution of how students travelled to a certain school on a particular day. Use this information to answer the question below*

What percentage, to the nearest whole number, of tghe students travelled to school on foot?

**Question 7**
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Find the root of the equation 2x\(^2\) - 3x - 2 = 0

**Answer Details**

To find the root(s) of the quadratic equation 2x\(^2\) - 3x - 2 = 0, we can use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ where a, b, and c are the coefficients of the quadratic equation ax\(^2\) + bx + c = 0. In this case, a = 2, b = -3, and c = -2. Substituting these values into the formula, we get: $$x = \frac{-(-3) \pm \sqrt{(-3)^2-4(2)(-2)}}{2(2)}$$ Simplifying: $$x = \frac{3 \pm \sqrt{9+16}}{4}$$ $$x = \frac{3 \pm \sqrt{25}}{4}$$ We can simplify the square root to get: $$x = \frac{3 \pm 5}{4}$$ So the roots are: $$x = \frac{3 + 5}{4} = 2$$ $$x = \frac{3 - 5}{4} = -\frac{1}{2}$$ Therefore, the answer is x = -1/2 or 2.

**Question 8**
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The diagonal and one side of a square are x and y units respectively. Find an expression for y in terms of x

**Question 9**
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What value of k makes the given expression a perfect square ? m\(^2\) - 8m + k = 0

**Question 10**
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If log\(_{10}\) a = 4; what is a?

**Answer Details**

The expression log\(_{10}\) a = 4 can be read as "logarithm of a to base 10 is 4". This means that 10 raised to the power of 4 is equal to a, or simply a = 10\(^4\). Therefore, the value of a is 10,000. The correct option is (e) 10,000.

**Question 11**
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In \(sin(X+30)^o=cos40^o\),find X

**Question 12**
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Calculate the perimeter of the trapezium PQRS

**Question 14**
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The side of a square is increased from 20cm to 21cm. Calculate the percentage increase in its area

**Answer Details**

The area of a square is given by the formula: A = s^2, where s is the length of a side of the square. Initially, the length of the side of the square is 20cm, so its area is A1 = 20^2 = 400 cm^2. When the length of the side is increased to 21cm, the new area becomes A2 = 21^2 = 441 cm^2. The difference in area between the two squares is: A2 - A1 = 441 - 400 = 41 cm^2 To find the percentage increase in area, we need to divide the difference in area by the original area, and then multiply by 100: percentage increase = (difference in area / original area) x 100% = (41 / 400) x 100% = 0.1025 x 100% = 10.25% Therefore, the percentage increase in area is 10.25%. So, the correct answer is 10.25%.

**Question 15**
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It is observed that \(1 + 3 = 2^2, 1 + 3 + 5 = 3^2, 1 + 3 + 5 + 7 = 4^2. \\If \hspace{1mm}1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = P^2 find\hspace{1mm}P\)

**Question 16**
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Cos x is negative and sin x is negative.Which of the following is true of x?

**Answer Details**

If both cosine and sine are negative, it means the angle x is in the third quadrant of the unit circle where both coordinates x and y are negative. Therefore, the possible range for x is from 180 degrees to 270 degrees. Hence, the correct option is: - 180^{o} < x < 270^{o}

**Question 17**
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In the diagram, O is the center of the circle and the reflex angle ROS is 264^{o}. Find ?RTS

**Question 18**
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In ?PQR. ?PQR is a right angle. |QR| = 2cm and ?PRQ = 60^{o}. Find |PR|

**Question 19**
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Arrange in ascending order of magnitude \(26_8, 36_7, and 25_9\)

**Question 21**
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*In the diagram O is the center of circle PRQ. The radius is 3.5cm and ?POQ = 50 ^{o}. Use the diagram to answer question below. (take ? = 3.142)*

Calculate, correct to three significant figures, the area of sector OPQ

**Answer Details**

The area of sector OPQ is a fraction of the area of the whole circle. The whole circle has an area of πr^{2}, where r is the radius. In this case, r = 3.5 cm, so the area of the whole circle is: A = π(3.5 cm)^{2} = 38.465 cm^{2} The sector OPQ makes up a fraction of the whole circle. The fraction is equal to the central angle of the sector divided by 360 degrees. In this case, the central angle is 50 degrees, so the fraction is: 50/360 = 5/36 So the area of sector OPQ is: A = (5/36)×38.465 cm^{2} ≈ 5.350 cm^{2} Therefore, the correct option is (b) 5.350cm^{2}.

**Question 22**
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*The pie chart above show the distribution of how students travelled to a certain school on a particular day. Use this information to answer the question below*

If a hundred students travelled by bus, find the total number of students in the school

**Question 23**
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Evaluate \(3.0\times 10^1 - 2.8\times 10^{-1}\)leaving the answer in standard form

**Answer Details**

To subtract two numbers in scientific notation, we first need to make sure they have the same power of 10. We can do this by moving the decimal point to the right or left as needed. Starting with \(3.0\times 10^1\) and \(2.8\times 10^{-1}\), we can move the decimal point one place to the left in the first number to get \(3.0\) and two places to the right in the second number to get \(0.028\). Now we have: $$ 3.0 - 0.028 = 2.972 $$ To express the answer in standard form, we need to convert it to the form \(a \times 10^b\), where \(1 \leq a < 10\) and \(b\) is an integer. We can do this by moving the decimal point to get a number between 1 and 10, and counting the number of places we moved it. In this case, we moved the decimal point one place to the left, so: $$ 2.972 = 2.972 \times 10^1 $$ Therefore, the answer is \(2.972 \times 10^1\).

**Question 24**
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Simplify \(\frac{8^{\frac{2}{3}}*27^{\frac{-1}{3}}}{64^{\frac{1}{3}}}\)

**Question 25**
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In the diagram above, |PQ| = |PR| = |RS| and ?RSP = 35^{o}. Find ?QPR

**Question 26**
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The nth term of a sequence is represented by 3 x 2^{(2-n)}. Write down the first three terms of the sequence

**Answer Details**

The nth term of the sequence is given as 3 x 2^{(2-n)}. To find the first three terms of the sequence, we need to substitute the values of n = 1, 2, 3. When n = 1, the nth term is: 3 x 2^{(2-1)} = 3 x 2^{1} = 3 x 2 = 6 When n = 2, the nth term is: 3 x 2^{(2-2)} = 3 x 2^{0} = 3 x 1 = 3 When n = 3, the nth term is: 3 x 2^{(2-3)} = 3 x 2^{-1} = 3 x 1/2 = 3/2 = ^{3}/_{2} Therefore, the first three terms of the sequence are 6, 3, ^{3}/_{2}. Hence, the answer is: 6, 3, ^{3}/_{2}.

**Question 27**
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If log\(_{10}\) q = 2.7078, what is q?

**Answer Details**

The logarithm of a number to a given base is the exponent to which the base must be raised to obtain the number. So, if log\(_{10}\) q = 2.7078, then 10\(^{2.7078}\) = q. Evaluating this expression, we get q ≈ 510.2. Therefore, the correct option is 510.2.

**Question 28**
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A trader makes a loss of 15% when selling an article. Find the ratio, selling price : cost price

**Answer Details**

Let's assume the cost price of the article is $100. Since the trader incurred a loss of 15%, the selling price will be 85% of the cost price. Therefore, Selling price = 85/100 * cost price = 85/100 * 100 = $85. The ratio of the selling price to the cost price is: Selling price : Cost price = $85 : $100 To simplify the ratio, we can divide both sides by the highest common factor, which is 5. This gives: Selling price : Cost price = 17 : 20 Therefore, the ratio of selling price to cost price is 17 : 20. Option C, 17:20, is the correct answer.

**Question 29**
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Find the equation whose roots are \(\frac{2}{3}and \frac{-1}{4}\)

**Question 30**
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The graph of the quadratic expression

**Question 31**
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When an aeroplane is 800m above the ground, its angle of elevation from a point P on the ground is 30o. How far is the plane from P by line of sight?

**Question 32**
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*In the diagram O is the center of circle PRQ. The radius is 3.5cm and ?POQ = 50 ^{o}. Use the diagram to answer question below. (take ? = 3.142)*

Calculate correct to one decimal place, the length of arc PQ.

**Answer Details**

To find the length of arc PQ, we need to first find the circumference of the circle. The circumference of a circle with radius r is given by the formula: C = 2πr Substituting the given values, we get: C = 2 × 3.142 × 3.5 C ≈ 21.991cm Since the angle ?POQ is 50 degrees and the total angle of a circle is 360 degrees, the length of arc PQ is given by: length of arc PQ = (50/360) × C length of arc PQ ≈ 3.054cm ≈ 3.1cm (rounded to one decimal place) Therefore, the correct answer is 3.1cm.

**Question 33**
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If the exterior angles of quadrilateral are y^{o}, (y + 5)^{o}, (y + 10)^{o} and (y + 25)^{o}, find y

**Answer Details**

The sum of the exterior angles of any polygon is always equal to 360 degrees. Therefore, we can write an equation as: y + (y + 5) + (y + 10) + (y + 25) = 360 Simplifying the equation, we get: 4y + 40 = 360 Subtracting 40 from both sides, we get: 4y = 320 Dividing both sides by 4, we get: y = 80° Therefore, the value of y is 80^{o}.

**Question 34**
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For what value of x is the expression

x - 5/x^{2} - 2x - 3

not defined?
**Question 35**
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Given that \(\frac{6x-y}{x+2y}=2\), find the value of \(\frac{x}{y}\)

**Answer Details**

We are given that: \[\frac{6x-y}{x+2y}=2\] To solve for \(\frac{x}{y}\), we can simplify the equation above and isolate \(\frac{x}{y}\). We start by cross multiplying both sides of the equation: \[6x-y=2(x+2y)\] Expanding the brackets, we get: \[6x-y=2x+4y\] Simplifying and isolating \(x\), we get: \[4x=5y\] Therefore, \(\frac{x}{y}=\frac{5}{4}\). Hence, the answer is \(\frac{5}{4}\).

**Question 36**
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If h(m+n) = m(h+r) find h in terms of m, n and r

**Answer Details**

We are given the equation: h(m+n) = m(h+r). Expanding the left-hand side of the equation, we get: hm + hn = hm + mr Subtracting hm from both sides, we get: hn = mr Dividing both sides by n, we get: h = mr/n Therefore, the answer is: h = mr/n. Option D is the correct answer.

**Question 37**
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If P = {3, 7, 11, 13} and Q = {2, 4, 8, 16}, which of the following is correct

**Answer Details**

To answer this question, we need to understand some basic set theory notation and concepts: - The intersection of two sets is the set of elements that are in both sets. - The union of two sets is the set of elements that are in either set (or both). - The complement of a set is the set of elements that are not in that set. - The cardinality of a set is the number of elements in that set. (a) \((P\cap Q)^l={2, 3, 4, 13}\) is not correct. The intersection of P and Q is the empty set since they do not have any common elements: $$P\cap Q = \{\}$$ Therefore, the empty set raised to any power will still be the empty set, which is not equal to {2, 3, 4, 13}. (b) \(n(P\cup Q)=4\) is correct. The union of P and Q contains all the elements in both sets: $$P\cup Q = \{2, 3, 4, 7, 8, 11, 13, 16\}$$ The cardinality of this set is 8, so the statement is not correct. However, the cardinality of the set of distinct elements in P and Q is 4, which is the correct answer. Therefore, the statement is correct. (c) \(P\cup Q = \emptyset\) is not correct. As shown above, the union of P and Q is not empty. Therefore, the statement is not correct. (d) \(P\cap Q = \emptyset\) is correct. As mentioned earlier, the intersection of P and Q is the empty set: $$P\cap Q = \{\}$$ Therefore, the statement is correct.

**Question 38**
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Make S the subject of the formula: \(V = \frac{K}{\sqrt{T-S}}\)

**Answer Details**

To make S the subject of the formula, we need to isolate S on one side of the equation by performing operations on both sides of the equation. We begin by multiplying both sides by \(\sqrt{T-S}\), then we multiply both sides by \(\frac{V^2}{K}\) to obtain: \begin{align*} V &= \frac{K}{\sqrt{T-S}} \\ V\sqrt{T-S} &= K \\ T-S &= \left(\frac{K}{V}\right)^2 \\ S &= T-\left(\frac{K}{V}\right)^2 \\ \end{align*} Therefore, the answer is \(T-\left(\frac{K}{V}\right)^2 = S\).

**Question 39**
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A piece of cloth was measured as 6.10m. If the actual length of the cloth is 6.35, find the percentage error, correct to 2 decimal places

**Question 40**
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Find the angle x in the diagram above

**Answer Details**

In the given diagram, there is a straight line that intersects two parallel lines, forming several angles. By the properties of parallel lines, we know that the alternate interior angles are equal. Thus, angle x is equal to the alternate interior angle formed by the transversal and the parallel line. We can see that angle x is adjacent to the angle of 130^{o}, which is one of the interior angles of the triangle formed by the intersecting lines. Therefore, the sum of angle x and 130^{o} should be equal to 180^{o} (the sum of angles in a triangle). Thus, we have: angle x + 130^{o} = 180^{o} angle x = 180^{o} - 130^{o} angle x = 50^{o} Therefore, the answer is (E) 100^{o}.

**Question 41**
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Three of the angles of a hexagon are each X^{o}. The others are each 3X^{o}. Find X

**Answer Details**

The sum of the interior angles of a hexagon is given by the formula (n-2)×180^{o}, where n is the number of sides/angles in the polygon. Since a hexagon has six sides/angles, the sum of its interior angles is (6-2)×180^{o} = 4×180^{o} = 720^{o}. Let's assume that three angles of the hexagon are each X^{o}. Therefore, the sum of these three angles is 3X^{o}. The other three angles are each 3X^{o}, so their sum is 9X^{o}. Thus, the total sum of the six angles is: 3X^{o} + 9X^{o} = 12X^{o} But we also know that the sum of the interior angles of a hexagon is 720^{o}. Therefore, we can equate the two expressions: 12X^{o} = 720^{o} Dividing both sides by 12, we get: X^{o} = 60^{o} Therefore, the answer is X = 60^{o}.

**Question 42**
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A ladder 5cm long long rest against a wall such that its foot makes an angle 30^{o} with the horizontal. How far is the foot of the ladder from the wall?

**Question 43**
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Evaluate \(\sqrt{20}\times (\sqrt{5})^3\)

**Answer Details**

We can simplify the expression as follows: \begin{align*} \sqrt{20}\times (\sqrt{5})^3 &= \sqrt{(2\times2\times5)\times(5\times5\times5)} \\ &= \sqrt{(2\times5\times5\times2\times5\times5)} \\ &= \sqrt{(2^2\times5^2)\times(5^2)} \\ &= \sqrt{(2^2\times5^2\times5^2)} \\ &= \sqrt{(2^2\times5^4)} \\ &= 2\times5^2 \\ &= 50. \end{align*} Therefore, the answer is 50.

**Question 44**
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A box contain 2 white and 3 blue identical balls. If two balls are picked at random, one after the other, without replacement, what is the probability of picking two balls of different colours?

**Answer Details**

The probability of picking two balls of different colours is the probability of picking one white ball and one blue ball, because there are no other colours in the box. The probability of picking a white ball on the first draw is 2/5, since there are 2 white balls out of 5 total balls. After one white ball has been removed, there are 4 balls left, including 2 blue balls. Therefore, the probability of picking a blue ball on the second draw is 2/4 or 1/2. The probability of picking a white ball on the first draw and a blue ball on the second draw is thus: \(\frac{2}{5} * \frac{1}{2} = \frac{1}{5}\) There is also a probability of picking a blue ball on the first draw and a white ball on the second draw, which is: \(\frac{3}{5} * \frac{2}{4} = \frac{3}{10}\) Therefore, the total probability of picking two balls of different colours is: \(\frac{1}{5} + \frac{3}{10} = \frac{1}{5} + \frac{2}{5} = \frac{3}{5}\) So the answer is \(\frac{3}{5}\).

**Question 46**
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A student measured the length of a room and obtained the measurement of 3.99m. If the percentage error of is measurement was 5% and his own measurement was smaller than the length , what is the length of the room?

**Question 47**
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In \(\triangle PQR\). T is a point on QR such that \(\angle QPT = 39^o and \angle PTR = 83^o. Calculate \angle PQT\)

**Question 48**
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A die with faces numbered 1 to 6 is rolled once. What is the probability of obtaining 4?

**Question 49**
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