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**Question 1**
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Find the next three terms of the sequence; 0, 1, 1, 2, 3, 5, 8...

**Answer Details**

The given sequence is the Fibonacci sequence, where the first two terms are 0 and 1, and each subsequent term is the sum of the two preceding it. Therefore, the next three terms are: - 13 (8 + 5) - 21 (13 + 8) - 34 (21 + 13) Hence, the answer is 13, 21, 34.

**Question 2**
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The figure is a pie chart which represents the expenditure of a family in a year. If the total income of the family was Le 10,800,000.00, how much was spent on food?

**Question 3**
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A straight line passes through the point P(1,2) and Q

(5,8). Calculate the length PQ

**Answer Details**

We can use the distance formula to find the length PQ, which is the distance between points P and Q on the line. The distance formula is: distance = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) where (x1, y1) = P and (x2, y2) = Q. Plugging in the values: distance = \(\sqrt{(5-1)^2 + (8-2)^2}\) = \(\sqrt{16 + 36}\) = \(\sqrt{52}\) = \(2\sqrt{13}\) Therefore, the length PQ is \(2\sqrt{13}\). Answer: \(2\sqrt{13}\).

**Question 4**
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A sphere of radius rcm has the same volume as cylinder of radius 3cm and height 4cm. Find the value of r

**Answer Details**

The volume of a sphere of radius r is given by the formula: \(\frac{4}{3}\pi r^3\). The volume of a cylinder of radius 3cm and height 4cm is given by the formula: \(\pi (3cm)^2(4cm) = 36\pi cm^3\). According to the question, the volume of the sphere is equal to the volume of the cylinder. Therefore: \[\frac{4}{3}\pi r^3 = 36\pi\] Dividing both sides by \(\frac{4}{3}\pi\), we get: \[r^3 = 27\] Taking the cube root of both sides, we get: \[r = 3\] Therefore, the value of r is 3cm. Answer is correct.

**Question 5**
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Find the lower quartile of the distribution illustrated by the cumulative frequency curve

**Answer Details**

To find the lower quartile, we need to identify the point on the cumulative frequency curve that corresponds to 25% of the total frequency. From the graph, we see that the total frequency is 40, and 25% of this is 10. The point on the curve that corresponds to a frequency of 10 is at a value of 19.0. Therefore, the lower quartile is 19.0. Note that the cumulative frequency curve shows the cumulative frequency of data values up to and including each data point on the horizontal axis. So, we can read off quartiles and other percentiles directly from the graph.

**Question 6**
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Simplify \(\frac{(p - r)^2 - r^2}{2p^2 - 4pr}\)

**Answer Details**

First, let's simplify the numerator: \begin{align*} (p-r)^2 - r^2 &= (p^2 - 2pr + r^2) - r^2 \\ &= p^2 - 2pr \end{align*} Now, let's factor the denominator: \begin{align*} 2p^2 - 4pr &= 2p(p - 2r) \end{align*} Substituting these results, we get: \begin{align*} \frac{(p-r)^2 - r^2}{2p^2 - 4pr} &= \frac{p^2 - 2pr}{2p(p - 2r)} \\ &= \frac{p(p-2r)}{2p(p-2r)} \\ &= \frac{1}{2} \end{align*} Therefore, the answer is: \boxed{\frac{1}{2}}.

**Question 7**
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The perimeter of a sector of a circle of radius 21cm is 64cm. Find the angle of the sector [Take \(\pi = \frac{22}{7}\)]

**Question 8**
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The marks of eight students in a test are: 3, 10, 4, 5, 14, 13, 16 and 7. Find the range

**Answer Details**

The range of a set of data is the difference between the maximum and minimum values in the set. In this case, the minimum mark is 3 and the maximum mark is 16. Therefore, the range is 16 - 3 = 13. Hence, the answer is 13.

**Question 9**
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If 23_{x} + 101_{x} = 130_{x}, find the value of x

**Question 10**
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In the diagram, TX is perpendicular to UW, |UX| = 1cm and |TX| = |WX| = \(\sqrt{3}\)cm. Find UTW

**Answer Details**

Using Pythagoras theorem, we can find that |UW| = 2cm. Next, we notice that \(\Delta\)UTW is an isosceles triangle since |TX| = |WX|. Therefore, o. Using the fact that the angles in a triangle add up to 180^{o}, we can find that o - o - (60^{o} + o. Therefore, we can substitute this into the previous equation to get: 90^{o} + o - (60^{o} + o = 2o, and since \(\Delta\)UTW is an isosceles triangle, we have that o. Hence, the answer is (C) 75^{o}.

**Question 11**
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Simplify: (\(\frac{3}{4} - \frac{2}{3}\)) x 1\(\frac{1}{5}\)

**Answer Details**

To simplify the expression (\(\frac{3}{4} - \frac{2}{3}\)) x 1\(\frac{1}{5}\), we first need to evaluate the subtraction inside the parentheses: \begin{align*} \frac{3}{4} - \frac{2}{3} &= \frac{9}{12} - \frac{8}{12} \\ &= \frac{1}{12} \end{align*} So now we have: \begin{align*} (\frac{3}{4} - \frac{2}{3})\times 1\frac{1}{5} &= \frac{1}{12} \times \frac{6}{5} \\ &= \frac{1 \times 6}{12 \times 5} \\ &= \frac{1}{10} \end{align*} Therefore, the answer is \(\frac{1}{10}\).

**Question 12**
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On a map, 1cm represent 5km. Find the area on the map that represents 100km^{2}.

**Answer Details**

If 1cm represents 5km on the map, then x cm will represent 100km^{2} on the map. To find x, we can use the formula for area of a square, which is A = s^{2}. In this case, we want to solve for s, where A = 100 and s represents the side length on the map in centimeters. So, s^{2} = A s^{2} = 100 s = √100 s = 10cm Therefore, 10cm on the map represents 100km^{2} on the ground. To find the area on the map that represents 100km^{2}, we need to find the area of a square with a side length of 10cm. Area = s^{2} Area = 10cm x 10cm Area = 100cm^{2} So, the area on the map that represents 100km^{2} is 100cm^{2}, which is equal to 1cm x 1cm, 4cm^{2}, or.

**Question 13**
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From the diagram, which of the following is true?

**Question 14**
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In the diagram MN is a chord of a circle KMN centre O and radius 10cm. If < MON = 140^{o}, find, correct to the nearest cm, the length of the chord MN.

**Question 15**
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Simplify; \(\frac{3^{n - 1} \times 27^{n + 1}}{81^{n}}\)

**Answer Details**

We can simplify this expression by using the properties of exponents and simplifying the numbers. First, we can write 27 as 3^{3} and 81 as 3^{4}. Next, we can simplify the numerator by using the distributive property of exponents: \begin{align*} \frac{3^{n-1} \times 27^{n+1}}{81^n} &= \frac{3^{n-1} \times 3^{3(n+1)}}{3^{4n}} \\ &= \frac{3^{n-1} \times 3^{3n+3}}{3^{4n}} \\ &= \frac{3^{4n-1}}{3^{4n}} \times 3^{3n+3} \\ &= 3^{-1} \times 3^{3n+3} \\ &= 3^{3n+2} \\ &= 3^2 \times 3^{3n} \\ &= 9 \times 3^n \\ \end{align*} Therefore, the simplified expression is 9 * 3^{n}. So, the answer is 9.

**Question 16**
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Make s the subject of the relation: P = S + \(\frac{sm^2}{nr}\)

**Answer Details**

To make s the subject of the relation P = S + \(\frac{sm^2}{nr}\), we need to isolate s on one side of the equation. First, we can start by moving the \(\frac{sm^2}{nr}\) term to the other side of the equation by subtracting it from both sides: P - \(\frac{sm^2}{nr}\) = S Next, we can solve for s by multiplying both sides of the equation by \(\frac{nr}{m^2}\): s = \(\frac{nr}{m^2}\)(P - \(\frac{sm^2}{nr}\)) Simplifying the right-hand side, we get: s = \(\frac{nrp}{m^2}\) - \(\frac{s}{m}\) To isolate s, we can add \(\frac{s}{m}\) to both sides of the equation: s + \(\frac{sm}{m^2}\) = \(\frac{nrp}{m^2}\) Simplifying the left-hand side, we get: s(\(\frac{m + 1}{m^2}\)) = \(\frac{nrp}{m^2}\) Finally, we can solve for s by dividing both sides of the equation by \(\frac{m+1}{m^2}\): s = \(\frac{nrp}{nr + m^2}\) Therefore, the answer is s = \(\frac{nrp}{nr + m^2}\).

**Question 17**
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Express 1975 correct to 2 significant figures

**Question 18**
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A straight line passes through the points P(1,2) and Q(5,8). Calculate the gradient of the line PQ

**Answer Details**

The gradient of a line is the measure of its steepness, or slope. To find the gradient of the line PQ, we need to use the formula: Gradient (m) = change in y / change in x We can calculate the change in y by subtracting the y-coordinate of point P from the y-coordinate of point Q: 8 - 2 = 6 Similarly, we can calculate the change in x by subtracting the x-coordinate of point P from the x-coordinate of point Q: 5 - 1 = 4 Therefore, the gradient of the line PQ is: Gradient (m) = change in y / change in x = 6 / 4 = 3/2 Therefore, the answer is option C: \(\frac{3}{2}\).

**Question 19**
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Kweku walked 8m up to slope and was 3m above the ground. If he walks 12m further up the slope, how far above the ground will he be?

**Question 20**
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In the diagram, \(\bar{YW}\) is a tangent to the circle at X, |UV| = |VX| and < VXW = 50o. Find the value of < UXY.

**Question 21**
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An object is 6m away from the base of a mast. If the angle of depression of the object from the top of the mast is 50^{o}, find, correct to 2 decimal places, the height of the mast.

**Question 22**
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\(\begin{array}{c|c}

Age(years) & 13 & 14 & 15 & 16 & 17 \\

\hline

Frequency & 10 & 24 & 8 & 5 & 3

\end{array}\)

Find the median age

**Answer Details**

To find the median, we need to first arrange the ages in order from lowest to highest. Then, we can determine which age lies in the middle of the list. Arranging the ages in order of increasing magnitude, we have: $$13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 17, 17, 17$$ There are a total of 10+24+8+5+3 = 50 ages in the list, which is an even number. To find the median, we need to take the average of the two middle ages. The two middle ages are the 25th and 26th ages in the list, which are both 14. Therefore, the median age is: $$(14 + 14)/2 = 14$$ So the correct answer is 14.

**Question 23**
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The volume of a pyramid with height 15cm is 90cm^{3}. If its base is a rectangle with dimension xcm by 6cm, find the value of x

**Answer Details**

The formula for the volume of a pyramid is given as: V = (1/3) * base_area * height Let the length of the rectangle be x, then the base area of the pyramid is given as: base_area = x * 6 From the question, we are told that the volume of the pyramid is 90cm^{3} and its height is 15cm. Substituting into the formula for the volume of a pyramid, we have: 90 = (1/3) * (x * 6) * 15 Multiplying both sides by 3 gives: 270 = 90 * 6x Dividing both sides by 90 gives: 3 = 2x Therefore, x = 3/2 = 1.5 Hence, the value of x is 3. Answer option A, 3, is the correct answer.

**Question 24**
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Find the values of y for which the expression \(\frac{y^2 - 9y + 18}{y^2 + 4y - 21}\) is undefined

**Answer Details**

The given expression will be undefined when its denominator is equal to zero since division by zero is undefined. So we need to find the values of y that make the denominator zero. \begin{align*} y^2 + 4y - 21 &= 0\\ (y+7)(y-3) &= 0 \end{align*} The denominator is equal to zero when either y+7=0 or y-3=0. Therefore, the expression is undefined when y=-7 or y=3. So, the answer is (c) 3, -7.

**Question 25**
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The roots of a quadratic equation are \(\frac{4}{3}\) and -\(\frac{3}{7}\). Find the equation

**Question 26**
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In the diagram, TS is a tangent to the circle at S. |PR| and < PQR = 177^{o}. Calculate < PST.

**Question 27**
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Halima is n years old. Her brother's age is 5 years more than half of her age. How old is her brother?

**Question 28**
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If 20(mod 9) is equivalent to y(mod 6), find y.

**Question 29**
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If cos \(\theta\) = x and sin 60^{o} = x + 0.5 0^{o} < \(\theta\) < 90^{o}, find, correct to the nearest degree, the value of \(\theta\)

**Question 30**
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If log_{2}(3x - 1) = 5, find x.

**Answer Details**

We are given that log_{2}(3x - 1) = 5. Using the definition of logarithms, we know that 2^{5} = 32 is equal to the expression inside the logarithm. That is, 3x - 1 = 32 Adding 1 to both sides, we get 3x = 33 Dividing by 3, we get x = 11 Therefore, the value of x is 11. Answer: 11.

**Question 31**
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The curve surface area of a cylinder, 5cm high is 110cm ^{2}. Find the radius of its base. [Take \(\pi = \frac{22}{7}\)]

**Answer Details**

The curved surface area of a cylinder is given by the formula: 2\(\pi\)rh. Given that the cylinder is 5cm high and its curved surface area is 110cm^{2}, we can write: 2\(\pi\)rh = 110, where h = 5 Substituting the value of \(\pi = \frac{22}{7}\) and h = 5, we get: 2 x \(\frac{22}{7}\) x r x 5 = 110 Simplifying this expression, we get: r = \(\frac{7}{2}\) r = 3.5cm (to one decimal place) Therefore, the radius of the cylinder is approximately 3.5cm. Hence, the answer is 3.5cm.

**Question 32**
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The diagram, MOPQ is a trapezium with QP||MO, MQ||NP, NQ||OP, |QP| = 9cm and the height of \(\Delta\) QNP = 6cm, calculate the area of the trapezium.

**Question 33**
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In the diagram, PR||SV||WY|, TX||QY|, < PQT = 48^{o} and < TXW = 60^{o}.Find < TQU.

**Question 34**
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A bag contains 5 red and 4 blue identical balls. Id two balls are selected at random from the bag, one after the other, with replacement, find the probability that the first is red and the second is blue

**Answer Details**

When two balls are drawn from the bag with replacement, there are a total of $9\times9=81$ possible outcomes, since there are 9 balls in the bag and we are replacing each ball after drawing. To find the probability that the first ball is red and the second ball is blue, we can use the multiplication rule of probability. The probability that the first ball is red is $\frac{5}{9}$, since there are 5 red balls out of 9 total balls in the bag. After replacing the first ball, there are still 9 balls in the bag, but now 4 of them are blue. So the probability that the second ball is blue, given that the first ball was red, is $\frac{4}{9}$. Therefore, the probability that the first ball is red and the second ball is blue is: $$\frac{5}{9} \times \frac{4}{9} = \frac{20}{81}$$ Hence the answer is $\frac{20}{81}$.

**Question 35**
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In the diagram, \(\bar{PF}\), \(\bar{QT}\), \(\bar{RG}\) intersect at S and PG||RG. If < SPQ = 113^{o} and < RSt = 220, find < PSQ

**Question 37**
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\(\begin{array}{c|c}

Age(years) & 13 & 14 & 15 & 16 & 17 \\

\hline

Frequency & 10 & 24 & 8 & 5 & 3

\end{array}\)

The table shows the ages of students in a club. How many students are in the club?

**Answer Details**

To find out how many students are in the club, we need to add up the frequencies in the table. So, Number of 13-year-olds = 10 Number of 14-year-olds = 24 Number of 15-year-olds = 8 Number of 16-year-olds = 5 Number of 17-year-olds = 3 Total number of students = 10 + 24 + 8 + 5 + 3 = 50 Therefore, there are 50 students in the club.

**Question 38**
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