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The probability of selecting a girl at random from the class is the ratio of the number of girls in the class to the total number of students in the class. Given that there are m boys and 12 girls in the class, the total number of students in the class is m + 12. Therefore, the probability of selecting a girl at random is: \begin{align*} \frac{12}{m+12} \end{align*} Therefore, the correct option is (c) \(\frac{12}{m+12}\).

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We can use the property of similar triangles that corresponding sides are proportional. First, let's find the ratio of corresponding sides between the two triangles. Ratio of corresponding sides: |XY|/|PR| = 5/8 |XZ|/|PQ| = 3.5/|PQ| Since the two triangles are similar, the ratio of corresponding sides must be equal. Therefore: 5/8 = 3.5/|PQ| Cross-multiplying: 5 x |PQ| = 8 x 3.5 |PQ| = 28/5 Simplifying: |PQ| = 5.6 cm Therefore, |PQ| is 5.6 cm.

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To calculate the average speed of the boy, we need to convert the distance and time to the same units. The boy walked 800m in 20 minutes. To convert minutes to hours, we divide by 60: 20 minutes ÷ 60 = 0.333 hours Now we can calculate the average speed: Speed = Distance ÷ Time Speed = 800m ÷ 0.333 hours Speed ≈ 2400m/h To convert meters per hour to kilometers per hour, we divide by 1000: Speed ≈ 2.4 km/h Therefore, the boy's average speed is 2.4 km/h. So the answer is (a) 2.4.

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First, we simplify the left-hand side of the equation: \begin{align*} \frac{3}{4}t + \frac{1}{3}(21-t) &= 11 \\ \frac{3}{4}t + 7 - \frac{1}{3}t &= 11 \\ \frac{5}{12}t &= 4 \\ t &= \frac{4 \times 12}{5} \\ t &= \frac{48}{5} \end{align*} Therefore, the value of t is \(9\frac{3}{5}\). So, the correct answer is (d).

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Option III is true, while options I and II are not true. Explanation: I. The locus of points equidistant from a straight line is actually a pair of parallel lines, not a circle. To see this, consider any point on one side of the line, and draw the perpendicular from that point to the line. The set of all points equidistant from the line will be the line that passes through all of those perpendiculars. II. The locus of points equidistant from two given points is actually the perpendicular bisector of the line segment joining the two points, not a circle. To see this, note that any point on the perpendicular bisector is equidistant from the two points by definition. Conversely, any point equidistant from the two points must lie on the perpendicular bisector. III. The locus of points equidistant from two points is indeed the perpendicular bisector of the line segment joining the two points. To see this, consider any point on the perpendicular bisector. By definition, it is equidistant from the two points. Conversely, any point equidistant from the two points must lie on the perpendicular bisector.

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We know that any number raised to the power of zero is equal to 1. Therefore, we can rewrite the equation \(\left(\frac{1}{4}\right)^{(2-y)} = 1\) as \(\left(\frac{1}{4}\right)^{(2-y)} = \left(\frac{1}{4}\right)^0\). Using the rule of exponents that states when we have the same base raised to different powers, we can multiply the bases and subtract the exponents. So, we get \(\left(\frac{1}{4}\right)^{(2-y)} = \left(\frac{1}{4}\right)^0 \Rightarrow \frac{1}{4^{(2-y)}} = \frac{1}{4^0}\). Since \(4^0 = 1\), we can simplify the right-hand side to 1. Therefore, we have \(\frac{1}{4^{(2-y)}} = 1\). Multiplying both sides by \(4^{(2-y)}\) gives us \(1 = 4^{(2-y)}\). We can rewrite the left-hand side as \(4^0\) because any number raised to the power of zero is equal to 1. So, we have \(4^0 = 4^{(2-y)}\). Using the rule of exponents again, we can set the exponents equal to each other, which gives us \(0 = 2 - y\). Solving for y, we get \(y = 2\). Therefore, the value of y that satisfies the equation \(\left(\frac{1}{4}\right)^{(2-y)} = 1\) is 2.

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The first four prime numbers greater than 10 are 11, 13, 17, and 19. To find their average, we add them up and divide by the total number of primes, which is 4. \begin{align*} \text{Average} &= \frac{11 + 13 + 17 + 19}{4} \\ &= \frac{60}{4} \\ &= 15 \end{align*} Therefore, the average of the first four prime numbers greater than 10 is 15. Answer: 15

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If the Cooperative Society charges an interest of 5¹/₂% per annum, then the total amount to be paid back by the member at the end of the year will be the principal amount plus 5¹/₂% of the principal amount. 5¹/₂% can be written as a fraction of 5/2 in its simplest form. Therefore, the interest on N125,000 will be (5/2) * N125,000/100 = N3125. The total amount to be paid back by the member after one year will be the sum of the principal and the interest, which is N125,000 + N3125 = N128,125. Therefore, the member will pay back N128,125 after one year. The option that matches this result is N131,875, which is not correct.

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The girl spends \(\frac{1}{4}\) of her pocket money on books and \(\frac{1}{3}\) on dress, which means she has spent a total of \(\frac{1}{4}+\frac{1}{3}\) of her money. To find the fraction that remains, we need to subtract this amount from 1 (since 1 represents the whole amount of money). \(\frac{1}{4}+\frac{1}{3} = \frac{3}{12}+\frac{4}{12} = \frac{7}{12}\) So, the fraction that remains is: \(1-\frac{7}{12} = \frac{12}{12}-\frac{7}{12} = \frac{5}{12}\) Therefore, the answer is \(\frac{5}{12}\).

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Let h be the height of the tower. We can form a right triangle with the tower, the ground point and the point directly below the top of the tower. Let's call the latter point P. Then, we have: - Angle of elevation of the top of the tower from point on ground = 30^o - Length of base of the right triangle (i.e., distance from the point on the ground to the foot of the tower) = 36m Therefore, we can use tangent function to find h: tan(30^o) = opposite / adjacent = h / 36 Solving for h, we have: h = 36 * tan(30^o) ≈ 20.78m Therefore, the height of the tower is approximately 20.78m. So, the correct option is (b) 20.78m.

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Since POQ is the diameter of the circle, we know that angle POR = 90 degrees. Therefore, angle QOR is also 90 degrees because it is vertically opposite to angle POR. Since QR is a chord of the circle, angle QRS is half of the angle subtended by the same chord at the circumference of the circle, which is angle QOR. Thus, angle QRS = 1/2 * angle QOR = 1/2 * 90 degrees = 45 degrees. Therefore, the answer is 45 degrees, which is 35^o.

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To solve the equation 10-3x-x² = 0, we can first rearrange it to get x²+3x-10=0. We can then factorize this quadratic equation as (x+5)(x-2)=0. This means that either x+5=0 or x-2=0, which gives us the solutions x=-5 and x=2. Therefore, the correct answer is x=2 or -5, which is the first option.

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Substitute the given values of p, q and r into the expression 3p² - q² - r³ to get: 3(2)² - (-5)² - (-4)³ = 12 - 25 + 64 = 51 Therefore, the answer is 51.

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We have the expression: \(\sqrt{128}+\sqrt{18}-\sqrt{K} = 7\sqrt{2}\). We can simplify the two radicals to get: \(\sqrt{128}=8\sqrt{2}\) and \(\sqrt{18}=3\sqrt{2}\). Substituting these values into the expression, we get: \(8\sqrt{2}+3\sqrt{2}-\sqrt{K} = 7\sqrt{2}\) Combining like terms on the left side, we have: \(11\sqrt{2}-\sqrt{K} = 7\sqrt{2}\) Isolating the square root on the left side, we get: \(-\sqrt{K} = -4\sqrt{2}\) Squaring both sides, we have: \(K = (-4\sqrt{2})^2\) Simplifying the expression, we get: \(K = 32\) Therefore, the value of K is 32.

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To find the mean of a set of numbers, we need to add up all the numbers and divide by the total number of numbers. Adding the given numbers, we get: 9 + 13 + 16 + 17 + 19 + 23 + 24 = 121 There are 7 numbers in the set, so to find the mean, we divide the sum by 7: Mean = 121/7 = 17.29 (correct to two decimal places) Therefore, the answer is (b) 17.29.

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We can start solving the inequality by first finding a common denominator for the fractions: \begin{align*} \frac{x+2}{4}-\frac{x+1}{3}&>\frac{1}{2}\\ \frac{3(x+2)}{12}-\frac{4(x+1)}{12}&>\frac{6}{12}\\ \frac{3x+6-4x-4}{12}&>\frac{1}{2}\\ -\frac{x-2}{12}&>\frac{1}{2}\\ \end{align*} Multiplying both sides by $-1$ changes the direction of the inequality: \begin{align*} \frac{x-2}{12}&<-\frac{1}{2}\\ \end{align*} Multiplying both sides by $12$ gives: \begin{align*} x-2&<-6\\ x&< -4\\ \end{align*} Therefore, the answer is: - x < -4

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The quadratic equation with roots \(-\frac{1}{2}\) and \(\frac{3}{4}\) can be written in factored form as: $$a(x+\frac{1}{2})(x-\frac{3}{4}) = 0$$ where a is a constant. Expanding this expression, we get: $$a(x+\frac{1}{2})(x-\frac{3}{4}) = ax^2+\frac{1}{8}a = 0$$ Simplifying the equation, we get: $$8ax^2 + a = 0$$ Now we can compare the coefficients of this equation with those in the given options to find the answer. Comparing the coefficients of the given options with our equation, we can see that only the equation: 8x² - 2x - 3 = 0 has the same coefficients, and therefore, the same roots, as the equation we derived. Therefore, the answer is: 8x² - 2x - 3 = 0.

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To solve this problem, we first need to find the values of x and y, and then substitute them into the expression \(\frac{y}{2}-3\). From the given equations, we can use the elimination method to solve for x and y. Multiplying the second equation by 2 and adding it to the first equation, we get: 2(3x-y) + (x+y) = 11x = 17 x = 17/11 Substituting x into the first equation, we get: y = 7 - x = 7 - 17/11 = 60/11 Now we can substitute these values into the expression \(\frac{y}{2}-3\): \(\frac{y}{2}-3 = \frac{60/11}{2} - 3 = \frac{30}{11} - \frac{33}{11} = -\frac{3}{11}\) Therefore, the answer is -1.

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There are two possible ways to get two balls of different colors: first picking a red ball, then picking a white ball, or first picking a white ball, then picking a red ball. Since we replace the first ball before picking the second one, the probability of picking a red ball on the first draw is $\frac{3}{5}$, and the probability of picking a white ball on the second draw is also $\frac{2}{5}$, hence the probability of getting a red ball followed by a white ball is $\frac{3}{5}\cdot \frac{2}{5} = \frac{6}{25}$. The probability of picking a white ball first and a red ball second is also $\frac{6}{25}$. Therefore, the probability of getting two balls of different colors is the sum of the probabilities of the two cases, which is $\frac{6}{25}+\frac{6}{25} = \frac{12}{25}$. So, the correct option is: - $\frac{12}{25}$

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To simplify the expression \(7\frac{1}{2}-\left(2\frac{1}{2}+3\right)\div16\frac{1}{2}\), we need to perform the arithmetic operations in the following order: division, addition, and subtraction. First, we need to simplify the expression inside the parentheses: \begin{align*} 2\frac{1}{2}+3 &= \frac{5}{2} + 3 \\ &= \frac{5}{2} + \frac{6}{2} \\ &= \frac{11}{2} \end{align*} Next, we need to divide $\frac{11}{2}$ by $16\frac{1}{2}$: \begin{align*} \frac{11}{2} \div 16\frac{1}{2} &= \frac{\frac{11}{2}}{\frac{33}{2}} \\ &= \frac{11}{33} \\ &= \frac{1}{3} \end{align*} Substituting $\frac{1}{3}$ into the original expression, we have: \begin{align*} 7\frac{1}{2}-\left(2\frac{1}{2}+3\right)\div16\frac{1}{2} &= 7\frac{1}{2}-\frac{1}{3} \\ &= \frac{22}{3}-\frac{1}{3} \\ &= \frac{21}{3} \\ &= 7 \end{align*} Therefore, the answer is 7, corrected to the nearest whole number. Answer is correct.

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The ratio of the sides of the two cubes is 2:5. Let us assume that the length of the sides of the first cube is 2x, then the length of the sides of the second cube will be 5x, since the ratio of their sides is 2:5. The volume of the first cube will be (2x)^3 = 8x^3, and the volume of the second cube will be (5x)^3 = 125x^3. The ratio of the volumes of the two cubes will be: Volume of the first cube : Volume of the second cube = 8x^3 : 125x^3 = 8 : 125 Therefore, the ratio of their volumes is 8:125. Hence, the correct answer is "8:125".

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To simplify the expression, we need to find a common denominator for the two fractions. We can use the difference of two squares identity \((a+b)(a-b) = a^2-b^2\) to find a common denominator. \begin{aligned} \frac{2}{a+b}-\frac{1}{a-b} &= \frac{2(a-b)}{(a+b)(a-b)} - \frac{1(a+b)}{(a-b)(a+b)}\\ &= \frac{2(a-b)-1(a+b)}{a^2-b^2}\\ &= \frac{2a-2b-a-b}{a^2-b^2}\\ &= \frac{a-3b}{a^2-b^2} \end{aligned} Therefore, the simplified expression is \(\frac{a-3b}{a^2-b^2}\), which corresponds to.

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To find the mean deviation, we first need to find the mean (average) of the given numbers. The mean is calculated by adding up all the numbers and then dividing by the total number of numbers: \[\text{Mean} = \frac{6 + 7 + 8 + 9 + 10}{5} = 8\] Next, we find the deviation of each number from the mean. To do this, we subtract the mean from each number: \[\begin{aligned} \text{Deviation of 6} &= 6 - 8 = -2 \\ \text{Deviation of 7} &= 7 - 8 = -1 \\ \text{Deviation of 8} &= 8 - 8 = 0 \\ \text{Deviation of 9} &= 9 - 8 = 1 \\ \text{Deviation of 10} &= 10 - 8 = 2 \end{aligned}\] Note that any negative deviations should be treated as positive values, so we need to ignore the negative signs when we calculate the mean deviation. To find the mean deviation, we add up all the deviations (ignoring any negative signs) and then divide by the total number of numbers: \[\begin{aligned} \text{Mean Deviation} &= \frac{|-2| + |-1| + |0| + |1| + |2|}{5} \\ &= \frac{2 + 1 + 0 + 1 + 2}{5} \\ &= \frac{6}{5} \\ &= 1.2 \end{aligned}\] Therefore, the mean deviation of 6, 7, 8, 9, 10 is 1.2.

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The volume of a cylinder can be calculated by multiplying the area of its base by its height. The base of this cylinder is a circle with diameter 7cm. The radius of the circle is half of the diameter, which is 3.5cm. Using the formula for the area of a circle, we can find that the area of the base is: \[\text{Area of base} = \pi r^2 = \frac{22}{7} \times 3.5^2 = 38.5\text{ cm}^2\] The height of the cylinder is also given as 7cm. Using the formula for the volume of a cylinder, we can find the volume of the solid: \[\text{Volume of cylinder} = \text{Area of base} \times \text{height} = 38.5\text{ cm}^2 \times 7\text{ cm} = 269.5\text{ cm}^3\] Therefore, the volume of the solid cylinder is 269.5cm³.

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If point X is on a bearing of 342^o from point Y, then point Y is on a bearing of 162^o from point X. This is because the bearing from X to Y is the opposite direction from the bearing from Y to X. Therefore, we subtract the original bearing of 342^o from 180^o to get 162^o.

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To find the number of sides of a regular polygon with interior angle 108°, we need to use the formula for the sum of interior angles of a polygon, which is (n-2) x 180°, where n is the number of sides. Since the polygon is regular, all its interior angles are equal, so we can use the fact that the sum of the interior angles of a polygon is also equal to the number of sides times the interior angle. Therefore, we have: (n-2) x 180° = n x 108° Simplifying this equation, we get: 180n - 360 = 108n 72n = 360 n = 5 Therefore, the regular polygon has 5 sides, and the answer is option A.

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Construction of Triangle PQR and the Required Loci

Given: Construct a triangle PQR such that

• PQ = 8 cm,
• PR = 7 cm,
• ∠QPR = 105°.

(a)(i) Construction of ΔPQR

1. Using a ruler, draw a line segment PQ = 8 cm.
2. At point P, construct an angle of 105°:
   • Construct a 90° angle at P by drawing a perpendicular to PQ.
   • Construct a 120° ray at P using equal arcs with the compasses.
   • Bisect the angle between the 90° ray and the 120° ray. The bisected angle is 105°.
3. On the 105° ray, use the compasses to mark a point R such that PR = 7 cm.
4. Join R to Q. This gives the required triangle ΔPQR.

(a)(ii) Locus L<sub>1</sub> of points equidistant from P and Q

The locus of points equidistant from P and Q is the perpendicular bisector of PQ.

1. With the compasses opened to more than half of PQ, draw arcs centred at P above and below PQ.
2. Without changing the compass width, draw arcs centred at Q to intersect the first arcs.
3. Join the two arc-intersection points with a ruler.

This straight line is labelled L<sub>1</sub>.

(a)(iii) Locus l<sub>2</sub> of points equidistant from Q and R

The locus of points equidistant from Q and R is the perpendicular bisector of QR.

1. Draw equal arcs centred at Q and R, using a radius greater than half of QR.
2. Join the two points where the arcs intersect.

This straight line is labelled l<sub>2</sub>.

(b)(i) Point of Intersection

Label the point where L<sub>1</sub> and l<sub>2</sub> intersect as T.

Conclusion: The intersection point T is the centre of the circle passing through P, Q and R.

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(a) Factor the first bracket:

\[\frac{x^{2}}{2}-x+\frac12=\frac12\left(x^{2}-2x+1\right)=\frac12(x-1)^{2}.\]

\[\therefore\ \frac12(x-1)^{2}\cdot\frac{1}{x-1}=\frac{x-1}{2}.\]

(b) \(P\) is 40 km from \(Q\) on bearing \(061^{\circ}\) (measured from north).

(i) North of Q: \(40\cos61^{\circ}=40(0.4848)=19.4\text{ km}\ (1\text{ d.p.}).\)

(ii) East of Q: \(40\sin61^{\circ}=40(0.8746)=35.0\text{ km}\ (1\text{ d.p.}).\)

(c) Let the son's share be \(s\). Each daughter gets \(\tfrac34 s\); three daughters get \(3\times\tfrac34 s=\tfrac94 s\).

\[s+\frac94 s=5720\Rightarrow\frac{13}{4}s=5720\Rightarrow s=5720\times\frac{4}{13}=\mathbf{N1760}.\]

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(a) For \(y=2x^2-7x-3\), the completed table is:

\(x\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)	\(5\)
\(y\)	\(19\)	\(6\)	\(-3\)	\(-8\)	\(-9\)	\(-6\)	\(1\)	\(12\)

(b) The points were plotted using the stated scales, and joined with a smooth curve. The straight line \(y=x+6\) and the tangent at \(x=1\) are included for parts (c) and (d).

(c)

(i) The lowest point of the curve gives the minimum value

\[\boxed{y\approx -9}\]

(ii) The tangent at \(x=1\) passes through \((1,-8)\). Using two points on the tangent, for example \((0,-5)\) and \((1,-8)\),

\[\text{gradient}=\frac{-8-(-5)}{1-0}=\boxed{-3}.\]

(d) Rearrange the equation in terms of the curve already drawn:

\[2x^2-7x-5=x+4\]

\[2x^2-7x-3=x+6.\]

Thus, draw the straight line \(y=x+6\). From its points of intersection with the parabola on the graph,

\[\boxed{x\approx -0.9\ \text{ or }\ x\approx 4.9}.\]

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(a) \(O\) is the centre, so \(|OA|=|OB|=r\). Perimeter of \(\triangle AOB\):

\[2r+|AB|=52.2\Rightarrow 2r+24.2=52.2\Rightarrow r=14\text{ cm}.\]

By the cosine rule in \(\triangle AOB\):

\[\cos(\angle AOB)=\frac{r^{2}+r^{2}-|AB|^{2}}{2r^{2}}=\frac{196+196-585.64}{392}=\frac{-193.64}{392}=-0.4939.\]

\[\angle AOB=\cos^{-1}(-0.4939)=119.6^{\circ}\approx\mathbf{120^{\circ}}\ (\text{nearest degree}).\]

(b) Volume of tank \(=60\times80\times100=480{,}000\text{ cm}^{3}\). Half filled:

\[\tfrac12\times480{,}000=240{,}000\text{ cm}^{3}=\frac{240{,}000}{1000}=\mathbf{240\text{ litres}}\quad(1000\text{ cm}^{3}=1\text{ litre}).\]

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(a) Simplify \(\sqrt{1001_{two}}\), leaving the answer in base two.

First convert \(1001_{two}\) to base ten:

Then \(\sqrt{9_{ten}}=3_{ten}\). Convert \(3\) back to base two:

Therefore \(\sqrt{1001_{two}}=\mathbf{11_{two}}\).

(b) Find \(z\) in terms of \(x\) and \(y\).

In the diagram, \(O\) is the centre, \(|OP|=x\) (a radius), \(|OK|=y\) and \(\angle OKP=90^\circ\). Since \(OK\) is drawn from the centre perpendicular to the chord \(PQ\), it bisects the chord, so \(K\) is the mid-point of \(PQ\):

Applying Pythagoras' theorem to right-angled triangle \(OKP\):

Therefore \(\displaystyle z=2\sqrt{x^{2}-y^{2}}\).

(c) \(P,Q,R,S\) lie on the circle centre \(O\), with \(\angle POQ=160^\circ\), \(\angle QSR=45^\circ\) and \(\angle PQS=40^\circ\).

(i) \(\angle QPS\): The chord \(PQ\) subtends the central angle \(\angle POQ=160^\circ\). The angle it subtends at the circumference (at \(S\)) is half of this:

In triangle \(PQS\), the three angles sum to \(180^\circ\):

(ii) \(\angle RQS\): Work out the arcs (central angles) from the given inscribed angles.

• Arc \(PQ\) corresponds to \(\angle POQ=160^\circ\).
• \(\angle PQS=40^\circ\) stands on chord \(PS\), so arc \(PS=2\times40^\circ=80^\circ\).
• \(\angle QSR=45^\circ\) stands on chord \(QR\), so arc \(QR=2\times45^\circ=90^\circ\).

The four arcs around the circle sum to \(360^\circ\):

\(\angle RQS\) stands at \(Q\) on chord \(RS\), so it equals half of arc \(RS\):

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(a) Write both sides in base 3: \(81=3^{4},\ 27=3^{3}\).

\[\frac{1}{81^{x-2}}=3^{-4(x-2)},\qquad 27^{1-x}=3^{3(1-x)}.\]

Equating indices:

\[-4(x-2)=3(1-x)\Rightarrow -4x+8=3-3x\Rightarrow -x=-5\Rightarrow x=5.\]

(b) Rationalise each term:

\[\frac{5}{\sqrt7-\sqrt3}=\frac{5(\sqrt7+\sqrt3)}{7-3}=\frac{5(\sqrt7+\sqrt3)}{4},\qquad \frac{1}{\sqrt7+\sqrt3}=\frac{\sqrt7-\sqrt3}{4}.\]

\[\text{Sum}=\frac{5\sqrt7+5\sqrt3+\sqrt7-\sqrt3}{4}=\frac{6\sqrt7+4\sqrt3}{4}=\frac{3\sqrt7+2\sqrt3}{2}.\]

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(a) Less-than cumulative frequency table

Marks	Class boundary	Frequency	Cumulative frequency
1–10	10.5	2	2
11–20	20.5	4	6
21–30	30.5	9	15
31–40	40.5	13	28
41–50	50.5	18	46
51–60	60.5	32	78
61–70	70.5	13	91
71–80	80.5	5	96
81–90	90.5	3	99
91–100	100.5	1	100

Plot cumulative frequency against the upper class boundaries, beginning with 0.5,0 . Draw a smooth increasing curve through the plotted points.

(b) Estimates from the ogive

The total frequency is 0.5+99.5=100 students.

1. Median position =rac{100}{2}=50 . From cumulative frequency 50, the corresponding mark is approximately 52 marks.
2. Lower quartile position =rac{100}{4}=25 . From cumulative frequency 25, the corresponding mark is approximately 39 marks.
3. 60th percentile position =rac{60}{100} imes100=60 . From cumulative frequency 60, the corresponding mark is approximately 55 marks.

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(a) Prepare the frequency table:

Mark, \(x\)	0	1	2	3	4	5	Total
Frequency, \(f\)	1	4	9	8	5	3	30
\(fx\)	0	4	18	24	20	15	81
Cumulative frequency	1	5	14	22	27	30

(i) Mean mark

\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{81}{30}=2.7\]

(ii) Median

There are \(30\) observations. The median lies between the 15th and 16th observations. From the cumulative frequencies, both the 15th and 16th observations have mark \(3\).

\[\text{Median}=3\]

(b) Pie chart

The sector angle for each mark is \(\dfrac{f}{30}\times360^\circ\). Thus, for mark \(2\):

\[\frac{9}{30}\times360^\circ=108^\circ\]

Therefore, the sectorial angle for mark \(2\) is \(108^\circ\).

The complete pie chart is shown below. Its sector angles are \(12^\circ,48^\circ,108^\circ,96^\circ,60^\circ\), and \(36^\circ\) for marks \(0,1,2,3,4\), and \(5\) respectively.

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(a) Substitute \(x=1,\ y=-2\):

\[p(1)+q(-2)=5\Rightarrow p-2q=5\quad(1)\]

\[q(1)+p(-2)=-10\Rightarrow -2p+q=-10\quad(2)\]

From (1), \(p=5+2q\). Substitute into (2): \(-2(5+2q)+q=-10\Rightarrow -10-4q+q=-10\Rightarrow -3q=0\Rightarrow q=0\).

Then \(p=5\). So \(\mathbf{p=5,\ q=0}\).

(b) Cross-multiply \(\dfrac{4r-3}{6r+1}=\dfrac{2r-1}{3r+4}\):

\[(4r-3)(3r+4)=(2r-1)(6r+1)\]

\[12r^{2}+7r-12=12r^{2}-4r-1\Rightarrow 7r-12=-4r-1\Rightarrow 11r=11\Rightarrow r=1.\]

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(a) Sides \(7:5:7\). The angle included between the two equal sides (each \(7k\)) is opposite the side \(5k\). By the cosine rule:

\[\cos\theta=\frac{7^{2}+7^{2}-5^{2}}{2(7)(7)}=\frac{49+49-25}{98}=\frac{73}{98}=0.7449.\]

\[\theta=\cos^{-1}(0.7449)=41.9^{\circ}\approx\mathbf{42^{\circ}}\ (\text{nearest degree}).\]

(b) Sum of interior angles \(=(n-2)\times180^{\circ}=1440^{\circ}\):

\[n-2=\frac{1440}{180}=8\Rightarrow \mathbf{n=10\text{ sides}}.\]

(ii) Each exterior angle \(=\dfrac{360^{\circ}}{n}=\dfrac{360^{\circ}}{10}=\mathbf{36^{\circ}}.\)

Antwoorddetails

(a) Area of the minor segment

The chord \(XY\) subtends \(\theta = 96^\circ\) at the centre and the radius is \(r = 5\text{ cm}\). The minor segment is the minor sector minus triangle \(OXY\).

Area of minor sector (with \(\pi = \tfrac{22}{7}\)):

\[\frac{\theta}{360}\pi r^2 = \frac{96}{360}\times\frac{22}{7}\times 5^2 = 20.95\text{ cm}^2\]

Area of triangle OXY:

\[\frac{1}{2}r^2\sin\theta = \frac{1}{2}\times 5^2\times\sin 96^\circ = 12.43\text{ cm}^2\]

Area of minor segment:

\[20.95 - 12.43 = 8.52\text{ cm}^2\]

Correct to three significant figures, the area of the minor segment is \(\mathbf{8.52\text{ cm}^2}\).

(b) Total area of the shaded portions

The square has side 14 cm, so the inscribed circle has radius \(r = 7\text{ cm}\).

Shaded sector of the circle (angle \(80^\circ\)):

\[\frac{80}{360}\times\frac{22}{7}\times 7^2 = 34.22\text{ cm}^2\]

Shaded portion of the square (the four corners between the square and the circle):

\[\text{Area of square} = 14^2 = 196\text{ cm}^2\]

\[\text{Area of circle} = \frac{22}{7}\times 7^2 = 154\text{ cm}^2\]

\[196 - 154 = 42\text{ cm}^2\]

Total shaded area:

\[42 + 34.22 = 76.22 \approx 76.2\text{ cm}^2\]