WAEC SSCE - General Mathematics - 2003

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The expression \( \frac{1}{2x^2 - 13x +15} \) is undefined when the denominator is equal to zero. Thus, we need to solve the equation: $$2x^2 - 13x + 15 = 0$$ We can factorize the quadratic as follows: $$2x^2 - 10x - 3x + 15 = 0$$ $$2x(x-5) - 3(x-5) = 0$$ $$(2x-3)(x-5) = 0$$ Thus, the values of x for which the denominator is zero (and the expression is undefined) are 3/2 and 5. Therefore, the correct answer is option A: 5 or 3/2.

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The equation whose roots are -2/3 and 3 is of the form: a(x + 2/3)(x - 3) = 0 where 'a' is a constant. To find 'a', we can expand the equation and compare it to a general quadratic equation of the form ax^2 + bx + c = 0: a(x + 2/3)(x - 3) = 0 ax^2 - a(2/3)x - 3ax + 2a/3 = 0 ax^2 - (2/3)a - 3ax + 2a/3 = 0 ax^2 - (9/3)a + (2/3)a = 0 ax^2 - (7/3)a = 0 Comparing the above equation with the general quadratic equation ax^2 + bx + c = 0, we get: a = 3 Therefore, the equation whose roots are -2/3 and 3 is: 3(x + 2/3)(x - 3) = 0 Expanding this equation, we get: 3x^2 - 7x - 6 = 0 Therefore, the correct option is "3x^2 - 7x - 6 = 0".

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We can start by expanding the brackets and simplifying the equation. 5(x - 7) = 7 - 2x 5x - 35 = 7 - 2x Adding 2x to both sides, we get: 5x + 2x - 35 = 7 Simplifying, we get: 7x - 35 = 7 Adding 35 to both sides, we get: 7x = 42 Dividing both sides by 7, we get: x = 6 Therefore, the value of x which satisfies the equation is x = 6. Answer: x = 6

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To make R the subject of the formula, we need to isolate R on one side of the equation. Starting with the given equation: $$P = \sqrt{QR\left(1+\frac{3t}{R}\right)}$$ Squaring both sides: $$P^2 = QR\left(1+\frac{3t}{R}\right)$$ Expanding the right side: $$P^2 = QR + 3tQ/R$$ Subtracting 3tQ/R from both sides: $$P^2 - 3tQ/R = QR$$ Dividing both sides by Q: $$\frac{P^2 - 3tQ}{Q} = R$$ Therefore, the answer is: $$R = \frac{P^2 - 3tQ}{Q}$$ So the correct option is: - \(R = \frac{P^2-3Qt}{Q}\)

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A regular polygon is a polygon with all sides and angles equal. Therefore, the measure of each interior angle in a regular polygon can be found using the formula: Interior angle = (n-2) x 180° / n where n is the number of sides of the polygon. Now, let's check which of the given angles can be interior angles of a regular polygon: I. 108° = (n-2) x 180° / n --> n = 5 II. 116° = (n-2) x 180° / n --> n ≈ 7.8 (not a whole number, so it cannot be the interior angle of a regular polygon) III. 120° = (n-2) x 180° / n --> n = 5 Therefore, the answer is (II only).

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A cyclic quadrilateral is a quadrilateral whose vertices all lie on a single circle. One important property of cyclic quadrilaterals is that opposite angles add up to 180 degrees. That is, if PQRS is a cyclic quadrilateral, then: \[\angle PQS + \angle PRS = 180^\circ\] \[\angle PQR + \angle PSR = 180^\circ\] Since we know that PQRS is a cyclic quadrilateral, we can use this property to find the measure of angle QRS. Let x be the measure of angle QRS. Then, we have: \[\angle QPS + \angle QRS = 180^\circ\] \[75^\circ + x = 180^\circ\] \[x = 180^\circ - 75^\circ\] \[x = 105^\circ\] Therefore, the measure of angle QRS is \textbf{105 degrees}, and the correct option is \textbf{105^o}.}

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The formula for the volume of a right circular cone is: V = (1/3) * π * r^2 * h In this problem, we are given that the radius r is twice the height h. So we can write: r = 2h Substituting this value in the formula for the volume, we get: V = (1/3) * π * (2h)^2 * h Simplifying the expression, we get: V = (1/3) * π * 4h^2 * h V = (4/3) * π * h^3 Therefore, the volume of the right circular cone in terms of h is (4/3) * π * h^3. Hence, the correct option is: - \(\frac{4}{3}\pi h^3\)

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Sum of interior angle of the diagram equals 360o

180o - 5yo + 136o + 180o + 180o - 3yo = 360o

-8yo + 136o = 0

-8yo = -136;

y = 17

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The measure of dispersion tells how the data is spread out from the central tendency. Mean and Standard deviation are both measures of dispersion. Range and Mean deviation are also measures of dispersion. Therefore, the answer is Mean.

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Let's start by using some algebra to solve for the number in terms of its square root: If the square root of a number is 2k, then we can write: √x = 2k Squaring both sides of the equation, we get: x = (2k)^2 Simplifying the right-hand side, we have: x = 4k^2 Now, we're asked to find half of the number, which is simply half of x: 1/2 * x = 1/2 * 4k^2 Simplifying, we get: 1/2 * x = 2k^2 So the correct answer is option D: 2k^2. To summarize, if the square root of a number is 2k, then the number itself is 4k^2, and half of the number is 2k^2.

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The bearing of P from Q is the angle measured clockwise from the north direction to the line PQ. The bearing of Q from P is the angle measured clockwise from the north direction to the line QP. Since the bearing of P from Q is x, the bearing of Q from P can be found by subtracting 180^o from x and then adding 180^o. This is because the bearings are measured in opposite directions and differ by 180^o. Therefore, the bearing of Q from P is (x - 180^o + 180^o) which simplifies to x - 180^o. Thus, the answer is (x - 180)^o.

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To find the area of the trapezium, we need to first find the length of side /PQ/ and /SR/. Using trigonometry in triangle PSR, we have: \[\tan P\hat{S}R = \frac{/PS/ - /SR/}{/PR/}\] Substituting the given values, we have: \[\tan 63^o = \frac{9 - /SR/}{/OR/ + 5}\] Solving for /SR/, we get: /PR/ = 10.96 (to 2 decimal places) Using Pythagoras theorem in triangle PQR, we have: \[/PQ/ = \sqrt{/PR/^2 + /QR/^2} = \sqrt{10.96^2 + 9^2} \approx 13.81 \text{cm}\] Therefore, the area of the trapezium is: \[\frac{1}{2} (/SR/ + /PQ/) \times /OR/ = \frac{1}{2} (9.19 + 13.81) \times 5 \approx 62.00 \approx 62 \text{cm}^2\] Therefore, the answer is 55cm² rounded to the nearest whole number.

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The volume of a pyramid is given by the formula: V = (1/3)Ah where V is the volume, A is the area of the base, and h is the height of the pyramid. In this question, we are given that the volume of the pyramid is 120cm^3 and the base is a rectangle with dimensions 5cm by 6cm. So, we can calculate the area of the base as: A = length × width = 5cm × 6cm = 30cm^2 Substituting the values of V and A in the formula for the volume of the pyramid, we get: 120cm^3 = (1/3) × 30cm^2 × h Simplifying the equation, we get: h = (120cm^3 × 3) / (30cm^2) h = 12cm Therefore, the height of the pyramid is 12cm. So, the correct option is "(3) 12cm".

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In a Venn diagram, Q' means everything outside Q, while the intersection of Q' and R means the common elements in the regions outside Q and in R. Thus, Q' ? R is the region(s) marked in R but not in Q. From the diagram, we see that only regions (c), (g), and (h) are in R but not in Q. Therefore, Q' ? R = (c, g, h). So the correct option is (c, g, h).

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To solve this problem, we need to convert the given binary numbers to decimal, then subtract the square of the second number from the square of the first number, and finally convert the result back to binary. (111_two)² in decimal is equal to (7)² = 49. (101_two)² in decimal is equal to (5)² = 25. Therefore, (111_two)² - (101_two)² = 49 - 25 = 24. Finally, we convert 24 to binary, which is 11000_two. Therefore, the answer is option (D) 11000_two.

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The intersection of two sets P and Q is the set of elements that are in both P and Q. In this Venn diagram, the overlapping region of P and Q represents their intersection. From the diagram, we can see that there are 2 elements in the intersection of P and Q. Therefore, the answer is 2.

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To solve this problem, we need to first understand what \(4P4_5\) means. Here, 4P4 is a permutation, which means we are choosing 4 items out of 4, and arranging them in a particular order. The subscript 5 means we are working in base 5. We can calculate the value of \(4P4_5\) as follows: \(4P4_5 = 4! = 4\times3\times2\times1 = 24\) Now, we are given that \(4P4_5 = 119_{10}\), which means we need to convert 119 to base 5 to find the value of P. 119 in base 5 is equal to 1444. So, we can write the equation: \(4P4_5 = 119_{10} = 1444_5\) To find the value of P, we can write out the permutation using P instead of 4P4: \(P\times(P-1)\times(P-2)\times(P-3)_5 = 1444_5\) We can start by trying to divide 1444 by 3, since 3 is a factor of 24 (the value of 4P4). \(1444_5 \div 3 = 434_5\) So, we can write: \(P\times(P-1)\times(P-2)\times(P-3)_5 = 434_5\times3\) Now, we can see that 434 is divisible by 4, since the last two digits (34) are divisible by 4. \(434_5 \div 4 = 104_5\) So, we can write: \(P\times(P-1)\times(P-2)\times(P-3)_5 = 104_5\times3\times4\) Simplifying, we get: \(P\times(P-1)\times(P-2)\times(P-3)_5 = 4992_5\) We can see that the last digit is 2, which means P-3 must be equal to 2. So, P = 5 - 2 = 3. Therefore, the value of P is 3.

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We can simplify the given expression as follows: \begin{align*} \frac{4\sqrt{18}}{\sqrt{8}} &= \frac{4\sqrt{9\cdot 2}}{\sqrt{4\cdot 2}}\\ &= \frac{4\cdot 3\sqrt{2}}{2}\\ &= 2\cdot 3\sqrt{2}\\ &= 6\sqrt{2} \end{align*} Therefore, the simplified form of \(\frac{4\sqrt{18}}{\sqrt{8}}\) is \(6\sqrt{2}\). So the correct option is (c) 6.

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Let the cost of a notebook be N and the cost of a pen be P. From the first transaction, we have: 3N + 1P = 35 ...(1) From the second transaction, we have: 2N + 2P = 30 ...(2) We want to find the value of P. We can rearrange equation (2) as follows: 2P = 30 - 2N P = (30 - 2N)/2 Substituting this value of P into equation (1), we get: 3N + 1[(30 - 2N)/2] = 35 Multiplying both sides by 2, we get: 6N + (30 - 2N) = 70 Simplifying the expression, we get: 4N = 40 N = 10 Substituting this value of N into equation (2), we get: 2(10) + 2P = 30 Simplifying the expression, we get: 2P = 10 P = 5 Therefore, the cost of a pen is N5.00. Answer: N5.00

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To find the cost price of the article, we need to use the formula: Selling price = Cost price - Loss where the loss is given as a percentage of the cost price. In this case, the loss is 15%. Let's first convert the loss percentage to a decimal: 15% = 0.15 Now we can substitute the given values into the formula: N595 = Cost price - 0.15(Cost price) Simplifying this equation, we get: N595 = Cost price - 0.15Cost price N595 = 0.85Cost price To solve for the cost price, we can divide both sides of the equation by 0.85: Cost price = N595 ÷ 0.85 Cost price = N700 Therefore, the cost price of the article is N700.00.

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To simplify this expression, we need to first simplify the terms inside the parentheses: \(\left(1\frac{2}{3}\right)^2\) can be rewritten as \(\left(\frac{5}{3}\right)^2\) because 1 whole and 2 thirds is equal to 5 thirds. Similarly, \(\left(\frac{2}{3}\right)^2\) can be simplified to \(\frac{4}{9}\). Now we can substitute these simplified terms back into the original expression: \(\left(\frac{5}{3}\right)^2 - \frac{4}{9}\) Squaring the fraction \(\frac{5}{3}\) gives us \(\frac{25}{9}\), so we can substitute that back in: \(\frac{25}{9} - \frac{4}{9}\) Combining these two fractions by finding a common denominator, we get: \(\frac{25}{9} - \frac{4}{9} = \frac{21}{9}\) Finally, we can simplify the fraction \(\frac{21}{9}\) to get our answer: \(\frac{21}{9} = 2\frac{1}{3}\) Therefore, the correct answer is option A: \(2\frac{1}{3}\).

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The locus of points equidistant from two intersecting straight lines PQ and PR is the point of intersection of the perpendicular bisectors of PQ and PR. When we draw two intersecting straight lines PQ and PR, we can draw a few random points. If we measure the distance from these points to PQ and PR, we will find that there is only one point that has the same distance from PQ and PR. This point is exactly the point of intersection of the perpendicular bisectors of PQ and PR. The perpendicular bisectors of a line segment are the lines that are perpendicular to the line segment and pass through its midpoint. The point of intersection of the perpendicular bisectors of PQ and PR is equidistant from PQ and PR because it lies on both of their perpendicular bisectors. Therefore, the locus of points equidistant from two intersecting straight lines PQ and PR is the point of intersection of the perpendicular bisectors of PQ and PR. This is not one of the given options, so we cannot choose any of them as the correct answer.

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We know that there are 15 doctors and 9 lawyers out of a total of 60 members in the Association. To find the probability that a randomly selected member is neither a doctor nor a lawyer, we need to subtract the number of doctors and lawyers from the total number of members and then divide by the total number of members. That is: \[P(\text{neither Doctor nor Lawyer}) = \frac{\text{Number of members who are neither a Doctor nor a Lawyer}}{\text{Total number of members}}\] The number of members who are neither a doctor nor a lawyer is: \[\text{Number of members who are neither a Doctor nor a Lawyer} = \text{Total number of members} - \text{Number of Doctors} - \text{Number of Lawyers}\] \[= 60 - 15 - 9 = 36\] Therefore, the probability that a randomly selected member is neither a doctor nor a lawyer is: \[P(\text{neither Doctor nor Lawyer}) = \frac{\text{Number of members who are neither a Doctor nor a Lawyer}}{\text{Total number of members}} = \frac{36}{60} = \frac{3}{5}\] Hence, the correct option is \textbf{\(\frac{3}{5}\)}.

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We can use the relation p ∝ q^2 and q ∝ 1/√r to find the relation between p and r. Substituting the value of q in terms of r in the relation p ∝ q^2, we get: p ∝ (1/√r)^2 p ∝ 1/r Therefore, p varies inversely as r. So, the answer is: "p varies inversely as r".

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Distance flown due west \(=\text{speed}\times\text{time}=600\times3=1800\) km, along the parallel of latitude \(50^{\circ}N\).

(a) Longitude of Q. Distance along a parallel \(=\dfrac{\theta}{360}\times2\pi R\cos(\text{lat})\), where \(\theta\) is the change in longitude.

\[1800=\frac{\theta}{360}\times2\times\frac{22}{7}\times6400\times\cos50^{\circ}\]

\(2\times\tfrac{22}{7}\times6400=40228.6\); \(\cos50^{\circ}=0.6428\); product \(=25859\).

\[\theta=\frac{1800\times360}{25859}\approx 25.1^{\circ}\]

Flying due west from long. \(60^{\circ}W\) increases the western longitude: \(60+25.1=85.1^{\circ}W\).

Longitude of Q \(\approx 85.1^{\circ}W\) (3 s.f.).

(b) Latitude of Y. Flying due south is along a meridian: distance \(=\dfrac{\phi}{360}\times2\pi R\).

\[500=\frac{\phi}{360}\times40228.6\Rightarrow \phi=\frac{500\times360}{40228.6}\approx 4.47^{\circ}\]

Moving south from \(50^{\circ}N\): latitude of \(Y=50-4.47\approx 45.5^{\circ}N\).

Latitude of Y \(\approx 45.5^{\circ}N\) (3 s.f.).

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(a) Write \(625=5^{4}\) and \(25=5^{2}\).

\[625^{\frac{3}{8}}\times5^{\frac{1}{2}}\div25=(5^{4})^{\frac{3}{8}}\times5^{\frac{1}{2}}\div5^{2}=5^{\frac{12}{8}}\times5^{\frac{1}{2}}\times5^{-2}\]

Add the indices: \(\dfrac{12}{8}+\dfrac{1}{2}-2=\dfrac{3}{2}+\dfrac{1}{2}-2=2-2=0\).

\[=5^{0}=1\]

\(=1\)

(b)(i) \(\tan(\theta+25)^{\circ}=5.145\).

\((\theta+25)^{\circ}=\tan^{-1}5.145=79.0^{\circ}\), so \(\theta=79.0-25=54.0^{\circ}\).

\(\theta\approx 54.0^{\circ}\)

(ii) \(5\cos\theta-1=0\Rightarrow \cos\theta=\dfrac{1}{5}=0.2\).

\(\theta=\cos^{-1}0.2=78.5^{\circ}\) (within \(0^{\circ}\le\theta\le90^{\circ}\)).

\(\theta\approx 78.5^{\circ}\)

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(a) Copy and complete the table

Using \(y = 11 - 2x - 2x^{2}\), build the value from its parts \(x^{2}\), \(11\), \(-2x\) and \(-2x^{2}\) for each \(x\):

\(x=-3:\ 11-2(-3)-2(9)=11+6-18=-1\)
\(x=-2:\ 11+4-8=7\)
\(x=-1:\ 11+2-2=11\)
\(x=1:\ 11-2-2=7\)
\(x=2:\ 11-4-8=-1\)
\(x=3:\ 11-6-18=-13\)

\(x\)	-4	-3	-2	-1	0	1	2	3
\(x^{2}\)	16	9	4	1	0	1	4	9
\(11\)	11	11	11	11	11	11	11	11
\(-2x\)	8	6	4	2	0	-2	-4	-6
\(-2x^{2}\)	-32	-18	-8	-2	0	-2	-8	-18
\(y\)	-13	-1	7	11	11	7	-1	-13

(b) The graph of \(y = 11 - 2x - 2x^{2}\)

Plotting the eight points \((-4,-13),(-3,-1),(-2,7),(-1,11),(0,11),(1,7),(2,-1),(3,-13)\) with a scale of 2 cm to 1 unit on the \(x\)-axis and 2 cm to 5 units on the \(y\)-axis, and joining them with a smooth curve, gives a downward (maximum) parabola with its peak midway between \(x=-1\) and \(x=0\), at about \((-0.5,\ 11.5)\).

(c) Using the graph

(i) Roots of \(11 - 2x - 2x^{2} = 0\). These are the \(x\)-values where the curve crosses the \(x\)-axis (\(y=0\)). Reading off the graph:

\[x \approx -2.9 \quad \text{or} \quad x \approx 1.9\]

(ii) Values of \(x\) for which \(3 - 2x - 2x^{2} = 0\). Subtracting this equation from the curve:

\[y - (3 - 2x - 2x^{2}) = (11 - 2x - 2x^{2}) - (3 - 2x - 2x^{2}) = 8\]

So where \(3 - 2x - 2x^{2}=0\) the curve has \(y = 8\). Drawing the horizontal line \(y = 8\) (shown dashed) and reading where it meets the curve:

\[x \approx -1.8 \quad \text{or} \quad x \approx 0.8\]

(iii) Gradient of the curve at \(x = 1\). Draw the tangent to the curve at the point \((1,\ 7)\) (shown as the straight line touching the curve). Taking two points on this tangent, e.g. \((0,\ 13)\) and \((2,\ 1)\):

\[\text{gradient} = \frac{1 - 13}{2 - 0} = \frac{-12}{2} = -6\]

Hence the gradient of the curve at \(x = 1\) is \(-6\).

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Given data.

No. of suitcases (x)	0	1	2	3	4	5
Travellers (f)	2	7	7	2	3	9

Total number of travellers: \(\sum f = 2+7+7+2+3+9 = 30\).

(a)(i) Median

With \(N = 30\) values (even), the median is the mean of the 15th and 16th values. Building the cumulative frequencies:

x	0	1	2	3	4	5
Cumulative f	2	9	16	18	21	30

Positions 10 to 16 correspond to \(x = 2\); hence both the 15th and 16th values equal 2.

\[\text{Median} = \frac{2+2}{2} = 2 \text{ suitcases.}\]

(a)(ii) Mean

\[\sum fx = (0)(2)+(1)(7)+(2)(7)+(3)(2)+(4)(3)+(5)(9) = 0+7+14+6+12+45 = 84.\]\[\bar{x} = \frac{\sum fx}{\sum f} = \frac{84}{30} = 2.8 \approx 3 \text{ (nearest whole number).}\]

(b) Bar chart

The number of suitcases (0 to 5) is placed on the horizontal axis and the number of travellers on the vertical axis. Each bar has equal width with equal gaps, and the heights are 2, 7, 7, 2, 3 and 9 respectively.

The tallest bar occurs at 5 suitcases, with a height of 9 travellers.

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The percentages account for the whole salary, and they sum to

\[5+25+40+10+12.5+7.5 = 100.\]

(a) Pie chart

The whole circle is \(360^\circ\), representing \(100\%\). Each item's sector angle is its percentage multiplied by \(\dfrac{360^\circ}{100} = 3.6^\circ\) per unit:

Item	Percentage (%)	Sector angle
Pension	5	\(\frac{5}{100}\times360^\circ = 18^\circ\)
Income Tax	25	\(\frac{25}{100}\times360^\circ = 90^\circ\)
Food	40	\(\frac{40}{100}\times360^\circ = 144^\circ\)
Transport	10	\(\frac{10}{100}\times360^\circ = 36^\circ\)
Rent	12.5	\(\frac{12.5}{100}\times360^\circ = 45^\circ\)
Others	7.5	\(\frac{7.5}{100}\times360^\circ = 27^\circ\)
Total	100	\(360^\circ\)

Check: \(18^\circ+90^\circ+144^\circ+36^\circ+45^\circ+27^\circ = 360^\circ.\) The completed pie chart is:

(b) Salary calculations (gross monthly salary \(=\) N10,800.00)

(i) Pension contribution \(= 5\%\) of N10,800, i.e. the \(18^\circ\) sector:

\[\frac{5}{100}\times 10800 \;=\; \frac{18}{360}\times 10800 \;=\; \text{N}540.00.\]

(ii) Income tax \(= 25\%\) of N10,800, i.e. the \(90^\circ\) sector:

\[\frac{25}{100}\times 10800 \;=\; \frac{90}{360}\times 10800 \;=\; \text{N}2{,}700.00.\]

(c) Take-home pay

Deduct the pension and income tax from the gross salary. Together they are \((5+25)\% = 30\%\), so the take-home pay is \(100\% - 30\% = 70\%\):

\[10800 - 540 - 2700 \;=\; \frac{70}{100}\times 10800 \;=\; \text{N}7{,}560.00.\]

The employee's take-home pay is N7,560.00.

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(a) Simplify \(\sqrt{50}-3\sqrt{2}(2\sqrt{2}-5)-5\sqrt{32}\).

\(\sqrt{50}=5\sqrt{2}\) and \(\sqrt{32}=4\sqrt{2}\), so \(5\sqrt{32}=20\sqrt{2}\).

Expand the bracket: \(3\sqrt{2}(2\sqrt{2}-5)=6(\sqrt{2}\cdot\sqrt{2})-15\sqrt{2}=6(2)-15\sqrt{2}=12-15\sqrt{2}\).

Combine:

\[5\sqrt{2}-(12-15\sqrt{2})-20\sqrt{2}=5\sqrt{2}-12+15\sqrt{2}-20\sqrt{2}=(5+15-20)\sqrt{2}-12=-12\]

\(=-12\)

(b) \(\tfrac{1}{2}\log_{10}\tfrac{25}{4}-2\log_{10}\tfrac{4}{5}+\log_{10}\tfrac{320}{125}\).

Bring the coefficients inside as powers: \(\tfrac{1}{2}\log\tfrac{25}{4}=\log\left(\tfrac{25}{4}\right)^{1/2}=\log\tfrac{5}{2}\); \(\;2\log\tfrac{4}{5}=\log\tfrac{16}{25}\).

\[\log\frac{5}{2}-\log\frac{16}{25}+\log\frac{320}{125}=\log\!\left(\frac{5}{2}\times\frac{25}{16}\times\frac{320}{125}\right)\]

\(\dfrac{5}{2}\times\dfrac{25}{16}=\dfrac{125}{32}\); then \(\dfrac{125}{32}\times\dfrac{320}{125}=\dfrac{320}{32}=10\).

\[=\log_{10}10=1\]

\(=1\)

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(a) \(p\propto r^{2}\Rightarrow p=kr^{2}\).

Using \(p=3.2,\ r=4\): \(3.2=k(16)\Rightarrow k=0.2\).

When \(r=6.5\): \(p=0.2(6.5)^{2}=0.2(42.25)=8.45\).

\(p=8.45\)

(b) Clear the fractions.

\(\dfrac{x}{2}+\dfrac{y}{4}=1\) \(\times4\Rightarrow 2x+y=4\) ... (1)

\(\dfrac{x}{3}-\dfrac{y}{4}=-\dfrac{1}{6}\) \(\times12\Rightarrow 4x-3y=-2\) ... (2)

From (1): \(y=4-2x\). Substitute into (2):

\[4x-3(4-2x)=-2\Rightarrow 4x-12+6x=-2\Rightarrow 10x=10\Rightarrow x=1\]

Then \(y=4-2(1)=2\).

\(x=1,\ y=2\)

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This is a ruler-and-compasses construction (the parallel-sides condition \(PQ\parallel XY\) fixes the shape). Keep all arcs visible.

(a) Construct quadrilateral PXYQ.

1. Draw \(PX=9.9\) cm.
2. At \(P\), construct \(\angle QPX=75^{\circ}\) (construct \(60^{\circ}\), then bisect the \(60^{\circ}\)-\(90^{\circ}\) gap to add \(15^{\circ}\)).
3. With centre \(X\), radius \(10.2\) cm, cut the \(75^{\circ}\) arm from \(P\) at \(Q\) (so \(QX=10.2\) cm).
4. Through \(X\) draw a line parallel to \(PQ\) (copy the angle at \(P\) to \(X\), or use equal-distance offsets). With centre \(Q\), radius \(10.4\) cm, cut this parallel line at \(Y\) (so \(QY=10.4\) cm).
5. Join \(QY\), \(XY\) and \(PQ\) to complete \(PXYQ\).

(b)(i) Locus \(l_{1}\): points equidistant from \(X\) and \(Y\) form the perpendicular bisector of \(XY\); construct it.

(ii) Locus \(l_{2}\): points equidistant from lines \(QY\) and \(YX\) lie on the bisector of \(\angle QYX\); construct it.

(c) Mark \(M\), the intersection of \(l_{1}\) and \(l_{2}\).

(d) Measure \(|PM|\) with the ruler and record the value (of the order of \(9\text{ - }11\) cm depending on your accurate drawing).

Antwortdetails

(a) P, Q, R, S lie on the circle, PQT is a straight line (T on PQ produced beyond Q), and \(SQ\parallel RT\). Join QR.

(i) Show that \(\angle RPS=\angle QRT\).

• \(\angle RPS\) and \(\angle RQS\) both stand on the same chord \(RS\) and P, Q are in the same segment, so \(\angle RPS=\angle RQS\) (angles in the same segment).
• \(SQ\parallel RT\) with QR as transversal gives \(\angle RQS=\angle QRT\) (alternate angles).

Therefore \(\angle RPS=\angle QRT\), as required.

(ii) Show that \(\angle PRS=\angle QTR\).

• \(\angle PRS\) and \(\angle PQS\) both stand on the same chord \(PS\) with R, Q in the same segment, so \(\angle PRS=\angle PQS\) (angles in the same segment).
• The straight line PQT is a transversal cutting the parallel lines \(SQ\) and \(RT\) at Q and T, so \(\angle PQS=\angle QTR\) (corresponding angles).

Therefore \(\angle PRS=\angle QTR\), as required.

(b) In triangle ABC, AB is produced to D and AC is produced to E, with \(\angle DBC=132^\circ\) and the exterior angle at C equal to \(96^\circ\).

\(\angle DBC\) is the exterior angle at B, and it is the supplement of the interior angle \(\angle ABC\) (angles on the straight line ABD):

\[\angle ABC=180^\circ-132^\circ=48^\circ\]

The exterior angle at C is the supplement of the interior angle \(\angle ACB\) (angles on the straight line ACE):

\[\angle ACB=180^\circ-96^\circ=84^\circ\]

The three interior angles of triangle ABC sum to \(180^\circ\):

\[\angle BAC=180^\circ-48^\circ-84^\circ=48^\circ\]

Hence \(\angle BAC=\angle ABC=48^\circ\). Since two angles are equal, the sides opposite them are equal, i.e. \(BC=AC\). Therefore triangle ABC is isosceles.

Antwortdetails

(a) Completed frequency distribution

Sorting the 40 marks into the given class boundaries and tallying each mark:

Class Boundaries	Tally	Frequency
59.5 - 64.5	||	2
64.5 - 69.5	|||	3
69.5 - 74.5	|||| |	6
74.5 - 79.5	|||| |||| |	11
79.5 - 84.5	|||| |||	8
84.5 - 89.5	|||| |	6
89.5 - 94.5	||	2
94.5 - 99.5	||	2
Total		40

(Check: \(2+3+6+11+8+6+2+2 = 40\) students.)

(b) Histogram

Since all class widths are equal (5 marks), the bar heights equal the frequencies. The class boundaries are plotted on the horizontal axis and frequency on the vertical axis, with the bars drawn adjoining (no gaps):

(c) Modal mark from the histogram

The modal class is 74.5 - 79.5, the tallest bar (frequency 11). To read the mode, straight lines are drawn inside the modal bar: one from its top-left corner to the top-left corner of the next bar, and one from its top-right corner to the top-right corner of the previous bar. The vertical through their point of intersection meets the mark axis at the mode (shown dashed above), giving mode \(\approx 77.6\).

This agrees with the standard formula \[\text{Mode} = L + \left(\frac{f_1 - f_0}{(f_1-f_0)+(f_1-f_2)}\right)c,\] where \(L = 74.5,\ f_1 = 11,\ f_0 = 6,\ f_2 = 8,\ c = 5\): \[\text{Mode} = 74.5 + \left(\frac{11-6}{(11-6)+(11-8)}\right)\times 5 = 74.5 + \frac{5}{8}\times 5 = 74.5 + 3.125 \approx 77.6.\] So the modal mark is approximately 77.6 (about 78 marks).

(d) Probability of a mark greater than 79

Marks greater than 79 fall in the four highest classes (79.5 and above), giving \[8 + 6 + 2 + 2 = 18 \text{ students.}\] With 40 students in all, \[P(\text{mark} > 79) = \frac{18}{40} = \frac{9}{20} = 0.45.\]

Antwortdetails

(a) Simplify the compound fraction

\[\frac{\tfrac{1}{3}c^{2} - \tfrac{2}{3}cd}{\tfrac{1}{2}d^{2} - \tfrac{1}{4}cd}\]

Factorise numerator and denominator.

\[\text{Numerator} = \tfrac{1}{3}c^{2} - \tfrac{2}{3}cd = \tfrac{1}{3}c\,(c - 2d)\]\[\text{Denominator} = \tfrac{1}{2}d^{2} - \tfrac{1}{4}cd = \tfrac{1}{4}d\,(2d - c)\]

Write \((c - 2d) = -(2d - c)\):

\[\frac{\tfrac{1}{3}c\,(c-2d)}{\tfrac{1}{4}d\,(2d-c)} = \frac{\tfrac{1}{3}c\,[-(2d-c)]}{\tfrac{1}{4}d\,(2d-c)} = -\frac{\tfrac{1}{3}c}{\tfrac{1}{4}d}\]\[= -\frac{1}{3}c \times \frac{4}{d} = \boxed{-\frac{4c}{3d}}\]

(b) Two right-angled triangles on the straight line \(YPF\)

From the diagram, \(YPF\) is a straight line, \(\angle XYP = \angle MFP = 90^\circ\), \(\angle XPY = 44^\circ\), \(\angle MPF = 46^\circ\), \(|XY| = 7\text{ cm}\) and \(|MP| = 9\text{ cm}\).

(i) Calculate \(|XM|\) and \(|YF|\).

Triangle \(XYP\) (right-angled at \(Y\)):

\[\tan 44^\circ = \frac{|XY|}{|YP|} \Rightarrow |YP| = \frac{7}{\tan 44^\circ} = \frac{7}{0.9657} = 7.249\text{ cm}\]\[\sin 44^\circ = \frac{|XY|}{|XP|} \Rightarrow |XP| = \frac{7}{\sin 44^\circ} = \frac{7}{0.6947} = 10.08\text{ cm}\]

Triangle \(MPF\) (right-angled at \(F\), hypotenuse \(MP = 9\)):

\[|PF| = 9\cos 46^\circ = 9(0.6947) = 6.252\text{ cm}\]

Length \(YF\): since \(Y\), \(P\), \(F\) are collinear,

\[|YF| = |YP| + |PF| = 7.249 + 6.252 = 13.5\text{ cm (3 s.f.)}\]

Length \(XM\): at \(P\) the angles on the straight line give

\[\angle XPM = 180^\circ - 44^\circ - 46^\circ = 90^\circ\]

so triangle \(XPM\) is right-angled at \(P\) with \(|XP| = 10.08\) and \(|PM| = 9\):

\[|XM| = \sqrt{|XP|^{2} + |PM|^{2}} = \sqrt{10.08^{2} + 9^{2}} = \sqrt{101.5 + 81} = \sqrt{182.5} = 13.5\text{ cm (3 s.f.)}\]

(ii) Find \(\angle XMP\). In the right-angled triangle \(XPM\) (right angle at \(P\)),

\[\tan(\angle XMP) = \frac{|XP|}{|PM|} = \frac{10.08}{9} = 1.1196\]\[\angle XMP = \tan^{-1}(1.1196) = 48.2^\circ\]

Hence \(|XM| \approx 13.5\text{ cm}\), \(|YF| \approx 13.5\text{ cm}\) and \(\angle XMP \approx 48.2^\circ\).

Antwortdetails

(a) \(U=\{1,2,3,4,5,6,7,8,9,10\}\), \(A=\{1,2,5,7\}\), \(B=\{1,3,6,7\}\).

(i) \(A'=U\setminus A=\{3,4,6,8,9,10\}\).

(ii) \(A\cap B=\{1,7\}\), so \((A\cap B)'=\{2,3,4,5,6,8,9,10\}\).

(iii) \(A\cup B=\{1,2,3,5,6,7\}\), so \((A\cup B)'=\{4,8,9,10\}\).

(iv) The three-element subsets of \(B=\{1,3,6,7\}\) are \(\{1,3,6\},\ \{1,3,7\},\ \{1,6,7\},\ \{3,6,7\}\).

(b) The sequence \(\dfrac{2}{1\times3},\dfrac{3}{2\times4},\dfrac{4}{3\times5},\dfrac{5}{4\times6},\ldots\) has \(n\)th term \(\dfrac{n+1}{n(n+2)}\).

\[\text{15th term}=\frac{15+1}{15(15+2)}=\frac{16}{15\times17}=\frac{16}{255}\]

(c) A.P. with \(a=3,\ d=4\).

(i) \(n\)th term \(=a+(n-1)d=3+(n-1)4=4n-1\).

(ii) Least term greater than 100: \(4n-1>100\Rightarrow 4n>101\Rightarrow n>25.25\), so \(n=26\).

\[T_{26}=4(26)-1=103\]

Least term above 100 is \(103\).