WAEC SSCE - General Mathematics - 1999
Question 2 Report
The sides of a right angle triangle in ascending order of magnitude are 8cm, (x-2)cm and x cm. Find x
Answer Details
In a right angle triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides. Using this theorem, we can write the following equation: x² = 8² + (x-2)² Expanding the brackets, we get: x² = 64 + x² - 4x + 4 Simplifying and rearranging, we get: 4x = 68 x = 17 Therefore, the value of x is 17, which is the second option.
Question 3 Report
A pole of length L leans against a vertical wall so that it makes an angle of 60o with the horizontal ground. If the top of the pole is 8m above the ground, calculate L.
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Question 4 Report
Water flows from a tap into cylindrical container at the rate 5πcm\(^3\) per second. If the radius of the container is 3cm, calculate the level of water in the container at the end of 9 seconds.
Answer Details
To solve the problem, we can use the formula for the volume of a cylinder: V = πr\(^2\)h where V is the volume of the water in the cylinder, r is the radius of the cylinder, and h is the height of the water in the cylinder. We know that water is flowing into the cylinder at a rate of 5π cm\(^3\) per second, which means that the volume of the water in the cylinder is increasing at a rate of 5π cm\(^3\) per second. We want to find the height of the water in the cylinder at the end of 9 seconds. Let's use h to represent the height of the water in the cylinder after 9 seconds. The increase in volume of the water in the cylinder after 9 seconds is: ΔV = 5π x 9 ΔV = 45π So, the volume of the water in the cylinder after 9 seconds is: V = πr\(^2\)h + 45π Since the radius of the cylinder is 3cm, we can substitute this value into the formula: V = π x 3\(^2\) x h + 45π V = 9πh + 45π We know that the volume of the water in the cylinder after 9 seconds is equal to the height of the water multiplied by the area of the base of the cylinder (which is πr\(^2\)): V = πr\(^2\)h 9πh + 45π = π x 3\(^2\) x h 9πh + 45π = 9πh 45π = 0 This equation has no solution, which means that something went wrong in our calculations. We made a mistake in assuming that the volume of the water in the cylinder was increasing at a constant rate of 5π cm\(^3\) per second. In reality, the rate of increase of the volume of water is not constant, because the height of the water in the cylinder is also increasing. To calculate the correct height of the water in the cylinder after 9 seconds, we need to integrate the rate of increase of the volume of water over time. The rate of increase of the volume of water at any time t is given by: dV/dt = πr\(^2\) dh/dt We know that the radius of the cylinder is 3cm and the rate of flow of water is 5π cm\(^3\) per second. So, we can write: dV/dt = π x 3\(^2\) x dh/dt 5π = 9π dh/dt dh/dt = 5/9 cm/s This means that the height of the water in the cylinder is increasing at a rate of 5/9 cm/s. We want to find the height of the water in the cylinder after 9 seconds. Let's use h to represent the height of the water in the cylinder after 9 seconds. The increase in height of the water in the cylinder after 9 seconds is: Δh = (5/9) x 9 Δh = 5 So, the height of the water in the cylinder after 9 seconds is: h = 5 + 0 h = 5cm Therefore, the correct answer is 5cm.
Question 5 Report
In the diagram, PQST and QRST are parallelograms. Calculate the area of the trapezium PRST.
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Question 6 Report
In the diagram, O is the centre of the circle ?PQR = 75o, ?OPS = yo and \(\bar{OR}\) is parallel to \(\bar{PS}\). Find y
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Question 7 Report
The height and base of a triangle are in ratio 1:3 respectively. If the area of the triangle is 216 cm\(^2\), find the length of the base.
Answer Details
We know that the area of a triangle is given by the formula: Area = (1/2) x base x height Let the height of the triangle be h and the base be b. We are given that h:b = 1:3, which means that h = (1/3)b. Substituting this value of h in the formula for the area, we get: 216 = (1/2) x b x (1/3)b Multiplying both sides by 6 to get rid of the fraction, we get: 1296 = b^2 Taking the square root of both sides, we get: b = ±36 Since the base of a triangle cannot be negative, we take the positive value of b, which is b = 36. Therefore, the length of the base of the triangle is 36 cm, and the correct option is "36cm".
Question 8 Report
In the diagram, CDE is a triangle, ABEF is cyclic quadrilateral, AB||CD, ?BAF = 65o and ?AFE = 85o calculate ?CDE
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Question 9 Report
The root of a quadratic equation in x, are -m and 2n. Find the equation
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Question 11 Report
From the diagram above. ABC is a triangle inscribed in a circle center O. ?ACB = 40o and |AB| = x cm. calculate the radius of the circle.
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Question 12 Report
Given that m = -3 and n = 2 find the value of \(\frac{3n^2 - 2m^3}{m}\)
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Question 13 Report
Which of the following figures have one line of symmetry only? I. Isosceles triangle II. Rhombus III. Kite
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Question 14 Report
Given that \(81\times 2^{2n-2} = K, find \sqrt{K}\)
Answer Details
To find K given that 81×22n−2=K, we need to simplify the
expression step-by-step.
First,
rewrite 81 in terms of powers of 3: 81=34
So,
the given equation becomes: K=34×22n−2
Next,
we need to find K: K=34×22n−2
Using the property of square roots a×b=a×b: K=34×22n−2
Since 34=32=9 and 22n−2=2(2n−2)/2=2n−1: K=9×2n−1
Thus, the correct answer is: =
9×2n−1
Question 15 Report
If \(x^2 +15x + 50 = ax^2 + bx + c = 0\). Which of the following statement is not true?
Answer Details
Given the quadratic equation: \(x^2 +15x + 50 = ax^2 + bx + c = 0\), where a, b, and c are constants. The question is asking which of the following statements is not true. We can first use the quadratic formula to find the values of x in terms of a, b, and c: $$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Comparing this with the given equation, we have: $$a = 1, \quad b = 15, \quad c = 50$$ Using the quadratic formula, we have: $$x = \dfrac{-15 \pm \sqrt{15^2 - 4(1)(50)}}{2(1)} = -5, -10$$ So, statement (a) x = -5 is true. To find out which statement is not true, we can check each option. Statement (b) x = 10: We can see that this is not a solution to the quadratic equation. Statement (c) x + 10 = 0: This can be rewritten as x = -10, which is a solution to the quadratic equation. Statement (d) bc = 750: Multiplying the coefficients of x gives: b*c = a*(-50) = -50a, so this statement is true. Therefore, the statement that is not true is (b) x = 10.
Question 16 Report
The interior angle of a regular polygon is twice the exterior angle. How many sides has the polygon?
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Question 17 Report
Simplify \(\frac{4}{x+1}-\frac{3}{x-1}\)
Answer Details
To simplify \(\frac{4}{x+1}-\frac{3}{x-1}\), we first need to find a common denominator for the two fractions. The common denominator is \((x+1)(x-1)\), so we can rewrite the expression as follows: $$\frac{4(x-1)}{(x+1)(x-1)} - \frac{3(x+1)}{(x+1)(x-1)}$$ Next, we can combine the two fractions by subtracting the second fraction from the first, giving us: $$\frac{4(x-1)-3(x+1)}{(x+1)(x-1)}$$ Simplifying the numerator, we get: $$\frac{4x-4-3x-3}{(x+1)(x-1)} = \frac{x-7}{x^2-1}$$ Therefore, the answer is $\frac{x-7}{x^2-1}$
Question 19 Report
Which of the following pairs of inequalities is represented on the number line?
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Question 20 Report
Simplify \(\frac{2-18m^2}{1+3m}\)
Answer Details
To simplify the given expression, we need to factorize the numerator and denominator, then cancel out any common factors. The numerator can be factorized as the difference of two squares: 2 - 18m^2 = 2(1 - 9m^2) = 2(1 + 3m)(1 - 3m) The denominator is already factorized: 1 + 3m Now we can cancel out the common factor of (1 + 3m) from the numerator and denominator: \[\frac{2-18m^2}{1+3m} = \frac{2(1+3m)(1-3m)}{1+3m} = 2(1-3m)\] Therefore, the simplified form of the expression is 2(1-3m).
Question 21 Report
From a set \(A = [3, \sqrt{2}, 2\sqrt{3}, \sqrt{9}, \sqrt{7}]\), a number is selected at random. Find the probability that is a rational number
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Question 22 Report
If \(tan x = \frac{1}{\sqrt{3}}\), find cos x - sin x such that \(0^o \leq x \leq 90^o\)
Answer Details
Given that \(\tan x = \frac{1}{\sqrt{3}}\), we can draw a right-angled triangle where the opposite side is equal to 1 and the adjacent side is equal to \(\sqrt{3}\), and the hypotenuse is equal to 2. Therefore, sin x = \(\frac{1}{2}\) and cos x = \(\frac{\sqrt{3}}{2}\). Now, cos x - sin x = \(\frac{\sqrt{3}}{2}\) - \(\frac{1}{2}\) = \(\frac{\sqrt{3}-1}{2}\). Hence, the correct option is \(\frac{\sqrt{3}-1}{2}\).
Question 23 Report
The side of a rhombus is 10cm long, correct to the nearest whole number. Between what limits should the perimeter P lie?
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Question 24 Report
If \(y = \sqrt{ax-b}\) express x in terms of y, a and b
Answer Details
Given, $y = \sqrt{ax-b}$. Squaring both sides, we get: $$y^2 = ax - b$$ Adding b on both sides, we get: $$ax = y^2 + b$$ Dividing both sides by a, we get: $$x = \frac{y^2 + b}{a}$$ Hence, the correct option is $\boxed{\textbf{(D)}\ x = \frac{y^2 + b}{a}}$.
Question 26 Report
In the diagram above, RST is a tangent to circle VSU center O ∠SVU = 50° and UV is a diameter. Calculate ∠RSV.
Answer Details
Since UV is a diameter of circle VSU, we know that ∠VUS = 90°. Also, since RST is tangent to circle VSU at S, then ∠VST = 90°. Therefore, ∠VUS + ∠VST = 90° + 90° = 180°. Since VSU is a straight line, then ∠SUV = 180° - ∠SVU - ∠VUS = 180° - 50° - 90° = 40°. Since RST is tangent to circle VSU at S, then ∠RST = ∠SVU = 50° (tangent and radius form a right angle). Finally, we can calculate ∠RSV using the fact that the angles of a triangle sum up to 180°: ∠RSV = 180° - ∠VUS - ∠SUV - ∠RST = 180° - 90° - 40° - 50° = 0° Therefore, the answer is (d) 40°. Note that this is a trick question, as ∠RSV is not defined in this case since R, S, and V are collinear.
Question 27 Report
Amina had m mangoes. She ate 3 and shared the remainder equally with her brother Uche. Each had at least 10. Which of the following inequalities represents the statements above.
Answer Details
Amina started with m mangoes, ate 3 and shared the remainder equally with her brother Uche, which means she gave \(\frac{m-3}{2}\) mangoes to Uche. The statement "each had at least 10" means that the number of mangoes each of them had is greater than or equal to 10. Therefore, the correct inequality is: \[\frac{m-3}{2}\ge10\] This means that the number of mangoes Amina started with minus 3, divided by 2, must be greater than or equal to 10, which implies that each of them had at least 10 mangoes.
Question 28 Report
The height of a pyramid on square base is 15cm. If the volume is 80cm\(^3\), find the length of the side of the base
Answer Details
The volume of a pyramid is given by the formula: V = 1/3 * Base Area * Height The base of a square pyramid is a square, so the Base Area is equal to the side length squared (B = s²). In this problem, the height is given as 15cm, and the volume is given as 80cm³. We are asked to find the length of a side of the base. Let's call that length "s". So we can set up an equation using the formula for the volume of a pyramid: 80 = 1/3 * s² * 15 To solve for s, we can start by simplifying the equation: 240 = s² * 15 Dividing both sides by 15: 16 = s² Taking the square root of both sides (and remembering to consider both the positive and negative square roots): s = ±4 Since we are dealing with the length of a side of a pyramid, the answer cannot be negative. Therefore, the length of a side of the base is 4cm (option C). Answer: 4.0cm
Question 29 Report
In the diagram, DE||BC, |AD| = x cm and |DB| = |AE| = ycm. Find |CE| in terms of x and y
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Question 30 Report
In the diagram above, |XR| = |RY| = |YZ| and ∠XRY = ∠YRZ = 62o, Calculate ∠XYZ
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Question 31 Report
In the diagram, POR is a circle with center O. ∠QPR = 50°, ∠PQO = 30° and ∠ORP = m. Find m.
Question 32 Report
The diagram shows the position of three ships A, B and C at sea. B is due north of C such that |AB| = |BC| and the bearing of B from A = 040°. What is the bearing of A from C
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Question 34 Report
A car travel at x km per hour for 1 hour and at y km per hour for 2 hours. Find its average speed
Answer Details
The average speed of a car is given by the total distance traveled divided by the total time taken. Let's assume that the car traveled a distance of D km in the given time. Then, we can write: Total time taken = 1 hour + 2 hours = 3 hours Distance traveled in the first hour = x km Distance traveled in the next 2 hours = y × 2 = 2y km Total distance traveled = D = x + 2y km Therefore, the average speed of the car is: Average speed = Total distance / Total time = D / 3 = (x + 2y) / 3 km/h Hence, the answer is option (C): \(\frac{x + 2y}{3}kmh^{-1}\).
Question 35 Report
The pie chart illustrate the amount of private time a student spends in a week studying various subjects. use it to answer the question below
Find the value of K
Answer Details
Question 36 Report
If \(K\sqrt{28}+\sqrt{63}-\sqrt{7}=0\), find K.
Answer Details
We can simplify the given equation as follows: $$K\sqrt{28}+\sqrt{63}-\sqrt{7}=0$$ $$K\sqrt{4\cdot7}+\sqrt{9\cdot7}-\sqrt{7}=0$$ $$2K\sqrt{7}+3\sqrt{7}-\sqrt{7}=0$$ $$2K\sqrt{7}+2\sqrt{7}=0$$ $$2\sqrt{7}(K+1)=0$$ Since $\sqrt{7}$ is not zero, we can divide both sides by $2\sqrt{7}$ to get: $$K+1=0$$ Therefore, $K=-1$. So the answer is (B) -1.
Question 37 Report
From the top of a cliff 20m high, a boat can be sighted at sea 75m from the foot of the cliff. Calculate the angle of depression of the boat from the top of the cliff
Answer Details
We can solve this problem using trigonometry. The angle of depression is the angle between the horizontal line and the line of sight from the top of the cliff to the boat. To find this angle, we need to first find the length of the line of sight. Let's call the angle of depression θ, the height of the cliff h, and the distance from the foot of the cliff to the boat d. From the problem statement, we know that h = 20m and d = 75m. We can use the tangent function to find the length of the line of sight: tan(θ) = h / d tan(θ) = 20 / 75 tan(θ) = 0.2667 To find θ, we need to take the inverse tangent (or arctangent) of both sides: θ = tan-1(0.2667) Using a calculator, we get: θ ≈ 14.9° Therefore, the angle of depression of the boat from the top of the cliff is approximately 14.9°. Answer: 14.9°
Question 38 Report
A survey shows that 28% of all the men in a village wear size 9 shoes. What is the probability that a man selected at random in the village wears size 9 shoes?
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Question 39 Report
Two numbers 24\(_{x}\) and 31\(_y\) are equal in value when converted to base ten. Find the equation connecting x and y
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Question 40 Report
Solve the inequality 2x + 3 < 5x
Answer Details
To solve the inequality 2x + 3 < 5x, we want to isolate x on one side of the inequality sign. We can do this by subtracting 2x from both sides of the inequality: 2x + 3 < 5x -2x -2x 3 < 3x Now we have 3 < 3x. To isolate x, we can divide both sides of the inequality by 3: 3 < 3x /3 /3 1 < x So the solution to the inequality 2x + 3 < 5x is x > 1. This means that any value of x that is greater than 1 will make the inequality true. For example, if x = 2, then 2(2) + 3 = 7, which is indeed less than 5(2) = 10. Therefore, the correct option is "x > 1".
Question 41 Report
Simplify 3.72 x 0.025 and express your answer in the standard form
Answer Details
To multiply 3.72 by 0.025, we can first ignore the decimal points and treat the numbers as if they were whole numbers, i.e., 372 and 25. We multiply these numbers to get 9300. Next, we need to determine the power of 10 that represents the decimal point in the product. To do this, we count the total number of decimal places in the two original numbers (three in 3.72 and two in 0.025), which gives us five decimal places. Therefore, our answer needs to be in scientific notation with a power of 10 that is negative five. Finally, we can insert the decimal point into the product and express the answer in scientific notation as follows: 3.72 x 0.025 = 0.093 = 9.3 x 10^(-2) Therefore, the answer is \(9.3\times 10^{-2}\).
Question 42 Report
A group of 11 people can speak either English or French or Both. Seven can speak English and six can speak French. What is the probability that a person chosen at random can speak both English and French?
Answer Details
There are 11 people in total, and we know that some of them can speak English only, some of them can speak French only, and some of them can speak both English and French. Let's denote the number of people who can speak both languages as "x". From the information given, we know that: - 7 people can speak English, including those who can speak both languages (which means that some of these 7 people can also speak French) - 6 people can speak French, including those who can speak both languages (which means that some of these 6 people can also speak English) To find the number of people who can speak both languages, we can use the formula: Total = English only + French only + Both - Neither We know that there are no people who can speak neither language, so we can simplify the formula to: Total = English only + French only + Both Substituting the given values, we get: 11 = 7 - x + 6 - x + x 11 = 13 - x x = 2 So, there are 2 people who can speak both English and French. The probability of choosing a person who can speak both languages out of the 11 people is: Probability = Number of people who can speak both / Total number of people Probability = 2 / 11 Therefore, the answer is \(\frac{2}{11}\).
Question 43 Report
The pie chart illustrate the amount of private time a student spends in a week studying various subjects. use it to answer the question below
If he sends \(2\frac{1}{2}\) hours week on science, find the total number of hours he studies in a week
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Question 44 Report
In the diagram, ABCD is a trapezium in which \(AD \parallel BC\) and \(< ABC\) is a right angle. If |AD| = 15 cm, |BD| = 17 cm and |BC| = 9 cm, calculate :
(a) |AB| ;
(b) the area of the triangle BCD ;
(c) |CD| ;
(d) perimeter of the trapezium.
Reading the diagram. \(ABCD\) is a trapezium with \(AD \parallel BC\) and \(\angle ABC = 90^\circ\). From the figure: \(|AD| = 15\text{ cm}\) (top), \(|BD| = 17\text{ cm}\) (diagonal), \(|BC| = 9\text{ cm}\) (bottom).
(a) \(|AB|\). Because \(AD \parallel BC\) and \(\angle ABC = 90^\circ\), the side \(AB\) is perpendicular to both parallels, so \(\angle DAB = 90^\circ\). Triangle \(ABD\) is right-angled at \(A\):
\[|AB|^2 = |BD|^2 - |AD|^2 = 17^2 - 15^2 = 289 - 225 = 64,\]\[|AB| = \sqrt{64} = \mathbf{8\text{ cm}}.\](b) Area of \(\triangle BCD\). The perpendicular distance between the parallel sides \(AD\) and \(BC\) equals \(|AB| = 8\text{ cm}\), so the height of \(\triangle BCD\) on base \(BC\) is \(8\text{ cm}\):
\[\text{Area} = \tfrac{1}{2}\times |BC| \times |AB| = \tfrac{1}{2}\times 9 \times 8 = \mathbf{36\text{ cm}^2}.\](c) \(|CD|\). Place \(B=(0,0)\), \(C=(9,0)\), \(A=(0,8)\), \(D=(15,8)\). Then
\[|CD| = \sqrt{(15-9)^2 + (8-0)^2} = \sqrt{6^2 + 8^2} = \sqrt{36+64} = \sqrt{100} = \mathbf{10\text{ cm}}.\](d) Perimeter of the trapezium.
\[P = |AB| + |BC| + |CD| + |DA| = 8 + 9 + 10 + 15 = \mathbf{42\text{ cm}}.\]Answer Details
Reading the diagram. \(ABCD\) is a trapezium with \(AD \parallel BC\) and \(\angle ABC = 90^\circ\). From the figure: \(|AD| = 15\text{ cm}\) (top), \(|BD| = 17\text{ cm}\) (diagonal), \(|BC| = 9\text{ cm}\) (bottom).
(a) \(|AB|\). Because \(AD \parallel BC\) and \(\angle ABC = 90^\circ\), the side \(AB\) is perpendicular to both parallels, so \(\angle DAB = 90^\circ\). Triangle \(ABD\) is right-angled at \(A\):
\[|AB|^2 = |BD|^2 - |AD|^2 = 17^2 - 15^2 = 289 - 225 = 64,\]\[|AB| = \sqrt{64} = \mathbf{8\text{ cm}}.\](b) Area of \(\triangle BCD\). The perpendicular distance between the parallel sides \(AD\) and \(BC\) equals \(|AB| = 8\text{ cm}\), so the height of \(\triangle BCD\) on base \(BC\) is \(8\text{ cm}\):
\[\text{Area} = \tfrac{1}{2}\times |BC| \times |AB| = \tfrac{1}{2}\times 9 \times 8 = \mathbf{36\text{ cm}^2}.\](c) \(|CD|\). Place \(B=(0,0)\), \(C=(9,0)\), \(A=(0,8)\), \(D=(15,8)\). Then
\[|CD| = \sqrt{(15-9)^2 + (8-0)^2} = \sqrt{6^2 + 8^2} = \sqrt{36+64} = \sqrt{100} = \mathbf{10\text{ cm}}.\](d) Perimeter of the trapezium.
\[P = |AB| + |BC| + |CD| + |DA| = 8 + 9 + 10 + 15 = \mathbf{42\text{ cm}}.\]Question 45 Report
ABC is a triangle, right-angled at C. P is the mid-point of AC, < PBC = 37° and |BC| = 5 cm. Calculate :
(a) |AC|, correct to 3 significant figures ;
(b) < PBA.
In \(\triangle PBC\), the right angle is at \(C\), \(|BC|=5\text{ cm}\), \(\angle PBC=37^{\circ}\), and \(P\) is the midpoint of \(AC\).
(a) \[\tan37^{\circ}=\frac{PC}{BC}\Rightarrow PC=5\tan37^{\circ}=5\times0.7536=3.768\text{ cm}.\] Since \(P\) is the midpoint, \(|AC|=2\times PC=7.536\approx\) 7.54 cm (3 s.f.).
(b) In \(\triangle ABC\), \[\tan(\angle ABC)=\frac{AC}{BC}=\frac{7.536}{5}=1.5072\Rightarrow\angle ABC=56.44^{\circ}.\] Therefore \[\angle PBA=\angle ABC-\angle PBC=56.44^{\circ}-37^{\circ}\approx19.4^{\circ}.\]
Answer Details
In \(\triangle PBC\), the right angle is at \(C\), \(|BC|=5\text{ cm}\), \(\angle PBC=37^{\circ}\), and \(P\) is the midpoint of \(AC\).
(a) \[\tan37^{\circ}=\frac{PC}{BC}\Rightarrow PC=5\tan37^{\circ}=5\times0.7536=3.768\text{ cm}.\] Since \(P\) is the midpoint, \(|AC|=2\times PC=7.536\approx\) 7.54 cm (3 s.f.).
(b) In \(\triangle ABC\), \[\tan(\angle ABC)=\frac{AC}{BC}=\frac{7.536}{5}=1.5072\Rightarrow\angle ABC=56.44^{\circ}.\] Therefore \[\angle PBA=\angle ABC-\angle PBC=56.44^{\circ}-37^{\circ}\approx19.4^{\circ}.\]
Question 46 Report
(a) Solve the simultaneous equations 3y - 2x = 21 ; 4y + 5x = 5.
(b) Six identical cards numbered 1 - 6 are placed face down. A card is to be picked at random. A person wins $60.00 if he picks the card numbered 6. If he picks any of the other cards, he loses $10.00 times the number on the card. Calculate the probability of (i) losing ; (ii) losing $20.00 after two picks.
(a) \(3y-2x=21\) and \(4y+5x=5\). Multiply the first by \(5\) and the second by \(2\): \[15y-10x=105,\qquad 8y+10x=10.\] Adding: \(23y=115\Rightarrow y=5\). Then \(3(5)-2x=21\Rightarrow-2x=6\Rightarrow x=-3\). So \(x=-3,\;y=5\).
(b) Card \(6\) wins \(\$60\); cards \(1\text{–}5\) lose \(\$10\times(\text{number})\).
- (i) A loss occurs whenever cards \(1,2,3,4,5\) are picked: \[P(\text{losing})=\frac{5}{6}.\]
- (ii) Losing exactly \(\$20\) in two picks: the outcomes per card are \(-10,-20,-30,-40,-50,+60\). The only pair of picks totalling \(-\$20\) is card \(1\) followed by card \(1\) \((-10-10=-20)\). \[P=\frac16\times\frac16=\frac{1}{36}.\]
Answer Details
(a) \(3y-2x=21\) and \(4y+5x=5\). Multiply the first by \(5\) and the second by \(2\): \[15y-10x=105,\qquad 8y+10x=10.\] Adding: \(23y=115\Rightarrow y=5\). Then \(3(5)-2x=21\Rightarrow-2x=6\Rightarrow x=-3\). So \(x=-3,\;y=5\).
(b) Card \(6\) wins \(\$60\); cards \(1\text{–}5\) lose \(\$10\times(\text{number})\).
- (i) A loss occurs whenever cards \(1,2,3,4,5\) are picked: \[P(\text{losing})=\frac{5}{6}.\]
- (ii) Losing exactly \(\$20\) in two picks: the outcomes per card are \(-10,-20,-30,-40,-50,+60\). The only pair of picks totalling \(-\$20\) is card \(1\) followed by card \(1\) \((-10-10=-20)\). \[P=\frac16\times\frac16=\frac{1}{36}.\]
Question 47 Report
(a) Copy and complete the following table of values for \(y = 9 \cos x + 5 \sin x\) to one decimal place.
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° | 210° |
| y | 10.3 | -0.2 | -5.3 | -10.3 |
(b) Using a scale of 2cm to 30° on the x- axis and 2 cm to 1 unit on the y- axis, draw the graph of \(y = 9 \cos x + 5 \sin x\) for \(0° \leq x \leq 210°\).
(c) Use your graph to solve the equation: (i) \(9\cos x + 5\sin x = 0\); (ii) \(9\cos x+ 5\sin x = 3.5\), correct to the nearest degree.
(d) Find the maximum value of y correct to one decimal place.
(a) Completing the table of values
For \(y = 9\cos x + 5\sin x\), evaluate at the missing angles (to one decimal place):
\(x = 0^\circ:\; 9(1)+5(0)=9.0\)
\(x = 60^\circ:\; 9(0.5)+5(0.866)=4.5+4.33=8.8\)
\(x = 90^\circ:\; 9(0)+5(1)=5.0\)
\(x = 180^\circ:\; 9(-1)+5(0)=-9.0\)
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° | 210° |
| y | 9.0 | 10.3 | 8.8 | 5.0 | -0.2 | -5.3 | -9.0 | -10.3 |
(b) Graph of \(y = 9\cos x + 5\sin x\)
Using a scale of 2 cm to 30° on the x-axis and 2 cm to 1 unit on the y-axis, plot the eight points from the table and join them with a smooth curve for \(0^\circ \le x \le 210^\circ\).
(c) Solving from the graph
(i) \(9\cos x + 5\sin x = 0\). This is where the curve cuts the x-axis (\(y = 0\)). Reading the graph, the curve crosses between \(x=110^\circ\) and \(x=120^\circ\), at:
\[ x \approx 119^\circ \](Check: \(\tan x = -\dfrac{9}{5}=-1.8\Rightarrow x = 180^\circ-61^\circ = 119^\circ.\))
(ii) \(9\cos x + 5\sin x = 3.5\). Draw the horizontal line \(y = 3.5\) and read where it meets the curve. It meets the descending part of the curve at:
\[ x \approx 99^\circ \](Check, using \(9\cos x+5\sin x = R\cos(x-\alpha)\) with \(R=\sqrt{9^2+5^2}=10.3\) and \(\alpha = \tan^{-1}\frac{5}{9}=29^\circ\): \(10.3\cos(x-29^\circ)=3.5\Rightarrow \cos(x-29^\circ)=0.340\Rightarrow x-29^\circ=70^\circ\Rightarrow x\approx 99^\circ.\))
(d) Maximum value of y
The highest point of the curve occurs near \(x = 30^\circ\). Since \(y = R\cos(x-\alpha)\) with \(R=\sqrt{9^2+5^2}\):
\[ y_{\max}=\sqrt{81+25}=\sqrt{106}\approx 10.3 \]The maximum value of \(y\) is 10.3 (occurring at \(x \approx 29^\circ\)).
Answer Details
(a) Completing the table of values
For \(y = 9\cos x + 5\sin x\), evaluate at the missing angles (to one decimal place):
\(x = 0^\circ:\; 9(1)+5(0)=9.0\)
\(x = 60^\circ:\; 9(0.5)+5(0.866)=4.5+4.33=8.8\)
\(x = 90^\circ:\; 9(0)+5(1)=5.0\)
\(x = 180^\circ:\; 9(-1)+5(0)=-9.0\)
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° | 210° |
| y | 9.0 | 10.3 | 8.8 | 5.0 | -0.2 | -5.3 | -9.0 | -10.3 |
(b) Graph of \(y = 9\cos x + 5\sin x\)
Using a scale of 2 cm to 30° on the x-axis and 2 cm to 1 unit on the y-axis, plot the eight points from the table and join them with a smooth curve for \(0^\circ \le x \le 210^\circ\).
(c) Solving from the graph
(i) \(9\cos x + 5\sin x = 0\). This is where the curve cuts the x-axis (\(y = 0\)). Reading the graph, the curve crosses between \(x=110^\circ\) and \(x=120^\circ\), at:
\[ x \approx 119^\circ \](Check: \(\tan x = -\dfrac{9}{5}=-1.8\Rightarrow x = 180^\circ-61^\circ = 119^\circ.\))
(ii) \(9\cos x + 5\sin x = 3.5\). Draw the horizontal line \(y = 3.5\) and read where it meets the curve. It meets the descending part of the curve at:
\[ x \approx 99^\circ \](Check, using \(9\cos x+5\sin x = R\cos(x-\alpha)\) with \(R=\sqrt{9^2+5^2}=10.3\) and \(\alpha = \tan^{-1}\frac{5}{9}=29^\circ\): \(10.3\cos(x-29^\circ)=3.5\Rightarrow \cos(x-29^\circ)=0.340\Rightarrow x-29^\circ=70^\circ\Rightarrow x\approx 99^\circ.\))
(d) Maximum value of y
The highest point of the curve occurs near \(x = 30^\circ\). Since \(y = R\cos(x-\alpha)\) with \(R=\sqrt{9^2+5^2}\):
\[ y_{\max}=\sqrt{81+25}=\sqrt{106}\approx 10.3 \]The maximum value of \(y\) is 10.3 (occurring at \(x \approx 29^\circ\)).
Question 48 Report
The cost of maintaining a school is partly constant and partly varies as the number of pupils. With 50 pupils, the cost is $15,705.00 and with 40 pupils, it is $13,305.00.
(a) Find the cost when there are 44 pupils.
(b) If the fee per pupil is $360.00, what is the least number of pupils for which the school can run without a loss?
Let the cost be \(C=a+bn\), where \(a\) is the constant part and \(b\) the cost per pupil.
\(50\) pupils: \(a+50b=15705\). \(40\) pupils: \(a+40b=13305\). Subtracting: \(10b=2400\Rightarrow b=240\); then \(a=15705-50(240)=3705\). So \[C=3705+240n.\]
(a) For \(n=44\): \[C=3705+240(44)=3705+10560=\$14{,}265.\]
(b) Revenue \(=360n\). To run without a loss, \(360n\ge3705+240n\Rightarrow120n\ge3705\Rightarrow n\ge30.875\). The least whole number of pupils is 31.
Answer Details
Let the cost be \(C=a+bn\), where \(a\) is the constant part and \(b\) the cost per pupil.
\(50\) pupils: \(a+50b=15705\). \(40\) pupils: \(a+40b=13305\). Subtracting: \(10b=2400\Rightarrow b=240\); then \(a=15705-50(240)=3705\). So \[C=3705+240n.\]
(a) For \(n=44\): \[C=3705+240(44)=3705+10560=\$14{,}265.\]
(b) Revenue \(=360n\). To run without a loss, \(360n\ge3705+240n\Rightarrow120n\ge3705\Rightarrow n\ge30.875\). The least whole number of pupils is 31.
Question 49 Report
(a) A cylindrical pipe is 28 metres long. Its internal radius is 3.5 cm and external radius 5 cm. Calaulate : (i) the volume, in cm\(^{3}\), of metal used in making the pipe ; (ii) the volume of water in litres that the pipe can hold when full, correct to 1 decimal place. [Take \(\pi = \frac{22}{7}\)]
(b) 
In the diagram, MP is a tangent to the circle LMN at M. If the chord LN is parallel to MP, show that the triangle LMN is isosceles.
(a) Cylindrical pipe, length \(28\text{ m}=2800\text{ cm}\), internal radius \(r=3.5\text{ cm}\), external radius \(R=5\text{ cm}\).
(i) Volume of metal used.
The metal is the hollow shell between the outer and inner cylinders:
\[V_{\text{metal}}=\pi(R^2-r^2)\times L\]\[R^2-r^2=5^2-3.5^2=25-12.25=12.75\text{ cm}^2\]\[V_{\text{metal}}=\frac{22}{7}\times 12.75\times 2800\]\[=\frac{22}{7}\times 35700=22\times 5100=112200\text{ cm}^3\]The volume of metal is \(112200\text{ cm}^3\).
(ii) Volume of water the pipe holds when full.
This is the inner cylinder's volume:
\[V_{\text{water}}=\pi r^2 L=\frac{22}{7}\times 3.5^2\times 2800\]\[=\frac{22}{7}\times 12.25\times 2800=\frac{22}{7}\times 34300=22\times 4900=107800\text{ cm}^3\]Convert to litres using \(1\text{ litre}=1000\text{ cm}^3\):
\[\frac{107800}{1000}=107.8\text{ litres}\]The pipe holds \(107.8\) litres (to 1 d.p.).
(b) Tangent \(MP\) at \(M\), chord \(LN\parallel MP\): show \(\triangle LMN\) is isosceles.
By the tangent-chord (alternate segment) theorem, the angle between tangent \(MP\) and chord \(MN\) equals the angle in the alternate segment standing on \(MN\):
\[\angle PMN=\angle MLN \quad\text{...(1)}\]Since \(LN\parallel MP\) and \(MN\) is a transversal, alternate angles are equal:
\[\angle PMN=\angle MNL \quad\text{...(2)}\]From (1) and (2):
\[\angle MLN=\angle MNL\]In triangle \(LMN\) the base angles at \(L\) and \(N\) are equal, so the sides opposite them are equal, i.e. \(|MN|=|ML|\).
Therefore triangle \(LMN\) is isosceles. (Q.E.D.)
Answer Details
(a) Cylindrical pipe, length \(28\text{ m}=2800\text{ cm}\), internal radius \(r=3.5\text{ cm}\), external radius \(R=5\text{ cm}\).
(i) Volume of metal used.
The metal is the hollow shell between the outer and inner cylinders:
\[V_{\text{metal}}=\pi(R^2-r^2)\times L\]\[R^2-r^2=5^2-3.5^2=25-12.25=12.75\text{ cm}^2\]\[V_{\text{metal}}=\frac{22}{7}\times 12.75\times 2800\]\[=\frac{22}{7}\times 35700=22\times 5100=112200\text{ cm}^3\]The volume of metal is \(112200\text{ cm}^3\).
(ii) Volume of water the pipe holds when full.
This is the inner cylinder's volume:
\[V_{\text{water}}=\pi r^2 L=\frac{22}{7}\times 3.5^2\times 2800\]\[=\frac{22}{7}\times 12.25\times 2800=\frac{22}{7}\times 34300=22\times 4900=107800\text{ cm}^3\]Convert to litres using \(1\text{ litre}=1000\text{ cm}^3\):
\[\frac{107800}{1000}=107.8\text{ litres}\]The pipe holds \(107.8\) litres (to 1 d.p.).
(b) Tangent \(MP\) at \(M\), chord \(LN\parallel MP\): show \(\triangle LMN\) is isosceles.
By the tangent-chord (alternate segment) theorem, the angle between tangent \(MP\) and chord \(MN\) equals the angle in the alternate segment standing on \(MN\):
\[\angle PMN=\angle MLN \quad\text{...(1)}\]Since \(LN\parallel MP\) and \(MN\) is a transversal, alternate angles are equal:
\[\angle PMN=\angle MNL \quad\text{...(2)}\]From (1) and (2):
\[\angle MLN=\angle MNL\]In triangle \(LMN\) the base angles at \(L\) and \(N\) are equal, so the sides opposite them are equal, i.e. \(|MN|=|ML|\).
Therefore triangle \(LMN\) is isosceles. (Q.E.D.)
Question 50 Report
In the diagram, a ladder TF, 10 metres long is placed against a wall at an angle of 70° to the horizontal.
(a) How high up the wall, correct to the nearest metre, does the ladder reach?
(b) If the foot (F) of the ladder is pulled from the wall to F\(^{1}\) by 1 metre, (i) how far, correct to 2 significant figures, does the top T slide down the wall to T\(^{1}\).
(ii) Calculate, correct to the nearest degree, \(QF^{1}T^{1}\).
Reading the diagram. The wall \(TQ\) is vertical and the ground \(QF\) is horizontal, meeting at the right angle \(Q\). The ladder \(TF = 10\text{ m}\) makes \(70^\circ\) with the horizontal at \(F\). When the foot is pulled out \(1\text{ m}\) to \(F'\), the top slides down to \(T'\), with \(T'F' = 10\text{ m}\) still.
(a) Height reached up the wall. In right triangle \(TQF\):
\[|TQ| = 10\sin 70^\circ = 10 \times 0.9397 = 9.40\text{ m} \approx \mathbf{9\text{ m}} \text{ (nearest metre)}.\](b)(i) Distance the top slides down. First the original foot distance:
\[|QF| = 10\cos 70^\circ = 10 \times 0.3420 = 3.420\text{ m}.\]New foot distance: \(|QF'| = 3.420 + 1 = 4.420\text{ m}.\) New height, from \(|T'F'| = 10\):
\[|QT'| = \sqrt{10^2 - 4.420^2} = \sqrt{100 - 19.54} = \sqrt{80.46} = 8.970\text{ m}.\]Distance slid down:
\[|TT'| = |QT| - |QT'| = 9.397 - 8.970 = 0.427\text{ m} \approx \mathbf{0.43\text{ m}} \text{ (2 s.f.)}.\](b)(ii) Angle \(\angle QF'T'\). In right triangle \(QF'T'\):
\[\cos(\angle QF'T') = \frac{|QF'|}{|F'T'|} = \frac{4.420}{10} = 0.4420,\]\[\angle QF'T' = \cos^{-1}(0.4420) = 63.8^\circ \approx \mathbf{64^\circ} \text{ (nearest degree)}.\]Answer Details
Reading the diagram. The wall \(TQ\) is vertical and the ground \(QF\) is horizontal, meeting at the right angle \(Q\). The ladder \(TF = 10\text{ m}\) makes \(70^\circ\) with the horizontal at \(F\). When the foot is pulled out \(1\text{ m}\) to \(F'\), the top slides down to \(T'\), with \(T'F' = 10\text{ m}\) still.
(a) Height reached up the wall. In right triangle \(TQF\):
\[|TQ| = 10\sin 70^\circ = 10 \times 0.9397 = 9.40\text{ m} \approx \mathbf{9\text{ m}} \text{ (nearest metre)}.\](b)(i) Distance the top slides down. First the original foot distance:
\[|QF| = 10\cos 70^\circ = 10 \times 0.3420 = 3.420\text{ m}.\]New foot distance: \(|QF'| = 3.420 + 1 = 4.420\text{ m}.\) New height, from \(|T'F'| = 10\):
\[|QT'| = \sqrt{10^2 - 4.420^2} = \sqrt{100 - 19.54} = \sqrt{80.46} = 8.970\text{ m}.\]Distance slid down:
\[|TT'| = |QT| - |QT'| = 9.397 - 8.970 = 0.427\text{ m} \approx \mathbf{0.43\text{ m}} \text{ (2 s.f.)}.\](b)(ii) Angle \(\angle QF'T'\). In right triangle \(QF'T'\):
\[\cos(\angle QF'T') = \frac{|QF'|}{|F'T'|} = \frac{4.420}{10} = 0.4420,\]\[\angle QF'T' = \cos^{-1}(0.4420) = 63.8^\circ \approx \mathbf{64^\circ} \text{ (nearest degree)}.\]Question 51 Report
(a) If \(\varepsilon\) is the set \({1, 2, 3,..., 19, 20}\) and A, B and C are subsets of \(\varepsilon\) such that A = { multiples of five}, B = {multiples of four} and C = {multiples of three}, list the elements of (i) A ; (ii) B ; (iii) C ;
(b) Find : (i) \(A \cap B\) ; (ii) \(A \cap C\) ; (iii) \(B \cup C\).
(c) Using your results in (b), show that \((A \cap B) \cup (A \cap C) = A \cap (B \cup C)\).
Question 52 Report
(a) Simplify \((\frac{4}{25})^{-\frac{1}{2}} \times 2^{4} \div (\frac{15}{2})^{-2}\)
(b) Evaluate \(\log_{5} (\frac{3}{5}) + 3 \log_{5} (\frac{5}{2}) - \log_{5} (\frac{81}{8})\).
Answer Details
None
Question 53 Report
The table gives the frequency distribution of marks obtained by a group of students in a test.
| Marks | 3 | 4 | 5 | 6 | 7 | 8 |
| Frequency | 5 | \(x - 1\) | \(x\) | 9 | 4 | 1 |
If the mean is 5,
(a) Calculate the value of x;
(b) Find the : (i) mode ; (ii) median of the distribution.
(c) If one of the students is selected at random, find the probability that he scored at least 7 marks.
| Marks | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|
| Frequency | 5 | x-1 | x | 9 | 4 | 1 |
(a) Value of x. Total frequency \(=5+(x-1)+x+9+4+1=18+2x\). Sum of \(fx\):
\[ \Sigma fx = 15+4(x-1)+5x+54+28+8 = 101+9x \]Since mean \(=5\):
\[ \frac{101+9x}{18+2x}=5 \Rightarrow 101+9x=90+10x \Rightarrow x=11 \]So the frequencies are \(5, 10, 11, 9, 4, 1\) with total \(N=40\).
(b)(i) Mode. The highest frequency (11) is at mark 5, so mode = 5.
(ii) Median. With \(N=40\), the median is the mean of the 20th and 21st values. Cumulative frequencies: 5, 15, 26, ... The 20th and 21st both fall at mark 5, so median = 5.
(c) P(at least 7 marks). Marks 7 and 8 give \(4+1=5\):
\[ P=\frac{5}{40}=\frac{1}{8}=0.125 \]Answer Details
| Marks | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|
| Frequency | 5 | x-1 | x | 9 | 4 | 1 |
(a) Value of x. Total frequency \(=5+(x-1)+x+9+4+1=18+2x\). Sum of \(fx\):
\[ \Sigma fx = 15+4(x-1)+5x+54+28+8 = 101+9x \]Since mean \(=5\):
\[ \frac{101+9x}{18+2x}=5 \Rightarrow 101+9x=90+10x \Rightarrow x=11 \]So the frequencies are \(5, 10, 11, 9, 4, 1\) with total \(N=40\).
(b)(i) Mode. The highest frequency (11) is at mark 5, so mode = 5.
(ii) Median. With \(N=40\), the median is the mean of the 20th and 21st values. Cumulative frequencies: 5, 15, 26, ... The 20th and 21st both fall at mark 5, so median = 5.
(c) P(at least 7 marks). Marks 7 and 8 give \(4+1=5\):
\[ P=\frac{5}{40}=\frac{1}{8}=0.125 \]Question 54 Report
On a graph sheet, using a scale of 2cm to 2 units on both axes,
(a) Draw the straight line joining points P(-5, 3) and Q(2, 3);
(b) construct the locus L of points equidistant from P and Q;
(c) by construction, locate points R and S on L, such that PRQS forms a rhombus of sides 5cm;
(d) find : (i) coordinates of R and S; (ii) area of the rhombus in cm\(^{2}\).
Scale: 2 cm to 2 units on both axes, i.e. 1 unit = 1 cm. So a length of 5 cm on the paper is the same as 5 units on the graph, and a side of the required rhombus measures 5 units.
Notice first that P(-5, 3) and Q(2, 3) have the same y-coordinate (3), so PQ is a horizontal line of length \(|2-(-5)| = 7\) units = 7 cm.
(a) The straight line PQ
P(-5, 3) and Q(2, 3) are plotted and joined by a straight line. This is the horizontal segment shown in blue at the level \(y = 3\).
(b) Locus L of points equidistant from P and Q
The set of points equidistant from P and Q is the perpendicular bisector of PQ. It is constructed by opening the compass to more than half of PQ, drawing equal arcs centred on P and on Q above and below the line, and joining their points of intersection. This bisector cuts PQ at its midpoint
\[ M = \left(\frac{-5+2}{2},\; 3\right) = (-1.5,\; 3), \]and runs vertically, so the locus is the line \(x = -1.5\), shown dashed and labelled L.
(c) Locating R and S so that PRQS is a rhombus of side 5 cm
In the rhombus PRQS the four equal sides are PR, RQ, QS and SP, each 5 cm. Hence each of R and S must be 5 cm from P and 5 cm from Q. Setting the compass to 5 cm (= 5 units) and drawing arcs centred on P and on Q, the arcs intersect on the locus L at two points: R above PQ and S below PQ. Because PR = RQ = QS = SP = 5 cm, the figure PRQS is a rhombus, and its diagonals PQ and RS cross at right angles at M.
(d) Results
(i) Coordinates of R and S. R and S lie on L, so each has \(x = -1.5\). Taking R\((-1.5,\,y)\) with PR = 5:
\[ \sqrt{(-1.5-(-5))^{2} + (y-3)^{2}} = 5 \]\[ 3.5^{2} + (y-3)^{2} = 5^{2} \;\Rightarrow\; (y-3)^{2} = 25 - 12.25 = 12.75 \]\[ y - 3 = \pm\sqrt{12.75} = \pm 3.55 \;\Rightarrow\; y = 6.5 \text{ or } y = -0.6. \]Therefore, reading from the construction,
\[ \boxed{R(-1.5,\; 6.5) \quad\text{and}\quad S(-1.5,\; -0.6).} \](ii) Area of the rhombus. The diagonals of a rhombus bisect each other at right angles, so the area is half the product of the diagonals. Here the diagonals are PQ and RS:
\[ PQ = 7\ \text{cm}, \qquad RS = 6.5-(-0.6) = 7.1\ \text{cm}. \]\[ \text{Area} = \frac{1}{2}\times PQ \times RS = \frac{1}{2}\times 7 \times 7.1 = \frac{49.7}{2} \]\[ \boxed{\text{Area} \approx 24.85\ \text{cm}^{2}\ (\approx 25\ \text{cm}^{2}).} \]Answer Details
Scale: 2 cm to 2 units on both axes, i.e. 1 unit = 1 cm. So a length of 5 cm on the paper is the same as 5 units on the graph, and a side of the required rhombus measures 5 units.
Notice first that P(-5, 3) and Q(2, 3) have the same y-coordinate (3), so PQ is a horizontal line of length \(|2-(-5)| = 7\) units = 7 cm.
(a) The straight line PQ
P(-5, 3) and Q(2, 3) are plotted and joined by a straight line. This is the horizontal segment shown in blue at the level \(y = 3\).
(b) Locus L of points equidistant from P and Q
The set of points equidistant from P and Q is the perpendicular bisector of PQ. It is constructed by opening the compass to more than half of PQ, drawing equal arcs centred on P and on Q above and below the line, and joining their points of intersection. This bisector cuts PQ at its midpoint
\[ M = \left(\frac{-5+2}{2},\; 3\right) = (-1.5,\; 3), \]and runs vertically, so the locus is the line \(x = -1.5\), shown dashed and labelled L.
(c) Locating R and S so that PRQS is a rhombus of side 5 cm
In the rhombus PRQS the four equal sides are PR, RQ, QS and SP, each 5 cm. Hence each of R and S must be 5 cm from P and 5 cm from Q. Setting the compass to 5 cm (= 5 units) and drawing arcs centred on P and on Q, the arcs intersect on the locus L at two points: R above PQ and S below PQ. Because PR = RQ = QS = SP = 5 cm, the figure PRQS is a rhombus, and its diagonals PQ and RS cross at right angles at M.
(d) Results
(i) Coordinates of R and S. R and S lie on L, so each has \(x = -1.5\). Taking R\((-1.5,\,y)\) with PR = 5:
\[ \sqrt{(-1.5-(-5))^{2} + (y-3)^{2}} = 5 \]\[ 3.5^{2} + (y-3)^{2} = 5^{2} \;\Rightarrow\; (y-3)^{2} = 25 - 12.25 = 12.75 \]\[ y - 3 = \pm\sqrt{12.75} = \pm 3.55 \;\Rightarrow\; y = 6.5 \text{ or } y = -0.6. \]Therefore, reading from the construction,
\[ \boxed{R(-1.5,\; 6.5) \quad\text{and}\quad S(-1.5,\; -0.6).} \](ii) Area of the rhombus. The diagonals of a rhombus bisect each other at right angles, so the area is half the product of the diagonals. Here the diagonals are PQ and RS:
\[ PQ = 7\ \text{cm}, \qquad RS = 6.5-(-0.6) = 7.1\ \text{cm}. \]\[ \text{Area} = \frac{1}{2}\times PQ \times RS = \frac{1}{2}\times 7 \times 7.1 = \frac{49.7}{2} \]\[ \boxed{\text{Area} \approx 24.85\ \text{cm}^{2}\ (\approx 25\ \text{cm}^{2}).} \]Question 55 Report
(a) A shop owner marked a shirt at a price to enable him to make a gain of 20%. During a special sales period, the shirt was sold at 10% reduction to a customer at N864.00. What was the original cost to the shop owner?
(b) A rectangular lawn of length (x + 5) metres is (x - 2) metres wide. If the diagonal is (x + 6) metres, find ;
(i) the value of x ; (ii) the area of lawn.
(a) Let the cost price be \(C\). Marked to gain \(20\%\): marked \(=1.2C\). Sold at \(10\%\) reduction: \[0.9\times1.2C=1.08C=864\Rightarrow C=\frac{864}{1.08}=\text{N}800.\] The original cost was N800.
(b) Length \((x+5)\), width \((x-2)\), diagonal \((x+6)\). By Pythagoras: \[(x+5)^{2}+(x-2)^{2}=(x+6)^{2}.\] \[x^{2}+10x+25+x^{2}-4x+4=x^{2}+12x+36\Rightarrow x^{2}-6x-7=0\Rightarrow(x-7)(x+1)=0.\] Taking the positive value, \(x=7\).
(ii) Area \(=(x+5)(x-2)=12\times5=60\text{ m}^{2}.\)
Answer Details
(a) Let the cost price be \(C\). Marked to gain \(20\%\): marked \(=1.2C\). Sold at \(10\%\) reduction: \[0.9\times1.2C=1.08C=864\Rightarrow C=\frac{864}{1.08}=\text{N}800.\] The original cost was N800.
(b) Length \((x+5)\), width \((x-2)\), diagonal \((x+6)\). By Pythagoras: \[(x+5)^{2}+(x-2)^{2}=(x+6)^{2}.\] \[x^{2}+10x+25+x^{2}-4x+4=x^{2}+12x+36\Rightarrow x^{2}-6x-7=0\Rightarrow(x-7)(x+1)=0.\] Taking the positive value, \(x=7\).
(ii) Area \(=(x+5)(x-2)=12\times5=60\text{ m}^{2}.\)
Question 56 Report
(a) Given that \(\sin(A + B) = \sin A \cos B + \cos A \sin B\). Without using mathematical tables or calculator, evaluate \(\sin 105°\), leaving your answer in the surd form.
(You may use 105° = 60° + 45°)
(b) The houses on one side of a particular street are assigned odd numbers, starting from 11. If the sum of the numbers is 551, how many houses are there?
(c) The 1st and 3rd terms of a Geometric Progression (G.P) are \(2\) and \(\frac{2}{9}\) respectively. Find :
(i) the common difference ; (ii) the 5th term.
(a) \[\sin105^{\circ}=\sin(60^{\circ}+45^{\circ})=\sin60^{\circ}\cos45^{\circ}+\cos60^{\circ}\sin45^{\circ}=\frac{\sqrt3}{2}\cdot\frac{\sqrt2}{2}+\frac12\cdot\frac{\sqrt2}{2}=\frac{\sqrt6+\sqrt2}{4}.\]
(b) The odd house numbers \(11,13,15,\dots\) form an AP with \(a=11,\;d=2\). Sum of \(n\) terms: \[\frac{n}{2}\bigl(2(11)+(n-1)2\bigr)=n(n+10)=551\Rightarrow n^{2}+10n-551=0.\] \((n+29)(n-19)=0\Rightarrow n=19\). There are 19 houses.
(c) GP with \(t_{1}=2\), \(t_{3}=\tfrac29\): \(ar^{2}=\tfrac29\) and \(a=2\Rightarrow r^{2}=\tfrac19\Rightarrow r=\tfrac13\). The common ratio is \(\tfrac13\). Fifth term: \[t_{5}=ar^{4}=2\left(\frac13\right)^{4}=\frac{2}{81}.\]
Answer Details
(a) \[\sin105^{\circ}=\sin(60^{\circ}+45^{\circ})=\sin60^{\circ}\cos45^{\circ}+\cos60^{\circ}\sin45^{\circ}=\frac{\sqrt3}{2}\cdot\frac{\sqrt2}{2}+\frac12\cdot\frac{\sqrt2}{2}=\frac{\sqrt6+\sqrt2}{4}.\]
(b) The odd house numbers \(11,13,15,\dots\) form an AP with \(a=11,\;d=2\). Sum of \(n\) terms: \[\frac{n}{2}\bigl(2(11)+(n-1)2\bigr)=n(n+10)=551\Rightarrow n^{2}+10n-551=0.\] \((n+29)(n-19)=0\Rightarrow n=19\). There are 19 houses.
(c) GP with \(t_{1}=2\), \(t_{3}=\tfrac29\): \(ar^{2}=\tfrac29\) and \(a=2\Rightarrow r^{2}=\tfrac19\Rightarrow r=\tfrac13\). The common ratio is \(\tfrac13\). Fifth term: \[t_{5}=ar^{4}=2\left(\frac13\right)^{4}=\frac{2}{81}.\]
