WAEC SSCE - General Mathematics - 2012
Frage 1 Bericht
The graph is that of y = 2x2 - 5x - 3. For what value of x will y be negative? For what value of x will y be negative?
Antwortdetails
To find when y is negative, we need to find the values of x for which 2x^2 - 5x - 3 is negative. One way to do this is to use the quadratic formula: x = (-b ± sqrt(b^2 - 4ac)) / 2a where a = 2, b = -5, and c = -3. Plugging these values in, we get: x = (-(-5) ± sqrt((-5)^2 - 4(2)(-3))) / 2(2) x = (5 ± sqrt(49)) / 4 x = (5 ± 7) / 4 So the solutions are x = -3/2 and x = 2. We can check that when x < -3/2 or x > 2, y is positive, and when -3/2 < x < 2, y is negative. Therefore, the answer is option (C) -\(\frac{1}{2} < x < 3\).
Frage 3 Bericht
Solve for x in the equation; \(\frac{1}{x} + \frac{2}{3x} = \frac{1}{3}\)
Antwortdetails
To solve the equation \(\frac{1}{x} + \frac{2}{3x} = \frac{1}{3}\), we can use the following steps: Step 1: Find the common denominator. The denominators in this equation are x and 3x. The common denominator is 3x. Step 2: Multiply both sides of the equation by the common denominator. Multiplying both sides by 3x gives: 3 + 2 = x Step 3: Simplify and solve for x. Simplifying the left side gives: 5 = x Therefore, the solution to the equation is x = 5. x = 5, is the correct answer. Options 2, 3, and 4 are incorrect because they do not equal 5.
Frage 4 Bericht
In 1995, the enrollment of two schools X and Y were 1,050 and 1,190 respectively. Find the ration of the enrollments of X and Y
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To find the ratio of enrollments of schools X and Y, we need to divide the enrollment of school X by the enrollment of school Y. Ratio of enrollments of X and Y = Enrollment of X / Enrollment of Y Enrollment of X = 1,050 Enrollment of Y = 1,190 Ratio of enrollments of X and Y = 1,050 / 1,190 Simplifying this fraction by dividing both numerator and denominator by 10 gives: Ratio of enrollments of X and Y = 105 / 119 Further simplifying by dividing both numerator and denominator by 7 gives: Ratio of enrollments of X and Y = 15 / 17 Therefore, the ratio of the enrollments of X and Y is 15:17. Answer option (B) is correct.
Frage 5 Bericht
In the diagram, MN//PO, < PMN = 112o, < PNO = 129oo and < MPN = yo. Find the value of y
Frage 6 Bericht
If x and y are variables and k is a constant, which of the following describes an inverse relationship between x and y?
Antwortdetails
The equation that describes an inverse relationship between x and y is y = k/x, which is the second option. An inverse relationship between two variables means that as one variable increases, the other variable decreases in a proportional manner. In other words, if we double the value of x, the value of y would be halved, and vice versa. In the equation y = k/x, as x increases, y decreases, and as x decreases, y increases. The constant k represents the scale of the relationship between x and y. The other options do not describe an inverse relationship between x and y. is a direct proportion where y increases as x increases. is a power relationship where y increases as the square root of x increases. is a linear relationship where y increases at a constant rate as x increases.
Frage 7 Bericht
Mr. Manu travelled from Accra to Pamfokromb a distance of 720km in 8 hours. What will be his speed in m/s?
Antwortdetails
To find the speed in meters per second (m/s), we need to convert the distance and time to meters and seconds respectively. Distance is given as 720km. 1km is equal to 1000m, so 720km = 720 x 1000m = 720,000m. Time taken is given as 8 hours. 1 hour is equal to 3600 seconds, so 8 hours = 8 x 3600 seconds = 28,800 seconds. Speed = Distance ÷ Time = 720,000m ÷ 28,800s = 25m/s. Therefore, Mr. Manu's speed is 25m/s. (25m/s) is the correct answer.
Frage 8 Bericht
Convert 3510 to number in base 2
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To convert a number from base 10 to base 2 (binary), we need to divide the decimal number by 2 repeatedly until the quotient is zero. Then, we write the remainders in reverse order. Let's convert 3510 to binary using this method: 35 / 2 = 17 remainder 1 17 / 2 = 8 remainder 1 8 / 2 = 4 remainder 0 4 / 2 = 2 remainder 0 2 / 2 = 1 remainder 0 1 / 2 = 0 remainder 1 The remainders, in reverse order, are 100011, which is the binary equivalent of 3510. Therefore, the correct answer is option (C) 100011. Each digit in a binary number represents a power of 2, starting from the rightmost digit, which represents 2^0 = 1. The next digit represents 2^1 = 2, the next represents 2^2 = 4, and so on. To convert a binary number to decimal, we multiply each digit by its corresponding power of 2 and add up the results.
Frage 9 Bericht
Express 302.10495 correct to five significant figures
Frage 10 Bericht
The sum of the interior angles of regular polygon is 1800o. How many sides has the polygon?
Antwortdetails
The sum of the interior angles of a polygon can be found by the formula: sum of interior angles = (n-2) x 180 degrees where n is the number of sides in the polygon. In this problem, we are given that the sum of the interior angles of the polygon is 1800 degrees. So we can set up an equation: 1800 = (n-2) x 180 Simplifying the equation: 10 = n - 2 n = 12 Therefore, the polygon has 12 sides.
Frage 11 Bericht
The diagram is a circle of radius |QR| = 4cm. \(\bar{R}\) is a tangent to the circle at R. If TPO = 120o, find |PQ|.
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Frage 12 Bericht
If x + 0.4y = 3 and y = \(\frac{1}{2}\)x, find the value of (x + y)
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We are given that x + 0.4y = 3 and y = \(\frac{1}{2}\)x. Substitute y in the first equation: x + 0.4 * (\(\frac{1}{2}\)x) = 3 x + 0.2x = 3 1.2x = 3 x = 2.5 Substitute x in the equation for y: y = \(\frac{1}{2}\) * 2.5 = 1.25 Then, x + y = 2.5 + 1.25 = 3.75. Therefore, the value of (x + y) is 3\(\frac{3}{4}\). Hence, the correct option is (c) 3\(\frac{3}{4}\).
Frage 13 Bericht
Factorise completely: 32x2y - 48x3y3
Antwortdetails
We can begin by factoring out the greatest common factor of the two terms, which is 16x2y. This gives: 32x2y - 48x3y3 = 16x2y(2 - 3xy2) So the correct option is 16x2y(2 - 3xy2).
Frage 14 Bericht
If p = {prime factors of 210} and Q = {prime less than 10}, find p \(\cap\) Q
Antwortdetails
The prime factors of 210 are 2, 3, 5, and 7. The prime numbers less than 10 are 2, 3, 5, and 7. The intersection (p ∩ Q) is the set of elements that are in both p and Q. Therefore, p ∩ Q is {2, 3, 5, 7}. So, the correct option is: - {2,3,5,7}
Frage 15 Bericht
Simplify: \(\frac{54k^2 - 6}{3k + 1}\)
Antwortdetails
To simplify the expression \(\frac{54k^2 - 6}{3k + 1}\), we can start by factoring out the numerator: \(\frac{54k^2 - 6}{3k + 1} = \frac{6(9k^2 - 1)}{3k + 1}\) We can further simplify the numerator by recognizing that \(9k^2 - 1\) is a difference of squares, and can be factored as \((3k + 1)(3k - 1)\): \(\frac{6(9k^2 - 1)}{3k + 1} = \frac{6(3k + 1)(3k - 1)}{3k + 1}\) We can then cancel out the common factor of \(3k + 1\) in the numerator and denominator, which gives: \(\frac{6(3k - 1)}{1} = 6(3k - 1)\) Therefore, the simplified expression is \(6(3k - 1)\). The correct option is: - 6(3k - 1)
Frage 16 Bericht
A bag contains 4 red and 6 black balls of the same size. If the balls are shuffled briskly and two balls are drawn one after the other without replacement, find the probability of picking balls of different colours
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Frage 18 Bericht
In the diagram, |SR| = |QR|. < SRP = 65o and < RPQ = 48o, find < PRQ
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Frage 19 Bericht
The diagram is a circle with centre P. PRST are points on the circle. Find the value of < PRS
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Frage 20 Bericht
The volume of a cuboid is 54cm3. If the length, width and height of the cuboid are in the ratio 2:1:1 respectively, find its total surface area
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To solve this problem, we first need to find the dimensions of the cuboid. Let the length, width and height be 2x, x and x respectively. Then we have: Volume of cuboid = length x width x height 54 = 2x * x * x 54 = 2x^3 x^3 = 27 x = 3 Therefore, the length of the cuboid is 2x = 6cm, the width is x = 3cm, and the height is x = 3cm. To find the total surface area, we need to find the area of each face and add them up. The total surface area is given by: Total surface area = 2lw + 2lh + 2wh = 2(6*3) + 2(6*3) + 2(3*3) = 36 + 36 + 18 = 90 cm^2 Therefore, the total surface area of the cuboid is 90cm^2. The correct answer is option B.
Frage 21 Bericht
The sum of 12 and one third of n is 1 more than twice n. Express the statement in the form of an equation
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Frage 22 Bericht
Is the diagram, MN, PQ and RS are three intersecting straight lines. Which of the following statements is/are true? i. t = y ii. x + y + z + m = 180o ii. x + m + n = 180o iv. x + n = m + z
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Frage 23 Bericht
The diagram is a circle centre O. If < SPR = 2m and < SQR = n, express m in terms of n
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Frage 24 Bericht
Solve (\(\frac{27}{125}\))-\(\frac{1}{3}\) x (\(\frac{4}{9}\))\(\frac{1}{2}\)
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Frage 25 Bericht
Alfred spent \(\frac{1}{4}\) of his money on food, \(\frac{1}{3}\) on clothing and save the rest. If he saved N72,20.00, how much did he spend on food?
Antwortdetails
Let A be the total amount of money that Alfred has. From the given information, he spent \(\frac{1}{4}\) of A on food and \(\frac{1}{3}\) of A on clothing, and he saved the rest. Therefore, we have: Amount spent on food = \(\frac{1}{4}\)A Amount spent on clothing = \(\frac{1}{3}\)A Amount saved = A - (\(\frac{1}{4}\)A + \(\frac{1}{3}\)A) Amount saved = A - \(\frac{7}{12}\)A Amount saved = \(\frac{5}{12}\)A We know that Alfred saved N72,20.00, so we can write: \(\frac{5}{12}\)A = N72,20.00 Multiplying both sides by \(\frac{12}{5}\), we get: A = N172,80.00 Therefore, the amount spent on food is: \(\frac{1}{4}\)A = \(\frac{1}{4}\) x N172,80.00 = N43,20.00 Hence, Alfred spent N43,20.00 on food. Answer is correct.
Frage 26 Bericht
The graph is that of y = 2x2 - 5x - 3. For what value of x will y be negative? What is the gradient of y = 2x2 - 5x - 3 at the point x = 4?
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Frage 27 Bericht
If x and y are variables and k is a constant, which of the following describes an inverse relationship between x and y?
Antwortdetails
An inverse relationship between two variables means that as one variable increases, the other variable decreases. Mathematically, an inverse relationship can be represented by an equation where one variable is multiplied or divided by a constant. Out of the given equations, the equation that represents an inverse relationship between x and y is y = \(\frac{k}{x}\). To see why this is the case, let's consider what happens to y as x increases. Suppose x doubles in value. Then, according to the equation y = \(\frac{k}{x}\), y will be halved in value. Similarly, if x triples in value, y will be divided by 3. This means that as x increases, y decreases, which is the characteristic of an inverse relationship. On the other hand, in the equation y = kx, as x increases, y also increases. This is not an inverse relationship but a direct relationship. The same applies to y = k\(\sqrt{x}\) and y = x + k. Therefore, the equation that describes an inverse relationship between x and y is y = \(\frac{k}{x}\).
Frage 28 Bericht
In the diagram, O is a circle centre of the circle PQRS and < PSR = 86o. If < PQR = xo, find x
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Frage 29 Bericht
Given that the mean of the scores 15, 21, 17, 26, 18 and 29 is 21, calculate the standard deviation of the scores
Antwortdetails
To calculate the standard deviation of the scores, we first need to find the variance, which is the average of the squared deviations from the mean. 1. Find the mean of the scores: Mean = (15 + 21 + 17 + 26 + 18 + 29)/6 = 126/6 = 21 2. Calculate the deviations from the mean for each score: 15 - 21 = -6 21 - 21 = 0 17 - 21 = -4 26 - 21 = 5 18 - 21 = -3 29 - 21 = 8 3. Square each deviation: (-6)^2 = 36 0^2 = 0 (-4)^2 = 16 5^2 = 25 (-3)^2 = 9 8^2 = 64 4. Find the average of the squared deviations: (36 + 0 + 16 + 25 + 9 + 64)/6 = 150/6 = 25 5. Take the square root of the variance to get the standard deviation: Standard deviation = sqrt(25) = 5 Therefore, the standard deviation of the scores is 5. The answer is 5.
Frage 30 Bericht
Simplify; \(\frac{3\sqrt{5} \times 4\sqrt{6}}{2 \sqrt{2} \times 3\sqrt{2}}\)
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Frage 31 Bericht
Find the coefficient of m in the expression of (\(\frac{m}{2} - 1 \frac{1}{2}\)) (m + \(\frac{2}{3}\))
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Frage 32 Bericht
If x + y = 2y - x + 1 = 5, find the value of x
Antwortdetails
We are given two equations, x + y = 5 and 2y - x + 1 = 5, and we need to find the value of x. From the first equation, we can rearrange it to get x = 5 - y. Substituting this value of x into the second equation, we get: 2y - (5 - y) + 1 = 5 Simplifying the equation, we get: 3y - 4 = 5 Adding 4 to both sides, we get: 3y = 9 Dividing both sides by 3, we get: y = 3 Substituting this value of y into x + y = 5, we get: x + 3 = 5 Subtracting 3 from both sides, we get: x = 2 Therefore, the value of x is 2.
Frage 33 Bericht
The bar chart shows the frequency distribution of marks scored by students in a class test. How many students are in the class?
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Frage 34 Bericht
In the diagram, MQ//RS, < TUV = 70o and < RLV = 30o. Find the value of x
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Frage 35 Bericht
The bar chart shows the frequency distribution of marks scored by students in a class test. Calculate the mean of the distribution.
Antwortdetails
To calculate the mean of a frequency distribution, we need to find the sum of all the values multiplied by their respective frequencies and divide by the total number of values. Looking at the bar chart, we can see that there are 5 values: 1, 2, 3, 4, and 5. The corresponding frequencies are 1, 2, 5, 6, and 1, respectively. To calculate the mean, we first need to calculate the sum of all the values multiplied by their respective frequencies: (1 x 1) + (2 x 2) + (3 x 5) + (4 x 6) + (5 x 1) = 1 + 4 + 15 + 24 + 5 = 49 Next, we need to calculate the total number of values, which is the sum of all the frequencies: 1 + 2 + 5 + 6 + 1 = 15 Finally, we divide the sum of the values multiplied by their frequencies by the total number of values: 49 / 15 = 3.27 (rounded to two decimal places) Therefore, the mean of the distribution is approximately 3.27. The correct answer is not provided in the options, but the closest one is option (C) 2.4.
Frage 36 Bericht
The lengths of the minor and major arcs 54cm and 126cm respectively. Calculate the angle of the major sector
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Frage 37 Bericht
The curved surface area of a cylindrical tin is 704cm2. If the radius of its base is 8cm, find the height. [Take \(\pi = \frac{22}{7}\)]`
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Frage 38 Bericht
Make p the subject of the relation: q = \(\frac{3p}{r} + \frac{s}{2}\)
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Frage 39 Bericht
If N2,500.00 amounted to N3,50.00 in 4 years at simple interest, find the rate at which the interest was charged
Antwortdetails
Simple interest is calculated by multiplying the principal amount, the interest rate, and the time in years. The formula for simple interest is given as: Simple Interest = (Principal x Rate x Time) / 100 In this problem, we know that the principal amount (P) is N2,500.00, the amount (A) after 4 years is N3,500.00. We need to find the rate (R) at which the interest was charged. Using the formula for simple interest, we can write: A = P + I where I is the interest amount. We can rearrange the formula to solve for I: I = A - P = N3,500.00 - N2,500.00 = N1,000.00 Now, we can substitute the values we know into the formula for simple interest and solve for R: I = (P x R x T) / 100 N1,000.00 = (N2,500.00 x R x 4) / 100 Simplifying, we get: R = (N1,000.00 x 100) / (N2,500.00 x 4) = 10% Therefore, the rate at which the interest was charged is 10%.
Frage 40 Bericht
A side and a diagonal of a rhombus are 10cm and 12cm respectively, Find its area
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Frage 41 Bericht
In \(\bigtriangleup\) XYZ, /XY/ = 8cm, /YZ/ = 10cm and /XZ/ = 6cm. Which of these relation is true?
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Frage 42 Bericht
The bar chart shows the frequency distribution of marks scored by students in a class test. What is the median of the distribution?
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Frage 43 Bericht
A kite flies on a taut string of length 50m inclined at tan angle 54o to the horizontal ground. The height of the kite above the ground is
Frage 44 Bericht
Solve the inequality: \(\frac{-m}{2} - \frac{5}{4} \geq \frac{5m}{12} - \frac{7}{6}\)
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Frage 46 Bericht
Express 3 - [\(\frac{x - y}{y}\)] as a single fraction
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To express 3 - [\(\frac{x - y}{y}\)] as a single fraction, we first need to simplify the expression inside the square brackets: 3 - [\(\frac{x - y}{y}\)] = 3 - (\(\frac{x}{y}\) - \(\frac{y}{y}\)) = 3 - \(\frac{x}{y}\) + 1 = 4 - \(\frac{x}{y}\) Therefore, 3 - [\(\frac{x - y}{y}\)] can be expressed as a single fraction: = 4 - \(\frac{x}{y}\) = \(\frac{4y - x}{y}\) Hence, the answer is \(\frac{4y - x}{y}\).
Frage 47 Bericht
In the diagram, |QR| = 10m, |SR| = 8m
< QPS = 30o, < QRP = 90o and |PS| = x, Find x
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Frage 48 Bericht
The position of three ships P,Q and R at sea are illustrated in the diagram. The arrows indicated the North direction. The bearing of Q from P is 050o and < PQR = 72o. Calculate the bearing of R and Q
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Frage 49 Bericht
The nth term of a sequence is Tn = 5 + (n - 1)2. Evaluate T4 - T6
Antwortdetails
To evaluate T4 - T6, we first need to find the values of T4 and T6. T4 = 5 + (4 - 1)2 = 5 + 9 = 14 T6 = 5 + (6 - 1)2 = 5 + 25 = 30 Therefore, T4 - T6 = 14 - 30 = -16 Hence, the answer is -16.
Frage 50 Bericht
| Class Interval |
Frequency |
| 60 - 64 | 2 |
| 65 - 69 | 3 |
| 70 - 74 | 6 |
| 75 - 79 | 11 |
| 80 - 84 | 8 |
| 85 - 89 | 7 |
| 90 - 94 | 2 |
| 95 - 99 | 1 |
The table shows the distribution of marks scored by students in an examination. Calculate, correct to 2 decimal places, the
(a) mean ; (b) standard deviation of the distribution.
Working table (midpoint \(x\), \(N = 40\)).
| Class | x | f | fx | fx² |
|---|---|---|---|---|
| 60-64 | 62 | 2 | 124 | 7688 |
| 65-69 | 67 | 3 | 201 | 13467 |
| 70-74 | 72 | 6 | 432 | 31104 |
| 75-79 | 77 | 11 | 847 | 65219 |
| 80-84 | 82 | 8 | 656 | 53792 |
| 85-89 | 87 | 7 | 609 | 52983 |
| 90-94 | 92 | 2 | 184 | 16928 |
| 95-99 | 97 | 1 | 97 | 9409 |
| Total | 40 | 3150 | 250590 |
(a) Mean.
\[\bar{x} = \frac{\sum fx}{\sum f} = \frac{3150}{40} = \mathbf{78.75}\](b) Standard deviation.
\[\sigma = \sqrt{\frac{\sum fx^{2}}{N} - \bar{x}^{2}} = \sqrt{\frac{250590}{40} - 78.75^{2}}\] \[= \sqrt{6264.75 - 6201.5625} = \sqrt{63.1875} = \mathbf{7.95}\]Antwortdetails
Working table (midpoint \(x\), \(N = 40\)).
| Class | x | f | fx | fx² |
|---|---|---|---|---|
| 60-64 | 62 | 2 | 124 | 7688 |
| 65-69 | 67 | 3 | 201 | 13467 |
| 70-74 | 72 | 6 | 432 | 31104 |
| 75-79 | 77 | 11 | 847 | 65219 |
| 80-84 | 82 | 8 | 656 | 53792 |
| 85-89 | 87 | 7 | 609 | 52983 |
| 90-94 | 92 | 2 | 184 | 16928 |
| 95-99 | 97 | 1 | 97 | 9409 |
| Total | 40 | 3150 | 250590 |
(a) Mean.
\[\bar{x} = \frac{\sum fx}{\sum f} = \frac{3150}{40} = \mathbf{78.75}\](b) Standard deviation.
\[\sigma = \sqrt{\frac{\sum fx^{2}}{N} - \bar{x}^{2}} = \sqrt{\frac{250590}{40} - 78.75^{2}}\] \[= \sqrt{6264.75 - 6201.5625} = \sqrt{63.1875} = \mathbf{7.95}\]Frage 51 Bericht
(a) (i) Using a scale of 2 cm to 1 unit on both axes, on the same graph sheet, draw the graphs of \(y - \frac{3x}{4} = 3\) and \(y + 2x = 6\).
(ii) From your graph, find the coordinates of the point of intersection of the two graphs.
(iii) Show, on the graph sheet, the region satisfied by the inequality \(y - \frac{3}{4}x \geq 3\).
(b) Given that \(x^{2} + bx + 18\) is factorized as \((x + 2)(x + c)\). Find the values of c and b.
(a)(i) Rearrange the equations:
\[y-\frac{3x}{4}=3\quad\Rightarrow\quad y=\frac34x+3\]
\[y+2x=6\quad\Rightarrow\quad y=6-2x\]
| Line | Points used for plotting |
|---|---|
| \(y=\frac34x+3\) | \((-4,0),\ (0,3),\ (4,6)\) |
| \(y=6-2x\) | \((0,6),\ (1,4),\ (3,0)\) |
Using the scale of 2 cm to 1 unit on both axes, the required graph is:
The boundary \(y-\frac34x=3\) is drawn as a solid line since the inequality includes equality. Region \(R\), the side above this line, is shaded.
(ii) The two lines intersect at approximately
\[\boxed{(1.1,\ 3.8)}\]
This agrees with the exact solution:
\[\frac34x+3=6-2x\]
\[\frac{11}{4}x=3\quad\Rightarrow\quad x=\frac{12}{11}\]
\[y=6-2\left(\frac{12}{11}\right)=\frac{42}{11}\]
Thus the exact coordinates are \(\left(\frac{12}{11},\frac{42}{11}\right)\), which are approximately \((1.1,3.8)\).
(iii) \[y-\frac34x\geq3\quad\Rightarrow\quad y\geq\frac34x+3.\]
Hence, the required region is the half-plane on and above \(y=\frac34x+3\), labelled \(R\) on the graph.
(b)
\[(x+2)(x+c)=x^2+(c+2)x+2c.\]
Comparing with \(x^2+bx+18\):
\[2c=18\Rightarrow c=9,\]
\[b=c+2=9+2=11.\]
\[\boxed{c=9,\quad b=11}\]
Antwortdetails
(a)(i) Rearrange the equations:
\[y-\frac{3x}{4}=3\quad\Rightarrow\quad y=\frac34x+3\]
\[y+2x=6\quad\Rightarrow\quad y=6-2x\]
| Line | Points used for plotting |
|---|---|
| \(y=\frac34x+3\) | \((-4,0),\ (0,3),\ (4,6)\) |
| \(y=6-2x\) | \((0,6),\ (1,4),\ (3,0)\) |
Using the scale of 2 cm to 1 unit on both axes, the required graph is:
The boundary \(y-\frac34x=3\) is drawn as a solid line since the inequality includes equality. Region \(R\), the side above this line, is shaded.
(ii) The two lines intersect at approximately
\[\boxed{(1.1,\ 3.8)}\]
This agrees with the exact solution:
\[\frac34x+3=6-2x\]
\[\frac{11}{4}x=3\quad\Rightarrow\quad x=\frac{12}{11}\]
\[y=6-2\left(\frac{12}{11}\right)=\frac{42}{11}\]
Thus the exact coordinates are \(\left(\frac{12}{11},\frac{42}{11}\right)\), which are approximately \((1.1,3.8)\).
(iii) \[y-\frac34x\geq3\quad\Rightarrow\quad y\geq\frac34x+3.\]
Hence, the required region is the half-plane on and above \(y=\frac34x+3\), labelled \(R\) on the graph.
(b)
\[(x+2)(x+c)=x^2+(c+2)x+2c.\]
Comparing with \(x^2+bx+18\):
\[2c=18\Rightarrow c=9,\]
\[b=c+2=9+2=11.\]
\[\boxed{c=9,\quad b=11}\]
Frage 52 Bericht
(a) A box contains 40 identical discs which are either red or white. If the probability of picking a red disc is \(\frac{1}{4}\); Calculate the number of (i) white discs ; (ii) red discs that should be added such that the probability of picking a red disc will be \(\frac{1}{3}\).
(b) A salesman bought some plates at N50.00 each. If he sold all of them for N600.00 and made a profit of 20% on the transaction, how many plates did he buy?
(a) Total discs \(= 40\), and \(P(\text{red}) = \dfrac14\), so number of red \(= \dfrac14\times 40 = 10\).
(i) Number of white discs \(= 40 - 10 = 30\).
(ii) Let \(x\) red discs be added. New red count \(= 10 + x\); new total \(= 40 + x\). Require \(P(\text{red}) = \dfrac13\):
\[\frac{10 + x}{40 + x} = \frac13 \;\Rightarrow\; 3(10 + x) = 40 + x \;\Rightarrow\; 30 + 3x = 40 + x\]
\[2x = 10 \;\Rightarrow\; x = 5\]
So \(5\) red discs should be added.
(b) Cost price per plate \(=\) N50. A profit of \(20\%\) on total cost means
\[\text{Selling price} = 1.20 \times \text{Cost price}\]
\[600 = 1.20 \times \text{Cost price} \;\Rightarrow\; \text{Cost price} = \frac{600}{1.20} = \text{N}500\]
\[\text{Number of plates} = \frac{500}{50} = 10\]
Antwortdetails
(a) Total discs \(= 40\), and \(P(\text{red}) = \dfrac14\), so number of red \(= \dfrac14\times 40 = 10\).
(i) Number of white discs \(= 40 - 10 = 30\).
(ii) Let \(x\) red discs be added. New red count \(= 10 + x\); new total \(= 40 + x\). Require \(P(\text{red}) = \dfrac13\):
\[\frac{10 + x}{40 + x} = \frac13 \;\Rightarrow\; 3(10 + x) = 40 + x \;\Rightarrow\; 30 + 3x = 40 + x\]
\[2x = 10 \;\Rightarrow\; x = 5\]
So \(5\) red discs should be added.
(b) Cost price per plate \(=\) N50. A profit of \(20\%\) on total cost means
\[\text{Selling price} = 1.20 \times \text{Cost price}\]
\[600 = 1.20 \times \text{Cost price} \;\Rightarrow\; \text{Cost price} = \frac{600}{1.20} = \text{N}500\]
\[\text{Number of plates} = \frac{500}{50} = 10\]
Frage 53 Bericht
Three towns X, Y and Z are such that Y is 20 km from X and 22 km from Z. Town X is 18 km from Z. A health centre is to be built by the government to serve the three towns. The centre is to be located such that patients from X and Y travel equal distance to access the health centre while patients from Z will travel exactly 10 km to reach the Health centre.
(a) Using a scale of 1 cm to 2 km, find the construction, using a pair of compasses and ruler only, the possible positions the Health centre can be located.
(b) In how many possible locations can the Health centre be built?
(c) Measure and record the distances of the location from town X.
(d) Which of these locations would be convenient for all three towns?
Frage 54 Bericht
Sonny is twice as old as Wale. Four years ago, he was four times as old as Wale. When will the sum of their ages be 66?
Let Wale's present age be \(w\). Since Sonny is twice as old, Sonny is \(2w\).
Four years ago Sonny was four times as old as Wale:
\[2w - 4 = 4(w - 4)\]
\[2w - 4 = 4w - 16 \;\Rightarrow\; 12 = 2w \;\Rightarrow\; w = 6\]
So presently Wale is \(6\) and Sonny is \(12\); their ages sum to \(18\).
Each year that passes adds \(2\) to the total (one year to each person). Let \(n\) be the number of years until the sum is \(66\):
\[18 + 2n = 66 \;\Rightarrow\; 2n = 48 \;\Rightarrow\; n = 24\]
The sum of their ages will be \(66\) in \(\mathbf{24}\) years' time (when Wale is 30 and Sonny is 36).
Antwortdetails
Let Wale's present age be \(w\). Since Sonny is twice as old, Sonny is \(2w\).
Four years ago Sonny was four times as old as Wale:
\[2w - 4 = 4(w - 4)\]
\[2w - 4 = 4w - 16 \;\Rightarrow\; 12 = 2w \;\Rightarrow\; w = 6\]
So presently Wale is \(6\) and Sonny is \(12\); their ages sum to \(18\).
Each year that passes adds \(2\) to the total (one year to each person). Let \(n\) be the number of years until the sum is \(66\):
\[18 + 2n = 66 \;\Rightarrow\; 2n = 48 \;\Rightarrow\; n = 24\]
The sum of their ages will be \(66\) in \(\mathbf{24}\) years' time (when Wale is 30 and Sonny is 36).
Frage 55 Bericht
(a) Copy and complete the table of values for \(y = 1 - 4\cos x\).
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° | 210° | 240° | 270° | 300° |
| y | -3.0 | 1.0 | 4.5 | -1.0 |
(b) Using a scale of 2cm to 30° on the x- axis and 2cm to 1 unit on the y- axis, draw the graph of \(y = 1 - 4\cos x\) for \(0° \leq x \leq 360°\).
(c) Use the graph to : (i) solve the equation \(1 - 4\cos x = 0\) ; (ii) find the value of y when x = 105° ; (iii) find x when y = 1.5.
(a) For each value of x, calculate \(y=1-4\cos x\), correct to 1 decimal place.
| \(x\) | \(0^\circ\) | \(30^\circ\) | \(60^\circ\) | \(90^\circ\) | \(120^\circ\) | \(150^\circ\) | \(180^\circ\) | \(210^\circ\) | \(240^\circ\) | \(270^\circ\) | \(300^\circ\) |
|---|---|---|---|---|---|---|---|---|---|---|---|
| \(y\) | -3.0 | -2.5 | -1.0 | 1.0 | 3.0 | 4.5 | 5.0 | 4.5 | 3.0 | 1.0 | -1.0 |
(b) The completed graph of \(y=1-4\cos x\), including \(y=-2.5\) at \(330^\circ\) and \(y=-3.0\) at \(360^\circ\), is shown below.
(c)
(i) The points where the curve cuts the \(x\)-axis give
\[x\approx75^\circ\quad\text{or}\quad285^\circ.\]
(ii) At \(x=105^\circ\), the ordinate of the curve is
\[y\approx2.0.\]
(iii) Drawing the horizontal line \(y=1.5\) and reading the two intersections with the curve gives
\[x\approx96^\circ\quad\text{or}\quad264^\circ.\]
Antwortdetails
(a) For each value of x, calculate \(y=1-4\cos x\), correct to 1 decimal place.
| \(x\) | \(0^\circ\) | \(30^\circ\) | \(60^\circ\) | \(90^\circ\) | \(120^\circ\) | \(150^\circ\) | \(180^\circ\) | \(210^\circ\) | \(240^\circ\) | \(270^\circ\) | \(300^\circ\) |
|---|---|---|---|---|---|---|---|---|---|---|---|
| \(y\) | -3.0 | -2.5 | -1.0 | 1.0 | 3.0 | 4.5 | 5.0 | 4.5 | 3.0 | 1.0 | -1.0 |
(b) The completed graph of \(y=1-4\cos x\), including \(y=-2.5\) at \(330^\circ\) and \(y=-3.0\) at \(360^\circ\), is shown below.
(c)
(i) The points where the curve cuts the \(x\)-axis give
\[x\approx75^\circ\quad\text{or}\quad285^\circ.\]
(ii) At \(x=105^\circ\), the ordinate of the curve is
\[y\approx2.0.\]
(iii) Drawing the horizontal line \(y=1.5\) and reading the two intersections with the curve gives
\[x\approx96^\circ\quad\text{or}\quad264^\circ.\]
Frage 56 Bericht
(a)
In the diagram, ABCD is a rectangular garden (3n - 1)m long and (2n + 1)m wide. A wire mesh 135m long is used to mark its boundary and to divide it into 8 equal plots. Find the value of n.
(b) A cylinder with base radius 14 cm has the same volume as a cube of side 22 cm. Calculate the ratio of the total surface area of the cylinder to that of the cube. [Take \(\pi = \frac{22}{7}\)]
(a) The diagram shows the rectangle divided into 4 columns and 2 rows, giving \(4 \times 2 = 8\) equal plots. Length \(AB = (3n-1)\ \text{m}\), width \(BC = (2n+1)\ \text{m}\).
The wire mesh forms the boundary and the internal dividing lines.
Horizontal lines (each of length \(3n-1\)): top, bottom and 1 internal line \(= 3\) lines.
Vertical lines (each of length \(2n+1\)): left, right and 3 internal lines \(= 5\) lines.
Total length of wire:
\[3(3n-1) + 5(2n+1) = 135\]\[9n - 3 + 10n + 5 = 135\]\[19n + 2 = 135\]\[19n = 133 \Rightarrow n = 7\]\(n = 7\)
(b) Take \(\pi = \frac{22}{7}\). Let the cylinder have radius \(r = 14\ \text{cm}\) and height \(h\).
Volume of cube (side \(22\ \text{cm}\)):
\[V = 22^3 = 10648\ \text{cm}^3\]Volume of cylinder equals this:
\[\pi r^2 h = \frac{22}{7} \times 14^2 \times h = 616h\]\[616h = 10648 \Rightarrow h = \frac{10648}{616} = \frac{121}{7}\ \text{cm} \;\left(=17\tfrac{2}{7}\right)\]Total surface area of cylinder:
\[2\pi r(r + h) = 2 \times \frac{22}{7} \times 14 \times \left(14 + \frac{121}{7}\right)\]\[= 88 \times \frac{98 + 121}{7} = 88 \times \frac{219}{7} = \frac{19272}{7}\ \text{cm}^2\]Total surface area of cube:
\[6 \times 22^2 = 6 \times 484 = 2904\ \text{cm}^2\]Ratio (cylinder : cube):
\[\frac{19272}{7} : 2904 = 19272 : 20328\]Dividing both by \(264\):
\[= 73 : 77\]Ratio \(= 73 : 77\)
Antwortdetails
(a) The diagram shows the rectangle divided into 4 columns and 2 rows, giving \(4 \times 2 = 8\) equal plots. Length \(AB = (3n-1)\ \text{m}\), width \(BC = (2n+1)\ \text{m}\).
The wire mesh forms the boundary and the internal dividing lines.
Horizontal lines (each of length \(3n-1\)): top, bottom and 1 internal line \(= 3\) lines.
Vertical lines (each of length \(2n+1\)): left, right and 3 internal lines \(= 5\) lines.
Total length of wire:
\[3(3n-1) + 5(2n+1) = 135\]\[9n - 3 + 10n + 5 = 135\]\[19n + 2 = 135\]\[19n = 133 \Rightarrow n = 7\]\(n = 7\)
(b) Take \(\pi = \frac{22}{7}\). Let the cylinder have radius \(r = 14\ \text{cm}\) and height \(h\).
Volume of cube (side \(22\ \text{cm}\)):
\[V = 22^3 = 10648\ \text{cm}^3\]Volume of cylinder equals this:
\[\pi r^2 h = \frac{22}{7} \times 14^2 \times h = 616h\]\[616h = 10648 \Rightarrow h = \frac{10648}{616} = \frac{121}{7}\ \text{cm} \;\left(=17\tfrac{2}{7}\right)\]Total surface area of cylinder:
\[2\pi r(r + h) = 2 \times \frac{22}{7} \times 14 \times \left(14 + \frac{121}{7}\right)\]\[= 88 \times \frac{98 + 121}{7} = 88 \times \frac{219}{7} = \frac{19272}{7}\ \text{cm}^2\]Total surface area of cube:
\[6 \times 22^2 = 6 \times 484 = 2904\ \text{cm}^2\]Ratio (cylinder : cube):
\[\frac{19272}{7} : 2904 = 19272 : 20328\]Dividing both by \(264\):
\[= 73 : 77\]Ratio \(= 73 : 77\)
Frage 57 Bericht
(a) A boy had M Dalasis (D). He spent D15 and shared the remainder equally with his sister. If the sister's share was equal to \(\frac{1}{3}\) of M, find the value of M.
(b) A number of tourists were interviewed on their choice of means of travel. Two- thirds said that they travelled by road, \(\frac{13}{30}\) by air and \(\frac{4}{15}\) by both air and road. If 20 tourists did not travel by either air or road ; (i) represent the information on a Venn diagram ; (ii) how many tourists (1) were interviewed ; (2) travelled by air only?
(a) The boy had \(M\) dalasis, spent D15, leaving \(M - 15\). This remainder is shared equally between him and his sister, so the sister's share is \(\dfrac{M - 15}{2}\). We are told this equals \(\dfrac13 M\):
\[\frac{M - 15}{2} = \frac{M}{3}\]
\[3(M - 15) = 2M \;\Rightarrow\; 3M - 45 = 2M \;\Rightarrow\; M = 45\]
So \(M = \text{D}45\).
(b) Let the total number of tourists be \(N\). Travelled by road \(= \tfrac23 N\), by air \(= \tfrac{13}{30}N\), by both \(= \tfrac{4}{15}N\).
(i) Venn diagram: two intersecting circles (Road, Air). Both region \(= \tfrac{4}{15}N\); road only \(= \tfrac23 N - \tfrac{4}{15}N = \tfrac{6}{15}N\); air only \(= \tfrac{13}{30}N - \tfrac{4}{15}N = \tfrac{5}{30}N\); outside both \(= 20\).
(ii)(1) Those using road or air:
\[\frac23 + \frac{13}{30} - \frac{4}{15} = \frac{20 + 13 - 8}{30} = \frac{25}{30} = \frac56\]
So the fraction using neither is \(1 - \tfrac56 = \tfrac16\), and \(\tfrac16 N = 20 \Rightarrow N = 120\). 120 tourists were interviewed.
(2) Air only \(= \tfrac{13}{30}N - \tfrac{4}{15}N = \tfrac{5}{30}\times 120 = 20\) tourists.
Antwortdetails
(a) The boy had \(M\) dalasis, spent D15, leaving \(M - 15\). This remainder is shared equally between him and his sister, so the sister's share is \(\dfrac{M - 15}{2}\). We are told this equals \(\dfrac13 M\):
\[\frac{M - 15}{2} = \frac{M}{3}\]
\[3(M - 15) = 2M \;\Rightarrow\; 3M - 45 = 2M \;\Rightarrow\; M = 45\]
So \(M = \text{D}45\).
(b) Let the total number of tourists be \(N\). Travelled by road \(= \tfrac23 N\), by air \(= \tfrac{13}{30}N\), by both \(= \tfrac{4}{15}N\).
(i) Venn diagram: two intersecting circles (Road, Air). Both region \(= \tfrac{4}{15}N\); road only \(= \tfrac23 N - \tfrac{4}{15}N = \tfrac{6}{15}N\); air only \(= \tfrac{13}{30}N - \tfrac{4}{15}N = \tfrac{5}{30}N\); outside both \(= 20\).
(ii)(1) Those using road or air:
\[\frac23 + \frac{13}{30} - \frac{4}{15} = \frac{20 + 13 - 8}{30} = \frac{25}{30} = \frac56\]
So the fraction using neither is \(1 - \tfrac56 = \tfrac16\), and \(\tfrac16 N = 20 \Rightarrow N = 120\). 120 tourists were interviewed.
(2) Air only \(= \tfrac{13}{30}N - \tfrac{4}{15}N = \tfrac{5}{30}\times 120 = 20\) tourists.
Frage 58 Bericht
(a) Simplify : \(\frac{1\frac{1}{4} + \frac{7}{9}}{1\frac{4}{9} - 2\frac{2}{3} \times \frac{9}{64}}\)
(b) Given that \(\sin x = \frac{2}{3}\), evaluate, leaving your answer in surd form and without using tables or calculator, \(\tan x - \cos x\).
(a) Convert to improper fractions.
Numerator: \(1\tfrac14 + \tfrac79 = \tfrac54 + \tfrac79 = \dfrac{45 + 28}{36} = \dfrac{73}{36}\).
Denominator: first \(2\tfrac23 \times \tfrac{9}{64} = \tfrac83 \times \tfrac{9}{64} = \dfrac{72}{192} = \dfrac38\). Then \(1\tfrac49 - \tfrac38 = \tfrac{13}{9} - \tfrac38 = \dfrac{104 - 27}{72} = \dfrac{77}{72}\).
\[\frac{73/36}{77/72} = \frac{73}{36}\times\frac{72}{77} = \frac{73 \times 2}{77} = \frac{146}{77} = 1\tfrac{69}{77}\]
(b) Given \(\sin x = \dfrac23\), the adjacent side is \(\sqrt{3^2 - 2^2} = \sqrt5\), so
\[\cos x = \frac{\sqrt5}{3}, \qquad \tan x = \frac{2}{\sqrt5} = \frac{2\sqrt5}{5}\]
\[\tan x - \cos x = \frac{2}{\sqrt5} - \frac{\sqrt5}{3} = \sqrt5\left(\frac{2}{5} - \frac{1}{3}\right) = \sqrt5\cdot\frac{6 - 5}{15} = \frac{\sqrt5}{15}\]
Antwortdetails
(a) Convert to improper fractions.
Numerator: \(1\tfrac14 + \tfrac79 = \tfrac54 + \tfrac79 = \dfrac{45 + 28}{36} = \dfrac{73}{36}\).
Denominator: first \(2\tfrac23 \times \tfrac{9}{64} = \tfrac83 \times \tfrac{9}{64} = \dfrac{72}{192} = \dfrac38\). Then \(1\tfrac49 - \tfrac38 = \tfrac{13}{9} - \tfrac38 = \dfrac{104 - 27}{72} = \dfrac{77}{72}\).
\[\frac{73/36}{77/72} = \frac{73}{36}\times\frac{72}{77} = \frac{73 \times 2}{77} = \frac{146}{77} = 1\tfrac{69}{77}\]
(b) Given \(\sin x = \dfrac23\), the adjacent side is \(\sqrt{3^2 - 2^2} = \sqrt5\), so
\[\cos x = \frac{\sqrt5}{3}, \qquad \tan x = \frac{2}{\sqrt5} = \frac{2\sqrt5}{5}\]
\[\tan x - \cos x = \frac{2}{\sqrt5} - \frac{\sqrt5}{3} = \sqrt5\left(\frac{2}{5} - \frac{1}{3}\right) = \sqrt5\cdot\frac{6 - 5}{15} = \frac{\sqrt5}{15}\]
Frage 59 Bericht
(a) 
In the diagram, /PQ/ = 6 cm, /QR/ = 13 cm, /RS/ = 5 cm and < RSQ is a right- angled triangle. Calculate, correct to one decimal place, /PS/.
(b) 
The diagram show a wooden structure in the form of a cone mounted on a hemispherical base. The vertical height of the cone is 24 cm and the base radius 7 cm. Calculate, correct to 3 significant figures, the surface area of the structure. [Take \(\pi = \frac{22}{7}\)].
(a) Calculating /PS/
From the diagram, \(P, Q, R\) lie on one straight line with \(PQ = 6\text{ cm}\) and \(QR = 13\text{ cm}\). The triangle \(QSR\) is right-angled at \(S\), with \(RS = 5\text{ cm}\) and hypotenuse \(QR = 13\text{ cm}\).
Step 1: Find QS using Pythagoras in triangle QSR.
\[ QS^2 = QR^2 - RS^2 = 13^2 - 5^2 = 169 - 25 = 144 \]
\[ QS = 12\text{ cm} \]
Step 2: Find angle RQS.
\[ \cos(\angle RQS) = \frac{QS}{QR} = \frac{12}{13} \]
Step 3: Use triangle PQS. Since \(P, Q, R\) are collinear, \(\angle PQS\) and \(\angle RQS\) are supplementary, so \(\cos(\angle PQS) = -\dfrac{12}{13}\).
Applying the cosine rule in triangle \(PQS\) with \(PQ = 6\), \(QS = 12\):
\[ PS^2 = PQ^2 + QS^2 - 2\,(PQ)(QS)\cos(\angle PQS) \]
\[ PS^2 = 6^2 + 12^2 - 2(6)(12)\left(-\tfrac{12}{13}\right) \]
\[ PS^2 = 36 + 144 + \frac{1728}{13} = 180 + 132.92 = 312.92 \]
\[ PS = \sqrt{312.92} \approx 17.7\text{ cm} \]
/PS/ \(\approx\) 17.7 cm.
(b) Surface area of the structure
The structure is a cone (height \(24\text{ cm}\), base radius \(7\text{ cm}\)) mounted on a hemisphere of the same radius \(7\text{ cm}\). The exposed surface is the curved surface of the cone plus the curved surface of the hemisphere (the joining circle is hidden).
Slant height of cone:
\[ l = \sqrt{h^2 + r^2} = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25\text{ cm} \]
Curved surface area of cone:
\[ \pi r l = \frac{22}{7} \times 7 \times 25 = 550\text{ cm}^2 \]
Curved surface area of hemisphere:
\[ 2\pi r^2 = 2 \times \frac{22}{7} \times 7^2 = 2 \times 22 \times 7 = 308\text{ cm}^2 \]
Total surface area:
\[ 550 + 308 = 858\text{ cm}^2 \]
Surface area \(= 858\text{ cm}^2\) (to 3 significant figures).
Antwortdetails
(a) Calculating /PS/
From the diagram, \(P, Q, R\) lie on one straight line with \(PQ = 6\text{ cm}\) and \(QR = 13\text{ cm}\). The triangle \(QSR\) is right-angled at \(S\), with \(RS = 5\text{ cm}\) and hypotenuse \(QR = 13\text{ cm}\).
Step 1: Find QS using Pythagoras in triangle QSR.
\[ QS^2 = QR^2 - RS^2 = 13^2 - 5^2 = 169 - 25 = 144 \]
\[ QS = 12\text{ cm} \]
Step 2: Find angle RQS.
\[ \cos(\angle RQS) = \frac{QS}{QR} = \frac{12}{13} \]
Step 3: Use triangle PQS. Since \(P, Q, R\) are collinear, \(\angle PQS\) and \(\angle RQS\) are supplementary, so \(\cos(\angle PQS) = -\dfrac{12}{13}\).
Applying the cosine rule in triangle \(PQS\) with \(PQ = 6\), \(QS = 12\):
\[ PS^2 = PQ^2 + QS^2 - 2\,(PQ)(QS)\cos(\angle PQS) \]
\[ PS^2 = 6^2 + 12^2 - 2(6)(12)\left(-\tfrac{12}{13}\right) \]
\[ PS^2 = 36 + 144 + \frac{1728}{13} = 180 + 132.92 = 312.92 \]
\[ PS = \sqrt{312.92} \approx 17.7\text{ cm} \]
/PS/ \(\approx\) 17.7 cm.
(b) Surface area of the structure
The structure is a cone (height \(24\text{ cm}\), base radius \(7\text{ cm}\)) mounted on a hemisphere of the same radius \(7\text{ cm}\). The exposed surface is the curved surface of the cone plus the curved surface of the hemisphere (the joining circle is hidden).
Slant height of cone:
\[ l = \sqrt{h^2 + r^2} = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25\text{ cm} \]
Curved surface area of cone:
\[ \pi r l = \frac{22}{7} \times 7 \times 25 = 550\text{ cm}^2 \]
Curved surface area of hemisphere:
\[ 2\pi r^2 = 2 \times \frac{22}{7} \times 7^2 = 2 \times 22 \times 7 = 308\text{ cm}^2 \]
Total surface area:
\[ 550 + 308 = 858\text{ cm}^2 \]
Surface area \(= 858\text{ cm}^2\) (to 3 significant figures).
Frage 60 Bericht
In the diagram, O is the centre of the circleand XY is a chord. If the radius is 5 cm and /XY/ = 6 cm, calculate, correct to 2 decimal places, the :
(a) angle which XY subtends at the centre O ;
(b) area of the shaded portion.
From the diagram, \(O\) is the centre, radius \(OX = OY = 5\ \text{cm}\), chord \(|XY| = 6\ \text{cm}\), and the shaded portion is the minor segment cut off by the chord \(XY\) (the region between the chord and the minor arc on the right).
(a) Angle subtended at the centre, \(\angle XOY\):
Drop the perpendicular from \(O\) to the midpoint \(M\) of \(XY\). Then \(XM = 3\ \text{cm}\) and:
\[\sin\left(\frac{\angle XOY}{2}\right) = \frac{XM}{OX} = \frac{3}{5} = 0.6\]\[\frac{\angle XOY}{2} = \sin^{-1}(0.6) = 36.8699^\circ\]\[\angle XOY = 2 \times 36.8699^\circ = 73.7398^\circ\]\(\angle XOY \approx 73.74^\circ\)
(b) Area of the shaded (minor) segment \(=\) area of sector \(XOY\) \(-\) area of triangle \(XOY\).
Area of sector:
\[\frac{\theta}{360^\circ} \times \pi r^2 = \frac{73.7398}{360} \times \frac{22}{7} \times 5^2\]\[= 0.204833 \times 78.5714 = 16.09\ \text{cm}^2\]Area of triangle \(XOY\):
\[\frac{1}{2} r^2 \sin\theta = \frac{1}{2} \times 25 \times \sin 73.7398^\circ = \frac{1}{2} \times 25 \times 0.96 = 12.00\ \text{cm}^2\]Area of shaded segment:
\[16.09 - 12.00 = 4.09\ \text{cm}^2\]Shaded area \(\approx 4.09\ \text{cm}^2\) (to 2 decimal places).
Antwortdetails
From the diagram, \(O\) is the centre, radius \(OX = OY = 5\ \text{cm}\), chord \(|XY| = 6\ \text{cm}\), and the shaded portion is the minor segment cut off by the chord \(XY\) (the region between the chord and the minor arc on the right).
(a) Angle subtended at the centre, \(\angle XOY\):
Drop the perpendicular from \(O\) to the midpoint \(M\) of \(XY\). Then \(XM = 3\ \text{cm}\) and:
\[\sin\left(\frac{\angle XOY}{2}\right) = \frac{XM}{OX} = \frac{3}{5} = 0.6\]\[\frac{\angle XOY}{2} = \sin^{-1}(0.6) = 36.8699^\circ\]\[\angle XOY = 2 \times 36.8699^\circ = 73.7398^\circ\]\(\angle XOY \approx 73.74^\circ\)
(b) Area of the shaded (minor) segment \(=\) area of sector \(XOY\) \(-\) area of triangle \(XOY\).
Area of sector:
\[\frac{\theta}{360^\circ} \times \pi r^2 = \frac{73.7398}{360} \times \frac{22}{7} \times 5^2\]\[= 0.204833 \times 78.5714 = 16.09\ \text{cm}^2\]Area of triangle \(XOY\):
\[\frac{1}{2} r^2 \sin\theta = \frac{1}{2} \times 25 \times \sin 73.7398^\circ = \frac{1}{2} \times 25 \times 0.96 = 12.00\ \text{cm}^2\]Area of shaded segment:
\[16.09 - 12.00 = 4.09\ \text{cm}^2\]Shaded area \(\approx 4.09\ \text{cm}^2\) (to 2 decimal places).
Frage 61 Bericht
A point H is 20 m away from the foot of a tower on the same horizontal ground. From the point H, the angle of elevation of the point P on the tower and the top (T) of the tower are 30° and 50° respectively. Calculate, correct to 3 significant figures :
(a) /PT/; (b) the distance between H and the top of the tower
(c) The position of H if the angle of depression of H from the top of the tower is to be 40°.
\(H\) is 20 m from the foot of the tower. From \(H\), \(P\) has elevation \(30^\circ\) and the top \(T\) has elevation \(50^\circ\).
Heights above ground of the two points:
\[\text{height of }P = 20\tan 30^\circ = 20(0.5774) = 11.547\text{ m}\]
\[\text{height of }T = 20\tan 50^\circ = 20(1.1918) = 23.835\text{ m}\]
(a) \(|PT| = 23.835 - 11.547 = 12.288 \approx 12.3\text{ m (3 s.f.)}\).
(b) Distance \(|HT|\) (the slant line to the top):
\[|HT| = \frac{20}{\cos 50^\circ} = \frac{20}{0.6428} = 31.1\text{ m (3 s.f.)}\]
(c) For the angle of depression of \(H\) from the top to be \(40^\circ\), the angle of elevation of \(T\) from \(H\) must be \(40^\circ\). With the top still \(23.835\) m high, the required horizontal distance \(x\) satisfies:
\[\tan 40^\circ = \frac{23.835}{x} \;\Rightarrow\; x = \frac{23.835}{0.8391} = 28.4\text{ m (3 s.f.)}\]
So \(H\) must be \(28.4\) m from the foot of the tower.
Antwortdetails
\(H\) is 20 m from the foot of the tower. From \(H\), \(P\) has elevation \(30^\circ\) and the top \(T\) has elevation \(50^\circ\).
Heights above ground of the two points:
\[\text{height of }P = 20\tan 30^\circ = 20(0.5774) = 11.547\text{ m}\]
\[\text{height of }T = 20\tan 50^\circ = 20(1.1918) = 23.835\text{ m}\]
(a) \(|PT| = 23.835 - 11.547 = 12.288 \approx 12.3\text{ m (3 s.f.)}\).
(b) Distance \(|HT|\) (the slant line to the top):
\[|HT| = \frac{20}{\cos 50^\circ} = \frac{20}{0.6428} = 31.1\text{ m (3 s.f.)}\]
(c) For the angle of depression of \(H\) from the top to be \(40^\circ\), the angle of elevation of \(T\) from \(H\) must be \(40^\circ\). With the top still \(23.835\) m high, the required horizontal distance \(x\) satisfies:
\[\tan 40^\circ = \frac{23.835}{x} \;\Rightarrow\; x = \frac{23.835}{0.8391} = 28.4\text{ m (3 s.f.)}\]
So \(H\) must be \(28.4\) m from the foot of the tower.
Frage 62 Bericht
(a) 
In the diagram, TU is tangent to the circle. < RVU = 100° and < URS = 36°. Calculate the value of angle STU.
(b) In triangle XYZ, |XY| = 5 cm, |YZ| = 8 cm and |XZ| = 6 cm. P is a point on the side XY such that |XP| = 2 cm and the line through P, parallel to YZ meets XZ at Q. Calculate |QZ|.
(a) Reading the diagram. \(R, V, S, U\) lie on the circle (\(R\) at top, \(V\) on the left, \(S\) on the right, \(U\) at the bottom). The line \(TU\) is the tangent that touches the circle at \(U\), and the secant through \(R\) and \(S\) is produced to meet the tangent at the external point \(T\). We are given \(\angle RVU = 100^\circ\) and \(\angle URS = 36^\circ\).
Step 1: Find arc \(SU\). \(\angle URS = 36^\circ\) is an inscribed angle at \(R\) standing on arc \(SU\) (the arc not containing \(R\)):
\[\text{arc } SU = 2\times 36^\circ = 72^\circ.\]
Step 2: Find arc \(RU\) on the far side. \(\angle RVU = 100^\circ\) is an inscribed angle at \(V\) standing on the arc \(RU\) that does not contain \(V\) (the arc going \(R\to S\to U\)):
\[\text{arc } RSU = \text{arc } RS + \text{arc } SU = 2\times 100^\circ = 200^\circ.\]
The remaining arc from \(R\) to \(U\) through \(V\) is therefore
\[\text{arc } RVU = 360^\circ - 200^\circ = 160^\circ.\]
Step 3: Apply the tangent-secant angle rule at \(T\). The angle between a tangent and a secant drawn from an external point equals half the difference of the two intercepted arcs. Here the far arc (between tangent point \(U\) and far point \(R\), through \(V\)) is \(160^\circ\) and the near arc (between \(U\) and near point \(S\)) is \(72^\circ\):
\[\angle STU = \tfrac{1}{2}\left(\text{arc } RVU - \text{arc } SU\right) = \tfrac{1}{2}(160^\circ - 72^\circ) = \tfrac{1}{2}(88^\circ) = 44^\circ.\]
Answer (a): \(\angle STU = 44^\circ\).
(b) Triangle \(XYZ\): \(|XY| = 5\), \(|YZ| = 8\), \(|XZ| = 6\text{ cm}\), with \(P\) on \(XY\), \(|XP| = 2\text{ cm}\), and \(PQ \parallel YZ\) with \(Q\) on \(XZ\).
Since \(PQ \parallel YZ\), triangle \(XPQ\) is similar to triangle \(XYZ\), so corresponding sides are proportional:
\[\frac{XP}{XY} = \frac{XQ}{XZ} \;\Rightarrow\; \frac{2}{5} = \frac{XQ}{6}.\]
\[XQ = \frac{2}{5}\times 6 = 2.4\text{ cm}.\]
\[|QZ| = |XZ| - |XQ| = 6 - 2.4 = 3.6\text{ cm}.\]
Answer (b): \(|QZ| = 3.6\text{ cm}\).
Antwortdetails
(a) Reading the diagram. \(R, V, S, U\) lie on the circle (\(R\) at top, \(V\) on the left, \(S\) on the right, \(U\) at the bottom). The line \(TU\) is the tangent that touches the circle at \(U\), and the secant through \(R\) and \(S\) is produced to meet the tangent at the external point \(T\). We are given \(\angle RVU = 100^\circ\) and \(\angle URS = 36^\circ\).
Step 1: Find arc \(SU\). \(\angle URS = 36^\circ\) is an inscribed angle at \(R\) standing on arc \(SU\) (the arc not containing \(R\)):
\[\text{arc } SU = 2\times 36^\circ = 72^\circ.\]
Step 2: Find arc \(RU\) on the far side. \(\angle RVU = 100^\circ\) is an inscribed angle at \(V\) standing on the arc \(RU\) that does not contain \(V\) (the arc going \(R\to S\to U\)):
\[\text{arc } RSU = \text{arc } RS + \text{arc } SU = 2\times 100^\circ = 200^\circ.\]
The remaining arc from \(R\) to \(U\) through \(V\) is therefore
\[\text{arc } RVU = 360^\circ - 200^\circ = 160^\circ.\]
Step 3: Apply the tangent-secant angle rule at \(T\). The angle between a tangent and a secant drawn from an external point equals half the difference of the two intercepted arcs. Here the far arc (between tangent point \(U\) and far point \(R\), through \(V\)) is \(160^\circ\) and the near arc (between \(U\) and near point \(S\)) is \(72^\circ\):
\[\angle STU = \tfrac{1}{2}\left(\text{arc } RVU - \text{arc } SU\right) = \tfrac{1}{2}(160^\circ - 72^\circ) = \tfrac{1}{2}(88^\circ) = 44^\circ.\]
Answer (a): \(\angle STU = 44^\circ\).
(b) Triangle \(XYZ\): \(|XY| = 5\), \(|YZ| = 8\), \(|XZ| = 6\text{ cm}\), with \(P\) on \(XY\), \(|XP| = 2\text{ cm}\), and \(PQ \parallel YZ\) with \(Q\) on \(XZ\).
Since \(PQ \parallel YZ\), triangle \(XPQ\) is similar to triangle \(XYZ\), so corresponding sides are proportional:
\[\frac{XP}{XY} = \frac{XQ}{XZ} \;\Rightarrow\; \frac{2}{5} = \frac{XQ}{6}.\]
\[XQ = \frac{2}{5}\times 6 = 2.4\text{ cm}.\]
\[|QZ| = |XZ| - |XQ| = 6 - 2.4 = 3.6\text{ cm}.\]
Answer (b): \(|QZ| = 3.6\text{ cm}\).
