WAEC SSCE - General Mathematics - 1992
Frage 1 Bericht
Write as a single fraction \(\frac{1}{1 - x} + \frac{2}{1 + x}\)
Antwortdetails
To add the fractions \(\frac{1}{1-x}\) and \(\frac{2}{1+x}\), we need to have a common denominator. We can obtain this by multiplying the first fraction by \((1+x)/(1+x)\) and the second fraction by \((1-x)/(1-x)\). This gives us: $$ \frac{1}{1-x} \cdot \frac{1+x}{1+x} + \frac{2}{1+x} \cdot \frac{1-x}{1-x} = \frac{1+x}{1-x^2} + \frac{2(1-x)}{1-x^2} $$ Combining the two fractions gives us: $$ \frac{1+x+2(1-x)}{1-x^2} = \frac{3-x}{1-x^2} $$ Therefore, the single fraction is \(\boxed{\frac{3-x}{1-x^2}}\).
Frage 2 Bericht
In the diagram above, ML//PQ and NP//QR, if ?LMN = 40° and ?MNP = 55°. Find ?QPR
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Frage 3 Bericht
If cos θ = 5/13, what is the value of tan \(\theta\) for 0 < θ < 90° ?
Antwortdetails
We can use the following trigonometric identity: \[\tan \theta = \frac{\sin \theta}{\cos \theta}\] We are given that $\cos \theta = 5/13$. To find $\sin \theta$, we can use the Pythagorean identity: \[\sin^2 \theta + \cos^2 \theta = 1\] Substituting $\cos \theta = 5/13$, we get: \[\sin^2 \theta + \left(\frac{5}{13}\right)^2 = 1\] Simplifying, we get: \[\sin^2 \theta = 1 - \frac{25}{169} = \frac{144}{169}\] Taking the square root of both sides, we get: \[\sin \theta = \frac{12}{13}\] Substituting into the formula for $\tan \theta$, we get: \[\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{12}{13}}{\frac{5}{13}} = \boxed{\frac{12}{5}}\]
Frage 4 Bericht
Construct a quadratic equation whose roots are \(-\frac{1}{2}\) and 2.
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Frage 5 Bericht
The nth term of a sequence is given by 3.2\(^{n-2}\). Write down the first three terms of the sequence.
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Frage 6 Bericht
Three balls are drawn one after the other with replacement, from a bag containing 5 red, 9white and 4 blue identical balls, What is the probability that they are one red, one white and one blue?
Antwortdetails
The probability of drawing one red ball out of 18 balls is 5/18. Similarly, the probability of drawing one white ball is 9/18 (or 1/2) and the probability of drawing one blue ball is 4/18 (or 2/9). Since we are drawing three balls, we can find the probability of getting one red, one white, and one blue by multiplying the probabilities of each event happening together. Since we are drawing with replacement, the probabilities do not change from draw to draw. Thus, the probability of getting one red, one white, and one blue is: (5/18) x (1/2) x (2/9) = 5/162 Therefore, the answer is option C: 5/162.
Frage 7 Bericht
In the diagram above, O is the center of the circle and ?POR = 126°. Find ?PQR
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Frage 8 Bericht
The angles marked in the diagram above are measured in degrees. Find x
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In the given diagram, we have a triangle with two sides of equal length, which means that the angles opposite these sides are also equal. Thus, we can conclude that: 4x - 6 = 3x + 12 (using the fact that angles in a triangle add up to 180 degrees) x = 18 Therefore, the value of x is 18 degrees. Answer: (b) 24°.
Frage 9 Bericht
The angle of a sector of a circle of radius 8cm is 240°. This sector is bent to form a cone. Find the radius of the base of the cone.
Antwortdetails
The sector of the circle with an angle of 240° forms a cone. To find the radius of the base of the cone, we need to use the formula relating the arc length of the sector and the circumference of the base of the cone. The arc length of the sector is (240/360) * 2π * 8 = (4/3) * π * 8 = (32/3)π cm. The circumference of the base of the cone is equal to the arc length of the sector, so it is (32/3)π cm. We know that the circumference of a circle is equal to 2πr, where r is the radius of the circle. Therefore, we can write: (32/3)π = 2πr Simplifying the equation, we get: r = (16/3) cm So, the radius of the base of the cone is 16/3 cm. Therefore, the correct option is: 16/3cm
Frage 10 Bericht
What must be added to the expression x\(^2\) - 18x to make it a perfect square?
Antwortdetails
To make the expression x\(^2\) - 18x a perfect square, we need to add a constant term. The constant term should be half of the coefficient of x squared, which is (18/2)\(^2\) = 81. Therefore, we need to add 81 to the expression to make it a perfect square. Adding 81 to the expression gives: x\(^2\) - 18x + 81 = (x - 9)\(^2\) Thus, the answer is 81.
Frage 11 Bericht
A cliff on the bank of a river is 300 metres
high. if the angle of depression of a point on the
opposite side of the river is 60° find the width of
the river.
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Frage 12 Bericht
A bricklayer measured the length of a wall and obtained 4.10m. If the actual length of the wall is 4.25m, find his percentage error.
Antwortdetails
To find the percentage error, we need to find the absolute difference between the measured value and actual value, divide it by the actual value, and then multiply by 100 to get a percentage. Absolute difference = |actual value - measured value| = |4.25 - 4.10| = 0.15 Percentage error = (|actual value - measured value| / actual value) x 100% Percentage error = (0.15 / 4.25) x 100% = 3.529% (rounded to 3 decimal places) Therefore, the percentage error is approximately 3.529%, which corresponds to 3 9/17%.
Frage 13 Bericht
In the diagram above, O is the center of the circle PQRS and ?QPS = 36°. Find ?QOS
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Frage 15 Bericht
In the diagram above, O is the center of the circle QOR is a diameter and ?PSR is 37°. Find ?PRQ
Antwortdetails
Since QOR is a diameter, it divides the circle into two equal halves, which means that ?QOR is a right angle (90°). Thus, we have: ?PQR = 180° - ?QOR (sum of angles in a triangle) ?PQR = 180° - 90° (since ?QOR is a right angle) ?PQR = 90° Now, we have a right triangle PQR with right angle at Q. Since we know that ?PSR is 37°, we can use the fact that the sum of angles in a triangle is 180° to find ?PRQ: ?PQR + ?QPR + ?PRQ = 180° 90° + ?QPR + 37° = 180° ?QPR = 180° - 90° - 37° ?QPR = 53° Therefore, the answer is 53°.
Frage 16 Bericht
| Marks | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| No of students | 1 | 3 | 2 | 0 | 1 | 6 | 1 | 0 | 1 | 0 |
The table above shows the scores of 15 students in a Physics test.
How many students scored at least 5?
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Frage 17 Bericht
The data below shows the number of worker?
employed in the various sections of a construction
company in Lagos.
Carpenters 24 Labourers 27
Plumbers 12 Plasterers 15
Painters 9 Messengers 3
Bricklayers 18
If one of the workers is absent on a certain day,
what is the probability that he is a bricklayer?
Antwortdetails
The total number of workers in the company is 24+27+12+15+9+3+18 = 108. The probability of an absent worker being a bricklayer is the ratio of the number of bricklayers to the total number of workers, which is 18/108 = 1/6. Therefore, the probability that the absent worker is a bricklayer is 1/6. Answer: 1/6.
Frage 18 Bericht
Sin 60° has the same value as I. Sin 120° II. cos 240° III. -sin 150° IV. cos 210° V. sin 240°
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Frage 19 Bericht
Solve the equation \(\frac{m}{3} + \frac{1}{2} = \frac{3}{4} + \frac{m}{4}\)
Antwortdetails
First, we simplify both sides of the equation by getting rid of the fractions. To do this, we can multiply both sides by the least common multiple (LCM) of the denominators, which is 12. So we have: \begin{align*} \frac{m}{3} + \frac{1}{2} &= \frac{3}{4} + \frac{m}{4}\\ 4 \cdot \left(\frac{m}{3} + \frac{1}{2}\right) &= 4 \cdot \left(\frac{3}{4} + \frac{m}{4}\right)\\ \frac{4m}{3} + 2 &= \frac{9}{4} + \frac{m}{4} \end{align*} Next, we can simplify further by moving all the terms involving m to one side of the equation, and all the constant terms to the other side: \begin{align*} \frac{4m}{3} - \frac{m}{4} &= \frac{9}{4} - 2\\ \frac{16m}{12} - \frac{3m}{12} &= \frac{9}{4} - \frac{8}{4}\\ \frac{13m}{12} &= \frac{1}{4}\\ 13m &= 3\\ m &= \frac{3}{13} \end{align*} Therefore, the solution to the equation is \(m = \frac{3}{13}\). So the correct answer is 3.
Frage 20 Bericht
The diagram above shows the shaded segment of a circle of radius 7cm. if the area of the triangle OXY is 12\(\frac{1}{4}\)cm\(^2\), calculate the area of the segment
[Take ? = 22/7]
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Frage 21 Bericht
A number is chosen at random from the set (1 ,2,3
....,9, 10). What is the probability that the number
is greater than or equal to 7?
Antwortdetails
There are 10 numbers in the set (1, 2, 3, ..., 9, 10). Out of these, the numbers that are greater than or equal to 7 are 7, 8, 9, and 10. So, there are 4 numbers that satisfy the given condition. Therefore, the probability of choosing a number that is greater than or equal to 7 is the number of numbers that satisfy the condition divided by the total number of numbers in the set, which is: 4/10 = 2/5 Hence, the probability that the number chosen at random from the set (1, 2, 3, ..., 9, 10) is greater than or equal to 7 is 2/5.
Frage 22 Bericht
Simplify (3/4 + 1/3) x 41/3 + 31/4
Antwortdetails
To simplify this expression, we need to follow the order of operations, which is Parentheses, Exponents, Multiplication and Division, and Addition and Subtraction (PEMDAS). First, we can simplify the expression inside the parentheses: (3/4 + 1/3) x 41/3 + 31/4 = (9/12 + 4/12) x 41/3 + 31/4 = 13/12 x 41/3 + 31/4 Next, we can simplify the exponent 41/3. To do this, we can convert the mixed number to an improper fraction: 41/3 = (4 x 3) + 1/3 = 13/3 Now we can substitute the value of 41/3 into the expression: 13/12 x 41/3 + 31/4 = 13/12 x 13/3 + 13/4 To simplify this, we can first simplify the multiplication of fractions: 13/12 x 13/3 = 169/36 Now we can add the two fractions: 169/36 + 13/4 = 169/36 + 39/36 = 208/36 = 13}{9 Therefore, the simplified expression is 13}{9, which corresponds to option (E).
Frage 23 Bericht
Points X and Y are respectively 12m Nonh and
5m of point Z. Calculate XY
Antwortdetails
To find the distance between points X and Y, we can use the Pythagorean theorem, which states that for a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In this case, we can consider points X, Y, and Z as the vertices of a right triangle, where the right angle is at point Z. Then, XY is the hypotenuse, and we know that XZ = 12m and YZ = 5m. Therefore, we have: XY^2 = XZ^2 + YZ^2 XY^2 = 12^2 + 5^2 XY^2 = 144 + 25 XY^2 = 169 XY = √169 XY = 13m Therefore, the distance between points X and Y is 13m. So the answer is option (C) 13m.
Frage 24 Bericht
Town P is on bearing 315o from town Q while town R is south of town P and west of town Q. lf town R is 60km away from Q, how far is R from P?
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Frage 25 Bericht
If x varies over the set of real numbers, which of the following is illustrated in the diagram above?
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Frage 26 Bericht
The value of tan 315°
Antwortdetails
To find the value of tan 315°, we can use the identity: tan (θ + 180°) = tan θ Since tan (315° + 180°) = tan 495°, and 495° is coterminal with 135°, we have: tan 315° = tan (135° + 180°) = tan 135° Using the fact that tan 45° = 1, and tan (θ + 90°) = -cot θ, we have: tan 135° = tan (45° + 90°) = -cot 45° = -1 Therefore, the value of tan 315° is -1.
Frage 27 Bericht
In the diagram above PQT is a tangent to the circle QRS at Q. Angle QTR = 48° and ?QRT = 95°. Find ?QRT
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Frage 28 Bericht
In the diagram above, PQT is an isosceles triangle.|PQ| = |QT|, ?SRQ = 75°, ?QPT = 25° and PQR is straight line. Find ?RST
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Frage 29 Bericht
Simplify: \((\frac{16}{81})^{\frac{1}{4}}\)
Antwortdetails
To simplify \((\frac{16}{81})^{\frac{1}{4}}\), we need to take the fourth root of \(\frac{16}{81}\). The fourth root of a number is the number that, when multiplied by itself four times, gives that number. So, to find the fourth root of \(\frac{16}{81}\), we need to find a number that, when multiplied by itself four times, gives \(\frac{16}{81}\). We can rewrite \(\frac{16}{81}\) as \((\frac{2}{3})^4\), since \(2^4=16\) and \(3^4=81\). Therefore, \((\frac{16}{81})^{\frac{1}{4}} = ((\frac{2}{3})^4)^{\frac{1}{4}} = \frac{2}{3}\), which is.
Frage 30 Bericht
Evaluate \(\log_{10} 25 + \log_{10} 32 - \log_{10} 8\)
Antwortdetails
We can use the following property of logarithms: - \(\log_{a}b + \log_{a}c = \log_{a}(bc)\) - \(\log_{a}b - \log_{a}c = \log_{a}\left(\dfrac{b}{c}\right)\) Using this property, we can rewrite the expression as: \begin{align*} \log_{10} 25 + \log_{10} 32 - \log_{10} 8 &= \log_{10}(25\cdot 32) - \log_{10} 8 \\ &= \log_{10}(800) - \log_{10} 8 \\ &= \log_{10}\left(\frac{800}{8}\right) \\ &= \log_{10}(100) \\ &= 2 \end{align*} Therefore, the value of the expression is 2. Thus, the answer is (2).
Frage 31 Bericht
The curved surface area of a cylindrical tin is 704cm\(^2\). Calculate the height when the radius is 8cm. [Take π = 22/7]
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Frage 32 Bericht
The data below shows the number of worker?
employed in the various sections of a construction
company in Lagos.
Carpenters 24 Labourers 27
Plumbers 12 Plasterers 15
Painters 9 Messengers 3
Bricklayers 18
If a worker is retrenched, what is the probability that he is a plumber or plasterer.
Antwortdetails
To calculate the probability that a retrenched worker is a plumber or plasterer, we need to first find the total number of workers in the company, and then find the number of workers who are plumbers or plasterers. The total number of workers in the company is: 24 (Carpenters) + 27 (Labourers) + 12 (Plumbers) + 15 (Plasterers) + 9 (Painters) + 3 (Messengers) + 18 (Bricklayers) = 108 The number of workers who are plumbers or plasterers is: 12 (Plumbers) + 15 (Plasterers) = 27 Therefore, the probability that a retrenched worker is a plumber or plasterer is: 27 (number of plumbers and plasterers) / 108 (total number of workers) = 1/4 So the correct answer is option D: 1/4.
Frage 33 Bericht
The data below shows the frequency distribution
of marks scored by a group of students in a class
test.
What is the modal score
Antwortdetails
We find the modal score by identifying the score that occurs most frequently in the data. Looking at the frequency distribution, we can see that the score 4 appears most frequently, with a frequency of 5. Therefore, the modal score is 4.
Frage 34 Bericht
In the diagram above, TR = TS and ?TRS = 60o. Find the value of x
Antwortdetails
Since TR = TS and ?TRS = 60o, then ?TSR = ?TRS = 60o. Therefore, the angles of triangle TSP add up to 180o: ?TPS + ?TSP + ?TSR = 180o. Substituting the known values: x + 60 + 60 = 180. Simplifying: x = 60. Hence, the value of x is 60.
Frage 35 Bericht
A regular polygon has each of its exterior angles as 18o. How many sides has the polygon
Antwortdetails
The exterior angles of a polygon add up to 360 degrees. If each exterior angle of a regular polygon is 18 degrees, then the polygon has 360/18 = 20 exterior angles. Since a regular polygon has the same number of sides as exterior angles, the polygon has 20 sides. Therefore, the answer is (d) 20.
Frage 36 Bericht
The area of a parallelogram is 513cm\(^2\) and the height is 19cm. Calculate the base.
Antwortdetails
The area of a parallelogram is given by the formula: Area = base × height In this problem, the area and height of the parallelogram are given as 513cm\(^2\) and 19cm respectively. We can rearrange the formula to solve for the base: base = Area ÷ height Substituting the given values, we get: base = 513cm\(^2\) ÷ 19cm = 27cm Therefore, the base of the parallelogram is 27cm. The answer is, 27cm.
Frage 38 Bericht
Find the median of the following set of scores:
65, 72, 55, 48, 78
Antwortdetails
To find the median of a set of scores, we need to arrange the scores in order of magnitude from smallest to largest. The scores in this set are: 48, 55, 65, 72, 78 There are five scores in this set, so the median will be the middle score. Since the scores are already in order, we can simply find the middle score, which is 65. Therefore, the median of this set of scores is 65.
Frage 39 Bericht
A ladder leans against a vertical wall at an angle
60° to the wall, if the foot of the ladder is 5 metres
away from the wall, calculate the length of the
ladder.
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Frage 41 Bericht
The data above shows the frequency distribution
of marks scored by a group of students in a class
test.
How many students took the test?
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Frage 42 Bericht
A group of students measured a certain angle
(to the nearest degree) and obtained the following
results: 75o, 76o, 72o, 73o, 74o, 79o, 72o, 72o, 77o,
72o, 71o, 70o, 78o, 73o. Find the mode
Antwortdetails
To find the mode of the given set of data, we need to identify the value that appears most frequently. From the given set of measurements, we can see that 72o appears 4 times, which is more than any other value. Therefore, the mode of the data set is 72o. So, the correct option is (e) 72o.
Frage 43 Bericht
Find the value of t in the diagram above
Antwortdetails
Using the Pythagorean theorem, we can find that the length of the hypotenuse of the right triangle on the left is: $$\sqrt{5^2+12^2} = 13$$ Since the triangle on the right is similar to the triangle on the left, we can set up the following proportion: $$\frac{5}{12} = \frac{t}{13}$$ Cross-multiplying, we get: $$12t = 65$$ Therefore: $$t = \frac{65}{12} \approx 5.4$$ So the answer is closest to 5.6cm. Therefore the correct option is (A).
Frage 44 Bericht
Let U = {1, 2, 3, 4}, P = {2, 3} and Q = {2, 4}. What is (P∩Q)'?
Antwortdetails
In set theory, the symbol ∩ denotes the intersection of two sets, meaning the set of elements that are in both sets. Therefore, P∩Q denotes the set of elements that are in both P and Q, which is {2}. The complement of a set A, denoted by A', is the set of elements that are not in A but are in the universal set U. Therefore, (P∩Q)' denotes the set of elements that are in U but are not in P∩Q, which is {1, 3, 4}. Thus, the correct answer is: (1, 3, 4).
Frage 45 Bericht
A fair die is rolled once. What is the probability of
obtaining 4 or 6?
Antwortdetails
When you roll a fair die, there are six possible outcomes, each with an equal chance of occurring. These outcomes are the numbers 1, 2, 3, 4, 5, and 6. To find the probability of obtaining 4 or 6, we need to count how many of the outcomes correspond to these numbers. In this case, there are two outcomes that satisfy the condition, namely 4 and 6. Therefore, the probability of obtaining 4 or 6 is 2 out of 6, or 2/6. We can simplify this fraction by dividing both the numerator and denominator by their greatest common factor, which is 2. This gives us 1/3. Therefore, the probability of obtaining 4 or 6 when rolling a fair die is 1/3.
Frage 46 Bericht
The volume of a cone of height 9cm is 1848cm\(^3\). Find its radius. [Take π = 22/7]
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Frage 47 Bericht
The data below shows the frequency distribution
of marks scored by a group of students in a class
test.
Find the mean mark.
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Frage 48 Bericht
What is the probability of throwing a number greater than 2 with a single fair die
Antwortdetails
The total number of outcomes when rolling a fair die is 6 since there are six sides on the die. The numbers that are greater than 2 are 3, 4, 5, and 6. So, there are four possible outcomes that are greater than 2. The probability of getting a number greater than 2 with a single fair die is the number of favorable outcomes divided by the total number of possible outcomes, which is 4/6 or 2/3. Therefore, the answer is (d) 2/3.
Frage 49 Bericht
(a) Solve the following pair of simultaneous equations: \(2x + 5y = 6\frac{1}{2} ; 5x - 2y = 9\)
(b) If \(\log_{10} (2x + 1) - \log_{10} (3x - 2) = 1\), find x.
(a) \(2x+5y=\tfrac{13}{2}\) and \(5x-2y=9\). Multiply the first by \(2\) and the second by \(5\): \[4x+10y=13,\qquad25x-10y=45.\] Adding: \(29x=58\Rightarrow x=2\). Then \(2(2)+5y=6\tfrac12\Rightarrow5y=2\tfrac12\Rightarrow y=\tfrac12\). So \(x=2,\;y=\tfrac12\).
(b) \[\log_{10}(2x+1)-\log_{10}(3x-2)=1\Rightarrow\log_{10}\frac{2x+1}{3x-2}=1\Rightarrow\frac{2x+1}{3x-2}=10.\] \[2x+1=30x-20\Rightarrow28x=21\Rightarrow x=\frac34.\] (Check: \(3x-2=\tfrac14>0\), valid.)
Antwortdetails
(a) \(2x+5y=\tfrac{13}{2}\) and \(5x-2y=9\). Multiply the first by \(2\) and the second by \(5\): \[4x+10y=13,\qquad25x-10y=45.\] Adding: \(29x=58\Rightarrow x=2\). Then \(2(2)+5y=6\tfrac12\Rightarrow5y=2\tfrac12\Rightarrow y=\tfrac12\). So \(x=2,\;y=\tfrac12\).
(b) \[\log_{10}(2x+1)-\log_{10}(3x-2)=1\Rightarrow\log_{10}\frac{2x+1}{3x-2}=1\Rightarrow\frac{2x+1}{3x-2}=10.\] \[2x+1=30x-20\Rightarrow28x=21\Rightarrow x=\frac34.\] (Check: \(3x-2=\tfrac14>0\), valid.)
Frage 50 Bericht
(a) The distribution of junior workers in an institution is as follows: Clerks - 78, Drivers - 36, Typists - 44, Messengers - 52, Others - 30. Represent the above information by a pie chart.
(b) The table below shows the frequency distribution of marks scored by 30 candidates in an aptitude test.
| Marks | 4 | 5 | 6 | 7 | 8 | 9 |
| No of candidates | 5 | 8 | 5 | 6 | 4 | 2 |
Find the mean score to the nearest whole number.
(a) Pie chart of junior workers
First find the total number of workers:
\[ 78+36+44+52+30 = 240 \]The angle of each sector is a fraction of the whole circle \(\left(360^\circ\right)\):
\[ \text{Sector angle} = \frac{\text{Number in category}}{240}\times 360^\circ \]| Category | Number | Sector angle |
|---|---|---|
| Clerks | 78 | \(\frac{78}{240}\times360^\circ=117^\circ\) |
| Drivers | 36 | \(\frac{36}{240}\times360^\circ=54^\circ\) |
| Typists | 44 | \(\frac{44}{240}\times360^\circ=66^\circ\) |
| Messengers | 52 | \(\frac{52}{240}\times360^\circ=78^\circ\) |
| Others | 30 | \(\frac{30}{240}\times360^\circ=45^\circ\) |
| Total | 240 | \(360^\circ\) |
The five sector angles add up to \(117^\circ+54^\circ+66^\circ+78^\circ+45^\circ=360^\circ\), confirming the calculation. Drawn to scale with a protractor, the pie chart is:
(b) Mean score of the aptitude test
Multiply each mark \(x\) by its frequency \(f\) to obtain \(fx\):
| Marks \((x)\) | 4 | 5 | 6 | 7 | 8 | 9 | Total |
|---|---|---|---|---|---|---|---|
| No. of candidates \((f)\) | 5 | 8 | 5 | 6 | 4 | 2 | 30 |
| \(fx\) | 20 | 40 | 30 | 42 | 32 | 18 | 182 |
To the nearest whole number, the mean score is 6 marks.
Antwortdetails
(a) Pie chart of junior workers
First find the total number of workers:
\[ 78+36+44+52+30 = 240 \]The angle of each sector is a fraction of the whole circle \(\left(360^\circ\right)\):
\[ \text{Sector angle} = \frac{\text{Number in category}}{240}\times 360^\circ \]| Category | Number | Sector angle |
|---|---|---|
| Clerks | 78 | \(\frac{78}{240}\times360^\circ=117^\circ\) |
| Drivers | 36 | \(\frac{36}{240}\times360^\circ=54^\circ\) |
| Typists | 44 | \(\frac{44}{240}\times360^\circ=66^\circ\) |
| Messengers | 52 | \(\frac{52}{240}\times360^\circ=78^\circ\) |
| Others | 30 | \(\frac{30}{240}\times360^\circ=45^\circ\) |
| Total | 240 | \(360^\circ\) |
The five sector angles add up to \(117^\circ+54^\circ+66^\circ+78^\circ+45^\circ=360^\circ\), confirming the calculation. Drawn to scale with a protractor, the pie chart is:
(b) Mean score of the aptitude test
Multiply each mark \(x\) by its frequency \(f\) to obtain \(fx\):
| Marks \((x)\) | 4 | 5 | 6 | 7 | 8 | 9 | Total |
|---|---|---|---|---|---|---|---|
| No. of candidates \((f)\) | 5 | 8 | 5 | 6 | 4 | 2 | 30 |
| \(fx\) | 20 | 40 | 30 | 42 | 32 | 18 | 182 |
To the nearest whole number, the mean score is 6 marks.
Frage 51 Bericht
(a) The sides PQ and PR of \(\Delta\) PQR are produced to T and S respectively, such that TQR = 131° and < QRS = 98°. Find < QPR.
(b) The circumference of a circular track is 400m. Find its radius, correct to the nearest metre. [Take \(\pi = \frac{22}{7}\)]
(a) \(PQ\) is produced to \(T\), so \(\angle PQR=180^{\circ}-131^{\circ}=49^{\circ}\). \(PR\) is produced to \(S\), so \(\angle PRQ=180^{\circ}-98^{\circ}=82^{\circ}\). The angles of \(\triangle PQR\) sum to \(180^{\circ}\): \[\angle QPR=180^{\circ}-49^{\circ}-82^{\circ}=49^{\circ}.\]
(b) Circumference \(=2\pi r=400\): \[r=\frac{400}{2\times\frac{22}{7}}=\frac{400\times7}{44}=\frac{2800}{44}=63.6\text{ m}\approx64\text{ m}.\]
Antwortdetails
(a) \(PQ\) is produced to \(T\), so \(\angle PQR=180^{\circ}-131^{\circ}=49^{\circ}\). \(PR\) is produced to \(S\), so \(\angle PRQ=180^{\circ}-98^{\circ}=82^{\circ}\). The angles of \(\triangle PQR\) sum to \(180^{\circ}\): \[\angle QPR=180^{\circ}-49^{\circ}-82^{\circ}=49^{\circ}.\]
(b) Circumference \(=2\pi r=400\): \[r=\frac{400}{2\times\frac{22}{7}}=\frac{400\times7}{44}=\frac{2800}{44}=63.6\text{ m}\approx64\text{ m}.\]
Frage 52 Bericht
(a) Copy and complete the following table for the relation \(y = \frac{5}{2} + x - 4x^{2}\)
| x | -2.0 | -1.5 | -1.0 | -0.5 | 0 | 0.5 | 1 | 1.5 | 2.0 |
| y | -15.5 | 1 | 2.5 |
(b) Using a scale of 2cm to 1 unit on the x- axis and 2cm to 5 units on the y- axis, draw the graph of the relation for \(-2.0 \leq x \leq 2.0\).
(c) What is the maximum value of y?
(d) From your graph, obtain the roots of the equation \(8x^{2} - 2x - 5 = 0\)
(a) Completing the table for \(y=\dfrac{5}{2}+x-4x^{2}\).
Each missing value is found by substitution, e.g.
\[ x=-1.5:\; y=2.5+(-1.5)-4(-1.5)^2 = 2.5-1.5-9 = -8.0 \] \[ x=1:\; y=2.5+1-4(1)^2 = 2.5+1-4 = -0.5 \] \[ x=1.5:\; y=2.5+1.5-4(1.5)^2 = 2.5+1.5-9 = -5.0 \] \[ x=2.0:\; y=2.5+2-4(2)^2 = 2.5+2-16 = -11.5 \]| x | -2.0 | -1.5 | -1.0 | -0.5 | 0 | 0.5 | 1 | 1.5 | 2.0 |
| y | -15.5 | -8.0 | -2.5 | 1.0 | 2.5 | 2.0 | -0.5 | -5.0 | -11.5 |
(b) Graph of \(y=\dfrac{5}{2}+x-4x^{2}\) for \(-2.0\le x\le 2.0\). Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, the nine points from the table are plotted and joined with a smooth curve (an inverted parabola).
(c) Maximum value of y. The curve turns at its highest point where \(x=-\dfrac{b}{2a}=-\dfrac{1}{2(-4)}=0.125\). Reading the top of the curve from the graph:
\[ y_{\max}=2.5+0.125-4(0.125)^2 = 2.5+0.125-0.0625 \approx 2.6 \]So the maximum value of \(y\) is approximately 2.6.
(d) Roots of \(8x^{2}-2x-5=0\). Divide the equation through by \(-2\):
\[ -4x^{2}+x+\tfrac{5}{2}=0 \quad\Longrightarrow\quad \tfrac{5}{2}+x-4x^{2}=0 \]This is exactly \(y=0\), so the required roots are the values of \(x\) where the graph cuts the x-axis. Reading these two intercepts from the curve:
\[ x \approx -0.7 \qquad\text{and}\qquad x \approx 0.9 \](Check by formula: \(x=\dfrac{1\pm\sqrt{1+40}}{8}=\dfrac{1\pm 6.40}{8}\), giving \(x=-0.68\) or \(x=0.93\), which agrees with the graph.)
Antwortdetails
(a) Completing the table for \(y=\dfrac{5}{2}+x-4x^{2}\).
Each missing value is found by substitution, e.g.
\[ x=-1.5:\; y=2.5+(-1.5)-4(-1.5)^2 = 2.5-1.5-9 = -8.0 \] \[ x=1:\; y=2.5+1-4(1)^2 = 2.5+1-4 = -0.5 \] \[ x=1.5:\; y=2.5+1.5-4(1.5)^2 = 2.5+1.5-9 = -5.0 \] \[ x=2.0:\; y=2.5+2-4(2)^2 = 2.5+2-16 = -11.5 \]| x | -2.0 | -1.5 | -1.0 | -0.5 | 0 | 0.5 | 1 | 1.5 | 2.0 |
| y | -15.5 | -8.0 | -2.5 | 1.0 | 2.5 | 2.0 | -0.5 | -5.0 | -11.5 |
(b) Graph of \(y=\dfrac{5}{2}+x-4x^{2}\) for \(-2.0\le x\le 2.0\). Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, the nine points from the table are plotted and joined with a smooth curve (an inverted parabola).
(c) Maximum value of y. The curve turns at its highest point where \(x=-\dfrac{b}{2a}=-\dfrac{1}{2(-4)}=0.125\). Reading the top of the curve from the graph:
\[ y_{\max}=2.5+0.125-4(0.125)^2 = 2.5+0.125-0.0625 \approx 2.6 \]So the maximum value of \(y\) is approximately 2.6.
(d) Roots of \(8x^{2}-2x-5=0\). Divide the equation through by \(-2\):
\[ -4x^{2}+x+\tfrac{5}{2}=0 \quad\Longrightarrow\quad \tfrac{5}{2}+x-4x^{2}=0 \]This is exactly \(y=0\), so the required roots are the values of \(x\) where the graph cuts the x-axis. Reading these two intercepts from the curve:
\[ x \approx -0.7 \qquad\text{and}\qquad x \approx 0.9 \](Check by formula: \(x=\dfrac{1\pm\sqrt{1+40}}{8}=\dfrac{1\pm 6.40}{8}\), giving \(x=-0.68\) or \(x=0.93\), which agrees with the graph.)
Frage 53 Bericht
(a) The angle of a sector of a circle radius 7cm is 108°. Calculate the perimeter of the sector. [Take \(\pi = \frac{22}{7}\)]
(b) A boat is on the same horizontal level as the foot of a cliff, and the angle of depression of the boat from the top of the cliff is 30°. If the boat is 120m away from the foot of the cliff, find the height of the cliff correct to three significant figures.
(a) Radius \(r=7\text{ cm}\), angle \(=108^{\circ}\). Arc length \[=\frac{108}{360}\times2\pi r=\frac{108}{360}\times2\times\frac{22}{7}\times7=\frac{3}{10}\times44=13.2\text{ cm}.\] Perimeter of sector \(=\) arc \(+2r=13.2+14=\) 27.2 cm.
(b) Angle of depression \(=30^{\circ}\) equals the angle of elevation of the cliff top from the boat. With the boat \(120\text{ m}\) from the foot: \[\tan30^{\circ}=\frac{h}{120}\Rightarrow h=120\tan30^{\circ}=120\times0.5774=69.28\text{ m}\approx69.3\text{ m}.\]
Antwortdetails
(a) Radius \(r=7\text{ cm}\), angle \(=108^{\circ}\). Arc length \[=\frac{108}{360}\times2\pi r=\frac{108}{360}\times2\times\frac{22}{7}\times7=\frac{3}{10}\times44=13.2\text{ cm}.\] Perimeter of sector \(=\) arc \(+2r=13.2+14=\) 27.2 cm.
(b) Angle of depression \(=30^{\circ}\) equals the angle of elevation of the cliff top from the boat. With the boat \(120\text{ m}\) from the foot: \[\tan30^{\circ}=\frac{h}{120}\Rightarrow h=120\tan30^{\circ}=120\times0.5774=69.28\text{ m}\approx69.3\text{ m}.\]
Frage 54 Bericht
(a) If \(9^{2x - 1} = \frac{81^{x - 2}}{3^{x}}\), find x.
(b) Without using Mathematical Tables, evaluate: \(\sqrt{\frac{0.81 \times 10^{-5}}{2.25 \times 10^{7}}}\)
(a) Write everything to base \(3\): \(9=3^{2}\), \(81=3^{4}\). \[9^{2x-1}=3^{4x-2},\qquad\frac{81^{x-2}}{3^{x}}=\frac{3^{4x-8}}{3^{x}}=3^{3x-8}.\] So \(4x-2=3x-8\Rightarrow x=-6\).
(b) \[\sqrt{\frac{0.81\times10^{-5}}{2.25\times10^{7}}}=\sqrt{\frac{0.81}{2.25}\times10^{-12}}=\sqrt{0.36\times10^{-12}}=0.6\times10^{-6}=6.0\times10^{-7}.\]
Antwortdetails
(a) Write everything to base \(3\): \(9=3^{2}\), \(81=3^{4}\). \[9^{2x-1}=3^{4x-2},\qquad\frac{81^{x-2}}{3^{x}}=\frac{3^{4x-8}}{3^{x}}=3^{3x-8}.\] So \(4x-2=3x-8\Rightarrow x=-6\).
(b) \[\sqrt{\frac{0.81\times10^{-5}}{2.25\times10^{7}}}=\sqrt{\frac{0.81}{2.25}\times10^{-12}}=\sqrt{0.36\times10^{-12}}=0.6\times10^{-6}=6.0\times10^{-7}.\]
Frage 55 Bericht
The table below shows the weekly profit in naira from a mini-market.
| Weekly profit (N) | 1-10 | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 |
| Freq | 6 | 6 | 12 | 11 | 10 | 5 |
(a) Draw the cumulative frequency curve of the data;
(b) From your graph, estimate the: (i) median; (ii) 80th percentile
(c) What is the modal weekly profit?
(a) Cumulative frequency table and curve (ogive)
The cumulative frequency is plotted against the upper class boundary of each class. The boundaries are obtained by adding \(0.5\) to each upper class limit.
| Weekly profit (N) | Upper class boundary | Frequency (f) | Cumulative frequency |
|---|---|---|---|
| 1 - 10 | 10.5 | 6 | 6 |
| 11 - 20 | 20.5 | 6 | 12 |
| 21 - 30 | 30.5 | 12 | 24 |
| 31 - 40 | 40.5 | 11 | 35 |
| 41 - 50 | 50.5 | 10 | 45 |
| 51 - 60 | 60.5 | 5 | 50 |
Starting the curve at the lower boundary \((0.5,\,0)\) and plotting each point \((\text{upper boundary},\ \text{cumulative frequency})\), then joining them with a smooth curve, gives the ogive below. Here \(N=50\).
(b) Estimates from the graph
(i) Median. The median corresponds to a cumulative frequency of \(\dfrac{N}{2}=\dfrac{50}{2}=25\). A horizontal line from cumulative frequency \(25\) meets the curve, and the vertical line down to the profit axis gives the median.
Reading from the graph (confirmed by interpolation between the points \((30.5,\,24)\) and \((40.5,\,35)\)):
\[ \text{Median}=30.5+\frac{25-24}{35-24}\times 10 = 30.5+0.9 \approx N\,31 \](ii) 80th percentile. This corresponds to a cumulative frequency of \(\dfrac{80}{100}\times 50 = 40\). A horizontal line from \(40\) meets the curve, and the vertical drop gives the 80th percentile.
Reading from the graph (confirmed by interpolation between \((40.5,\,35)\) and \((50.5,\,45)\)):
\[ P_{80}=40.5+\frac{40-35}{45-35}\times 10 = 40.5+5 = N\,45.5 \](c) Modal weekly profit
The modal class is the class with the highest frequency, i.e. 21 - 30 (frequency \(12\)). Using
\[ \text{Mode}=L+\frac{(f-f_1)}{(f-f_1)+(f-f_2)}\times c \]where \(L=20.5\) (lower boundary of the modal class), \(f=12\), \(f_1=6\) (frequency of the class before), \(f_2=11\) (frequency of the class after) and \(c=10\):
\[ \text{Mode}=20.5+\frac{(12-6)}{(12-6)+(12-11)}\times 10 = 20.5+\frac{60}{7} = 20.5+8.57 \]\[ \boxed{\text{Modal weekly profit}\approx N\,29.07} \]Antwortdetails
(a) Cumulative frequency table and curve (ogive)
The cumulative frequency is plotted against the upper class boundary of each class. The boundaries are obtained by adding \(0.5\) to each upper class limit.
| Weekly profit (N) | Upper class boundary | Frequency (f) | Cumulative frequency |
|---|---|---|---|
| 1 - 10 | 10.5 | 6 | 6 |
| 11 - 20 | 20.5 | 6 | 12 |
| 21 - 30 | 30.5 | 12 | 24 |
| 31 - 40 | 40.5 | 11 | 35 |
| 41 - 50 | 50.5 | 10 | 45 |
| 51 - 60 | 60.5 | 5 | 50 |
Starting the curve at the lower boundary \((0.5,\,0)\) and plotting each point \((\text{upper boundary},\ \text{cumulative frequency})\), then joining them with a smooth curve, gives the ogive below. Here \(N=50\).
(b) Estimates from the graph
(i) Median. The median corresponds to a cumulative frequency of \(\dfrac{N}{2}=\dfrac{50}{2}=25\). A horizontal line from cumulative frequency \(25\) meets the curve, and the vertical line down to the profit axis gives the median.
Reading from the graph (confirmed by interpolation between the points \((30.5,\,24)\) and \((40.5,\,35)\)):
\[ \text{Median}=30.5+\frac{25-24}{35-24}\times 10 = 30.5+0.9 \approx N\,31 \](ii) 80th percentile. This corresponds to a cumulative frequency of \(\dfrac{80}{100}\times 50 = 40\). A horizontal line from \(40\) meets the curve, and the vertical drop gives the 80th percentile.
Reading from the graph (confirmed by interpolation between \((40.5,\,35)\) and \((50.5,\,45)\)):
\[ P_{80}=40.5+\frac{40-35}{45-35}\times 10 = 40.5+5 = N\,45.5 \](c) Modal weekly profit
The modal class is the class with the highest frequency, i.e. 21 - 30 (frequency \(12\)). Using
\[ \text{Mode}=L+\frac{(f-f_1)}{(f-f_1)+(f-f_2)}\times c \]where \(L=20.5\) (lower boundary of the modal class), \(f=12\), \(f_1=6\) (frequency of the class before), \(f_2=11\) (frequency of the class after) and \(c=10\):
\[ \text{Mode}=20.5+\frac{(12-6)}{(12-6)+(12-11)}\times 10 = 20.5+\frac{60}{7} = 20.5+8.57 \]\[ \boxed{\text{Modal weekly profit}\approx N\,29.07} \]Frage 56 Bericht
(a) 
Calculate the area of the shaded segment of the circle shown in the diagram [Take \(\pi = \frac{22}{7}\)]
(b) A tin has radius 3cm and height 6cm. Find the (i) total surface area of the tin ; (ii) volume, in litres, that will fill the tin to capacity, correct to two decimal places.
[Take \(\pi = \frac{22}{7}\)]
Antwortdetails
None
Frage 57 Bericht
(a) If \(17x = 375^{2} - 356^{2}\), find the exact value of x.
(b) If \(4^{x} = 2^{\frac{1}{2}} \times 8\), find x.
(c) The sum of the first 9 terms of an A.P is 72 and the sum of the next 4 terms is 71, find the A.P.
(a) Using difference of two squares: \[375^{2}-356^{2}=(375-356)(375+356)=19\times731=13889.\] So \(17x=13889\Rightarrow x=\dfrac{13889}{17}=817\).
(b) \(4^{x}=2^{2x}\) and \(2^{\frac12}\times8=2^{\frac12}\times2^{3}=2^{\frac72}\). Thus \(2x=\tfrac72\Rightarrow x=\tfrac74\).
(c) \(S_{9}=72\Rightarrow\tfrac92(2a+8d)=72\Rightarrow a+4d=8\). Sum of next \(4\) terms \(=S_{13}-S_{9}=71\Rightarrow S_{13}=143\Rightarrow\tfrac{13}{2}(2a+12d)=143\Rightarrow a+6d=11\). Subtracting: \(2d=3\Rightarrow d=1.5\), then \(a=2\). The AP is \[2,\;3\tfrac12,\;5,\;6\tfrac12,\;8,\;\dots\]
Antwortdetails
(a) Using difference of two squares: \[375^{2}-356^{2}=(375-356)(375+356)=19\times731=13889.\] So \(17x=13889\Rightarrow x=\dfrac{13889}{17}=817\).
(b) \(4^{x}=2^{2x}\) and \(2^{\frac12}\times8=2^{\frac12}\times2^{3}=2^{\frac72}\). Thus \(2x=\tfrac72\Rightarrow x=\tfrac74\).
(c) \(S_{9}=72\Rightarrow\tfrac92(2a+8d)=72\Rightarrow a+4d=8\). Sum of next \(4\) terms \(=S_{13}-S_{9}=71\Rightarrow S_{13}=143\Rightarrow\tfrac{13}{2}(2a+12d)=143\Rightarrow a+6d=11\). Subtracting: \(2d=3\Rightarrow d=1.5\), then \(a=2\). The AP is \[2,\;3\tfrac12,\;5,\;6\tfrac12,\;8,\;\dots\]
Frage 58 Bericht
(a) Using a ruler and a pair of compasses only, construct: (i) a triangle ABC such that |AB| = 5cm, |AC| = 7.5cm and < CAB = 120°; (ii) the locus \(l_{1}\) of points equidistant from A and B; (iii) the locus \(l_{2}\) of points equidistant from AB and AC which passes through triangle ABC .
(b) Label the point P where \(l_{1}\) and \(l_{2}\) intersect.
(c) Measure |CP|.
(a) Construction (ruler and a pair of compasses only, drawn to scale 1 cm to 1 cm):
Steps.
- Draw a base line and mark \(|AB| = 5\text{ cm}\).
- Construct \(\angle CAB = 120^{\circ}\) at \(A\). With centre \(A\) draw an arc cutting \(AB\); step the radius round the arc twice to obtain \(60^{\circ}\) and then \(120^{\circ}\), and rule the arm \(AX\) through the \(120^{\circ}\) mark.
- With centre \(A\) and radius \(7.5\text{ cm}\) cut arm \(AX\) at \(C\), so \(|AC| = 7.5\text{ cm}\). Join \(BC\); triangle \(ABC\) is complete.
- Locus \(l_{1}\): the perpendicular bisector of \(AB\). With centre \(A\), then centre \(B\), and the same radius (a little more than half of \(|AB|\)), draw arcs that meet above and below \(AB\); the line through the two meeting points is \(l_{1}\), the locus of points equidistant from \(A\) and \(B\).
- Locus \(l_{2}\): the bisector of \(\angle CAB\). With centre \(A\) draw an arc cutting both \(AB\) and \(AC\); from those two cuts draw equal arcs meeting inside the angle, and rule \(l_{2}\) from \(A\) through that meeting point. \(l_{2}\) is the locus of points equidistant from lines \(AB\) and \(AC\).
(b) \(l_{1}\) and \(l_{2}\) intersect at the point marked \(P\) inside the triangle.
(c) Measurement. Measuring directly from the construction, \(|CP| = 6.7\text{ cm}\) (accept \(6.5\text{ cm}\) to \(6.7\text{ cm}\)).
This agrees with a coordinate check. Taking \(A(0,0)\) and \(B(5,0)\), then \(C = 7.5(\cos120^{\circ},\ \sin120^{\circ}) = (-3.75,\ 6.50)\). The perpendicular bisector of \(AB\) is \(x = 2.5\), and the bisector of the \(120^{\circ}\) angle rises at \(60^{\circ}\), so \(P = (2.5,\ 2.5\tan60^{\circ}) = (2.5,\ 4.33)\). Hence
\[|CP| = \sqrt{(2.5-(-3.75))^{2} + (4.33-6.50)^{2}} = \sqrt{6.25^{2} + 2.17^{2}} = \sqrt{43.75} \approx 6.7\text{ cm}.\]Antwortdetails
(a) Construction (ruler and a pair of compasses only, drawn to scale 1 cm to 1 cm):
Steps.
- Draw a base line and mark \(|AB| = 5\text{ cm}\).
- Construct \(\angle CAB = 120^{\circ}\) at \(A\). With centre \(A\) draw an arc cutting \(AB\); step the radius round the arc twice to obtain \(60^{\circ}\) and then \(120^{\circ}\), and rule the arm \(AX\) through the \(120^{\circ}\) mark.
- With centre \(A\) and radius \(7.5\text{ cm}\) cut arm \(AX\) at \(C\), so \(|AC| = 7.5\text{ cm}\). Join \(BC\); triangle \(ABC\) is complete.
- Locus \(l_{1}\): the perpendicular bisector of \(AB\). With centre \(A\), then centre \(B\), and the same radius (a little more than half of \(|AB|\)), draw arcs that meet above and below \(AB\); the line through the two meeting points is \(l_{1}\), the locus of points equidistant from \(A\) and \(B\).
- Locus \(l_{2}\): the bisector of \(\angle CAB\). With centre \(A\) draw an arc cutting both \(AB\) and \(AC\); from those two cuts draw equal arcs meeting inside the angle, and rule \(l_{2}\) from \(A\) through that meeting point. \(l_{2}\) is the locus of points equidistant from lines \(AB\) and \(AC\).
(b) \(l_{1}\) and \(l_{2}\) intersect at the point marked \(P\) inside the triangle.
(c) Measurement. Measuring directly from the construction, \(|CP| = 6.7\text{ cm}\) (accept \(6.5\text{ cm}\) to \(6.7\text{ cm}\)).
This agrees with a coordinate check. Taking \(A(0,0)\) and \(B(5,0)\), then \(C = 7.5(\cos120^{\circ},\ \sin120^{\circ}) = (-3.75,\ 6.50)\). The perpendicular bisector of \(AB\) is \(x = 2.5\), and the bisector of the \(120^{\circ}\) angle rises at \(60^{\circ}\), so \(P = (2.5,\ 2.5\tan60^{\circ}) = (2.5,\ 4.33)\). Hence
\[|CP| = \sqrt{(2.5-(-3.75))^{2} + (4.33-6.50)^{2}} = \sqrt{6.25^{2} + 2.17^{2}} = \sqrt{43.75} \approx 6.7\text{ cm}.\]Frage 59 Bericht
Three towns P, Q and R are such that the distance between P and Q is 50km and the distance between P and R is 90km. If the bearing of Q from P is 075° and the bearing of R from P is 310°, find the :
(a) distance between Q and R ;
(b) baering of R from Q.
Setting up. With P as the vertex, the bearing of Q is 075° and the bearing of R is 310°. The angle turned from PQ to PR is \(310^\circ - 75^\circ = 235^\circ\), so the actual angle inside triangle PQR is
\[\angle QPR = 360^\circ - 235^\circ = 125^\circ.\]
(a) Distance QR (cosine rule).
\[QR^2 = PQ^2 + PR^2 - 2(PQ)(PR)\cos\angle QPR\]
\[QR^2 = 50^2 + 90^2 - 2(50)(90)\cos 125^\circ = 2500 + 8100 - 9000(-0.5736)\]
\[QR^2 = 10600 + 5162.2 = 15762.2,\qquad QR = 125.5\text{ km}.\]
(b) Bearing of R from Q. First find \(\angle PQR\) by the sine rule:
\[\frac{\sin\angle PQR}{PR} = \frac{\sin\angle QPR}{QR}\Rightarrow \sin\angle PQR = \frac{90\sin125^\circ}{125.5} = \frac{90(0.8192)}{125.5} = 0.5873.\]
\[\angle PQR = 36.0^\circ.\]
The bearing of P from Q is \(075^\circ + 180^\circ = 255^\circ\). Point R lies to the west of QP, so the bearing of R from Q is
\[255^\circ + 36^\circ = 291^\circ.\]
Answers: (a) \(QR \approx 125.5\text{ km}\); (b) bearing of R from Q \(\approx 291^\circ\).
Antwortdetails
Setting up. With P as the vertex, the bearing of Q is 075° and the bearing of R is 310°. The angle turned from PQ to PR is \(310^\circ - 75^\circ = 235^\circ\), so the actual angle inside triangle PQR is
\[\angle QPR = 360^\circ - 235^\circ = 125^\circ.\]
(a) Distance QR (cosine rule).
\[QR^2 = PQ^2 + PR^2 - 2(PQ)(PR)\cos\angle QPR\]
\[QR^2 = 50^2 + 90^2 - 2(50)(90)\cos 125^\circ = 2500 + 8100 - 9000(-0.5736)\]
\[QR^2 = 10600 + 5162.2 = 15762.2,\qquad QR = 125.5\text{ km}.\]
(b) Bearing of R from Q. First find \(\angle PQR\) by the sine rule:
\[\frac{\sin\angle PQR}{PR} = \frac{\sin\angle QPR}{QR}\Rightarrow \sin\angle PQR = \frac{90\sin125^\circ}{125.5} = \frac{90(0.8192)}{125.5} = 0.5873.\]
\[\angle PQR = 36.0^\circ.\]
The bearing of P from Q is \(075^\circ + 180^\circ = 255^\circ\). Point R lies to the west of QP, so the bearing of R from Q is
\[255^\circ + 36^\circ = 291^\circ.\]
Answers: (a) \(QR \approx 125.5\text{ km}\); (b) bearing of R from Q \(\approx 291^\circ\).
Frage 60 Bericht
(a) In a game, a fair die is rolled once and two unbiased coins are tossed at once. What is the probability of obtaining 3 and a tail?
(b) A box contains 10 marbles, 7 of which are black and 3 are red. Two marbles are drawn one after the other without replacement. Find the probability of getting:
(i) a red, then a black marble ; (ii) two black marbles.
