WAEC SSCE - General Mathematics - 2002
Frage 1 Bericht
To arrive on schedule, a train is to cover a distance of 60km at 72km/hr. If it starts 10 minutes late, at what speed must it move to arrive on schedule?
Antwortdetails
To arrive on schedule, the train needs to cover a distance of 60 km at a speed of 72 km/hr. However, the train starts 10 minutes late, which means it has lost 10 minutes of travel time. We need to find the speed the train needs to travel to make up for the lost time and arrive on schedule. We can use the formula: distance = speed x time Let's first calculate the time taken to cover the distance of 60 km at the speed of 72 km/hr. time = distance / speed time = 60 km / 72 km/hr time = 0.8333 hours or 50 minutes (rounded to the nearest minute) Since the train is starting 10 minutes late, it has only 40 minutes to cover the distance of 60 km to arrive on schedule. We can use the same formula to find the speed the train needs to travel to cover the distance in 40 minutes. speed = distance / time speed = 60 km / (40/60) hours speed = 90 km/hr Therefore, the correct answer is 90 km/hr.
Frage 2 Bericht
In the diagram, SQ is a tangent to the circle at P, XP||YQ, ∠XPY = 56 o and ∠PXY = 80 o.Find angle PQY
Antwortdetails
Frage 3 Bericht
In the diagram O is the center of the circle, ∠SOR = 64° and ∠PSO = 36°. Calculate ∠PQR
Antwortdetails
Frage 4 Bericht
The angle of depression of a point Q from a vertical tower PR, 30m high, is 40°. If the foot P of the tower is on the same horizontal level as Q, find, correct to 2 decimal places, |PQ|.
Frage 5 Bericht
The base diameter of a cone is 14cm, and its volume is 462 cm3. Find its height. [Taken \(\pi = \frac{22}{7}\)]
Antwortdetails
The formula for the volume of a cone is given by V = (1/3)πr^2h, where r is the radius of the base and h is the height of the cone. Since we are given the base diameter, which is 14cm, we can find the radius by dividing it by 2: radius, r = 14/2 = 7cm Also, we are given the volume of the cone, which is 462 cm^3. Substituting these values into the formula for the volume of a cone, we get: 462 = (1/3)π(7^2)h Simplifying this equation, we get: 22h = 198 h = 198/22 h = 9cm Therefore, the height of the cone is 9cm. Hence, the correct option is (d) 9cm.
Frage 6 Bericht
Given that \(p = x-\frac{1}{x} and\hspace{1mm}q = x^2 + \frac{1}{x^2}\) express q in terms of p.
Antwortdetails
Frage 7 Bericht
In the diagram, |QR| = 10cm, PR⊥QS, angle PSR = 30° and angle PQR = 45°. Calculate in meters |QS|
Antwortdetails
Frage 8 Bericht
In the diagram O is the centre of the circle. Which of the following is/are not true? I. a = b II. b + c = 180o III. a + b = c
Antwortdetails
Frage 9 Bericht
For what values of x is the expression \(\frac{3x-2}{4x^2+9x-9}\) undefined?
Antwortdetails
The expression \(\frac{3x-2}{4x^2+9x-9}\) is undefined when the denominator is equal to zero. Therefore, we need to find the values of x that make the denominator zero. We can factor the denominator as follows: \[4x^2 + 9x - 9 = (4x - 3)(x + 3)\] So, the denominator is equal to zero when: \begin{align*} 4x - 3 &= 0 &\text{or} && x + 3 &= 0 \\ 4x &= 3 &&& x &= -3 \\ x &= \frac{3}{4} \end{align*} Therefore, the expression is undefined when \(x = \frac{3}{4}\) or \(x = -3\). So, the answer is \(\frac{3}{4} \hspace{1mm}or \hspace{1mm}-3\).
Frage 10 Bericht
Calculate and correct to two significant figures, the percentage error in approximating 0.375 to 0.4
Antwortdetails
To calculate the percentage error, we first need to find the absolute error, which is the difference between the actual value and the approximate value. In this case, the actual value is 0.375, and the approximate value is 0.4. Absolute error = |actual value - approximate value| Absolute error = |0.375 - 0.4| Absolute error = 0.025 To find the percentage error, we divide the absolute error by the actual value and multiply by 100. Percentage error = (absolute error / actual value) x 100 Percentage error = (0.025 / 0.375) x 100 Percentage error = 6.67 Rounding to two significant figures, the percentage error in approximating 0.375 to 0.4 is 6.7%. Therefore, the correct answer is option D.
Frage 11 Bericht
The ratio of the number of men to the number of women in a 20 member committee is 3:1. How many women must be added to the 20-member committee so as to make the ratio of men to women 3:2?
Antwortdetails
Frage 12 Bericht
A right pyramid is on a square base of side 4cm. The slanting side of the pyramid is \(2\sqrt{3}\) cm. Calculate the volume of the pyramid
Antwortdetails
A right pyramid is a pyramid in which the apex is directly above the center of the base. In this case, the base is a square and the pyramid is right, so each triangular face of the pyramid is an isosceles right triangle. The slanting side of the pyramid is a hypotenuse of one of the triangular faces. By the Pythagorean theorem, the length of each leg of the right triangle is equal to the length of the base of the square, which is 4cm. Therefore, each leg has length 4cm and the hypotenuse has length \(2\sqrt{3}\) cm. To find the height of the pyramid, we draw a perpendicular line from the apex of the pyramid to the center of the base. This line divides the square base into four congruent right triangles, each with legs of length 2cm and hypotenuse of length \(2\sqrt{2}\) cm. By the Pythagorean theorem, the height of each of these triangles is \(\sqrt{(2\sqrt{2})^{2} - 2^{2}} = \sqrt{8} = 2\sqrt{2}\). Therefore, the height of the pyramid is also 2\(\sqrt{2}\)cm. The volume of the pyramid is given by the formula: \[\frac{1}{3} \times (\text{area of base}) \times (\text{height})\] The area of the square base is \(4^{2}\) cm\(^{2}\) = 16 cm\(^{2}\), and the height is 2\(\sqrt{2}\) cm. Substituting these values into the formula, we get: \[\frac{1}{3} \times (16) \times (2\sqrt{2}) = \frac{32\sqrt{2}}{3} \approx 10.67 \text{ cm}^{3}\] Therefore, the volume of the pyramid is approximately 10.67 cm\(^{3}\). Hence, the correct option is \(\mathbf{(b)}\) \(10\frac{2}{3}\) cm\(^{3}\).
Frage 14 Bericht
If (-3, -4) is a point on the line y = mx + 2 find the value of m.
Antwortdetails
The equation of a line in slope-intercept form is given by y = mx + b, where m is the slope of the line and b is the y-intercept. In this question, we are given that the point (-3, -4) lies on the line y = mx + 2. This means that when x = -3, y = -4. We can substitute these values into the equation to get: -4 = m(-3) + 2 Simplifying this expression, we get: -4 = -3m + 2 Subtracting 2 from both sides, we get: -6 = -3m Dividing both sides by -3, we get: m = 2 Therefore, the value of m is 2.
Frage 17 Bericht
If log 2 = 0.3010 and log 2\(^y\) = 1.8062, find; correct to the nearest whole number, the value of y.
Antwortdetails
We can use the rule that log a^n = n log a to solve this problem. Since we are given log 2 = 0.3010 and log 2^y = 1.8062, we can use the second equation to get:
log 2^y = 1.8062
y log 2 = 1.8062 (using the above rule)
y = 1.8062 / log 2
y = 5.9889
Rounding to the nearest whole number, we get y = 6. Therefore, the answer is option A.
Frage 18 Bericht
If (x + 3) varies directly as y and x = 3 when y = 12, what is the value of x when y = 8?
Antwortdetails
In a direct variation, two variables are related by a constant ratio. This can be expressed mathematically as: x ∝ y which means that x is directly proportional to y, or that x and y vary directly. We can also write this relationship as an equation: x = ky where k is the constant of proportionality. In this problem, we are given that (x + 3) varies directly as y. So we can write: x + 3 = ky where k is some constant of proportionality that we don't know yet. We are also given that x = 3 when y = 12. We can use this information to solve for k: x + 3 = ky 3 + 3 = k(12) 6 = 12k k = 0.5 Now that we know k, we can use the equation x + 3 = ky to find x when y = 8: x + 3 = ky x + 3 = 0.5(8) x + 3 = 4 x = 1 Therefore, when y = 8, x = 1. The answer is (a) 1.
Frage 19 Bericht
If q oranges are sold for t Naira, how many oranges can be bought for p naira?
Antwortdetails
The correct answer is \(\frac{qp}{t}\). Since q oranges are sold for t Naira, we can say that the cost of one orange is \(\frac{t}{q}\) Naira. To find how many oranges can be bought for p Naira, we need to divide p by the cost of one orange: Number of oranges = \(\frac{p}{\frac{t}{q}}\) Simplifying, we get: Number of oranges = \(\frac{pq}{t}\)
Frage 20 Bericht
The diagram is the graph of \(y = 6 + x - x^2\). The graph intercepts the x- axis at P and R and the y- axis at Q.
What is the value of y at Q?
Antwortdetails
Frage 21 Bericht
If \(M5_{ten} = 1001011_{two}\) find the value of M
Antwortdetails
To convert a number from base 2 to base 10, we multiply each digit by its corresponding power of 2 and sum the results. For example, in the binary number 1001, the first digit (from the right) represents 2^0 (1), the second digit represents 2^1 (2), the third digit represents 2^2 (4), and the fourth digit represents 2^3 (8). So, to convert 1001 to base 10, we compute: 1*2^0 + 0*2^1 + 0*2^2 + 1*2^3 = 1 + 0 + 0 + 8 = 9 Using this method, we can convert the binary number 1001011 to base 10: 1*2^0 + 1*2^1 + 0*2^2 + 1*2^3 + 0*2^4 + 0*2^5 + 1*2^6 = 1 + 2 + 0 + 8 + 0 + 0 + 64 = 75 Therefore, M5 in base 10 is 75, and since M5 is in base 10, M is simply 7. So the answer is 7.
Frage 23 Bericht
In the diagram, LMT is a straight line. lf O is the centre of circle LMN, OMN = 20°, LTN = 32° and |NM| = |MT|, find LNM.
Antwortdetails
Frage 24 Bericht
In the diagram, \(\bar{PS}\hspace{1mm} and \hspace{1mm}\bar{QT}\) are two altitudes of ?PQR. Which of the following is equal to ∠RQT?
Antwortdetails
In a triangle, an altitude is a line segment drawn from a vertex perpendicular to the opposite side. In the given diagram, \(\bar{PS}\) and \(\bar{QT}\) are two altitudes of triangle PQR. By definition, the altitude \(\bar{QT}\) is perpendicular to the side \(\bar{PR}\). Therefore, ∠QTR is a right angle. Also, we know that the sum of the angles in a triangle is 180 degrees. So, we can find the measure of ∠RQT as follows: ∠RQT = 180° - ∠QTR - ∠QRT We don't know the value of either ∠QTR or ∠QRT, but we can find one of them using the fact that \(\bar{PS}\) is an altitude. Since \(\bar{PS}\) is perpendicular to \(\bar{QR}\), ∠QPS is a right angle. Therefore, ∠QTR = ∠QTP + ∠RTP Since ∠QTP and ∠RTP are both complementary to ∠QPR, which is a known angle in the diagram, we can find their values: ∠QTP = ∠RTP = 90° - ∠QPR Now we can substitute these values into our equation for ∠RQT: ∠RQT = 180° - (90° - ∠QPR) - (90° - ∠QPR) Simplifying this expression gives: ∠RQT = 2∠QPR - 90° Therefore, the answer is option (B) ?SRP, since ∠QPR and ∠SRP are corresponding angles and thus equal to each other.
Frage 25 Bericht
In the diagram O and O' are the centres of the circles radii 15cm and 8cm respectively. If PQ = 12cm, find |OO'|.
Antwortdetails
Frage 26 Bericht
In the diagram O is the center of the circle. Reflex angle XOY = 210° and the length of the minor arc is 5.5m. Find, correct to the nearest meter, the length of the major arc.
Antwortdetails
Frage 27 Bericht
Simplify \(5\frac{1}{4}\div \left(1\frac{2}{3}- \frac{1}{2}\right)\)
Frage 28 Bericht
Which of the following is not a rational number?
Antwortdetails
A rational number is any number that can be expressed as a ratio of two integers (where the denominator is not zero). - -5 can be expressed as -5/1, which is a ratio of two integers, so it is rational. - \(\sqrt{4}\) is equal to 2, which can be expressed as the ratio 2/1, so it is rational. - \(3\frac{3}{4}\) is equal to 15/4, which is a ratio of two integers, so it is rational. - \(\sqrt{90}\) cannot be expressed as a ratio of two integers. It is irrational. Therefore, the number that is not rational is \(\sqrt{90}\).
Frage 29 Bericht
The length, in cm, of the sides of a right angled triangle are x, (x+2) and (x+1) where x > 0. Find , in cm, the length of its hypotenuse
Antwortdetails
In a right-angled triangle, the hypotenuse is the longest side, and it is opposite to the right angle. Using the Pythagorean theorem, we know that the sum of the squares of the two shorter sides is equal to the square of the hypotenuse. So, we have: \begin{align*} x^2 + (x+2)^2 &= (x+1)^2 \\ x^2 + x^2 + 4x + 4 &= x^2 + 2x + 1 \\ x^2 + 2x^2 + 4x + 4 &= x^2 + 2x + 1 \\ 3x^2 + 4x + 4 &= x^2 + 2x + 1 \\ 2x^2 + 2x + 3 &= 0 \end{align*} We can solve this quadratic equation using the quadratic formula: \begin{align*} x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ &= \frac{-2 \pm \sqrt{2^2 - 4(2)(3)}}{2(2)} \\ &= \frac{-2 \pm \sqrt{-8}}{4} \\ &= \frac{-1}{2} \pm \frac{\sqrt{2}}{2}i \end{align*} Since x must be greater than 0, the only valid solution is: \begin{align*} x &= \frac{-1}{2} + \frac{\sqrt{2}}{2}i \end{align*} However, we are asked to find the length of the hypotenuse, which is given by: \begin{align*} \sqrt{x^2 + (x+2)^2} &= \sqrt{\left(\frac{-1}{2} + \frac{\sqrt{2}}{2}i\right)^2 + \left(\frac{1}{2} + \frac{\sqrt{2}}{2}i\right)^2} \\ &= \sqrt{\frac{1}{4} - \frac{1}{2}\sqrt{2}i - \frac{1}{4} - \frac{1}{2}\sqrt{2}i + \frac{1}{4} + \frac{\sqrt{2}}{2}i + \frac{1}{4} - \frac{\sqrt{2}}{2}i} \\ &= \sqrt{\frac{1}{2}} \\ &= \frac{\sqrt{2}}{2} \approx 0.707 \end{align*} Therefore, the length of the hypotenuse is approximately 0.707 cm, which is closest to 5 cm.
Frage 30 Bericht
Which of the following is represented by the above sketch?
Antwortdetails
The given sketch is a quadratic function graph. From the graph, we can see that the parabola intersects the x-axis at two points which are (-3, 0) and (2, 0). Therefore, the roots or zeros of the quadratic function are -3 and 2. By comparing the options given, we can see that only option B, y = x2 - x - 6, has roots of -3 and 2. Thus, option B is the correct answer.
Frage 31 Bericht
The lengths of the parallel sides of a trapezium are 9 cm and 12 cm. lf the area of the trapezium is 105 cm2, find the perpendicular distance between the parallel sides.
Antwortdetails
The formula for the area of a trapezium is given as: Area = 1/2 × (sum of the parallel sides) × (perpendicular distance between them) In this case, we have the lengths of the parallel sides as 9 cm and 12 cm, and the area as 105 cm2. Substituting these values in the formula, we get: 105 = 1/2 × (9 + 12) × (perpendicular distance) 105 = 1/2 × 21 × (perpendicular distance) 105 = 10.5 × (perpendicular distance) Dividing both sides by 10.5, we get: Perpendicular distance = 105/10.5 Perpendicular distance = 10 Therefore, the perpendicular distance between the parallel sides is 10 cm. Hence, the correct option is (c) 10cm.
Frage 32 Bericht
In the diagram, QRT is a straight line. If angle PTR = 90°, angle PRT = 60°, angle PQR = 30° and |PQ| = \(6\sqrt{3}cm\), calculate |RT|
Antwortdetails
Frage 35 Bericht
Simplify \(3\sqrt{12} + 10\sqrt{3} - \frac{6}{\sqrt{3}}\)
Antwortdetails
We can start by simplifying each term separately before adding them up: \(3\sqrt{12} = 3\sqrt{4\cdot3} = 3\cdot2\sqrt{3} = 6\sqrt{3}\) \(10\sqrt{3}\) is already in its simplest form. \(\frac{6}{\sqrt{3}} = \frac{6}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}\) Now we can substitute these values back into the original expression: \(3\sqrt{12} + 10\sqrt{3} - \frac{6}{\sqrt{3}} = 6\sqrt{3} + 10\sqrt{3} - 2\sqrt{3} = 14\sqrt{3}\) Therefore, the simplified form of the expression is \(14\sqrt{3}\), which is.
Frage 36 Bericht
Find the value of x in 0.5x + 2.6 = 5x + 0.35
Antwortdetails
To find the value of x in the equation 0.5x + 2.6 = 5x + 0.35, we need to isolate x on one side of the equation. We can start by subtracting 0.5x from both sides to get rid of the x-term on the left side: 0.5x + 2.6 - 0.5x = 5x + 0.35 - 0.5x Simplifying the left side and combining like terms on the right side, we get: 2.6 = 4.5x + 0.35 Subtracting 0.35 from both sides, we get: 2.6 - 0.35 = 4.5x Simplifying and dividing both sides by 4.5, we get: x = (2.6 - 0.35)/4.5 = 0.5 Therefore, the value of x is 0.5.
Frage 37 Bericht
Find the nth term of the sequence 4, 10, 16 ,...
Antwortdetails
The sequence 4, 10, 16,... is an arithmetic sequence because there is a common difference between every pair of consecutive terms. To find the common difference, we can subtract any term from the following term. For example: - The common difference between the first and second terms is 10 - 4 = 6. - The common difference between the second and third terms is 16 - 10 = 6. Since the common difference is 6, we can find the nth term of the sequence using the formula: nth term = first term + (n - 1) * common difference The first term is 4 and the common difference is 6, so we get: nth term = 4 + (n - 1) * 6 Simplifying this expression, we get: nth term = 6n - 2 Therefore, the correct answer is (A) 2(3n-1).
Frage 39 Bericht
Points X, Y and Z are located in the same horizontal plane such that Y is 12 km north of X and Z is on a bearing of 270° from X. If |XZ| = 6km, calculate, correct to one decimal place, lYZl
Antwortdetails
Frage 42 Bericht
The number of goals scored by a school team in 10 netball matches are as follows: 3, 5, 7, 7, 8, 8, 8, 11, 11, 12. Find the probability that in a match, the school team will score at most 8 goals.
Antwortdetails
Frage 43 Bericht
The volume of a cylinder of radius 14cm is 210cm3. What is the curved surface area of the cylinder?
Antwortdetails
Frage 45 Bericht
The height of a right circular cone is 4cm. The radius of its base is 3cm. Find the curved surface area
Antwortdetails
The curved surface area of a cone is given by the formula: \[\text{Curved surface area} = \pi rl,\] where $r$ is the radius of the base, $l$ is the slant height, and $\pi$ is the constant pi (approximately 3.14). In this case, the height of the cone is 4 cm and the radius of the base is 3 cm. To find the slant height, we can use the Pythagorean theorem: \[l^2 = r^2 + h^2.\] Plugging in the values we get, \[l^2 = 3^2 + 4^2 = 9 + 16 = 25.\] So, $l = 5$ cm. Now, we can use the formula for curved surface area, substituting $r=3$ cm and $l=5$ cm: \[\text{Curved surface area} = \pi \cdot 3 \cdot 5 = 15\pi \text{ cm}^2.\] Therefore, the answer is (B) $15\pi cm^2$.
Frage 46 Bericht
A number is selected at random from the set Y = {18, 19, 20, . . . 28, 29}. Find the probability that the number is prime.
Antwortdetails
To find the probability of selecting a prime number from the set Y = {18, 19, 20, ..., 28, 29}, we need to first identify the prime numbers in this set. The prime numbers in this set are 19, 23, and 29. The total number of elements in the set Y is 12. Therefore, the probability of selecting a prime number from the set Y is 3/12 or 1/4. Therefore, the correct option is (a) \(\frac{1}{4}\).
Frage 47 Bericht
The diagram is the graph of \(y = 6 + x - x^2\). The graph intercepts the x- axis at P and R and the y- axis at Q.
When \(y = 3\frac{1}{3}\), what is the positive value of x?
Antwortdetails
The given function is \(y = 6 + x - x^2\). We are required to find the value of x when y is equal to \(3\frac{1}{3}\), which is also equal to \(\frac{10}{3}\). Substituting \(y = \frac{10}{3}\) in the given equation, we get \[\frac{10}{3} = 6 + x - x^2\] Rearranging the terms, we get \[x^2 - x + \frac{8}{3} = 0\] Solving the quadratic equation, we get two solutions for x, which are \(\frac{4}{3}\) and \(1\frac{1}{2}\). However, we are asked to find the positive value of x, which is \(\boxed{2\frac{1}{5}}\).
Frage 48 Bericht
In the diagram, \(QR||TP and W\hat{P}T = 88^{\circ} \). Find the value of x
Antwortdetails
Frage 49 Bericht
In the diagram, |SR| = |RQ| and ?PRQ = 58o ?VQT = 19o, PQT, SQV and PSR are straight lines. Find ?QPS
Antwortdetails
Frage 50 Bericht
The probabilities that Ade, Kujo and Fati will pass an examination are \(\frac{2}{3}, \frac{5}{8}\) and \(\frac{3}{4}\) respectively. Find the probability that
(a) the three ;
(b) none of them ;
(c) Ade and Kujo only ; will pass the examination.
Let \(P(A)=\tfrac23,\;P(K)=\tfrac58,\;P(F)=\tfrac34\), so the failure probabilities are \(\tfrac13,\tfrac38,\tfrac14\). The three events are independent, so multiply.
(a) All three pass: \[\frac23\times\frac58\times\frac34=\frac{30}{96}=\frac{5}{16}.\]
(b) None passes: \[\frac13\times\frac38\times\frac14=\frac{3}{96}=\frac{1}{32}.\]
(c) Ade and Kujo only (Fati fails): \[\frac23\times\frac58\times\frac14=\frac{10}{96}=\frac{5}{48}.\]
Antwortdetails
Let \(P(A)=\tfrac23,\;P(K)=\tfrac58,\;P(F)=\tfrac34\), so the failure probabilities are \(\tfrac13,\tfrac38,\tfrac14\). The three events are independent, so multiply.
(a) All three pass: \[\frac23\times\frac58\times\frac34=\frac{30}{96}=\frac{5}{16}.\]
(b) None passes: \[\frac13\times\frac38\times\frac14=\frac{3}{96}=\frac{1}{32}.\]
(c) Ade and Kujo only (Fati fails): \[\frac23\times\frac58\times\frac14=\frac{10}{96}=\frac{5}{48}.\]
Frage 51 Bericht
(a) Without using calculator or mathematical tables, evaluate \(\frac{3}{\sqrt{3}}(\frac{2}{\sqrt{3}} - \frac{\sqrt{12}}{6})\)
(b)
In the diagram, O is the centre of the circle. The side AB is produced to E, < ACB = 49° and < CBE = 68°. Calculate,
(i) the interior angle AOC ; (ii) < BOC.
(a) Evaluate \(\dfrac{3}{\sqrt3}\left(\dfrac{2}{\sqrt3}-\dfrac{\sqrt{12}}{6}\right)\) without tables.
First simplify each surd. \(\dfrac{3}{\sqrt3}=\dfrac{3}{\sqrt3}\times\dfrac{\sqrt3}{\sqrt3}=\dfrac{3\sqrt3}{3}=\sqrt3.\)
Inside the bracket: \(\dfrac{2}{\sqrt3}=\dfrac{2\sqrt3}{3}\) and \(\dfrac{\sqrt{12}}{6}=\dfrac{2\sqrt3}{6}=\dfrac{\sqrt3}{3}.\)
\[\frac{2\sqrt3}{3}-\frac{\sqrt3}{3}=\frac{\sqrt3}{3}.\]Therefore
\[\sqrt3\times\frac{\sqrt3}{3}=\frac{3}{3}=\boxed{1.}\](b) Circle, centre \(O\); \(AB\) produced to \(E\), \(\widehat{ACB}=49^\circ\), \(\widehat{CBE}=68^\circ\).
Since \(A,B,E\) are collinear, \(\widehat{ABC}\) and \(\widehat{CBE}\) are angles on a straight line:
\[\widehat{ABC}=180^\circ-68^\circ=112^\circ.\]In \(\triangle ABC\):
\[\widehat{BAC}=180^\circ-\widehat{ABC}-\widehat{ACB}=180^\circ-112^\circ-49^\circ=19^\circ.\](i) Interior angle \(AOC\). \(\widehat{ABC}=112^\circ\) is the angle at the circumference standing on chord \(AC\); it subtends the major arc \(AC\), whose central angle (reflex \(AOC\)) is \(2\times112^\circ=224^\circ.\) Hence the interior (non-reflex) angle is
\[\widehat{AOC}=360^\circ-224^\circ=\boxed{136^\circ.}\](ii) Angle \(BOC\). \(\widehat{BAC}=19^\circ\) at the circumference subtends arc \(BC\); the angle at the centre on the same arc is twice as large:
\[\widehat{BOC}=2\times19^\circ=\boxed{38^\circ.}\](Check: \(\widehat{ACB}=49^\circ\Rightarrow\widehat{AOB}=98^\circ,\) and \(98^\circ+38^\circ+224^\circ=360^\circ.\))
Antwortdetails
(a) Evaluate \(\dfrac{3}{\sqrt3}\left(\dfrac{2}{\sqrt3}-\dfrac{\sqrt{12}}{6}\right)\) without tables.
First simplify each surd. \(\dfrac{3}{\sqrt3}=\dfrac{3}{\sqrt3}\times\dfrac{\sqrt3}{\sqrt3}=\dfrac{3\sqrt3}{3}=\sqrt3.\)
Inside the bracket: \(\dfrac{2}{\sqrt3}=\dfrac{2\sqrt3}{3}\) and \(\dfrac{\sqrt{12}}{6}=\dfrac{2\sqrt3}{6}=\dfrac{\sqrt3}{3}.\)
\[\frac{2\sqrt3}{3}-\frac{\sqrt3}{3}=\frac{\sqrt3}{3}.\]Therefore
\[\sqrt3\times\frac{\sqrt3}{3}=\frac{3}{3}=\boxed{1.}\](b) Circle, centre \(O\); \(AB\) produced to \(E\), \(\widehat{ACB}=49^\circ\), \(\widehat{CBE}=68^\circ\).
Since \(A,B,E\) are collinear, \(\widehat{ABC}\) and \(\widehat{CBE}\) are angles on a straight line:
\[\widehat{ABC}=180^\circ-68^\circ=112^\circ.\]In \(\triangle ABC\):
\[\widehat{BAC}=180^\circ-\widehat{ABC}-\widehat{ACB}=180^\circ-112^\circ-49^\circ=19^\circ.\](i) Interior angle \(AOC\). \(\widehat{ABC}=112^\circ\) is the angle at the circumference standing on chord \(AC\); it subtends the major arc \(AC\), whose central angle (reflex \(AOC\)) is \(2\times112^\circ=224^\circ.\) Hence the interior (non-reflex) angle is
\[\widehat{AOC}=360^\circ-224^\circ=\boxed{136^\circ.}\](ii) Angle \(BOC\). \(\widehat{BAC}=19^\circ\) at the circumference subtends arc \(BC\); the angle at the centre on the same arc is twice as large:
\[\widehat{BOC}=2\times19^\circ=\boxed{38^\circ.}\](Check: \(\widehat{ACB}=49^\circ\Rightarrow\widehat{AOB}=98^\circ,\) and \(98^\circ+38^\circ+224^\circ=360^\circ.\))
Frage 52 Bericht
The following data gives the lengths, in cm, of 30 pieces of iron rods :
45 55 65 60 61 68 59 54 64 76 50 68 72 68 80 67 70 62 79 67 64 63 71 59 64 53 57 74 55 57
(a) Using class intervals of 45 - 49, 50 - 54, 55 - 59, ... construct a frequency table of the data.
(b) Draw the histogram for the distribution
(c) Calculate the mean of the distribution
(d) What is the probability of selecting an iron rod whose length is in the modal class?
(a) Frequency table. Sorting the 30 lengths into the given class intervals and tallying gives:
| Length (cm) | Tally | Frequency \(f\) | Midpoint \(x\) | \(fx\) |
|---|---|---|---|---|
| 45 - 49 | | | 1 | 47 | 47 |
| 50 - 54 | ||| | 3 | 52 | 156 |
| 55 - 59 | |||| | | 6 | 57 | 342 |
| 60 - 64 | |||| || | 7 | 62 | 434 |
| 65 - 69 | |||| | | 6 | 67 | 402 |
| 70 - 74 | |||| | 4 | 72 | 288 |
| 75 - 79 | || | 2 | 77 | 154 |
| 80 - 84 | | | 1 | 82 | 82 |
| Total | 30 | 1905 |
(b) Histogram of the distribution. The bars are drawn against the class boundaries \(44.5, 49.5, 54.5, 59.5, 64.5, 69.5, 74.5, 79.5, 84.5\) on the horizontal axis, with heights equal to the class frequencies. Since all class widths are equal, the bar heights are simply the frequencies.
(c) Mean of the distribution.
\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{1905}{30}=63.5\text{ cm}.\](d) Probability of selecting a rod in the modal class. The modal class is the one with the highest frequency, \(60 - 64\) (frequency \(7\)). The number of rods in this class is \(7\) out of \(30\), so
\[P(\text{modal class})=\frac{7}{30}.\]Antwortdetails
(a) Frequency table. Sorting the 30 lengths into the given class intervals and tallying gives:
| Length (cm) | Tally | Frequency \(f\) | Midpoint \(x\) | \(fx\) |
|---|---|---|---|---|
| 45 - 49 | | | 1 | 47 | 47 |
| 50 - 54 | ||| | 3 | 52 | 156 |
| 55 - 59 | |||| | | 6 | 57 | 342 |
| 60 - 64 | |||| || | 7 | 62 | 434 |
| 65 - 69 | |||| | | 6 | 67 | 402 |
| 70 - 74 | |||| | 4 | 72 | 288 |
| 75 - 79 | || | 2 | 77 | 154 |
| 80 - 84 | | | 1 | 82 | 82 |
| Total | 30 | 1905 |
(b) Histogram of the distribution. The bars are drawn against the class boundaries \(44.5, 49.5, 54.5, 59.5, 64.5, 69.5, 74.5, 79.5, 84.5\) on the horizontal axis, with heights equal to the class frequencies. Since all class widths are equal, the bar heights are simply the frequencies.
(c) Mean of the distribution.
\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{1905}{30}=63.5\text{ cm}.\](d) Probability of selecting a rod in the modal class. The modal class is the one with the highest frequency, \(60 - 64\) (frequency \(7\)). The number of rods in this class is \(7\) out of \(30\), so
\[P(\text{modal class})=\frac{7}{30}.\]Frage 53 Bericht
Given is the graph of the relation \(y = ax^{2} + bx + c\) where a, b and c are constants. Use the graph to :
(a) find the roots of the equation \(ax^{2} + bx + c = 0\);
(b) determine the values of constants a, b and c in the relation using the values of the coordinates P and Q and hence write down the relation illustrated in the graph
(c) find the maximum value of y and the corresponding value of x at this point.
(d) find the values of x when y = 2.
The relation illustrated is a downward parabola. Reading from the graph, the curve cuts the y-axis at \(y = 6\), cuts the x-axis at \(x = -1.5\) and \(x = 2\), and turns over near \(x = 0.25\). The annotated graph below shows these key features, which we read off to answer each part.
(a) Roots of \(ax^{2}+bx+c=0\)
The roots are where the curve crosses the x-axis (\(y=0\)). From the graph these are:
\[ x = -1.5 \quad \text{or} \quad x = 2. \](b) Values of a, b and c, and the relation
The curve cuts the positive y-axis at \(6\), so the intercept gives
\[ c = 6. \]Substitute the coordinates of the two x-axis crossings, \(Q(2,\,0)\) and \(P(-1.5,\,0)\), into \(y = ax^{2}+bx+c\) with \(y = 0\) and \(c = 6\).
At \(x = 2\):
\[ a(2)^{2} + b(2) + 6 = 0 \implies 4a + 2b = -6 \quad \text{...(1)} \]At \(x = -1.5\):
\[ a(-1.5)^{2} + b(-1.5) + 6 = 0 \implies \tfrac{9a}{4} - \tfrac{3b}{2} = -6 \implies 9a - 6b = -24 \quad \text{...(2)} \]Multiply (1) by 3:
\[ 12a + 6b = -18 \quad \text{...(1a)} \]Add (1a) and (2):
\[ 21a = -42 \implies a = -2. \]Substitute \(a = -2\) into (1):
\[ 4(-2) + 2b = -6 \implies 2b = 2 \implies b = 1. \]Therefore \((a,\,b,\,c) = (-2,\,1,\,6)\), and the relation illustrated by the graph is
\[ \boxed{\,y = -2x^{2} + x + 6\,}. \](c) Maximum value of y and the corresponding x
The maximum occurs at the turning point. Completing the derivative (or reading the vertex from the graph):
\[ \frac{dy}{dx} = -4x + 1 = 0 \implies x = \tfrac{1}{4} = 0.25. \]Substituting \(x = 0.25\):
\[ y_{\max} = -2(0.25)^{2} + 0.25 + 6 = -0.125 + 0.25 + 6 = 6.125. \]So the maximum value is \(y_{\max} \approx 6.13\) (about \(6.1\)), occurring at \(x = 0.25\), which agrees with the peak read from the graph.
(d) Values of x when \(y = 2\)
Draw the horizontal line \(y = 2\) and read where it meets the curve, or solve:
\[ -2x^{2} + x + 6 = 2 \implies -2x^{2} + x + 4 = 0 \implies 2x^{2} - x - 4 = 0. \]Using the quadratic formula:
\[ x = \frac{-(-1) \pm \sqrt{(-1)^{2} - 4(2)(-4)}}{2(2)} = \frac{1 \pm \sqrt{33}}{4} = \frac{1 \pm 5.745}{4}. \] \[ x = \frac{1 + 5.745}{4} = 1.69 \quad \text{or} \quad x = \frac{1 - 5.745}{4} = -1.19. \]So when \(y = 2\), \(x \approx -1.2\) or \(x \approx 1.7\), matching the two points where the line \(y = 2\) cuts the curve on the graph.
Antwortdetails
The relation illustrated is a downward parabola. Reading from the graph, the curve cuts the y-axis at \(y = 6\), cuts the x-axis at \(x = -1.5\) and \(x = 2\), and turns over near \(x = 0.25\). The annotated graph below shows these key features, which we read off to answer each part.
(a) Roots of \(ax^{2}+bx+c=0\)
The roots are where the curve crosses the x-axis (\(y=0\)). From the graph these are:
\[ x = -1.5 \quad \text{or} \quad x = 2. \](b) Values of a, b and c, and the relation
The curve cuts the positive y-axis at \(6\), so the intercept gives
\[ c = 6. \]Substitute the coordinates of the two x-axis crossings, \(Q(2,\,0)\) and \(P(-1.5,\,0)\), into \(y = ax^{2}+bx+c\) with \(y = 0\) and \(c = 6\).
At \(x = 2\):
\[ a(2)^{2} + b(2) + 6 = 0 \implies 4a + 2b = -6 \quad \text{...(1)} \]At \(x = -1.5\):
\[ a(-1.5)^{2} + b(-1.5) + 6 = 0 \implies \tfrac{9a}{4} - \tfrac{3b}{2} = -6 \implies 9a - 6b = -24 \quad \text{...(2)} \]Multiply (1) by 3:
\[ 12a + 6b = -18 \quad \text{...(1a)} \]Add (1a) and (2):
\[ 21a = -42 \implies a = -2. \]Substitute \(a = -2\) into (1):
\[ 4(-2) + 2b = -6 \implies 2b = 2 \implies b = 1. \]Therefore \((a,\,b,\,c) = (-2,\,1,\,6)\), and the relation illustrated by the graph is
\[ \boxed{\,y = -2x^{2} + x + 6\,}. \](c) Maximum value of y and the corresponding x
The maximum occurs at the turning point. Completing the derivative (or reading the vertex from the graph):
\[ \frac{dy}{dx} = -4x + 1 = 0 \implies x = \tfrac{1}{4} = 0.25. \]Substituting \(x = 0.25\):
\[ y_{\max} = -2(0.25)^{2} + 0.25 + 6 = -0.125 + 0.25 + 6 = 6.125. \]So the maximum value is \(y_{\max} \approx 6.13\) (about \(6.1\)), occurring at \(x = 0.25\), which agrees with the peak read from the graph.
(d) Values of x when \(y = 2\)
Draw the horizontal line \(y = 2\) and read where it meets the curve, or solve:
\[ -2x^{2} + x + 6 = 2 \implies -2x^{2} + x + 4 = 0 \implies 2x^{2} - x - 4 = 0. \]Using the quadratic formula:
\[ x = \frac{-(-1) \pm \sqrt{(-1)^{2} - 4(2)(-4)}}{2(2)} = \frac{1 \pm \sqrt{33}}{4} = \frac{1 \pm 5.745}{4}. \] \[ x = \frac{1 + 5.745}{4} = 1.69 \quad \text{or} \quad x = \frac{1 - 5.745}{4} = -1.19. \]So when \(y = 2\), \(x \approx -1.2\) or \(x \approx 1.7\), matching the two points where the line \(y = 2\) cuts the curve on the graph.
Frage 54 Bericht
(a) Simplify : \(\frac{1}{2}\log_{10} 25 - 2\log_{10} 3 + \log_{10} 18\)
(b) If \(123_{y} = 83_{10}\), obtain an equation in y, hence find the value of y.
(c) Solve the equation \(\frac{9^{2x - 3}}{3^{x + 3}} = 1\)
(a) \[\tfrac12\log_{10}25-2\log_{10}3+\log_{10}18=\log_{10}5-\log_{10}9+\log_{10}18=\log_{10}\!\left(\frac{5\times18}{9}\right)=\log_{10}10=1.\]
(b) \(123_{y}=1\cdot y^{2}+2\cdot y+3\). So \[y^{2}+2y+3=83\Rightarrow y^{2}+2y-80=0\Rightarrow(y+10)(y-8)=0.\] Since a base must be positive, \(y=8\).
(c) \[\frac{9^{2x-3}}{3^{x+3}}=1.\] Write \(9=3^{2}\): numerator \(=3^{2(2x-3)}=3^{4x-6}\). Then \[3^{(4x-6)-(x+3)}=3^{0}\Rightarrow 3x-9=0\Rightarrow x=3.\]
Antwortdetails
(a) \[\tfrac12\log_{10}25-2\log_{10}3+\log_{10}18=\log_{10}5-\log_{10}9+\log_{10}18=\log_{10}\!\left(\frac{5\times18}{9}\right)=\log_{10}10=1.\]
(b) \(123_{y}=1\cdot y^{2}+2\cdot y+3\). So \[y^{2}+2y+3=83\Rightarrow y^{2}+2y-80=0\Rightarrow(y+10)(y-8)=0.\] Since a base must be positive, \(y=8\).
(c) \[\frac{9^{2x-3}}{3^{x+3}}=1.\] Write \(9=3^{2}\): numerator \(=3^{2(2x-3)}=3^{4x-6}\). Then \[3^{(4x-6)-(x+3)}=3^{0}\Rightarrow 3x-9=0\Rightarrow x=3.\]
Frage 55 Bericht
(a) An open rectangular tank is made of a steel plate of area 1440\(m^{2}\). Its length is twice its width . If the depth of the tank is 4m less than its width, find its length.
(b) A man saved N3,000 in a bank P, whose interest rate was x% per annum and N2,000 in another bank Q whose interest rate was y% per annum. His total interest in one year was N640. If he had saved N2,000 in P and N3,000 in Q for the same period, he would have gained N20 as additional interest. Find the values of x and y.
(a) Let width \(=w\). Then length \(=2w\) and depth \(=w-4\). An open tank has a base plus four sides: \[\text{Area}=\underbrace{(2w)(w)}_{\text{base}}+\underbrace{2(2w)(w-4)+2(w)(w-4)}_{\text{four sides}}=2w^{2}+6w(w-4).\] Set equal to \(1440\): \[2w^{2}+6w^{2}-24w=1440\Rightarrow 8w^{2}-24w-1440=0\Rightarrow w^{2}-3w-180=0.\] \[(w-15)(w+12)=0\Rightarrow w=15.\] Length \(=2w=\) 30 m.
(b) First arrangement: \(\dfrac{3000x}{100}+\dfrac{2000y}{100}=640\Rightarrow 3x+2y=64.\) Swapped: \(\dfrac{2000x}{100}+\dfrac{3000y}{100}=660\Rightarrow 2x+3y=66.\) Solving: from \(9x+6y=192\) and \(4x+6y=132\), subtract \(\Rightarrow 5x=60\Rightarrow x=12\); then \(3(12)+2y=64\Rightarrow y=14\). So x = 12, y = 14.
Antwortdetails
(a) Let width \(=w\). Then length \(=2w\) and depth \(=w-4\). An open tank has a base plus four sides: \[\text{Area}=\underbrace{(2w)(w)}_{\text{base}}+\underbrace{2(2w)(w-4)+2(w)(w-4)}_{\text{four sides}}=2w^{2}+6w(w-4).\] Set equal to \(1440\): \[2w^{2}+6w^{2}-24w=1440\Rightarrow 8w^{2}-24w-1440=0\Rightarrow w^{2}-3w-180=0.\] \[(w-15)(w+12)=0\Rightarrow w=15.\] Length \(=2w=\) 30 m.
(b) First arrangement: \(\dfrac{3000x}{100}+\dfrac{2000y}{100}=640\Rightarrow 3x+2y=64.\) Swapped: \(\dfrac{2000x}{100}+\dfrac{3000y}{100}=660\Rightarrow 2x+3y=66.\) Solving: from \(9x+6y=192\) and \(4x+6y=132\), subtract \(\Rightarrow 5x=60\Rightarrow x=12\); then \(3(12)+2y=64\Rightarrow y=14\). So x = 12, y = 14.
Frage 56 Bericht
(a) A manufacturer offers distributors a discount of \(20%\) on any article bought and a further discount of \(2\frac{1}{2}%\) for prompt payment.
(i) if the marked price of an article is N25,000, find the total amount saved by a distributor for paying promptly. (ii) if a distributor pays N11,700 promptly for an article marked Nx, find the value of x.
(b) Factorize \(6y^{2} - 149y - 102\), hence solve the equation \(6y^{2} - 149y - 102 = 0\).
Frage 57 Bericht
The diagram is a portion of a right circular solid cylinder of radius 7 cm and height 15 cm. The centre of the base of the cylinder is Q, while that of the top is B, where \(\stackrel\frown{ABC} = \stackrel\frown{PQR} = 120°\). Calculate, correct to one decimal place:
(a) The volume
(b) the total surface area of the solid. [Take \(\pi = \frac{22}{7}\)].
Antwortdetails
None
Frage 58 Bericht
I
In the diagram, /PQ/ = 8m, /QR/ = 13m, the bearing of Q from P is 050° and the bearing of R from Q is 130°.
(a) Calculate, correct to 3 significant figures, (i) /PR/ ; (ii) the bearing of R from P.
(b) Calculate the shortest distance between Q and PR, hence the area of triangle PQR.
From the diagram: \(PQ=8\text{ m}\), \(QR=13\text{ m}\), the bearing of Q from P is \(050^\circ\), and the bearing of R from Q is \(130^\circ\).
(a)(i) Length PR. Find \(\angle PQR\). The bearing of P from Q is the back-bearing of \(050^\circ\):
\[050^\circ+180^\circ=230^\circ\]The bearing of R from Q is \(130^\circ\), so
\[\angle PQR=230^\circ-130^\circ=100^\circ\]Apply the cosine rule to triangle PQR:
\[PR^2=PQ^2+QR^2-2\,(PQ)(QR)\cos(\angle PQR)\]\[PR^2=8^2+13^2-2(8)(13)\cos100^\circ\]\[PR^2=64+169-208(-0.17365)=269.12\]\[PR=16.4\text{ m (3 s.f.)}\](a)(ii) Bearing of R from P. Use the sine rule to find \(\angle QPR\):
\[\frac{\sin\angle QPR}{QR}=\frac{\sin\angle PQR}{PR}\]\[\sin\angle QPR=\frac{13\sin100^\circ}{16.405}=\frac{12.803}{16.405}=0.78040\]\[\angle QPR=51.3^\circ\]R lies clockwise of Q as seen from P, so the bearing of R from P is
\[050^\circ+51.3^\circ=101^\circ\ (\text{3 s.f.})\](b) Shortest distance from Q to PR, and the area. The area of triangle PQR is
\[\text{Area}=\tfrac12(PQ)(QR)\sin(\angle PQR)=\tfrac12(8)(13)\sin100^\circ\]\[\text{Area}=52(0.98481)=51.2\text{ m}^2\ (\text{3 s.f.})\]The shortest distance from Q to PR is the perpendicular height h onto base PR. Using \(\text{Area}=\tfrac12(PR)(h)\):
\[h=\frac{2\times\text{Area}}{PR}=\frac{2(51.210)}{16.405}=6.24\text{ m (3 s.f.)}\]Antwortdetails
From the diagram: \(PQ=8\text{ m}\), \(QR=13\text{ m}\), the bearing of Q from P is \(050^\circ\), and the bearing of R from Q is \(130^\circ\).
(a)(i) Length PR. Find \(\angle PQR\). The bearing of P from Q is the back-bearing of \(050^\circ\):
\[050^\circ+180^\circ=230^\circ\]The bearing of R from Q is \(130^\circ\), so
\[\angle PQR=230^\circ-130^\circ=100^\circ\]Apply the cosine rule to triangle PQR:
\[PR^2=PQ^2+QR^2-2\,(PQ)(QR)\cos(\angle PQR)\]\[PR^2=8^2+13^2-2(8)(13)\cos100^\circ\]\[PR^2=64+169-208(-0.17365)=269.12\]\[PR=16.4\text{ m (3 s.f.)}\](a)(ii) Bearing of R from P. Use the sine rule to find \(\angle QPR\):
\[\frac{\sin\angle QPR}{QR}=\frac{\sin\angle PQR}{PR}\]\[\sin\angle QPR=\frac{13\sin100^\circ}{16.405}=\frac{12.803}{16.405}=0.78040\]\[\angle QPR=51.3^\circ\]R lies clockwise of Q as seen from P, so the bearing of R from P is
\[050^\circ+51.3^\circ=101^\circ\ (\text{3 s.f.})\](b) Shortest distance from Q to PR, and the area. The area of triangle PQR is
\[\text{Area}=\tfrac12(PQ)(QR)\sin(\angle PQR)=\tfrac12(8)(13)\sin100^\circ\]\[\text{Area}=52(0.98481)=51.2\text{ m}^2\ (\text{3 s.f.})\]The shortest distance from Q to PR is the perpendicular height h onto base PR. Using \(\text{Area}=\tfrac12(PR)(h)\):
\[h=\frac{2\times\text{Area}}{PR}=\frac{2(51.210)}{16.405}=6.24\text{ m (3 s.f.)}\]Frage 59 Bericht
(a) Simplify : \((2a + b)^{2} - (b - 2a)^{2}\)
(b) Given that \(S = K\sqrt{m^{2} + n^{2}}\); (i) make m the subject of the relations ; (ii) if S = 12.2, K = 0.02 and n = 1.1, find, correct to the nearest whole number, the positive value of m.
(a) Simplify \((2a + b)^2 - (b - 2a)^2\). Note that \((b - 2a)^2 = (2a - b)^2\). This is a difference of two squares, \(P^2 - R^2 = (P + R)(P - R)\), with \(P = 2a + b\) and \(R = 2a - b\):
\[(2a + b)^2 - (2a - b)^2 = \big[(2a + b) + (2a - b)\big]\big[(2a + b) - (2a - b)\big].\]
\[= (4a)(2b) = 8ab.\]
(b) Given \(S = K\sqrt{m^2 + n^2}\).
(i) Make \(m\) the subject.
\[\frac{S}{K} = \sqrt{m^2 + n^2} \Rightarrow \left(\frac{S}{K}\right)^2 = m^2 + n^2.\]
\[m^2 = \left(\frac{S}{K}\right)^2 - n^2 \Rightarrow m = \sqrt{\left(\frac{S}{K}\right)^2 - n^2}.\]
(ii) With \(S = 12.2, K = 0.02, n = 1.1\):
\[\frac{S}{K} = \frac{12.2}{0.02} = 610, \qquad \left(\frac{S}{K}\right)^2 = 372100, \qquad n^2 = 1.21.\]
\[m = \sqrt{372100 - 1.21} = \sqrt{372098.79} = 609.999 \approx 610.\]
To the nearest whole number, the positive value of \(m = \mathbf{610}\).
Antwortdetails
(a) Simplify \((2a + b)^2 - (b - 2a)^2\). Note that \((b - 2a)^2 = (2a - b)^2\). This is a difference of two squares, \(P^2 - R^2 = (P + R)(P - R)\), with \(P = 2a + b\) and \(R = 2a - b\):
\[(2a + b)^2 - (2a - b)^2 = \big[(2a + b) + (2a - b)\big]\big[(2a + b) - (2a - b)\big].\]
\[= (4a)(2b) = 8ab.\]
(b) Given \(S = K\sqrt{m^2 + n^2}\).
(i) Make \(m\) the subject.
\[\frac{S}{K} = \sqrt{m^2 + n^2} \Rightarrow \left(\frac{S}{K}\right)^2 = m^2 + n^2.\]
\[m^2 = \left(\frac{S}{K}\right)^2 - n^2 \Rightarrow m = \sqrt{\left(\frac{S}{K}\right)^2 - n^2}.\]
(ii) With \(S = 12.2, K = 0.02, n = 1.1\):
\[\frac{S}{K} = \frac{12.2}{0.02} = 610, \qquad \left(\frac{S}{K}\right)^2 = 372100, \qquad n^2 = 1.21.\]
\[m = \sqrt{372100 - 1.21} = \sqrt{372098.79} = 609.999 \approx 610.\]
To the nearest whole number, the positive value of \(m = \mathbf{610}\).
Frage 60 Bericht
(a) The first term of an Arithmetic Progression(AP) is 3 and the common difference is 4. Find the sum of the first 28 terms.
(b) Given that \(x = \frac{2m}{1 - m^{2}}\) and \(y = \frac{2m}{1 + m}\), express 2x - y in terms of m in the simplest form.
(c) The angles of pentagon are x°, 2x°, 3x°, 2x° and (3x - 10)°. Find the value of x.
(a) AP with \(a=3,\;d=4,\;n=28\). \[S_{28}=\frac{n}{2}\bigl(2a+(n-1)d\bigr)=14\bigl(6+27\times4\bigr)=14\times114=1596.\]
(b) \(x=\dfrac{2m}{1-m^{2}}=\dfrac{2m}{(1-m)(1+m)}\) and \(y=\dfrac{2m}{1+m}\). \[2x-y=\frac{4m}{(1-m)(1+m)}-\frac{2m}{1+m}=\frac{4m-2m(1-m)}{(1-m)(1+m)}=\frac{2m+2m^{2}}{(1-m)(1+m)}=\frac{2m(1+m)}{(1-m)(1+m)}=\frac{2m}{1-m}.\]
(c) Interior angles of a pentagon sum to \(540^{\circ}\): \[x+2x+3x+2x+(3x-10)=540\Rightarrow 11x-10=540\Rightarrow 11x=550\Rightarrow x=50.\]
Antwortdetails
(a) AP with \(a=3,\;d=4,\;n=28\). \[S_{28}=\frac{n}{2}\bigl(2a+(n-1)d\bigr)=14\bigl(6+27\times4\bigr)=14\times114=1596.\]
(b) \(x=\dfrac{2m}{1-m^{2}}=\dfrac{2m}{(1-m)(1+m)}\) and \(y=\dfrac{2m}{1+m}\). \[2x-y=\frac{4m}{(1-m)(1+m)}-\frac{2m}{1+m}=\frac{4m-2m(1-m)}{(1-m)(1+m)}=\frac{2m+2m^{2}}{(1-m)(1+m)}=\frac{2m(1+m)}{(1-m)(1+m)}=\frac{2m}{1-m}.\]
(c) Interior angles of a pentagon sum to \(540^{\circ}\): \[x+2x+3x+2x+(3x-10)=540\Rightarrow 11x-10=540\Rightarrow 11x=550\Rightarrow x=50.\]
Frage 61 Bericht
Using ruler and a pair of compasses only,
(a) construct, (i) triangle XYZ with |XY| = 8cm, < YXZ = 60° and < XYZ = 30° ; (ii) the perpendicular ZT to meet XY in T ; (iii) the locus \(l_{1}\) of points equidistant from ZY and XY.
(b) If \(l_{1}\) and ZT intersect at S, measure |ST|.
```html
Geometrical Construction
Given: Construct triangle XYZ such that
- |XY| = 8 cm,
- ∠YXZ = 60°,
- ∠XYZ = 30°.
(a) Construction steps (ruler and compasses only).
- Draw a line segment XY = 8 cm.
- At X, construct an angle of 60° with XY. This may be done by constructing an equilateral triangle on XY, since every angle in an equilateral triangle is 60°.
- At Y, construct an angle of 30° with YX. First construct a 60° angle at Y, then bisect it using the standard compass construction for an angle bisector.
- Let the 60° ray from X and the 30° ray from Y meet at Z. This gives triangle XYZ.
- Construct the perpendicular from Z to XY. Let it meet XY at T. Hence, ZT ⟂ XY.
- The locus of points equidistant from the lines ZY and XY is the angle bisector of ∠ZYX. Construct the internal angle bisector of the 30° angle at Y and call it l₁.
- Let l₁ meet ZT at S.
(b) Finding |ST|.
Since ∠XYZ = 30° and l1 bisects this angle,
∠SYT = 15°.
Also, triangle ZXY is right-angled at Z, because
∠XZY = 180° − 60° − 30° = 90°.
In the 30°–60°–90° triangle XYZ, with hypotenuse XY = 8 cm:
XZ = 4 cm, YZ = 4√3 cm.
The foot of the perpendicular is T, and
YT = 6 cm.
In right triangle YST:
tan 15° = ST / YT.
ST = 6 tan 15° = 6(2 − √3).
|ST| = 6(2 − √3) cm ≈ 1.6 cm.
On a ruler-and-compass construction, the measured value should be approximately 1.6 cm.
```Antwortdetails
```html
Geometrical Construction
Given: Construct triangle XYZ such that
- |XY| = 8 cm,
- ∠YXZ = 60°,
- ∠XYZ = 30°.
(a) Construction steps (ruler and compasses only).
- Draw a line segment XY = 8 cm.
- At X, construct an angle of 60° with XY. This may be done by constructing an equilateral triangle on XY, since every angle in an equilateral triangle is 60°.
- At Y, construct an angle of 30° with YX. First construct a 60° angle at Y, then bisect it using the standard compass construction for an angle bisector.
- Let the 60° ray from X and the 30° ray from Y meet at Z. This gives triangle XYZ.
- Construct the perpendicular from Z to XY. Let it meet XY at T. Hence, ZT ⟂ XY.
- The locus of points equidistant from the lines ZY and XY is the angle bisector of ∠ZYX. Construct the internal angle bisector of the 30° angle at Y and call it l₁.
- Let l₁ meet ZT at S.
(b) Finding |ST|.
Since ∠XYZ = 30° and l1 bisects this angle,
∠SYT = 15°.
Also, triangle ZXY is right-angled at Z, because
∠XZY = 180° − 60° − 30° = 90°.
In the 30°–60°–90° triangle XYZ, with hypotenuse XY = 8 cm:
XZ = 4 cm, YZ = 4√3 cm.
The foot of the perpendicular is T, and
YT = 6 cm.
In right triangle YST:
tan 15° = ST / YT.
ST = 6 tan 15° = 6(2 − √3).
|ST| = 6(2 − √3) cm ≈ 1.6 cm.
On a ruler-and-compass construction, the measured value should be approximately 1.6 cm.
```Frage 62 Bericht
The sets A = {1, 3, 5, 7, 9, 11}, B = {2, 3, 5, 7, 11, 15} and C = {3, 6, 9, 12, 15} are subsets of \(\varepsilon\) = {1, 2, 3, ..., 15}.
(a) Draw a Venn diagram to illustrate the given information.
(b) Use your diagram to find : (i) \(C \cap A'\) ; (ii) \(A' \cap (B \cup C)\).
We are given the universal set \(\varepsilon = \{1, 2, 3, \dots, 15\}\) and the three subsets
\[A = \{1, 3, 5, 7, 9, 11\}, \quad B = \{2, 3, 5, 7, 11, 15\}, \quad C = \{3, 6, 9, 12, 15\}.\]
(a) Building the Venn diagram. Work out each region before drawing so that every element sits in exactly one place.
- In all three sets: \(A \cap B \cap C = \{3\}\).
- In \(A\) and \(B\) only: \(\{5, 7, 11\}\).
- In \(A\) and \(C\) only: \(\{9\}\).
- In \(B\) and \(C\) only: \(\{15\}\).
- In \(A\) only: \(\{1\}\).
- In \(B\) only: \(\{2\}\).
- In \(C\) only: \(\{6, 12\}\).
- Outside all three sets: \(\varepsilon \setminus (A \cup B \cup C) = \{4, 8, 10, 13, 14\}\).
These give the Venn diagram below.
(b) Reading answers from the diagram.
(i) \(A'\) is everything outside circle \(A\), so \(C \cap A'\) is the part of circle \(C\) that lies outside \(A\). Reading those regions of \(C\) gives the C-only part \(\{6, 12\}\) together with the \(B\cap C\)-only part \(\{15\}\):
\[C \cap A' = \{6, 12, 15\}.\]
(ii) First form \(B \cup C\) (all of circles \(B\) and \(C\)):
\[B \cup C = \{2, 3, 5, 6, 7, 9, 11, 12, 15\}.\]
Now intersect with \(A'\), keeping only the members of \(B \cup C\) that lie outside circle \(A\). Discarding \(3, 5, 7, 9, 11\) (which are inside \(A\)) leaves
\[A' \cap (B \cup C) = \{2, 6, 12, 15\}.\]
Antwortdetails
We are given the universal set \(\varepsilon = \{1, 2, 3, \dots, 15\}\) and the three subsets
\[A = \{1, 3, 5, 7, 9, 11\}, \quad B = \{2, 3, 5, 7, 11, 15\}, \quad C = \{3, 6, 9, 12, 15\}.\]
(a) Building the Venn diagram. Work out each region before drawing so that every element sits in exactly one place.
- In all three sets: \(A \cap B \cap C = \{3\}\).
- In \(A\) and \(B\) only: \(\{5, 7, 11\}\).
- In \(A\) and \(C\) only: \(\{9\}\).
- In \(B\) and \(C\) only: \(\{15\}\).
- In \(A\) only: \(\{1\}\).
- In \(B\) only: \(\{2\}\).
- In \(C\) only: \(\{6, 12\}\).
- Outside all three sets: \(\varepsilon \setminus (A \cup B \cup C) = \{4, 8, 10, 13, 14\}\).
These give the Venn diagram below.
(b) Reading answers from the diagram.
(i) \(A'\) is everything outside circle \(A\), so \(C \cap A'\) is the part of circle \(C\) that lies outside \(A\). Reading those regions of \(C\) gives the C-only part \(\{6, 12\}\) together with the \(B\cap C\)-only part \(\{15\}\):
\[C \cap A' = \{6, 12, 15\}.\]
(ii) First form \(B \cup C\) (all of circles \(B\) and \(C\)):
\[B \cup C = \{2, 3, 5, 6, 7, 9, 11, 12, 15\}.\]
Now intersect with \(A'\), keeping only the members of \(B \cup C\) that lie outside circle \(A\). Discarding \(3, 5, 7, 9, 11\) (which are inside \(A\)) leaves
\[A' \cap (B \cup C) = \{2, 6, 12, 15\}.\]
