WAEC SSCE - General Mathematics - 2002
Question 1 Report
Calculate and correct to two significant figures, the percentage error in approximating 0.375 to 0.4
Answer Details
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.
Question 2 Report
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?
Answer Details
Question 3 Report
Find the value of x in 0.5x + 2.6 = 5x + 0.35
Answer Details
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.
Question 4 Report
In the diagram O is the center of the circle, ∠SOR = 64° and ∠PSO = 36°. Calculate ∠PQR
Answer Details
Question 5 Report
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
Answer Details
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\(^{3}\).
Question 6 Report
Which of the following is not a rational number?
Answer Details
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 \(\sqrt{90}\).
Question 7 Report
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|.
Question 8 Report
In the diagram, \(QR||TP and W\hat{P}T = 88^{\circ} \). Find the value of x
Answer Details
Question 9 Report
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.
Answer Details
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.
Question 10 Report
The height of a right circular cone is 4cm. The radius of its base is 3cm. Find the curved surface area
Answer Details
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) $15\pi cm^2$.
Question 11 Report
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?
Answer Details
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.
Question 12 Report
In the diagram, |SR| = |RQ| and ?PRQ = 58o ?VQT = 19o, PQT, SQV and PSR are straight lines. Find ?QPS
Answer Details
Question 13 Report
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.
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Question 14 Report
In the diagram above, ∠PQU=36°, ∠QRT = 29°, PQ||RT. Find ∠PQR
Question 15 Report
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
Question 17 Report
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?
Answer Details
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 \(\boxed{2\frac{1}{5}}\).
Question 18 Report
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.
Answer Details
Question 20 Report
If \(M5_{ten} = 1001011_{two}\) find the value of M
Answer Details
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.
Question 21 Report
Which of the following is represented by the above sketch?
Answer Details
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.
Question 22 Report
If (x + 3) varies directly as y and x = 3 when y = 12, what is the value of x when y = 8?
Answer Details
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.
Question 23 Report
Given that \(p = x-\frac{1}{x} and\hspace{1mm}q = x^2 + \frac{1}{x^2}\) express q in terms of p.
Answer Details
Question 24 Report
If log 2 = 0.3010 and log 2\(^y\) = 1.8062, find; correct to the nearest whole number, the value of y.
Answer Details
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.
Question 25 Report
The base diameter of a cone is 14cm, and its volume is 462 cm3. Find its height. [Taken \(\pi = \frac{22}{7}\)]
Answer Details
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.
Question 26 Report
If q oranges are sold for t Naira, how many oranges can be bought for p naira?
Answer Details
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 = \(\frac{pq}{t}\)
Question 27 Report
The volume of a cylinder of radius 14cm is 210cm3. What is the curved surface area of the cylinder?
Question 28 Report
Find the nth term of the sequence 4, 10, 16 ,...
Answer Details
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).
Question 29 Report
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|
Answer Details
Question 31 Report
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?
Answer Details
Question 32 Report
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
Question 33 Report
Simplify \(5\frac{1}{4}\div \left(1\frac{2}{3}- \frac{1}{2}\right)\)
Question 34 Report
Find the value of x in the diagram
Question 36 Report
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.
Answer Details
Question 39 Report
A number is selected at random from the set Y = {18, 19, 20, . . . 28, 29}. Find the probability that the number is prime.
Answer Details
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) \(\frac{1}{4}\).
Question 40 Report
In the diagram, |QR| = 10cm, PR⊥QS, angle PSR = 30° and angle PQR = 45°. Calculate in meters |QS|
Question 41 Report
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
Answer Details
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.
Question 42 Report
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?
Answer Details
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.
Question 43 Report
In the diagram O and O' are the centres of the circles radii 15cm and 8cm respectively. If PQ = 12cm, find |OO'|.
Answer Details
Question 44 Report
For what values of x is the expression \(\frac{3x-2}{4x^2+9x-9}\) undefined?
Answer Details
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 \(\frac{3}{4} \hspace{1mm}or \hspace{1mm}-3\).
Question 47 Report
In the diagram, SQ is a tangent to the circle at P, XP||YQ, ∠XPY = 56 o and ∠PXY = 80 o.Find angle PQY
Answer Details
Question 48 Report
Simplify \(3\sqrt{12} + 10\sqrt{3} - \frac{6}{\sqrt{3}}\)
Answer Details
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.
Question 49 Report
If (-3, -4) is a point on the line y = mx + 2 find the value of m.
Answer Details
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.
Question 50 Report
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): \[\frac23\times\frac58\times\frac14=\frac{10}{96}=\frac{5}{48}.\]
Answer Details
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): \[\frac23\times\frac58\times\frac14=\frac{10}{96}=\frac{5}{48}.\]
Question 51 Report
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|.
Construction using ruler and a pair of compasses only
(a)(i) Triangle XYZ. Draw a straight line and mark X and Y on it with \(|XY| = 8\,\text{cm}\). At X, use the compasses to construct \(\angle YXZ = 60^{\circ}\) (construct a \(60^{\circ}\) angle by stepping the radius round an arc). At Y, construct \(\angle XYZ = 30^{\circ}\) (first construct \(60^{\circ}\) at Y, then bisect it to obtain \(30^{\circ}\)). The two ray lines meet at Z. Since \(\angle X + \angle Y = 60^{\circ}+30^{\circ}=90^{\circ}\), the remaining angle \(\angle XZY = 180^{\circ}-90^{\circ}=90^{\circ}\), so the triangle is right-angled at Z.
(a)(ii) Perpendicular ZT. With centre Z, draw an arc cutting XY at two points; from each of these two points draw equal arcs to intersect below (or on) XY. The line joining Z to that intersection is the perpendicular from Z; it meets XY at T. T is the foot of the perpendicular.
(a)(iii) Locus \(l_{1}\). The set of points equidistant from the two lines ZY and XY is the bisector of the angle between them. These lines meet at Y (\(\angle XYZ = 30^{\circ}\)), so bisect \(\angle XYZ\): the bisector makes \(15^{\circ}\) with XY at Y. Produce this bisector into the triangle. This straight line is \(l_{1}\).
The completed construction is shown below.
(b) Point S and measurement of |ST|. \(l_{1}\) (the bisector) meets the perpendicular ZT at S. Placing the base on coordinates with \(X(0,0)\) and \(Y(8,0)\):
Vertex \(Z\): \(|XZ| = |XY|\cos 60^{\circ} = 8\times 0.5 = 4\,\text{cm}\), giving \(Z = (4\cos 60^{\circ},\,4\sin 60^{\circ}) = (2.00,\,3.46)\). Hence the perpendicular ZT is the line \(x = 2\) and \(T = (2.00,\,0)\).
The bisector \(l_{1}\) leaves \(Y(8,0)\) at \(15^{\circ}\) above XY. Its gradient is \(\tan 15^{\circ} = 0.2679\), so\[y = 0.2679\,(8 - x).\]At \(S\), \(x = 2\):\[|ST| = y = 0.2679\,(8 - 2) = 0.2679 \times 6 = 1.61\,\text{cm}.\]
Check (equidistance at S): distance from \(S(2,\,1.61)\) to line XY is \(1.61\,\text{cm}\); distance from \(S\) to line ZY is also \(1.61\,\text{cm}\), confirming \(S\) lies on \(l_{1}\).
\(\therefore\ |ST| \approx 1.6\,\text{cm}.\)
Answer Details
Construction using ruler and a pair of compasses only
(a)(i) Triangle XYZ. Draw a straight line and mark X and Y on it with \(|XY| = 8\,\text{cm}\). At X, use the compasses to construct \(\angle YXZ = 60^{\circ}\) (construct a \(60^{\circ}\) angle by stepping the radius round an arc). At Y, construct \(\angle XYZ = 30^{\circ}\) (first construct \(60^{\circ}\) at Y, then bisect it to obtain \(30^{\circ}\)). The two ray lines meet at Z. Since \(\angle X + \angle Y = 60^{\circ}+30^{\circ}=90^{\circ}\), the remaining angle \(\angle XZY = 180^{\circ}-90^{\circ}=90^{\circ}\), so the triangle is right-angled at Z.
(a)(ii) Perpendicular ZT. With centre Z, draw an arc cutting XY at two points; from each of these two points draw equal arcs to intersect below (or on) XY. The line joining Z to that intersection is the perpendicular from Z; it meets XY at T. T is the foot of the perpendicular.
(a)(iii) Locus \(l_{1}\). The set of points equidistant from the two lines ZY and XY is the bisector of the angle between them. These lines meet at Y (\(\angle XYZ = 30^{\circ}\)), so bisect \(\angle XYZ\): the bisector makes \(15^{\circ}\) with XY at Y. Produce this bisector into the triangle. This straight line is \(l_{1}\).
The completed construction is shown below.
(b) Point S and measurement of |ST|. \(l_{1}\) (the bisector) meets the perpendicular ZT at S. Placing the base on coordinates with \(X(0,0)\) and \(Y(8,0)\):
Vertex \(Z\): \(|XZ| = |XY|\cos 60^{\circ} = 8\times 0.5 = 4\,\text{cm}\), giving \(Z = (4\cos 60^{\circ},\,4\sin 60^{\circ}) = (2.00,\,3.46)\). Hence the perpendicular ZT is the line \(x = 2\) and \(T = (2.00,\,0)\).
The bisector \(l_{1}\) leaves \(Y(8,0)\) at \(15^{\circ}\) above XY. Its gradient is \(\tan 15^{\circ} = 0.2679\), so\[y = 0.2679\,(8 - x).\]At \(S\), \(x = 2\):\[|ST| = y = 0.2679\,(8 - 2) = 0.2679 \times 6 = 1.61\,\text{cm}.\]
Check (equidistance at S): distance from \(S(2,\,1.61)\) to line XY is \(1.61\,\text{cm}\); distance from \(S\) to line ZY is also \(1.61\,\text{cm}\), confirming \(S\) lies on \(l_{1}\).
\(\therefore\ |ST| \approx 1.6\,\text{cm}.\)
Question 52 Report
(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 \(6y^{2} - 149y - 102 = 0\).
Question 53 Report
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
\[P(\text{modal class})=\frac{7}{30}.\]Answer Details
(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
\[P(\text{modal class})=\frac{7}{30}.\]Question 54 Report
(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 \(98^\circ+38^\circ+224^\circ=360^\circ.\))
Answer Details
(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 \(98^\circ+38^\circ+224^\circ=360^\circ.\))
Question 55 Report
(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 \(540^{\circ}\): \[x+2x+3x+2x+(3x-10)=540\Rightarrow 11x-10=540\Rightarrow 11x=550\Rightarrow x=50.\]
Answer Details
(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 \(540^{\circ}\): \[x+2x+3x+2x+(3x-10)=540\Rightarrow 11x-10=540\Rightarrow 11x=550\Rightarrow x=50.\]
Question 56 Report
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
\[A' \cap (B \cup C) = \{2, 6, 12, 15\}.\]
Answer Details
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
\[A' \cap (B \cup C) = \{2, 6, 12, 15\}.\]
Question 57 Report
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.
Answer Details
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.
Question 58 Report
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 \(\text{Area}=\tfrac12(PR)(h)\):
\[h=\frac{2\times\text{Area}}{PR}=\frac{2(51.210)}{16.405}=6.24\text{ m (3 s.f.)}\]Answer Details
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 \(\text{Area}=\tfrac12(PR)(h)\):
\[h=\frac{2\times\text{Area}}{PR}=\frac{2(51.210)}{16.405}=6.24\text{ m (3 s.f.)}\]Question 59 Report
(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 \(m = \mathbf{610}\).
Answer Details
(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 \(m = \mathbf{610}\).
Question 60 Report
(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.
Answer Details
(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.
Question 61 Report
(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 \[3^{(4x-6)-(x+3)}=3^{0}\Rightarrow 3x-9=0\Rightarrow x=3.\]
Answer Details
(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 \[3^{(4x-6)-(x+3)}=3^{0}\Rightarrow 3x-9=0\Rightarrow x=3.\]
Question 62 Report
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 \(\pi = \frac{22}{7}\)].
Answer Details
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