WAEC SSCE - General Mathematics - 1990
Vraag 1 Verslag
A cylindrical container closed at both ends, has a radius of 7cm and height 5cm [Take π = 22/7]. What is the volume of the container?
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To find the volume of a cylindrical container, we use the formula V = πr^2h, where V is the volume, r is the radius, and h is the height of the cylinder. In this problem, the radius of the container is given as 7cm, and the height is given as 5cm. Substituting these values into the formula, we get: V = πr^2h V = (22/7)(7^2)(5) V = (22/7)(49)(5) V = 22(5)(7) V = 770 Therefore, the volume of the container is 770 cm^3. So, the correct answer is option (E) 770cm^3.
Vraag 2 Verslag
Solve the equation 7y\(^2\) = 3y
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To solve the equation 7y\(^2\) = 3y, we can start by rearranging it to get 7y\(^2\) - 3y = 0. We can then factor out y to get y(7y - 3) = 0. From here, we have two solutions: y = 0 or 7y - 3 = 0, which gives us y = 3/7. Therefore, the solutions to the equation 7y\(^2\) = 3y are y = 0 or y = 3/7. Therefore, the answer is: y = 0 or 3/7.
Vraag 3 Verslag
If 3\(^{2x}\) = 27, what is x?
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We can solve for x by using the laws of exponents and taking the logarithm of both sides. First, we can rewrite 3²x as (3²)ⁿ, where n = 2x. So, we have: (3²)ⁿ = 27 3ⁿ² = 27 Now, we can take the logarithm of both sides of the equation. Let's use the natural logarithm, denoted as ln, which is the logarithm to the base e: ln(3ⁿ²) = ln(27) Using the power rule of logarithms, we can simplify the left-hand side: n² ln(3) = ln(27) Now, we can solve for n: n² = ln(27) / ln(3) n² = 3 Taking the square root of both sides, we get: n = ± √3 But we know that n = 2x, so we can substitute back: 2x = ± √3 Solving for x, we get: x = ± (1/2) √3 Since the question is asking for a real value of x, we can take the positive square root: x = (1/2) √3 ≈ 0.866 Therefore, x is approximately 0.866, which is option (B) 1.5 rounded to one decimal place.
Vraag 4 Verslag
A cylindrical container closed at both ends, has a radius of 7cm and height 5cm [Take π = 22/7]
Find the total surface area of the container
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The total surface area of the cylindrical container is the sum of the areas of its top, bottom, and lateral surface. The top and bottom of the container are both circles, each with an area of πr², where r is the radius of the container. So the total area of the top and bottom is 2πr². The lateral surface of the container is a rectangle that has been rolled into a cylinder. The length of the rectangle is the circumference of the circle, which is 2πr. The height of the rectangle is the height of the container, which is given as 5cm. So the area of the lateral surface is 2πrh. Therefore, the total surface area of the container is: 2πr² + 2πrh Substituting the given values for r and h, we get: 2 × (22/7) × 7² + 2 × (22/7) × 7 × 5 = 2 × (22/7) × 49 + 2 × (22/7) × 35 = (22/7) × 2 × (49 + 35) = (22/7) × 168 = 528 cm² Therefore, the total surface area of the container is 528 cm². So, the correct option is (D) 528cm².
Vraag 6 Verslag
The table shows that the amount of money (in naira)collected through voluntary donations in a secondary school.
What is the median of distributionAntwoorddetails
To find the median of a distribution, we first need to arrange the values in order from smallest to largest. In this case, the data is already presented in order, so we can simply identify the middle value of the data set. Since there are 5 values, the middle value will be the 3rd value. So, the median of the distribution is N12.00, which is the value in the middle of the data set. Therefore, the answer is (c) N12.00.
Vraag 8 Verslag
Simplify 125\(^{\frac{-1}{3}}\) x 49\(^{\frac{-1}{2}}\) x 10\(^0\)
Vraag 9 Verslag
The annual salary of Mr. Johnson Mohammed for 1989 was N12,000.00. He spent this on agriculture projects, education of his children, food items, saving , maintenance and miscellaneous items as shown in the pie chart
How much did he spend on food items?
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Vraag 10 Verslag
A sector of a circle of radius 7cm has an area of 44cm2. Calculate the angle of the sector correct to the nearest degree [Take π = 22/7]
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The formula for the area of a sector is: A = (θ/360)πr2 where A is the area of the sector, θ is the angle of the sector in degrees, r is the radius of the circle, and π is the mathematical constant pi. We are given that the radius of the circle is 7cm and the area of the sector is 44cm2. Substituting these values into the formula above, we get: 44 = (θ/360) × (22/7) × 72 Simplifying this equation, we get: 44 = (θ/360) × 22 × 7 44 = (θ/360) × 154 Multiplying both sides by 360, we get: θ = (44 × 360) / 154 θ = 102.597 Rounding to the nearest degree, we get: θ ≈ 103o Therefore, the angle of the sector correct to the nearest degree is 103o. Answer is correct.
Vraag 11 Verslag
The annual salary of Mr. Johnson Mohammed for 1989 was N12,000.00. He spent this on agriculture projects, education of his children, food items, saving , maintenance and miscellaneous items as shown in the pie chart
How much money did he invest in agriculture?
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Vraag 12 Verslag
While doing his physics practical, Idowu recorded a reading as 1.12cm instead of 1.21cm. Calculate his percentage error
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Vraag 13 Verslag
If log x = \(\bar{2}.3675\) and log y = 0.9750, what is the value of x + y? Correct to three significant figures
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Vraag 14 Verslag
If the radius of the parallel of latitude 30°N is equal to the radius of the parallel of latitude θ°S, what is the value of θ?
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Vraag 15 Verslag
Find the 4th term of an A.P, whose first term is 2 and the common difference is 0.5
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In an arithmetic progression (A.P.), the terms increase or decrease by a constant difference called the common difference. In this problem, the first term is 2, and the common difference is 0.5. Therefore, the second term would be 2 + 0.5 = 2.5, the third term would be 2.5 + 0.5 = 3, and the fourth term would be 3 + 0.5 = 3.5. Therefore, the answer is option (C) 3.5.
Vraag 16 Verslag
Express 0.00562 in standard form
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To convert a decimal number to standard form, we need to express it in the form of a x 10n, where a is a number between 1 and 10 and n is an integer. Starting with 0.00562, we need to move the decimal point to the right until we obtain a number between 1 and 10. This means moving the decimal point four places to the right: 0.00562 = 5.62 x 10-3 Therefore, the answer is option A: 5.62 x 10-3.
Vraag 17 Verslag
The bearing of a point X from a point Y is 074°. What is the bearing of Y from X?
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If the bearing of point X from point Y is 074°, then the bearing of Y from X is the opposite direction, which is 180° away from 074°. To find the bearing of Y from X, we can add 180° to 74°, which gives us 254°. Therefore, the correct answer is 254°.
Vraag 18 Verslag
In the diagram above, O is the center of the circle with radius 10cm, and ?ABC = 30°. Calculate, correct to 1 decimal place, the length of arc AC [Take ? = 22/7]
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Vraag 20 Verslag
In the diagram above, |PQ| = |PR| = |RS| and ?RPS = 32°. Find the value of ?QPR
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Vraag 21 Verslag
In the diagram below, O is the center of the circle if ?QOR = 290o, find the size ?QPR
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Vraag 22 Verslag
calculate the surface area of a sphere of radius 7cm [Take π = 22/7]
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The surface area of a sphere can be calculated using the formula: Surface area = 4πr2 where r is the radius of the sphere. Substituting r=7cm and π = 22/7, we get: Surface area = 4 x (22/7) x 72 cm2 Surface area = 4 x (22/7) x 49 cm2 Surface area = 616 cm2 Therefore, the surface area of the sphere is 616cm2. So the answer is 616cm2.
Vraag 24 Verslag
Given that 1/3log10 P = 1, find the value of P
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We can solve for P by isolating it on one side of the equation. Starting with: 1/3log10 P = 1 We can multiply both sides by 3 to get rid of the 1/3: log10 P = 3 Now, we can rewrite this equation in exponential form: 103 = P Simplifying: P = 1000 Therefore, the value of P is 1000.
Vraag 25 Verslag
Solve the equation 2a\(^2\) - 3a - 27 = 0
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To solve the equation 2a\(^2\) - 3a - 27 = 0, we can use the quadratic formula which is: a = (-b ± sqrt(b\(^2\) - 4ac)) / 2a In this case, we have a = 2, b = -3, and c = -27. Substituting these values into the formula, we get: a = (-(-3) ± sqrt((-3)\(^2\) - 4(2)(-27))) / 2(2) Simplifying the expression under the square root, we get: a = (-(-3) ± sqrt(225)) / 4 which gives us: a = (3 ± 15) / 4 Therefore, we have two solutions: a = 3 and a = -6/2 = -3/2 Hence, the correct option is -3, 9/2.
Vraag 26 Verslag
Which angle is equal to ?VWZ?
Vraag 27 Verslag
In a class of 80 students, every students had to study economics or geography, or both economics and geography, if 65 students studied economics and 50 studied geography, how many studied both subjects?
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To find out how many students studied both economics and geography, we need to use the principle of inclusion-exclusion. First, we add the number of students who studied economics and geography, as some students could have studied both subjects. Let's call this number "x." Then, we subtract x from the total number of students (80), which gives us the number of students who studied only economics or only geography. We know that 65 students studied economics, so the number of students who studied only economics is 65 - x. Similarly, the number of students who studied only geography is 50 - x. Since every student had to study at least one subject, the total number of students who studied only economics or only geography is equal to the sum of the students who studied only economics and the students who studied only geography: (65 - x) + (50 - x) Simplifying this expression, we get: 115 - 2x But we know that this number is equal to the total number of students who studied only economics or only geography, which is 80 minus the number of students who studied both subjects: 80 - x Therefore, we can set up an equation: 115 - 2x = 80 - x Solving for x, we get: x = 35 So 35 students studied both economics and geography.
Vraag 28 Verslag
From the top of a building 10m high, the angle of depression of a stone lying on the horizontal ground is 69o. Calculate ,correct to one decimal place, the distance of the stone from the foot of the building
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Vraag 29 Verslag
In the diagram above PS||RQ, |RQ| = 6.4cm and perpendicular PH = 3.2cm. Find the area of SQR
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Vraag 30 Verslag
Which of the sketches above gives a correct method for constructed an angle of 120o at the point P?
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Vraag 31 Verslag
In the diagram above, PQRS s a cyclic quadrilateral, ?PSR = 86o and ?QPR = 38o. Calculate PRQ
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In a cyclic quadrilateral, the opposite angles add up to 180 degrees. Therefore, we can find the value of angle PQR as follows: angle PSR + angle PQR = 180 (since PQRS is a cyclic quadrilateral) 86 + angle PQR = 180 angle PQR = 180 - 86 = 94 degrees We are given angle QPR as 38 degrees, and since angles in a triangle add up to 180 degrees, we can find angle PRQ as: angle PRQ = 180 - angle PQR - angle QPR angle PRQ = 180 - 94 - 38 angle PRQ = 48 degrees Therefore, the answer is option (C) 48o.
Vraag 32 Verslag
Use mathematical table to evaluate (cos40° - sin30°)
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To evaluate (cos40° - sin30°), we first need to find the values of cos40° and sin30°. Using a mathematical table (such as a trigonometric table), we can look up the values of cos40° and sin30°: - cos40° = 0.7660 - sin30° = 0.5000 Now we can substitute these values into the expression: (cos40° - sin30°) = (0.7660 - 0.5000) = 0.2660 Therefore, the answer is option (D) 0.2660.
Vraag 33 Verslag
Two groups of male students cast their vote on a particular proposal. The results are as follows:
| In favor | Against | |
| Group A | 128 | 32 |
| Group B | 96 | 48 |
If a student is chosen at random, what is probability that he is against the proposal?
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To find the probability that a student chosen at random is against the proposal, we need to add up the number of students from Group A and Group B who are against the proposal, and divide that by the total number of students. The number of students against the proposal in Group A is 32, and in Group B it is 48. So the total number of students against the proposal is: 32 + 48 = 80 The total number of students in both groups is: 128 + 32 + 96 + 48 = 304 So the probability that a student chosen at random is against the proposal is: 80 / 304 = 5/19 Therefore, the answer is option (C) 5/19.
Vraag 34 Verslag
The table shows that the amount of money (in naira)collected through voluntary donations in a secondary school. What is the mode?
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Vraag 36 Verslag
Two groups of male students cast their votes on a particular proposal. The result are as follows:
| In favor | Against | |
| Group A | 128 | 32 |
| Group B | 96 | 48 |
If a student in favor of the proposal is selected for a post, what is the probability that he is from group A?
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Vraag 37 Verslag
Factorize 2e\(^2\) - 3e + 1
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To factorize 2e\(^2\) - 3e + 1, we can use the quadratic formula: e = (-b ± √(b² - 4ac)) / 2a where a = 2, b = -3, and c = 1. Plugging in the values, we get: e = (3 ± √(9 - 8)) / 4 e = (3 ± 1) / 4 So the roots are e = 1 and e = 1/2. Therefore, we can factorize 2e\(^2\) - 3e + 1 as: 2e\(^2\) - 3e + 1 = 2(e - 1)(e - 1/2) Simplifying this expression, we get: 2(e - 1)(2e - 1) Therefore, the factorization of 2e\(^2\) - 3e + 1 is (2e-1) (e-1), which corresponds to option (A).
Vraag 38 Verslag
If the shadow of a pole 7m high is 1/2 its length what is the angle of elevation of the sun, correct to the nearest degree?
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Vraag 40 Verslag
In the diagram above, WXYZ is a rhombus and ?WYX = 20°. What is the value of ?XZY
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Vraag 41 Verslag
Simplify \(\frac{\log \sqrt{8}}{\log 8}\)
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We can simplify the expression using the laws of logarithms. First, we can simplify the numerator by using the fact that the square root of 8 is equal to 8 raised to the power of 1/2: log√8 = log(81/2) = 1/2 log 8 Now, we can substitute this into the original expression: \(\frac{\log \sqrt{8}}{\log 8} = \frac{1/2 \log 8}{\log 8}\) We can simplify this by canceling out the factor of log 8 in the numerator and denominator: \(\frac{1/2 \log 8}{\log 8} = \frac{1}{2}\) Therefore, the simplified expression is 1/2, which corresponds to option (B).
Vraag 43 Verslag
An arc of length 22cm subtends an angle of θ at the center of the circle. What is the value of θ if the radius of the circle is 15cm?[Take π = 22/7]
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To find the value of θ, we can use the formula: θ = (arc length / radius) In this case, the arc length is given as 22cm, and the radius is given as 15cm. So we have: θ = (22 / 15) θ = 1.47 (approx) However, the answer options are in degrees, so we need to convert radians to degrees. We can use the formula: degrees = (radians × 180) / π Substituting θ = 1.47 and π = 22/7, we get: degrees = (1.47 × 180) / (22/7) degrees = 89.14 (approx) Therefore, the answer is closest to option (B) 84o.
Vraag 44 Verslag
In the diagram above, PQ is a tangent at T to the circle ABT. ABC is a straight line and TC bisects ?BTO. Find x.
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Vraag 45 Verslag
Let J be the set of positive integers, If H = {x: x∈J, x\(^2\) < 3 and x ≠ 0}, then
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The set H consists of all positive integers x such that x² is less than 3 and x is not equal to zero. The only positive integer that satisfies this condition is 1, because 1²=1 which is less than 3. Therefore, the set H contains only one element which is 1, so the answer is H = {1}. Option (a) is correct, and options (b), (c), (d), and (e) are incorrect. It is important to note that the symbol ∈ means "is an element of", and the symbol ≠ means "is not equal to". The symbol J ≤ H is not a valid statement because J is not a subset of H.
Vraag 46 Verslag
The area shaded with horizontal lines is the solution set of the inequalities;
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Vraag 47 Verslag
the area shaded with vertical lines is the solution set of the inequalities;
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Vraag 48 Verslag
(a) Find the volume of a right solid cone of base radius 4cm and perpendicular height 6cm. [\(\pi = 3.142\)]
(b) A hemispherical tank of diameter which is 10m is filled by water issuing from a pipe of radius 20cm at 2m per second. Calculate, correct to three significant figures, the time, in minutes, it takes to fill the tank.
(a) Volume of a right circular cone \(= \dfrac{1}{3}\pi r^2 h\), with \(r = 4\) cm, \(h = 6\) cm, \(\pi = 3.142\):
\[ V = \frac{1}{3} \times 3.142 \times 4^2 \times 6 = \frac{1}{3} \times 3.142 \times 16 \times 6 = 3.142 \times 32 = 100.544\ \text{cm}^3. \]\(V \approx 100.5\ \text{cm}^3\).
(b) The tank is a hemisphere of diameter 10 m, so radius \(R = 5\) m.
\[ \text{Volume of tank} = \frac{2}{3}\pi R^3 = \frac{2}{3} \times 3.142 \times 5^3 = \frac{2}{3} \times 3.142 \times 125 = 261.83\ \text{m}^3. \]The pipe has radius \(20\ \text{cm} = 0.2\ \text{m}\) and water flows at 2 m/s. Volume delivered per second:
\[ \text{Flow rate} = \pi (0.2)^2 \times 2 = 3.142 \times 0.04 \times 2 = 0.25136\ \text{m}^3/\text{s}. \]Time to fill:
\[ t = \frac{261.83}{0.25136} = 1041.7\ \text{s} = \frac{1041.7}{60} = 17.4\ \text{minutes (to 3 s.f.).} \]Antwoorddetails
(a) Volume of a right circular cone \(= \dfrac{1}{3}\pi r^2 h\), with \(r = 4\) cm, \(h = 6\) cm, \(\pi = 3.142\):
\[ V = \frac{1}{3} \times 3.142 \times 4^2 \times 6 = \frac{1}{3} \times 3.142 \times 16 \times 6 = 3.142 \times 32 = 100.544\ \text{cm}^3. \]\(V \approx 100.5\ \text{cm}^3\).
(b) The tank is a hemisphere of diameter 10 m, so radius \(R = 5\) m.
\[ \text{Volume of tank} = \frac{2}{3}\pi R^3 = \frac{2}{3} \times 3.142 \times 5^3 = \frac{2}{3} \times 3.142 \times 125 = 261.83\ \text{m}^3. \]The pipe has radius \(20\ \text{cm} = 0.2\ \text{m}\) and water flows at 2 m/s. Volume delivered per second:
\[ \text{Flow rate} = \pi (0.2)^2 \times 2 = 3.142 \times 0.04 \times 2 = 0.25136\ \text{m}^3/\text{s}. \]Time to fill:
\[ t = \frac{261.83}{0.25136} = 1041.7\ \text{s} = \frac{1041.7}{60} = 17.4\ \text{minutes (to 3 s.f.).} \]Vraag 49 Verslag
The feet of two vertical poles of height 3m and 7m are in line with a point P on the ground, the smaller pole being between the taller pole and P and at a distance of 20m from P. The angle of elevation of the top (T) of the taller pole from the top (R) of the smaller pole is 30°. Calculate the :
(i) distance RT ; (ii) distance of the foot of the taller pole from P, correct to three significant figures ; (iii) angle of elevation of T from P, correct to one decimal place.
Let the taller pole (7 m) stand at foot F and the shorter pole (3 m) at foot S, with S between F and P. The distance SP = 20 m. R is the top of the short pole and T is the top of the tall pole.
Draw a horizontal line from R to the tall pole meeting it at N. Then RN is horizontal and NT is vertical, with
\[ NT = 7 - 3 = 4\ \text{m}. \]The angle of elevation of T from R is \(30^\circ\), so in the right-angled triangle RNT:
(i) \[ \sin 30^\circ = \frac{NT}{RT} \implies RT = \frac{4}{\sin 30^\circ} = \frac{4}{0.5} = 8\ \text{m}. \]
(ii) The horizontal distance is \[ RN = SF = \frac{NT}{\tan 30^\circ} = \frac{4}{0.5774} = 6.928\ \text{m}. \] So the foot of the taller pole is \[ FP = SP + SF = 20 + 6.928 = 26.928 \approx 26.9\ \text{m from P.} \]
(iii) The top T is 7 m high and its foot is 26.928 m from P. The angle of elevation of T from P is \(\theta\) where
\[ \tan \theta = \frac{7}{26.928} = 0.2600 \implies \theta = 14.6^\circ \ (\text{to 1 d.p.}). \]Antwoorddetails
Let the taller pole (7 m) stand at foot F and the shorter pole (3 m) at foot S, with S between F and P. The distance SP = 20 m. R is the top of the short pole and T is the top of the tall pole.
Draw a horizontal line from R to the tall pole meeting it at N. Then RN is horizontal and NT is vertical, with
\[ NT = 7 - 3 = 4\ \text{m}. \]The angle of elevation of T from R is \(30^\circ\), so in the right-angled triangle RNT:
(i) \[ \sin 30^\circ = \frac{NT}{RT} \implies RT = \frac{4}{\sin 30^\circ} = \frac{4}{0.5} = 8\ \text{m}. \]
(ii) The horizontal distance is \[ RN = SF = \frac{NT}{\tan 30^\circ} = \frac{4}{0.5774} = 6.928\ \text{m}. \] So the foot of the taller pole is \[ FP = SP + SF = 20 + 6.928 = 26.928 \approx 26.9\ \text{m from P.} \]
(iii) The top T is 7 m high and its foot is 26.928 m from P. The angle of elevation of T from P is \(\theta\) where
\[ \tan \theta = \frac{7}{26.928} = 0.2600 \implies \theta = 14.6^\circ \ (\text{to 1 d.p.}). \]Vraag 50 Verslag
(a) ABCD is a trapezium in which AB // DC, |AB| = 8cm, < ABC = 60°, |BC| = 5.5cm and |BD| = 8.3cm. Using a ruler and a pair of compasses only, construct:
(i) the trapezium ABCD ; (ii) a rectangle PQCD, where P, Q are two points AB;
(b) Measure |AB| and |QB|.
(a) Construction
- Draw \(AB=8.0\text{ cm}\).
- At \(B\), construct \(\angle ABC=60^\circ\). On this ray, mark \(C\) such that \(BC=5.5\text{ cm}\).
- With centre \(B\) and radius \(8.3\text{ cm}\), draw an arc.
- Through \(C\), construct a line parallel to \(AB\). Let it meet the arc at \(D\), choosing the intersection to the left of \(C\). Join \(AD\) and \(BD\) to complete trapezium \(ABCD\).
- Construct perpendiculars from \(D\) and \(C\) to the straight line \(AB\), meeting it at \(P\) and \(Q\) respectively. Join \(P\) to \(Q\). Thus \(PQCD\) is a rectangle.
(b) Measurements from the construction
\[|AC|=7.0\text{ cm},\qquad |QB|=2.9\text{ cm}.\]
Antwoorddetails
(a) Construction
- Draw \(AB=8.0\text{ cm}\).
- At \(B\), construct \(\angle ABC=60^\circ\). On this ray, mark \(C\) such that \(BC=5.5\text{ cm}\).
- With centre \(B\) and radius \(8.3\text{ cm}\), draw an arc.
- Through \(C\), construct a line parallel to \(AB\). Let it meet the arc at \(D\), choosing the intersection to the left of \(C\). Join \(AD\) and \(BD\) to complete trapezium \(ABCD\).
- Construct perpendiculars from \(D\) and \(C\) to the straight line \(AB\), meeting it at \(P\) and \(Q\) respectively. Join \(P\) to \(Q\). Thus \(PQCD\) is a rectangle.
(b) Measurements from the construction
\[|AC|=7.0\text{ cm},\qquad |QB|=2.9\text{ cm}.\]
Vraag 51 Verslag
In a class of 40 students, 25 speak Hausa, 16 speak Igbo, 21 speak Yoruba and each of the students speak at least one of the these three languages. If 8 speak Hausa and Igbo, 11 speak Hausa and Yoruba and 6 speak Igbo and Yoruba.
(a) Draw a Venn diagram to illustrate the information, using x to represent the number of students that speak all three languages.
(b) calculate the value of x.
(a) Let \(H\), \(I\) and \(Y\) represent the sets of students who speak Hausa, Igbo and Yoruba respectively. Since the given pairwise intersections include those who speak all three languages, the Venn diagram is:
(b) Since every student speaks at least one of the languages,
\[n(H\cup I\cup Y)=40.\]
Using the inclusion-exclusion principle,
\[40=25+16+21-8-11-6+x\]
\[40=37+x\]
\[x=3.\]
Therefore, 3 students speak all three languages.
Hence the numerical entries in the Venn diagram are:
\[H\cap I\cap Y=3,\quad H\cap I\text{ only}=5,\quad H\cap Y\text{ only}=8,\quad I\cap Y\text{ only}=3,\]
\[H\text{ only}=9,\quad I\text{ only}=5,\quad Y\text{ only}=7.\]
Antwoorddetails
(a) Let \(H\), \(I\) and \(Y\) represent the sets of students who speak Hausa, Igbo and Yoruba respectively. Since the given pairwise intersections include those who speak all three languages, the Venn diagram is:
(b) Since every student speaks at least one of the languages,
\[n(H\cup I\cup Y)=40.\]
Using the inclusion-exclusion principle,
\[40=25+16+21-8-11-6+x\]
\[40=37+x\]
\[x=3.\]
Therefore, 3 students speak all three languages.
Hence the numerical entries in the Venn diagram are:
\[H\cap I\cap Y=3,\quad H\cap I\text{ only}=5,\quad H\cap Y\text{ only}=8,\quad I\cap Y\text{ only}=3,\]
\[H\text{ only}=9,\quad I\text{ only}=5,\quad Y\text{ only}=7.\]
Vraag 52 Verslag
(a) Prove that the sum of the angles in a triangle is 2 right angles.
(b) The side AB of a triangle ABC is produced to a point D. The bisector of ACB cuts AB at E. Prove that < CAE + < CBD = 2 < CEB.
(a) The angles of a triangle sum to two right angles.
Let triangle ABC have interior angles at A, B and C. Through C draw a line XY parallel to AB, with X and Y on opposite sides of C.
- \(\angle XCA = \angle CAB\) (alternate angles, XY \(\parallel\) AB, transversal CA).
- \(\angle YCB = \angle CBA\) (alternate angles, XY \(\parallel\) AB, transversal CB).
The angles on the straight line XY at C add to two right angles:
\[ \angle XCA + \angle ACB + \angle YCB = 180^\circ. \]Replacing the two alternate angles gives
\[ \angle CAB + \angle ACB + \angle CBA = 180^\circ, \]so the three interior angles of the triangle sum to two right angles. \(\blacksquare\)
(b) Let \(\angle ACE = \angle ECB = c\) (CE bisects \(\angle ACB\)), and write \(\angle CAB = A\) and \(\angle CBA = B\).
\(\angle CAE = A\). The exterior angle at B, \(\angle CBD\), equals the sum of the two remote interior angles:
\[ \angle CBD = A + \angle ACB = A + 2c. \]Therefore
\[ \angle CAE + \angle CBD = A + (A + 2c) = 2A + 2c. \tag{1} \]Now \(\angle CEB\) is the exterior angle of triangle ACE at E, so it equals the sum of the two remote interior angles \(\angle CAE\) and \(\angle ACE\):
\[ \angle CEB = A + c \implies 2\,\angle CEB = 2A + 2c. \tag{2} \]From (1) and (2),
\[ \angle CAE + \angle CBD = 2\,\angle CEB. \ \blacksquare \]Antwoorddetails
(a) The angles of a triangle sum to two right angles.
Let triangle ABC have interior angles at A, B and C. Through C draw a line XY parallel to AB, with X and Y on opposite sides of C.
- \(\angle XCA = \angle CAB\) (alternate angles, XY \(\parallel\) AB, transversal CA).
- \(\angle YCB = \angle CBA\) (alternate angles, XY \(\parallel\) AB, transversal CB).
The angles on the straight line XY at C add to two right angles:
\[ \angle XCA + \angle ACB + \angle YCB = 180^\circ. \]Replacing the two alternate angles gives
\[ \angle CAB + \angle ACB + \angle CBA = 180^\circ, \]so the three interior angles of the triangle sum to two right angles. \(\blacksquare\)
(b) Let \(\angle ACE = \angle ECB = c\) (CE bisects \(\angle ACB\)), and write \(\angle CAB = A\) and \(\angle CBA = B\).
\(\angle CAE = A\). The exterior angle at B, \(\angle CBD\), equals the sum of the two remote interior angles:
\[ \angle CBD = A + \angle ACB = A + 2c. \]Therefore
\[ \angle CAE + \angle CBD = A + (A + 2c) = 2A + 2c. \tag{1} \]Now \(\angle CEB\) is the exterior angle of triangle ACE at E, so it equals the sum of the two remote interior angles \(\angle CAE\) and \(\angle ACE\):
\[ \angle CEB = A + c \implies 2\,\angle CEB = 2A + 2c. \tag{2} \]From (1) and (2),
\[ \angle CAE + \angle CBD = 2\,\angle CEB. \ \blacksquare \]Vraag 53 Verslag
(a) Simplify \(\frac{0.016 \times 0.084}{0.48}\) [Leave your answer in standard form].
(b) Eight wooden poles are to be used for pillars and the lengths of the poles form an Arithmetic Progression (A.P). If the second pole is 2m and the sixth is 5m, give the lengths of the poles, in order.
(a) Multiply out the numerator and divide:
\[ \frac{0.016 \times 0.084}{0.48} = \frac{0.001344}{0.48} = 0.0028. \]In standard form:
\[ 0.0028 = 2.8 \times 10^{-3}. \](b) Let the lengths form an A.P. with first term \(a\) and common difference \(d\).
\[ T_2 = a + d = 2, \qquad T_6 = a + 5d = 5. \]Subtracting: \(4d = 3 \implies d = 0.75\ \text{m}\), and \(a = 2 - 0.75 = 1.25\ \text{m}\).
The eight poles, in order, are:
\[ 1.25,\ 2,\ 2.75,\ 3.5,\ 4.25,\ 5,\ 5.75,\ 6.5 \text{ (all in metres).} \]Antwoorddetails
(a) Multiply out the numerator and divide:
\[ \frac{0.016 \times 0.084}{0.48} = \frac{0.001344}{0.48} = 0.0028. \]In standard form:
\[ 0.0028 = 2.8 \times 10^{-3}. \](b) Let the lengths form an A.P. with first term \(a\) and common difference \(d\).
\[ T_2 = a + d = 2, \qquad T_6 = a + 5d = 5. \]Subtracting: \(4d = 3 \implies d = 0.75\ \text{m}\), and \(a = 2 - 0.75 = 1.25\ \text{m}\).
The eight poles, in order, are:
\[ 1.25,\ 2,\ 2.75,\ 3.5,\ 4.25,\ 5,\ 5.75,\ 6.5 \text{ (all in metres).} \]Vraag 54 Verslag
Show on a graph, the area which gives the solution set of the inequalities: \(y - 2x \leq 4 ; 3y + x \geq 6 ; y \geq 7x - 9\).
Rewrite the inequalities in terms of y:
Draw the three boundary lines as solid lines, since equality is included. Their pairwise intersections are:
The required solution set is the common region below \(y=2x+4\) and above both \(y=2-\frac{x}{3}\) and \(y=7x-9\). It is the shaded triangular region shown.
Hence, the solution region is the closed triangle with vertices
Antwoorddetails
Rewrite the inequalities in terms of y:
Draw the three boundary lines as solid lines, since equality is included. Their pairwise intersections are:
The required solution set is the common region below \(y=2x+4\) and above both \(y=2-\frac{x}{3}\) and \(y=7x-9\). It is the shaded triangular region shown.
Hence, the solution region is the closed triangle with vertices
Vraag 55 Verslag
(a) If a number is chosen at random from the integers 5 to 25 inclusive, find the probability that the number is a multiple of 5 or 3.
(b) A bag contains 10 balls that differ only in colour; 4 are blue and 6 are red. Two balls are picked one after the other, with replacement. What is the probability that:
(i) both are red? (ii) both are the same colour?
Vraag 56 Verslag
The following is an incomplete table for the relation \(y = 2x^{2} - 5x + 1\)
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
| y | 8 | 1 | -1 | 26 |
(a) Copy and complete the table.
(b) Using a scale of 2cm to 1 unit on the x- axis and 2cm to 10 units on the y- axis, draw the graph of the relation \(y = 2x^{2} - 5x + 1\) for \(-3 \leq x \leq 5\).
(c) Using the same scale and axes, draw the graph of \(y = x + 6\).
(d) Estimate from your graphs, correct to one decimal place : (i) the least value of y and the value of x for which it occurs ; (ii) the solution of the equation \(2x^{2} - 5x + 1 = x + 6\).
(a) Table of values for \(y = 2x^{2} - 5x + 1\)
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
| y | 34 | 19 | 8 | 1 | -2 | -1 | 4 | 13 | 26 |
(b) Plot the points
\[
(-3,34),\,(-2,19),\,(-1,8),\,(0,1),\,(1,-2),\,(2,-1),\,(3,4),\,(4,13),\,(5,26)
\]
using the given scales, and join them with a smooth curve.
(c) Table of values for \(y=x+6\)
| x | -2 | 0 | 2 |
| y | 4 | 6 | 8 |
Plot the points \((-2,4)\), \((0,6)\) and \((2,8)\), then draw a straight line through them.
(d)
(i) Least value of \(y = -2.1\), occurring at \(x = 1.3\), correct to one decimal place.
(ii) The solutions of \(2x^{2}-5x+1=x+6\) are:
\[ x=-0.7 \quad \text{and} \quad x=3.7 \]
Antwoorddetails
(a) Table of values for \(y = 2x^{2} - 5x + 1\)
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
| y | 34 | 19 | 8 | 1 | -2 | -1 | 4 | 13 | 26 |
(b) Plot the points
\[
(-3,34),\,(-2,19),\,(-1,8),\,(0,1),\,(1,-2),\,(2,-1),\,(3,4),\,(4,13),\,(5,26)
\]
using the given scales, and join them with a smooth curve.
(c) Table of values for \(y=x+6\)
| x | -2 | 0 | 2 |
| y | 4 | 6 | 8 |
Plot the points \((-2,4)\), \((0,6)\) and \((2,8)\), then draw a straight line through them.
(d)
(i) Least value of \(y = -2.1\), occurring at \(x = 1.3\), correct to one decimal place.
(ii) The solutions of \(2x^{2}-5x+1=x+6\) are:
\[ x=-0.7 \quad \text{and} \quad x=3.7 \]
Vraag 57 Verslag
The weights to the nearest kilogram, of a group of 50 students in a College of Technology are given below:
65, 70, 60, 46, 51, 55, 59, 63, 68, 53, 47, 53, 72, 53, 67, 62, 64, 70, 57, 56, 73, 56, 48, 51, 58, 63, 65, 62, 49, 64, 53, 59, 63, 50, 48, 72, 67, 56, 61, 64, 66, 52, 49, 62, 71, 58, 53, 69, 63, 59.
(a) Prepare a grouped fraquency table with class intervals 45 - 49, 50 - 54, 55 - 59 etc.
(b) Using an assumed mean of 62 or otherwise, calculate the mean and standard deviation of the grouped data, correct to one decimal place.
(a) Grouped frequency table
| Column 1 | Column 2 | Column 3 |
|---|---|---|
| Data | Data | Data |
| Data | Data | Data |
| Data | Data | Data |
| Class Interval | Tally | Frequency, \(f\) |
|---|---|---|
| 45 - 49 | ||||| | | 6 |
| 50 - 54 | ||||| |||| | 9 |
| 55 - 59 | ||||| ||||| | 10 |
| 60 - 64 | ||||| ||||| || | 12 |
| 65 - 69 | ||||| || | 7 |
| 70 - 74 | ||||| | | 6 |
| Total | 50 |
(b) Let the assumed mean, \(A = 62\).
| Class Interval | Mid-value, \(x\) | \(d=x-62\) | \(d^2\) | \(f\) | \(fd\) | \(fd^2\) | ||||||||||||||||||||||||||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 45 - 49 | 47 | -15 | 225 | 6 | -90 | 1350 | ||||||||||||||||||||||||||||||||||||||||||||||||||
| 50 - 54 | 52 | -10 | 100 | 9 | -90 | 900 | ||||||||||||||||||||||||||||||||||||||||||||||||||
| Column 1 | Column 2 | Column 3 |
|---|---|---|
| Data | Data | Data |
| Data | Data | Data |
| Data | Data | Data |
| Class Interval | Tally | Frequency, \(f\) |
|---|---|---|
| 45 - 49 | ||||| | | 6 |
| 50 - 54 | ||||| |||| | 9 |
| 55 - 59 | ||||| ||||| | 10 |
| 60 - 64 | ||||| ||||| || | 12 |
| 65 - 69 | ||||| || | 7 |
| 70 - 74 | ||||| | | 6 |
| Total | 50 |
(b) Let the assumed mean, \(A = 62\).
| Class Interval | Mid-value, \(x\) | \(d=x-62\) | \(d^2\) | \(f\) | \(fd\) | \(fd^2\) |
|---|---|---|---|---|---|---|
| 45 - 49 | 47 | -15 | 225 | 6 | -90 | 1350 |
| 50 - 54 | 52 | -10 | 100 | 9 | -90 | 900 |
