WAEC SSCE - General Mathematics - 1993
Tambaya 1 Rahoto
A plane tiles 90km on a bearing 030° and then flies 150km due east. How far east of the starting point is the plane?
Bayanin Amsa
Tambaya 3 Rahoto
The angle of elevation of the top of a tower from a point on the horizontal ground, 40m away from the foot of the tower is 30o. Find the height of the tower.
Tambaya 4 Rahoto
A die is rolled 200 times the outcomes obtained are shown in the table below.
Find the probability of obtaining 2.
Bayanin Amsa
To find the probability of obtaining 2, we need to look at the frequency of occurrence of 2 in the outcomes. From the table, we see that 2 appears 30 times out of 200. Therefore, the probability of obtaining 2 on a single roll of the die is: Probability of obtaining 2 = (number of times 2 appears) / (total number of rolls) Probability of obtaining 2 = 30 / 200 Probability of obtaining 2 = 0.15 Therefore, the probability of obtaining 2 is 0.15, which is.
Tambaya 5 Rahoto
The angle of a sector of a circle radius 10.5cm is 120°. Find the perimeter of the sector [Take π = 22/7]
Bayanin Amsa
Tambaya 7 Rahoto
The area and a diagonal of a rhombus are 60 cm\(^2\) and 12 cm respectively. Calculate the length of the other diagonal.
Bayanin Amsa
The area of a rhombus is half the product of the lengths of its diagonals. Therefore, if we know the area of the rhombus and one of its diagonals, we can use this formula to find the length of the other diagonal. Let's call the length of the first diagonal 12 cm, and let's call the length of the other diagonal d cm. The area of the rhombus is given as 60 cm\(^2\). We can write the formula for the area of a rhombus as: Area = (1/2) x (diagonal 1) x (diagonal 2) Substituting the values we know, we get: 60 = (1/2) x 12 x d Simplifying this equation, we get: 60 = 6d Dividing both sides by 6, we get: d = 10 Therefore, the length of the other diagonal is 10 cm. So, the answer is option C - 10cm.
Tambaya 8 Rahoto
A sector of a circle of radius 9cm subtends angle 120° at the centre of the circle. Find the area of the sector to the nearest cm\(^2\) [Take π = 22/7]
Bayanin Amsa
To find the area of a sector, we need to use the formula: Area of sector = (θ/360) x πr² where θ is the angle subtended at the centre of the circle, r is the radius of the circle, and π is a mathematical constant approximately equal to 22/7. In this case, the radius is given as 9cm and the angle is given as 120°. So, substituting these values into the formula, we get: Area of sector = (120/360) x (22/7) x 9² = (1/3) x (22/7) x 81 = 754/7 ≈ 107.71 cm² (rounded to the nearest cm²) Therefore, the correct answer is: 85 cm² (option C).
Tambaya 9 Rahoto
Simplify: \(\frac{\log \sqrt{27}}{\log \sqrt{81}}\)
Tambaya 10 Rahoto
Write as a single fraction: \(\frac{5}{6r} - \frac{3}{4r}\)
Bayanin Amsa
To add or subtract fractions, we need to have a common denominator. Here, we have denominators of 6r and 4r. The least common multiple (LCM) of 6r and 4r is 12r. So, we need to convert each fraction so that it has a denominator of 12r. \(\frac{5}{6r} - \frac{3}{4r} = \frac{5 \cdot 4}{6r \cdot 4} - \frac{3 \cdot 3}{4r \cdot 3}\) Simplifying the resulting fractions, we get: \(\frac{20}{24r} - \frac{9}{12r} = \frac{20}{24r} - \frac{18}{24r}\) Now, we can combine the fractions by subtracting the numerators and keeping the same denominator: \(\frac{20}{24r} - \frac{18}{24r} = \frac{2}{24r}\) Finally, we can simplify the resulting fraction by dividing the numerator and the denominator by 2: \(\frac{2}{24r} = \frac{1}{12r}\) Therefore, the answer is:
Tambaya 12 Rahoto
In the diagram above, |PQ| = |QR|, |PS| = |RS|, ?PSR = 30o and ?PQR = 80o. Find ?SPQ.
Bayanin Amsa
Tambaya 13 Rahoto
At a point 500m from the base of a water tank the angle of elevation of the top of the tank is 45o. Find the height of the tank,
Bayanin Amsa
We can use the trigonometric ratio tangent to solve this problem. Let h be the height of the tank. Then, we have: tan(45°) = h/500 Since tan(45°) = 1, we can simplify to: h/500 = 1 h = 500 Therefore, the height of the tank is 500 meters. Note that the answer is not in the options provided.
Tambaya 14 Rahoto
Simplify: \(\frac{4^{-\frac{1}{2}} \times 16^{\frac{3}{4}}}{4^{\frac{1}{2}}}\)
Tambaya 15 Rahoto
What is the probability that an integer selected from the set of integers (20, 21, ...., 30) is a prime number?
Bayanin Amsa
The set of integers from 20 to 30 is {20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30}. We need to find the probability of selecting a prime number from this set. Prime numbers are the numbers that are only divisible by 1 and themselves. The prime numbers in this set are 23 and 29. So, out of the 11 numbers in the set, only 2 are prime. Therefore, the probability of selecting a prime number from the set is 2/11. Hence, the answer is (a) 2/11.
Tambaya 16 Rahoto
The table above shows the scores of a group of 40 students in a physics test.
If the mode is m and the median is n, then (m,n) is
Tambaya 17 Rahoto
The nth term of a sequence is given by (-1)\(^{n-2}\) x 2\(^{n+1}\). Find the sum of the second and third terms.
Bayanin Amsa
Tambaya 18 Rahoto
Convert 89\(_{10}\) to a number in base two.
Bayanin Amsa
To convert a decimal number to a binary number, we need to repeatedly divide the decimal number by 2 and take note of the remainder. The binary number is obtained by arranging the remainders in the reverse order. Here's the process for converting 89\(_{10}\) to binary: 1. Divide 89 by 2. The quotient is 44 and the remainder is 1. 2. Divide 44 by 2. The quotient is 22 and the remainder is 0. 3. Divide 22 by 2. The quotient is 11 and the remainder is 0. 4. Divide 11 by 2. The quotient is 5 and the remainder is 1. 5. Divide 5 by 2. The quotient is 2 and the remainder is 1. 6. Divide 2 by 2. The quotient is 1 and the remainder is 0. 7. Divide 1 by 2. The quotient is 0 and the remainder is 1. The remainders, taken in reverse order, are 1011001, which is the binary equivalent of 89\(_{10}\). Therefore, the correct answer is option (B) 1011001.
Tambaya 19 Rahoto
A cone is 14cm deep and the base radius is 41/2cm. Calculate the volume of water that is exactly half the volume of the cone.[Take π = 22/7]
Bayanin Amsa
The volume of a cone is given by the formula: V = (1/3)πr2h, where r is the radius of the base and h is the height of the cone. The cone in the question has height 14 cm and base radius 4.5 cm. Therefore, its volume is: V = (1/3) × (22/7) × (4.5)2 × 14 = 346.5 cm3 Half of this volume is: V/2 = 346.5/2 = 173.25 cm3 Therefore, the volume of water that is exactly half the volume of the cone is 173.25 cm3. The answer closest to this value is 148.5 cm3, so the correct option is: - 148.5cm3
Tambaya 20 Rahoto
A stick of length 1.75m was measured by a boy as 1.80m. Find the percentage error in his measurement
Bayanin Amsa
To find the percentage error, we need to first calculate the absolute error which is the difference between the measured value and the actual value. Absolute error = |1.80 - 1.75| = 0.05m Then, we can calculate the percentage error using the formula: Percentage error = (Absolute error / Actual value) x 100% Percentage error = (0.05 / 1.75) x 100% = 2.86% Therefore, the percentage error in the boy's measurement is approximately 2.86%. The closest option to this value is 26/7%.
Tambaya 21 Rahoto
From a point on the edge of the sea, one ship is 24km away on a bearing S 50oE and another ship is 7km away on a bearing S40oW. How far apart are the ships?
Tambaya 22 Rahoto
Solve the simultaneous equations y = 3x; 4y - 5x =14
Bayanin Amsa
The first equation can be rearranged as x = y/3. Then substituting this value of x into the second equation, we get: 4y - 5(y/3) = 14 Multiplying both sides by 3, we get: 12y - 5y = 42 7y = 42 y = 6 Substituting y = 6 into the first equation, we get: x = y/3 = 2 Therefore, the solution to the simultaneous equations is x = 2 and y = 6. So the answer is 2,6.
Tambaya 24 Rahoto
Factorize 2x\(^2\) - 21x + 45
Bayanin Amsa
To factorize 2x\(^2\) - 21x + 45, we need to find two numbers that multiply to give 2\(\times\)45 = 90 and add to give -21. Let's factorize 90 to find such numbers: 90 = 1\(\times\)90 = 2\(\times\)45 = 3\(\times\)30 = 5\(\times\)18 = 6\(\times\)15 Out of these, the pair that adds up to -21 is 6 and 15. So, we can rewrite the expression as: 2x\(^2\) - 21x + 45 = 2x\(^2\) - 6x - 15x + 45 Now, we can group the first two terms and the last two terms separately and factorize them using the distributive law. That gives us: 2x\(^2\) - 6x - 15x + 45 = 2x(x - 3) - 15(x - 3) Notice that we have a common factor of (x - 3) in both terms. We can factor it out to get the final answer: 2x\(^2\) - 21x + 45 = (x - 3)(2x - 15) Therefore, the correct answer is (x - 3)(2x - 15).
Tambaya 26 Rahoto
Solve the inequality: 3m + 3 > 9
Bayanin Amsa
The inequality 3m + 3 > 9 can be solved by isolating m on one side of the inequality sign. First, subtract 3 from both sides to get 3m > 6. Then, divide both sides by 3 to get m > 2. Therefore, the solution to the inequality is m > 2, which means any value of m that is greater than 2 would make the inequality true. The correct label for the solution is "m > 2".
Tambaya 27 Rahoto
A fair die is rolled once. What is the probability of obtaining a number less than 3?
Bayanin Amsa
A fair die has six possible outcomes, which are the numbers 1, 2, 3, 4, 5, and 6. Since we want to find the probability of obtaining a number less than 3, we need to count how many of these outcomes satisfy this condition. There are only two possible outcomes that satisfy this condition: 1 and 2. Therefore, the probability of obtaining a number less than 3 is 2 out of 6 possible outcomes or 2/6, which simplifies to 1/3. So, the answer is: - Probability = 1/3
Tambaya 28 Rahoto
In the diagram above, |QR| = 12cm and |QS| = 10cm. If ?PQR = 90°, ?RSQ = 90° and PSQ is a straight line, find |PS|
Bayanin Amsa
Tambaya 30 Rahoto
Find the median of the set of numbers 12, 15, 13, 14, 12, 12.
Bayanin Amsa
To find the median, we need to first arrange the set of numbers in order from least to greatest: 12, 12, 12, 13, 14, 15 The median is the middle number in the set. Since there are six numbers in this set, the middle number will be the average of the two middle numbers. In this case, the two middle numbers are 12 and 13, so their average is: (12 + 13) / 2 = 12.5 Therefore, the median of this set of numbers is 12.5. Option (B) is the correct answer.
Tambaya 31 Rahoto
Two chords PQ and RS of a circle when produced meet at K. If ∠KPS = 31o and ∠PKR = 42o, find ∠KQR
Bayanin Amsa
Tambaya 32 Rahoto
| Number | 1 | 2 | 3 | 4 | 5 | 6 |
| No of times | 25 | 30 | 45 | 28 | 40 | 32 |
A die rolled 200 times. The outcome obtained are shown in the table above.
What is the probability of obtaining a number less than 3 ?
Bayanin Amsa
To find the probability of obtaining a number less than 3, we need to add up the number of times the die rolled a 1 or a 2, since these are the numbers less than 3. From the table, we see that the number of times the die rolled a 1 is 25 and the number of times it rolled a 2 is 30. So the total number of times the die rolled a number less than 3 is: 25 + 30 = 55 Therefore, the probability of obtaining a number less than 3 is: 55/200 = 0.275 So the answer is option C: 0.275.
Tambaya 33 Rahoto
Solve the equation x\(^2\) - 2x - 3 = 0
Bayanin Amsa
To solve the equation x\(^2\) - 2x - 3 = 0, we can use the quadratic formula, which is given by: x = (-b ± sqrt(b\^2 - 4ac)) / 2a where a, b, and c are the coefficients of the quadratic equation ax\^2 + bx + c = 0. Using the coefficients of the given equation, a=1, b=-2, and c=-3, we get: x = (-(-2) ± sqrt((-2)\^2 - 4(1)(-3))) / 2(1) x = (2 ± sqrt(16)) / 2 x = (2 ± 4) / 2 So the solutions of the equation are: x = 3 or x = -1 Therefore, the correct answer is: (-1, 3).
Tambaya 34 Rahoto
The sum of an interior angles of a regular polygon is 30 right angles. How many sides has the polygon?
Bayanin Amsa
The sum of the interior angles of a polygon with n sides is given by the formula (n - 2) x 180 degrees. Since a right angle is equal to 90 degrees, 30 right angles is equal to 30 x 90 = 2700 degrees. Thus, we can set up the equation: (n - 2) x 180 = 2700 Solving for n: n - 2 = 15 n = 17 Therefore, the polygon has 17 sides, so the answer is (d) 17 sides.
Tambaya 35 Rahoto
Calculate the total surface area of a cone of height 12cm and base radius 5cm. [Take π = 22/7]
Bayanin Amsa
To find the total surface area of a cone, we need to add the area of the base and the curved surface area. Given that the base radius is 5cm, we can find the base area by using the formula for the area of a circle: Area of base = πr² = π(5)² = 25π To find the curved surface area, we need to first find the slant height of the cone. We can use the Pythagorean theorem to find this value: l = √(r² + h²) = √(5² + 12²) = √169 = 13cm Now, we can find the curved surface area using the formula: Curved surface area = πrl Curved surface area = π(5)(13) = 65π Therefore, the total surface area of the cone is: Total surface area = Base area + Curved surface area Total surface area = 25π + 65π = 90π Using the approximation π = 22/7, we get: Total surface area = 90π = 90(22/7) = 282.8571... cm² Rounding off to one decimal place, we get: Total surface area ≈ 282.9 cm² Therefore, the correct option is 4) 282 6/7cm².
Tambaya 37 Rahoto
If sin x = cos 50o, then x equals
Bayanin Amsa
Since sin x = cos 90 - x for any angle x, we can write: sin x = cos (90 - x) Given sin x = cos 50°, we can substitute to get: cos (90 - x) = cos 50° Taking the inverse cosine of both sides, we get: 90 - x = 50° Solving for x, we get: x = 40° Therefore, the answer is option A: 40°.
Tambaya 38 Rahoto
What is the probability of having an even number in a single toss of a fair die?
Bayanin Amsa
In a single toss of a fair die, there are six possible outcomes: 1, 2, 3, 4, 5, and 6. Out of these outcomes, three of them (2, 4, and 6) are even numbers. Therefore, the probability of getting an even number in a single toss of a fair die is 3 out of 6 or 1 out of 2. This can be expressed as a fraction of 1/2 or a decimal of 0.5. So, the correct answer is 1/2.
Tambaya 40 Rahoto
State the fifth and seventh terms of the sequence \(-2, -3, -4\frac{1}{2}, ...\)
Bayanin Amsa
Tambaya 41 Rahoto
What is the probability that the total sum of seven would appear in toss of a fair die?
Bayanin Amsa
When a fair die is rolled, there are 6 possible outcomes, each with an equal probability of 1/6. The only way to obtain a sum of 7 is by rolling a combination of (1,6), (2,5), (3,4), (4,3), (5,2), or (6,1). That means there are 6 possible ways to get a sum of 7 out of 36 possible outcomes. Therefore, the probability of obtaining a sum of 7 is 6/36, which simplifies to 1/6. So the answer is: 1/6.
Tambaya 42 Rahoto
Simplify \(\frac{2.25}{0.015}\) leaving your answer in standard form
Tambaya 43 Rahoto
The distribution by state of 840 students in the Faculty of Science of a Nigerian University in a certain session is as follows:
| Bendel | 45 |
| kwara | 410 |
| Ogun | 105 |
| Ondo | 126 |
| Oyo | 154 |
In a pie chart drawn to represent this distribution, the angle subtended by Ondo is
Bayanin Amsa
To find the angle subtended by Ondo in the pie chart, we need to first calculate the total number of students in the Faculty of Science. Total number of students = 45 + 410 + 105 + 126 + 154 = 840 Next, we need to find the percentage of students from Ondo: Percentage of students from Ondo = (Number of students from Ondo / Total number of students) x 100 = (126/840) x 100 = 15 Since there are five states in total, and the pie chart represents 360 degrees, we can find the angle subtended by Ondo as follows: Angle subtended by Ondo = (Percentage of students from Ondo / 100) x 360 = (15/100) x 360 = 54 degrees Therefore, the answer is (e) 54o.
Tambaya 44 Rahoto
In the diagram, PQR is a tangent to the circle QST at Q. If |QT| = |ST| and ?SQR = 68°, find ?PQT.
Bayanin Amsa
Tambaya 45 Rahoto
The table above shows the scores of a group of 40 students in a physics test
What is the mean of the distribution?
Tambaya 46 Rahoto
S = {1, 2, 3, 4, 5, 6}, T = {2,4,5,7} and R = {1,4, 5}, and (S∩T) ∪ R
Bayanin Amsa
The given problem involves set theory and requires finding the union and intersection of sets. We are given three sets: S, T, and R. Set S contains six elements, {1, 2, 3, 4, 5, 6}, set T contains four elements, {2, 4, 5, 7}, and set R contains three elements, {1, 4, 5}. First, we need to find the intersection of sets S and T, denoted as S∩T, which represents the elements that are common to both sets. The elements common to sets S and T are 2, 4, and 5. Next, we need to find the union of the intersection of sets S and T with set R, denoted as (S∩T) ∪ R, which represents all the elements that are in either set. The elements in (S∩T) ∪ R are {1, 2, 4, 5}. Therefore, the answer is {1, 2, 4, 5}.
Tambaya 47 Rahoto
If sin x = 12/13, where 0° < x < 90°, find the value of 1 - cos\(^2\)x
Bayanin Amsa
We know that sin x = opposite/hypotenuse = 12/13, therefore, the adjacent side of the right-angled triangle is √(13^2 - 12^2) = 5. Now, we need to find the value of 1 - cos^2x. Using the identity sin^2x + cos^2x = 1, we can write: cos^2x = 1 - sin^2x Substituting sin x = 12/13, we get: cos^2x = 1 - (12/13)^2 cos^2x = 1 - 144/169 cos^2x = 25/169 Therefore, 1 - cos^2x = 1 - 25/169 = 144/169. Hence, the answer is (d) 144/169.
Tambaya 48 Rahoto
Find the median of the following set of numbers: 28, 29, 39, 38, 33, 37, 26, 20, 15, 25
Bayanin Amsa
Tambaya 49 Rahoto
The universal set \(\varepsilon\) is the set of all integers and the subset P, Q, R of \(\varepsilon\) are given by:
\(P = {x : x < 0} ; Q = {... , -5, -3, -1, 1, 3, 5} ; R = {x : -2 \leq x < 7}\)
(a) Find \(Q \cap R\).
(b) Find \(R'\) where R' is the complement of R with respect to \(\varepsilon\).
(c) Find \(P' \cup R'\)
(d) List the members of \((P \cap Q)'\).
Tambaya 50 Rahoto
(a) 
In the diagram, BA is parallel to DE. Find the value of x.
(b) Illustrate graphically and shade the region in which inequalities \(y - 2x < 5 ; 2y + x \geq 4 ; y + 2x \leq 10\) are satisfied.
Bayanin Amsa
None
Tambaya 51 Rahoto
A box contains identical balls of which 12 are red, 16 white and 8 blue. Three balls are drawn from the box one after the other without replacement. Find the probability that :
(a) three are red;
(b) the first is blue and the other two are red;
(c) two are white and one is blue.
Tambaya 52 Rahoto
(a) Prove that the angle which an arc of a circle subtends at the centre is twice that which it subtends at any point on the remaining part of the circumference.
(b) 
In the diagram, O is the centre of the circle ACDB. If < CAO = 26° and < AOB = 130°. Calculate : (i) < OBC ; (ii) < COB.
Tambaya 54 Rahoto
(a)(i) Given that \(\log_{10} 5 = 0.699\) and \(\log_{10} 3 = 0.477\), find \(\log_{10} 45\), without using Mathematical tables.
(ii) Hence, solve \(x^{0.8265} = 45\).
(b) Use Mathematical tables to evaluate \(\sqrt{\frac{2.067}{0.0348 \times 0.538}}\)
Tambaya 55 Rahoto
A simple measuring device is used at points X and Y on the same horizontal level to measure the angles of elevation of the peak P of a certain mountain. If X is known to 5,200m above sea level, /XY/ = 4,000m and the measurements of the angles of elevation of P at X and Y are 15° and 35° respectively, find the height of the mountain. (Take \(\tan 15 = 0.3\) and \(\tan 35 = 0.7\))
Bayanin Amsa
None
Tambaya 56 Rahoto
(a) Simplify \(\frac{3}{m + 2n} - \frac{2}{m - 3n}\)
(b) A number is made up of two digits. The sum of the digits is 11. If the digits are interchanged, the original number is increased by 9. Find the number.
Tambaya 57 Rahoto
The table below shows the frequency distribution of the marks scored by fifty students in an examination.
| Marks (%) | 0-9 | 10-19 | 20-29 | 30-39 | 40-49 | 50-59 | 60-69 | 70-79 | 80-89 | 90-99 |
| Freq | 2 | 3 | 4 | 6 | 13 | 10 | 5 | 3 | 2 | 2 |
(a) Draw the cumulative frequency curve for the distribution.
(b) Use your curve to estimate the : (i) upper quartile; (ii) pass mark if 60% of the students passed.
Tambaya 58 Rahoto
P and Q are two points on latitude 55°N and their longitudes are 33°W and 20°E respectively. Calculate the distance between P and Q measured along
(a) the parallel of latitude ;
(b) a great circle.
[Take \(\pi = \frac{22}{7}\) and radius of the earth = 6400km].
Tambaya 59 Rahoto
(a) Simplify, without using Mathematical tables: \(\log_{10} (\frac{30}{16}) - 2 \log_{10} (\frac{5}{9}) + \log_{10} (\frac{400}{243})\)
(b) Without using Mathematical tables, calculate \(\sqrt{\frac{P}{Q}}\) where \(P = 3.6 \times 10^{-3}\) and \(Q = 2.25 \times 10^{6}\), leaving your answer in standard form.
Bayanin Amsa
None
Tambaya 60 Rahoto
(a) Copy and complete the following table of values for \(y = 3\sin 2\theta - \cos \theta\).
| \(\theta\) | 0° | 30° | 60° | 90° | 120° | 150° | 180° |
| y | -1.0 | 0 | 1.0 |
(b) Using a scale of 2cm to 30° on the \(\theta\) axis and 2cm to 1 unit on the y- axis, draw the graph of \(y = 3 \sin 2\theta - \cos \theta\) for \(0° \leq \theta \leq 180°\).
(c) Use your graph to find the : (i) solution of the equation \(3 \sin 2\theta - \cos \theta = 0\), correct to the nearest degree; (ii) maximum value of y, correct to one decimal place.
Bayanin Amsa
None
