WAEC SSCE - General Mathematics - 2015
Ajụjụ 1 Ripọtì
A letter is selected from the letters of the English alphabet. What is the probability that the letter selected is from the word MATHEMATICS?
Akọwa Nkọwa
The word MATHEMATICS contains the letters A, C, E, H, I, M, S, and T. Since there are 26 letters in the English alphabet, the probability of selecting any letter from the alphabet is 1 out of 26. Out of the letters in the word MATHEMATICS, there are 8 letters, so the probability of selecting a letter from the word MATHEMATICS is 8 out of 26. Therefore, the probability of selecting a letter from the word MATHEMATICS is: $$\frac{8}{26}=\frac{4}{13}$$ Therefore, the correct answer is option (C).
Ajụjụ 3 Ripọtì
A trader bought an engine for $15,000.00 outside Nigeria. If the exchange rate is $0.075 to N1.00, how much did the engine cost in Naira?
Akọwa Nkọwa
Ajụjụ 4 Ripọtì
In a circle radius rcm, a chord 16\(\sqrt{3}cm\) long is 10cmfrom the centre of the circle. Find, correct to the nearest cm, the value of r
Akọwa Nkọwa
In a circle, a radius is a line segment that connects the center of the circle to any point on the circle. A chord is a line segment that connects two points on a circle. Given that a chord of length 16\(\sqrt{3}cm\) is 10cm from the center of the circle, we can use the Pythagorean theorem to find the length of the radius. First, we draw a line from the center of the circle perpendicular to the chord, creating a right triangle with the radius, the perpendicular line, and half the chord length (which is \(8\sqrt{3}\) cm). Using the Pythagorean theorem: \begin{align*} r^2 &= (8\sqrt{3})^2 + 10^2 \\ r^2 &= 192 + 100 \\ r^2 &= 292 \\ r &\approx \sqrt{292} \\ r &\approx 17 \text{ cm} \end{align*} Rounding to the nearest whole number, we get the answer of 17cm. Therefore, the correct option is (b) 17cm.
Ajụjụ 5 Ripọtì
If \(\frac{27^x \times 3^{1 - x}}{9^{2x}} = 1\), find the value of x.
Ajụjụ 7 Ripọtì
If m = 4, n = 9 and r = 16., evaluate \(\frac{m}{n}\) - 1\(\frac{7}{9}\) + \(\frac{n}{r}\)
Ajụjụ 8 Ripọtì
If \(\frac{\sqrt{2} + \sqrt{3}}{\sqrt{3}}\) is simplified as m + n\(\sqrt{6}\), find the value of (m + n)
Akọwa Nkọwa
Ajụjụ 9 Ripọtì
Find the equation whose roots are \(\frac{3}{4}\) and -4
Akọwa Nkọwa
If a quadratic equation has roots p and q, then it can be expressed in the factored form as:
(x - p)(x - q) = 0
Expanding this equation, we get:
x2 - (p + q)x + pq = 0
In this case, the roots are given as \(\frac{3}{4}\) and -4. Thus, the equation can be expressed as:
(x - \(\frac{3}{4}\))(x + 4) = 0
Expanding this equation, we get:
x2 + (4 - \(\frac{3}{4}\))x - 3 = 0
Multiplying throughout by 4 to eliminate fractions, we get:
4x2 + 13x - 12 = 0
Therefore, the equation whose roots are \(\frac{3}{4}\) and -4 is 4x2 + 13x - 12 = 0.
Hence, the answer is 4x2 + 13x - 12 = 0.
Ajụjụ 11 Ripọtì
In the diagram, O is the centre, \(\bar{RT}\) is a diameter, < PQT = 33\(^o\) and <TOS = 76\(^o\). Using the diagram, calculate the value of angle PTR.
Akọwa Nkọwa
Ajụjụ 12 Ripọtì
\(\begin{array}{c|c}
x & 0 & 1\frac{1}{4} & 2 & 4\\
\hline
y & 3 & 5\frac{1}{2} & &
\end{array}\)
The table given shows some values for a linear graph. Find the gradient of the line
Akọwa Nkọwa
To find the gradient of a line, we need to calculate the change in y divided by the change in x between two points on the line. Let's choose two points on the line from the given table. We can select the points (0,3) and (2,y), where y is the value of y for x = 2. The change in y is y - 3, and the change in x is 2 - 0 = 2. Therefore, the gradient of the line is: gradient = change in y / change in x = (y - 3) / 2 We don't know the value of y yet, but we can use the other point on the line to find it. Using the points (1.25, 5.5) and (2, y), we get: gradient = (y - 5.5) / (2 - 1.25) = (y - 5.5) / 0.75 Setting the two expressions for the gradient equal to each other, we get: (y - 3) / 2 = (y - 5.5) / 0.75 Solving for y gives: y = 7 Therefore, the value of y for x = 2 is 7, and the gradient of the line is: gradient = (y - 3) / 2 = (7 - 3) / 2 = 2 So the answer is 2, and the gradient of the line is 2. In simple terms, the gradient of a line tells us how steep the line is. If the gradient is positive, the line is sloping upwards from left to right. If the gradient is negative, the line is sloping downwards from left to right. The larger the absolute value of the gradient, the steeper the line.
Ajụjụ 13 Ripọtì
If {X: 2 d- x d- 19; X integer} and 7 + x = 4 (mod 9), find the highest value of x
Akọwa Nkọwa
To solve this problem, we need to use the given conditions to find the possible values of x and then determine the highest value among them. From the first condition, we know that x is an integer between 2 and 19 (inclusive) and can be represented as {X: 2 d- x d- 19; X integer}. This means that x can take on any of the following values: 2, 3, 4, ..., 18, 19. The second condition tells us that 7 + x is congruent to 4 modulo 9, which can be written as: 7 + x ≡ 4 (mod 9) To solve for x, we can subtract 7 from both sides of the congruence: x ≡ -3 (mod 9) Since we want x to be between 2 and 19, we can add or subtract multiples of 9 to -3 until we get a value within the range of possible values for x. Doing so, we obtain: x ≡ -3 (mod 9) x ≡ 15 (mod 9) Since 15 is the largest value of x that satisfies both conditions, our answer is 15.
Ajụjụ 14 Ripọtì
Ada draws the graph of y = x2 - x - 2 and y = 2x - 1 on the same axes. Which of these equations is she solving?
Akọwa Nkọwa
Ajụjụ 15 Ripọtì
The scores of twenty students in a test are as follows: 44, 47, 48, 49, 50, 51, 52, 53, 53, 54, 58, 59, 60, 61, 63, 65, 67, 70, 73, 75. Find the third quartile.
Ajụjụ 16 Ripọtì
A trader bought 100 oranges at 5 for N40.00 and 20 for N120.00. Find the profit or loss percent
Ajụjụ 18 Ripọtì
A farmer uses \(\frac{2}{5}\) of his land to grow cassava, \(\frac{1}{3}\) of the remaining for yam and the rest for maize. Find the part of the land used for maize
Akọwa Nkọwa
Let's assume the farmer has 15 parts of land. The fraction of land used for cassava is \(\frac{2}{5}\) of 15, which is 6 parts. The remaining land is 15 - 6 = 9 parts. The fraction of the remaining land used for yam is \(\frac{1}{3}\) of 9, which is 3 parts. Therefore, the part of the land used for maize is the remaining land after cassava and yam are planted, which is 9 - 3 = 6 parts. So the fraction of the land used for maize is \(\frac{6}{15}\), which simplifies to \(\frac{2}{5}\). Therefore, the answer is option (B) \(\frac{2}{5}\).
Ajụjụ 19 Ripọtì
In the diagram, O is the centre. If PQ//RS and ∠ONS = 140°, find the size of ∠POM.
Akọwa Nkọwa
Ajụjụ 21 Ripọtì
The rate of consumption of petrol by a vehicle varies directly as the square of the distance covered. If 4 litres of petrol is consumed on a distance of 15km. how far would the vehicle go on 9 litres of petrol?
Akọwa Nkọwa
Since the rate of consumption of petrol varies directly as the square of the distance covered, let's assume that the rate is k\(\frac{\text{litres}}{\text{km}^2}\). Then we can write the relationship as: rate of consumption = k(distance covered)\(^2\) We are given that when the distance covered is 15 km, the rate of consumption is 4 litres. Substituting these values in the above equation, we get: 4 = k(15)\(^2\) k = \(\frac{4}{15^2}\) Now, we need to find how far the vehicle can go on 9 litres of petrol. Let's assume that the vehicle can travel a distance of x km on 9 litres of petrol. Using the same equation as before, we get: 9 = \(\frac{4}{15^2}\)(x)\(^2\) Simplifying this equation, we get: x\(^2\) = \(\frac{9}{\frac{4}{15^2}}\) x\(^2\) = 506.25 x = 22.5 Therefore, the vehicle can travel 22.5 km on 9 litres of petrol. So the answer is 22\(\frac{1}{2}\)km.
Ajụjụ 22 Ripọtì
The area of a rhombus is 110 cm\(^2\). If the diagonals are 20 cm and (2x + 1) cm long, find the value of x.
Ajụjụ 23 Ripọtì
Factorize completely: 6ax - 12by - 9ay + 8bx
Ajụjụ 24 Ripọtì
Simplify: \(\frac{3x - y}{xy} - \frac{2x + 3y}{2xy} + \frac{1}{2}\)
Akọwa Nkọwa
To simplify the expression, we first need to find a common denominator for the fractions. The lowest common multiple of the denominators xy and 2xy is 2xy. \(\frac{3x - y}{xy} - \frac{2x + 3y}{2xy} + \frac{1}{2} = \frac{3x\cdot2 - y\cdot2}{xy\cdot2} - \frac{2x\cdot1 + 3y\cdot1}{2xy\cdot1} + \frac{1\cdot xy}{2\cdot xy}\) Simplifying further, we get: \(\frac{6x - 2y}{2xy} - \frac{2x + 3y}{2xy} + \frac{xy}{2xy}\) Combining like terms, we get: \(\frac{4x - 5y + xy}{2xy}\) Therefore, the answer is option D: \(\frac{4x - 5y + xy}{2xy}\).
Ajụjụ 26 Ripọtì
The angle of elevation of an aircraft from a point K on the horizontal ground 30\(\alpha\). If the aircraft is 800m above the ground, how far is it from K?
Akọwa Nkọwa
Ajụjụ 27 Ripọtì
The dimension of a rectangular tank are 2m by 7m by 11m. If its volume is equal to that of a cylindrical tank of height 4cm, calculate the base radius of the cylindrical tank. [Take \(\pi = \frac{22}{7}\)]
Akọwa Nkọwa
Ajụjụ 28 Ripọtì
The population of students in a school is 810. If this is represented on a pie chart, calculate the sectoral angle for a class of 7 students
Akọwa Nkọwa
To find the sectoral angle for a class of 7 students, we first need to calculate what percentage of the total population they represent: Percentage of class of 7 students = (Number of students in the class / Total population) x 100% Percentage of class of 7 students = (7 / 810) x 100% Percentage of class of 7 students = 0.864% (rounded to three decimal places) Next, we need to convert this percentage to an angle of the pie chart. We know that a full circle is 360 degrees, and the percentage of the class of 7 students represents a certain portion of this circle. We can use the following formula to calculate the sectoral angle: Sectoral angle = Percentage of class of 7 students x 360 degrees Plugging in the percentage we calculated earlier, we get: Sectoral angle = 0.864% x 360 degrees Sectoral angle ≈ 3.11 degrees (rounded to two decimal places) Therefore, the sectoral angle for a class of 7 students in the pie chart is approximately 3.11 degrees. So, the correct option is (a) 32o.
Ajụjụ 29 Ripọtì
Tom will be 25 years old in n years' time. If he is 5 years younger than Bade's present age.
Akọwa Nkọwa
Ajụjụ 30 Ripọtì
Given that logx 64 = 3, evaluate x log\(_2\)8
Ajụjụ 31 Ripọtì
In the diagram, O is the centre, \(\bar{RT}\) is a diameter, < PQT = 33\(^o\) and < TOS = 76\(^o\). Using the diagram, find the size of angle PRS.
Akọwa Nkọwa
Ajụjụ 34 Ripọtì
Two angles of a pentagon are in the ratio 2:3. The others are 60o each. Calculate the smaller of the two angles
Akọwa Nkọwa
Let's call the two angles that are in the ratio 2:3 as "2x" and "3x". The sum of the interior angles of a pentagon is given by the formula (5-2) x 180 degrees = 540 degrees. Since the other three angles are all 60 degrees each, we can find the sum of those angles by multiplying 60 by 3, which gives us 180 degrees. Now we can set up an equation to solve for the unknown angle, which is 2x + 3x. 2x + 3x + 180 = 540 Simplifying this equation, we get: 5x + 180 = 540 Subtracting 180 from both sides, we get: 5x = 360 Dividing both sides by 5, we get: x = 72 So the smaller angle, which is 2x, is equal to 2 times 72, which is 144 degrees. Therefore, the correct answer is 144 degrees.
Ajụjụ 36 Ripọtì
In the diagram, PTR is a tangent to the centre O. If angles TON = 108°, Calculate the size of angle PTN
Akọwa Nkọwa
Ajụjụ 37 Ripọtì
If 2n = y, Find 2\(^{(2 + \frac{n}{3})}\)
Akọwa Nkọwa
We know that 2\(^n\) = y. Therefore, 2\(^{(2 + \frac{n}{3})}\) can be written as 2\(^2\) × 2\(^{\frac{n}{3}}\) = 4 × 2\(^{\frac{n}{3}}\). Substituting 2\(^n\) = y in the above equation, we get: 2\(^{(2 + \frac{n}{3})}\) = 4y\(^\frac{1}{3}\) Therefore, the answer is (a) 4y\(^\frac{1}{3}\).
Ajụjụ 38 Ripọtì
Adding 42 to a given positive number gives the same result as squaring the number. Find the number
Akọwa Nkọwa
Let's call the unknown number "x". According to the problem, adding 42 to x gives the same result as squaring x. We can write this as an equation: x + 42 = x^2 To solve for x, we can rearrange this equation to get it in standard quadratic form: x^2 - x - 42 = 0 Now we can factor this quadratic equation: (x - 7)(x + 6) = 0 This gives us two possible solutions: x = 7 and x = -6. However, we were given that x is a positive number, so we can discard the negative solution. Therefore, the answer is x = 7. To check our answer, we can substitute x = 7 into the original equation: 7 + 42 = 49 7^2 = 49 So adding 42 to 7 does indeed give the same result as squaring 7.
Ajụjụ 39 Ripọtì
In the diagram, the shaded part is carpet laid in a room with dimensions 3.5m by 2.2m leaving a margin of 0.5m round it. Find area of the margin
Ajụjụ 40 Ripọtì
In the diagram, \(\bar{OX}\) bisects < YXZ and \(\bar{OZ}\) bisects < YZX. If < XYZ = 68o, calculate the value of < XOZ
Ajụjụ 41 Ripọtì
Calculate the mean deviation of 5, 3, 0, 7, 2, 1
Akọwa Nkọwa
To find the mean deviation, we need to first calculate the mean of the given set of numbers: Mean = (5 + 3 + 0 + 7 + 2 + 1) / 6 = 18 / 6 = 3 Next, we need to find the absolute deviation of each number from the mean: |5 - 3| = 2 |3 - 3| = 0 |0 - 3| = 3 |7 - 3| = 4 |2 - 3| = 1 |1 - 3| = 2 To calculate the mean deviation, we add up these absolute deviations and divide by the number of numbers: Mean deviation = (2 + 0 + 3 + 4 + 1 + 2) / 6 = 12 / 6 = 2 Therefore, the answer is option (B) 2.0.
Ajụjụ 42 Ripọtì
The probability that kebba, Ebou and Omar will hit a target are \(\frac{2}{3}\), \(\frac{3}{4}\) and \(\frac{4}{5}\) respectively. Find the probability that only Kebba will hit the target.
Akọwa Nkọwa
The probability that only Kebba will hit the target means that Kebba hits the target and the other two miss the target. We can find this probability by multiplying the probability of Kebba hitting the target with the probability of Ebou missing the target and the probability of Omar missing the target. Probability that only Kebba hits the target = Probability of Kebba hitting the target × Probability of Ebou missing the target × Probability of Omar missing the target = \(\frac{2}{3}\) × \(\frac{1}{4}\) × \(\frac{1}{5}\) = \(\frac{2}{3\times4\times5}\) = \(\frac{1}{30}\) Therefore, the probability that only Kebba will hit the target is \(\frac{1}{30}\).
Ajụjụ 43 Ripọtì
In the given diagram, \(\bar{QT}\) and \(\bar{PR}\) are straight lines, < ROS = (3n - 20), < SOT = n, < POL = m and < QOL is a right angle. Find the value of n.
Akọwa Nkọwa
Since < QOL is a right angle, we have: < QOT + < SOT = 90o Since < QOT = 180o - < POL, we have: < QOT = 180o - m Substituting into the first equation, we get: 180o - m + n = 90o n - m = 90o - 180o = -90o Also, from the diagram, we have: < QOT + < ROS + < SOT = 180o Substituting < QOT = 180o - < POL and simplifying, we get: 180o - m + (3n - 20) + n = 180o 4n - m = 20 We now have a system of two equations: n - m = -90o 4n - m = 20 Solving simultaneously, we get: n = 35o
Ajụjụ 44 Ripọtì
\(\begin{array}{c|c}
Scores & 0 - 4 & 5 - 9 & 10 - 14\\
\hline Frequency & 2 & 1 & 2\end{array}\)
The table shows the distribution of the scores of some students in a test. Calculate the mean scores.
Ajụjụ 45 Ripọtì
Describe the shaded portion in the diagram
Akọwa Nkọwa
The shaded portion in the diagram represents the set of elements that belong to the complement of P and the intersection of Q and R. In other words, it represents the set of elements that are not in P, but are in both Q and R. Therefore, the correct answer is P' \(\cap\) Q \(\cap\) R'.
Ajụjụ 47 Ripọtì
The table shows the marks scored by some candidates in an examination.
| Marks (%) | 0-9 | 10-19 | 20-29 | 30-39 | 40-49 | 50-59 | 60-69 | 70-79 | 80-89 | 90-99 |
| Frequency | 7 | 11 | 17 | 20 | 29 | 34 | 30 | 25 | 21 | 6 |
(a) Construct a cumulative frequency table for the distribution and draw a cumulative frequency curve.
(b) Use the curve to estimate, correct to one decimal place, the :
(i) Lowest mark for distinction if 5% of the candidates passed with distinction ; (ii) probability of selecting a candidate who scored at most 45%.
None
Akọwa Nkọwa
None
Ajụjụ 48 Ripọtì
(a) Solve the inequality : \(4 + \frac{3}{4}(x + 2) \leq \frac{3}{8}x + 1\)
(b) 
The diagram shows a rectangle PQRS from which a square of side x cm has been cut. If the area of the shaded portion is 484\(cm^{2}\), find the values of x.
Akọwa Nkọwa
None
Ajụjụ 49 Ripọtì
(a) The first term of an Arithmetic Progression (AP) is 8, the ratio of the 7th term to the 9th term is 5 : 8, find the common difference of the AP.
(b) A trader bought 30 baskets of pawpaw and 100 baskets of mangoes for N2,450.00. She sold the pawpaw at a profit of 40% and the mangoes at a profit of 30%. If her profit on the entire transaction was N855.00, find the (i) cost price of a basket of pawpaw ; (ii) selling price of the 100 baskets of mangoes.
Ajụjụ 50 Ripọtì
(a) The ratio of the interior angle to the exterior angle of a regular polygon is 5 : 2, Find the number of sides of the polygon.
(b) 
The diagram shows a circle PQRS with centre O, < UQR = 68°, < TPS = 74° and < QSR = 40°. Calculate the value of < PRS.
Akọwa Nkọwa
None
Ajụjụ 51 Ripọtì
(a) Without using Mathematical tables or calculators, simplify : \(\frac{2\tan 60° + \cos 30°}{\sin 60°}\)
(b) From an aeroplane in the air and at a horizontal distance of 1050m, the angles of depression of the top and base of a control tower at an instance are 36° and 41° respectively. Calculate, correct to the nearest meter, the :
(i) height of the control tower ; (ii) shortest distance between the aeroplane and the base of the control tower.
Akọwa Nkọwa
None
Ajụjụ 52 Ripọtì
(a) (i) Illustrate the following statements in a Venn diagram : All good Literature students in a school are in the General Arts class.
(ii) Use ths diagram to determine whether or not the following are valid conclusions from the given statement.
(1) Vivian is in the General Arts class therefore she is a good Literature student.
(2) Audu is not a good Literature student therefore he is not in the General Arts class;
(3) Kweku is not in the General Arts class therefore he is not a good Literature student.
(b) The cost (c) of producing n bricks is the sum of a fixed amount, h, and a variable amount, y, where y varies directly as n. If it costs GH¢950.00 to produce 600 bricks and GH¢ 1,030.00 to produce 1000 bricks,
(i) Find the relationship between c, h and n ; (ii) Calculate the cost of producing 500 bricks.
None
Akọwa Nkọwa
None
Ajụjụ 53 Ripọtì
A water reservoir in the form of a cone mounted on a hemisphere is built such that the plane face of the hemisphere fits exactly to the base of the cone and the height of the cone is 6 times thr radius of its base.
(a) Illustrate this information in a diagram.
(b) If the volume of the reservoir is \(333\frac{1}{3}\pi m^{3}\), calculate, correct to the nearest whole number, the :
(I) volume of the hemisphere ; (II) Total surface area of the reservoir. [Take \(\pi = \frac{22}{7}\)].
None
Akọwa Nkọwa
None
Ajụjụ 54 Ripọtì
(a) Without using Mathematical tables or calculators, simplify:
\(3\frac{4}{9} \div (5\frac{1}{3} - 2\frac{3}{4}) + 5\frac{9}{10}\)
(b) A number is selected at random from each of the sets {2, 3, 4} and {1, 3, 5}. Find the probability that the sum of the two numbers is greater than 3 and less than 7.
Akọwa Nkọwa
None
Ajụjụ 56 Ripọtì
The table is for the relation \(y = px^{2} - 5x + q\).
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
| y | 21 | 6 | -12 | 0 | 13 |
(a)(i) Use the table to find the values of p and q.
(ii) Copy and complete the table.
(b) Using scales of 2cm to 1 unit on the x- axis and 2 cm to 5 units on the y- axis, draw the graph of the relation for \(-3 \leq x \leq 5\).
(c) Use the graph to find :
(i) y when x = 1.8 ; (ii) x when y = -8.
Ajụjụ 57 Ripọtì
(a) Make m the subject of the relations \(h = \frac{mt}{d(m + p)}\).
(b) 
In the diagram, WY and WZ are straight lines, O is the centre of circle WXM and < XWM = 48°. Calculate the value of < WYZ.
(c) An operation \(\star\) is defind on the set X = {1, 3, 5, 6} by \(m \star n = m + n + 2 (mod 7)\) where \(m, n \in X\).
(i) Draw a table for the operation.
(ii) Using the table, find the truth set of : (I) \(3 \star n = 3\) ; (II) \(n \star n = 3\).
Akọwa Nkọwa
None
Ajụjụ 58 Ripọtì
(a) By how much is the sum of \(3\frac{2}{3}\) and \(2\frac{1}{5}\) less than 7?
(b) The height, h m, of a dock above sea level is given by \(h = 6 + 4\cos (15p)°, 0 < p < 6\). Find :
(i) the value of h when p = 4 ; (ii) correct to two significant figures, the value of p when h = 9 m.
Akọwa Nkọwa
None
