WAEC SSCE - General Mathematics - 1989
Ajụjụ 1 Ripọtì
In the diagram above O is the center of the circle, if ∠PAQ = 75o, what is the value of ∠PBQ?
Akọwa Nkọwa
Ajụjụ 2 Ripọtì
In an A.P the first term is 2, and the sum of the 1st and the 6th term is 161/2. What is the 4th term
Akọwa Nkọwa
Ajụjụ 3 Ripọtì
Given that logp = 2 logx + 3logq, which of the following expresses p in terms of x and q?
Akọwa Nkọwa
We are given the equation: log p = 2 log x + 3 log q. We know that log a + log b = log(ab) and log a - log b = log(a/b). Using these properties, we can rewrite the given equation as follows: log p = log x^2 + log q^3 log p = log (x^2q^3) Now we can write the exponential form of this equation as: p = x^2q^3 Therefore, the expression for p in terms of x and q is p = x^2q^3. Hence, the correct answer is "p = x^2q^3".
Ajụjụ 4 Ripọtì
Find the volume of a cone of radius 3.5cm and vertical height 12cm [Take π = 22/7]
Akọwa Nkọwa
Ajụjụ 5 Ripọtì
In the diagram above, O is the center of the circle. Calculate the length of the chord AB if |OA| = 5cm, |OD| = 3cm and ?AOD = ?BOD
Ajụjụ 6 Ripọtì
With reference to the graph above, which f the following solution sets is represented by ΔOPQ?
Akọwa Nkọwa
Ajụjụ 7 Ripọtì
In the diagram above ABDE and FCDE are parallelograms. If |FC| = 12.2cm and the height |PC| = 4.0cm, calculate the area of the parallelogram ABDE.
Akọwa Nkọwa
We know that ABDE and FCDE are parallelograms which means their opposite sides are equal and parallel. Therefore, |AB| = |DE| and |AE| = |BD|. We can calculate the area of parallelogram ABDE as follows: Area of ABDE = |AB| x |PC| We need to calculate |AB| and we can do this by finding the length of |DE|. In parallelogram FCDE, the opposite sides |DE| and |FC| are equal. Therefore, |DE| = |FC| = 12.2cm. Also, we know that FCDE is a parallelogram, so the height |PC| is equal to the height of parallelogram ABDE. Now, we can find the length of |AB| using the Pythagorean theorem. |AB|2 = |AE|2 + |DE|2 |AB|2 = (4.0cm)2 + (12.2cm)2 |AB|2 = 149.24cm2 |AB| = 12.2cm Therefore, the area of ABDE is: Area of ABDE = |AB| x |PC| = 12.2cm x 4.0cm = 48.8cm2 Therefore, the answer is 48.8cm2.
Ajụjụ 9 Ripọtì
The angle of elevation of X from Y is 30o. If XY = 40m, how high is X above the level of Y?
Akọwa Nkọwa
Ajụjụ 10 Ripọtì
A sector is cut off from a circle radius 8.2cm to form a cone, if the radius of the resulting cone is 3.5cm, calculate the curved surface area of the cone. [take π = 22/7]
Akọwa Nkọwa
Ajụjụ 11 Ripọtì
The angle of depression of a point on the ground from the top of a building is 20.3o. If the foot of the building is 40m, calculate the height of the building, correct to one decimal place
Akọwa Nkọwa
In this problem, we can use trigonometry to determine the height of the building. Let h be the height of the building and d be the distance from the point on the ground to the foot of the building. We can see that the angle of depression is the angle between the line of sight from the top of the building to the point on the ground and the horizontal line. Therefore, we can draw a right-angled triangle with the height of the building, the distance from the point on the ground to the foot of the building, and the line of sight from the top of the building to the point on the ground as the sides. From the triangle, we can use the tangent function since we have the opposite and adjacent sides. Therefore: tan(20.3°) = h/40 h = 40 tan(20.3°) ≈ 14.8m Therefore, the height of the building is approximately 14.8m. The correct answer is 14.8m.
Ajụjụ 12 Ripọtì
The graph above is that of a quadratic function y = 2x2 - x - 6
The roots of the equation are
Akọwa Nkọwa
To find the roots of a quadratic function, we need to set y to zero and solve for x. Therefore, we have: 2x² - x - 6 = 0 We can factorize this quadratic equation by splitting the middle term: 2x² - 4x + 3x - 6 = 0 2x(x - 2) + 3(x - 2) = 0 (2x + 3)(x - 2) = 0 Setting each factor to zero, we get: 2x + 3 = 0 or x - 2 = 0 Solving for x in each equation, we get: x = -3/2 or x = 2 Therefore, the roots of the equation are -3/2 and 2. So the correct answer is (B) -1.5, 2.
Ajụjụ 13 Ripọtì
If the second and fourth term of a G.P are 8 and 32 respectively,what is the sum of the first four terms?
Ajụjụ 14 Ripọtì
If logax = p, express x in terms of a and p
Akọwa Nkọwa
We know that logax = p This means that ap = x So, we can express x in terms of a and p as x = ap. Therefore, the answer is x = ap.
Ajụjụ 15 Ripọtì
sinθ = 1/2 and cosθ = -√3/2, what is the value of θ?
Akọwa Nkọwa
We can use the fact that sin2θ + cos2θ = 1, to find the value of sinθ or cosθ given the other one. Since sinθ = 1/2, we know that sin2θ + cos2θ = 1 becomes: (1/2)2 + cos2θ = 1 Simplifying the equation, we get: 1/4 + cos2θ = 1 cos2θ = 3/4 Taking the square root of both sides, we get: cosθ = ±√3/2 Since we know that cosθ = -√3/2, we can conclude that θ = 150o (or 5π/6 radians) as this is the angle in the fourth quadrant that has a cosine of -√3/2. Therefore, the answer is option E: 150o.
Ajụjụ 16 Ripọtì
For a class of 30 students, the scores in a Mathematics test out of 10 marks were as follows;
4,5,7,2,3,6,5,5,8,9,5,4,2,3,7,9,8,7,7,7,3,4,5,5,2,3,6,7,7,2
What is the mode of the score
Akọwa Nkọwa
The mode of a set of data is the value that appears most frequently. To find the mode of the scores in the Mathematics test, we need to determine which score occurs the most number of times. From the given data, we can see that the score "7" appears 6 times, which is more than any other score. Therefore, the mode of the score is 7. So the answer is (e) 7.
Ajụjụ 18 Ripọtì
An arc of a circle radius 7cm is 14cm long. What angle does the arc subtend at the center of circle?
[Take π = 22/7]
Akọwa Nkọwa
Ajụjụ 19 Ripọtì
Simplify \(\frac{9^{-\frac{1}{2}}}{27^{\frac{2}{3}}}\)
Akọwa Nkọwa
We can simplify this expression using the rules of exponents and radicals. First, we can rewrite 9 as \(3^2\) and 27 as \(3^3\): \(\frac{9^{-\frac{1}{2}}}{27^{\frac{2}{3}}} = \frac{(3^2)^{-\frac{1}{2}}}{(3^3)^{\frac{2}{3}}}\) Next, we can simplify the exponents using the product rule of exponents: \(\frac{(3^2)^{-\frac{1}{2}}}{(3^3)^{\frac{2}{3}}} = \frac{3^{-1}}{3^2} = \frac{1}{3^{1+2}} = \frac{1}{3^3} = \frac{1}{27}\) Therefore, the simplified expression is 1/27. The answer is option E.
Ajụjụ 21 Ripọtì
Which of the following gives the point of intersection of the graph y = x2 and y = x + 6 shown above?
Akọwa Nkọwa
The problem is asking us to find the point of intersection of the two graphs y = x2 and y = x + 6. This can be done by solving the equations simultaneously. We need to find the values of x and y that satisfy both equations. We can do this by substituting y = x + 6 for y in the equation y = x2, giving us: x + 6 = x2 Rearranging this equation gives us: x2 - x - 6 = 0 We can factor this quadratic equation to obtain: (x - 3)(x + 2) = 0 Thus, the solutions are x = 3 or x = -2. To find the corresponding values of y, we can substitute these values of x into either of the original equations. For example, if we use y = x + 6, we get: When x = 3, y = 3 + 6 = 9, giving us the point (3, 9). When x = -2, y = -2 + 6 = 4, giving us the point (-2, 4). Therefore, the point of intersection of the two graphs is (3, 9) and (-2, 4).
Ajụjụ 22 Ripọtì
Points X and Y are respectively 20km North and 9km East of a point O. What is the bearing of Y from X? Correct to the nearest degree
Akọwa Nkọwa
Ajụjụ 23 Ripọtì
What percentage of the population are married? correct to one decimal place
Akọwa Nkọwa
Ajụjụ 25 Ripọtì
In the diagram O is the center of the circle, if ?QRS = 62o, find the value of ?SQR.
Akọwa Nkọwa
Ajụjụ 26 Ripọtì
Evaluate \(\frac{0.009}{0.012}\), leaving your answer in standard form.
Akọwa Nkọwa
To evaluate \(\frac{0.009}{0.012}\) in standard form, we need to simplify the fraction and express the result in the form of \(a \times 10^b\), where \(1 \leq a < 10\) and \(b\) is an integer. We can simplify the fraction by dividing both the numerator and the denominator by 0.003: \[\frac{0.009}{0.012} = \frac{0.009 \div 0.003}{0.012 \div 0.003} = \frac{3}{4}\] Now we can express 3/4 in standard form: \[\frac{3}{4} = 0.75 = 7.5 \times 10^{-1}\] Therefore, the answer is option (C) 7.5 x 10-1.
Ajụjụ 27 Ripọtì
In the diagram, O is the center of the circle, If ?POQ = 80o and ?PRQ = 5x, find the value of x.
Akọwa Nkọwa
In a circle, the angle subtended by an arc at the center is twice the angle subtended by the same arc at any point on the circumference of the circle. Using this property, we can find the value of ?PRQ: ?POQ = 80o (given) Therefore, ?PRQ = 1/2 ?POQ = 1/2 x 80o = 40o Now, we can use the value of ?PRQ to find the value of x: ?PRQ = 5x (given) Substituting the value of ?PRQ, we get: 40o = 5x Simplifying this equation, we get: x = 8 Therefore, the value of x is 8. The answer is option B.
Ajụjụ 28 Ripọtì
If twice a certain integer is subtracted from 5 times the integer, the result is 63. Find the integer.
Akọwa Nkọwa
Ajụjụ 29 Ripọtì
For a class of 30 students, the scores in a Mathematics test out of 10 marks were as follows;
4,5,7,2,3,6,5,5,8,9,5,4,2,3,7,9,8,7,7,7,3,4,5,5,2,3,6,7,7,2
What is the range of the distribution?
Akọwa Nkọwa
Range of a distribution refers to the difference between the highest and lowest value in a dataset. To find the range of the given Mathematics test scores, we need to determine the highest and lowest values. The lowest score in the test is 2, while the highest score is 9. Therefore, the range of the distribution is 9 - 2 = 7. Hence, the answer is 7.
Ajụjụ 30 Ripọtì
Instead of recording the number 1.23cm of the radius of a tube, a student recorded 1.32cm. Find the percentage of error, correct to one decimal place.
Akọwa Nkọwa
To find the percentage of error, we first need to find the absolute error, which is the difference between the measured value and the true value. In this case, the true value is 1.23 cm, and the measured value is 1.32 cm. Absolute error = |measured value - true value| = |1.32 - 1.23| = 0.09 cm Next, we can find the relative error, which is the ratio of the absolute error to the true value. Relative error = (absolute error / true value) x 100% Relative error = (0.09 / 1.23) x 100% = 7.32% Therefore, the percentage of error, correct to one decimal place, is 7.3%. The answer is option B.
Ajụjụ 31 Ripọtì
Evaluate, using logarithm tables \(\frac{5.34 \times 67.4}{2.7}\)
Akọwa Nkọwa
To evaluate this expression using logarithm tables, we can use the following steps: 1. Take the logarithm (base 10) of each of the numbers in the expression. 2. Add the logarithms of the two factors in the numerator. 3. Subtract the logarithm of the denominator from the result in step 2. 4. Take the antilogarithm (base 10) of the result in step 3. Using these steps, we get: log(5.34) = 0.727 log(67.4) = 1.829 log(2.7) = 0.431 log(5.34 x 67.4) = log(5.34) + log(67.4) = 0.727 + 1.829 = 2.556 log(5.34 x 67.4) - log(2.7) = 2.556 - 0.431 = 2.125 antilog(2.125) = 133.2 Therefore, the value of the expression is 133.2. The answer is option C.
Ajụjụ 33 Ripọtì
In the diagram above, O is the center of the circle QRT and PT is a tangent to the circle at T. Calculate the angle x
Akọwa Nkọwa
Ajụjụ 35 Ripọtì
In the diagram above, O is the center of the circle. If ?POR = 114o, calculate ?PQR
Akọwa Nkọwa
Ajụjụ 36 Ripọtì
If 3\(^y\) = 243, find the value of y.
Akọwa Nkọwa
Since 3 raised to what power gives 243, we need to find the exponent that makes this equation true: 3\(^y\) = 243 We can rewrite 243 as 3\(^5\), so the equation becomes: 3\(^y\) = 3\(^5\) Since the bases are equal, the exponents must also be equal. Therefore: y = 5 So the value of y is 5.
Ajụjụ 37 Ripọtì
Calculate and correct to three significant figures, the length of an arc subtends an angle of 70o at the center of the circle radius 4cm. [Take π = 22/7]
Akọwa Nkọwa
Ajụjụ 38 Ripọtì
If A = {a,b,c}, B = {a,b,c,d,e} and C = {a,b,c,d,e,f}. Find (A∪B)∩(A∪C)
Akọwa Nkọwa
The union of sets A and B (A ∪ B) is the set of elements that are in either set A, set B or both. Similarly, the union of sets A and C (A ∪ C) is the set of elements that are in either set A, set C or both. The intersection of these two sets ((A ∪ B) ∩ (A ∪ C)) is the set of elements that are common to both sets. Therefore, we can first find the union of A and B which gives us {a, b, c, d, e}. Then, we can find the union of A and C which gives us {a, b, c, d, e, f}. The intersection of these two sets is the set of elements that are in both sets, which is {a, b, c, d, e}. Therefore, (A∪B)∩(A∪C) = {a, b, c, d, e}. So the correct answer is: {a,b,c,d,e}.
Ajụjụ 39 Ripọtì
What is the probability that 3 customers waiting in a bank will be served in the sequence of their arrival at the bank
Akọwa Nkọwa
The probability that 3 customers waiting in a bank will be served in the sequence of their arrival at the bank can be calculated using the permutation formula. Since there are three customers waiting in line, the total number of possible arrangements is 3! = 6. However, only one arrangement results in the customers being served in the sequence of their arrival at the bank. Therefore, the probability of this happening is 1/6. Hence, the correct answer is 1/6.
Ajụjụ 40 Ripọtì
Calculate the surface area of a hallow cylinder which is closed at one end, if the base radius is 3.5cm and the height 8cm.[take π = 22/7]
Akọwa Nkọwa
Ajụjụ 41 Ripọtì
Using the above graph, if 10o < x < 60o, what is the value of x for which sin x = cos x?
Akọwa Nkọwa
Since sin x = cos x, we have: sin x = cos x Dividing both sides by cos x, we get: tan x = 1 Taking the inverse tangent (tan-1) of both sides, we get: x = 45o Therefore, the value of x for which sin x = cos x is 45o. Therefore, the correct option is (d) 45o.
Ajụjụ 43 Ripọtì
Solve for x: (x2 + 2x + 1) = 25
Akọwa Nkọwa
We are given the equation: x2 + 2x + 1 = 25 To solve for x, we first simplify the left-hand side of the equation by subtracting 24 from both sides: x2 + 2x - 24 = 0 We can then factor this quadratic equation as follows: (x + 6)(x - 4) = 0 Using the zero product property, we know that this equation is true if either (x + 6) = 0 or (x - 4) = 0. Solving for x in each case, we get: x + 6 = 0 or x - 4 = 0 x = -6 or x = 4 Therefore, the solutions to the original equation are x = -6 and x = 4. So the answer is: -6, 4
Ajụjụ 44 Ripọtì
The venn diagram below shows the number of students who studied Physics, Chemistry, and Mathematics in a certain school. How many students took at least two of the three subjects?
Akọwa Nkọwa
Ajụjụ 45 Ripọtì
For a class of 30 students, the scores in a Mathematics test out of 10 marks were as follows;
4,5,7,2,3,6,5,5,8,9,5,4,2,3,7,9,8,7,7,7,3,4,5,5,2,3,6,7,7,2
What is the median score?
Akọwa Nkọwa
Ajụjụ 46 Ripọtì
Factorize the following expression; 2x2 + x - 15
Ajụjụ 48 Ripọtì
The table above gives the scores of a group of students in an English Language test
If a student is chosen at random from the group, what is the probability that the scored at least 6 marks?
Ajụjụ 49 Ripọtì
Using a ruler and a pair of compasses only, construct a triangle ABC, given that |AB| = 8.4cm, |BC| = 6.5cm and < ABC = 30°. Construct the locus:
(a) \(l_{1}\) of points equidistant from AB and BC, and within the angle ABC;
(b) \(l_{2}\) of points equidistant from B and C. Locate the point of intersection P of \(l_{1}\) and \(l_{2}\). Measure |AP|.
Ajụjụ 50 Ripọtì
(a) 
In the diagram, O is the centre of the circle radius 3.2cm. If < PRQ = 42°, calculate, correct to two decimal places, the area of the:
(i) minor sector POQ ; (ii) shaded part.
(b) If the sector POQ in (a) is used to form the curved surface of a cone with vertex O, calculate the base radius of the cone, correct to one decimal place.
Ajụjụ 51 Ripọtì
(a) (i) 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.
(ii) In the diagram above, O is the centre of the circle and PT is a diameter. If < PTQ = 22° and < TOR = 98°, calculate < QRS.
(b) ABCD is a cyclic quadrilateral and the diagonals AC and BD intersect at H. If < DAC = 41° and < AHB = 70°, calculate < ABC.
None
Akọwa Nkọwa
None
Ajụjụ 52 Ripọtì
(a) In an A.P, the difference between the 8th and 4th terms is 20 and the 8th term is \(1\frac{1}{2}\) times the 4th term. What is the:
(i) common difference ; (ii) first term of the sequence?
(b) The value of a machine depreciates each year by 5% of its value at the beginning of that year. If its value when new on 1st January 1980 was N10,250.00, what was its value in January 1989 when it was 9 years old? Give your answer correct to three significant figures.
Akọwa Nkọwa
None
Ajụjụ 53 Ripọtì
(a) Triangle PQR is right-angled at Q. PQ = 3a cm and QR = 4a cm. Determine PR in terms of a.
(b) Ayo travels a distance of 24km from X on a bearing of 060° to Y. He then travels a distance of 18km to a point Z and Z is 30km from X.
(i) Draw the diagram to show the positions of X, Y and Z ; (ii) What is the bearing of Z from Y ; (iii) Calculate the bearing of X from Z.
Akọwa Nkọwa
None
Ajụjụ 54 Ripọtì
(a) If \(\cos \alpha = 0.6717\), use mathematical tables to find (i) \(\alpha\) ; (ii) \(\sin \alpha\)
(b) The angle of depression of a point P on the ground, from the top T of a building is 23.6°. If the distance of P from the foot of the building is 50m, calculate the height of the building, correct to the nearest metre.
Akọwa Nkọwa
None
Ajụjụ 55 Ripọtì
The diagram shows a wooden structure in the form of a cone, mounted on a hemispherical base. The vertical height of the cone is 24cm and the base height 7cm. Calculate, correct to three significant figures, the surface area of the structure. [Take \(\pi = \frac{22}{7}\)].
Ajụjụ 56 Ripọtì
When a stone is thrown vertically upwards, its distance d metres after t seconds is given by the formula \(d = 60t - 10t^{2}\). Draw the graph of \(d = 60t - 10t^{2}\) for values of t from 1 to 5 seconds using 2cm to 1 unit on the t- axis and 2cm to 20 units on the d- axis.
(a) Using your graph, (i) how long does it take to reach a height of 70 metres? (ii) determine the height of the stone after 5 seconds. (iii) after how many seconds does it reach its maximum height.
(b) Determine the slope of the graph when t = 4 seconds.
Ajụjụ 57 Ripọtì
(a) Without using Mathematical tables, find x, given that \(6 \log (x + 4) = \log 64\)
(b) If \(U = {1, 2, 3,4, 5, 6, 7, 8, 9, 10}, X = {1, 2, 4, 6, 7, 8, 9}, Y = {1, 2, 3, 4, 7, 9}\) and \(Z = {2, 3, 4, 7, 9}\). What is \(X \cap Y \cap Z' \)?
Akọwa Nkọwa
None
Ajụjụ 58 Ripọtì
The table shows the weights, to the nearest kilogram, of twelve students in a Further Mathematics class
| Weight in kg | 55 | 57 | 59 | 61 | 63 |
| No of students | 2 | 1 | 2 | 4 | 3 |
(a) Draw a bar chart to illustrate the above information;
(b) What is (i) the mode; (ii) the median of the distribution?
(c) Calculate the mean weight correct to the nearest kilogram.
None
Akọwa Nkọwa
None
Ajụjụ 59 Ripọtì
(a) A pair of fair dice each numbered 1 to 6 is tossed. Find the probability of getting a sum of at least 9.
(b) If the probability that a civil servant owns a car is \(\frac{1}{6}\), find the probability that:
(i) two civil servants, A and B, selected at random each owns a car ; (ii) of two civil servants, C and D selected at random, only one owns a car ; (iii) of three civil servants, X, Y and Z, selected at random, only one owns a car.
Ajụjụ 60 Ripọtì
(a) Derive the smallest equation whose coefficients are integers and which has roots of \(\frac{1}{2}\) and -7.
(b) Three years ago, a father was four times as old as his daughter is now. The product of their present ages is 430. Calculate the ages of the father and daughter.
Akọwa Nkọwa
None
