WAEC SSCE - General Mathematics - 2009
Ajụjụ 1 Ripọtì
If M and N are two fixed points in a plane. Find the locus L = [P : PM = PN]
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Ajụjụ 2 Ripọtì
If a positive integer, list the values of x which satisfy the equation 3x - 4 < 6 and x - 1 > 0
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Ajụjụ 3 Ripọtì
In the diagram, \SQ\ = 4cm, \PT\ = 7cm. /TR/ = 5cm and ST//OR. If /SP/ = xcm, find the value of x
Ajụjụ 4 Ripọtì
If log 2 = x, log 3 = y and log 7 = z, find, in terms of x, y and z, the value of log (\(\frac{28}{3}\))
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We can use logarithmic rules to solve this question. Let's first express log(\(\frac{28}{3}\)) in terms of x, y, and z. log(\(\frac{28}{3}\)) = log(28) - log(3) We know that: log(28) = log(4 × 7) = log(4) + log(7) = 2x + z log(3) = y Therefore: log(\(\frac{28}{3}\)) = 2x + z - y So the correct answer is 2x + z - y.
Ajụjụ 6 Ripọtì
If /XY/ = 50m, how far cast of X is Y?
Ajụjụ 7 Ripọtì
In an examination, Kofi scored x% in Physics, 50% in Chemistry and 70% in Biology. If his mean score for the three subjects was 55%, find x
Akọwa Nkọwa
Kofi's mean score for the three subjects was 55%, so the total percentage score for the three subjects is 3 x 55 = 165%. Let's assume that Kofi scored x% in Physics. Then, the total percentage score for Physics, Chemistry and Biology would be: x + 50 + 70 = 165 Simplifying the equation, we get: x = 165 - 50 - 70 x = 45 Therefore, Kofi scored 45% in Physics. So the correct option is (B) 45.
Ajụjụ 8 Ripọtì
Factorize x2 - 2x - 3xy completely
Ajụjụ 9 Ripọtì
The diameter of a bicycle wheel is 42cm. If the wheel makes 16 complete revolution, what will be the total distance covered by the wheel? [Take \(\pi \frac{22}{7}\)
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The distance covered by the wheel is equal to the circumference of the wheel multiplied by the number of revolutions made. The circumference of the wheel can be calculated using the formula: circumference = diameter x pi circumference = 42cm x 22/7 circumference = 132cm Therefore, the distance covered by the wheel in 16 complete revolutions is: distance = circumference x number of revolutions distance = 132cm x 16 distance = 2112cm So, the answer is (c) 2112cm.
Ajụjụ 10 Ripọtì
Simplify: (3\(\frac{1}{2} + 4\frac{1}{3}) \div (\frac{5}{6} - \frac{2}{3}\))
Akọwa Nkọwa
We will start by simplifying the expression inside the parenthesis first. 3\(\frac{1}{2}\) + 4\(\frac{1}{3}\) = (7/2) + (13/3) To add these two fractions, we need a common denominator. Multiplying the denominators together gives us 6, so: (7/2) + (13/3) = (21/6) + (26/6) = 47/6 Now, let's simplify the expression in the denominator: \(\frac{5}{6} - \frac{2}{3}\) = \(\frac{5}{6} - \frac{4}{6}\) = \(\frac{1}{6}\) Finally, we can substitute these values into the original expression: (3\(\frac{1}{2}\) + 4\(\frac{1}{3}\)) ÷ (\(\frac{5}{6}\) - \(\frac{2}{3}\)) = (47/6) ÷ (1/6) When dividing fractions, we can multiply the first fraction by the reciprocal of the second: (47/6) ÷ (1/6) = (47/6) x (6/1) = 47 Therefore, the answer is 47.
Ajụjụ 11 Ripọtì
Each of the interior angles of a regular polygon is 140o. Calculate the sum of all the interior angles of the polygon
Akọwa Nkọwa
In a regular polygon with n sides, each interior angle measures: \begin{align*} 180^\circ - \frac{360^\circ}{n} \end{align*} Since each interior angle of this polygon is 140o, we can equate the above formula to 140o: \begin{align*} 180^\circ - \frac{360^\circ}{n} &= 140^\circ\\ \frac{360^\circ}{n} &= 40^\circ\\ n &= \frac{360^\circ}{40^\circ}\\ n &= 9 \end{align*} Therefore, the given polygon is a nonagon (a polygon with nine sides). The sum of the interior angles of a polygon with n sides can be calculated using the formula: \begin{align*} S &= (n-2)180^\circ \end{align*} Substituting n=9 into this formula, we get: \begin{align*} S &= (9-2)180^\circ\\ &= 7\cdot180^\circ\\ &= 1260^\circ \end{align*} Therefore, the sum of all the interior angles of the given polygon is 1260o. Hence, the correct option is: - 1260o
Ajụjụ 12 Ripọtì
If 85% of x is N3230, what is the value of x?
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We can solve this problem using a proportion. If 85% of x is N3230, that means we can write: 0.85x = N3230 To solve for x, we need to isolate x on one side of the equation. We can do this by dividing both sides by 0.85: x = N3230 ÷ 0.85 Using a calculator, we get: x ≈ N3800.00 Therefore, the value of x is N3800.00.
Ajụjụ 13 Ripọtì
in the diagram, angle 20o is subtended at the centre of the circle, find the value of x
Akọwa Nkọwa
In the given diagram, we have a circle with center O and angle 20o subtended at the center. We need to find the value of x. Firstly, we know that the angle subtended at the center of a circle is twice the angle subtended at the circumference by the same arc. Therefore, angle AOB = 2 × 20o = 40o Also, we know that angle in a semicircle is a right angle. So, angle AOC = 90o. Using the fact that the angles in a triangle add up to 180o, we can find angle BOC as follows: angle BOC = 180o - angle AOB - angle AOC = 180o - 40o - 90o = 50o Since angle BOC is an angle at the circumference that subtends the arc BC, which is equal to x degrees, we have: x = angle BOC = 50o Therefore, the value of x is 50o. Answer is correct.
Ajụjụ 15 Ripọtì
The capacity of a water tank is 1,800 litres. If the tank is in form of a cuboid with base 600cm by 150 cm. Find the height of the tank
Ajụjụ 16 Ripọtì
The following is the graph of a quadratic friction, find the value of x when y = 0
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Ajụjụ 17 Ripọtì
In the diagram, /TP/ = 12cm and it is 6cm from O, the centre of the circle, Calculate < TOP
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Ajụjụ 18 Ripọtì
Find the value to which N3000.00 will amount in 5 years at 6% per annum simple interest
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Simple interest is calculated as the product of the principal, the rate of interest, and the time duration. From the question, we have a principal of N3000.00, an interest rate of 6%, and a duration of 5 years. Using the formula for simple interest, we can find the interest accrued over the 5 years as: Interest = (P * R * T) / 100 = (3000 * 6 * 5) / 100 = N900.00 The total value to which N3000.00 will amount to after 5 years is the sum of the principal and the interest, which is: Total = Principal + Interest = 3000 + 900 = N3900.00 Therefore, N3000.00 will amount to N3900.00 after 5 years at 6% per annum simple interest. So the correct answer is option A, N3,900.00.
Ajụjụ 19 Ripọtì
From the diagram which of the following statements are correct? i. XQ is a radius of a circle centre Q. ii. /XQ/ = /QY/. iii. /QX/ = /XY/
Ajụjụ 20 Ripọtì
In a particular year, the exchange rate of naira (N) varies directly with the Dollars{$}. If N1122 is equivalent to $8, find the Naira equivalent of $36
Akọwa Nkọwa
The exchange rate of naira (N) varies directly with the Dollars($). This means that the exchange rate is constant, and we can set up a proportion to solve the problem. If N1122 is equivalent to $8, we can write: N1122/$8 = x/$36 where x is the Naira equivalent of $36. To solve for x, we can cross-multiply and simplify: N1122 x 36 = $8 x x 40392 = 8x x = 40392/8 = 5049 Therefore, the Naira equivalent of $36 is N5049. So, the correct option is (B) N5049.
Ajụjụ 21 Ripọtì
A bucket holds 10 litres of water. How many buckets of water will fill a reservoir of size 8m x 7m x 5m.(1 litre = 1000cm3)`
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The volume of the reservoir can be found by multiplying its dimensions: 8m x 7m x 5m = 280 m^3 Since 1 liter is equal to 1000 cubic centimeters (cm^3), 1 cubic meter is equal to 1,000,000 cubic centimeters. Therefore, the reservoir has a volume of: 280 m^3 x 1,000,000 cm^3/m^3 = 280,000,000 cm^3 Each bucket can hold 10 liters of water or 10,000 cubic centimeters (cm^3) of water since 1 liter is equal to 1000 cm^3. Therefore, the number of buckets needed to fill the reservoir is: 280,000,000 cm^3 ÷ 10,000 cm^3/bucket = 28,000 buckets Therefore, the answer is 28,000.
Ajụjụ 22 Ripọtì
The angles of triangle are (x + 10)o, (2x - 40)o and (3x - 90)o. Which of the following accurately describes the triangle?
Ajụjụ 23 Ripọtì
solve \(\frac{2x + 1}{6} - \frac{3x - 1}{4}\) = 0
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To solve the equation \(\frac{2x + 1}{6} - \frac{3x - 1}{4} = 0\), we need to simplify the left-hand side and solve for x. First, we need to find a common denominator for the two fractions. The smallest common multiple of 6 and 4 is 12, so we can rewrite the equation as: \[\frac{2x+1}{6}\cdot \frac{2}{2} - \frac{3x-1}{4}\cdot \frac{3}{3} = 0\] Simplifying, we get: \[\frac{4x+2}{12} - \frac{9x-3}{12} = 0\] Combining the fractions, we get: \[\frac{4x+2-(9x-3)}{12} = 0\] Simplifying further, we get: \[\frac{-5x+5}{12} = 0\] Multiplying both sides by 12, we get: \[-5x+5=0\] Adding 5 to both sides, we get: \[-5x=-5\] Dividing both sides by -5, we get: \[x=1\] Therefore, the solution to the equation is x = 1.
Ajụjụ 25 Ripọtì
Given that sin 60o = \(\frac{\sqrt{3}}{2}\) and cos 60o = \(\frac{1}{2}\), evaluate \(\frac{1 - sin 60^o}{1 + cos 60^o}\)
Ajụjụ 27 Ripọtì
Express 2.7864 x 10-3 to 2 significant figures
Akọwa Nkọwa
To express 2.7864 x 10-3 to 2 significant figures, we need to round off the number to the second significant digit from the left. Since the first significant digit is 2, and the second significant digit is 7, we will look at the third digit, which is 8. Since 8 is greater than or equal to 5, we round up the second significant digit (7) by 1. Therefore, the number becomes 0.0028. So, the answer is 0.0028.
Ajụjụ 28 Ripọtì
Using the diagram, find the bearing of X from Y
Ajụjụ 29 Ripọtì
The interior angles of a pentagon are (2x + 5)o, (x + 20)o, xo, (3x - 20)o and (x + 15)o. Find the value of x
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Ajụjụ 30 Ripọtì
PQRS is a trapezuim. QR//PS, /PQ/ = 5cm, /OR/ = 6cm, /PS/ = 10cm and angle QPS = 42o. Calculate the perpendicular distance between the parallel sides
Ajụjụ 32 Ripọtì
Express the square root of 0.000144 in the standard form
Ajụjụ 33 Ripọtì
If (2x + 3)3 = 125, find the value of x
Akọwa Nkọwa
We are given that (2x + 3)3 = 125. We can rewrite 125 as 53. Therefore, we have: (2x + 3)3 = 53 Taking the cube root of both sides, we get: 2x + 3 = 5 Solving for x, we get: 2x = 2 x = 1 Therefore, the value of x is 1. Answer: (a) 1.
Ajụjụ 34 Ripọtì
Find the quadratic equation whose roots are -\(\frac{1}{2}\) and 3
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To find the quadratic equation given its roots, we use the fact that for a quadratic equation of the form ax2 + bx + c = 0, the roots are given by the formula: x = (-b ± √(b2 - 4ac)) / 2a If the roots are given as α and β, then the quadratic equation can be written as: (x - α)(x - β) = 0 Expanding the above equation gives: x2 - (α + β)x + αβ = 0 Therefore, to find the quadratic equation whose roots are -\(\frac{1}{2}\) and 3, we substitute α = -\(\frac{1}{2}\) and β = 3 into the equation: x2 - (α + β)x + αβ = 0 x2 - (-\(\frac{1}{2}\) + 3)x + (-\(\frac{1}{2}\) × 3) = 0 Simplifying the above equation, we get: 2x2 - 5x - 3 = 0 Therefore, the quadratic equation whose roots are -\(\frac{1}{2}\) and 3 is 2x2 - 5x - 3 = 0. The correct option is (C) 2x2 - 5x - 3 = 0.
Ajụjụ 35 Ripọtì
A box contains black, white and red identical balls. The probability of picking a black ball at random from the box is \(\frac{3}{10}\) and the probability of picking a white ball at random is \(\frac{2}{5}\). If there are 30 balls in the box, how many of them are red?
Akọwa Nkọwa
The probability of picking a black ball at random from the box is \(\frac{3}{10}\) and the probability of picking a white ball at random is \(\frac{2}{5}\). Let's assume that there are x red balls in the box. Since there are a total of 30 balls, we can write: number of black balls + number of white balls + number of red balls = total number of balls \(\frac{3}{10}(30) + \frac{2}{5}(30) + x = 30\) Simplifying the above equation gives: 9 + 12 + x = 30 x = 30 - 9 - 12 x = 9 Therefore, there are 9 red balls in the box.
Ajụjụ 37 Ripọtì
The following is the graph of a quadratic friction, find the co-ordinates of point P
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Ajụjụ 38 Ripọtì
Two sets are disjoint if
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Two sets are said to be disjoint if their intersection is an empty set. In other words, if there are no common elements between the sets, they are said to be disjoint. For example, the sets {1,2,3} and {4,5,6} are disjoint since their intersection is an empty set {} or ∅. On the other hand, the sets {1,2,3} and {2,3,4} are not disjoint since they have elements in common, namely 2 and 3.
Ajụjụ 39 Ripọtì
An arc of a circle subtends an angle of 60o at the centre. If the radius of the circle is 3cm, find , in terms of \(\pi\), the length of the arc
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When an angle in degrees is subtended at the center of a circle, the length of the arc it cuts out is given by: $$\text{Length of arc} = \frac{\text{angle}}{360^\circ} \times 2\pi r$$ where r is the radius of the circle. In this case, the angle is 60° and the radius is 3 cm. Substituting these values into the formula above, we get: $$\text{Length of arc} = \frac{60}{360^\circ} \times 2\pi (3\text{ cm}) = \frac{1}{6} \times 6\pi = \pi \text{ cm}$$ Therefore, the length of the arc, in terms of π, is π cm. Hence, the correct option is: \(\pi\)cm.
Ajụjụ 40 Ripọtì
What is the median of the following scores: 22 35 41 63 74 82
Akọwa Nkọwa
To find the median of a set of numbers, we first need to arrange them in ascending or descending order. In this case, arranging the numbers in ascending order gives us: 22, 35, 41, 63, 74, 82 The median is the middle number in the set when the numbers are arranged in order. Since we have an even number of numbers in this set, there are two middle numbers: 41 and 63. To find the median, we take the average of these two numbers: Median = (41 + 63) / 2 = 104 / 2 = 52 Therefore, the median of the scores is 52.
Ajụjụ 41 Ripọtì
The pie chart shows the distribution of 4320 students who graduated from four departments in a university. If a student is picked at random from the four departments, what id the probability that he is not from the education department?
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To find the probability that a student picked at random is not from the education department, we need to find the total number of students who are not from the education department and divide it by the total number of students in all departments. From the pie chart, we can see that the education department has 30% of the total students. Therefore, the remaining three departments have a total of 70% of the total students. To find the probability of picking a student who is not from the education department, we divide the percentage of students who are not in the education department by 100%: Probability = \(\frac{70}{100}\) = \(\frac{7}{10}\) Therefore, the probability that a student picked at random is not from the education department is \(\frac{7}{10}\). The correct option is: \(\frac{7}{10}\).
Ajụjụ 42 Ripọtì
Find the value of x in the diagram
Ajụjụ 45 Ripọtì
Arrange the following numbers in descending orders of magnitude: 22three, 34five, 21six
Akọwa Nkọwa
To compare these numbers, we need to convert all of them to the same base. Let's convert them all to base 10 for simplicity: - 22three = 2\*31 + 2\*30 = 6 + 2 = 8 - 34five = 3\*51 + 4\*50 = 15 + 4 = 19 - 21six = 2\*61 + 1\*60 = 12 + 1 = 13 So, in base 10, the numbers are 8, 19, and 13. To arrange them in descending order of magnitude, we simply sort them from largest to smallest: - 19, 13, 8 Therefore, the correct answer is: 34five, 21six, 22three
Ajụjụ 46 Ripọtì
The pie chart shows the distribution of 4320 students who graduated from four departments in a university. How many students graduated from the science department?
Akọwa Nkọwa
To find the number of students who graduated from the science department, we need to look at the pie chart and determine the percentage of students that belong to that department. From the pie chart, we can see that the science department occupies 20% of the entire circle. To find the number of students in the science department, we need to calculate 20% of 4320, which can be done by multiplying 4320 by 0.20. 20% of 4320 = 0.20 x 4320 = 864 Therefore, the number of students who graduated from the science department is 864. So, the correct answer is option B: 864.
Ajụjụ 47 Ripọtì
The table shows the number of children per family in a community.
| No of children | 0 | 1 | 2 | 3 | 4 | 5 |
| No of families | 3 | 5 | 7 | 4 | 3 | 2 |
(a) Find the : (i) mode ; (ii) third quartile ; (iii) probability that a family has at least 2 children.
(b) If a pie chart were to be drawn for the data, what would be the sectorial angle representing families with one child?
None
Akọwa Nkọwa
None
Ajụjụ 48 Ripọtì
(a) Given that \((\sqrt{3} - 5\sqrt{2})(\sqrt{3} + \sqrt{2}) = a + b\sqrt{6}\), find a and b.
(b) If \(\frac{2^{1 - y} \times 2^{y - 1}}{2^{y + 2}} = 8^{2 - 3y}\), find y.
Akọwa Nkọwa
None
Ajụjụ 49 Ripọtì
(a) Using ruler and a pair of compasses only, construct : (i) quadrilateral PQRS such that /PQ/ = 10 cm, /QR/ = 8 cm, /PS/ = 6 cm, < PQR = 60° and < QPS = 75° ;
(ii) the locus \(l_{1}\) of points equidistant from QR and RS ; (iii) locus \(l_{2}\) of points equidistant from R and S ;
(b) Measure /RS/.
Ajụjụ 50 Ripọtì
(a) Copy and complete the table of values for \(y = \sin x + 2 \cos x\), correct to one decimal place.
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° | 210° | 240° |
| y | 2.2 | -1.2 | -2.0 | -1.9 |
(b) Using a scale of 2 cm to 30° on the x- axis and 2 cm to 0.5 units on the y- axis, draw the graph of \(y = \sin x + 2\cos x\) for \(0° \leq x \leq 240°\).
(c) Use your graph to solve the equation : (i) \(\sin x + 2 \cos x = 0\) ; (ii) \(\sin x = 2.1 - 2\cos x\).
(d) From the graph, find y when x = 171°.
Ajụjụ 52 Ripọtì
The marks scored by 50 students in a Geography examination are as follows :
60 54 40 67 53 73 37 55 62 43 44 69 39 32 45 58 48 67 39 51 46 59 40 52 61 48 23 60 59 47 65 58 74 47 40 59 68 51 50 50 71 51 26 36 38 70 46 40 51 42.
(a) Using class intervals 21 - 30, 31 - 40, ..., prepare a frequency distribution table.
(b) Calculate the mean mark of the distribution.
(c) What percentage of the students scored more than 60%?
Akọwa Nkọwa
None
Ajụjụ 53 Ripọtì
(a) How many numbers between 75 and 500 are divisible by 7?
(b) The 8th term of an Arithmetic Progression (A.P) is 5 times the 3rd term while the 7th term is 9 greater than the 4th term. Write the first 5 terms of the A.P.
Ajụjụ 54 Ripọtì
(a) A woman looking out from the window of a building at a height of 30m, observed that the angle of depression of the top of a flag pole was 44°. If the foot of the pole is 25m from the foot of the building and on the same horizontal ground, find, correct to the nearest whole number, the (i) angle of depression of the foot of the pole from the woman ; (ii) height of the flag pole.
(b)
In the diagram, O is the centre of the circle, < OQR = 32° and < TPQ = 15°. Calculate, (i) < QPR ; (ii) < TQo.
None
Akọwa Nkọwa
None
Ajụjụ 55 Ripọtì
(a) A circle is inscribed in a square. If the sum of the perimeter of the square and the circumference of the circle is 100 cm, calculate the radius of the circle. [Take \(\pi = \frac{22}{7}\)].
(b) A rope 60cm long is made to form a rectangle. If the length is 4 times its breadth, calculate, correct to one decimal place, the :
(i) length ; (ii) diagonal of the rectangle.
None
Akọwa Nkọwa
None
Ajụjụ 56 Ripọtì
(a) Simplify \(\frac{x + 2}{x - 2} - \frac{x + 3}{x - 1}\)
(b) The graph of the equation \(y = Ax^{2} + Bx + C\) passes through the point (0, 0), (1, 4) and (2, 10). Find the :
(i) value of C ; (ii) values of A and B ; (iii) co-ordinates of the other point where the graph cuts the x- axis.
Akọwa Nkọwa
None
Ajụjụ 57 Ripọtì
(a) Out of 30 candidates applying for a post, 17 have degrees, 15 have diplomas and 4 neither degree nor diploma. How many of them have both?
(b) In triangle PQR, M and N are points on the side PQ and PR respectively such that MN is parallel to QR. If < PRQ = 75°, PN = QN and < PNQ = 125°, determine :
(i) < NQR ; (ii) < NPM.
Ajụjụ 58 Ripọtì
(a) If \(9 \cos x - 7 = 1\) and \(0° \leq x \leq 90°\), find x.
(b) Given that x is an integer, find the three greatest values of x which satisfy the inequality \(7x < 2x - 13\).
Akọwa Nkọwa
None
Ajụjụ 59 Ripọtì
(a) If \(\log 5 = 0.6990, \log 7 = 0.8451\) and \(\log 8 = 0.9031\), evaluate \(\log (\frac{35 \times 49}{40 \div 56})\).
(b) For a musical show, x children were present. There were 60 more adults than children. An adult paid D5 and a child D2. If a total of D1280 was collected, calculate the
(i) value of x ; (ii) ratio of the number of children to the number of adults ; (iii) average amount paid per person ; (iv) percentage gain if the organisers spent D720 on the show.
Akọwa Nkọwa
None
