WAEC SSCE - General Mathematics - 2012 (Objective)
Vraag 1 Verslag
If x and y are variables and k is a constant, which of the following describes an inverse relationship between x and y?
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An inverse relationship between two variables means that as one variable increases, the other variable decreases. Mathematically, an inverse relationship can be represented by an equation where one variable is multiplied or divided by a constant. Out of the given equations, the equation that represents an inverse relationship between x and y is y = \(\frac{k}{x}\). To see why this is the case, let's consider what happens to y as x increases. Suppose x doubles in value. Then, according to the equation y = \(\frac{k}{x}\), y will be halved in value. Similarly, if x triples in value, y will be divided by 3. This means that as x increases, y decreases, which is the characteristic of an inverse relationship. On the other hand, in the equation y = kx, as x increases, y also increases. This is not an inverse relationship but a direct relationship. The same applies to y = k\(\sqrt{x}\) and y = x + k. Therefore, the equation that describes an inverse relationship between x and y is y = \(\frac{k}{x}\).
Vraag 2 Verslag
The diagram is a circle with centre P. PRST are points on the circle. Find the value of < PRS
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Vraag 3 Verslag
Simplify; \(\frac{3\sqrt{5} \times 4\sqrt{6}}{2 \sqrt{2} \times 3\sqrt{2}}\)
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Vraag 4 Verslag
Simplify: \(\frac{54k^2 - 6}{3k + 1}\)
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To simplify the expression \(\frac{54k^2 - 6}{3k + 1}\), we can start by factoring out the numerator: \(\frac{54k^2 - 6}{3k + 1} = \frac{6(9k^2 - 1)}{3k + 1}\) We can further simplify the numerator by recognizing that \(9k^2 - 1\) is a difference of squares, and can be factored as \((3k + 1)(3k - 1)\): \(\frac{6(9k^2 - 1)}{3k + 1} = \frac{6(3k + 1)(3k - 1)}{3k + 1}\) We can then cancel out the common factor of \(3k + 1\) in the numerator and denominator, which gives: \(\frac{6(3k - 1)}{1} = 6(3k - 1)\) Therefore, the simplified expression is \(6(3k - 1)\). The correct option is: - 6(3k - 1)
Vraag 5 Verslag
In the diagram, |SR| = |QR|. < SRP = 65o and < RPQ = 48o, find < PRQ
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Vraag 6 Verslag
The diagram is a circle of radius |QR| = 4cm. \(\bar{R}\) is a tangent to the circle at R. If TPO = 120o, find |PQ|.
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Vraag 7 Verslag
Is the diagram, MN, PQ and RS are three intersecting straight lines. Which of the following statements is/are true? i. t = y ii. x + y + z + m = 180o ii. x + m + n = 180o iv. x + n = m + z
Vraag 8 Verslag
The nth term of a sequence is Tn = 5 + (n - 1)2. Evaluate T4 - T6
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To evaluate T4 - T6, we first need to find the values of T4 and T6. T4 = 5 + (4 - 1)2 = 5 + 9 = 14 T6 = 5 + (6 - 1)2 = 5 + 25 = 30 Therefore, T4 - T6 = 14 - 30 = -16 Hence, the answer is -16.
Vraag 9 Verslag
The diagram is a circle centre O. If < SPR = 2m and < SQR = n, express m in terms of n
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Vraag 10 Verslag
Alfred spent \(\frac{1}{4}\) of his money on food, \(\frac{1}{3}\) on clothing and save the rest. If he saved N72,20.00, how much did he spend on food?
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Let A be the total amount of money that Alfred has. From the given information, he spent \(\frac{1}{4}\) of A on food and \(\frac{1}{3}\) of A on clothing, and he saved the rest. Therefore, we have: Amount spent on food = \(\frac{1}{4}\)A Amount spent on clothing = \(\frac{1}{3}\)A Amount saved = A - (\(\frac{1}{4}\)A + \(\frac{1}{3}\)A) Amount saved = A - \(\frac{7}{12}\)A Amount saved = \(\frac{5}{12}\)A We know that Alfred saved N72,20.00, so we can write: \(\frac{5}{12}\)A = N72,20.00 Multiplying both sides by \(\frac{12}{5}\), we get: A = N172,80.00 Therefore, the amount spent on food is: \(\frac{1}{4}\)A = \(\frac{1}{4}\) x N172,80.00 = N43,20.00 Hence, Alfred spent N43,20.00 on food. Answer is correct.
Vraag 11 Verslag
Find the coefficient of m in the expression of (\(\frac{m}{2} - 1 \frac{1}{2}\)) (m + \(\frac{2}{3}\))
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Vraag 12 Verslag
If x + 0.4y = 3 and y = \(\frac{1}{2}\)x, find the value of (x + y)
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We are given that x + 0.4y = 3 and y = \(\frac{1}{2}\)x. Substitute y in the first equation: x + 0.4 * (\(\frac{1}{2}\)x) = 3 x + 0.2x = 3 1.2x = 3 x = 2.5 Substitute x in the equation for y: y = \(\frac{1}{2}\) * 2.5 = 1.25 Then, x + y = 2.5 + 1.25 = 3.75. Therefore, the value of (x + y) is 3\(\frac{3}{4}\). Hence, the correct option is (c) 3\(\frac{3}{4}\).
Vraag 13 Verslag
If x + y = 2y - x + 1 = 5, find the value of x
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We are given two equations, x + y = 5 and 2y - x + 1 = 5, and we need to find the value of x. From the first equation, we can rearrange it to get x = 5 - y. Substituting this value of x into the second equation, we get: 2y - (5 - y) + 1 = 5 Simplifying the equation, we get: 3y - 4 = 5 Adding 4 to both sides, we get: 3y = 9 Dividing both sides by 3, we get: y = 3 Substituting this value of y into x + y = 5, we get: x + 3 = 5 Subtracting 3 from both sides, we get: x = 2 Therefore, the value of x is 2.
Vraag 14 Verslag
The graph is that of y = 2x2 - 5x - 3. For what value of x will y be negative? What is the gradient of y = 2x2 - 5x - 3 at the point x = 4?
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Vraag 16 Verslag
If x and y are variables and k is a constant, which of the following describes an inverse relationship between x and y?
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The equation that describes an inverse relationship between x and y is y = k/x, which is the second option. An inverse relationship between two variables means that as one variable increases, the other variable decreases in a proportional manner. In other words, if we double the value of x, the value of y would be halved, and vice versa. In the equation y = k/x, as x increases, y decreases, and as x decreases, y increases. The constant k represents the scale of the relationship between x and y. The other options do not describe an inverse relationship between x and y. is a direct proportion where y increases as x increases. is a power relationship where y increases as the square root of x increases. is a linear relationship where y increases at a constant rate as x increases.
Vraag 17 Verslag
The sum of 12 and one third of n is 1 more than twice n. Express the statement in the form of an equation
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Vraag 18 Verslag
The curved surface area of a cylindrical tin is 704cm2. If the radius of its base is 8cm, find the height. [Take \(\pi = \frac{22}{7}\)]`
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Vraag 19 Verslag
The sum of the interior angles of regular polygon is 1800o. How many sides has the polygon?
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The sum of the interior angles of a polygon can be found by the formula: sum of interior angles = (n-2) x 180 degrees where n is the number of sides in the polygon. In this problem, we are given that the sum of the interior angles of the polygon is 1800 degrees. So we can set up an equation: 1800 = (n-2) x 180 Simplifying the equation: 10 = n - 2 n = 12 Therefore, the polygon has 12 sides.
Vraag 21 Verslag
Solve the inequality: \(\frac{-m}{2} - \frac{5}{4} \geq \frac{5m}{12} - \frac{7}{6}\)
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Vraag 22 Verslag
The graph is that of y = 2x2 - 5x - 3. For what value of x will y be negative? For what value of x will y be negative?
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To find when y is negative, we need to find the values of x for which 2x^2 - 5x - 3 is negative. One way to do this is to use the quadratic formula: x = (-b ± sqrt(b^2 - 4ac)) / 2a where a = 2, b = -5, and c = -3. Plugging these values in, we get: x = (-(-5) ± sqrt((-5)^2 - 4(2)(-3))) / 2(2) x = (5 ± sqrt(49)) / 4 x = (5 ± 7) / 4 So the solutions are x = -3/2 and x = 2. We can check that when x < -3/2 or x > 2, y is positive, and when -3/2 < x < 2, y is negative. Therefore, the answer is option (C) -\(\frac{1}{2} < x < 3\).
Vraag 23 Verslag
Solve (\(\frac{27}{125}\))-\(\frac{1}{3}\) x (\(\frac{4}{9}\))\(\frac{1}{2}\)
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Vraag 24 Verslag
In 1995, the enrollment of two schools X and Y were 1,050 and 1,190 respectively. Find the ration of the enrollments of X and Y
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To find the ratio of enrollments of schools X and Y, we need to divide the enrollment of school X by the enrollment of school Y. Ratio of enrollments of X and Y = Enrollment of X / Enrollment of Y Enrollment of X = 1,050 Enrollment of Y = 1,190 Ratio of enrollments of X and Y = 1,050 / 1,190 Simplifying this fraction by dividing both numerator and denominator by 10 gives: Ratio of enrollments of X and Y = 105 / 119 Further simplifying by dividing both numerator and denominator by 7 gives: Ratio of enrollments of X and Y = 15 / 17 Therefore, the ratio of the enrollments of X and Y is 15:17. Answer option (B) is correct.
Vraag 25 Verslag
If p = {prime factors of 210} and Q = {prime less than 10}, find p \(\cap\) Q
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The prime factors of 210 are 2, 3, 5, and 7. The prime numbers less than 10 are 2, 3, 5, and 7. The intersection (p ∩ Q) is the set of elements that are in both p and Q. Therefore, p ∩ Q is {2, 3, 5, 7}. So, the correct option is: - {2,3,5,7}
Vraag 26 Verslag
The position of three ships P,Q and R at sea are illustrated in the diagram. The arrows indicated the North direction. The bearing of Q from P is 050o and < PQR = 72o. Calculate the bearing of R and Q
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Vraag 27 Verslag
A bag contains 4 red and 6 black balls of the same size. If the balls are shuffled briskly and two balls are drawn one after the other without replacement, find the probability of picking balls of different colours
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Vraag 28 Verslag
In \(\bigtriangleup\) XYZ, /XY/ = 8cm, /YZ/ = 10cm and /XZ/ = 6cm. Which of these relation is true?
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Vraag 29 Verslag
If N2,500.00 amounted to N3,50.00 in 4 years at simple interest, find the rate at which the interest was charged
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Simple interest is calculated by multiplying the principal amount, the interest rate, and the time in years. The formula for simple interest is given as: Simple Interest = (Principal x Rate x Time) / 100 In this problem, we know that the principal amount (P) is N2,500.00, the amount (A) after 4 years is N3,500.00. We need to find the rate (R) at which the interest was charged. Using the formula for simple interest, we can write: A = P + I where I is the interest amount. We can rearrange the formula to solve for I: I = A - P = N3,500.00 - N2,500.00 = N1,000.00 Now, we can substitute the values we know into the formula for simple interest and solve for R: I = (P x R x T) / 100 N1,000.00 = (N2,500.00 x R x 4) / 100 Simplifying, we get: R = (N1,000.00 x 100) / (N2,500.00 x 4) = 10% Therefore, the rate at which the interest was charged is 10%.
Vraag 30 Verslag
Express 3 - [\(\frac{x - y}{y}\)] as a single fraction
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To express 3 - [\(\frac{x - y}{y}\)] as a single fraction, we first need to simplify the expression inside the square brackets: 3 - [\(\frac{x - y}{y}\)] = 3 - (\(\frac{x}{y}\) - \(\frac{y}{y}\)) = 3 - \(\frac{x}{y}\) + 1 = 4 - \(\frac{x}{y}\) Therefore, 3 - [\(\frac{x - y}{y}\)] can be expressed as a single fraction: = 4 - \(\frac{x}{y}\) = \(\frac{4y - x}{y}\) Hence, the answer is \(\frac{4y - x}{y}\).
Vraag 31 Verslag
Make p the subject of the relation: q = \(\frac{3p}{r} + \frac{s}{2}\)
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Vraag 32 Verslag
Solve for x in the equation; \(\frac{1}{x} + \frac{2}{3x} = \frac{1}{3}\)
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To solve the equation \(\frac{1}{x} + \frac{2}{3x} = \frac{1}{3}\), we can use the following steps: Step 1: Find the common denominator. The denominators in this equation are x and 3x. The common denominator is 3x. Step 2: Multiply both sides of the equation by the common denominator. Multiplying both sides by 3x gives: 3 + 2 = x Step 3: Simplify and solve for x. Simplifying the left side gives: 5 = x Therefore, the solution to the equation is x = 5. x = 5, is the correct answer. Options 2, 3, and 4 are incorrect because they do not equal 5.
Vraag 33 Verslag
Convert 3510 to number in base 2
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To convert a number from base 10 to base 2 (binary), we need to divide the decimal number by 2 repeatedly until the quotient is zero. Then, we write the remainders in reverse order. Let's convert 3510 to binary using this method: 35 / 2 = 17 remainder 1 17 / 2 = 8 remainder 1 8 / 2 = 4 remainder 0 4 / 2 = 2 remainder 0 2 / 2 = 1 remainder 0 1 / 2 = 0 remainder 1 The remainders, in reverse order, are 100011, which is the binary equivalent of 3510. Therefore, the correct answer is option (C) 100011. Each digit in a binary number represents a power of 2, starting from the rightmost digit, which represents 2^0 = 1. The next digit represents 2^1 = 2, the next represents 2^2 = 4, and so on. To convert a binary number to decimal, we multiply each digit by its corresponding power of 2 and add up the results.
Vraag 34 Verslag
Factorise completely: 32x2y - 48x3y3
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We can begin by factoring out the greatest common factor of the two terms, which is 16x2y. This gives: 32x2y - 48x3y3 = 16x2y(2 - 3xy2) So the correct option is 16x2y(2 - 3xy2).
Vraag 35 Verslag
The volume of a cuboid is 54cm3. If the length, width and height of the cuboid are in the ratio 2:1:1 respectively, find its total surface area
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To solve this problem, we first need to find the dimensions of the cuboid. Let the length, width and height be 2x, x and x respectively. Then we have: Volume of cuboid = length x width x height 54 = 2x * x * x 54 = 2x^3 x^3 = 27 x = 3 Therefore, the length of the cuboid is 2x = 6cm, the width is x = 3cm, and the height is x = 3cm. To find the total surface area, we need to find the area of each face and add them up. The total surface area is given by: Total surface area = 2lw + 2lh + 2wh = 2(6*3) + 2(6*3) + 2(3*3) = 36 + 36 + 18 = 90 cm^2 Therefore, the total surface area of the cuboid is 90cm^2. The correct answer is option B.
Vraag 37 Verslag
In the diagram, MQ//RS, < TUV = 70o and < RLV = 30o. Find the value of x
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Vraag 38 Verslag
The bar chart shows the frequency distribution of marks scored by students in a class test. What is the median of the distribution?
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Vraag 39 Verslag
In the diagram, O is a circle centre of the circle PQRS and < PSR = 86o. If < PQR = xo, find x
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Vraag 40 Verslag
Mr. Manu travelled from Accra to Pamfokromb a distance of 720km in 8 hours. What will be his speed in m/s?
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To find the speed in meters per second (m/s), we need to convert the distance and time to meters and seconds respectively. Distance is given as 720km. 1km is equal to 1000m, so 720km = 720 x 1000m = 720,000m. Time taken is given as 8 hours. 1 hour is equal to 3600 seconds, so 8 hours = 8 x 3600 seconds = 28,800 seconds. Speed = Distance ÷ Time = 720,000m ÷ 28,800s = 25m/s. Therefore, Mr. Manu's speed is 25m/s. (25m/s) is the correct answer.
Vraag 41 Verslag
Given that the mean of the scores 15, 21, 17, 26, 18 and 29 is 21, calculate the standard deviation of the scores
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To calculate the standard deviation of the scores, we first need to find the variance, which is the average of the squared deviations from the mean. 1. Find the mean of the scores: Mean = (15 + 21 + 17 + 26 + 18 + 29)/6 = 126/6 = 21 2. Calculate the deviations from the mean for each score: 15 - 21 = -6 21 - 21 = 0 17 - 21 = -4 26 - 21 = 5 18 - 21 = -3 29 - 21 = 8 3. Square each deviation: (-6)^2 = 36 0^2 = 0 (-4)^2 = 16 5^2 = 25 (-3)^2 = 9 8^2 = 64 4. Find the average of the squared deviations: (36 + 0 + 16 + 25 + 9 + 64)/6 = 150/6 = 25 5. Take the square root of the variance to get the standard deviation: Standard deviation = sqrt(25) = 5 Therefore, the standard deviation of the scores is 5. The answer is 5.
Vraag 42 Verslag
The bar chart shows the frequency distribution of marks scored by students in a class test. Calculate the mean of the distribution.
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To calculate the mean of a frequency distribution, we need to find the sum of all the values multiplied by their respective frequencies and divide by the total number of values. Looking at the bar chart, we can see that there are 5 values: 1, 2, 3, 4, and 5. The corresponding frequencies are 1, 2, 5, 6, and 1, respectively. To calculate the mean, we first need to calculate the sum of all the values multiplied by their respective frequencies: (1 x 1) + (2 x 2) + (3 x 5) + (4 x 6) + (5 x 1) = 1 + 4 + 15 + 24 + 5 = 49 Next, we need to calculate the total number of values, which is the sum of all the frequencies: 1 + 2 + 5 + 6 + 1 = 15 Finally, we divide the sum of the values multiplied by their frequencies by the total number of values: 49 / 15 = 3.27 (rounded to two decimal places) Therefore, the mean of the distribution is approximately 3.27. The correct answer is not provided in the options, but the closest one is option (C) 2.4.
Vraag 43 Verslag
In the diagram, MN//PO, < PMN = 112o, < PNO = 129oo and < MPN = yo. Find the value of y
Vraag 44 Verslag
The bar chart shows the frequency distribution of marks scored by students in a class test. How many students are in the class?
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Vraag 46 Verslag
A kite flies on a taut string of length 50m inclined at tan angle 54o to the horizontal ground. The height of the kite above the ground is
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Vraag 47 Verslag
The lengths of the minor and major arcs 54cm and 126cm respectively. Calculate the angle of the major sector
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Vraag 48 Verslag
A side and a diagonal of a rhombus are 10cm and 12cm respectively, Find its area
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Vraag 49 Verslag
In the diagram, |QR| = 10m, |SR| = 8m
< QPS = 30o, < QRP = 90o and |PS| = x, Find x
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