WAEC SSCE - General Mathematics - 2016

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The range of a set of data is the difference between the maximum and minimum values in the set. In this case, the minimum mark is 3 and the maximum mark is 16. Therefore, the range is 16 - 3 = 13. Hence, the answer is 13.

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Using Pythagoras theorem, we can find that |UW| = 2cm. Next, we notice that \(\Delta\)UTW is an isosceles triangle since |TX| = |WX|. Therefore, o. Using the fact that the angles in a triangle add up to 180^o, we can find that o - o - (60^o + o. Therefore, we can substitute this into the previous equation to get: 90^o + o - (60^o + o = 2o, and since \(\Delta\)UTW is an isosceles triangle, we have that o. Hence, the answer is (C) 75^o.

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To find the lower quartile, we need to identify the point on the cumulative frequency curve that corresponds to 25% of the total frequency. From the graph, we see that the total frequency is 40, and 25% of this is 10. The point on the curve that corresponds to a frequency of 10 is at a value of 19.0. Therefore, the lower quartile is 19.0. Note that the cumulative frequency curve shows the cumulative frequency of data values up to and including each data point on the horizontal axis. So, we can read off quartiles and other percentiles directly from the graph.

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We know that the relation y = x² + 2x + k passes through the point (2,0), which means that when x=2, y=0. So, substituting these values in the relation, we get: 0 = 2² + 2(2) + k 0 = 4 + 4 + k 0 = 8 + k Therefore, k = -8. Hence, the value of k is -8.

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When two balls are drawn from the bag with replacement, there are a total of $9\times9=81$ possible outcomes, since there are 9 balls in the bag and we are replacing each ball after drawing. To find the probability that the first ball is red and the second ball is blue, we can use the multiplication rule of probability. The probability that the first ball is red is $\frac{5}{9}$, since there are 5 red balls out of 9 total balls in the bag. After replacing the first ball, there are still 9 balls in the bag, but now 4 of them are blue. So the probability that the second ball is blue, given that the first ball was red, is $\frac{4}{9}$. Therefore, the probability that the first ball is red and the second ball is blue is: $$\frac{5}{9} \times \frac{4}{9} = \frac{20}{81}$$ Hence the answer is $\frac{20}{81}$.

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The problem can be solved using the formula for direct variation: d = kt^2 where d is the distance, t is the time, and k is the constant of variation. We can solve for k using the given information that the stone falls 45cm in 3 seconds: 45 = k(3)^2 45 = 9k k = 5 Now that we know k, we can use the formula to find how far the stone will fall in 6 seconds: d = 5(6)^2 d = 5(36) d = 180cm Therefore, the answer is 180cm.

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The given expression will be undefined when its denominator is equal to zero since division by zero is undefined. So we need to find the values of y that make the denominator zero. \begin{align*} y^2 + 4y - 21 &= 0\\ (y+7)(y-3) &= 0 \end{align*} The denominator is equal to zero when either y+7=0 or y-3=0. Therefore, the expression is undefined when y=-7 or y=3. So, the answer is (c) 3, -7.

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The gradient of a line is the measure of its steepness, or slope. To find the gradient of the line PQ, we need to use the formula: Gradient (m) = change in y / change in x We can calculate the change in y by subtracting the y-coordinate of point P from the y-coordinate of point Q: 8 - 2 = 6 Similarly, we can calculate the change in x by subtracting the x-coordinate of point P from the x-coordinate of point Q: 5 - 1 = 4 Therefore, the gradient of the line PQ is: Gradient (m) = change in y / change in x = 6 / 4 = 3/2 Therefore, the answer is option C: \(\frac{3}{2}\).

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If 1cm represents 5km on the map, then x cm will represent 100km² on the map. To find x, we can use the formula for area of a square, which is A = s². In this case, we want to solve for s, where A = 100 and s represents the side length on the map in centimeters. So, s² = A s² = 100 s = √100 s = 10cm Therefore, 10cm on the map represents 100km² on the ground. To find the area on the map that represents 100km², we need to find the area of a square with a side length of 10cm. Area = s² Area = 10cm x 10cm Area = 100cm² So, the area on the map that represents 100km² is 100cm², which is equal to 1cm x 1cm, 4cm², or.

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To find the median, we need to first arrange the ages in order from lowest to highest. Then, we can determine which age lies in the middle of the list. Arranging the ages in order of increasing magnitude, we have: $$13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 17, 17, 17$$ There are a total of 10+24+8+5+3 = 50 ages in the list, which is an even number. To find the median, we need to take the average of the two middle ages. The two middle ages are the 25th and 26th ages in the list, which are both 14. Therefore, the median age is: $$(14 + 14)/2 = 14$$ So the correct answer is 14.

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We can use the distance formula to find the length PQ, which is the distance between points P and Q on the line. The distance formula is: distance = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) where (x1, y1) = P and (x2, y2) = Q. Plugging in the values: distance = \(\sqrt{(5-1)^2 + (8-2)^2}\) = \(\sqrt{16 + 36}\) = \(\sqrt{52}\) = \(2\sqrt{13}\) Therefore, the length PQ is \(2\sqrt{13}\). Answer: \(2\sqrt{13}\).

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To find out how many students are in the club, we need to add up the frequencies in the table. So, Number of 13-year-olds = 10 Number of 14-year-olds = 24 Number of 15-year-olds = 8 Number of 16-year-olds = 5 Number of 17-year-olds = 3 Total number of students = 10 + 24 + 8 + 5 + 3 = 50 Therefore, there are 50 students in the club.

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The volume of a sphere of radius r is given by the formula: \(\frac{4}{3}\pi r^3\). The volume of a cylinder of radius 3cm and height 4cm is given by the formula: \(\pi (3cm)^2(4cm) = 36\pi cm^3\). According to the question, the volume of the sphere is equal to the volume of the cylinder. Therefore: \[\frac{4}{3}\pi r^3 = 36\pi\] Dividing both sides by \(\frac{4}{3}\pi\), we get: \[r^3 = 27\] Taking the cube root of both sides, we get: \[r = 3\] Therefore, the value of r is 3cm. Answer is correct.

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First, let's simplify the numerator: \begin{align*} (p-r)^2 - r^2 &= (p^2 - 2pr + r^2) - r^2 \\ &= p^2 - 2pr \end{align*} Now, let's factor the denominator: \begin{align*} 2p^2 - 4pr &= 2p(p - 2r) \end{align*} Substituting these results, we get: \begin{align*} \frac{(p-r)^2 - r^2}{2p^2 - 4pr} &= \frac{p^2 - 2pr}{2p(p - 2r)} \\ &= \frac{p(p-2r)}{2p(p-2r)} \\ &= \frac{1}{2} \end{align*} Therefore, the answer is: \boxed{\frac{1}{2}}.

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The valid conclusion from the premise "Nigeria footballers are good footballers" is "Joseph is a Nigerian footballer therefore he is a good footballer". This is because the premise establishes that Nigeria footballers are good, so anyone who is a Nigeria footballer can be inferred to be a good footballer. Therefore, Joseph, who is a Nigerian footballer, can be concluded to be a good footballer. The other options are not valid because they do not follow logically from the given premise.

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The curved surface area of a cylinder is given by the formula: 2\(\pi\)rh. Given that the cylinder is 5cm high and its curved surface area is 110cm², we can write: 2\(\pi\)rh = 110, where h = 5 Substituting the value of \(\pi = \frac{22}{7}\) and h = 5, we get: 2 x \(\frac{22}{7}\) x r x 5 = 110 Simplifying this expression, we get: r = \(\frac{7}{2}\) r = 3.5cm (to one decimal place) Therefore, the radius of the cylinder is approximately 3.5cm. Hence, the answer is 3.5cm.

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The given sequence is the Fibonacci sequence, where the first two terms are 0 and 1, and each subsequent term is the sum of the two preceding it. Therefore, the next three terms are: - 13 (8 + 5) - 21 (13 + 8) - 34 (21 + 13) Hence, the answer is 13, 21, 34.

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To make s the subject of the relation P = S + \(\frac{sm^2}{nr}\), we need to isolate s on one side of the equation. First, we can start by moving the \(\frac{sm^2}{nr}\) term to the other side of the equation by subtracting it from both sides: P - \(\frac{sm^2}{nr}\) = S Next, we can solve for s by multiplying both sides of the equation by \(\frac{nr}{m^2}\): s = \(\frac{nr}{m^2}\)(P - \(\frac{sm^2}{nr}\)) Simplifying the right-hand side, we get: s = \(\frac{nrp}{m^2}\) - \(\frac{s}{m}\) To isolate s, we can add \(\frac{s}{m}\) to both sides of the equation: s + \(\frac{sm}{m^2}\) = \(\frac{nrp}{m^2}\) Simplifying the left-hand side, we get: s(\(\frac{m + 1}{m^2}\)) = \(\frac{nrp}{m^2}\) Finally, we can solve for s by dividing both sides of the equation by \(\frac{m+1}{m^2}\): s = \(\frac{nrp}{nr + m^2}\) Therefore, the answer is s = \(\frac{nrp}{nr + m^2}\).

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To simplify the expression (\(\frac{3}{4} - \frac{2}{3}\)) x 1\(\frac{1}{5}\), we first need to evaluate the subtraction inside the parentheses: \begin{align*} \frac{3}{4} - \frac{2}{3} &= \frac{9}{12} - \frac{8}{12} \\ &= \frac{1}{12} \end{align*} So now we have: \begin{align*} (\frac{3}{4} - \frac{2}{3})\times 1\frac{1}{5} &= \frac{1}{12} \times \frac{6}{5} \\ &= \frac{1 \times 6}{12 \times 5} \\ &= \frac{1}{10} \end{align*} Therefore, the answer is \(\frac{1}{10}\).

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The formula for the volume of a pyramid is given as: V = (1/3) * base_area * height Let the length of the rectangle be x, then the base area of the pyramid is given as: base_area = x * 6 From the question, we are told that the volume of the pyramid is 90cm³ and its height is 15cm. Substituting into the formula for the volume of a pyramid, we have: 90 = (1/3) * (x * 6) * 15 Multiplying both sides by 3 gives: 270 = 90 * 6x Dividing both sides by 90 gives: 3 = 2x Therefore, x = 3/2 = 1.5 Hence, the value of x is 3. Answer option A, 3, is the correct answer.

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To simplify the expression \(\frac{2}{1-x} - \frac{1}{x}\), we need to first find a common denominator. The denominator of the first fraction is \((1-x)\) and the denominator of the second fraction is \(x\). The common denominator of these two fractions is \(x(1-x)\). Now, we need to rewrite each fraction with this common denominator. For the first fraction, we can multiply the numerator and denominator by \(x\), giving us \(\frac{2x}{x(1-x)}\). For the second fraction, we can multiply the numerator and denominator by \((1-x)\), giving us \(\frac{-(1-x)}{x(1-x)}\). Putting these two fractions together, we get: \[\frac{2x}{x(1-x)} - \frac{1-x}{x(1-x)} = \frac{2x - (1-x)}{x(1-x)} = \frac{3x-1}{x(1-x)}\] Therefore, the answer is \(\frac{3x-1}{x(1-x)}\).

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We are given that log₂(3x - 1) = 5. Using the definition of logarithms, we know that 2⁵ = 32 is equal to the expression inside the logarithm. That is, 3x - 1 = 32 Adding 1 to both sides, we get 3x = 33 Dividing by 3, we get x = 11 Therefore, the value of x is 11. Answer: 11.

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We can simplify this expression by using the properties of exponents and simplifying the numbers. First, we can write 27 as 3³ and 81 as 3⁴. Next, we can simplify the numerator by using the distributive property of exponents: \begin{align*} \frac{3^{n-1} \times 27^{n+1}}{81^n} &= \frac{3^{n-1} \times 3^{3(n+1)}}{3^{4n}} \\ &= \frac{3^{n-1} \times 3^{3n+3}}{3^{4n}} \\ &= \frac{3^{4n-1}}{3^{4n}} \times 3^{3n+3} \\ &= 3^{-1} \times 3^{3n+3} \\ &= 3^{3n+2} \\ &= 3^2 \times 3^{3n} \\ &= 9 \times 3^n \\ \end{align*} Therefore, the simplified expression is 9 * 3ⁿ. So, the answer is 9.

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