WAEC SSCE - General Mathematics - 2015

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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) \(\frac{2}{5}\).

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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 \(\frac{1}{30}\).

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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.

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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.

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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.

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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.

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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\(^\frac{1}{3}\).

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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.

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Since < QOL is a right angle, we have: < QOT + < SOT = 90^o Since < QOT = 180^o - < POL, we have: < QOT = 180^o - m Substituting into the first equation, we get: 180^o - m + n = 90^o n - m = 90^o - 180^o = -90^o Also, from the diagram, we have: < QOT + < ROS + < SOT = 180^o Substituting < QOT = 180^o - < POL and simplifying, we get: 180^o - m + (3n - 20) + n = 180^o 4n - m = 20 We now have a system of two equations: n - m = -90^o 4n - m = 20 Solving simultaneously, we get: n = 35^o

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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) 32^o.

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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).

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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.

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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: \(\frac{4x - 5y + xy}{2xy}\).

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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'.

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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.

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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.

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(a) Brackets first:

(b) One number from \(\{2, 3, 4\}\) and one from \(\{1, 3, 5\}\): \(3 \times 3 = 9\) equally likely pairs.

Pairs whose sum satisfies \(3 < \text{sum} < 7\): \((2,3) = 5,\ (3,1) = 4,\ (3,3) = 6,\ (4,1) = 5\), giving 4 outcomes.

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(a) Use exact surd values: \(\tan 60° = \sqrt{3}\), \(\cos 30° = \dfrac{\sqrt{3}}{2}\), \(\sin 60° = \dfrac{\sqrt{3}}{2}\).

Numerator: \(2\sqrt{3} + \dfrac{\sqrt{3}}{2} = \dfrac{4\sqrt{3} + \sqrt{3}}{2} = \dfrac{5\sqrt{3}}{2}\).

\(\dfrac{2\tan 60° + \cos 30°}{\sin 60°} = \dfrac{5\sqrt{3}/2}{\sqrt{3}/2} = \mathbf{5}\)

(b) Horizontal distance to the tower \(= 1050\text{ m}\). The top has depression \(36°\) and the base \(41°\). The vertical drop from the aeroplane's level to each point is \(1050\tan(\text{angle})\).

(i) Height of tower \(=\) (drop to base) \(-\) (drop to top):

\(= 1050\tan 41° - 1050\tan 36° = 1050(0.8693 - 0.7265)\)

\(= 1050 \times 0.1428 = 149.9 \approx \mathbf{150\text{ m}}\).

(ii) Shortest distance (line of sight to the base):

\(= \dfrac{1050}{\cos 41°} = \dfrac{1050}{0.7547} = 1391.3 \approx \mathbf{1391\text{ m}}\).

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(a)(i) Given that \(y=px^2-5x+q\).

When \(x=0\), \(y=-12\):

\[q=-12.\]

When \(x=4\), \(y=0\):

\[0=16p-5(4)-12\]

\[0=16p-32\]

\[p=2.\]

Therefore,

\[\boxed{y=2x^2-5x-12}.\]

(a)(ii) The completed table is:

\(x\)	\(-3\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)	\(5\)
\(y\)	\(21\)	\(6\)	\(-5\)	\(-12\)	\(-15\)	\(-14\)	\(-9\)	\(0\)	\(13\)

(b) The graph of \(y=2x^2-5x-12\), plotted using the stated scales, is shown below.

(c)

(i) From the graph, when \(x=1.8\),

\[\boxed{y\approx -14.5}.\]

(ii) Drawing the horizontal line \(y=-8\) and reading the two points of intersection with the curve gives

\[\boxed{x\approx -0.65\text{ or }x\approx 3.15}.\]

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(a) Sum \(= 3\tfrac{2}{3} + 2\tfrac{1}{5} = \tfrac{11}{3} + \tfrac{11}{5} = \tfrac{55 + 33}{15} = \tfrac{88}{15}\).

It is less than 7 by \(7 - \tfrac{88}{15} = \tfrac{105 - 88}{15} = \tfrac{17}{15} = 1\tfrac{2}{15}\).

(b)(i) With \(p = 4\): \(15p = 60^{\circ}\) and \(\cos 60^{\circ} = 0.5\).

(b)(ii) With \(h = 9\):

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(a)(i) Venn diagram: Draw a rectangle for the school. Inside it draw a circle for the General Arts class \((A)\). Entirely within that circle draw a smaller circle for good Literature students \((L)\). This shows \(L \subset A\): every good Literature student is in the General Arts class, but not every General Arts student is a good Literature student.

(ii) Testing the conclusions:

1. Vivian is in General Arts, therefore she is a good Literature student. NOT valid: being in \(A\) does not place her inside the smaller circle \(L\).
2. Audu is not a good Literature student, therefore he is not in General Arts. NOT valid: he could be in \(A\) but outside \(L\).
3. Kweku is not in General Arts, therefore he is not a good Literature student. VALID: since \(L \subset A\), anyone outside \(A\) must be outside \(L\).

(b) Cost \(c = h + y\) where \(y = kn\) (\(y\) varies directly as \(n\)), so \(c = h + kn\).

\(950 = h + 600k \quad(1)\)

\(1030 = h + 1000k \quad(2)\)

\((2) - (1):\ 80 = 400k \ \Rightarrow\ k = 0.2\). Then \(h = 950 - 600(0.2) = 830\).

(i) \(\mathbf{c = h + 0.2n}\), that is \(c = 830 + 0.2n\) (with \(h = \text{GH¢}830\)).

(ii) For \(n = 500\): \(c = 830 + 0.2(500) = 830 + 100 = \mathbf{\text{GH¢}930.00}\).

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(a) Area. The perpendicular distance between the parallel sides is 40 cm.

(b) Angle QRS. Place S at \((0, 0)\) and R at \((60, 0)\). P is 40 cm above RS; since \(SP = 50\), its horizontal offset is \(\sqrt{50^2 - 40^2} = 30\), so P is at \((30, 40)\). As \(PQ \parallel SR\) with \(PQ = 20\), Q is at \((50, 40)\).

At R, \(\vec{RQ} = (-10, 40)\) and \(\vec{RS} = (-60, 0)\):

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(a) Cumulative frequency table

Marks (%)	Frequency, \(f\)	Upper class boundary	Cumulative frequency
0–9	7	9.5	7
10–19	11	19.5	18
20–29	17	29.5	35
30–39	20	39.5	55
40–49	29	49.5	84
50–59	34	59.5	118
60–69	30	69.5	148
70–79	25	79.5	173
80–89	21	89.5	194
90–99	6	99.5	200

Total number of candidates \(=200\).

Plot the upper class boundaries against the cumulative frequencies, including \((-0.5,0)\), and draw a smooth cumulative frequency curve.

(b)(i) The number who passed with distinction is

\[5\%\text{ of }200=10.\]

Hence the lowest distinction mark has cumulative frequency

\[200-10=190.\]

From the cumulative frequency curve, a cumulative frequency of \(190\) gives a mark of approximately \(87.5\%\).

Lowest mark for distinction \(=87.5\%\).

(b)(ii) From the curve, the cumulative frequency at \(45\%\) is approximately \(69\).

\[P(\text{score at most }45\%)=\frac{69}{200}=0.345\approx 0.3\text{ (to 1 d.p.)}.\]

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(a) Solving the inequality

\[4 + \frac{3}{4}(x+2) \le \frac{3}{8}x + 1\]

Multiply every term by 8 (the LCM of the denominators) to clear fractions:

\[8(4) + 8\cdot\frac{3}{4}(x+2) \le 8\cdot\frac{3}{8}x + 8(1)\]

\[32 + 6(x+2) \le 3x + 8\]

\[32 + 6x + 12 \le 3x + 8\]

\[6x + 44 \le 3x + 8\]

Collect like terms:

\[6x - 3x \le 8 - 44\]

\[3x \le -36\]

\[x \le -12\]

(b) Finding x from the rectangle

From the diagram, rectangle PQRS has height \(|PS| = 20\ \text{cm}\). Along the base SR a square of side x cm is cut out, leaving 10 cm on the left and 10 cm on the right. So the full length of the rectangle is

\[|SR| = 10 + x + 10 = (20 + x)\ \text{cm}\]

Area of the whole rectangle:

\[(20 + x)\times 20 = 400 + 20x\]

Area of the square removed:

\[x \times x = x^2\]

The shaded (remaining) area is 484 cm\(^2\):

\[(400 + 20x) - x^2 = 484\]

\[-x^2 + 20x + 400 - 484 = 0\]

\[-x^2 + 20x - 84 = 0\]

\[x^2 - 20x + 84 = 0\]

Solve using factorisation (\(-6\) and \(-14\) multiply to 84 and add to \(-20\)):

\[(x - 6)(x - 14) = 0\]

\[x = 6 \quad\text{or}\quad x = 14\]

Both values are geometrically valid (the cut height stays within the 20 cm rectangle), so

\[x = 6\ \text{cm} \quad\text{or}\quad x = 14\ \text{cm}\]

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(a) Number of sides of the polygon

At each vertex the interior and exterior angles are supplementary, so they add up to \(180^\circ\). The ratio interior : exterior is \(5:2\), giving \(5+2 = 7\) equal parts.

For any regular polygon the exterior angles sum to \(360^\circ\), so the number of sides is

The polygon has 7 sides (a regular heptagon).

(b) Finding \(\angle PRS\)

From the diagram, \(TPQU\) is a straight line through the two points \(P\) and \(Q\) on the circle, and \(PQRS\) is a cyclic quadrilateral. The given angles are \(\angle TPS = 74^\circ\), \(\angle UQR = 68^\circ\) and \(\angle QSR = 40^\circ\).

Step 1: Interior angles of the cyclic quadrilateral. Since \(TPQ\) is a straight line,

Since \(PQU\) is a straight line,

Step 2: Use the opposite-angles property of a cyclic quadrilateral.

Step 3: Split \(\angle SRQ\) using angles in the same segment. The angle \(\angle PRQ\) stands on chord \(PQ\), as does \(\angle PSQ\); angles in the same segment are equal, so \(\angle PRQ = \angle PSQ\).

Now find \(\angle PSQ\). We have \(\angle PSR = 180^\circ - \angle PQR = 180^\circ - 112^\circ = 68^\circ\), and this splits as \(\angle PSR = \angle PSQ + \angle QSR\). Therefore

Step 4: Obtain \(\angle PRS\). Since \(\angle SRQ = \angle PRS + \angle PRQ\),

Hence \(\angle PRS = 46^\circ\).

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(a) Diagram: A cone sitting on top of a hemisphere, sharing the same circular base of radius \(r\). The flat face of the hemisphere coincides with the base of the cone, and the cone height is \(h = 6r\).

(b) Let the common radius be \(r\), so \(h = 6r\).

\(V_{\text{cone}} = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\pi r^2(6r) = 2\pi r^3\)

\(V_{\text{hemisphere}} = \tfrac{2}{3}\pi r^3\)

Total: \(2\pi r^3 + \tfrac{2}{3}\pi r^3 = \tfrac{8}{3}\pi r^3 = 333\tfrac{1}{3}\pi = \tfrac{1000}{3}\pi\).

\(8r^3 = 1000 \ \Rightarrow\ r^3 = 125 \ \Rightarrow\ r = 5\text{ m}\) (so \(h = 30\text{ m}\)).

(I) Volume of hemisphere: \(\tfrac{2}{3}\pi r^3 = \tfrac{2}{3}\times\tfrac{22}{7}\times 125 = \dfrac{5500}{21} = 261.9 \approx \mathbf{262\text{ m}^3}\).

(II) Total surface area \(=\) curved surface of cone \(+\) curved surface of hemisphere (the joining faces are internal).

Slant height \(l = \sqrt{r^2 + h^2} = \sqrt{25 + 900} = \sqrt{925} = 30.41\text{ m}\).

\(\pi r l = \tfrac{22}{7}\times 5\times 30.41 = 477.9\text{ m}^2\); \(2\pi r^2 = 2\times\tfrac{22}{7}\times 25 = 157.1\text{ m}^2\).

Total \(= 477.9 + 157.1 = 635.0 \approx \mathbf{635\text{ m}^2}\).

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(a) Construction

1. Draw the straight line WX = 8.0 cm.
2. At X, construct an angle of 60° with XW. On this ray, mark Y such that XY = 5.5 cm.
3. Through Y, construct a line parallel to WX. On it, mark Z such that YZ = 8.3 cm, on the side shown. Join Z to W to complete trapezium WXYZ.
4. From Y and Z, construct perpendiculars to the line WX, meeting it at Q and P respectively. Join PQ. Thus, PQYZ is a rectangle.

(b) Measurements from the construction

(i) \(|QX| = 2.7\text{ cm}\).

(ii) \(\angle XWZ = 75^\circ\).

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