WAEC SSCE - General Mathematics - 2012

Akọwa Nkọwa

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\(\frac{3}{4}\).

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

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.

Akọwa Nkọwa

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.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

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.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

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.

Akọwa Nkọwa

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.

Akọwa Nkọwa

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.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

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 = \(\frac{k}{x}\).

Akọwa Nkọwa

Akọwa Nkọwa

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.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

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.

Akọwa Nkọwa

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.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

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) -\(\frac{1}{2} < x < 3\).

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

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)

Akọwa Nkọwa

Akọwa Nkọwa

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.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

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).

Akọwa Nkọwa

Akọwa Nkọwa

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 \(\frac{4y - x}{y}\).

Akọwa Nkọwa

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%.

Akọwa Nkọwa

Akọwa Nkọwa

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.

Akọwa Nkọwa

Akọwa Nkọwa

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}

Akọwa Nkọwa

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.

Akọwa Nkọwa

(a) The boy had \(M\) dalasis, spent D15, leaving \(M - 15\). This remainder is shared equally between him and his sister, so the sister's share is \(\dfrac{M - 15}{2}\). We are told this equals \(\dfrac13 M\):

\[\frac{M - 15}{2} = \frac{M}{3}\]

\[3(M - 15) = 2M \;\Rightarrow\; 3M - 45 = 2M \;\Rightarrow\; M = 45\]

So \(M = \text{D}45\).

(b) Let the total number of tourists be \(N\). Travelled by road \(= \tfrac23 N\), by air \(= \tfrac{13}{30}N\), by both \(= \tfrac{4}{15}N\).

(i) Venn diagram: two intersecting circles (Road, Air). Both region \(= \tfrac{4}{15}N\); road only \(= \tfrac23 N - \tfrac{4}{15}N = \tfrac{6}{15}N\); air only \(= \tfrac{13}{30}N - \tfrac{4}{15}N = \tfrac{5}{30}N\); outside both \(= 20\).

(ii)(1) Those using road or air:

\[\frac23 + \frac{13}{30} - \frac{4}{15} = \frac{20 + 13 - 8}{30} = \frac{25}{30} = \frac56\]

So the fraction using neither is \(1 - \tfrac56 = \tfrac16\), and \(\tfrac16 N = 20 \Rightarrow N = 120\). 120 tourists were interviewed.

(2) Air only \(= \tfrac{13}{30}N - \tfrac{4}{15}N = \tfrac{5}{30}\times 120 = 20\) tourists.

Akọwa Nkọwa

(a) Calculating /PS/

From the diagram, \(P, Q, R\) lie on one straight line with \(PQ = 6\text{ cm}\) and \(QR = 13\text{ cm}\). The triangle \(QSR\) is right-angled at \(S\), with \(RS = 5\text{ cm}\) and hypotenuse \(QR = 13\text{ cm}\).

Step 1: Find QS using Pythagoras in triangle QSR.

\[ QS^2 = QR^2 - RS^2 = 13^2 - 5^2 = 169 - 25 = 144 \]

\[ QS = 12\text{ cm} \]

Step 2: Find angle RQS.

\[ \cos(\angle RQS) = \frac{QS}{QR} = \frac{12}{13} \]

Step 3: Use triangle PQS. Since \(P, Q, R\) are collinear, \(\angle PQS\) and \(\angle RQS\) are supplementary, so \(\cos(\angle PQS) = -\dfrac{12}{13}\).

Applying the cosine rule in triangle \(PQS\) with \(PQ = 6\), \(QS = 12\):

\[ PS^2 = PQ^2 + QS^2 - 2\,(PQ)(QS)\cos(\angle PQS) \]

\[ PS^2 = 6^2 + 12^2 - 2(6)(12)\left(-\tfrac{12}{13}\right) \]

\[ PS^2 = 36 + 144 + \frac{1728}{13} = 180 + 132.92 = 312.92 \]

\[ PS = \sqrt{312.92} \approx 17.7\text{ cm} \]

/PS/ \(\approx\) 17.7 cm.

(b) Surface area of the structure

The structure is a cone (height \(24\text{ cm}\), base radius \(7\text{ cm}\)) mounted on a hemisphere of the same radius \(7\text{ cm}\). The exposed surface is the curved surface of the cone plus the curved surface of the hemisphere (the joining circle is hidden).

Slant height of cone:

\[ l = \sqrt{h^2 + r^2} = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25\text{ cm} \]

Curved surface area of cone:

\[ \pi r l = \frac{22}{7} \times 7 \times 25 = 550\text{ cm}^2 \]

Curved surface area of hemisphere:

\[ 2\pi r^2 = 2 \times \frac{22}{7} \times 7^2 = 2 \times 22 \times 7 = 308\text{ cm}^2 \]

Total surface area:

\[ 550 + 308 = 858\text{ cm}^2 \]

Surface area \(= 858\text{ cm}^2\) (to 3 significant figures).

Akọwa Nkọwa

Let Wale's present age be \(w\). Since Sonny is twice as old, Sonny is \(2w\).

Four years ago Sonny was four times as old as Wale:

\[2w - 4 = 4(w - 4)\]

\[2w - 4 = 4w - 16 \;\Rightarrow\; 12 = 2w \;\Rightarrow\; w = 6\]

So presently Wale is \(6\) and Sonny is \(12\); their ages sum to \(18\).

Each year that passes adds \(2\) to the total (one year to each person). Let \(n\) be the number of years until the sum is \(66\):

\[18 + 2n = 66 \;\Rightarrow\; 2n = 48 \;\Rightarrow\; n = 24\]

The sum of their ages will be \(66\) in \(\mathbf{24}\) years' time (when Wale is 30 and Sonny is 36).

Akọwa Nkọwa

(a) The diagram shows the rectangle divided into 4 columns and 2 rows, giving \(4 \times 2 = 8\) equal plots. Length \(AB = (3n-1)\ \text{m}\), width \(BC = (2n+1)\ \text{m}\).

The wire mesh forms the boundary and the internal dividing lines.

Horizontal lines (each of length \(3n-1\)): top, bottom and 1 internal line \(= 3\) lines.

Vertical lines (each of length \(2n+1\)): left, right and 3 internal lines \(= 5\) lines.

Total length of wire:

\(n = 7\)

(b) Take \(\pi = \frac{22}{7}\). Let the cylinder have radius \(r = 14\ \text{cm}\) and height \(h\).

Volume of cube (side \(22\ \text{cm}\)):

Volume of cylinder equals this:

Total surface area of cylinder:

Total surface area of cube:

Ratio (cylinder : cube):

Dividing both by \(264\):

Ratio \(= 73 : 77\)

Akọwa Nkọwa

Working table (midpoint \(x\), \(N = 40\)).

(a) Mean.

(b) Standard deviation.

Akọwa Nkọwa

\(H\) is 20 m from the foot of the tower. From \(H\), \(P\) has elevation \(30^\circ\) and the top \(T\) has elevation \(50^\circ\).

Heights above ground of the two points:

\[\text{height of }P = 20\tan 30^\circ = 20(0.5774) = 11.547\text{ m}\]

\[\text{height of }T = 20\tan 50^\circ = 20(1.1918) = 23.835\text{ m}\]

(a) \(|PT| = 23.835 - 11.547 = 12.288 \approx 12.3\text{ m (3 s.f.)}\).

(b) Distance \(|HT|\) (the slant line to the top):

\[|HT| = \frac{20}{\cos 50^\circ} = \frac{20}{0.6428} = 31.1\text{ m (3 s.f.)}\]

(c) For the angle of depression of \(H\) from the top to be \(40^\circ\), the angle of elevation of \(T\) from \(H\) must be \(40^\circ\). With the top still \(23.835\) m high, the required horizontal distance \(x\) satisfies:

\[\tan 40^\circ = \frac{23.835}{x} \;\Rightarrow\; x = \frac{23.835}{0.8391} = 28.4\text{ m (3 s.f.)}\]

So \(H\) must be \(28.4\) m from the foot of the tower.

Akọwa Nkọwa

(a) For each value of x, calculate \(y=1-4\cos x\), correct to 1 decimal place.

\(x\)	\(0^\circ\)	\(30^\circ\)	\(60^\circ\)	\(90^\circ\)	\(120^\circ\)	\(150^\circ\)	\(180^\circ\)	\(210^\circ\)	\(240^\circ\)	\(270^\circ\)	\(300^\circ\)
\(y\)	-3.0	-2.5	-1.0	1.0	3.0	4.5	5.0	4.5	3.0	1.0	-1.0

(b) The completed graph of \(y=1-4\cos x\), including \(y=-2.5\) at \(330^\circ\) and \(y=-3.0\) at \(360^\circ\), is shown below.

(c)

(i) The points where the curve cuts the \(x\)-axis give
\[x\approx75^\circ\quad\text{or}\quad285^\circ.\]

(ii) At \(x=105^\circ\), the ordinate of the curve is
\[y\approx2.0.\]

(iii) Drawing the horizontal line \(y=1.5\) and reading the two intersections with the curve gives
\[x\approx96^\circ\quad\text{or}\quad264^\circ.\]

Akọwa Nkọwa

(a)(i) Rearrange the equations:

\[y-\frac{3x}{4}=3\quad\Rightarrow\quad y=\frac34x+3\]

\[y+2x=6\quad\Rightarrow\quad y=6-2x\]

Line	Points used for plotting
\(y=\frac34x+3\)	\((-4,0),\ (0,3),\ (4,6)\)
\(y=6-2x\)	\((0,6),\ (1,4),\ (3,0)\)

Using the scale of 2 cm to 1 unit on both axes, the required graph is:

The boundary \(y-\frac34x=3\) is drawn as a solid line since the inequality includes equality. Region \(R\), the side above this line, is shaded.

(ii) The two lines intersect at approximately

\[\boxed{(1.1,\ 3.8)}\]

This agrees with the exact solution:

\[\frac34x+3=6-2x\]

\[\frac{11}{4}x=3\quad\Rightarrow\quad x=\frac{12}{11}\]

\[y=6-2\left(\frac{12}{11}\right)=\frac{42}{11}\]

Thus the exact coordinates are \(\left(\frac{12}{11},\frac{42}{11}\right)\), which are approximately \((1.1,3.8)\).

(iii) \[y-\frac34x\geq3\quad\Rightarrow\quad y\geq\frac34x+3.\]

Hence, the required region is the half-plane on and above \(y=\frac34x+3\), labelled \(R\) on the graph.

(b)

\[(x+2)(x+c)=x^2+(c+2)x+2c.\]

Comparing with \(x^2+bx+18\):

\[2c=18\Rightarrow c=9,\]

\[b=c+2=9+2=11.\]

\[\boxed{c=9,\quad b=11}\]

Akọwa Nkọwa

(a) Reading the diagram. \(R, V, S, U\) lie on the circle (\(R\) at top, \(V\) on the left, \(S\) on the right, \(U\) at the bottom). The line \(TU\) is the tangent that touches the circle at \(U\), and the secant through \(R\) and \(S\) is produced to meet the tangent at the external point \(T\). We are given \(\angle RVU = 100^\circ\) and \(\angle URS = 36^\circ\).

Step 1: Find arc \(SU\). \(\angle URS = 36^\circ\) is an inscribed angle at \(R\) standing on arc \(SU\) (the arc not containing \(R\)):

\[\text{arc } SU = 2\times 36^\circ = 72^\circ.\]

Step 2: Find arc \(RU\) on the far side. \(\angle RVU = 100^\circ\) is an inscribed angle at \(V\) standing on the arc \(RU\) that does not contain \(V\) (the arc going \(R\to S\to U\)):

\[\text{arc } RSU = \text{arc } RS + \text{arc } SU = 2\times 100^\circ = 200^\circ.\]

The remaining arc from \(R\) to \(U\) through \(V\) is therefore

\[\text{arc } RVU = 360^\circ - 200^\circ = 160^\circ.\]

Step 3: Apply the tangent-secant angle rule at \(T\). The angle between a tangent and a secant drawn from an external point equals half the difference of the two intercepted arcs. Here the far arc (between tangent point \(U\) and far point \(R\), through \(V\)) is \(160^\circ\) and the near arc (between \(U\) and near point \(S\)) is \(72^\circ\):

\[\angle STU = \tfrac{1}{2}\left(\text{arc } RVU - \text{arc } SU\right) = \tfrac{1}{2}(160^\circ - 72^\circ) = \tfrac{1}{2}(88^\circ) = 44^\circ.\]

Answer (a): \(\angle STU = 44^\circ\).

(b) Triangle \(XYZ\): \(|XY| = 5\), \(|YZ| = 8\), \(|XZ| = 6\text{ cm}\), with \(P\) on \(XY\), \(|XP| = 2\text{ cm}\), and \(PQ \parallel YZ\) with \(Q\) on \(XZ\).

Since \(PQ \parallel YZ\), triangle \(XPQ\) is similar to triangle \(XYZ\), so corresponding sides are proportional:

\[\frac{XP}{XY} = \frac{XQ}{XZ} \;\Rightarrow\; \frac{2}{5} = \frac{XQ}{6}.\]

\[XQ = \frac{2}{5}\times 6 = 2.4\text{ cm}.\]

\[|QZ| = |XZ| - |XQ| = 6 - 2.4 = 3.6\text{ cm}.\]

Answer (b): \(|QZ| = 3.6\text{ cm}\).

Akọwa Nkọwa

None

Akọwa Nkọwa

From the diagram, \(O\) is the centre, radius \(OX = OY = 5\ \text{cm}\), chord \(|XY| = 6\ \text{cm}\), and the shaded portion is the minor segment cut off by the chord \(XY\) (the region between the chord and the minor arc on the right).

(a) Angle subtended at the centre, \(\angle XOY\):

Drop the perpendicular from \(O\) to the midpoint \(M\) of \(XY\). Then \(XM = 3\ \text{cm}\) and:

\(\angle XOY \approx 73.74^\circ\)

(b) Area of the shaded (minor) segment \(=\) area of sector \(XOY\) \(-\) area of triangle \(XOY\).

Area of sector:

Area of triangle \(XOY\):

Area of shaded segment:

Shaded area \(\approx 4.09\ \text{cm}^2\) (to 2 decimal places).

Akọwa Nkọwa

(a) Convert to improper fractions.

Numerator: \(1\tfrac14 + \tfrac79 = \tfrac54 + \tfrac79 = \dfrac{45 + 28}{36} = \dfrac{73}{36}\).

Denominator: first \(2\tfrac23 \times \tfrac{9}{64} = \tfrac83 \times \tfrac{9}{64} = \dfrac{72}{192} = \dfrac38\). Then \(1\tfrac49 - \tfrac38 = \tfrac{13}{9} - \tfrac38 = \dfrac{104 - 27}{72} = \dfrac{77}{72}\).

\[\frac{73/36}{77/72} = \frac{73}{36}\times\frac{72}{77} = \frac{73 \times 2}{77} = \frac{146}{77} = 1\tfrac{69}{77}\]

(b) Given \(\sin x = \dfrac23\), the adjacent side is \(\sqrt{3^2 - 2^2} = \sqrt5\), so

\[\cos x = \frac{\sqrt5}{3}, \qquad \tan x = \frac{2}{\sqrt5} = \frac{2\sqrt5}{5}\]

\[\tan x - \cos x = \frac{2}{\sqrt5} - \frac{\sqrt5}{3} = \sqrt5\left(\frac{2}{5} - \frac{1}{3}\right) = \sqrt5\cdot\frac{6 - 5}{15} = \frac{\sqrt5}{15}\]

Akọwa Nkọwa

(a) Total discs \(= 40\), and \(P(\text{red}) = \dfrac14\), so number of red \(= \dfrac14\times 40 = 10\).

(i) Number of white discs \(= 40 - 10 = 30\).

(ii) Let \(x\) red discs be added. New red count \(= 10 + x\); new total \(= 40 + x\). Require \(P(\text{red}) = \dfrac13\):

\[\frac{10 + x}{40 + x} = \frac13 \;\Rightarrow\; 3(10 + x) = 40 + x \;\Rightarrow\; 30 + 3x = 40 + x\]

\[2x = 10 \;\Rightarrow\; x = 5\]

So \(5\) red discs should be added.

(b) Cost price per plate \(=\) N50. A profit of \(20\%\) on total cost means

\[\text{Selling price} = 1.20 \times \text{Cost price}\]

\[600 = 1.20 \times \text{Cost price} \;\Rightarrow\; \text{Cost price} = \frac{600}{1.20} = \text{N}500\]

\[\text{Number of plates} = \frac{500}{50} = 10\]