WAEC SSCE - General Mathematics - 2017

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To calculate the variance of a set of data, you need to follow these steps: 1. Find the mean of the data set. 2. For each data point, subtract the mean and square the result. 3. Add up all the squared differences. 4. Divide the sum of squared differences by the total number of data points minus one. Here are the steps for the given data set: 1. Find the mean: (2 + 3 + 3 + 4 + 5 + 5 + 5 + 7 + 7 + 9) / 10 = 5 2. Subtract the mean and square the result for each data point: (2 - 5)^2 = 9 (3 - 5)^2 = 4 (3 - 5)^2 = 4 (4 - 5)^2 = 1 (5 - 5)^2 = 0 (5 - 5)^2 = 0 (5 - 5)^2 = 0 (7 - 5)^2 = 4 (7 - 5)^2 = 4 (9 - 5)^2 = 16 3. Add up all the squared differences: 9 + 4 + 4 + 1 + 0 + 0 + 0 + 4 + 4 + 16 = 42 4. Divide the sum of squared differences by the total number of data points minus one: 42 / (10 - 1) = 4.67 Therefore, the variance of the given data set is approximately 4.67. The closest option is 4.2.

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The given problem states that x varies inversely with y, which means that as y increases, x decreases and vice versa. Similarly, it also states that y varies directly with z, which means that as z increases, y increases and vice versa. Now, to find the relationship between x and z, we need to combine these two statements. Since y is the common variable, we can rewrite the second statement as y \(\alpha\) z. Substituting this value of y in the first statement, we get x \(\alpha\) \(\frac{1}{y}\), which can be further simplified as x \(\alpha\) \(\frac{1}{z}\). Hence, the relationship between x and z is x \(\alpha\) \(\frac{1}{z}\). is the correct answer.

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To find the perimeter of a square, we need to know the length of one of its sides. Let's call the length of one side of the square "x". Since the square has four equal sides, we can say that the perimeter of the square is equal to 4 times x, or 4x. We are given that the diagonal of the square is 60 cm. We know that the diagonal of a square forms a right triangle with two sides that are equal to the length of one side of the square. We can use the Pythagorean theorem to find the length of one side of the square. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the longest side) is equal to the sum of the squares of the lengths of the other two sides. In this case, the hypotenuse is the diagonal of the square, which we know is 60 cm. We can let one side of the square be "x", as we mentioned earlier. The other side of the square is also "x" because the square has four equal sides. So we have: 60^2 = x^2 + x^2 Simplifying this equation, we get: 3600 = 2x^2 Dividing both sides by 2, we get: 1800 = x^2 Taking the square root of both sides, we get: x = sqrt(1800) We can simplify this answer by factoring out the square of a perfect square: x = sqrt(900 * 2) x = sqrt(900) * sqrt(2) x = 30 * sqrt(2) Now that we know the length of one side of the square, we can find its perimeter by multiplying by 4: perimeter = 4x = 4(30 * sqrt(2)) = 120 * sqrt(2) Therefore, the answer is 120\(\sqrt{2}\).

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To evaluate this expression, we need to follow the order of operations, which is a set of rules for the order in which we perform arithmetic operations. The order of operations is: 1. Parentheses 2. Exponents 3. Multiplication and division (performed from left to right) 4. Addition and subtraction (performed from left to right) In this expression, there are no parentheses or exponents, so we can start by performing the multiplication and division. First, we need to convert the mixed numbers to improper fractions, so that we can easily multiply and divide them. \(\frac{3\frac{1}{4} \times 1\frac{3}{5}}{11\frac{1}{3} - 5 \frac{1}{3}} = \frac{\frac{13}{4} \times \frac{8}{5}}{\frac{32}{3}}\) Next, we can simplify the expression by canceling out common factors. \(\frac{\frac{13}{4} \times \frac{8}{5}}{\frac{32}{3}} = \frac{13 \times 2}{4 \times 5} = \frac{26}{20} = \frac{13}{10}\) Therefore, the answer is \(\frac{13}{10}\), which is equivalent to \(\frac{26}{20}\) and can be simplified to \(\frac{13}{15}\). So the correct option is: \(\frac{13}{15}\).

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There are three types of bottles in the crate: Coke, Sprite, and not Coke (which includes Sprite bottles). To find the probability that the first bottle is Coke and the second bottle is not Coke, we need to multiply two probabilities: the probability of selecting a Coke bottle first and the probability of selecting a not Coke bottle (i.e., a Sprite bottle) second. The probability of selecting a Coke bottle first is 8/24 because there are 8 Coke bottles in the crate out of a total of 24 bottles (8 Coke + 12 Sprite + 4 Sprite = 24). After selecting the first bottle, there will be 23 bottles left in the crate. If the first bottle was a Coke bottle, then there will be 7 Coke bottles and 12 Sprite bottles left in the crate. Therefore, the probability of selecting a not Coke (i.e., a Sprite) bottle second is 12/23. Multiplying these probabilities, we get: (8/24) * (12/23) = 96/552 ≈ 0.174 Therefore, the probability that the first bottle is Coke and the second bottle is not Coke is approximately 0.174, which is closest to option (C) 2/9.

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To calculate the length of PQ, we need to use the distance formula, which is: distance = square root of [(x2 - x1)^2 + (y2 - y1)^2] Here, P is at (2,3) and Q is at (2,5). So, x1=2, y1=3, x2=2, and y2=5. Plugging these values into the formula, we get: distance = square root of [(2 - 2)^2 + (5 - 3)^2] distance = square root of [0 + 4] distance = square root of 4 distance = 2 Therefore, the length of PQ is 2 units.

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The statement "It is raining if and only if it is cold" can be written as "q if and only if \(\sim\)p". This is because the "if and only if" connector implies that both statements are true, so it is equivalent to saying "if it is raining, then it is cold" and "if it is cold, then it is raining". We can represent this statement using the "if and only if" symbol (\(\iff\)) as: q \(\iff\) \(\sim\)p Therefore, the correct option is (D) q \(\iff\) \(\sim\)p.

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We can start by using the laws of exponents to rewrite \(2^{x+1}\) in terms of \(2^{-x}\), which is given by \(t\). Recall that \(a^{m+n} = a^m \times a^n\) and \(a^{-n} = \frac{1}{a^n}\). Therefore, we have: $$2^{x+1} = 2^x \times 2^1 = 2^x \times 2 = 2 \times 2^x$$ Next, we can substitute \(2^{-x} = t\) into this expression: $$2 \times 2^x = 2 \times \frac{1}{t} = \frac{2}{t}$$ So the answer is: \(\frac{2}{t}\).

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To simplify this expression, we can use the distributive property of multiplication and simplify the terms inside the parentheses first: \(\sqrt{2}(\sqrt{6} + 2\sqrt{2}) - 2\sqrt{3}\) = \(\sqrt{2} \times \sqrt{6} + \sqrt{2} \times 2\sqrt{2} - 2\sqrt{3}\) = \(\sqrt{12} + 2\sqrt{4} - 2\sqrt{3}\) = \(2\sqrt{3} + 4 - 2\sqrt{3}\) = \(4\) Therefore, the simplified expression is 4. In summary, we can simplify this expression by first using the distributive property to multiply the \(\sqrt{2}\) by the terms inside the parentheses, then simplifying the resulting expression by combining like terms.

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The simple interest on a certain amount of money saved in a bank for 5 years at 2\(\frac{1}{2}\)% per annum is N500.00. We can use the formula for simple interest to find the principal amount, which is the initial amount of money saved. Simple Interest (SI) = (Principal * Rate * Time) / 100 In this case, we know that the rate is 2\(\frac{1}{2}\)% per annum, which is equivalent to 0.025 as a decimal. We also know that the time is 5 years, and the simple interest is N500.00. We can plug these values into the formula and solve for the principal amount: 500 = (P * 0.025 * 5) / 100 P = 500 * 100 / (0.025 * 5) P = 4000 So the principal amount is N4,000.00. Now we need to calculate the total amount due after 6 years at the same rate. We can again use the formula for simple interest, but this time the time is 6 years instead of 5. Simple Interest (SI) = (Principal * Rate * Time) / 100 SI = (4000 * 0.025 * 6) / 100 SI = N600.00 So the simple interest for 6 years is N600.00. To find the total amount due after 6 years, we need to add the simple interest to the principal amount: Total amount due = Principal + Simple Interest Total amount due = 4000 + 600 Total amount due = N4,600.00 Therefore, the total amount due after 6 years at the same rate is N4,600.00.

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We can use the logarithmic identities to simplify the expression for log 35 in terms of a and b. Firstly, we can write 35 as the product of 7 and 5: 35 = 7 x 5 Next, we can use the logarithmic identity: log (a x b) = log a + log b to express log 35 in terms of log 7 and log 5 as follows: log 35 = log (7 x 5) = log 7 + log 5 Now, we need to express log 5 in terms of a and b. We can write 5 as the product of 2 and 2.5: 5 = 2 x 2.5 Using the logarithmic identity, we get: log 5 = log (2 x 2.5) = log 2 + log 2.5 We can express log 2.5 in terms of log 10 (which is equal to 1) and log 2 as follows: log 2.5 = log (2.5/1) = log (5/2) = log 5 - log 2 Substituting this into our expression for log 35, we get: log 35 = log 7 + log 5 = log 7 + (log 5 - log 2) Finally, we can substitute the given values of a and b into the expression above to obtain: log 35 = a + (log 5 - b) Simplifying further, we get: log 35 = a + log 5 - b Therefore, the answer is (c) a - b + 1.

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When expressing a number to a certain number of significant figures, you need to look at the digits that are significant and those that are not significant. Significant digits are those that carry meaning, while non-significant digits are just placeholders. In this case, the number 0.0000407 has 4 significant digits. To express it to 2 significant figures, you need to round it to the nearest value with 2 significant digits. The third significant digit in 0.0000407 is 7. Since it is greater than 5, we round up the second significant digit (which is 4) to 5. Therefore, the final answer, rounded to 2 significant figures, is 0.000041.

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We are asked to evaluate the expression xy - $\frac{y}{x}$, where x = $\frac{2}{3}$ and y = -6. First, we can find xy: xy = $\frac{2}{3}$ * -6 = -4 Next, we can find $\frac{y}{x}$: $\frac{y}{x}$ = -6 * $\frac{3}{2}$ = -9 Finally, we can subtract $\frac{y}{x}$ from xy: xy - $\frac{y}{x}$ = -4 - (-9) = -4 + 9 = 5 So the expression xy - $\frac{y}{x}$ = 5.

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To find the area of the path around the circular pond, we need to subtract the area of the inner circle from the area of the outer circle. The radius of the outer circle is the sum of the radius of the pond and the width of the path. Therefore, the radius of the outer circle is: 4m + 2.5m = 6.5m The area of the outer circle is given by: A_outer = π * r_outer^2 A_outer = π * (6.5m)^2 A_outer = 132.73\(m^2\) The area of the inner circle is simply: A_inner = π * r_inner^2 A_inner = π * (4m)^2 A_inner = 50.27\(m^2\) Therefore, the area of the path is: A_path = A_outer - A_inner A_path = 132.73\(m^2\) - 50.27\(m^2\) A_path = 82.46\(m^2\) Rounding to two decimal places, the area of the path is approximately 82.50\(m^2\). Therefore, the correct option is: - 82.50\(m^2\)

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Let's first understand what M and N represent. M is a set of numbers x that are greater than or equal to 3 but less than 8, while N is a set of numbers x that are greater than 8 but less than or equal to 12. i. 8 is not in the set M because M only includes numbers less than 8. Similarly, 8 is not in the set N because N only includes numbers greater than 8. Therefore, 8 is not in the intersection of M and N, making option i false. ii. The union of M and N includes all the numbers in both sets. Therefore, the union of M and N would include all numbers greater than or equal to 3 but less than or equal to 12. 8 is included in this range, so it must also be in the union of M and N. Therefore, option ii is true. iii. The intersection of M and N includes all the numbers that are in both sets. However, M and N do not overlap since there are no numbers that are both greater than or equal to 3 and less than or equal to 12. Therefore, the intersection of M and N is an empty set, making option iii true. Therefore, the correct answer is option iii only.

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The expression can be simplified as follows: First, we can factor out the common factor of \(ab\) from the numerator: \begin{align*} \frac{a^2 b^4 - b^2 a^4}{ab(a + b)} &= \frac{ab(a^2 b^3 - b^2 a^3)}{ab(a + b)} \\ &= \frac{ab(a^2 b^3 - b^2 a^3)}{ab(a + b)} \\ &= \frac{a^2 b^3 - b^2 a^3}{a + b}. \end{align*} Next, we can simplify the numerator: \begin{align*} a^2 b^3 - b^2 a^3 &= (a^2)(b^3) - (b^2)(a^3) \\ &= a^2b^2 (b - a) \\ &= ab^2 (a - b)(b + a) \\ &= ab^2 (a^2 - b^2). \end{align*} Therefore, the expression simplifies to: \[\frac{ab^2 (a^2 - b^2)}{a + b} = ab (a - b) = \boxed{ab^2 - a^2b}.\]

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The sum of all exterior angles of a polygon is 360 degrees. Exterior angles are formed by extending one side of a polygon to the next. So, the sum of the three exterior angles that we know is 30 + 40 + 60 = 130 degrees. And, if each of the remaining exterior angles measures 46 degrees, then there must be 360 - 130 = 230 degrees left. Dividing this by 46, we get 230/46 = 5. So, the polygon has a total of 5 + 3 = 8 sides. Therefore, the polygon is an octagon.

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The gradient or slope of a line is a measure of how steep it is, and is defined as the ratio of the vertical change between two points to the horizontal change between the same two points. To find the gradient of the line that passes through the points (-1, 1) and (2, -2), we need to calculate the difference in the y-coordinates (the vertical change) divided by the difference in the x-coordinates (the horizontal change). So, the gradient is: (y2 - y1) / (x2 - x1) = (-2 - 1) / (2 - (-1)) = -3 / 3 = -1. Therefore, the gradient of the line that passes through the points (-1, 1) and (2, -2) is -1.

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The total amount of money shared among 5 boys and 4 girls is N18,100.00. If each boy takes N20.00 more than each girl, then the difference between a boy's share and a girl's share is N20.00. We can use this information to find the total share for both the boys and girls. Let's call the girl's share "x". If a boy takes N20.00 more than a girl, then the boy's share is "x + N20.00". The total amount of money for the girls is 4 * x, and the total amount of money for the boys is 5 * (x + N20.00). We can use this information to set up an equation: 4 * x + 5 * (x + N20.00) = N18,100.00 Expanding the right side of the equation: 4 * x + 5 * x + 5 * N20.00 = N18,100.00 Combining like terms: 9 * x + 5 * N20.00 = N18,100.00 Subtracting 5 * N20.00 from both sides: 9 * x = N18,100.00 - 5 * N20.00 Calculating N18,100.00 - 5 * N20.00: 9 * x = N18,100.00 - 5 * N20.00 = N18,100.00 - N100.00 = N18,000.00 Dividing both sides by 9: x = N18,000.00 / 9 = N2,000.00 So each girl's share is N2,000.00, and each boy's share is N2,000.00 + N20.00 = N2,020.00.

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To find the new average age of the group, we need to calculate the sum of the ages of all the girls in the group after the new girl joins, and then divide by the total number of girls in the group. The sum of the ages of the original 25 girls is 25 * 10 = 250 years. When the new girl joins the group, the total number of girls becomes 26, and the sum of their ages becomes 250 + 12.33 (12 years and 4 months converted to decimal form) = 262.33 years. To find the new average age, we divide the sum of the ages by the number of girls: New average age = 262.33 / 26 ≈ 10.1 years Therefore, the new average age of the group is 10.1 years.

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To calculate the volume of water in the tank, we need to first calculate the total volume of the tank and then divide it by 2 (since the tank is half-filled with water). The volume of the tank can be calculated by multiplying its length, width, and height. Therefore, the total volume of the tank is: Volume of tank = Length × Width × Height = 13cm × 10cm × 70cm = 9100 cubic centimeters (or cm³) Since we want to find the volume of water in litres, we need to convert the volume from cubic centimeters to litres. There are 1000 cubic centimeters in one litre, so we can convert the volume by dividing it by 1000. Volume of tank in litres = Volume of tank in cubic centimeters ÷ 1000 = 9100 ÷ 1000 = 9.1 litres Finally, since the tank is half-filled with water, we need to divide the total volume by 2 to get the volume of water in the tank: Volume of water in tank = Volume of tank ÷ 2 = 9.1 ÷ 2 = 4.55 litres Therefore, the volume of water in the tank is 4.55 litres, which is option A.

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(a) Processing \(\text{N}87.00\) worth of raw materials needs 3 hours of labour. For \(\text{N}30{,}450.00\) worth:

\[\text{Hours} = \frac{30{,}450}{87}\times 3 = 350 \times 3 = 1050\text{ hours}.\]

Budget at \(\text{N}18.25\) per hour:

\[1050 \times 18.25 = \text{N}19{,}162.50.\]

(b) Let \(\text{N}x\) be invested at 6% and \(\text{N}y\) at 8%, with \(x + y = 8{,}000{,}000\) ... (1)

Simple interest over 4 years: \(I = \dfrac{P R T}{100}\).

• Bank M: \(\dfrac{x(6)(4)}{100} = 0.24x\)
• Bank N: \(\dfrac{y(8)(4)}{100} = 0.32y\)

Total interest: \(0.24x + 0.32y = 2{,}320{,}000\) ... (2)

From (1), \(x = 8{,}000{,}000 - y\). Substitute into (2):

\[0.24(8{,}000{,}000 - y) + 0.32y = 2{,}320{,}000.\]

\[1{,}920{,}000 + 0.08y = 2{,}320{,}000 \Rightarrow 0.08y = 400{,}000 \Rightarrow y = 5{,}000{,}000.\]

Then \(x = 8{,}000{,}000 - 5{,}000{,}000 = 3{,}000{,}000\).

(i) \(x = \text{N}3{,}000{,}000\) and \(y = \text{N}5{,}000{,}000\).

(ii) Interest paid by the second bank (bank N) \(= 0.32y = 0.32 \times 5{,}000{,}000 = \text{N}1{,}600{,}000\).

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(a) Finding m. Form the totals in terms of m.

\[\sum f=(m+2)+(m-1)+(2m-3)+(m+5)+(3m-4)=8m-1.\]

\[\sum fx=1(m+2)+2(m-1)+3(2m-3)+4(m+5)+5(3m-4)=28m-9.\]

The mean is \(3\frac{6}{23}=\frac{75}{23}\), so

\[\frac{28m-9}{8m-1}=\frac{75}{23}.\]

\[23(28m-9)=75(8m-1)\Rightarrow 644m-207=600m-75\Rightarrow 44m=132\Rightarrow m=3.\]

Substituting \(m=3\) gives the frequencies:

(b)(i) Interquartile range. \(N=23\). \(Q_1\) is the \(\frac{N+1}{4}=6\text{th}\) value \(=2\); \(Q_3\) is the \(\frac{3(N+1)}{4}=18\text{th}\) value \(=4\).

\[\text{IQR}=Q_3-Q_1=4-2=2.\]

(b)(ii) Probability of at least 4 marks.

\[P(\ge4)=\frac{8+5}{23}=\frac{13}{23}.\]

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(a) Solve the equation.

Multiply every term by \(15\) (the LCM of \(3\) and \(5\)):

(b) Find \(m+n\).

From the diagram, \(P,Q,R\) lie on a straight line and \(P,U,T,S\) lie on a straight line, with \(\angle STQ=m\), \(\angle TUQ=80^{\circ}\), \(\angle UPQ=r\), \(\angle PQU=n\) and \(\angle RQT=88^{\circ}\).

At \(U\): \(P,U,T\) are collinear, so \(\angle QUP\) and \(\angle QUT\) are angles on a straight line:

At \(Q\): \(P,Q,R\) are collinear, so the three angles on that line satisfy

Triangle \(UQT\):

At \(T\): \(U,T,S\) are collinear, so \(\angle UTQ\) and \(\angle STQ(=m)\) are angles on a straight line:

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(a) Given \((y - 1)\log_{10}4 = y\log_{10}16\).

Since \(16 = 4^2\), we have \(\log_{10}16 = 2\log_{10}4\). Substituting:

\[(y - 1)\log_{10}4 = y(2\log_{10}4).\]

Divide both sides by \(\log_{10}4\) (which is not zero):

\[y - 1 = 2y \Rightarrow -1 = y \Rightarrow y = -1.\]

(b) Let the distance from house to office be \(d\) km. Time at 4 km/h is \(\dfrac{d}{4}\) h; time at 5 km/h is \(\dfrac{d}{5}\) h. Walking at the slower speed takes 30 minutes \(\left(= \tfrac{1}{2}\text{ h}\right)\) longer:

\[\frac{d}{4} - \frac{d}{5} = \frac{1}{2}.\]

\[d\left(\frac{5 - 4}{20}\right) = \frac{1}{2} \Rightarrow \frac{d}{20} = \frac{1}{2} \Rightarrow d = 10.\]

The distance between the house and the office is \(\mathbf{10\text{ km}}\).

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(a) Completing the table for \(y = 2x^{2}-7x-9\).

(b) Graph.

Using the scales of 2 cm to 1 unit on the x-axis and 2 cm to 4 units on the y-axis, plot the points from the table and join them with a smooth curve.

(c)

(i) Since

\[ 2x^{2}-7x=26 \Rightarrow y+9=26 \Rightarrow y=17, \]

draw the line \(y=17\) on the graph. The roots are approximately:

\[ x=-2.2 \quad \text{and} \quad x=5.8. \]

(ii) The minimum point of the curve is approximately:

\[ (1.8,\,-15). \]

(iii)

\[ 2x^{2}-7x<9 \]

is equivalent to

\[ 2x^{2}-7x-9<0, \]

that is, \(y<0\). The curve lies below the x-axis between \(x=-1\) and \(x=4.5\). Hence,

\[ -1<x<4.5. \]

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Let \(b\) be the number who bought bags and \(s\) the number who bought shoes. Every customer bought at least one item, so \(n(\text{bags} \cup \text{shoes}) = 120\), with \(n(\text{both}) = 45\).

Using \(n(B \cup S) = b + s - n(\text{both})\):

\[120 = b + s - 45 \Rightarrow b + s = 165.\]

Also 11 more bought shoes than bags: \(s = b + 11\).

(a) Diagram. A two-circle Venn diagram (Bags and Shoes) with the overlap \(= 45\), bags-only \(= b - 45\), shoes-only \(= s - 45\).

(b) Substitute \(s = b + 11\) into \(b + s = 165\):

\[b + (b + 11) = 165 \Rightarrow 2b = 154 \Rightarrow b = 77, \quad s = 88.\]

The number who bought shoes is \(\mathbf{88}\).

(c) The number who bought bags is 77, so

\[P(\text{bought bags}) = \frac{77}{120}.\]

Checking the regions: bags-only \(= 32\), both \(= 45\), shoes-only \(= 43\); total \(= 120\).

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(a) Solve \(3y^2 - 5y + 2 = 0\) by completing the square. Divide through by 3:

\[y^2 - \frac{5}{3}y + \frac{2}{3} = 0 \Rightarrow y^2 - \frac{5}{3}y = -\frac{2}{3}.\]

Half the coefficient of \(y\) is \(\dfrac{5}{6}\); add \(\left(\dfrac{5}{6}\right)^2 = \dfrac{25}{36}\) to both sides:

\[\left(y - \frac{5}{6}\right)^2 = -\frac{2}{3} + \frac{25}{36} = \frac{-24 + 25}{36} = \frac{1}{36}.\]

\[y - \frac{5}{6} = \pm\frac{1}{6} \Rightarrow y = \frac{5}{6} + \frac{1}{6} = 1 \quad\text{or}\quad y = \frac{5}{6} - \frac{1}{6} = \frac{2}{3}.\]

Hence \(y = 1.00\) or \(y = 0.67\) (2 d.p.).

(b) Let \(N = \begin{pmatrix} m & x \\ n & y \end{pmatrix}\). Then

\[MN = \begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix}\begin{pmatrix} m & x \\ n & y \end{pmatrix} = \begin{pmatrix} m + 2n & x + 2y \\ 4m + 3n & 4x + 3y \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix}.\]

First column: \(m + 2n = 2\) and \(4m + 3n = 3\). From the first, \(m = 2 - 2n\); substituting: \(4(2 - 2n) + 3n = 3 \Rightarrow 8 - 5n = 3 \Rightarrow n = 1,\; m = 0\).

Second column: \(x + 2y = 1\) and \(4x + 3y = 4\). From the first, \(x = 1 - 2y\); substituting: \(4(1 - 2y) + 3y = 4 \Rightarrow 4 - 5y = 4 \Rightarrow y = 0,\; x = 1\).

\[N = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.\]

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(a) The number of tanks filled varies jointly with the number of students and the time worked: tanks \(= k \times (\text{students})\times(\text{time})\).

From the given data, 8 students in \(\tfrac{2}{3}\) h fill 12 tanks:

\[12 = k \times 8 \times \frac{2}{3} = \frac{16k}{3} \Rightarrow k = \frac{36}{16} = 2.25.\]

For 4 students in \(\tfrac{1}{3}\) h:

\[\text{tanks} = 2.25 \times 4 \times \frac{1}{3} = 2.25 \times \frac{4}{3} = 3.\]

So \(\mathbf{3}\) tanks are filled.

(b) A chord of length 20 cm is 12 cm from the centre. The perpendicular from the centre bisects the chord, giving a right triangle with legs 12 and 10.

Radius: \(r = \sqrt{12^2 + 10^2} = \sqrt{244} = 15.62\text{ cm}\).

(i) If \(\theta\) is the angle subtended at the centre, then in the right triangle \(\tan\left(\tfrac{\theta}{2}\right) = \dfrac{10}{12}\):

\[\frac{\theta}{2} = 39.8^\circ \Rightarrow \theta = 79.6^\circ.\]

(ii) The perimeter of the minor segment \(=\) chord \(+\) arc length. Arc length \(= \dfrac{\theta}{360^\circ}\times 2\pi r\):

\[\text{arc} = \frac{79.6}{360}\times 2(3.142)(15.62) = 0.2211 \times 98.16 = 21.7\text{ cm}.\]

\[\text{Perimeter} = 20 + 21.7 = 41.7\text{ cm}.\]

Awọn alaye Idahun

For an A.P. with first term \(a\) and common difference \(d\):

Sixth term: \(a + 5d = 37\) ... (1)

Sum of first six terms: \(S_6 = \dfrac{6}{2}(2a + 5d) = 3(2a + 5d) = 147\), so \(2a + 5d = 49\) ... (2)

(a) Subtract (1) from (2):

\[(2a + 5d) - (a + 5d) = 49 - 37 \Rightarrow a = 12.\]

The first term is \(a = \mathbf{12}\).

From (1): \(12 + 5d = 37 \Rightarrow 5d = 25 \Rightarrow d = 5\).

(b) Sum of the first fifteen terms:

\[S_{15} = \frac{15}{2}\big(2a + 14d\big) = \frac{15}{2}\big(2(12) + 14(5)\big) = \frac{15}{2}(24 + 70) = \frac{15}{2}(94) = 705.\]

Therefore \(S_{15} = \mathbf{705}\).

Awọn alaye Idahun

(a) Given \(\sin x = \dfrac{5}{13}\) with \(0^\circ < x < 90^\circ\). Using a 5-12-13 right triangle (\(\sqrt{13^2 - 5^2} = 12\)):

\[\cos x = \frac{12}{13}, \qquad \tan x = \frac{5}{12}.\]

Now evaluate \(\dfrac{\cos x - 2\sin x}{2\tan x}\):

\[\cos x - 2\sin x = \frac{12}{13} - \frac{10}{13} = \frac{2}{13}, \qquad 2\tan x = \frac{10}{12} = \frac{5}{6}.\]

\[\frac{\cos x - 2\sin x}{2\tan x} = \frac{2/13}{5/6} = \frac{2}{13}\times\frac{6}{5} = \frac{12}{65}.\]

(b) The point \(L\) is 9.6 m up the vertical pole, with foot \(F\). Ladder \(LB = 12\) m reaches the ground at \(B\); \(A\) and \(B\) are 10 m apart on opposite sides of the pole.

Horizontal distance \(FB\):

\[FB = \sqrt{12^2 - 9.6^2} = \sqrt{144 - 92.16} = \sqrt{51.84} = 7.2\text{ m}.\]

Since \(A\) and \(B\) are on opposite sides, \(FA + FB = 10\), so \(FA = 10 - 7.2 = 2.8\text{ m}\).

(i) \(LA = \sqrt{FA^2 + 9.6^2} = \sqrt{2.8^2 + 9.6^2} = \sqrt{7.84 + 92.16} = \sqrt{100} = 10\text{ m}\) (2 s.f.).

(ii) Let \(\theta\) be the angle \(LA\) makes with the ground:

\[\tan\theta = \frac{9.6}{2.8} = 3.4286 \Rightarrow \theta = 73.7^\circ \approx 74^\circ.\]

Awọn alaye Idahun

(a) The operation is \(x * y = \dfrac{x + y}{2}\).

(i) \(3 * \dfrac{2}{5} = \dfrac{3 + \tfrac{2}{5}}{2} = \dfrac{\tfrac{17}{5}}{2} = \dfrac{17}{10} = 1\tfrac{7}{10}\).

(ii) \(8 * y = 8\tfrac{1}{4}\) means \(\dfrac{8 + y}{2} = \dfrac{33}{4}\).

\[8 + y = \frac{33}{2} = 16.5 \Rightarrow y = 8.5 = 8\tfrac{1}{2}.\]

(b) Take \(A\) as the reference point. Then \(\overline{AB} = \begin{pmatrix} -4 \\ 6 \end{pmatrix}\) and \(\overline{AC} = \begin{pmatrix} 3 \\ -8 \end{pmatrix}\).

\(P\) is the midpoint of \(\overline{AB}\), so

\[\overline{AP} = \frac{1}{2}\overline{AB} = \begin{pmatrix} -2 \\ 3 \end{pmatrix}.\]

Then

\[\overline{CP} = \overline{AP} - \overline{AC} = \begin{pmatrix} -2 \\ 3 \end{pmatrix} - \begin{pmatrix} 3 \\ -8 \end{pmatrix} = \begin{pmatrix} -5 \\ 11 \end{pmatrix}.\]

Therefore \(\overline{CP} = \begin{pmatrix} -5 \\ 11 \end{pmatrix}\).

Awọn alaye Idahun

(a) Tangent \(PQ\) touches the circle \(RST\) at \(S\), with \(PRT\) a straight secant, \(\angle TPS=34^\circ\) and \(\angle TSQ=65^\circ\).

(i) Diagram.

(ii)(a) Finding \(\angle RTS\). Since \(P\), \(S\) and \(Q\) lie on the tangent line, \(\angle TSQ=65^\circ\) is the exterior angle of triangle \(PST\) at \(S\). An exterior angle equals the sum of the two interior opposite angles, \(\angle TPS\) and \(\angle PTS\). Because \(R\) lies on \(PT\), the angle \(\angle PTS\) is the same as \(\angle RTS\):

\[ \angle TSQ=\angle TPS+\angle RTS \]

\[ 65^\circ=34^\circ+\angle RTS\ \Rightarrow\ \angle RTS=31^\circ. \]

(ii)(b) Finding \(\angle SRP\). By the alternate segment theorem, the tangent-chord angle \(\angle TSQ\) equals the inscribed angle in the alternate segment standing on chord \(ST\), which is \(\angle SRT\):

\[ \angle SRT=\angle TSQ=65^\circ. \]

Since \(PRT\) is a straight line, \(\angle SRP\) and \(\angle SRT\) are angles on a straight line:

\[ \angle SRP=180^\circ-\angle SRT=180^\circ-65^\circ=115^\circ. \]

(b) In the given diagram \(VZ=YZ\), \(\angle YXZ=20^\circ\) and \(\angle ZVY=52^\circ\); find \(\angle WYZ\).

Triangle \(VZY\) has \(VZ=YZ\), so it is isosceles and its base angles at \(V\) and \(Y\) are equal:

\[ \angle ZYV=\angle ZVY=52^\circ. \]

The inscribed angle \(\angle ZVY=52^\circ\) stands on chord \(ZY\), so the arc \(ZY\) it cuts off is

\[ \text{arc } ZY=2\times52^\circ=104^\circ. \]

\(X\) is an external point with the two secants \(XVY\) and \(XWZ\). The angle between two secants from an external point equals half the difference of the two intercepted arcs:

\[ \angle YXZ=\tfrac{1}{2}\big(\text{arc } YZ-\text{arc } VW\big) \]

\[ 20^\circ=\tfrac{1}{2}\big(104^\circ-\text{arc } VW\big)\ \Rightarrow\ \text{arc } VW=64^\circ. \]

Equal chords cut off equal arcs, so \(VZ=YZ\) gives \(\text{arc } VWZ=\text{arc } YZ=104^\circ\). Since \(W\) lies on the arc between \(V\) and \(Z\),

\[ \text{arc } WZ=\text{arc } VWZ-\text{arc } VW=104^\circ-64^\circ=40^\circ. \]

Finally, \(\angle WYZ\) is the inscribed angle at \(Y\) standing on chord \(WZ\), so it is half of arc \(WZ\):

\[ \angle WYZ=\tfrac{1}{2}\times40^\circ=20^\circ. \]

Awọn alaye Idahun

(a) Let the building have height \(h\) and foot \(B\), with \(P\) on the ground \(50\ \text{m}\) from \(B\). The angle of depression from the top \(T\) to \(P\) equals the angle of elevation from \(P\) to \(T\) (alternate angles), so this angle at \(P\) is \(23.6^\circ\).

Height \(\approx 22\ \text{m}\) (to the nearest metre).

(b) From the diagram, \(PT \parallel SU\) and \(QS \parallel TR\), with \(S\), \(R\), \(U\) on the bottom line where \(|SR| = 6\ \text{cm}\) and \(|RU| = 10\ \text{cm}\).

Find the height between the parallel lines. Triangle \(TRU\) has base \(RU = 10\ \text{cm}\) on the bottom line, and apex \(T\) on the top line, so its height is the perpendicular distance \(H\) between \(PT\) and \(SU\):

Find \(QT\). Since \(QT \parallel SR\) (both on the parallel top and bottom lines) and \(QS \parallel TR\), the figure \(QTRS\) is a parallelogram, so:

Length of \(SU\):

Area of trapezium \(QTUS\) (parallel sides \(QT = 6\ \text{cm}\) and \(SU = 16\ \text{cm}\), height \(H = 9\ \text{cm}\)):

Area of trapezium \(QTUS = 99\ \text{cm}^2\)