WAEC SSCE - General Mathematics - 2009

Akọwa Nkọwa

In the given diagram, we have a circle with center O and angle 20^o subtended at the center. We need to find the value of x. Firstly, we know that the angle subtended at the center of a circle is twice the angle subtended at the circumference by the same arc. Therefore, angle AOB = 2 × 20^o = 40^o Also, we know that angle in a semicircle is a right angle. So, angle AOC = 90^o. Using the fact that the angles in a triangle add up to 180^o, we can find angle BOC as follows: angle BOC = 180^o - angle AOB - angle AOC = 180^o - 40^o - 90^o = 50^o Since angle BOC is an angle at the circumference that subtends the arc BC, which is equal to x degrees, we have: x = angle BOC = 50^o Therefore, the value of x is 50^o. Answer is correct.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Two sets are said to be disjoint if their intersection is an empty set. In other words, if there are no common elements between the sets, they are said to be disjoint. For example, the sets {1,2,3} and {4,5,6} are disjoint since their intersection is an empty set {} or ∅. On the other hand, the sets {1,2,3} and {2,3,4} are not disjoint since they have elements in common, namely 2 and 3.

Akọwa Nkọwa

When an angle in degrees is subtended at the center of a circle, the length of the arc it cuts out is given by: $$\text{Length of arc} = \frac{\text{angle}}{360^\circ} \times 2\pi r$$ where r is the radius of the circle. In this case, the angle is 60° and the radius is 3 cm. Substituting these values into the formula above, we get: $$\text{Length of arc} = \frac{60}{360^\circ} \times 2\pi (3\text{ cm}) = \frac{1}{6} \times 6\pi = \pi \text{ cm}$$ Therefore, the length of the arc, in terms of π, is π cm. Hence, the correct option is: \(\pi\)cm.

Akọwa Nkọwa

We will start by simplifying the expression inside the parenthesis first. 3\(\frac{1}{2}\) + 4\(\frac{1}{3}\) = (7/2) + (13/3) To add these two fractions, we need a common denominator. Multiplying the denominators together gives us 6, so: (7/2) + (13/3) = (21/6) + (26/6) = 47/6 Now, let's simplify the expression in the denominator: \(\frac{5}{6} - \frac{2}{3}\) = \(\frac{5}{6} - \frac{4}{6}\) = \(\frac{1}{6}\) Finally, we can substitute these values into the original expression: (3\(\frac{1}{2}\) + 4\(\frac{1}{3}\)) ÷ (\(\frac{5}{6}\) - \(\frac{2}{3}\)) = (47/6) ÷ (1/6) When dividing fractions, we can multiply the first fraction by the reciprocal of the second: (47/6) ÷ (1/6) = (47/6) x (6/1) = 47 Therefore, the answer is 47.

Akọwa Nkọwa

Akọwa Nkọwa

The volume of the reservoir can be found by multiplying its dimensions: 8m x 7m x 5m = 280 m^3 Since 1 liter is equal to 1000 cubic centimeters (cm^3), 1 cubic meter is equal to 1,000,000 cubic centimeters. Therefore, the reservoir has a volume of: 280 m^3 x 1,000,000 cm^3/m^3 = 280,000,000 cm^3 Each bucket can hold 10 liters of water or 10,000 cubic centimeters (cm^3) of water since 1 liter is equal to 1000 cm^3. Therefore, the number of buckets needed to fill the reservoir is: 280,000,000 cm^3 ÷ 10,000 cm^3/bucket = 28,000 buckets Therefore, the answer is 28,000.

Akọwa Nkọwa

To find the quadratic equation given its roots, we use the fact that for a quadratic equation of the form ax² + bx + c = 0, the roots are given by the formula: x = (-b ± √(b² - 4ac)) / 2a If the roots are given as α and β, then the quadratic equation can be written as: (x - α)(x - β) = 0 Expanding the above equation gives: x² - (α + β)x + αβ = 0 Therefore, to find the quadratic equation whose roots are -\(\frac{1}{2}\) and 3, we substitute α = -\(\frac{1}{2}\) and β = 3 into the equation: x² - (α + β)x + αβ = 0 x² - (-\(\frac{1}{2}\) + 3)x + (-\(\frac{1}{2}\) × 3) = 0 Simplifying the above equation, we get: 2x² - 5x - 3 = 0 Therefore, the quadratic equation whose roots are -\(\frac{1}{2}\) and 3 is 2x² - 5x - 3 = 0. The correct option is (C) 2x² - 5x - 3 = 0.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Simple interest is calculated as the product of the principal, the rate of interest, and the time duration. From the question, we have a principal of N3000.00, an interest rate of 6%, and a duration of 5 years. Using the formula for simple interest, we can find the interest accrued over the 5 years as: Interest = (P * R * T) / 100 = (3000 * 6 * 5) / 100 = N900.00 The total value to which N3000.00 will amount to after 5 years is the sum of the principal and the interest, which is: Total = Principal + Interest = 3000 + 900 = N3900.00 Therefore, N3000.00 will amount to N3900.00 after 5 years at 6% per annum simple interest. So the correct answer is option A, N3,900.00.

Akọwa Nkọwa

To find the number of students who graduated from the science department, we need to look at the pie chart and determine the percentage of students that belong to that department. From the pie chart, we can see that the science department occupies 20% of the entire circle. To find the number of students in the science department, we need to calculate 20% of 4320, which can be done by multiplying 4320 by 0.20. 20% of 4320 = 0.20 x 4320 = 864 Therefore, the number of students who graduated from the science department is 864. So, the correct answer is option B: 864.

Akọwa Nkọwa

To express 2.7864 x 10^-3 to 2 significant figures, we need to round off the number to the second significant digit from the left. Since the first significant digit is 2, and the second significant digit is 7, we will look at the third digit, which is 8. Since 8 is greater than or equal to 5, we round up the second significant digit (7) by 1. Therefore, the number becomes 0.0028. So, the answer is 0.0028.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

In a regular polygon with n sides, each interior angle measures: \begin{align*} 180^\circ - \frac{360^\circ}{n} \end{align*} Since each interior angle of this polygon is 140^o, we can equate the above formula to 140^o: \begin{align*} 180^\circ - \frac{360^\circ}{n} &= 140^\circ\\ \frac{360^\circ}{n} &= 40^\circ\\ n &= \frac{360^\circ}{40^\circ}\\ n &= 9 \end{align*} Therefore, the given polygon is a nonagon (a polygon with nine sides). The sum of the interior angles of a polygon with n sides can be calculated using the formula: \begin{align*} S &= (n-2)180^\circ \end{align*} Substituting n=9 into this formula, we get: \begin{align*} S &= (9-2)180^\circ\\ &= 7\cdot180^\circ\\ &= 1260^\circ \end{align*} Therefore, the sum of all the interior angles of the given polygon is 1260^o. Hence, the correct option is: - 1260^o

Akọwa Nkọwa

To find the median of a set of numbers, we first need to arrange them in ascending or descending order. In this case, arranging the numbers in ascending order gives us: 22, 35, 41, 63, 74, 82 The median is the middle number in the set when the numbers are arranged in order. Since we have an even number of numbers in this set, there are two middle numbers: 41 and 63. To find the median, we take the average of these two numbers: Median = (41 + 63) / 2 = 104 / 2 = 52 Therefore, the median of the scores is 52.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

To compare these numbers, we need to convert all of them to the same base. Let's convert them all to base 10 for simplicity: - 22_three = 2\*3¹ + 2\*3⁰ = 6 + 2 = 8 - 34_five = 3\*5¹ + 4\*5⁰ = 15 + 4 = 19 - 21_six = 2\*6¹ + 1\*6⁰ = 12 + 1 = 13 So, in base 10, the numbers are 8, 19, and 13. To arrange them in descending order of magnitude, we simply sort them from largest to smallest: - 19, 13, 8 Therefore, the correct answer is: 34_five, 21_six, 22_three

Akọwa Nkọwa

We can use logarithmic rules to solve this question. Let's first express log(\(\frac{28}{3}\)) in terms of x, y, and z. log(\(\frac{28}{3}\)) = log(28) - log(3) We know that: log(28) = log(4 × 7) = log(4) + log(7) = 2x + z log(3) = y Therefore: log(\(\frac{28}{3}\)) = 2x + z - y So the correct answer is 2x + z - y.

Akọwa Nkọwa

Kofi's mean score for the three subjects was 55%, so the total percentage score for the three subjects is 3 x 55 = 165%. Let's assume that Kofi scored x% in Physics. Then, the total percentage score for Physics, Chemistry and Biology would be: x + 50 + 70 = 165 Simplifying the equation, we get: x = 165 - 50 - 70 x = 45 Therefore, Kofi scored 45% in Physics. So the correct option is (B) 45.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

To solve the equation \(\frac{2x + 1}{6} - \frac{3x - 1}{4} = 0\), we need to simplify the left-hand side and solve for x. First, we need to find a common denominator for the two fractions. The smallest common multiple of 6 and 4 is 12, so we can rewrite the equation as: \[\frac{2x+1}{6}\cdot \frac{2}{2} - \frac{3x-1}{4}\cdot \frac{3}{3} = 0\] Simplifying, we get: \[\frac{4x+2}{12} - \frac{9x-3}{12} = 0\] Combining the fractions, we get: \[\frac{4x+2-(9x-3)}{12} = 0\] Simplifying further, we get: \[\frac{-5x+5}{12} = 0\] Multiplying both sides by 12, we get: \[-5x+5=0\] Adding 5 to both sides, we get: \[-5x=-5\] Dividing both sides by -5, we get: \[x=1\] Therefore, the solution to the equation is x = 1.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

We can solve this problem using a proportion. If 85% of x is N3230, that means we can write: 0.85x = N3230 To solve for x, we need to isolate x on one side of the equation. We can do this by dividing both sides by 0.85: x = N3230 ÷ 0.85 Using a calculator, we get: x ≈ N3800.00 Therefore, the value of x is N3800.00.

Akọwa Nkọwa

Akọwa Nkọwa

The distance covered by the wheel is equal to the circumference of the wheel multiplied by the number of revolutions made. The circumference of the wheel can be calculated using the formula: circumference = diameter x pi circumference = 42cm x 22/7 circumference = 132cm Therefore, the distance covered by the wheel in 16 complete revolutions is: distance = circumference x number of revolutions distance = 132cm x 16 distance = 2112cm So, the answer is (c) 2112cm.

Akọwa Nkọwa

Akọwa Nkọwa

We are given that (2x + 3)³ = 125. We can rewrite 125 as 5³. Therefore, we have: (2x + 3)³ = 5³ Taking the cube root of both sides, we get: 2x + 3 = 5 Solving for x, we get: 2x = 2 x = 1 Therefore, the value of x is 1. Answer: (a) 1.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

The exchange rate of naira (N) varies directly with the Dollars($). This means that the exchange rate is constant, and we can set up a proportion to solve the problem. If N1122 is equivalent to $8, we can write: N1122/$8 = x/$36 where x is the Naira equivalent of $36. To solve for x, we can cross-multiply and simplify: N1122 x 36 = $8 x x 40392 = 8x x = 40392/8 = 5049 Therefore, the Naira equivalent of $36 is N5049. So, the correct option is (B) N5049.

Akọwa Nkọwa

Akọwa Nkọwa

The probability of picking a black ball at random from the box is \(\frac{3}{10}\) and the probability of picking a white ball at random is \(\frac{2}{5}\). Let's assume that there are x red balls in the box. Since there are a total of 30 balls, we can write: number of black balls + number of white balls + number of red balls = total number of balls \(\frac{3}{10}(30) + \frac{2}{5}(30) + x = 30\) Simplifying the above equation gives: 9 + 12 + x = 30 x = 30 - 9 - 12 x = 9 Therefore, there are 9 red balls in the box.

Akọwa Nkọwa

Akọwa Nkọwa

To find the probability that a student picked at random is not from the education department, we need to find the total number of students who are not from the education department and divide it by the total number of students in all departments. From the pie chart, we can see that the education department has 30% of the total students. Therefore, the remaining three departments have a total of 70% of the total students. To find the probability of picking a student who is not from the education department, we divide the percentage of students who are not in the education department by 100%: Probability = \(\frac{70}{100}\) = \(\frac{7}{10}\) Therefore, the probability that a student picked at random is not from the education department is \(\frac{7}{10}\). The correct option is: \(\frac{7}{10}\).

Akọwa Nkọwa

(a) For each value of x, evaluate \(y=\sin x+2\cos x\), correct to one decimal place.

\(x\)	0°	30°	60°	90°	120°	150°	180°	210°	240°
\(y\)	2.0	2.2	1.9	1.0	−0.1	−1.2	−2.0	−2.2	−1.9

(b) The plotted graph is shown below. The points are joined with a smooth curve.

(c)

(i) The x-intercept of the curve gives

\[x\approx117^\circ.\]

(ii) Draw the horizontal line \(y=2.1\). Its intersections with the curve give

\[x\approx9^\circ\quad\text{or}\quad x\approx45^\circ.\]

(d) Reading the ordinate at \(x=171^\circ\) from the graph gives

\[y\approx-1.75.\]

Akọwa Nkọwa

(a) The woman is at a window \(30\text{ m}\) above the ground. The foot of the flag pole is \(25\text{ m}\) horizontally from the foot of the building, on the same level ground.

(i) Angle of depression of the foot of the pole. From the window, the horizontal distance to the pole is \(25\text{ m}\) and the vertical drop to the foot of the pole is the full window height \(30\text{ m}\):

(ii) Height of the flag pole. The angle of depression of the top of the pole is \(44^\circ\). Over the horizontal distance \(25\text{ m}\), the vertical drop from window level down to the top of the pole is

So the top of the pole is \(30-24.14=5.86\text{ m}\) above the ground. Since the pole stands on the same ground, its height is

(b) \(O\) is the centre of the circle, \(\angle OQR=32^\circ\) and \(\angle TPQ=15^\circ\).

(i) \(\angle QPR\): \(OQ\) and \(OR\) are radii, so triangle \(OQR\) is isosceles and

\(\angle QOR\) is the angle subtended by chord \(QR\) at the centre, and \(\angle QPR\) is subtended by the same chord at the circumference, so it is half:

(ii) \(\angle TQO\): \(\angle TPQ=15^\circ\) stands on chord \(TQ\), so the central angle on the same chord is

\(OT\) and \(OQ\) are radii, so triangle \(TOQ\) is isosceles:

Akọwa Nkọwa

(a) A circle inscribed in a square has diameter equal to the side, so side \(= 2r.\) Then \[\text{perimeter of square} + \text{circumference} = 4(2r) + 2\pi r = 100.\] \[8r + 2\times\frac{22}{7}r = 100 \Rightarrow 8r + \frac{44}{7}r = 100 \Rightarrow \frac{56r + 44r}{7} = 100.\] \[\frac{100r}{7} = 100 \Rightarrow r = \mathbf{7\,\text{cm}}.\]

(b) Perimeter of rectangle \(= 60\,\text{cm}\), with length \(= 4\times\)breadth. \[2(l + b) = 60 \Rightarrow l + b = 30,\quad l = 4b \Rightarrow 5b = 30 \Rightarrow b = 6.\]

(i) Length \(l = 4(6) = \mathbf{24\,\text{cm}}.\)

(ii) Diagonal \(= \sqrt{l^2 + b^2} = \sqrt{24^2 + 6^2} = \sqrt{576 + 36} = \sqrt{612} \approx \mathbf{24.7\,\text{cm}}.\)

Akọwa Nkọwa

(a) Construction

1. Draw PQ = 10 cm.
2. At Q, construct an angle of 60° to QP. On this ray, mark R such that QR = 8 cm.
3. At P, construct an angle of 75° to PQ. On this ray, mark S such that PS = 6 cm.
4. Join R to S to complete quadrilateral PQRS.
5. Bisect the angle between QR and RS at R. The straight line through R obtained is the locus l₁ of points equidistant from QR and RS.
6. Construct the perpendicular bisector of RS. This is the locus l₂ of points equidistant from R and S.

(b) From the construction,

\[|RS| = 4.4\text{ cm}.\]

Akọwa Nkọwa

(a)

Taking the LCM, \, \((x-2)(x-1)\),

\[\begin{aligned} \frac{x+2}{x-2}-\frac{x+3}{x-1} &=\frac{(x+2)(x-1)-(x+3)(x-2)}{(x-2)(x-1)}\\ &=\frac{(x^2+x-2)-(x^2+x-6)}{(x-2)(x-1)}\\ &=\frac{4}{(x-2)(x-1)}. \end{aligned}\]

Hence, \(\displaystyle \frac{x+2}{x-2}-\frac{x+3}{x-1}=\frac{4}{(x-2)(x-1)}\), where \(x\ne 1,2\).

(b)

Given \(y=Ax^2+Bx+C\) and that the graph passes through \((0,0)\):

\[0=A(0)^2+B(0)+C,\]

therefore,

\[\boxed{C=0}.\]

Using \((1,4)\) and \((2,10)\):

\[\begin{aligned} A+B&=4 \qquad \text{(1)}\\ 4A+2B&=10 \qquad \text{(2)} \end{aligned}\]

Twice (1) gives \(2A+2B=8\). Subtracting this from (2),

\[2A=2\quad\Rightarrow\quad A=1.\]

From \(A+B=4\),

\[1+B=4\quad\Rightarrow\quad B=3.\]

Thus the equation of the graph is

\[\boxed{y=x^2+3x}.\]

The plotted curve is shown below.

To find the other point where the graph cuts the \(x\)-axis, put \(y=0\):

\[\begin{aligned} x^2+3x&=0\\ x(x+3)&=0. \end{aligned}\]

Hence \(x=0\) or \(x=-3\). Since \((0,0)\) is already one intercept, the other point is

\[\boxed{(-3,0)}.\]

Akọwa Nkọwa

(a) \(9\cos x-7=1\Rightarrow 9\cos x=8\Rightarrow \cos x=\dfrac{8}{9}=0.8889\).

For \(0^{\circ}\le x\le 90^{\circ}\): \(x=\cos^{-1}(0.8889)\approx \mathbf{27.3^{\circ}}\).

(b) \(7x<2x-13 \Rightarrow 7x-2x<-13 \Rightarrow 5x<-13 \Rightarrow x<-2.6\).

Since \(x\) is an integer, the values are \(\ldots,-5,-4,-3\). The three greatest are \(\mathbf{-3,\ -4,\ -5}\).

Akọwa Nkọwa

(a) \(n=30\); degrees \(=17\), diplomas \(=15\), neither \(=4\), so \(30-4=26\) have at least one.

\(|D\cup P|=|D|+|P|-|D\cap P| \Rightarrow 26=17+15-|D\cap P| \Rightarrow |D\cap P|=32-26=\mathbf{6}\).

Six candidates have both a degree and a diploma.

(b) In \(\triangle PQR\), \(M\) on \(PQ\), \(N\) on \(PR\), \(MN\parallel QR\); \(\angle PRQ=75^{\circ}\), \(PN=QN\), \(\angle PNQ=125^{\circ}\).

Triangle \(PNQ\) is isosceles (\(PN=QN\)), so \(\angle NPQ=\angle NQP\).

\(\angle NPQ+\angle NQP+\angle PNQ=180^{\circ}\Rightarrow 2\angle NPQ=180^{\circ}-125^{\circ}=55^{\circ}\Rightarrow \angle NPQ=\angle NQP=27.5^{\circ}\).

In \(\triangle PQR\): \(\angle P=27.5^{\circ}\), \(\angle R=75^{\circ}\), so \(\angle PQR=180^{\circ}-27.5^{\circ}-75^{\circ}=77.5^{\circ}\).

(i) \(\angle NQR=\angle PQR-\angle NQP=77.5^{\circ}-27.5^{\circ}=\mathbf{50^{\circ}}\).

(ii) \(\angle NPM\) is the angle at \(P\) between \(PM\) (on \(PQ\)) and \(PN\) (on \(PR\)), i.e. \(\angle QPR=\mathbf{27.5^{\circ}}\).

Akọwa Nkọwa

(a) Frequency distribution table

(b) Mean \[\bar{x} = \frac{\sum fx}{\sum f} = \frac{2535}{50} = \mathbf{50.7}.\]

(c) Marks more than \(60\): the values greater than \(60\) are \(61, 62, 65, 67, 67, 68, 69, 70, 71, 73, 74\) - that is \(11\) students. \[\text{percentage} = \frac{11}{50}\times 100 = \mathbf{22\%}.\]

Akọwa Nkọwa

(a) Expand \((\sqrt{3}-5\sqrt{2})(\sqrt{3}+\sqrt{2})\):

\(=\sqrt{3}\cdot\sqrt{3}+\sqrt{3}\cdot\sqrt{2}-5\sqrt{2}\cdot\sqrt{3}-5\sqrt{2}\cdot\sqrt{2}\)

\(=3+\sqrt{6}-5\sqrt{6}-10=-7-4\sqrt{6}\).

Comparing with \(a+b\sqrt{6}\): \(\mathbf{a=-7,\ b=-4}\).

(b) \(\dfrac{2^{1-y}\times 2^{y-1}}{2^{y+2}}=8^{2-3y}\).

Numerator: \(2^{(1-y)+(y-1)}=2^{0}=1\). Left side \(=2^{-(y+2)}\).

Right side: \(8^{2-3y}=2^{3(2-3y)}=2^{6-9y}\).

\(-(y+2)=6-9y \Rightarrow -y-2=6-9y \Rightarrow 8y=8 \Rightarrow y=\mathbf{1}\).

Akọwa Nkọwa

The solid is a triangular prism with congruent right-angled triangular ends \(ABC\) and \(DEF\) (\(\angle ABC = \angle DEF = 90^\circ\)). The given lengths are \(|AB| = 24\) cm, \(|BC| = 7\) cm, and the prism length \(|CD| = 40\) cm.

(a) Length AC. Triangle \(ABC\) is right-angled at \(B\), so by Pythagoras:

\[ |AC| = \sqrt{|AB|^2 + |BC|^2} = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25\ \text{cm} \]

(b) Total surface area. The prism has two triangular ends and three rectangular faces. The prism length is \(40\) cm.

The two triangular ends:

\[ 2 \times \left(\tfrac{1}{2}\times 24 \times 7\right) = 2 \times 84 = 168\ \text{cm}^2 \]

The three rectangles have widths equal to the sides of the triangle \((24, 7, 25)\) and common length \(40\):

\[ (24 + 7 + 25)\times 40 = 56 \times 40 = 2240\ \text{cm}^2 \]

Total surface area:

\[ 168 + 2240 = 2408\ \text{cm}^2 \]

Answers: (a) \(|AC| = 25\) cm; (b) total surface area \(= 2408\ \text{cm}^2\).

Akọwa Nkọwa

Total number of families:

\[N=3+5+7+4+3+2=24.\]

Number of children, \(x\)	0	1	2	3	4	5
Number of families, \(f\)	3	5	7	4	3	2
Cumulative frequency	3	8	15	19	22	24

(a)(i) Mode

The greatest frequency is \(7\), which corresponds to 2 children.

\[\boxed{\text{Mode}=2}\]

(a)(ii) Third quartile

\[Q_3=\frac{3}{4}\times24=18.\]

The 18th observation lies between cumulative frequencies 15 and 19, corresponding to 3 children.

\[\boxed{Q_3=3}\]

(a)(iii) Probability of at least 2 children

\[P(X\geq2)=\frac{7+4+3+2}{24}=\frac{16}{24}=\boxed{\frac{2}{3}}.\]

(b) Pie chart

The sector angle for each category is \(\dfrac{f}{24}\times360^\circ\). The completed pie chart is shown below.

For families with one child:

\[\frac{5}{24}\times360^\circ=\boxed{75^\circ}.\]

Akọwa Nkọwa

(a) Simplify inside first. \(40 \div 56 = \dfrac{40}{56} = \dfrac{5}{7}\), so \[\frac{35\times 49}{40\div 56} = \frac{35\times 49}{5/7} = 35\times 49\times\frac{7}{5} = 7\times 49\times 7 = 7^4 = 2401.\] \[\log 2401 = \log 7^4 = 4\log 7 = 4(0.8451) = \mathbf{3.3804}.\]

(b) Let there be \(x\) children; adults \(= x + 60\). Adult pays \(D5\), child \(D2\), total \(D1280\): \[5(x + 60) + 2x = 1280 \Rightarrow 7x + 300 = 1280 \Rightarrow 7x = 980 \Rightarrow x = 140.\]

(i) \(x = \mathbf{140}\) children (adults \(= 200\)).

(ii) children : adults \(= 140 : 200 = \mathbf{7 : 10}.\)

(iii) Total people \(= 340\); average \(= \dfrac{1280}{340} = \mathbf{D3.76}\) per person.

(iv) Gain \(= 1280 - 720 = D560\); percentage gain \(= \dfrac{560}{720}\times 100 = \mathbf{77.8\%}.\)

Akọwa Nkọwa

(a) The multiples of \(7\) between \(75\) and \(500\): the first is \(77 = 7\times 11\) and the last is \(497 = 7\times 71.\) \[\text{count} = 71 - 11 + 1 = \mathbf{61}.\]

(b) Let the first term be \(a\) and common difference \(d.\)

8th term is \(5\) times the 3rd: \(a + 7d = 5(a + 2d) \Rightarrow a + 7d = 5a + 10d \Rightarrow 4a + 3d = 0.\)

7th term is \(9\) greater than the 4th: \((a + 6d) - (a + 3d) = 9 \Rightarrow 3d = 9 \Rightarrow d = 3.\)

Then \(4a + 3(3) = 0 \Rightarrow 4a = -9 \Rightarrow a = -\tfrac{9}{4} = -2.25.\)

First five terms (adding \(d = 3\) each time): \[\mathbf{-2\tfrac{1}{4},\ \tfrac{3}{4},\ 3\tfrac{3}{4},\ 6\tfrac{3}{4},\ 9\tfrac{3}{4}}.\]