WAEC SSCE - General Mathematics - 2020

Akọwa Nkọwa

Akọwa Nkọwa

The value of y can only lie in the third quadrant. In the third quadrant, the sine values are negative and the tangent values are positive. This means that if sine is negative and tangent is positive, the angle must be in the third quadrant. It's important to remember that the sine and tangent functions have specific signs in each quadrant, which can be used to determine the location of an angle in the coordinate plane.

Akọwa Nkọwa

The volume of a pyramid can be calculated using the formula: V = (base area) * (height) / 3 We know the base area (9 cm x 5 cm = 45 cm\(^2\)), and the volume (105 cm\(^3\)), so we can rearrange the formula to solve for the height: height = 3 * (volume) / (base area) Plugging in the given values: height = 3 * (105) / (45) height = 7 cm So the height of the pyramid is 7 cm.

Akọwa Nkọwa

Standard deviation is a measure of how spread out the data is from the average (mean). It tells us how much the individual data points deviate from the mean. The standard deviation is expressed in the same units as the data. To calculate the standard deviation, we first find the difference between each data point and the mean. We square these differences to eliminate any negative values, and then take the average of the squared differences (this is known as the variance). Finally, we take the square root of the variance to find the standard deviation. For the numbers 15, 21, 17, 26, 18, and 29, with a mean of 21, the standard deviation is approximately 5.24. So, the answer is closest to: 5.

Akọwa Nkọwa

The gradient of a line describes how steep the line is, and can be calculated as the change in y (rise) divided by the change in x (run). To find the gradient of the line in this case, we need to find the difference in y between two points and divide it by the difference in x between the same points. For example, we can choose the first two points: (6.20, 3.90) and (6.85, 5.20). The change in y is 5.20 - 3.90 = 1.30, and the change in x is 6.85 - 6.20 = 0.65. The gradient is then 1.30 / 0.65 = 2. Therefore, the gradient of the line is 2.

Akọwa Nkọwa

There are 12 balls in the box, 5 of which are red, 4 blue, and the remaining ones are green. If we select a ball at random, the probability of selecting a red ball is 5/12, since there are 5 red balls out of 12 total. If we replace the ball and select again, the probability of selecting a red ball is still 5/12, since we put back the ball we selected and the number of red balls remains the same. To find the probability of selecting two red balls in a row, we need to multiply the probability of selecting a red ball on the first draw (5/12) by the probability of selecting a red ball on the second draw (also 5/12), since the events are independent. Therefore, the probability of selecting two red balls in a row with replacement is (5/12) x (5/12) = 25/144. Hence, the correct option is (A) \(\frac{25}{144}\).

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

The two buses are moving away from each other, so we can add their speeds to find the relative speed between them. The relative speed is 72 km/h + 48 km/h = 120 km/h. We want to know when the buses will be 240 km apart, which is twice the distance they cover in the first hour, since they are moving away from each other at a relative speed of 120 km/h. To cover a distance of 240 km at a relative speed of 120 km/h, it will take them (240 km) / (120 km/h) = 2 hours. Since they started at 9:00 am, they will be 240 km apart at 9:00 am + 2 hours = 11:00 am. Therefore, the correct option is (C) 11:00 am.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

The "SSA" (Side-Side-Angle) condition is not a sufficient condition for two triangles to be congruent. This is because two triangles can have the same lengths for two sides and the same measure for an angle that is not included between those sides, but still have different shapes and sizes. In this case, the triangles would not be congruent, but they would have what is called a "SSA" similarity, meaning they have two sides and a non-included angle in common. On the other hand, "SSS" (Side-Side-Side), "SAS" (Side-Angle-Side), and "AAS" (Angle-Angle-Side) are all sufficient conditions for two triangles to be congruent. "SSS" means that all three sides of one triangle are congruent to the three sides of another triangle. "SAS" means that two sides and the included angle of one triangle are congruent to the corresponding two sides and included angle of another triangle. "AAS" means that two angles and a non-included side of one triangle are congruent to the corresponding two angles and non-included side of another triangle.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Given the diagram, if PQ is parallel to SR, that means their corresponding angles are equal, and in this case, the angle marked "x" is equal in both triangles PQR and SRT. Since the two triangles are similar, we can use the fact that corresponding sides are proportional to find the value of x. Let's say the length of PQ is a and the length of SR is b. Since the two triangles are similar, we have the following proportion: a/PQ = b/SR Since we know the value of a and b, we can use this proportion to find the value of x. For example, if a = 8 and b = 10, we have: 8/PQ = 10/SR PQ = 10 * (8/10) PQ = 8 So, the length of PQ is 8 and the length of SR is 10, and since the angles marked x are equal in both triangles, we can conclude that x is equal to 46 degrees. Therefore, the value of x is 46.

Akọwa Nkọwa

The sum of the interior angles of a polygon can be found using the formula: (n-2) x 180 degrees, where "n" represents the number of sides of the polygon. In this case, the polygon is a pentagon, which has five sides. Substituting "n" with 5, we get: (5-2) x 180 degrees = 3 x 180 degrees = 540 degrees. Therefore, the sum of the interior angles of a pentagon is 540 degrees. Hence, the correct option is 540\(^o\).

Akọwa Nkọwa

Akọwa Nkọwa

The probability of an event can be calculated by dividing the number of favorable outcomes by the number of possible outcomes. In this case, we want to find the probability of selecting a green ball from the box. There are 3 green balls out of a total of 12 balls in the box, so the probability is 3/12 = 1/4. So, the answer is: The probability of selecting a green ball from the box is 1/4.

Akọwa Nkọwa

Akọwa Nkọwa

To solve the two equations 3x - 2y = 10 and x + 3y = 7 simultaneously, we need to find the values of x and y that satisfy both equations. To do this, we'll use a method called substitution. First, we'll solve one of the equations for one of the variables. Let's solve the second equation, x + 3y = 7, for x: x = 7 - 3y Next, we'll substitute this expression for x into the first equation: 3x - 2y = 10 3(7 - 3y) - 2y = 10 21 - 9y - 2y = 10 21 - 11y = 10 11y = 11 y = 1 Now that we have y = 1, we can substitute this value back into the expression we found for x: x = 7 - 3y x = 7 - 3(1) x = 4 So the solution to the two equations is x = 4 and y = 1.

Akọwa Nkọwa

Akọwa Nkọwa

To find out how much money Fati receives from the sales of 10 tins of milk, we need to know two things: the cost of each tin of milk and the profit she makes from selling each tin. Let's call the cost of each tin of milk "x" and the profit she makes from selling each tin "y". So, if she buys each tin of milk for ₦x and sells it at a profit of ₦y, then the total amount of money she receives for each tin of milk is ₦x + ₦y. Now, if she sells 10 tins of milk, then the total amount of money she receives from the sales of these 10 tins is 10 * (₦x + ₦y) = ₦10(x + y). So, the answer is ₦10(x + y).

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

To determine the value of P, we need to use the factor theorem, which states that if (x + a) is a factor of a polynomial, then the polynomial evaluated at -a will equal zero. In this case, we know that (x + 2) is a factor of x\(^2\) + px - 10, so we can set x = -2 and solve for P as follows: (-2)\(^2\) + P(-2) - 10 = 0 4 - 2P - 10 = 0 -2P - 6 = 0 -2P = 6 P = -3 Therefore, the value of P is -3.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

To find X \(\cap\) Y, we need to determine the common elements that are present in both sets X and Y. Set X is defined as the set of all numbers x, such that x is less than 7. Therefore, X = {1, 2, 3, 4, 5, 6}. Set Y is defined as the set of all factors of 24. The factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24. Therefore, Y = {1, 2, 3, 4, 6, 8, 12, 24}. The intersection of sets X and Y, denoted as X \(\cap\) Y, is the set of all elements that are present in both sets X and Y. Therefore, X \(\cap\) Y = {x : x is an element of X and x is an element of Y} From the sets X and Y, we see that the common elements are 1, 2, 3, 4, 6. Therefore, X \(\cap\) Y = {1, 2, 3, 4, 6} Thus, the correct answer is {1, 2, 3, 4, 6}.

Akọwa Nkọwa

The expression 2(x\(^2\) - y\(^2\)) means 2 times the difference of x squared and y squared. First, we need to calculate x\(^2\) and y\(^2\) x\(^2\) = 3\(^2\) = 9 y\(^2\) = (-1)\(^2\) = 1 Next, we subtract y\(^2\) from x\(^2\) to get the difference x\(^2\) - y\(^2\) = 9 - 1 = 8 Finally, we multiply the difference by 2 2(x\(^2\) - y\(^2\)) = 2 * 8 = 16 So, the answer is 16.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

The formula to calculate simple interest is: Simple Interest = Principal x Rate x Time where Rate is the interest rate per year and Time is the number of years. In this case, we are given the amount and the interest rate, and we need to find the principal. We know that: Amount = Principal + Simple Interest Substituting the values given: N550,000 = x + (x x 0.02 x 5) Simplifying: N550,000 = x + 0.1x N550,000 = 1.1x Dividing both sides by 1.1: x = N500,000 Therefore, the value of the principal is N500,000. Hence, the correct answer is N500,000.

Akọwa Nkọwa

The expression \(\frac{5x + 3}{6x (x + 1)}\) will be undefined when the denominator of the fraction equals zero. In this case, the denominator is \(6x(x+1)\), which will equal zero if either \(x=0\) or \(x=-1\). Therefore, the expression will be undefined when \(x\) is equal to either 0 or -1. To understand why this is the case, we can think of the denominator as representing the area of a rectangle with a width of \(6x\) and a height of \(x+1\). If either the width or the height of the rectangle is equal to zero, then the rectangle has zero area, and the fraction becomes undefined.

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

Akọwa Nkọwa

To solve this problem, we can use trigonometry and the concept of similar triangles. Let's draw a diagram to visualize the situation. We have a fence that is 2.4 meters tall and a tree that is 16 meters tall. The tree is 10 meters away from the fence. We want to find the angle of elevation of the top of the tree from the top of the fence, which is the angle between the horizontal line passing through the top of the fence and the line connecting the top of the fence and the top of the tree. To start, we can create a right triangle using the height of the fence and the distance between the fence and the tree as the base and the hypotenuse, respectively. Let's call this triangle ABC, where A is the top of the fence, B is the bottom of the fence, and C is the point on the ground directly below the top of the tree. The angle between AB and AC is 90 degrees. Next, we can create another right triangle using the height of the tree and the same distance as the base and the hypotenuse, respectively. Let's call this triangle ACD, where D is the top of the tree. The angle between AD and AC is also 90 degrees. Since triangles ABC and ACD share the angle at A, they are similar triangles. This means that their corresponding angles are equal, and their corresponding sides are proportional. In particular, we can use the ratio of their heights to find the tangent of the angle we're looking for. Let's call the angle of elevation we're looking for x. Then, we have: tan(x) = CD/AB = (16-2.4)/10 = 1.36 To solve for x, we take the inverse tangent of both sides: x = tan^-1(1.36) ≈ 53.67° Therefore, the angle of elevation of the top of the tree from the top of the fence is approximately 53.67 degrees.

Akọwa Nkọwa

The interquartile range (IQR) is the difference between the upper quartile and the lower quartile of a distribution. We are given that the IQR is 7, which means the upper quartile (Q3) is 7 units away from the lower quartile (Q1). We are also given that the 25th percentile (Q1) is 16. The 25th percentile means that 25% of the data is below 16, which is the same as saying that the first quartile (Q1) is 16. To find the upper quartile (Q3), we need to add the IQR to Q1. So, Q3 = Q1 + IQR = 16 + 7 = 23. Therefore, the correct option is (C) 23.

Akọwa Nkọwa

To find the equation of a line that is parallel to another line and passes through a specific point, we can use the slope-point form of a line. The slope-point form of a line is given by: y - y1 = m(x - x1) where m is the slope of the line and (x1, y1) is a point on the line. The first step is to find the slope of the line 2y = 3(x - 2), which can be found by rearranging the equation into slope-intercept form (y = mx + b). To do this, we first isolate y on one side of the equation: 2y = 3x - 6 y = (3/2)x - 3 The slope of the line is (3/2). Next, we use the point (2, 3) and the slope of the line to write the equation in slope-point form: y - 3 = (3/2)(x - 2) This is the equation of the line that is parallel to 2y = 3(x - 2) and passes through the point (2, 3). So, the correct answer is y = (3/2)x - 3.

Akọwa Nkọwa

Akọwa Nkọwa

Given: Total surface area of a cone = 224π cm², slant height = 1 cm, and r : 1 = 2.5

(a) To find the value of r, we can use the formula for the total surface area of a cone:

Total surface area = πrℓ + πr², where ℓ is the slant height of the cone.

Substituting the given values, we get:

224π = πr√(1 + r²) + πr²

Simplifying and rearranging the terms, we get:

√(1 + r²) = 224/r - r

Squaring both sides, we get:

1 + r² = 50176/r² - 448 + r²

Simplifying and rearranging, we get:

r⁴ - 448r² - 50175 = 0

This is a quadratic equation in terms of r², so we can use the quadratic formula:

r² = (448 ± √(448² + 4(1)(50175)))/2

Simplifying, we get:

r² = 225 or r² = 22400

Since r : 1 = 2.5, we can choose the positive value of r², which is 225. Therefore, r = 15 (correct to one decimal place).

(b) To find the volume of the cone, we can use the formula:

Volume = (1/3)πr²ℓ

Substituting the given values, we get:

Volume = (1/3)(22/7)(15)²(1)

Simplifying, we get:

Volume = 350 cm³ (correct to the nearest whole number)

Therefore, the volume of the cone is approximately 350 cm³.

Akọwa Nkọwa

(a) Completing the Table of Values for $\mathrm{<span\; style="font-family:\; Ubuntu;\; font-size:\; 16px;">y</span>}<span\; style="font-family:\; Ubuntu;\; font-size:\; 16px;">=</span>\mathrm{<span\; style="font-family:\; Ubuntu;\; font-size:\; 16px;">3</span>}\mathrm{<span\; style="font-family:\; Ubuntu;\; font-size:\; 16px;">sin</span>}<span\; style="font-family:\; Ubuntu;\; font-size:\; 16px;"></span>\mathrm{<span\; style="font-family:\; Ubuntu;\; font-size:\; 16px;">2</span>}\mathrm{<span\; style="font-family:\; Ubuntu;\; font-size:\; 16px;">x</span>}$y = 3 \sin 2xsinx

To complete the table, we need to calculate the values of $yy$ for given $xx$ values using the equation $y=3\sin 2xy\; =\; 3\; \backslash sin\; 2x$.

Let's calculate the values for each $xx$:

• $x=0^{\circ }x\; =\; 0^\backslash circ$

  $y=3\sin 2(0^{\circ })=3\sin 0=0y\; =\; 3\; \backslash sin\; 2(0^\backslash circ)\; =\; 3\; \backslash sin\; 0\; =\; 0$

• $x=15^{\circ }x\; =\; 15^\backslash circ$

  $y=3\sin 2(15^{\circ })=3\sin 30^{\circ }=3\times 0.5=1.5y\; =\; 3\; \backslash sin\; 2(15^\backslash circ)\; =\; 3\; \backslash sin\; 30^\backslash circ\; =\; 3\; \backslash times\; 0.5\; =\; 1.5$

• $x=30^{\circ }x\; =\; 30^\backslash circ$

  $y=3\sin 2(30^{\circ })=3\sin 60^{\circ }=3\times \frac{\sqrt{3}}{2}\approx 2.598y\; =\; 3\; \backslash sin\; 2(30^\backslash circ)\; =\; 3\; \backslash sin\; 60^\backslash circ\; =\; 3\; \backslash times\; \backslash frac\{\backslash sqrt\{3\}\}\{2\}\; \backslash approx\; 2.598$

• $x=45^{\circ }x\; =\; 45^\backslash circ$

  $y=3\sin 2(45^{\circ })=3\sin 90^{\circ }=3\times 1=3y\; =\; 3\; \backslash sin\; 2(45^\backslash circ)\; =\; 3\; \backslash sin\; 90^\backslash circ\; =\; 3\; \backslash times\; 1\; =\; 3$

• $x=60^{\circ }x\; =\; 60^\backslash circ$

  $y=3\sin 2(60^{\circ })=3\sin 120^{\circ }=3\times \frac{\sqrt{3}}{2}\approx 2.598y\; =\; 3\; \backslash sin\; 2(60^\backslash circ)\; =\; 3\; \backslash sin\; 120^\backslash circ\; =\; 3\; \backslash times\; \backslash frac\{\backslash sqrt\{3\}\}\{2\}\; \backslash approx\; 2.598$

• $x=75^{\circ }x\; =\; 75^\backslash circ$

  $y=3\sin 2(75^{\circ })=3\sin 150^{\circ }=3\times 0.5=1.5y\; =\; 3\; \backslash sin\; 2(75^\backslash circ)\; =\; 3\; \backslash sin\; 150^\backslash circ\; =\; 3\; \backslash times\; 0.5\; =\; 1.5$

• $x=90^{\circ }x\; =\; 90^\backslash circ$

  $y=3\sin 2(90^{\circ })=3\sin 180^{\circ }=3\times 0=0y\; =\; 3\; \backslash sin\; 2(90^\backslash circ)\; =\; 3\; \backslash sin\; 180^\backslash circ\; =\; 3\; \backslash times\; 0\; =\; 0$

• $x=105^{\circ }x\; =\; 105^\backslash circ$

  $y=3\sin 2(105^{\circ })=3\sin 210^{\circ }=3\times (-0.5)=-1.5y\; =\; 3\; \backslash sin\; 2(105^\backslash circ)\; =\; 3\; \backslash sin\; 210^\backslash circ\; =\; 3\; \backslash times\; (-0.5)\; =\; -1.5$

• $x=120^{\circ }x\; =\; 120^\backslash circ$

  $y=3\sin 2(120^{\circ })=3\sin 240^{\circ }=3\times (-\frac{\sqrt{3}}{2})\approx -2.598y\; =\; 3\; \backslash sin\; 2(120^\backslash circ)\; =\; 3\; \backslash sin\; 240^\backslash circ\; =\; 3\; \backslash times\; (-\backslash frac\{\backslash sqrt\{3\}\}\{2\})\; \backslash approx\; -2.598$

• $x=135^{\circ }x\; =\; 135^\backslash circ$

  $y=3\sin 2(135^{\circ })=3\sin 270^{\circ }=3\times (-1)=-3y\; =\; 3\; \backslash sin\; 2(135^\backslash circ)\; =\; 3\; \backslash sin\; 270^\backslash circ\; =\; 3\; \backslash times\; (-1)\; =\; -3$

• $x=150^{\circ }x\; =\; 150^\backslash circ$

  $y=3\sin 2(150^{\circ })=3\sin 300^{\circ }=3\times (-\frac{\sqrt{3}}{2})\approx -2.598y\; =\; 3\; \backslash sin\; 2(150^\backslash circ)\; =\; 3\; \backslash sin\; 300^\backslash circ\; =\; 3\; \backslash times\; (-\backslash frac\{\backslash sqrt\{3\}\}\{2\})\; \backslash approx\; -2.598$

Here is the completed table:

b)

c)

i) 113

(ii) 3o ${}^{\mathrm{<br>}}$ and 82o

Akọwa Nkọwa

a)

(i) To make Q the subject of the formula, we need to isolate Q on one side of the equation.

P = \(\left(\frac{rk}{Q} - ms\right)^{\frac{2}{3}}\)

Take the reciprocal of both sides:

\(\frac{1}{P}\) = \(\frac{Q}{rk}\) - \(\frac{ms}{rk}\)

Add \(\frac{ms}{rk}\) to both sides:

\(\frac{ms}{rk}\) + \(\frac{1}{P}\) = \(\frac{Q}{rk}\)

Multiply both sides by \(\frac{rk}{ms + Prk}\):

Q = \(\frac{rk}{ms + Prk}\)

(ii) Now that we have the formula for Q, we can substitute the given values and calculate:

Q = \(\frac{(10)(4)}{(0.2) + (3)(15)(4)}\)

Q = \(\frac{40}{182}\)

Q = 0.22 (correct to two decimal places)

Therefore, when P = 3, m = 15, s = 0.2, k = 4, and r = 10, Q is approximately equal to 0.22.

b) To find x : y, we can use algebraic manipulation to isolate one of the variables in terms of the other:

\(\frac{x + 2y}{5}\) = x - 2y

Distribute the 5 on the right side of the equation:

\(\frac{x + 2y}{5}\) = x - \(\frac{10y}{5}\)

Simplify:

\(\frac{x + 2y}{5}\) = x - 2y

x + 2y = 5x - 10y

12y = 4x

y = \(\frac{1}{3}\)x

Therefore, x : y is 3 : 1.

Akọwa Nkọwa

(a) Finding ${\mathrm{<span\; style="font-family:\; Ubuntu;\; font-size:\; 16px;">A</span>}}^{\mathrm{<span\; style="font-family:\; Ubuntu;\; font-size:\; 16px;">\prime </span>}}<span\; style="font-family:\; Ubuntu;\; font-size:\; 16px;">\cap </span>{\mathrm{<span\; style="font-family:\; Ubuntu;\; font-size:\; 16px;">B</span>}}^{\mathrm{<span\; style="font-family:\; Ubuntu;\; font-size:\; 16px;">\prime </span>}}<span\; style="font-family:\; Ubuntu;\; font-size:\; 16px;">\cap </span>{\mathrm{<span\; style="font-family:\; Ubuntu;\; font-size:\; 16px;">C</span>}}^{\mathrm{<span\; style="font-family:\; Ubuntu;\; font-size:\; 16px;">\prime </span>}}$A' \cap B' \cap C'

Given:

• $\mu =\{x:1\le x\le 10\}=\{1,2,3,4,5,6,7,8,9,10\}\backslash mu\; =\; \backslash \{\; x\; :\; 1\; \backslash leq\; x\; \backslash leq\; 10\; \backslash \}\; =\; \backslash \{1,\; 2,\; 3,\; 4,\; 5,\; 6,\; 7,\; 8,\; 9,\; 10\backslash \}$
• $A=\{\text{multiples of 2}\}=\{2,4,6,8,10\}A\; =\; \backslash \{\; \backslash text\{multiples\; of\; 2\}\; \backslash \}\; =\; \backslash \{2,\; 4,\; 6,\; 8,\; 10\backslash \}$
• $B=\{\text{multiples of 3}\}=\{3,6,9\}B\; =\; \backslash \{\; \backslash text\{multiples\; of\; 3\}\; \backslash \}\; =\; \backslash \{3,\; 6,\; 9\backslash \}$
• $C=\{\text{factors of 6}\}=\{1,2,3,6\}C\; =\; \backslash \{\; \backslash text\{factors\; of\; 6\}\; \backslash \}\; =\; \backslash \{1,\; 2,\; 3,\; 6\backslash \}$

First, find the complements $A^{\mathrm{\prime }},B^{\mathrm{\prime }},A\text{'},\; B\text{'},$ and $C^{\mathrm{\prime }}C\text{'}$:

• $A^{\mathrm{\prime }}=\mu -A=\{1,3,5,7,9\}A\text{'}\; =\; \backslash mu\; -\; A\; =\; \backslash \{1,\; 3,\; 5,\; 7,\; 9\backslash \}$
• $B^{\mathrm{\prime }}=\mu -B=\{1,2,4,5,7,8,10\}B\text{'}\; =\; \backslash mu\; -\; B\; =\; \backslash \{1,\; 2,\; 4,\; 5,\; 7,\; 8,\; 10\backslash \}$
• $C^{\mathrm{\prime }}=\mu -C=\{4,5,7,8,9,10\}C\text{'}\; =\; \backslash mu\; -\; C\; =\; \backslash \{4,\; 5,\; 7,\; 8,\; 9,\; 10\backslash \}$

Now, find the intersection $A^{\mathrm{\prime }}\cap B^{\mathrm{\prime }}\cap C^{\mathrm{\prime }}A\text{'}\; \backslash cap\; B\text{'}\; \backslash cap\; C\text{'}$:

$A^{\mathrm{\prime }}\cap B^{\mathrm{\prime }}=\{1,3,5,7,9\}\cap \{1,2,4,5,7,8,10\}=\{1,5,7\}A\text{'}\; \backslash cap\; B\text{'}\; =\; \backslash \{1,\; 3,\; 5,\; 7,\; 9\backslash \}\; \backslash cap\; \backslash \{1,\; 2,\; 4,\; 5,\; 7,\; 8,\; 10\backslash \}\; =\; \backslash \{1,\; 5,\; 7\backslash \}$

$A^{\mathrm{\prime }}\cap B^{\mathrm{\prime }}\cap C^{\mathrm{\prime }}=\{1,5,7\}\cap \{4,5,7,8,9,10\}=\{5,7\}A\text{'}\; \backslash cap\; B\text{'}\; \backslash cap\; C\text{'}\; =\; \backslash \{1,\; 5,\; 7\backslash \}\; \backslash cap\; \backslash \{4,\; 5,\; 7,\; 8,\; 9,\; 10\backslash \}\; =\; \backslash \{5,\; 7\backslash \}$

So, $A^{\mathrm{\prime }}\cap B^{\mathrm{\prime }}\cap C^{\mathrm{\prime }}=\{5,7\}A\text{'}\; \backslash cap\; B\text{'}\; \backslash cap\; C\text{'}\; =\; \backslash \{5,\; 7\backslash \}$.

(b) Calculating the Savings Per Person

Given:

• Individual ticket price: $18.50
• Bulk purchase price for 5 tickets: $80.00

First, calculate the total cost of 5 individual tickets: $5\times 18.50=92.505\; \backslash times\; 18.50\; =\; 92.50$

The cost difference between individual tickets and the bulk purchase price: $92.50-80.00=12.5092.50\; -\; 80.00\; =\; 12.50$

Each person would save: $\frac{12.50}{5}=2.50\backslash frac\{12.50\}\{5\}\; =\; 2.50$

So, if the 4 gentlemen get a fifth person to join them and share the bulk purchase price equally, each person would save $2.50.

Akọwa Nkọwa

Akọwa Nkọwa

a) To construct a cumulative frequency table, we add up the frequencies for each interval to find the cumulative frequency for each interval. Here's what the cumulative frequency table would look like:

b)

(c)

(i) Median Mark = 54.5 

(ii) Lowest mark, if 5% of the students passed = 14

Akọwa Nkọwa

a) (i) To find Ms. Maureen's monthly income:

To find Ms. Maureen's monthly income, we can set up an equation using the information given. Let's call her monthly income "M".

Ms. Maureen spent 1/4 of her monthly income at the shopping mall, so: M * 1/4 = M/4

She spent 1/3 of her monthly income at the open market, so: M * 1/3 = M/3

The remaining amount after spending at the shopping mall and open market is M - M/4 - M/3 = M * (3/4) - M/3.

She then spent 2/5 of this remaining amount at the mechanic workshop, so: (M * (3/4) - M/3) * 2/5 = 2/5 * M * (3/4) - 2/5 * M/3.

We know that she had N222,000 left, so we can set this equal to the amount spent at the mechanic workshop: N222,000 = 2/5 * M * (3/4) - 2/5 * M/3.

Solving for M, we can find her monthly income:

M = N222,000 * 5/2 * 4/3 / (3/4 - 2/5) = N222,000 * 5/2 * 4/3 / (5/4 - 6/5) = N222,000 * 20/3 / (25/20 - 24/20) = N222,000 * 20/3 / (5/20) = N222,000 * 20/3 * 20/5 = N222,000 * 4 = N888,000.

So Ms. Maureen's monthly income is N888,000.

(ii) To find the amount spent at the open market:

To find the amount spent at the open market, we use the equation M * 1/3 = M/3 and substitute M = N888,000:

N888,000 * 1/3 = N888,000/3

N888,000/3 = N296,000

So the amount spent at the open market is N296,000.

b) The third term of an arithmetic progression (A.P.) is 4m - 2n, and the ninth term is 2m - 8n. To find the common difference:

To find the common difference, we can subtract the third term from the ninth term and divide by 6, since there are 6 terms between the third and ninth terms:

(2m - 8n) - (4m - 2n) = -2m + 6n

The common difference is -2m + 6n / 6 = -m + n.

So the common difference in terms of m and n is -m + n.

Akọwa Nkọwa

(a) The diagram can be drawn by placing town Q at the origin and drawing the bearings of 049° and 319°. Then, draw lines from town Q at those bearings to represent the paths of cyclists X and Y.

(b)

1. (i) To find the distance between cyclist X and Y, we need to use the cosine rule. Let a be the distance that cyclist X travels and b be the distance that cyclist Y travels. Then, we have:
• a² = (2 × 5)² = 100
• b² = (2 × 9)² = 324
• c² = a² + b² - 2ab cos(128°) ? 302.73
• c ? 17.4 km (nearest whole number)
2. (ii) To find the bearing of cyclist X from Y, we need to use the sine rule. Let C be the angle between cyclist X's path and the line connecting X and Y, and let D be the angle between cyclist Y's path and the same line. Then, we have:
• sin(C) / a = sin(128°) / c
• sin(D) / b = sin(33°) / c
Solving for C and D, we get:
• C ? 97.2°
• D ? 35.7°
Since cyclist X is to the right of cyclist Y, the bearing of cyclist X from Y is:
• 360° - 319° + 97.2° ? 138.2°

(c) To find the average speed at which cyclist X will get to Y in 4 hours, we need to find the total distance between them and divide by the time taken. Using the distance formula from part (b)(i), we have:

d = ?302.73 ? 17.4 km
time taken = 4 hours

Therefore, the average speed at which cyclist X will get to Y in 4 hours is:

17.4 km / 4 hours = 4.35 km/h