WAEC SSCE - General Mathematics - 2010

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To round off 0.002473 to 3 significant figures, we count the number of digits from the first non-zero digit to the third significant figure. In this case, the first non-zero digit is 2, and the third significant figure is 4. So, we count 1, 2, 3 digits from the first non-zero digit, which means the third digit is the last significant figure. The fourth digit, which is 7 in this case, is greater than or equal to 5, so we round up the last significant figure by adding 1 to it. Therefore, the final answer is 0.00247, which is the third option.

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The distance covered in one complete lap around a circular track of diameter 100m is equal to the circumference of the circle. The circumference of a circle is given by the formula C = 2πr, where r is the radius of the circle. The diameter of the circle is 100m, so the radius is half of that, which is 50m. Thus, the circumference of the circle is C = 2π(50) = 100π m. To cover a distance of 1000m, the man needs to run around the track 1000/100π = 10/π times. To the nearest whole number, this is equal to 3. So the man will need to run around the track 3 times to cover a distance of 1000m. Therefore, the answer is 3.

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The shaded portion in the diagram represents the region where the values of x and y satisfy the inequality x + y < 3. This inequality represents a boundary line passing through (3, 0) and (0, 3) and the shaded region lies below this line. Therefore, the answer is x + y < 3.

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An exterior angle of a polygon is the angle between any side of the polygon and its adjacent extended side. The sum of the exterior angles of any polygon, regardless of the number of sides, is always equal to 360 degrees. The sum of the interior angles of a polygon can be found using the formula (n-2) x 180, where n is the number of sides of the polygon. So if we have an n-sided polygon, the sum of its interior angles is (n-2) x 180. According to the problem, the sum of the exterior angles is half the sum of the interior angles. Mathematically, we can express this as: Sum of exterior angles = 1/2 x sum of interior angles Substituting the formulas we have for the sum of exterior and interior angles, we get: 360 = 1/2 x (n-2) x 180 Simplifying, we get: 2 x 360 = (n-2) x 180 720 = (n-2) x 180 Dividing both sides by 180, we get: 4 = n-2 n = 6 Therefore, the number of sides of the polygon is 6.

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Let the total age of all R men in the club be T. Then, we know that the mean age of R men is 50 years. Therefore, we have: T/R = 50 When two men aged 55 and 63 leave the club, the total age of the remaining men becomes T - 55 - 63, and the total number of remaining men becomes R - 2. We also know that the new mean age is reduced by 1 year. Therefore, we have: (T - 55 - 63)/(R - 2) = 49 Expanding this equation and substituting T/R = 50, we get: (50R - 118)/(R - 2) = 49 Multiplying both sides by (R - 2), we get: 50R - 118 = 49(R - 2) 50R - 118 = 49R - 98 R = 20 Therefore, there were originally 20 men in the club.

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We can approach this problem using a well-known algebraic identity: (a + b)^2 = a^2 + 2ab + b^2 If we let a = x and b = \(\frac{4}{3}\), we can write: x^2 + kx + \(\frac{16}{9}\) = (x + \(\frac{4}{3}\))^2 Expanding the right side of the equation, we get: x^2 + 2x(\(\frac{4}{3}\)) + \(\frac{16}{9}\) = x^2 + kx + \(\frac{16}{9}\) Simplifying, we get: 2x(\(\frac{4}{3}\)) = kx k = 2(\(\frac{4}{3}\)) = \(\frac{8}{3}\) Therefore, the value of k is \(\frac{8}{3}\).

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Let the two consecutive whole numbers be x and x+1. Then, according to the problem statement: x + (x+1) = \(\frac{5}{6}\)(x(x+1)) 2x+1 = \(\frac{5}{6}\)(x² + x) Multiplying both sides by 6 to remove the fraction: 12x + 6 = 5x² + 5x 5x² - 7x - 6 = 0 Solving the quadratic equation above by factoring or using the quadratic formula, we get: x = -1 or x = 1.2 Since we're looking for consecutive whole numbers, x must be 1. Therefore, the two consecutive whole numbers are 1 and 2. So the answer is (2, 3).

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To simplify the expression, we need to find a common denominator first. The denominators are \((2+x)\) and \((2-x)\), whose product is \((2+x)(2-x)=4-x^2\). So, we can write: $$ \frac{2}{2 + x} + \frac{2}{2 - x} = \frac{2(2-x)}{(2+x)(2-x)} + \frac{2(2+x)}{(2+x)(2-x)} = \frac{4 - 2x + 4 + 2x}{4 - x^2} = \frac{8}{4 - x^2} $$ Therefore, the simplified expression is \(\frac{8}{4 - x^2}\), which is option (B).

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Since tan x = 1, we can determine that x = 45 degrees or \(\frac{\pi}{4}\) radians. Using the identity \(\tan^2x + 1 = \sec^2x\), we can determine that \(\sec x = \sqrt{2}\). Next, we can use the identity \(\sin^2x + \cos^2x = 1\) to determine that \(\cos x = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}\). Now we can substitute the values of \(\sec x\) and \(\cos x\) into the expression for \(\frac{1 - \sin^2 x}{\cos x}\) to get: \[\frac{1 - \sin^2 x}{\cos x} = \frac{1 - \frac{1}{2}}{\frac{\sqrt{2}}{2}} = \frac{\frac{1}{2}}{\frac{\sqrt{2}}{2}} = \frac{1}{\sqrt{2}} = \boxed{\frac{\sqrt{2}}{2}}\]

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Let's start by finding an equation that relates the length and width of the rectangle to its perimeter. We know that the perimeter of a rectangle is the sum of the lengths of all its sides, which in this case is given as 16cm. The perimeter, P = sum of all sides of rectangle = 2(length + width) Therefore, we can write the equation as: 2(x + (x - 1)) = 16 Simplifying the equation, we have: 2x + 2(x - 1) = 16 4x - 2 = 16 4x = 18 x = 4.5cm Therefore, the value of x is 4\(\frac{1}{2}\)cm, which is.

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Using the identity sin(90° - θ) = cos θ, we can write cos 2y as sin (90° - 2y), and so sin 3y = sin (90° - 2y). Since the sine function is periodic with a period of 360°, we can write: 3y = 90° - 2y + 360°k or 3y = 270° - 2y + 360°k for some integer k. Solving for y in each equation gives: 5y = 90° + 360°k or 5y = 270° + 360°k Dividing both sides by 5 gives: y = 18° + 72°k or y = 54° + 72°k Since the problem specifies that 0° ≤ y ≤ 90°, the only solution that works is y = 18°. Therefore, the answer is (a) 18°.

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To determine how many students passed the test, we need to calculate the total number of students who scored 50 or more. From the bar chart, we can see that the number of students who scored 50 or more in the test is the sum of the heights of the bars for the grades C, B, and A. The height of the bar for grade C is 20, the height of the bar for grade B is 30, and the height of the bar for grade A is 35. Therefore, the total number of students who scored 50 or more is 20 + 30 + 35 = 85. Hence, the answer is 85, and 85 students passed the test.

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If Bola sold the article for N6,900.00 and made a profit of 15%, then the cost price of the article would be: Cost price = (100 / (100 + 15)) x N6,900.00 = N6,000.00 If Bola sold the same article for N6,600.00, he would be selling it at a loss. To find out the percentage of loss, we need to calculate the selling price that would result in a profit of 0%. Let's call the new selling price "S". We know that the cost price is N6,000.00 and the desired profit is 0%, so: S - N6,000.00 = 0 S = N6,000.00 This means that if Bola sells the article for N6,000.00, he will break even and make no profit or loss. But Bola is actually selling the article for N6,600.00, which is N600.00 more than the break-even price. To find the percentage of loss, we need to calculate what percentage of the cost price N600.00 represents: Percentage loss = (600 / 6000) x 100% = 10% Therefore, if Bola sells the article for N6,600.00, he would make a loss of 10%. So the answer is "loss of 10%".

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We can begin by calculating the volume of water that has been poured into the tank. We know that the volume of a rectangular tank is given by the formula V = lwh, where l is the length, w is the width, and h is the height. In this case, we are given that l = 250cm, w = 120cm, and V = 180 litres. However, we need to convert the volume to cubic centimeters, since the dimensions of the tank are given in centimeters. To do this, we can multiply 180 litres by 1000 cubic centimeters per liter, which gives us: V = 180 x 1000 = 180000 cubic centimeters Now we can use the formula for the volume of a rectangular tank to solve for the height of the water. We rearrange the formula to get: h = V / lw Substituting the given values, we get: h = 180000 / (250 x 120) = 6 cm Therefore, the height of the water in the tank is 6cm. The answer is 6.0cm.

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We are given a linear equation y = (1/2)x + 1 and asked to find the value of x when y = 5. Substituting y = 5 into the equation, we get: 5 = (1/2)x + 1 Subtracting 1 from both sides, we get: 4 = (1/2)x Multiplying both sides by 2, we get: 8 = x Therefore, when y = 5, x = 8. Hence, the answer is (a) 8.

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To convert from km/h to m/s, we need to multiply by a conversion factor. We know that 1 km = 1000 m and 1 hour = 3600 seconds. So, 1 km/h = \(\frac{1000\text{m}}{3600\text{s}}\) = \(\frac{5}{18}\) m/s. Therefore, to convert x km/h to m/s, we multiply by \(\frac{5}{18}\) to get y. So, y = \(\frac{5}{18}\)x. Therefore, option D, \(\frac{5}{18}\)x, is the correct answer.

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If 65% of the casting is Zinc, then the percentage of Copper in the casting is (100% - 65%) = 35%. Let's assume that the mass of the casting is x grams. Then, the mass of Zinc in the casting is 0.65x grams and the mass of Copper in the casting is 0.35x grams. We know that the mass of Copper in the casting is 147g, so we can write the following equation: 0.35x = 147 Solving for x, we get: x = 147 / 0.35 = 420 Therefore, the mass of the casting is 420g. Hence, the answer is 420g.

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To simplify 0.000215 x 0.000028, we multiply the two numbers to obtain: 0.000215 x 0.000028 = 0.00000000602 To express this in standard form, we move the decimal point 9 places to the right, since the number is less than 1. This gives: 0.00000000602 = 6.02 x 10^-9 Therefore, the answer is 6.02 x 10^-9.

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Since MN is tangent to the circle, it must be perpendicular to the radius OP drawn to the point of contact. Therefore, angle MOP is 90 degrees. Let x be the measure of angle RPS. Then, angle RPO is also x degrees because RP is a radius and angles at the center are twice those at the circumference. Since MN is tangent to the circle, angle MPN is also 90 degrees. Since angle MPN and angle NPS form a linear pair (as they add up to 180 degrees) and angle MPN is 90 degrees, we have that angle NPS is 90 - x degrees. Finally, since angle MNQ and angle NPS are alternate angles, they are equal. Thus, angle MNQ is also 90 - x degrees. Now, we have a triangle MPN with angles 90, 90 - x, and 55 degrees. The angles of a triangle add up to 180 degrees, so we can write: 90 + (90 - x) + 55 = 180 Simplifying this equation gives: x = 35 Therefore, < RPS = 35 degrees.

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The intersection of two sets, denoted by A ∩ B, is the set of elements that are in both set A and set B. In this case, P ∩ Q represents the set of integers that are between 1 and 6, and also between 2 and 10. The smallest integer in the intersection is 2, and the largest is 6. Therefore, the set P ∩ Q is {2, 3, 4, 5, 6}. The number of elements in this set is 5. Hence, the answer is: n(P ∩ Q) = 5.

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We can use the values in the table to find the values of m and c. Since y = mx + c, we can write: When x = 0, y = c, so c = 1 When x = 2, y = 2m + 1 When x = 4, y = 4m + 1 When x = 6, y = 6m + 1 To find the value of m, we can use the values for x = 2 and x = 4: 2m + 1 = 2 4m + 1 = 3 Solving these equations simultaneously gives m = 1/2. Substituting this value of m and c = 1 into y = mx + c gives: y = (1/2)x + 1 So the equation of the line described in the table is y = (1/2)x + 1. Therefore, the answer is (d) y = \(\frac{1}{2}x + 1\).

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The given information can be written in the form of an equation as follows: 6 + (1/3)x = 2x + 1 To solve for x, we need to isolate x on one side of the equation. First, we can subtract 6 from both sides: (1/3)x = 2x - 5 Next, we can subtract 2x from both sides: (1/3)x - 2x = -5 Simplifying the left side: (1/3 - 6/3)x = -5 -5/3x = -5 Multiplying both sides by -3/5: x = 3 Therefore, the solution is x = 3.

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Since triangles HKL and HIJ are similar, their corresponding sides are proportional. We know that LH and JH are corresponding sides. Therefore, the ratio \(\frac{LH}{JH}\) can be expressed as the ratio of the corresponding sides of the two triangles that form the fraction. In other words, we need to find the ratio of the side lengths of the triangles that share the same vertex as L and J, respectively. That would be the ratio of KL to JI. Thus, the answer is (a) \(\frac{KL}{JI}\).

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Let the length and width of the rectangular garden be L and W, respectively. The perimeter is the sum of all sides of the rectangle, so we have: P = 2L + 2W = 32cm The area is given by: A = LW = 39cm² We can use the first equation to solve for one of the variables in terms of the other. For example, solving for L we get: L = 16 - W Substituting this expression for L in the equation for the area we obtain: (16 - W)W = 39 Expanding and rearranging, we get a quadratic equation: W² - 16W + 39 = 0 Solving for W using the quadratic formula, we get: W = 3 or W = 13 Since the length cannot be smaller than the width, we discard the solution W = 3. Therefore, W = 13cm, and L = 16 - W = 3cm. Therefore, the length of the rectangular garden is 13cm. So the answer is (c) 13cm.

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To find the range of values of x, we first need to find the mean of the numbers 2, 5, 2x, and 7, and then solve for x. The mean of the four numbers is given by: \[\frac{2 + 5 + 2x + 7}{4} = \frac{14 + 2x}{4} = \frac{7 + x}{2}\] Since the mean is less than or equal to 5, we can write: \[\frac{7 + x}{2} \leq 5\] Multiplying both sides by 2, we get: \[7 + x \leq 10\] Subtracting 7 from both sides, we get: \[x \leq 3\] Therefore, the range of values of x that satisfy the condition is: \[x \leq 3\] So the answer is: x ≤ 3.

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The car uses 1 litre of petrol for every 14km, so for N63.00 (the cost of one litre of petrol), the car can travel 14km. We can use this information to find how far the car can go with N900.00 worth of petrol. To do this, we need to find how many litres of petrol N900.00 can buy. We can set up a proportion using the cost of petrol and the amount of petrol used per kilometre: 1 litre of petrol costs N63.00, so x litres cost N900.00 1 litre of petrol takes the car 14km, so x litres of petrol will take the car 14x km. Putting these together, we get: 1/63 = x/900 x = (1/63) * 900 x = 14.29 So, N900.00 can buy 14.29 litres of petrol. To find how far the car can go, we multiply the amount of petrol by the distance travelled per litre: 14.29 litres of petrol * 14km per litre = 200.06 km Therefore, the car can go approximately 200km with N900.00 worth of petrol. The closest option is 200km.

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None

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The universal set is \(U = \{2,3,4,5,6,7,8,9\}\), with \(A = \{2,4,6,8\}\), \(B = \{2,3,7,9\}\) and \(C = \{x : 3 < x < 9\} = \{4,5,6,7,8\}\).

Complements within \(U\):

• \(B' = U \setminus B = \{4,5,6,8\}\)
• \(C' = U \setminus C = \{2,3,9\}\)

(a) \(B' \cap C' = \{4,5,6,8\} \cap \{2,3,9\} = \varnothing\).

Therefore \(A \cap (B' \cap C') = A \cap \varnothing = \varnothing\) (the empty set).

(b) First form the unions:

• \(A \cup B = \{2,3,4,6,7,8,9\}\)
• \(B \cup C = \{2,3,4,5,6,7,8,9\}\)

\[(A \cup B) \cap (B \cup C) = \{2,3,4,6,7,8,9\}.\]

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Setting up the angle at B.

The bearing of A from B is \(310°\), so \(BA\) lies \(360°-310°=50°\) west of north at B (the diagram marks \(40°\) between the west arm and \(BA\), and \(90°-50°=40°\)).

The bearing of B from C is \(230°\), so the bearing of C from B is \(230°-180°=050°\); thus \(BC\) lies \(50°\) east of north at B (marked \(50°\) at C).

\[\angle ABC = 50°+50° = 100°.\]

(a) Distance AC. By the cosine rule,

\[AC^2 = AB^2+BC^2-2\,AB\cdot BC\cos\angle ABC\]

\[AC^2 = 8^2+13^2-2(8)(13)\cos100° = 233-208(-0.1736)=269.1\]

\[AC=\sqrt{269.1}=16.4\text{ km (3 s.f.)}\]

(b) Bearing of C from A. By the sine rule,

\[\frac{\sin\angle BAC}{BC}=\frac{\sin\angle ABC}{AC}\Rightarrow \sin\angle BAC=\frac{13\sin100°}{16.4}=0.7804\]

\[\angle BAC=51.3°.\]

The bearing of B from A is \(310°-180°=130°\). C lies to the north of the line AB, nearer A's north arm, so the bearing of C from A is

\[130°-51.3°=078.7°\approx 079°.\]

(c) How far east of B is C. C is on bearing \(050°\) from B at a distance of \(13\) km, so the eastward component is

\[13\sin50°=13(0.7660)=9.96\text{ km (3 s.f.)}\]

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(a) For a G.P. with first term \(a\) and common ratio \(r\):

Third term: \(ar^2 = 24\) ... (1)

Seventh term: \(ar^6 = 4\tfrac{20}{27} = \dfrac{128}{27}\) ... (2)

Divide (2) by (1):

\[r^4 = \frac{128/27}{24} = \frac{128}{648} = \frac{16}{81}.\]

\[r^2 = \sqrt{\frac{16}{81}} = \frac{4}{9} \Rightarrow r = \frac{2}{3}.\]

From (1): \(a\left(\dfrac{4}{9}\right) = 24 \Rightarrow a = 24 \times \dfrac{9}{4} = 54\).

The first term is \(a = \mathbf{54}\).

(b) \(y\) varies directly as \(x\) and inversely as the square of \(z\):

\[y = \frac{kx}{z^2}.\]

When \(y = 4, x = 3, z = 1\): \(4 = \dfrac{k(3)}{1^2} \Rightarrow k = \dfrac{4}{3}\).

When \(x = 3, z = 2\):

\[y = \frac{\tfrac{4}{3}(3)}{2^2} = \frac{4}{4} = 1.\]

Therefore \(y = \mathbf{1}\).

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(a) The observer is at the mid-point of a vertical cliff. Let the mid-point be at height \(h\) above the foot, so the full cliff height is \(2h\). The boat is 120 m from the foot, and the angle of depression from the mid-point is \(35^\circ\).

The angle of depression equals the angle of elevation from the boat to the mid-point:

\[\tan 35^\circ = \frac{h}{120} \Rightarrow h = 120\tan 35^\circ = 120(0.7002) = 84.03\text{ m}.\]

Height of the cliff \(= 2h = 2(84.03) = 168\text{ m}\) (to 3 s.f.).

(b) Let the distance be \(x\) km. Time at 60 km/h is \(\dfrac{x}{60}\) h; time at 80 km/h is \(\dfrac{x}{80}\) h. The faster motorist arrives 30 minutes \(\left(\tfrac{1}{2}\text{ h}\right)\) earlier:

\[\frac{x}{60} - \frac{x}{80} = \frac{1}{2}.\]

\[x\left(\frac{4 - 3}{240}\right) = \frac{1}{2} \Rightarrow \frac{x}{240} = \frac{1}{2} \Rightarrow x = 120.\]

Therefore \(x = \mathbf{120\text{ km}}\).

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Given quadrilateral \(PQRS\): \(|PS| = 6\text{ cm}\), \(\angle RSP = 90^\circ\), \(|SR| = 9\text{ cm}\), \(|QR| = 8.4\text{ cm}\) and \(|PQ| = 5.4\text{ cm}\).

(a)(i) Constructing quadrilateral \(PQRS\)

1. Draw the base line \(PS = 6\text{ cm}\).
2. At \(S\), erect a perpendicular to \(SP\) (construct \(\angle RSP = 90^\circ\) with ruler and compasses). Along this perpendicular mark off \(|SR| = 9\text{ cm}\) to fix \(R\).
3. With centre \(P\) and radius \(5.4\text{ cm}\) draw an arc; with centre \(R\) and radius \(8.4\text{ cm}\) draw a second arc. The two arcs cut at \(Q\).
4. Join \(PQ\), \(QR\) and \(RS\) to complete quadrilateral \(PQRS\).

(a)(ii) Bisectors of \(\angle RSP\) and \(\angle SPQ\) meeting at \(X\)

1. Bisect \(\angle RSP\): with centre \(S\) draw an arc cutting \(SR\) and \(SP\); from those two intersections draw equal arcs that cross, and join \(S\) to the crossing point. This bisector runs at \(45^\circ\) to \(PS\) (since \(\angle RSP = 90^\circ\)).
2. Bisect \(\angle SPQ\) in the same way, with centre \(P\).
3. Produce both bisectors until they intersect; label this point \(X\).

(a)(iii) The perpendicular \(XT\)

From \(X\), drop a perpendicular to \(PS\) (with centre \(X\) mark two points on \(PS\), then bisect the distance between them). The foot of this perpendicular on \(PS\) is \(T\).

(b) Measurement

Measuring the perpendicular from the accurate construction:

Check by calculation

Take \(S\) as origin with \(SP\) along the horizontal and \(SR\) vertical (\(\angle RSP = 90^\circ\)):

Point \(Q\) lies \(5.4\text{ cm}\) from \(P\) and \(8.4\text{ cm}\) from \(R\), giving \(Q=(7.49,\,5.19)\). \(X\) is equidistant from lines \(PS\), \(SR\) and \(PQ\) (it lies on both bisectors), and this common distance is the required perpendicular:

This agrees with the measured value, so \(|XT| = 3.4\text{ cm}\).

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(a) Area of quadrilateral \(QRST\). From the diagram, \(P, T, S\) lie on a straight top line with \(PS = 14.4\text{ cm}\); \(TQ = 6\text{ cm}\) is perpendicular to \(PS\) at \(T\) (\(\angle PTQ = 90^\circ\)); \(SR\) is perpendicular to \(PS\) at \(S\) (\(\angle PSR = 90^\circ\)); and \(P, Q, R\) are collinear along the slant \(PR\), with \(PQ = 10\text{ cm}\).

Step 1 - the small triangle \(PTQ\). It is right-angled at \(T\), so by Pythagoras

Step 2 - similar triangles. Triangles \(PTQ\) and \(PSR\) share \(\angle P\) and each has a right angle, so they are similar with scale factor

Hence

Step 3 - subtract the areas. The quadrilateral \(QRST\) is the large triangle \(PSR\) with the small triangle \(PTQ\) removed:

(b) Ratio of the rectangle's area to the square's. Let the square have side \(s\), so its area is \(s^2\).

Two opposite sides are decreased by \(10\%\) to \(0.9s\); the other two are increased by \(15\%\) to \(1.15s\). The rectangle's area is

Therefore

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(a) Cost of the fish \(= \text{GH¢}400.00\).

Import duty \(= 15\%\) of 400 \(= 0.15 \times 400 = 60\). Total cost including duty \(= 400 + 60 = 460\).

Sales tax \(= 10\%\) of 460 \(= 46\). Total cost \(= 460 + 46 = 506\).

Selling price \(= 660\), so profit \(= 660 - 506 = 154\).

\[\text{Percentage profit} = \frac{154}{506}\times 100 = 30.4\%\ (\text{to 1 d.p.}).\]

(b) Let the number of girls be \(g\). There are 1000 boys, so the total number of students is \(1000 + g\).

The number that passed is \(48\%\) of the total, and this equals \(50\%\) of the boys plus \(40\%\) of the girls:

\[0.48(1000 + g) = 0.50(1000) + 0.40g.\]

\[480 + 0.48g = 500 + 0.40g \Rightarrow 0.08g = 20 \Rightarrow g = 250.\]

There are \(\mathbf{250}\) girls in the school.

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(a) Completing the table of values for \(y=-x^{2}+x+2\).

Evaluate the relation at each missing value of \(x\):

\(x=-3:\ y=-(-3)^{2}+(-3)+2=-9-3+2=-10\)

\(x=-1:\ y=-(-1)^{2}+(-1)+2=-1-1+2=0\)

\(x=1:\ y=-(1)^{2}+(1)+2=-1+1+2=2\)

\(x=2:\ y=-(2)^{2}+(2)+2=-4+2+2=0\)

The completed table is:

(b) Graph of \(y=-x^{2}+x+2\).

Using a scale of 2 cm to 1 unit on the \(x\)-axis and 2 cm to 2 units on the \(y\)-axis, the seven points \((-3,-10),(-2,-4),(-1,0),(0,2),(1,2),(2,0),(3,-4)\) are plotted and joined with a smooth curve. Because the coefficient of \(x^{2}\) is negative, the curve opens downward with its turning point (a maximum) near \(x=0.5\).

(c) Readings from the graph.

(i) Turning value of \(y\). The curve turns at \(x=\tfrac{1}{2}\), where

\[y=-(0.5)^{2}+0.5+2=-0.25+0.5+2=2.25.\]

Since the parabola opens downward, this turning point is a maximum, so the turning value read from the graph is \(y=2.25\) at \(x=0.5\). Over the drawn range \(-3\le x\le 3\) the least (lowest) value reached is \(y=-10\) at \(x=-3\).

(ii) Roots of \(x^{2}-x-2=0\). Since \(-x^{2}+x+2=-(x^{2}-x-2)\), the equation \(x^{2}-x-2=0\) is equivalent to \(y=0\). The curve cuts the \(x\)-axis where \(y=0\), and from the graph these crossings are at

\[x=-1\quad\text{and}\quad x=2.\]

(iii) Gradient of the curve at \(x=-0.5\). A tangent is drawn to the curve at the point \((-0.5,\,1.25)\). Taking a convenient right-angled triangle along the tangent, the change in \(y\) divided by the change in \(x\) gives

\[\text{gradient}=\frac{\Delta y}{\Delta x}\approx 2.\]

(This agrees with the exact slope: \(\dfrac{dy}{dx}=-2x+1\), so at \(x=-0.5\), \(\dfrac{dy}{dx}=-2(-0.5)+1=1+1=2\).) The gradient of the curve at \(x=-0.5\) is therefore approximately \(2\).

Awọn alaye Idahun

Two fair dice are thrown, giving 36 equally likely ordered outcomes.

(a) Elements.

\(M\) = "sum of the scores is 10":

\[M = \{(4,6),\ (5,5),\ (6,4)\}.\]

\(N\) = "difference between the scores is 3":

\[N = \{(1,4),\ (4,1),\ (2,5),\ (5,2),\ (3,6),\ (6,3)\}.\]

So \(n(M) = 3\) and \(n(N) = 6\).

(b) P(M or N). Checking for common outcomes: the pairs in \(M\) have differences 2, 0, 2, so \(M \cap N = \varnothing\). Hence

\[P(M \cup N) = P(M) + P(N) = \frac{3}{36} + \frac{6}{36} = \frac{9}{36} = \frac{1}{4}.\]

(c) Yes, \(M\) and \(N\) are mutually exclusive, because they have no outcome in common (\(M \cap N = \varnothing\)); no single throw can give a sum of 10 and a difference of 3 at the same time.

Awọn alaye Idahun

(a) Cumulative frequency table.

Total number of participants \(N = 10+18+22+25+16+9 = 100\). The cumulative frequency is plotted against the upper class boundary of each class.

(b) Cumulative frequency curve (ogive).

The curve begins at the lower boundary of the first class, \((19.5,\,0)\), and passes through each point \((\text{upper boundary},\ \text{cumulative frequency})\); the points are joined with a smooth curve.

(c) Estimates read from the curve.

(i) Median \((Q_2)\). The median is the weight corresponding to \(\dfrac{N}{2} = \dfrac{100}{2} = 50\) on the cumulative-frequency axis. Drawing a horizontal line from cumulative frequency \(50\) to the curve and dropping down to the weight axis:

(ii) Semi-interquartile range. The lower quartile \(Q_1\) is read at \(\dfrac{N}{4} = \dfrac{100}{4} = 25\), and the upper quartile \(Q_3\) at \(\dfrac{3N}{4} = \dfrac{300}{4} = 75\). From the curve:

(iii) Probability of weighing at least 60 kg. A weight of at least \(60\) kg corresponds to the class boundary \(59.5\) kg. From the table the cumulative frequency at \(59.5\) kg is \(75\), so the number weighing at least \(60\) kg is \(100 - 75 = 25\).