WAEC SSCE - General Mathematics - 2011
Question 1 Report
From the equation whose roots are x = \(\frac{1}{2}\) and -\(\frac{2}{3}\)
Answer Details
When a quadratic equation has roots at x = a and x = b, it can be written in factored form as (x-a)(x-b) = 0. Therefore, from the given roots, the factors are (x - \(\frac{1}{2}\)) and (x + \(\frac{2}{3}\)). To get the quadratic equation, we can expand the factors by multiplying them together, which gives us: (x - \(\frac{1}{2}\))(x + \(\frac{2}{3}\)) = x2 - \(\frac{1}{2}\)x + \(\frac{2}{3}\)x - \(\frac{1}{2}\)\(\frac{2}{3}\) = x2 + \(\frac{1}{6}\)x - \(\frac{1}{3}\) Therefore, the correct option is 6x2 + x - 2 = 0.
Question 2 Report
The perpendicular bisectors of the sides of an acute-angled triangle are drawn. Which of these statements is correct? They intersect
Answer Details
The perpendicular bisectors of the sides of an acute-angled triangle intersect at a point inside the triangle. This point is called the circumcenter, which is equidistant from the three vertices of the triangle. To see why this is true, consider two sides of the triangle, and let their perpendicular bisectors intersect at a point O. Since O lies on the perpendicular bisector of each of the two sides, it is equidistant from the endpoints of each of those sides. Therefore, O is equidistant from two vertices of the triangle. Similarly, O is equidistant from the third vertex, so it must be the circumcenter of the triangle. Since the triangle is acute-angled, the circumcenter lies inside the triangle.
Question 3 Report
Simplify \(\frac{\log \sqrt{27}}{\log \sqrt{81}}\)
Answer Details
Using the property that \(\log_{a}b = \frac{\log{b}}{\log{a}}\), we can simplify the given expression as follows: \[\frac{\log \sqrt{27}}{\log \sqrt{81}} = \frac{\log 27^{\frac{1}{2}}}{\log 81^{\frac{1}{2}}} = \frac{\frac{1}{2}\log 27}{\frac{1}{2}\log 81} = \frac{\log 3^3}{\log 3^4} = \frac{3\log 3}{4\log 3} = \frac{3}{4}\] Therefore, the simplified form of \(\frac{\log \sqrt{27}}{\log \sqrt{81}}\) is \(\frac{3}{4}\), and the correct option is (D).
Question 6 Report
In the diagram /Pq//TS//TU, reflex angle QPS = 245o angle PST = 115o, , STU = 65o and < RPS = x. Find the value of x
Answer Details
Question 7 Report
If a number is chosen at random from the set {x: 4 \(\leq x \leq 15\)}. Find the probability that it is a multiple of 3 or a multiple of 4
Question 8 Report
Factorize the expression: am + bn - an - bm
Answer Details
We can begin by grouping the first two terms and the last two terms together: am + bn = a(m) + b(n) = (a+b)n - bn an + bm = a(n) + b(m) = (a+b)m - am Now, we can substitute these expressions back into the original equation: am + bn - an - bm = [(a+b)n - bn] - [(a+b)m - am] We can simplify this expression by combining like terms: am + bn - an - bm = (a+b)n - bn - (a+b)m + am am + bn - an - bm = (a+b)n - (a+b)m + am - bn Finally, we can factor out the common factor of (a+b) from the first two terms and the common factor of (-1) from the last two terms: am + bn - an - bm = (a+b)(n-m) - (b-a)(n-m) Therefore, the answer is (a+b)(n-m) - (b-a)(n-m), which can be further simplified to (a-b)(m-n). Thus, the correct option is (a - b)(m - n).
Question 9 Report
The diagram shows a cyclic quadrilateral PQRS with its diagonals intersecting at K. Which of the following triangles is similar to triangle QKR?
Answer Details
Question 10 Report
G varies directly as the square of H, If G is 4 when H is 3, find H when G = 100
Answer Details
In this problem, we are given that G varies directly as the square of H. This means that if H is multiplied by some factor, then G will be multiplied by the square of that factor. Mathematically, we can write this as: G ∝ H^2 where the symbol "∝" means "varies directly as". We are also given that G is 4 when H is 3. Using this information, we can write: 4 ∝ 3^2 To find H when G = 100, we can use the same relationship: G ∝ H^2 If we let the constant of proportionality be k, we can write: G = kH^2 To solve for k, we can use the initial condition where G is 4 when H is 3: 4 = k(3^2) Simplifying, we get: k = 4/9 Now we can use this value of k to find H when G is 100: 100 = (4/9)H^2 Multiplying both sides by 9/4, we get: 225 = H^2 Taking the square root of both sides, we get: H = 15 Therefore, the correct answer is (a) 15. In summary, we used the direct variation relationship between G and H^2 to find the constant of proportionality, and then used that constant and the given value of G to solve for H.
Question 11 Report
Find the smaller value of x that satisfies the equation x2 + 7x + 10 = 0
Answer Details
We are given a quadratic equation x2 + 7x + 10 = 0 and we need to find the smaller value of x that satisfies the equation. To solve the equation, we can factorize it by finding two numbers whose product is 10 and whose sum is 7. We can see that the two numbers are 2 and 5, since 2 × 5 = 10 and 2 + 5 = 7. So, we can write the equation as (x + 2)(x + 5) = 0. For this equation to be true, either (x + 2) = 0 or (x + 5) = 0. Therefore, we get x = -2 or x = -5. Since we are asked to find the smaller value of x, we choose x = -5 as the answer. Hence, the smaller value of x that satisfies the equation x2 + 7x + 10 = 0 is -5.
Question 12 Report
The venn diagram shows the number of students in a class who like reading(R), dancing(D) and swimming(S). How many students like dancing and swimming?
Question 13 Report
in a quiz competition, a student answers n questions correctly and was given D(n + 50) for each question correctly answered. If he gets D600.00 altogether, how many questions did he answer correctly?
Answer Details
Question 14 Report
The graph represents the relation y = x\(^2\) - 3x - 3. What is the equation of the line of symmetry of the graph?
Answer Details
To find the equation of the line of symmetry of the graph, we need to identify the axis of symmetry. The axis of symmetry is a vertical line that passes through the vertex of the parabola. The vertex of the parabola is the point where the parabola changes direction, and it can be found by using the formula: x = -b / (2a) where a and b are the coefficients of the quadratic equation y = ax\(^2\) + bx + c. In the given equation y = x\(^2\) - 3x - 3, a = 1, b = -3, and c = -3. Substituting these values in the formula, we get: x = -(-3) / (2*1) = 3/2 = 1.5 Therefore, the line of symmetry is a vertical line passing through x = 1.5. So, the correct answer is (C) x = 1.5.
Question 15 Report
In the diagram, /MN/, /OP/, /QOP/ = 125o. What is the size of < MQR?
Question 16 Report
Simplify \(\frac{m}{n} + \frac{(m - 1)}{5n} = \frac{(m - 2)}{10n}\) where n \(\neq\) 0
Answer Details
Question 17 Report
The cross section section of a uniform prism is a right-angled triangle with sides 3cm. 4cm and 5cm. If its length is 10cm. Calculate the total surface area
Answer Details
Question 18 Report
Esther was facing S 20° W. She turned 90° in the clock wise direction. What direction is she facing?
Answer Details
If Esther was initially facing S 20° W and then turned 90° clockwise, she would end up facing in a new direction. To determine the new direction, we can add 90° to her initial direction. When we add 90° to S 20° W, we rotate the direction clockwise by 90°, which means the new direction will be to the right of the initial direction. To find the new direction, we need to subtract the initial angle from 90°: 90° - 20° = 70° Therefore, Esther is facing in the direction of N 70° W after turning 90° clockwise from her initial direction of S 20° W.
Question 19 Report
John pours 96 litres of red oil into a rectangular container with length 220cm and breadth 40cm. Calculate, correct to the nearest cm, the height of the oil in the container
Answer Details
To solve this problem, we need to use the formula for the volume of a rectangular container: Volume = Length x Breadth x Height We are given the length and breadth of the container, as well as the volume of the oil. We need to find the height of the oil. First, we need to convert the volume of the oil from litres to cubic centimeters, since the dimensions of the container are given in centimeters. 1 litre = 1000 cubic centimeters Therefore, 96 litres = 96 x 1000 = 96,000 cubic centimeters Now, we can plug in the values we have into the formula for the volume of the container: Volume = Length x Breadth x Height 96,000 = 220 x 40 x Height Simplifying, we get: Height = 96,000 / (220 x 40) Height = 2.18 To round off to the nearest centimeter, we need to look at the first decimal place. If the value in the first decimal place is 5 or more, we round up. If it is less than 5, we round down. In this case, the value in the first decimal place is 1, which is less than 5. Therefore, we round down to 2. So, the height of the oil in the container is approximately 2cm. Therefore, the correct answer is (a) 11cm.
Question 20 Report
What is the value of 3 in the number 42.7531?
Answer Details
The number 42.7531 can be written in expanded form as: 42 + 0.7 + 0.05 + 0.003 + 0.0001 The digit 3 is located in the thousandths place, which represents the decimal value of 0.001. So, the value of 3 in the number 42.7531 is equal to: 3 x 0.001 = 0.003 Therefore, the correct answer is option (A), which is 3 divided by 10000.
Question 21 Report
If N112.00 exchanges for D14.95, calculate the value of D1.00 in naira
Answer Details
To calculate the value of D1.00 in naira, we can use the given exchange rate of N112.00 to D14.95. We can find the value of 1 D in Naira by dividing N112.00 by the equivalent value of D14.95. So, 1 D = N112.00/D14.95 To simplify this, we can first convert D14.95 to its decimal equivalent by dividing by 100: D14.95 = 14.95/100 = 0.1495 Now we can substitute this value into the equation: 1 D = N112.00/0.1495 Simplifying this expression, we get: 1 D = N748.16 Therefore, the value of D1.00 in Naira is N748.16. Answer: 7.49.
Question 22 Report
In the diagram, PO and OR are radii, |PQ| = |QR| and reflex < PQR is 240o. Calculate the value x
Answer Details
In the given diagram, we have a circle with center O and radii OP and OR. The reflex angle PQR is 240° and |PQ| = |QR|. We need to find the value of x. Since |PQ| = |QR|, we know that triangle PQR is an isosceles triangle. Therefore, the angles opposite to PQ and QR are equal. Let's denote the angle PQR by y. Then we have: 2y + 60° = 360° (sum of angles in a triangle) 2y = 300° y = 150° Therefore, each of the angles opposite to PQ and QR is equal to (180° - 150°)/2 = 15°. Now, consider the triangle OQP. We know that the sum of angles in a triangle is 180°. Therefore: ∠OQP + ∠QOP + ∠OPQ = 180° Since OP and OQ are radii, ∠QOP = ∠OPQ. Let's denote this angle by z. Then we have: z + z + 15° = 180° 2z = 165° z = 82.5° Finally, consider the triangle OXR. We know that the sum of angles in a triangle is 180°. Therefore: ∠OXR + ∠ORX + ∠ROX = 180° Since OR and OX are radii, ∠ORX = ∠ROX. Let's denote this angle by x. Then we have: x + x + 60° = 180° 2x = 120° x = 60° Therefore, the value of x is 60°. Answer: 60°.
Question 24 Report
Given that cos xo = \(\frac{1}{r}\), express tan x in terms of r
Answer Details
We know that: cos x = adjacent side/hypotenuse So, if cos x = 1/r, then adjacent side = 1 and hypotenuse = r. Using the Pythagorean theorem, we can find the opposite side: opposite side = √(hypotenuse^2 - adjacent side^2) = √(r^2 - 1) Finally, we can find the value of tan x: tan x = opposite side/adjacent side = √(r^2 - 1)/1 = √(r^2 - 1) Therefore, the answer is (d) \(\sqrt{r^2 - 1}\).
Question 25 Report
from the diagram, Which of the following statements are true? i. m = q ii. n = q iii. n + p = 180o iv. p + m = 180o
Answer Details
In the given diagram, we can see that lines n and q are parallel and m is a transversal cutting them. Therefore, angles n and q are alternate interior angles and are equal, i.e., statement i is true. Also, we can see that lines n and p are parallel and q is a transversal cutting them. Therefore, angles n and p are corresponding angles and are equal. As the sum of the corresponding angles is equal to 180 degrees, we have n + p = 180 degrees, i.e., statement iii is also true. However, we cannot determine whether statement ii and iv are true or not based on the given information and the diagram. Therefore, the correct answer is (a) i and iii.
Question 26 Report
The length of a piece of stick is 1.75m. A girl measured it as 1.80m. Find the percentage error
Answer Details
The actual length of the stick is 1.75m and the measured length is 1.80m. The error is the difference between the actual and measured length: 1.80m - 1.75m = 0.05m To find the percentage error, we divide the error by the actual length and multiply by 100%: \frac{0.05}{1.75} \times 100\% \approx 2.857\% \approx \frac{20}{7}\% Therefore, the percentage error is approximately \frac{20}{7}\%.
Question 27 Report
Given that n(p) = 19, m(P \(\cup\) Q) = 38 and n(P \(\cap\) Q) = 7, Find n(C)
Answer Details
Question 28 Report
A rectangular garden measures 18.6m by 12.5m. Calculate, correct to three significant figures, the area of the garden
Answer Details
The area of a rectangle is given by multiplying the length by the width. Therefore, the area of the garden is: Area = length × width Area = 18.6m × 12.5m Area = 232.5m2 Rounding to three significant figures gives 233m2. Therefore, the answer is (d) 233m2.
Question 29 Report
Which of these angles can be constructed using ruler and a pair of compasses only?
Answer Details
Question 30 Report
What is the value of 3 in the number 42.7531?
Answer Details
The digit 3 in the number 42.7531 is in the thousandth place (the digit after the decimal point and three places to the right of it). Therefore, its value is \(\frac{3}{1000}\).
Question 31 Report
What must be added to (2x - 3y) to get (x - 2y)?
Answer Details
To get from (2x - 3y) to (x - 2y), we need to subtract x from 2x and add 2y to -3y. Therefore, we need to add (x - 2y) - (2x - 3y) to (2x - 3y) to get (x - 2y). Simplifying (x - 2y) - (2x - 3y), we have: (x - 2y) - (2x - 3y) = x - 2y - 2x + 3y = -x + y Therefore, we need to add (-x + y) to (2x - 3y) to get (x - 2y). Simplifying (2x - 3y) + (-x + y), we have: (2x - 3y) + (-x + y) = 2x - 3y - x + y = x - 2y So, we need to add (-x + y) to (2x - 3y) to get (x - 2y). Therefore, the answer is (B) y - x.
Question 32 Report
The histogram shows the age distribution of members of a club. How many members are in the club?
Answer Details
Question 33 Report
One of the factors of (mn - nq - n2 + mq) is (m - n). The other factor is?
Answer Details
Question 34 Report
The height of a cylinder is equal to its radius. If the volume is 0.216 \(\pi\) m\(^3\). Calculate the radius.
Answer Details
Let's denote the radius of the cylinder as r and its height as h. We are given that the height of the cylinder is equal to its radius, so h = r. We also know the volume of the cylinder, which is given by: V = \(\pi\)r\(^2\)h Substituting h = r, we get: V = \(\pi\)r\(^2\)r = \(\pi\)r\(^3\) We are given that the volume of the cylinder is 0.216 \(\pi\) m\(^3\). So, we can solve for r as follows: 0.216 \(\pi\) = \(\pi\)r\(^3\) r\(^3\) = 0.216 Taking the cube root of both sides, we get: r = 0.6 Therefore, the radius of the cylinder is 0.6 meters. So, the answer is 0.60m.
Question 35 Report
The perimeter of a sector of a circle of radius 4cm is (\(\pi + 8\))cm. Calculate the anle of the sector
Question 36 Report
Solve for x in the equation; \(\frac{3}{5}\)(2x - 1) = \(\frac{1}{4}\)(5x - 3)
Answer Details
Question 37 Report
In the diagram /Pq//TS//TU, reflex angle QPS = 245o angle PST = 115o, , STU = 65o and < RPS = x. Find the value of x
Question 38 Report
A regular polygon of n sides has each exterior angle to 45o. Find the value of n
Answer Details
In a regular polygon with n sides, each exterior angle measures 360/n degrees. We are given that in this polygon, each exterior angle is 45 degrees. Therefore, we can set up an equation: 360/n = 45 To solve for n, we can cross-multiply and simplify: 360 = 45n n = 360/45 n = 8 Therefore, the regular polygon in question has 8 sides. Answer: 8.
Question 39 Report
Using the cumulative frequency curve, estimate the median of the data represented on the graph.
Answer Details
Question 40 Report
A cylindrical container has a base radius of 14cm and height 18cm. How many litres of liquid can it hold? correct to the nearest litre [Take \(\pi = \frac{22}{7}\)]
Answer Details
The volume of a cylinder can be calculated using the formula: V = πr2h, where r is the radius of the base and h is the height of the cylinder. Substituting the given values, we have: V = π(14)2(18) V = 11088 cm3 To convert cm3 to litres, we divide by 1000: V = 11088/1000 V = 11.088 litres Rounding to the nearest litre, we get: V ≈ 11 litres Therefore, the answer is 11.
Question 41 Report
If 27x = 9y. Find the value of \(\frac{x}{y}\)
Answer Details
If we can find the value of x and y, then we can calculate x/y by dividing x by y. Given: 27x = 9y We can rewrite 27 as 33 and 9 as 32 to get: (33)x = (32)y Applying the power of a power rule, we get: 33x = 32y For two exponential expressions to be equal, their bases must be equal. Therefore: 33x = 32y implies 3x = 2y Dividing both sides by y, we get: \(\frac{x}{y} = \frac{2}{3}\) Therefore, the answer is \(\frac{2}{3}\).
Question 42 Report
In the diagram, STUV is a straight line. < TSY = < UXY = 40o and < VUW = 110o. Calculate < TYW
Question 43 Report
The graph represents the relation y = xo2 - 3x - 3. Find the value of x for which x2 - 3x = 7
Answer Details
Question 44 Report
The height of a cylinder is equal to its radius. If the volume is 0.216 \(\pi m^3\) Calculate the radius.
Answer Details
Question 45 Report
a boy looks through a window of a building and sees a mango fruit on the ground 50m away from the foot of the building. If the window is 9m from the ground, calculate, correct to the nearest degree, the angle of depression of the mango from the window
Question 46 Report
John pours 96 litres of red oil into a rectangular container with length 220cm and breadth 40cm. Calculate, correct to the nearest cm, the height of the oil in the container
Answer Details
To calculate the height of the oil in the container, we need to use the formula for the volume of a rectangular prism: Volume = length x breadth x height First, we need to convert the given volume from liters to cubic centimeters, since the dimensions of the container are in centimeters. 96 liters = 96,000 cubic centimeters Next, we can plug in the given values into the formula: 96,000 = 220 x 40 x height Solving for height, we get: height = 96,000 / (220 x 40) height ≈ 11.0 cm (rounded to the nearest cm) Therefore, the height of the oil in the container is approximately 11 cm. Note: When working with volume, it's important to make sure the units are consistent throughout the problem. In this case, we converted liters to cubic centimeters to match the dimensions of the container.
Question 47 Report
If \(\sqrt{72} + \sqrt{32} - 3 \sqrt{18} = x \sqrt{8}\), Find the value of x
Answer Details
Question 48 Report
Solve the equation; 3x - 2y = 7, x + 2y = -3
Answer Details
To solve this system of equations, we can use the method of elimination. We will add the two equations together, which will eliminate the y variable: (3x - 2y) + (x + 2y) = 7 + (-3) Simplifying, we get: 4x = 4 Dividing both sides by 4, we get: x = 1 Substituting x = 1 into one of the equations, we get: 1 + 2y = -3 Solving for y, we get: y = -2 Therefore, the solution to the system of equations is x = 1, y = -2. So, the correct option is (a) x = 1, y = -2.
Question 49 Report
The histogram shows the age distribution of members of a club. What is their modal age?
Answer Details
Question 50 Report
(a) Simplify : \(\frac{\frac{1}{2} of \frac{1}{4} \div \frac{1}{3}}{\frac{1}{6} - \frac{3}{4} + \frac{1}{2}}\).
(b) Given that \(\sqrt{x} = 10^{\bar{1}.6741}\), without using calculators, find the value of x.
(a) Numerator: \(\tfrac{1}{2}\text{ of }\tfrac{1}{4} \div \tfrac{1}{3} = \tfrac{1}{8}\div\tfrac{1}{3} = \tfrac{1}{8}\times3 = \tfrac{3}{8}.\)
Denominator: \(\tfrac{1}{6} - \tfrac{3}{4} + \tfrac{1}{2} = \tfrac{2 - 9 + 6}{12} = -\tfrac{1}{12}.\)
\[\frac{\tfrac{3}{8}}{-\tfrac{1}{12}} = \tfrac{3}{8}\times(-12) = -\tfrac{9}{2} = -4\tfrac{1}{2}.\]
(b) The bar denotes a negative characteristic, so \(\bar{1}.6741 = -1 + 0.6741 = -0.3259\). Squaring \(\sqrt{x}\):
\[x = \left(10^{\bar{1}.6741}\right)^2 = 10^{\,2(-0.3259)} = 10^{-0.6518} = 10^{\bar{1}.3482}.\]
The antilog of \(0.3482\) is \(2.229\), and the characteristic \(\bar{1}\) places one zero after the point:
\[x = 0.2229 \approx 0.223.\]
Answer Details
(a) Numerator: \(\tfrac{1}{2}\text{ of }\tfrac{1}{4} \div \tfrac{1}{3} = \tfrac{1}{8}\div\tfrac{1}{3} = \tfrac{1}{8}\times3 = \tfrac{3}{8}.\)
Denominator: \(\tfrac{1}{6} - \tfrac{3}{4} + \tfrac{1}{2} = \tfrac{2 - 9 + 6}{12} = -\tfrac{1}{12}.\)
\[\frac{\tfrac{3}{8}}{-\tfrac{1}{12}} = \tfrac{3}{8}\times(-12) = -\tfrac{9}{2} = -4\tfrac{1}{2}.\]
(b) The bar denotes a negative characteristic, so \(\bar{1}.6741 = -1 + 0.6741 = -0.3259\). Squaring \(\sqrt{x}\):
\[x = \left(10^{\bar{1}.6741}\right)^2 = 10^{\,2(-0.3259)} = 10^{-0.6518} = 10^{\bar{1}.3482}.\]
The antilog of \(0.3482\) is \(2.229\), and the characteristic \(\bar{1}\) places one zero after the point:
\[x = 0.2229 \approx 0.223.\]
Question 51 Report
In a class of 40 students, 18 passed Mathematics, 19 passed Accounts, 16 passed Economics, 5 passed Mathematics and Accounts only, 6 Mathematics only, 9 Accounts only, 2 Accounts and Economics only. If each student offered at least one of the subjects,
(a) how many students failed in all subjects?
(b) find the percentage number that failed in at least one of Economics and Mathematics
(c) calculate the probability that a student picked at random failed in Accounts?
Answer Details
None
Question 52 Report
A library received $1,300 grant. It spends 10% of the grant on magazine subscriptions, 35% on new books, 15% to repair damaged books, 30% to buy new furniture and 10% to train library staff.
(a) Represent this information on a pie chart.
(b) Calculate, correct to the nearest whole number, the percentage increase of the amount for buying books over that of new furniture.
(a) Pie chart
The whole grant of \($1{,}300\) is represented by the full circle, \(360^{\circ}\). Each item's sector angle is found from \(\text{angle}=\dfrac{\text{percentage}}{100}\times360^{\circ}\), which is the same as \(\dfrac{\text{amount}}{1300}\times360^{\circ}\).
| Item | Percentage | Amount ($) | Sector angle |
|---|---|---|---|
| Magazine subscriptions | 10% | 130 | \(\frac{130}{1300}\times360^{\circ}=36^{\circ}\) |
| New books | 35% | 455 | \(\frac{455}{1300}\times360^{\circ}=126^{\circ}\) |
| Repair of damaged books | 15% | 195 | \(\frac{195}{1300}\times360^{\circ}=54^{\circ}\) |
| New furniture | 30% | 390 | \(\frac{390}{1300}\times360^{\circ}=108^{\circ}\) |
| Staff training | 10% | 130 | \(\frac{130}{1300}\times360^{\circ}=36^{\circ}\) |
| Total | 100% | 1300 | \(360^{\circ}\) |
Drawing a circle and marking each sector in turn with the angles above gives the following pie chart:
(b) Percentage increase of books over furniture
Amount for new books \(=35\%\) of \(1300 = \$455\).
Amount for new furniture \(=30\%\) of \(1300 = \$390\).
The increase in the amount is \(455-390=\$65\), and this is expressed as a percentage of the furniture amount:
\[\text{Percentage increase}=\frac{455-390}{390}\times100\%=\frac{65}{390}\times100\%=16.67\%\approx 17\%\]The books allocation is about \(17\%\) more than the furniture allocation (to the nearest whole number).
Answer Details
(a) Pie chart
The whole grant of \($1{,}300\) is represented by the full circle, \(360^{\circ}\). Each item's sector angle is found from \(\text{angle}=\dfrac{\text{percentage}}{100}\times360^{\circ}\), which is the same as \(\dfrac{\text{amount}}{1300}\times360^{\circ}\).
| Item | Percentage | Amount ($) | Sector angle |
|---|---|---|---|
| Magazine subscriptions | 10% | 130 | \(\frac{130}{1300}\times360^{\circ}=36^{\circ}\) |
| New books | 35% | 455 | \(\frac{455}{1300}\times360^{\circ}=126^{\circ}\) |
| Repair of damaged books | 15% | 195 | \(\frac{195}{1300}\times360^{\circ}=54^{\circ}\) |
| New furniture | 30% | 390 | \(\frac{390}{1300}\times360^{\circ}=108^{\circ}\) |
| Staff training | 10% | 130 | \(\frac{130}{1300}\times360^{\circ}=36^{\circ}\) |
| Total | 100% | 1300 | \(360^{\circ}\) |
Drawing a circle and marking each sector in turn with the angles above gives the following pie chart:
(b) Percentage increase of books over furniture
Amount for new books \(=35\%\) of \(1300 = \$455\).
Amount for new furniture \(=30\%\) of \(1300 = \$390\).
The increase in the amount is \(455-390=\$65\), and this is expressed as a percentage of the furniture amount:
\[\text{Percentage increase}=\frac{455-390}{390}\times100\%=\frac{65}{390}\times100\%=16.67\%\approx 17\%\]The books allocation is about \(17\%\) more than the furniture allocation (to the nearest whole number).
Question 53 Report
Using ruler and a pair of compasses only,
(a) construct a rhombus PQRS of side 7 cm and < PQR = 60°;
(b) locate point X such that X lies on the locus of points equidistant from PQ and QR and also equidistant from Q and R ;
(c) measure |XR|.
Construction (ruler and a pair of compasses only).
(a) Rhombus PQRS of side 7 cm with ∠PQR = 60°.
- Draw the base QR = 7 cm.
- At Q construct an angle of 60°: with centre Q and a convenient radius draw an arc cutting QR; with the same radius and centre at that cut, step one arc across the first to fix the 60° direction. Along this direction mark off QP = 7 cm.
- With centre P, radius 7 cm, and centre R, radius 7 cm, draw two arcs meeting at S. Join RS and SP. PQRS is the required rhombus (all four sides 7 cm).
(b) Locating the point X.
- Bisect ∠PQR. This bisector from Q is the locus of points equidistant from the lines PQ and QR.
- Construct the perpendicular bisector of QR. This is the locus of points equidistant from Q and R.
- The two loci intersect at X (shown as dashed construction lines below).
(c) Measuring |XR|.
Because X lies on the bisector of the 60° angle at Q, the line QX makes 30° with QR. Because X also lies on the perpendicular bisector of QR, the foot M of that bisector is the midpoint of QR, so \(QM = \tfrac{7}{2} = 3.5\text{ cm}\). In right-angled triangle QMX (right angle at M):
\[ QX = \frac{QM}{\cos 30^\circ} = \frac{3.5}{0.8660} = 4.04\text{ cm}. \]Since X is equidistant from Q and R, \(|XR| = |XQ|\). Therefore
\[ |XR| \approx 4.0\text{ cm}. \]Measured directly on the accurate construction, \(|XR| \approx 4.0\text{ cm}\).
Answer Details
Construction (ruler and a pair of compasses only).
(a) Rhombus PQRS of side 7 cm with ∠PQR = 60°.
- Draw the base QR = 7 cm.
- At Q construct an angle of 60°: with centre Q and a convenient radius draw an arc cutting QR; with the same radius and centre at that cut, step one arc across the first to fix the 60° direction. Along this direction mark off QP = 7 cm.
- With centre P, radius 7 cm, and centre R, radius 7 cm, draw two arcs meeting at S. Join RS and SP. PQRS is the required rhombus (all four sides 7 cm).
(b) Locating the point X.
- Bisect ∠PQR. This bisector from Q is the locus of points equidistant from the lines PQ and QR.
- Construct the perpendicular bisector of QR. This is the locus of points equidistant from Q and R.
- The two loci intersect at X (shown as dashed construction lines below).
(c) Measuring |XR|.
Because X lies on the bisector of the 60° angle at Q, the line QX makes 30° with QR. Because X also lies on the perpendicular bisector of QR, the foot M of that bisector is the midpoint of QR, so \(QM = \tfrac{7}{2} = 3.5\text{ cm}\). In right-angled triangle QMX (right angle at M):
\[ QX = \frac{QM}{\cos 30^\circ} = \frac{3.5}{0.8660} = 4.04\text{ cm}. \]Since X is equidistant from Q and R, \(|XR| = |XQ|\). Therefore
\[ |XR| \approx 4.0\text{ cm}. \]Measured directly on the accurate construction, \(|XR| \approx 4.0\text{ cm}\).
Question 54 Report
(a) P varies directly as Q and inversely as the square of R. If P = 1 when Q = 8 and R = 2, find the value of Q when P = 3 and R = 5.
(b) An aeroplane flies from town A(20°N, 60°E) to town B(20°N, 20°E). (i) if the journey takes 6 hours, calculate, correct to 3 significant figures, the average speed of the aeroplane. (ii) if it then flies due North from town B to town C, 420 km away, calculate correct to the nearest degree, the latitude of town C. [Take radius of the earth = 6400 km and \(\pi\) = 3.142].
(a) \(P\propto\dfrac{Q}{R^{2}}\Rightarrow P=\dfrac{kQ}{R^{2}}\).
Using \(P=1,\ Q=8,\ R=2\): \(1=\dfrac{8k}{4}=2k\Rightarrow k=\tfrac{1}{2}\).
When \(P=3,\ R=5\): \(3=\dfrac{\tfrac{1}{2}\,Q}{25}=\dfrac{Q}{50}\Rightarrow Q=150\).
\(Q=150\)
(b)(i) \(A(20^{\circ}N,60^{\circ}E)\) and \(B(20^{\circ}N,20^{\circ}E)\) are on the same parallel of latitude, so the flight is along the parallel \(20^{\circ}N\). Difference in longitude \(=60-20=40^{\circ}\).
\[\text{Distance}=\frac{40}{360}\times2\pi R\cos20^{\circ}=\frac{40}{360}\times2(3.142)(6400)(0.9397)\approx 4199\text{ km}\]\[\text{Average speed}=\frac{4199}{6}\approx 700\text{ km/h (3 s.f.)}\](ii) Flying due North from \(B\) is along a meridian, where distance \(=\dfrac{\theta}{360}\times2\pi R\), \(\theta\) the change in latitude.
\[420=\frac{\theta}{360}\times2(3.142)(6400)\Rightarrow \theta=\frac{420\times360}{40217.6}\approx 3.76^{\circ}\]Starting at \(20^{\circ}N\) and moving north, latitude of \(C=20+3.76\approx 24^{\circ}N\).
Latitude of C \(\approx 24^{\circ}N\).
Answer Details
(a) \(P\propto\dfrac{Q}{R^{2}}\Rightarrow P=\dfrac{kQ}{R^{2}}\).
Using \(P=1,\ Q=8,\ R=2\): \(1=\dfrac{8k}{4}=2k\Rightarrow k=\tfrac{1}{2}\).
When \(P=3,\ R=5\): \(3=\dfrac{\tfrac{1}{2}\,Q}{25}=\dfrac{Q}{50}\Rightarrow Q=150\).
\(Q=150\)
(b)(i) \(A(20^{\circ}N,60^{\circ}E)\) and \(B(20^{\circ}N,20^{\circ}E)\) are on the same parallel of latitude, so the flight is along the parallel \(20^{\circ}N\). Difference in longitude \(=60-20=40^{\circ}\).
\[\text{Distance}=\frac{40}{360}\times2\pi R\cos20^{\circ}=\frac{40}{360}\times2(3.142)(6400)(0.9397)\approx 4199\text{ km}\]\[\text{Average speed}=\frac{4199}{6}\approx 700\text{ km/h (3 s.f.)}\](ii) Flying due North from \(B\) is along a meridian, where distance \(=\dfrac{\theta}{360}\times2\pi R\), \(\theta\) the change in latitude.
\[420=\frac{\theta}{360}\times2(3.142)(6400)\Rightarrow \theta=\frac{420\times360}{40217.6}\approx 3.76^{\circ}\]Starting at \(20^{\circ}N\) and moving north, latitude of \(C=20+3.76\approx 24^{\circ}N\).
Latitude of C \(\approx 24^{\circ}N\).
Question 55 Report
(a) Divide \(\frac{x^{2} - 4}{x^{2} + x}\) by \(\frac{x^{2} - 4x + 4}{x + 1}\).
(b) The diagram below shows the graphs of \(y = ax^{2} + bx + c\) and \(y = mx + k\) where a, b, c and m are constants. Use the graph(s) to :
(i) find the roots of the equation \(ax^{2} + bx + c = mx + k\);
(ii) determine the values of a, b and c using the coordinates of points L, M and N and hence write down the equation of the curve;
(iii) determine the line of symmetry of the curve \(y = ax^{2} + bx + c\).
(a) Divide the algebraic fractions.
Dividing by a fraction is the same as multiplying by its reciprocal:
\[\frac{x^{2}-4}{x^{2}+x}\div\frac{x^{2}-4x+4}{x+1}=\frac{x^{2}-4}{x^{2}+x}\times\frac{x+1}{x^{2}-4x+4}\]Factorise each expression fully:
\[x^{2}-4=(x-2)(x+2),\qquad x^{2}+x=x(x+1),\qquad x^{2}-4x+4=(x-2)^{2}\]Substitute the factorised forms:
\[=\frac{(x-2)(x+2)}{x(x+1)}\times\frac{x+1}{(x-2)^{2}}\]Cancel the common factors \((x+1)\) and one \((x-2)\):
\[=\frac{x+2}{x(x-2)}=\frac{x+2}{x^{2}-2x}\](b) Using the graph.
Reading the marked points from the diagram, the curve \(y=ax^{2}+bx+c\) cuts the x-axis at \(L(-1,0)\) and \(N(2,0)\) and cuts the y-axis at \(M(0,2)\). The straight line \(y=mx+k\) passes through \(L(-1,0)\) and cuts the curve again at \(H(1.5,\,1.25)\). The reconstructed graph below shows these features.
(ii) Values of \(a\), \(b\), \(c\) and the equation of the curve. (Solved first, as the roots in part (i) depend on this equation.)
Substitute the coordinates of \(L\), \(M\) and \(N\) into \(y=ax^{2}+bx+c\):
\[L(-1,0):\;\;a(-1)^{2}+b(-1)+c=0\;\Rightarrow\;a-b+c=0\quad(1)\]\[M(0,2):\;\;a(0)^{2}+b(0)+c=2\;\Rightarrow\;c=2\quad(2)\]\[N(2,0):\;\;a(2)^{2}+b(2)+c=0\;\Rightarrow\;4a+2b+c=0\quad(3)\]Put \(c=2\) from (2) into (1) and (3):
\[a-b=-2\quad(1a),\qquad 4a+2b=-2\quad(3a)\]From (1a), \(b=a+2\). Substitute into (3a):
\[4a+2(a+2)=-2\;\Rightarrow\;6a+4=-2\;\Rightarrow\;6a=-6\;\Rightarrow\;a=-1\]Then \(b=a+2=1\) and \(c=2\). Therefore
\[a=-1,\qquad b=1,\qquad c=2,\]and the equation of the curve is
\[y=-x^{2}+x+2.\](i) Roots of \(ax^{2}+bx+c=mx+k\).
The solutions are the x-coordinates of the points where the line meets the curve, namely \(L\) and \(H\). First find the line \(y=mx+k\) through \(L(-1,0)\) and \(H(1.5,1.25)\):
\[m=\frac{1.25-0}{1.5-(-1)}=\frac{1.25}{2.5}=0.5\]Using \(L(-1,0)\): \(0=0.5(-1)+k\Rightarrow k=0.5\), so \(y=0.5x+0.5\).
Now equate the curve and the line:
\[-x^{2}+x+2=0.5x+0.5\]\[-x^{2}+0.5x+1.5=0\;\Rightarrow\;x^{2}-0.5x-1.5=0\]Solve using the formula \(x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) with \(a=1,\,b=-0.5,\,c=-1.5\):
\[x=\frac{0.5\pm\sqrt{(-0.5)^{2}-4(1)(-1.5)}}{2}=\frac{0.5\pm\sqrt{0.25+6}}{2}=\frac{0.5\pm\sqrt{6.25}}{2}=\frac{0.5\pm2.5}{2}\]\[x=\frac{0.5+2.5}{2}=1.5\qquad\text{or}\qquad x=\frac{0.5-2.5}{2}=-1\]The roots are \(x=-1\) and \(x=1.5\), which are exactly the x-coordinates of \(L\) and \(H\) read from the graph.
(iii) Line of symmetry of the curve.
The line of symmetry of \(y=ax^{2}+bx+c\) is \(x=-\dfrac{b}{2a}\). With \(a=-1\) and \(b=1\):
\[x=-\frac{1}{2(-1)}=\frac{1}{2}\]The line of symmetry is \(x=0.5\), the vertical line midway between the roots \(-1\) and \(2\) (shown dashed on the graph, passing through the maximum point \((0.5,\,2.25)\)).
Answer Details
(a) Divide the algebraic fractions.
Dividing by a fraction is the same as multiplying by its reciprocal:
\[\frac{x^{2}-4}{x^{2}+x}\div\frac{x^{2}-4x+4}{x+1}=\frac{x^{2}-4}{x^{2}+x}\times\frac{x+1}{x^{2}-4x+4}\]Factorise each expression fully:
\[x^{2}-4=(x-2)(x+2),\qquad x^{2}+x=x(x+1),\qquad x^{2}-4x+4=(x-2)^{2}\]Substitute the factorised forms:
\[=\frac{(x-2)(x+2)}{x(x+1)}\times\frac{x+1}{(x-2)^{2}}\]Cancel the common factors \((x+1)\) and one \((x-2)\):
\[=\frac{x+2}{x(x-2)}=\frac{x+2}{x^{2}-2x}\](b) Using the graph.
Reading the marked points from the diagram, the curve \(y=ax^{2}+bx+c\) cuts the x-axis at \(L(-1,0)\) and \(N(2,0)\) and cuts the y-axis at \(M(0,2)\). The straight line \(y=mx+k\) passes through \(L(-1,0)\) and cuts the curve again at \(H(1.5,\,1.25)\). The reconstructed graph below shows these features.
(ii) Values of \(a\), \(b\), \(c\) and the equation of the curve. (Solved first, as the roots in part (i) depend on this equation.)
Substitute the coordinates of \(L\), \(M\) and \(N\) into \(y=ax^{2}+bx+c\):
\[L(-1,0):\;\;a(-1)^{2}+b(-1)+c=0\;\Rightarrow\;a-b+c=0\quad(1)\]\[M(0,2):\;\;a(0)^{2}+b(0)+c=2\;\Rightarrow\;c=2\quad(2)\]\[N(2,0):\;\;a(2)^{2}+b(2)+c=0\;\Rightarrow\;4a+2b+c=0\quad(3)\]Put \(c=2\) from (2) into (1) and (3):
\[a-b=-2\quad(1a),\qquad 4a+2b=-2\quad(3a)\]From (1a), \(b=a+2\). Substitute into (3a):
\[4a+2(a+2)=-2\;\Rightarrow\;6a+4=-2\;\Rightarrow\;6a=-6\;\Rightarrow\;a=-1\]Then \(b=a+2=1\) and \(c=2\). Therefore
\[a=-1,\qquad b=1,\qquad c=2,\]and the equation of the curve is
\[y=-x^{2}+x+2.\](i) Roots of \(ax^{2}+bx+c=mx+k\).
The solutions are the x-coordinates of the points where the line meets the curve, namely \(L\) and \(H\). First find the line \(y=mx+k\) through \(L(-1,0)\) and \(H(1.5,1.25)\):
\[m=\frac{1.25-0}{1.5-(-1)}=\frac{1.25}{2.5}=0.5\]Using \(L(-1,0)\): \(0=0.5(-1)+k\Rightarrow k=0.5\), so \(y=0.5x+0.5\).
Now equate the curve and the line:
\[-x^{2}+x+2=0.5x+0.5\]\[-x^{2}+0.5x+1.5=0\;\Rightarrow\;x^{2}-0.5x-1.5=0\]Solve using the formula \(x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) with \(a=1,\,b=-0.5,\,c=-1.5\):
\[x=\frac{0.5\pm\sqrt{(-0.5)^{2}-4(1)(-1.5)}}{2}=\frac{0.5\pm\sqrt{0.25+6}}{2}=\frac{0.5\pm\sqrt{6.25}}{2}=\frac{0.5\pm2.5}{2}\]\[x=\frac{0.5+2.5}{2}=1.5\qquad\text{or}\qquad x=\frac{0.5-2.5}{2}=-1\]The roots are \(x=-1\) and \(x=1.5\), which are exactly the x-coordinates of \(L\) and \(H\) read from the graph.
(iii) Line of symmetry of the curve.
The line of symmetry of \(y=ax^{2}+bx+c\) is \(x=-\dfrac{b}{2a}\). With \(a=-1\) and \(b=1\):
\[x=-\frac{1}{2(-1)}=\frac{1}{2}\]The line of symmetry is \(x=0.5\), the vertical line midway between the roots \(-1\) and \(2\) (shown dashed on the graph, passing through the maximum point \((0.5,\,2.25)\)).
Question 56 Report
(a) Make q the subject of the relation \(t = \sqrt{\frac{pq}{r} - r^{2}q}\).
(b) If \(9^{(1 - x)} = 27^{y}\) and \(x - y = -1\frac{1}{2}\), find the value of x and y.
(a) Square both sides to remove the root:
\[t^2 = \frac{pq}{r} - r^2 q = q\left(\frac{p}{r} - r^2\right) = q\left(\frac{p - r^3}{r}\right).\]
Make q the subject:
\[q = \frac{r\,t^2}{p - r^3}.\]
(b) Write both sides to base 3: \(9^{(1-x)} = 3^{2(1-x)}\) and \(27^{y} = 3^{3y}\). Equating indices,
\[2(1 - x) = 3y \Rightarrow 2 - 2x = 3y. \quad (1)\]
Also \(x - y = -\tfrac{3}{2}\Rightarrow x = y - \tfrac{3}{2}. \quad (2)\)
Substitute (2) into (1):
\[2 - 2\left(y - \tfrac{3}{2}\right) = 3y \Rightarrow 2 - 2y + 3 = 3y \Rightarrow 5 = 5y \Rightarrow y = 1.\]
\[x = 1 - \tfrac{3}{2} = -\tfrac{1}{2}.\]
Answer: \(x = -\tfrac{1}{2},\ y = 1\).
Answer Details
(a) Square both sides to remove the root:
\[t^2 = \frac{pq}{r} - r^2 q = q\left(\frac{p}{r} - r^2\right) = q\left(\frac{p - r^3}{r}\right).\]
Make q the subject:
\[q = \frac{r\,t^2}{p - r^3}.\]
(b) Write both sides to base 3: \(9^{(1-x)} = 3^{2(1-x)}\) and \(27^{y} = 3^{3y}\). Equating indices,
\[2(1 - x) = 3y \Rightarrow 2 - 2x = 3y. \quad (1)\]
Also \(x - y = -\tfrac{3}{2}\Rightarrow x = y - \tfrac{3}{2}. \quad (2)\)
Substitute (2) into (1):
\[2 - 2\left(y - \tfrac{3}{2}\right) = 3y \Rightarrow 2 - 2y + 3 = 3y \Rightarrow 5 = 5y \Rightarrow y = 1.\]
\[x = 1 - \tfrac{3}{2} = -\tfrac{1}{2}.\]
Answer: \(x = -\tfrac{1}{2},\ y = 1\).
Question 57 Report
A sector of a circle with radius 21 cm has an area of 280\(cm^{2}\).
(a) Calculate, correct to 1 decimal place, the perimeter of the sector.
(b) If the sector is bent such that its straight edges coincide to form a cone, calculate, correct to the nearest degree, the vertical angle of the cone. [Take \(\pi = \frac{22}{7}\)].
(a) Perimeter of the sector. First find the angle \(\theta\) from the area:
\[\text{Area} = \frac{\theta}{360}\pi r^2 \Rightarrow 280 = \frac{\theta}{360}\times\frac{22}{7}\times21^2.\]
\[\frac{22}{7}\times441 = 1386,\qquad \frac{\theta}{360} = \frac{280}{1386} = 0.2020.\]
Arc length \(= \dfrac{\theta}{360}\times2\pi r = 0.2020\times\left(2\times\tfrac{22}{7}\times21\right) = 0.2020\times132 = 26.7\text{ cm}.\)
\[\text{Perimeter} = \text{arc} + 2r = 26.7 + 42 = 68.7\text{ cm}.\]
(b) Vertical angle of the cone. When bent, the arc becomes the base circumference and the sector radius becomes the slant height \(l = 21\) cm. Base radius R:
\[2\pi R = 26.7 \Rightarrow R = \frac{26.7}{2\times\tfrac{22}{7}} = 4.24\text{ cm}.\]
Half the vertical angle \(\alpha\) satisfies \(\sin\alpha = \dfrac{R}{l} = \dfrac{4.24}{21} = 0.2020\), so \(\alpha = 11.66^\circ\).
\[\text{Vertical angle} = 2\alpha = 23.3^\circ \approx 23^\circ.\]
Answer Details
(a) Perimeter of the sector. First find the angle \(\theta\) from the area:
\[\text{Area} = \frac{\theta}{360}\pi r^2 \Rightarrow 280 = \frac{\theta}{360}\times\frac{22}{7}\times21^2.\]
\[\frac{22}{7}\times441 = 1386,\qquad \frac{\theta}{360} = \frac{280}{1386} = 0.2020.\]
Arc length \(= \dfrac{\theta}{360}\times2\pi r = 0.2020\times\left(2\times\tfrac{22}{7}\times21\right) = 0.2020\times132 = 26.7\text{ cm}.\)
\[\text{Perimeter} = \text{arc} + 2r = 26.7 + 42 = 68.7\text{ cm}.\]
(b) Vertical angle of the cone. When bent, the arc becomes the base circumference and the sector radius becomes the slant height \(l = 21\) cm. Base radius R:
\[2\pi R = 26.7 \Rightarrow R = \frac{26.7}{2\times\tfrac{22}{7}} = 4.24\text{ cm}.\]
Half the vertical angle \(\alpha\) satisfies \(\sin\alpha = \dfrac{R}{l} = \dfrac{4.24}{21} = 0.2020\), so \(\alpha = 11.66^\circ\).
\[\text{Vertical angle} = 2\alpha = 23.3^\circ \approx 23^\circ.\]
Question 58 Report
The table shows the scores obtained when a fair die was thrown a number of times.
| Score | 1 | 2 | 3 | 4 | 5 | 6 |
| Frequency | 2 | 5 | x | 11 | 9 | 10 |
If the probability of obtaining a 3 is 0.26, find the (a) median
(b) standard deviation of the distribution.
| Score | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Frequency | 2 | 5 | x | 11 | 9 | 10 |
Find x. Total \(N=37+x\). Since \(P(3)=0.26\):
\[ \frac{x}{37+x}=0.26 \Rightarrow x=9.62+0.26x \Rightarrow 0.74x=9.62 \Rightarrow x=13 \]So frequencies are \(2,5,13,11,9,10\) with \(N=50\).
(a) Median. With \(N=50\), the median is the mean of the 25th and 26th values. Cumulative frequencies: 2, 7, 20, 31, ... Both the 25th and 26th fall at score 4, so median = 4.
(b) Standard deviation. First the mean:
\[ \Sigma fx = 2+10+39+44+45+60 = 200,\quad \bar{x}=\frac{200}{50}=4 \] \[ \Sigma fx^2 = 2+20+117+176+225+360 = 900 \] \[ \text{Variance}=\frac{\Sigma fx^2}{N}-\bar{x}^2=\frac{900}{50}-4^2=18-16=2 \] \[ \text{SD}=\sqrt{2}\approx 1.41 \]Answer Details
| Score | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Frequency | 2 | 5 | x | 11 | 9 | 10 |
Find x. Total \(N=37+x\). Since \(P(3)=0.26\):
\[ \frac{x}{37+x}=0.26 \Rightarrow x=9.62+0.26x \Rightarrow 0.74x=9.62 \Rightarrow x=13 \]So frequencies are \(2,5,13,11,9,10\) with \(N=50\).
(a) Median. With \(N=50\), the median is the mean of the 25th and 26th values. Cumulative frequencies: 2, 7, 20, 31, ... Both the 25th and 26th fall at score 4, so median = 4.
(b) Standard deviation. First the mean:
\[ \Sigma fx = 2+10+39+44+45+60 = 200,\quad \bar{x}=\frac{200}{50}=4 \] \[ \Sigma fx^2 = 2+20+117+176+225+360 = 900 \] \[ \text{Variance}=\frac{\Sigma fx^2}{N}-\bar{x}^2=\frac{900}{50}-4^2=18-16=2 \] \[ \text{SD}=\sqrt{2}\approx 1.41 \]Question 59 Report
(a) Given that \(\sin x = 0.6, 0° \leq x \leq 90°\), evaluate \(2\cos x + 3\sin x\), leaving your answer in the form \(\frac{m}{n}\), where m and n are integers.
(b) 
In the diagram, a semi-circle WXYZ with centre O is inscribed in an isosceles triangle ABC. If /AC/ = /BC/, |OC| = 30 cm and < ACB = 130°, calculate, correct to one decimal place, the (i) radius of the circle ; (ii) area oc the shaded portion. [Take \(\pi = \frac{22}{7}\)].
(a) Evaluating \(2\cos x + 3\sin x\)
Given \(\sin x = 0.6 = \dfrac{3}{5}\) with \(0^\circ \le x \le 90^\circ\), x is acute, so \(\cos x\) is positive.
\[\cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 - (0.6)^2} = \sqrt{1 - 0.36} = \sqrt{0.64} = 0.8 = \frac{4}{5}\]
Then
\[2\cos x + 3\sin x = 2\left(\frac{4}{5}\right) + 3\left(\frac{3}{5}\right) = \frac{8}{5} + \frac{9}{5} = \frac{17}{5}\]
(b) Semi-circle inscribed in isosceles triangle ABC
From the diagram, the diameter WZ lies along the top side AB, the centre O is on AB, and the semicircular arc is tangent to the two equal sides. The apex C is below, with \(|OC| = 30\ \text{cm}\) measured along the axis of symmetry, and \(\angle ACB = 130^\circ\).
By symmetry (AC = BC), the line CO bisects angle C, so the half-angle at C is
\[\frac{130^\circ}{2} = 65^\circ\]
(i) Radius of the circle. The radius drawn to the point where the arc touches a side is perpendicular to that side. In the right-angled triangle formed by O, C and the tangent point, OC is the hypotenuse and the radius r is opposite the \(65^\circ\) angle:
\[r = |OC|\sin 65^\circ = 30 \times 0.9063 = 27.19\]
\[r \approx 27.2\ \text{cm}\]
(ii) Area of the shaded portion. The shaded region is the part of triangle ABC lying outside the semicircle:
\[\text{Shaded} = \text{Area of } \triangle ABC - \text{Area of semicircle}\]
The centre O lies on AB, so OC = 30 cm is the height of the triangle from C to AB. The half-base is
\[\frac{1}{2}|AB| = 30\tan 65^\circ = 30 \times 2.1445 = 64.34\ \text{cm}\]
\[|AB| = 128.67\ \text{cm}\]
\[\text{Area of } \triangle ABC = \frac{1}{2}\times |AB| \times \text{height} = \frac{1}{2}\times 128.67 \times 30 = 1930.1\ \text{cm}^2\]
Area of the semicircle, with \(\pi = \dfrac{22}{7}\) and \(r = 27.2\):
\[\frac{1}{2}\pi r^2 = \frac{1}{2}\times \frac{22}{7}\times (27.2)^2 = \frac{1}{2}\times \frac{22}{7}\times 739.84 = 1162.6\ \text{cm}^2\]
Therefore
\[\text{Shaded area} = 1930.1 - 1162.6 = 767.5\ \text{cm}^2\]
Answer Details
(a) Evaluating \(2\cos x + 3\sin x\)
Given \(\sin x = 0.6 = \dfrac{3}{5}\) with \(0^\circ \le x \le 90^\circ\), x is acute, so \(\cos x\) is positive.
\[\cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 - (0.6)^2} = \sqrt{1 - 0.36} = \sqrt{0.64} = 0.8 = \frac{4}{5}\]
Then
\[2\cos x + 3\sin x = 2\left(\frac{4}{5}\right) + 3\left(\frac{3}{5}\right) = \frac{8}{5} + \frac{9}{5} = \frac{17}{5}\]
(b) Semi-circle inscribed in isosceles triangle ABC
From the diagram, the diameter WZ lies along the top side AB, the centre O is on AB, and the semicircular arc is tangent to the two equal sides. The apex C is below, with \(|OC| = 30\ \text{cm}\) measured along the axis of symmetry, and \(\angle ACB = 130^\circ\).
By symmetry (AC = BC), the line CO bisects angle C, so the half-angle at C is
\[\frac{130^\circ}{2} = 65^\circ\]
(i) Radius of the circle. The radius drawn to the point where the arc touches a side is perpendicular to that side. In the right-angled triangle formed by O, C and the tangent point, OC is the hypotenuse and the radius r is opposite the \(65^\circ\) angle:
\[r = |OC|\sin 65^\circ = 30 \times 0.9063 = 27.19\]
\[r \approx 27.2\ \text{cm}\]
(ii) Area of the shaded portion. The shaded region is the part of triangle ABC lying outside the semicircle:
\[\text{Shaded} = \text{Area of } \triangle ABC - \text{Area of semicircle}\]
The centre O lies on AB, so OC = 30 cm is the height of the triangle from C to AB. The half-base is
\[\frac{1}{2}|AB| = 30\tan 65^\circ = 30 \times 2.1445 = 64.34\ \text{cm}\]
\[|AB| = 128.67\ \text{cm}\]
\[\text{Area of } \triangle ABC = \frac{1}{2}\times |AB| \times \text{height} = \frac{1}{2}\times 128.67 \times 30 = 1930.1\ \text{cm}^2\]
Area of the semicircle, with \(\pi = \dfrac{22}{7}\) and \(r = 27.2\):
\[\frac{1}{2}\pi r^2 = \frac{1}{2}\times \frac{22}{7}\times (27.2)^2 = \frac{1}{2}\times \frac{22}{7}\times 739.84 = 1162.6\ \text{cm}^2\]
Therefore
\[\text{Shaded area} = 1930.1 - 1162.6 = 767.5\ \text{cm}^2\]
Question 60 Report
(a) The area of trapezium PQRS is 60\(cm^{2}\). PQ // RS, /PQ/ = 15 cm, /RS/ = 25 cm and < PSR = 60°. Calculate the : (i) perpendicular height of PQRS ; (ii) |PS|.
(b) Ade received \(\frac{3}{5}\) of a sum of money, Nelly \(\frac{1}{3}\) of the remainder while Austin took the rest. If Austin's share is greater than Nelly's share by N3,000, how much did Ade get?
Question 61 Report
(a) 
In the diagram, PQRST is a quadrilateral. PT // QS, < PTQ = 42°, < TSQ = 38° and < QSR = 30°. If < QTS = x and < POT = y, find: (i) x ; (ii) y.
(b) 
In the diagram, PQRS is a circle centre O. If POQ = 150°, < QSR = 40° and < SQP = 45°, calculate < RQS.
(a) In the quadrilateral \(PT \parallel QS\), with \(\angle PTQ = 42^{\circ}\), \(\angle TSQ = 38^{\circ}\) and \(\angle QSR = 30^{\circ}\).
(i) Find \(x = \angle QTS\).
Since \(PT \parallel QS\), \(\angle PTS = \angle TSQ = 38^{\circ}\) (alternate angles). The angles at \(T\) on the straight line give:
\[ 42^{\circ} + x + 38^{\circ} = 180^{\circ}. \]
\[ 80^{\circ} + x = 180^{\circ} \quad\Rightarrow\quad x = 100^{\circ}. \]
(ii) Find \(y = \angle POT\).
Since \(PT \parallel QS\), \(\angle SQT = \angle PTQ = 42^{\circ}\) (alternate angles). In triangle \(QRS\), \(\angle SQR = 180^{\circ} - 90^{\circ} - 30^{\circ} = 60^{\circ}\) (sum of angles in a triangle). The angles on the straight line at \(Q\) give:
\[ y + 42^{\circ} + 60^{\circ} = 180^{\circ}. \]
\[ y + 102^{\circ} = 180^{\circ} \quad\Rightarrow\quad y = 78^{\circ}. \]
(b) \(PQRS\) is a circle centre \(O\), with \(\angle POQ = 150^{\circ}\), \(\angle QSR = 40^{\circ}\) and \(\angle SQP = 45^{\circ}\). Find \(\angle RQS\).
The angle at the centre is twice the angle at the circumference on the same arc \(PQ\):
\[ \angle QSP = \tfrac{1}{2}\times 150^{\circ} = 75^{\circ}. \]
In triangle \(SPQ\), the sum of the angles is \(180^{\circ}\):
\[ \angle QPS + 75^{\circ} + 45^{\circ} = 180^{\circ} \quad\Rightarrow\quad \angle QPS = 60^{\circ}. \]
\(PQRS\) is a cyclic quadrilateral, so opposite angles are supplementary:
\[ \angle QRS = 180^{\circ} - \angle QPS = 180^{\circ} - 60^{\circ} = 120^{\circ}. \]
In triangle \(QRS\), the sum of the angles is \(180^{\circ}\):
\[ \angle RQS = 180^{\circ} - (\angle QRS + \angle QSR) = 180^{\circ} - (120^{\circ} + 40^{\circ}) = 20^{\circ}. \]
\(\angle RQS = 20^{\circ}.\)
Answer Details
(a) In the quadrilateral \(PT \parallel QS\), with \(\angle PTQ = 42^{\circ}\), \(\angle TSQ = 38^{\circ}\) and \(\angle QSR = 30^{\circ}\).
(i) Find \(x = \angle QTS\).
Since \(PT \parallel QS\), \(\angle PTS = \angle TSQ = 38^{\circ}\) (alternate angles). The angles at \(T\) on the straight line give:
\[ 42^{\circ} + x + 38^{\circ} = 180^{\circ}. \]
\[ 80^{\circ} + x = 180^{\circ} \quad\Rightarrow\quad x = 100^{\circ}. \]
(ii) Find \(y = \angle POT\).
Since \(PT \parallel QS\), \(\angle SQT = \angle PTQ = 42^{\circ}\) (alternate angles). In triangle \(QRS\), \(\angle SQR = 180^{\circ} - 90^{\circ} - 30^{\circ} = 60^{\circ}\) (sum of angles in a triangle). The angles on the straight line at \(Q\) give:
\[ y + 42^{\circ} + 60^{\circ} = 180^{\circ}. \]
\[ y + 102^{\circ} = 180^{\circ} \quad\Rightarrow\quad y = 78^{\circ}. \]
(b) \(PQRS\) is a circle centre \(O\), with \(\angle POQ = 150^{\circ}\), \(\angle QSR = 40^{\circ}\) and \(\angle SQP = 45^{\circ}\). Find \(\angle RQS\).
The angle at the centre is twice the angle at the circumference on the same arc \(PQ\):
\[ \angle QSP = \tfrac{1}{2}\times 150^{\circ} = 75^{\circ}. \]
In triangle \(SPQ\), the sum of the angles is \(180^{\circ}\):
\[ \angle QPS + 75^{\circ} + 45^{\circ} = 180^{\circ} \quad\Rightarrow\quad \angle QPS = 60^{\circ}. \]
\(PQRS\) is a cyclic quadrilateral, so opposite angles are supplementary:
\[ \angle QRS = 180^{\circ} - \angle QPS = 180^{\circ} - 60^{\circ} = 120^{\circ}. \]
In triangle \(QRS\), the sum of the angles is \(180^{\circ}\):
\[ \angle RQS = 180^{\circ} - (\angle QRS + \angle QSR) = 180^{\circ} - (120^{\circ} + 40^{\circ}) = 20^{\circ}. \]
\(\angle RQS = 20^{\circ}.\)
Question 62 Report
(a) The total surface area of two spheres are in the ratio 9 : 49. If the radius of the smaller sphere is 12 cm, find, correct to the nearest \(cm^{3}\), the volume of the bigger sphere.
(b) A cyclist starts from a point X and rides 3 km due West to a point Y. At Y, he changes direction and rides 5 km North- West to a point Z.
(i) How far is he from the starting point, correct to the nearest km? ; (ii) Find the bearing of Z from X, to the nearest degree.
(a) Surface areas of similar spheres are in the ratio of the squares of their radii:
\[\frac{r_{1}^{2}}{r_{2}^{2}}=\frac{9}{49}\Rightarrow \frac{r_{1}}{r_{2}}=\frac{3}{7}\]The smaller radius \(r_{1}=12\) cm corresponds to the ratio part 3, so one part \(=\dfrac{12}{3}=4\) cm, and the bigger radius \(r_{2}=7\times4=28\) cm.
\[V=\frac{4}{3}\pi r_{2}^{3}=\frac{4}{3}(3.142)(28)^{3}=\frac{4}{3}(3.142)(21952)\approx 91964\text{ cm}^{3}\]Volume of bigger sphere \(\approx 91{,}964\text{ cm}^{3}\).
(b) Take \(X\) as origin, West as the negative \(x\)-direction. \(Y=(-3,0)\). North-West is the bearing \(315^{\circ}\), i.e. direction \((-\sin45^{\circ},\cos45^{\circ})\), so
\[Z=(-3,0)+5(-\sin45^{\circ},\cos45^{\circ})=(-3-3.536,\ 3.536)=(-6.536,\ 3.536)\](i) Distance from \(X\):
\[XZ=\sqrt{6.536^{2}+3.536^{2}}=\sqrt{42.72+12.50}=\sqrt{55.22}\approx 7.4\ \text{km}\approx 7\text{ km}\](ii) \(Z\) lies to the North and West of \(X\). The angle west of due North is \(\tan^{-1}\dfrac{6.536}{3.536}=\tan^{-1}1.848=61.6^{\circ}\).
\[\text{Bearing of }Z\text{ from }X=360^{\circ}-61.6^{\circ}\approx 298^{\circ}\]Bearing \(\approx 298^{\circ}\).
Answer Details
(a) Surface areas of similar spheres are in the ratio of the squares of their radii:
\[\frac{r_{1}^{2}}{r_{2}^{2}}=\frac{9}{49}\Rightarrow \frac{r_{1}}{r_{2}}=\frac{3}{7}\]The smaller radius \(r_{1}=12\) cm corresponds to the ratio part 3, so one part \(=\dfrac{12}{3}=4\) cm, and the bigger radius \(r_{2}=7\times4=28\) cm.
\[V=\frac{4}{3}\pi r_{2}^{3}=\frac{4}{3}(3.142)(28)^{3}=\frac{4}{3}(3.142)(21952)\approx 91964\text{ cm}^{3}\]Volume of bigger sphere \(\approx 91{,}964\text{ cm}^{3}\).
(b) Take \(X\) as origin, West as the negative \(x\)-direction. \(Y=(-3,0)\). North-West is the bearing \(315^{\circ}\), i.e. direction \((-\sin45^{\circ},\cos45^{\circ})\), so
\[Z=(-3,0)+5(-\sin45^{\circ},\cos45^{\circ})=(-3-3.536,\ 3.536)=(-6.536,\ 3.536)\](i) Distance from \(X\):
\[XZ=\sqrt{6.536^{2}+3.536^{2}}=\sqrt{42.72+12.50}=\sqrt{55.22}\approx 7.4\ \text{km}\approx 7\text{ km}\](ii) \(Z\) lies to the North and West of \(X\). The angle west of due North is \(\tan^{-1}\dfrac{6.536}{3.536}=\tan^{-1}1.848=61.6^{\circ}\).
\[\text{Bearing of }Z\text{ from }X=360^{\circ}-61.6^{\circ}\approx 298^{\circ}\]Bearing \(\approx 298^{\circ}\).
