WAEC SSCE - General Mathematics - 2011
Tambaya 1 Rahoto
What is the value of 3 in the number 42.7531?
Bayanin Amsa
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.
Tambaya 2 Rahoto
Given that n(p) = 19, m(P \(\cup\) Q) = 38 and n(P \(\cap\) Q) = 7, Find n(C)
Bayanin Amsa
Tambaya 3 Rahoto
Simplify \(\frac{\log \sqrt{27}}{\log \sqrt{81}}\)
Bayanin Amsa
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).
Tambaya 4 Rahoto
G varies directly as the square of H, If G is 4 when H is 3, find H when G = 100
Bayanin Amsa
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.
Tambaya 5 Rahoto
The graph represents the relation y = xo2 - 3x - 3. Find the value of x for which x2 - 3x = 7
Bayanin Amsa
Tambaya 6 Rahoto
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 \(\pi = \frac{22}{7}\)]
Bayanin Amsa
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.
Tambaya 8 Rahoto
One of the factors of (mn - nq - n2 + mq) is (m - n). The other factor is?
Bayanin Amsa
Tambaya 9 Rahoto
What must be added to (2x - 3y) to get (x - 2y)?
Bayanin Amsa
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.
Tambaya 10 Rahoto
If 27x = 9y. Find the value of \(\frac{x}{y}\)
Bayanin Amsa
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 \(\frac{2}{3}\).
Tambaya 11 Rahoto
In the diagram, PO and OR are radii, |PQ| = |QR| and reflex < PQR is 240o. Calculate the value x
Bayanin Amsa
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°.
Tambaya 12 Rahoto
In the diagram /Pq//TS//TU, reflex angle QPS = 245o angle PST = 115o, , STU = 65o and < RPS = x. Find the value of x
Bayanin Amsa
Tambaya 14 Rahoto
In the diagram /Pq//TS//TU, reflex angle QPS = 245o angle PST = 115o, , STU = 65o and < RPS = x. Find the value of x
Bayanin Amsa
Tambaya 15 Rahoto
The height of a cylinder is equal to its radius. If the volume is 0.216 \(\pi m^3\) Calculate the radius.
Bayanin Amsa
Tambaya 16 Rahoto
Which of these angles can be constructed using ruler and a pair of compasses only?
Tambaya 17 Rahoto
Solve for x in the equation; \(\frac{3}{5}\)(2x - 1) = \(\frac{1}{4}\)(5x - 3)
Bayanin Amsa
Tambaya 18 Rahoto
If \(\sqrt{72} + \sqrt{32} - 3 \sqrt{18} = x \sqrt{8}\), Find the value of x
Bayanin Amsa
Tambaya 19 Rahoto
The histogram shows the age distribution of members of a club. How many members are in the club?
Bayanin Amsa
Tambaya 20 Rahoto
The perpendicular bisectors of the sides of an acute-angled triangle are drawn. Which of these statements is correct? They intersect
Bayanin Amsa
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.
Tambaya 21 Rahoto
Factorize the expression: am + bn - an - bm
Bayanin Amsa
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).
Tambaya 22 Rahoto
Find the smaller value of x that satisfies the equation x2 + 7x + 10 = 0
Bayanin Amsa
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.
Tambaya 24 Rahoto
from the diagram, Which of the following statements are true? i. m = q ii. n = q iii. n + p = 180o iv. p + m = 180o
Bayanin Amsa
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.
Tambaya 25 Rahoto
If N112.00 exchanges for D14.95, calculate the value of D1.00 in naira
Bayanin Amsa
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.
Tambaya 26 Rahoto
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
Bayanin Amsa
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.
Tambaya 27 Rahoto
Solve the equation; 3x - 2y = 7, x + 2y = -3
Bayanin Amsa
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.
Tambaya 28 Rahoto
In the diagram, STUV is a straight line. < TSY = < UXY = 40o and < VUW = 110o. Calculate < TYW
Bayanin Amsa
Tambaya 29 Rahoto
A rectangular garden measures 18.6m by 12.5m. Calculate, correct to three significant figures, the area of the garden
Bayanin Amsa
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.
Tambaya 30 Rahoto
The length of a piece of stick is 1.75m. A girl measured it as 1.80m. Find the percentage error
Bayanin Amsa
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}\%.
Tambaya 31 Rahoto
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?
Bayanin Amsa
Tambaya 32 Rahoto
A regular polygon of n sides has each exterior angle to 45o. Find the value of n
Bayanin Amsa
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.
Tambaya 33 Rahoto
What is the value of 3 in the number 42.7531?
Bayanin Amsa
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 \(\frac{3}{1000}\).
Tambaya 34 Rahoto
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
Bayanin Amsa
Tambaya 35 Rahoto
The diagram shows a cyclic quadrilateral PQRS with its diagonals intersecting at K. Which of the following triangles is similar to triangle QKR?
Bayanin Amsa
Tambaya 36 Rahoto
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
Bayanin Amsa
Tambaya 38 Rahoto
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
Tambaya 39 Rahoto
The graph represents the relation y = x\(^2\) - 3x - 3. What is the equation of the line of symmetry of the graph?
Bayanin Amsa
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.
Tambaya 40 Rahoto
Esther was facing S 20° W. She turned 90° in the clock wise direction. What direction is she facing?
Bayanin Amsa
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.
Tambaya 41 Rahoto
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?
Tambaya 43 Rahoto
The height of a cylinder is equal to its radius. If the volume is 0.216 \(\pi\) m\(^3\). Calculate the radius.
Bayanin Amsa
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.
Tambaya 44 Rahoto
The histogram shows the age distribution of members of a club. What is their modal age?
Tambaya 45 Rahoto
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
Bayanin Amsa
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.
Tambaya 46 Rahoto
Given that cos xo = \(\frac{1}{r}\), express tan x in terms of r
Bayanin Amsa
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) \(\sqrt{r^2 - 1}\).
Tambaya 47 Rahoto
The perimeter of a sector of a circle of radius 4cm is (\(\pi + 8\))cm. Calculate the anle of the sector
Bayanin Amsa
Tambaya 48 Rahoto
Simplify \(\frac{m}{n} + \frac{(m - 1)}{5n} = \frac{(m - 2)}{10n}\) where n \(\neq\) 0
Tambaya 49 Rahoto
From the equation whose roots are x = \(\frac{1}{2}\) and -\(\frac{2}{3}\)
Bayanin Amsa
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.
Tambaya 50 Rahoto
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?
Bayanin Amsa
None
Tambaya 51 Rahoto
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.
Tambaya 53 Rahoto
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 \(\pi = \frac{22}{7}\)].
None
Bayanin Amsa
None
Tambaya 54 Rahoto
(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].
Bayanin Amsa
None
Tambaya 55 Rahoto
(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.
Bayanin Amsa
None
Tambaya 56 Rahoto
(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 \(y = ax^{2} + bx + c\).
Bayanin Amsa
None
Tambaya 57 Rahoto
(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 \(\pi = \frac{22}{7}\)].
Bayanin Amsa
None
Tambaya 58 Rahoto
(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.
Bayanin Amsa
None
Tambaya 59 Rahoto
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.
Bayanin Amsa
None
Tambaya 60 Rahoto
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|.
Bayanin Amsa
None
Tambaya 61 Rahoto
(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?
Bayanin Amsa
None
Tambaya 62 Rahoto
(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.
None
Bayanin Amsa
None
