WAEC SSCE - General Mathematics - 2014
Ajụjụ 1 Ripọtì
F x varies inversely as y and y varies directly as Z, what is the relationship between x and z?
Akọwa Nkọwa
The statement "F x varies inversely as y" can be mathematically written as x = k/y, where k is the constant of proportionality. Similarly, "y varies directly as Z" can be written as y = kz, where k is again the constant of proportionality. Substituting y in the first equation, we get x = k/(kz), which simplifies to x = 1/z. Therefore, we can conclude that x varies inversely with z, which is mathematically represented as x \(\alpha\) 1/z. So, the correct answer is: x \(\alpha \frac{1}{z}\).
Ajụjụ 2 Ripọtì
In the diagram, O is the centre of the circle of the circle, PR is a tangent to the circle at Q < SOQ = 86o. Calculate the value of < SQR.
Akọwa Nkọwa
Ajụjụ 3 Ripọtì
Each exterior angle of a polygon is 30o. Calculate the sum of the interior angles
Akọwa Nkọwa
The sum of the exterior angles of a polygon is always 360 degrees. Therefore, if each exterior angle is 30 degrees, the polygon must have 360/30 = 12 sides. The formula to find the sum of interior angles of a polygon is (n-2) x 180 degrees, where n is the number of sides. Substituting n = 12 into the formula, we get: (12-2) x 180 = 10 x 180 = 1800 degrees Therefore, the sum of the interior angles of this polygon is 1800 degrees. So the correct option is 1800o.
Ajụjụ 4 Ripọtì
approximate 0.0033780 to 3 significant figures
Akọwa Nkọwa
To approximate 0.0033780 to 3 significant figures, we need to count the first three non-zero digits, then round off the remaining digits. The first three non-zero digits are 3, 3, and 7, so we keep them. The next digit is 8, which is greater than or equal to 5, so we round up the last digit (0) to 1. Therefore, the answer is 0.00338 (option c).
Ajụjụ 5 Ripọtì
The graph given is for the relation y = 2x2 + x - 1. Find the minimum value of y
Akọwa Nkọwa
To find the minimum value of y for the given relation y = 2x^2 + x - 1, we need to locate the vertex of the parabolic graph. The vertex of a parabola with equation y = ax^2 + bx + c is given by the coordinates (-b/2a, c - b^2/4a). In this case, a = 2, b = 1, and c = -1. Substituting these values into the formula, we get: Vertex x-coordinate = -b/2a = -1/(2*2) = -1/4 Vertex y-coordinate = 2(-1/4)^2 + (1/4) - 1 = -1.25 Therefore, the minimum value of y is -1.25. So the correct option is (C) -1.25.
Ajụjụ 6 Ripọtì
If \(\frac{2}{x - 3} - \frac{3}{x - 2}\) is equal to \(\frac{p}{(x - 3)(x - 2)}\), find p
Akọwa Nkọwa
Ajụjụ 7 Ripọtì
In the figures, PQ is a tangent to the circle at R and UT is parallel to PQ. if < TRQ = xo, find < URT in terms of x
Akọwa Nkọwa
Ajụjụ 9 Ripọtì
The volume of a cube is 512cm3. Find the length of its side
Akọwa Nkọwa
To find the length of a side of a cube, we need to take the cube root of its volume. Given that the volume of the cube is 512cm³, we have: side³ = volume side³ = 512cm³ Taking the cube root of both sides, we get: side = ∛(512cm³) side = 8cm Therefore, the length of the side of the cube is 8cm. Answer: Option (C) 8cm
Ajụjụ 10 Ripọtì
If a number is selected at random from each of the sets p = {1, 2, 3} and Q = {2, 3, 5}, find the probability that the sum of the numbers is prime
Akọwa Nkọwa
To find the probability that the sum of the numbers is prime, we need to first determine all the possible sums of numbers from sets P and Q. The elements in set P are {1, 2, 3}, and the elements in set Q are {2, 3, 5}. To find all the possible sums, we can create a table of sums by adding each element of set P to each element of set Q: | | 2 | 3 | 5 | |---|---|---|---| | 1 | 3 | 4 | 6 | | 2 | 4 | 5 | 7 | | 3 | 5 | 6 | 8 | From this table, we see that there are nine possible sums. To determine which of these sums are prime, we can simply check each one: - 3: Not prime - 4: Not prime - 5: Prime - 6: Not prime - 7: Prime - 8: Not prime - 9: Not prime - 10: Not prime - 11: Prime So, there are three possible sums that are prime: 5, 7, and 11. Therefore, the probability that the sum of the numbers is prime is: $$\frac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}} = \frac{3}{9} = \frac{1}{3}$$ Therefore, the answer is option (C) \(\frac{1}{3}\).
Ajụjụ 11 Ripọtì
Find the number of term in the Arithmetic Progression(A.P) 2, -9, -20,...-141.
Akọwa Nkọwa
Ajụjụ 12 Ripọtì
The probability of an event P happening is \(\frac{1}{5}\) and that of event Q is \(\frac{1}{4}\). If the events are independent, what is the probability that neither of them happens?
Akọwa Nkọwa
Ajụjụ 14 Ripọtì
The bar chart shows the scores of some students in a test. If one students is selected at random, find the probability that he/she scored at most 2 marks
Ajụjụ 17 Ripọtì
Subtract \(\frac{1}{2}\)(a - b - c) from the sum of \(\frac{1}{2}\)(a - b + c) and \(\frac{1}{2}\)
(a + b - c)
Akọwa Nkọwa
We have: \begin{align*} &\frac{1}{2}(a-b+c)+\frac{1}{2}(a+b-c)-\frac{1}{2}(a-b-c)\\ &=\frac{1}{2}(2a)\\ &=a \end{align*} Therefore, the expression simplifies to $a$. So, the answer is: $\frac{1}{2}(a+b+c)$.
Ajụjụ 18 Ripọtì
Which of the following is a measure of dispersion?
Akọwa Nkọwa
A measure of dispersion describes how spread out or varied a set of data is. Among the options given, only the range is a measure of dispersion. Range is calculated by subtracting the minimum value from the maximum value in a dataset. It provides information about the spread of the data from the lowest to the highest value. Percentile, median, and quartile are measures of central tendency, which describe the typical or central value in a dataset.
Ajụjụ 20 Ripọtì
A chord, 7cm long, is drawn in a circle with radius 3.7cm. Calculate the distance of the chord from the centre of the circle
Akọwa Nkọwa
To find the distance of the chord from the center of the circle, we can use the formula: distance from center = √(r^2 - (c/2)^2) where r is the radius of the circle and c is the length of the chord. In this case, r = 3.7 cm and c = 7 cm. Therefore: distance from center = √(3.7^2 - (7/2)^2) = √(13.69 - 12.25) = √1.44 = 1.2 cm Therefore, the distance of the chord from the center of the circle is 1.2 cm. The logic behind the formula is that the distance from the center of the circle to the chord is the perpendicular distance from the center to the line that contains the chord. If we draw radii from the center of the circle to the endpoints of the chord, we can form a right triangle with the chord as the hypotenuse. The distance from the center to the chord is the height of this triangle, which can be found using the Pythagorean theorem.
Ajụjụ 22 Ripọtì
The graph given is for the relation y = 2x2 + x - 1.What are the coordinates of the point S?
Akọwa Nkọwa
To find the coordinates of point S on the graph, we need to use the equation of the graph, which is given as y = 2x^2 + x - 1. From the graph, we can see that the point S has x-coordinate of 1. To find the y-coordinate, we can substitute x=1 into the equation and solve for y: y = 2(1)^2 + (1) - 1 y = 2 + 1 - 1 y = 2 Therefore, the coordinates of point S are (1, 2). So, the answer is (c) (1, 2.0).
Ajụjụ 23 Ripọtì
If x = {0, 2, 4, 6}, y = {1, 2, 3, 4} and z = {1, 3} are subsets of u = {x:0 \(\geq\) x \(\geq\) 6}, find x \(\cap\) (Y' \(\cup\) Z)
Akọwa Nkọwa
To solve this problem, we need to follow the given operations in order of precedence, i.e., first we have to find the complement of y, then union it with z and finally find the intersection of x with the resulting set. The complement of y (denoted by Y') is the set of all elements in u that are not in y. So, Y' = {0, 5, 6}. The union of Y' and Z (denoted by Y' \(\cup\) Z) is the set of all elements that are in Y' or Z or both. So, Y' \(\cup\) Z = {0, 1, 3, 5, 6}. Finally, the intersection of x with Y' \(\cup\) Z (denoted by x \(\cap\) (Y' \(\cup\) Z)) is the set of all elements that are in both x and Y' \(\cup\) Z. So, x \(\cap\) (Y' \(\cup\) Z) = {0, 6}. Therefore, the answer is {0, 6}.
Ajụjụ 25 Ripọtì
The bar chart shows the scores of some students in a test. How many students took the test?
Ajụjụ 26 Ripọtì
The radii of the base of two cylindrical tins, P and Q are r and 2r respectively. If the water level in p is 10cm high, would be the height of the same quantity of water in Q?
Akọwa Nkọwa
The volume of a cylinder is given by the formula V = πr²h, where r is the radius of the base of the cylinder and h is the height of the cylinder. The volume of the water in tin P is VP = πr²hP, where r is the radius of the base of tin P and hP is the height of the water in tin P. The volume of the same quantity of water in tin Q is VQ = π(2r)²hQ = 4πr²hQ, where 2r is the radius of the base of tin Q and hQ is the height of the water in tin Q. Since the same quantity of water is in both tins, VP = VQ. Substituting the expression for VP and VQ into the equation VP = VQ gives πr²hP = 4πr²hQ, which simplifies to hQ = hP/4. Therefore, the height of the water in tin Q is one-fourth of the height of the water in tin P. Since the height of the water in tin P is 10cm, the height of the water in tin Q would be (10/4) cm = 2.5 cm. Therefore, the height of the same quantity of water in tin Q is 2.5 cm. Answer option (A) is the correct answer.
Ajụjụ 27 Ripọtì
Make u the subject of formula, E = \(\frac{m}{2g}\)(v2 - u2)
Akọwa Nkọwa
The formula we want to solve for u is E = (m/2g)(v^2 - u^2).
To make u the subject of the formula, we need to isolate u on one side of the equation by performing the necessary algebraic operations.
First, we'll simplify the right side of the equation:
E = (m/2g)(v^2 - u^2)
2gE/m = v^2 - u^2 // Multiply both sides by 2g/m
Next, we'll isolate u^2 by adding it to both sides of the equation:
2gE/m + u^2 = v^2
Finally, we'll solve for u by taking the square root of both sides of the equation:
u = sqrt(v^2 - 2gE/m)
Therefore, the value of u as the subject of the formula is:
u = sqrt(v^2 - 2gE/m)
Option A is the correct answer.
Ajụjụ 29 Ripọtì
The coordinates of points P and Q are (4, 3) and (2, -1) respectively. Find the shortest distance between P and Q.
Ajụjụ 30 Ripọtì
Given that cos x = \(\frac{12}{13}\), evaluate \(\frac{1 - \tan x}{\tan x}\)
Akọwa Nkọwa
We can use trigonometric identities to evaluate the expression. We know that: - $\tan x = \frac{\sin x}{\cos x}$. - $\sin^2x + \cos^2x = 1$. We can use the second identity to find $\sin x$: $$\begin{aligned} \sin^2x + \cos^2x &= 1 \\ \sin^2x &= 1 - \cos^2x \\ \sin x &= \sqrt{1 - \cos^2x} \\ &= \sqrt{1 - \left(\frac{12}{13}\right)^2} \\ &= \frac{5}{13} \end{aligned}$$ Now we can substitute $\sin x$ and $\cos x$ into the expression we need to evaluate: $$\begin{aligned} \frac{1 - \tan x}{\tan x} &= \frac{1 - \frac{\sin x}{\cos x}}{\frac{\sin x}{\cos x}} \\ &= \frac{\cos x - \sin x}{\sin x} \\ &= \frac{\frac{12}{13} - \frac{5}{13}}{\frac{5}{13}} \\ &= \frac{7}{5} \end{aligned}$$ Therefore, the answer is $\frac{7}{5}$.
Ajụjụ 31 Ripọtì
A man's eye level is 1.7m above the horizontal ground and 13m from a vertical pole. If the pole is 8.3m high, calculate, correct to the nearest degree, the angle of elevation of the top of the pole from his eyes.
Akọwa Nkọwa
Ajụjụ 32 Ripọtì
A ship sails x km due east to a point E and continues x km due north to F. Find the bearing the bearing of f from the starting point.
Akọwa Nkọwa
Let's draw a diagram to visualize the situation:
N
|
F
|
W-------+-------E
|
|
S
The ship starts at the point marked "W", then sails east to reach the point marked "E". The distance between W and E is x km. Then the ship sails north from E to reach the point marked "F". The distance between E and F is also x km.
We want to find the bearing of F from W. This is the angle that the line segment WF makes with the north-south line, measured in a clockwise direction.
Let's call the point where the north-south line intersects the line WE as point G. Then we have a right triangle WGE, where WG is the distance travelled east by the ship, and GE is the distance travelled north. The angle WGE is the bearing we're looking for.
From the right triangle WGE, we can use trigonometry to find the angle WGE:
tan(WGE) = opposite / adjacent = GE / WG = x / x = 1
Taking the arctan of both sides, we get:
WGE = arctan(1) = 45 degrees
Therefore, the bearing of F from W is 045 degrees. Answer: (a) 045o.
Ajụjụ 33 Ripọtì
The area of a sector a circle with diameter 12cm is 66cm2. If the sector is folded to form a cone, calculate the radius of the base of the cone [Take \(\pi = \frac{22}{7}\)]
Akọwa Nkọwa
The area of a sector of a circle is given by the formula: A = (θ/360)πr^2 where A is the area of the sector, θ is the angle of the sector in degrees, r is the radius of the circle. Let's first find the radius of the circle. The diameter of the circle is given as 12 cm. Therefore, the radius is half of the diameter, i.e., r = 12/2 = 6 cm. The area of the sector is given as 66 cm^2. Therefore, 66 = (θ/360)π(6^2) 66 = (θ/360)π(36) 66 = (θ/10)π θ = (660/π) × 10/360 θ = 18.86° (approx.) Now, we can use the radius of the circle and the angle of the sector to find the slant height (l) of the cone: l = √(r^2 + h^2), where h is the height of the cone. The lateral area of the cone (i.e., the curved surface area) is half the area of the sector: L = (1/2)A = (1/2) × 66 = 33 cm^2 The slant height of the cone is given by: l = L/r = 33/6 = 5.5 cm We can use the Pythagorean theorem to find the height of the cone: h = √(l^2 - r^2) = √(5.5^2 - 6^2) = √(30.25 - 36) = √5.75 Therefore, the height of the cone is h = 2.4 cm (approx.). Finally, we can use the formula for the volume of a cone to find the radius of the base: V = (1/3)πr^2h We know the volume of the cone is equal to the area of the sector: V = A = (θ/360)πr^2 = (18.86/360) × π × r^2 Therefore, (1/3)πr^2h = (18.86/360) × π × r^2 h = (18.86/360) × 3 r^2 = (3/π) × (360/18.86) × 66 r = √((3/π) × (360/18.86) × 66) = 3.5 cm (approx.) Hence, the radius of the base of the cone is 3.5 cm. Therefore, the correct option is (b) 3.5cm.
Ajụjụ 35 Ripọtì
A box contains 13 currency notes, all of which are either N50 or N20 notes. The total value of the currency notes is N530. How many N50 notes are in the box?
Akọwa Nkọwa
Let's assume that there are x N50 notes and y N20 notes in the box. From the problem, we know that the total number of currency notes is 13, so we have: x + y = 13 ----(1) We also know that the total value of the currency notes is N530. If we express this in terms of the number of N50 and N20 notes, we get: 50x + 20y = 530 ----(2) We now have two equations (1) and (2) with two variables (x and y), which we can solve simultaneously. Multiplying equation (1) by 20, we get: 20x + 20y = 260 Subtracting this equation from equation (2), we get: 30x = 270 Dividing both sides by 30, we get: x = 9 Therefore, there are 9 N50 notes in the box. Answer: 9
Ajụjụ 36 Ripọtì
Given that x > y and 3 < y, which of the following is/are true? i. y > 3 ii. x < 3 iii. x > y > 3
Akọwa Nkọwa
Ajụjụ 38 Ripọtì
In a cumulative frequency graph, the lower quartile is 18 years while the 60th percentile is 48 years. What percentage of the distribution is at most 18 years or greater than 48 years?
Akọwa Nkọwa
Ajụjụ 40 Ripọtì
If (x - a) is a factor pf bx - ax + x2, find the other factor.
Ajụjụ 41 Ripọtì
If 23x = 325, find the value of x
Akọwa Nkọwa
To find the value of x, we need to determine what base value the number 23x represents. We are told that 23x is equal to 325. In other words, the value represented by 23x is equal to the value represented by 325. We can convert 325 to base 10 to get: 325 = (3 x 51) + (2 x 50) = 15 + 2 = 17 So, we have: 23x = 17 To find the value of x, we need to determine what base value raised to what power is equal to 17. We can do this by trying different values of x. If x is 2, then 232 = (2 x 21) + (3 x 20) = 4 + 3 = 7 If x is 3, then 233 = (2 x 31) + (3 x 30) = 6 + 3 = 9 If x is 4, then 234 = (2 x 41) + (3 x 40) = 8 + 3 = 11 If x is 5, then 235 = (2 x 51) + (3 x 50) = 10 + 3 = 13 If x is 6, then 236 = (2 x 61) + (3 x 60) = 12 + 3 = 15 If x is 7, then 237 = (2 x 71) + (3 x 70) = 14 + 3 = 17 So, we have found that if x is 7, then 23x = 17, which matches the value of 325. Therefore, the value of x is 7. Answer: 7
Ajụjụ 42 Ripọtì
Find the gradient of the line joining the points (2, -3) and 2, 5)
Akọwa Nkọwa
To find the gradient of the line joining two points, we use the formula: Gradient = (change in y) / (change in x) We can calculate the change in y by subtracting the y-coordinate of one point from the y-coordinate of the other point, and similarly, we can calculate the change in x by subtracting the x-coordinate of one point from the x-coordinate of the other point. In this case, the two points are (2, -3) and (2, 5). The change in y is: 5 - (-3) = 8 The change in x is: 2 - 2 = 0 Since the change in x is 0, we cannot divide by it, and the gradient is undefined. This means that the line joining these two points is vertical and has no slope. Therefore, the correct answer is: undefined.
Ajụjụ 43 Ripọtì
Three quarters of a number added to two and a half of the number gives 13. Find the number
Akọwa Nkọwa
Let's assume the number we are looking for to be 'x'. According to the problem statement, three quarters of the number can be written as \(\frac{3}{4}x\) and two and a half of the number can be written as \(2.5x\). It is given that the sum of these two quantities gives 13. So we can write it as: \(\frac{3}{4}x + 2.5x = 13\) Now, we can simplify this equation by combining the two terms on the left-hand side: \(\frac{3}{4}x + 2.5x = 13\) \(\Rightarrow \frac{3}{4}x + \frac{10}{4}x = 13\) \(\Rightarrow \frac{13}{4}x = 13\) \(\Rightarrow x = \frac{13 \times 4}{13}\) \(\Rightarrow x = 4\) Therefore, the number we are looking for is 4. Hence, the answer is 4.
Ajụjụ 46 Ripọtì
\(\begin{array}{c|c}height & 2 & 3 & 4 & 5 & 6 \\ \hline frequency & 2 & 4 & 5 & 3 & 1
\end{array}\)
The table shows the distribution of the height of plants in a nursery. Calculate the mean height of the plants.
Akọwa Nkọwa
To calculate the mean height of the plants, we need to first find the sum of all the heights and divide it by the total number of plants. We can do this using the formula: mean = (sum of all heights) / (total number of plants) To find the sum of all the heights, we need to multiply each height by its corresponding frequency and add up the results. So, the sum of all heights = (2 x 2) + (3 x 4) + (4 x 5) + (5 x 3) + (6 x 1) = 4 + 12 + 20 + 15 + 6 = 57 The total number of plants is the sum of all the frequencies: 2 + 4 + 5 + 3 + 1 = 15. Therefore, the mean height of the plants = 57 / 15 = 3.8. Hence, the answer is 3.8.
Ajụjụ 47 Ripọtì
In parallelogram PQRS, QR is produced to M such that |QR| = |RM|. What fraction of the area of PQMS is the area of PRMS?
Ajụjụ 48 Ripọtì
In the diagram, PQR is a straight line, (m + n) = 120o and (n + r) = 100o. Find (m + r)
Akọwa Nkọwa
Let's start by using the fact that PQR is a straight line, which means that m + n + r = 180°. We can then use the given information to form two equations and solve for m + r. From (m + n) = 120°, we can substitute n = 120° - m to get: (m + 120° - m + r) = 180° r + 120° = 180° r = 60° From (n + r) = 100°, we can substitute n = 120° - m and r = 60° to get: (120° - m + 60°) = 100° m = 80° Finally, we can find m + r by adding m and r: m + r = 80° + 60° = 140° Therefore, the answer is 140°.
Ajụjụ 49 Ripọtì
Simplify 10\(\frac{2}{5} - 6 \frac{2}{3} + 3\)
Akọwa Nkọwa
To simplify the expression 10\(\frac{2}{5} - 6 \frac{2}{3} + 3\), we first need to convert the mixed numbers to improper fractions. 10\(\frac{2}{5}\) can be written as \(\frac{52}{5}\) and 6\(\frac{2}{3}\) can be written as \(\frac{20}{3}\). So, the expression becomes: \begin{align*} 10\frac{2}{5} - 6 \frac{2}{3} + 3 &= \frac{52}{5} - \frac{20}{3} + 3\\ &= \frac{156}{15} - \frac{100}{15} + \frac{45}{15}\\ &= \frac{101}{15} \end{align*} Therefore, the simplified expression is 6\(\frac{11}{15}\). Hence, the correct option is: \boxed{2. 6\frac{11}{15}}
Ajụjụ 50 Ripọtì
(a) Copy and complete the table of values for the relation \(y = 2 \sin x + 1\)
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° | 210° | 240° | 270° |
| y | 1.0 | 2.7 | 0.0 | -0.7 |
(b) Using scales of 2 cm to 30° on the x- axis and 2 cm to 1 unit on the y- axis, draw the graph of \(y = 2 \sin x + 1, 0° \leq x \leq 270°\).
(c) Use the graph to find the values of x for which \(\sin x = \frac{1}{4}\).
Akọwa Nkọwa
None
Ajụjụ 51 Ripọtì
(a) Two functions, f and g, are defined by \(f : x \to 2x^{2} - 1\) and \(g : x \to 3x + 2\) where x is a real number.
(i) If \(f(x - 1) - 7 = 0\), find the values of x.
(ii) Evaluate : \(\frac{f(-\frac{1}{2}) . g(3)}{f(4) - g(5)}\).
(b) An operation, \((\ast)\) is defined on the set R, of real numbers, by \(m \ast n = \frac{-n}{m^{2} + 1}\), where \(m, n \in R\). If \(-3, -10 \in R\), show whether or not \(\ast\) is commutative.
Ajụjụ 52 Ripọtì
(a) If \(\frac{3}{2p - \frac{1}{2}} = \frac{\frac{1}{3}}{\frac{1}{4}p + 1}\), find p.
(b) A television set was marked for sale at GH¢ 760.00 in order to make a profit of 20%. The television set was actually sold at a discount of 5%. Calculate, correct to 2 significant figures, the actual percentage profit.
Akọwa Nkọwa
None
Ajụjụ 53 Ripọtì
(a) 
In the diagram, O is the centre of the circle radius r cm and < XOY = 90°.If the area of the shaded part is 504\(cm^{2}\), calculate the value of r. [Take \(\pi = \frac{22}{7}\)].
(b) Two isosceles triangles PQR and PQS are drawn on opposite sides of a common base PQ. If \(< PQR = 66°\) and \(< PSQ = 109°\), calculate the value of \(< RQS\).
None
Akọwa Nkọwa
None
Ajụjụ 54 Ripọtì
A building contractor tendered for two independent contracts, X and Y. The probabilities that he will win contract X is 0.5 and not win contract Y is 0.3, What is the probability that he will win :
(a) both contracts ;
(b) exactly one of the contracts ;
(c) neither of the contracts?
Ajụjụ 55 Ripọtì
(a) Given that \(5 \cos (x + 8.5)° - 1 = 0, 0° \leq x \leq 90°\), calculate, correct to the nearest degree, the value of x.
(b) The bearing of Q from P is 0150° and the bearing of P from R is 015°. If Q and R are 24km and 32km respectively from P : (i) represent this information in a diagram;
(ii) calculate the distance between Q and R, correct to two decimal places ; (iii) find the bearing of R from Q, correct to the nearest degree.
Ajụjụ 56 Ripọtì
(a) Solve : \((x - 2)(x - 3) = 12\).
(b) 
In the diagram, M and N are the centres of two circles of equal radii 7cm. The circle intercept at P and Q. If < PMQ = < PNQ = 60°, calculate, correct to the nearest whole number, the area of the shaded portion. [Take \(\pi = \frac{22}{7}\)].
Akọwa Nkọwa
None
Ajụjụ 57 Ripọtì
(a) Simplify : \(3\sqrt{75} - \sqrt{12} + \sqrt{108}\), leaving the answer in surd form (radicals).
(b) If \(124_{n} = 232_{five}\), find n.
Akọwa Nkọwa
None
Ajụjụ 58 Ripọtì
(a) Without using tables or calculator, simplify : \(\frac{0.6 \times 32 \times 0.004}{1.2 \times 0.008 \times 0.16}\), leaving the answer in standard form (scientific notation).
(b) 
In the diagram, \(\overline{EF}\) is parallel to \(\overline{GH}\). If \(< AEF = 3x°, < ABC = 120°\) and \(< CHG = 7x°\), find the value of \(< GHB\).
Akọwa Nkọwa
None
Ajụjụ 59 Ripọtì
(a) 
In the Venn diagram, P, Qand R are subsets of the universal set U. If n(U) = 125, find : (i) the value of x ; (ii) n(\(P \cup Q \cap R'\)).
(b)
In the diagram, O is the centre of the circle. If WX is parallel to YZ and < WXY = 50°, find the value of (i) , WYZ
(ii) < YEZ.
Akọwa Nkọwa
None
Ajụjụ 60 Ripọtì
| Scores | 1 | 2 | 3 | 4 | 5 | 6 |
| Frequency | 2 | 5 | 13 | 11 | 9 | 10 |
The table shows the distribution of outcomes when a die is thrown 50 times. Calculate the :
(a) Mean deviation of the distribution ; (b) probability that a score selected at random is at least a 4.
Akọwa Nkọwa
None
Ajụjụ 61 Ripọtì
(a) Copy and complete the following table for multiplication modulo 11.
| \(\otimes\) | 1 | 5 | 9 | 10 |
| 1 | 1 | 5 | 9 | 10 |
| 5 | 5 | |||
| 9 | 9 | |||
| 10 | 10 |
Use the table to : (i) evaluate \((9 \otimes 5) \otimes (10 \otimes 10)\);
(ii) find the truth set of :(1) \(10 \otimes m = 2\); (2) \(n \otimes n = 4\)
(b) When a fraction is reduced to its lowest term, it is equal to \(\frac{3}{4}\). The numerator of the fraction when doubled would be 34 greater than the denominator. Find the fraction.
Ajụjụ 62 Ripọtì
(a) Solve the simultaneous equation : \(\frac{1}{x} + \frac{1}{y} = 5 ; \frac{1}{y} - \frac{1}{x} = 1\).
(b) A man drives from Ibadan to Oyo, a distance of 48km in 45 minutes. If he drives at 72 km/h where the surface is good and 48 km/h where it is bad, find the number of kilometers of good surface.
None
Akọwa Nkọwa
None
