WAEC SSCE - General Mathematics - 2018
Ajụjụ 1 Ripọtì
The solution of x + 2 \(\geq\) 2x + 1 is illustrated
Ajụjụ 2 Ripọtì
The equation of the line through the points (4,2) and (-8, -2) is 3y = px + q, where p and q are constants. Find the value of p.
Akọwa Nkọwa
To find the value of p, we need to determine the slope of the line passing through the two given points. The slope of a line passing through two points (x1, y1) and (x2, y2) is given by the formula: slope = (y2 - y1) / (x2 - x1) Using the coordinates of the given points, we get: slope = (-2 - 2) / (-8 - 4) = -4 / (-12) = 1/3 Since the equation of the line is given as 3y = px + q, we can rewrite this equation in slope-intercept form, y = (p/3)x + (q/3), by dividing both sides by 3. The slope of the line in slope-intercept form is then (p/3). Since we know the slope of the line passing through the two given points is 1/3, we can set these two expressions equal to each other and solve for p: (p/3) = 1/3 Multiplying both sides by 3, we get: p = 1 Therefore, the value of p is 1.
Ajụjụ 3 Ripọtì
The table shows the distribution of goals scored by 25 teams in a football competition. Calculate the probability that a team selected at random scored at most 3 goals.
Akọwa Nkọwa
Ajụjụ 4 Ripọtì
The angles of a polygon are x, 2x, 2x, (x + \(30^o\)), (x + \(20^o\)) and (x - \(10^o\)). Find the value of x
Ajụjụ 5 Ripọtì
Donations during the launching of a church project were sent in sealed envolopes. The table shows the distribution of the amount of money in the envelope. How much was the donation?
Akọwa Nkọwa
Ajụjụ 6 Ripọtì
Find the median of 2, 1, 0, 3, 1, 1, 4, 0, 1 and 2
Akọwa Nkọwa
To find the median of a set of numbers, we first need to arrange them in order from smallest to largest or largest to smallest. In this case, the numbers are: 0, 0, 1, 1, 1, 2, 2, 3, 4 Next, we determine the middle value(s) of the set. If the set has an odd number of values, there will be one middle value. If the set has an even number of values, there will be two middle values, and the median is the average of these two values. In this case, the set has an even number of values (10), so there are two middle values: 1 and 1. To find the median, we take the average of these two values: (1 + 1)/2 = 1 Therefore, the median of the set {2, 1, 0, 3, 1, 1, 4, 0, 1, 2} is 1. So the answer is 1.0.
Ajụjụ 7 Ripọtì
In the diagram, WXYZ is a rectangle with diamension 8cm by 6cm. P, Q, R and S are the midpoints of the rectangle as shown. Using this information calculate the area of the part of the rectangle that is not shaded
Akọwa Nkọwa
The area of the rectangle WXYZ is 8cm * 6cm = 48 cm\(^2\). The shaded part is made up of four congruent triangles, each of which has an area of 1/2 * base * height = 1/2 * 4cm * 3cm = 6 cm\(^2\). So the total area of the shaded part is 4 * 6 cm\(^2\) = 24 cm\(^2\). The area of the part of the rectangle that is not shaded is 48 cm\(^2\) - 24 cm\(^2\) = 24 cm\(^2\).
Ajụjụ 9 Ripọtì
In the diagram, PS and RS are tangents to the circle centre O,
Ajụjụ 10 Ripọtì
Simplify; \(\frac{2 - 18m^2}{1 + 3m}\)
Akọwa Nkọwa
To simplify \(\frac{2-18m^2}{1+3m}\), we need to factor the numerator and denominator. First, we can factor out a 2 from the numerator: \[\frac{2-18m^2}{1+3m} = \frac{2(1-9m^2)}{1+3m}\] Next, we can factor the numerator further using the difference of squares formula: \[\frac{2(1-9m^2)}{1+3m} = \frac{2(1-3m)(1+3m)}{1+3m}\] Finally, we can cancel out the common factor of \((1+3m)\) in the numerator and denominator: \[\frac{2(1-3m)(1+3m)}{1+3m} = 2(1-3m)\] Therefore, the simplified form of \(\frac{2-18m^2}{1+3m}\) is \(2(1-3m)\).
Ajụjụ 11 Ripọtì
M and N are two subsets of the universal set (U). If n(U) = 48, n(M) = 20, n(N) = 30 and n(MUN) = 40, find n(M \(\cap\) N)
Akọwa Nkọwa
Ajụjụ 12 Ripọtì
If \(\log_{10}\)(6x - 4) - \(\log_{10}\)2 = 1, solve for x.
Akọwa Nkọwa
To solve for x, we can use the following steps: 1. Rewrite the equation using exponential form: 10^(log(6x - 4) - log(2)) = 10^1 2. Use the fact that log(a) - log(b) = log(a/b) to simplify the equation: 10^(log((6x - 4) / 2)) = 10^1 3. Use the definition of logarithms to simplify the equation: (6x - 4) / 2 = 10 4. Solve for x by multiplying both sides of the equation by 2 and then subtracting 4 from both sides: 6x - 4 = 20, 6x = 24, and x = 4 So, the solution to the equation is x = 4.
Ajụjụ 13 Ripọtì
Find the \(n^{th}\) term of the sequence 2 x 3, 4 x 6, 8 x 9, 16 x 12...
Akọwa Nkọwa
Ajụjụ 14 Ripọtì
The total surface area of a hemispher is 75\(\pi cm^2\). Find the radius.
Ajụjụ 15 Ripọtì
The mean of 1, 3, 5, 7 and x is 4. Find the value of x
Akọwa Nkọwa
To find the value of x, we need to use the given information that the mean of 1, 3, 5, 7, and x is 4. The mean of a set of numbers is found by adding up all the numbers in the set and then dividing by the total number of numbers. So, we can write an equation: (1 + 3 + 5 + 7 + x) / 5 = 4 To solve for x, we can first simplify the equation by multiplying both sides by 5: 1 + 3 + 5 + 7 + x = 20 Then, we can solve for x by subtracting the sum of the known numbers from both sides: x = 20 - (1 + 3 + 5 + 7) x = 20 - 16 x = 4 Therefore, the value of x is 4.
Ajụjụ 16 Ripọtì
A curve is such that when y = 0, x = -2 or x = 3. Find the equation of the curve.
Akọwa Nkọwa
We are given that the curve passes through the points \((-2, 0)\) and \((3, 0)\). Since the curve is a function of \(x\), we can assume that the equation of the curve is of the form \(y = f(x)\). If the curve passes through the point \((-2, 0)\), then we can substitute \(x = -2\) and \(y = 0\) into the equation of the curve to get: $$ 0 = f(-2) $$ Similarly, if the curve passes through the point \((3, 0)\), then we can substitute \(x = 3\) and \(y = 0\) into the equation of the curve to get: $$ 0 = f(3) $$ Therefore, the curve must have at least two roots, namely \(x = -2\) and \(x = 3\). This suggests that the curve is a quadratic function. We can use the information about the roots of the curve to write the equation of the curve in factored form: $$ y = A(x + 2)(x - 3) $$ where \(A\) is a constant. Since the coefficient of \(x^2\) in the equation is 1, we can simplify this equation to: $$ y = Ax^2 + Bx + C $$ where \(A = 1\), and \(B\) and \(C\) are constants. To find the values of \(B\) and \(C\), we can substitute the points \((-2, 0)\) and \((3, 0)\) into the equation: $$ 0 = A(-2)^2 + B(-2) + C \quad \text{and} \quad 0 = A(3)^2 + B(3) + C $$ Simplifying these equations, we get: $$ 4A - 2B + C = 0 \quad \text{and} \quad 9A + 3B + C = 0 $$ We can use these two equations to solve for \(B\) and \(C\). Adding the two equations, we get: $$ 13A + B = 0 $$ Substituting this expression for \(B\) into one of the previous equations, we get: $$ 4A - 2(-13A) + C = 0 $$ Simplifying, we get: $$ 20A + C = 0 $$ Solving these two equations simultaneously, we get: $$ A = \frac{1}{13}, \quad B = -\frac{4}{13}, \quad C = 0 $$ Therefore, the equation of the curve is: $$ y = \frac{1}{13}(x + 2)(x - 3) = \frac{1}{13}(x^2 - x - 6) $$ So the correct answer is \(y = x^2 - x - 6\).
Ajụjụ 17 Ripọtì
If tan x = \(\frac{4}{3}\), 0\(^o\) < x < 90\(^o\), find the value of sin x - cos x
Akọwa Nkọwa
Ajụjụ 20 Ripọtì
The volume of a cylindrical tank, 10m high is 385 m\(^2\). Find the diameter of the tank. [Take \(\pi = \frac{22}{7}\)]
Akọwa Nkọwa
The volume of a cylinder is given by the formula V = πr²h, where V is the volume, r is the radius and h is the height of the cylinder. In this question, we are given that the height of the cylindrical tank is 10m and its volume is 385 m². Therefore, we can find the radius of the cylinder as follows: 385 = πr² × 10 r² = 38.5/π r = √(38.5/π) = 3.5m (approximately) Finally, we can find the diameter of the cylinder by doubling the radius: Diameter = 2r = 2 × 3.5 = 7m Therefore, the diameter of the tank is 7m.
Ajụjụ 21 Ripọtì
Simplify; 2\(\frac{1}{4} \times 3\frac{1}{2} \div 4 \frac{3}{8}\)
Akọwa Nkọwa
To simplify this expression, we need to follow the order of operations (PEMDAS): 1. First, we perform the multiplication of the mixed numbers: 2\(\frac{1}{4} \times 3\frac{1}{2} = \frac{9}{4} \times \frac{7}{2} = \frac{63}{8}\) 2. Then, we perform the division of the mixed numbers: 4 \(\frac{3}{8} = \frac{35}{8}\) \(\frac{63}{8} \div \frac{35}{8} = \frac{63}{8} \times \frac{8}{35} = \frac{9}{5} =\) 1\(\frac{4}{5}\) Therefore, the answer is option (D), 1\(\frac{4}{5}\).
Ajụjụ 22 Ripọtì
Find the value of t in the diagram
Akọwa Nkọwa
To find the value of t, we need to use the information given in the diagram. In a triangle, the sum of all angles is 180 degrees. So, if we know two angles, we can find the third one by subtracting their sum from 180. In this case, we know the angles labeled x and y. So, to find t, we can add x and y and then subtract the sum from 180. That is, t = 180 - (x + y). So, the value of t is 126 degrees.
Ajụjụ 23 Ripọtì
There are 250 boys and 150 girls in a school, if 60% of the boys and 40% of the girls play football, what percentage of the school play football?
Akọwa Nkọwa
We are given that there are 250 boys and 150 girls in a school. We are also given that 60% of the boys play football, which means that 0.6 x 250 = 150 boys play football. Similarly, 40% of the girls play football, which means that 0.4 x 150 = 60 girls play football. Therefore, the total number of students who play football is 150 + 60 = 210. To find the percentage of the school that plays football, we need to divide the number of students who play football by the total number of students in the school and multiply by 100. The total number of students in the school is 250 + 150 = 400. So, the percentage of the school that plays football is (210/400) x 100 = 52.5%. Therefore, the correct option is 52.5% (option D).
Ajụjụ 27 Ripọtì
If x : y = \(\frac{1}{4} : \frac{3}{8}\) and y : z = \(\frac{1}{3} : \frac{4}{9}\), find x : z
Akọwa Nkọwa
To find x : z, we need to have a common ratio between x, y, and z. We can use the given ratios to find a common ratio involving all three. Since x : y = 1/4 : 3/8, we can simplify this ratio by multiplying both terms by 8 to get: x : y = 2 : 3 Similarly, since y : z = 1/3 : 4/9, we can simplify this ratio by multiplying both terms by 3 to get: y : z = 1 : 4/3 Now we have a common ratio of y between the two ratios. We can use this common ratio to find x : z by multiplying the two simplified ratios: x : y = 2 : 3 y : z = 1 : 4/3 Multiplying these ratios gives: x : z = (2/3) * (1/(4/3)) = 2/4 = 1/2 Therefore, x : z = 1 : 2. In summary, x : z = 1 : 2.
Ajụjụ 28 Ripọtì
In the diagram, PQ is a straight line, (m + n) = 110\(^o\) and (n + r) = 130\(^o\) and (m + r) = 120\(^o\). Find the ratio of m : n : r
Akọwa Nkọwa
Ajụjụ 29 Ripọtì
Expression 0.612 in the form \(\frac{x}{y}\), where x and y are integers and y \(\neq\) 0
Akọwa Nkọwa
Ajụjụ 30 Ripọtì
The surface area of a sphere is \(\frac{792}{7} cm^2\). Find, correct to the nearest whole number, its volume. [Take \(\pi = \frac{22}{7}\)]
Akọwa Nkọwa
The formula for the surface area of a sphere is \(4\pi r^2\), where \(r\) is the radius of the sphere. Given that the surface area of the sphere is \(\frac{792}{7} cm^2\) and \(\pi = \frac{22}{7}\), we can set up an equation: $$ 4\left(\frac{22}{7}\right)r^2 = \frac{792}{7} $$ Simplifying this equation by canceling out the \(7\)s, we get: $$ 4\left(\frac{22}{1}\right)r^2 = 792 $$ Multiplying both sides by \(\frac{1}{4}\) to isolate \(r^2\), we get: $$ \left(\frac{22}{1}\right)r^2 = 198 $$ Dividing both sides by \(\frac{22}{1}\), we get: $$ r^2 = 9 $$ Taking the square root of both sides, we get: $$ r = 3 $$ Therefore, the radius of the sphere is 3 cm. The formula for the volume of a sphere is \(\frac{4}{3}\pi r^3\). Substituting the value of \(r\) into this formula, we get: $$ \frac{4}{3}\left(\frac{22}{7}\right)(3)^3 \approx 113 $$ Therefore, the volume of the sphere is approximately 113\(cm^3\). So the correct answer is 113\(cm^3\).
Ajụjụ 32 Ripọtì
Find the value of x for which \(32_{four} = 22_x\)
Akọwa Nkọwa
In this question, we are asked to find the value of x for which \(32_{four} = 22_x\). We know that \(32_{four}\) means 3 fours plus 2 ones, which is equal to 14 in the decimal system. So, we have: $$14_{10} = 2\times x^1 + 2\times x^0 = 2x+2$$ Now we can solve for x: $$14_{10} = 2x+2$$ $$12_{10} = 2x$$ $$x = 6_{10}$$ Therefore, \(32_{four} = 22_6\) and the value of x is 6.
Ajụjụ 33 Ripọtì
If M and N are the points (-3, 8) and (5, -7) respectively, find |MN|
Akọwa Nkọwa
To find the distance between two points M(-3, 8) and N(5, -7), we use the distance formula: |MN| = √[(x₂ - x₁)² + (y₂ - y₁)²] where (x₁, y₁) and (x₂, y₂) are the coordinates of M and N, respectively. Substituting the values, we get: |MN| = √[(5 - (-3))² + (-7 - 8)²] |MN| = √[(5 + 3)² + (-15)²] |MN| = √[8² + 15²] |MN| = √(64 + 225) |MN| = √289 |MN| = 17 Therefore, the distance |MN| between the two points is 17 units. So, the correct option is "17 units".
Ajụjụ 34 Ripọtì
If F = \(\frac{9}{5}\)C + 32, find C when F = 98.6
Akọwa Nkọwa
The given equation relates the temperature in degrees Celsius (C) to the temperature in degrees Fahrenheit (F). To find C when F = 98.6, we can simply substitute F = 98.6 into the equation and solve for C. So, we have: F = \(\frac{9}{5}\)C + 32 98.6 = \(\frac{9}{5}\)C + 32 (substituting F = 98.6) Subtracting 32 from both sides, we get: 66.6 = \(\frac{9}{5}\)C Multiplying both sides by 5/9, we get: C ≈ 37 Therefore, C is approximately equal to 37 degrees Celsius (not exactly any of the options provided).
Ajụjụ 35 Ripọtì
Find the inter-quartile range of 1, 3, 4, 5, 8, 9, 10, 11, 12, 14, 16
Ajụjụ 36 Ripọtì
The graph of y = x\(^2\) and y = x intersect at which of these points?
Akọwa Nkọwa
The graph of y = x\(^2\) and y = x intersect at two points where the two curves have the same y-value. To find these points, we need to find the x-values where x\(^2\) = x. We can start by subtracting x from both sides: x\(^2\) - x = 0 Next, we can factor out x: x (x - 1) = 0 This equation tells us that either x = 0 or x - 1 = 0. So, the two points of intersection are (0, 0) and (1, 1).
Ajụjụ 37 Ripọtì
In the diagram, WXYZ is a rectangle with dimension 8cm by 6cm. P, Q, R and S are the midpoints of the sides of the rectangle as shown. Using this information, what type of quadrilateral is the shaded region?
Ajụjụ 38 Ripọtì
If y + 2x = 4 and y - 3x = -1, find the value of (x + y)
Akọwa Nkọwa
To find the value of (x + y), we need to solve for x and y first. We can do this by using two equations: y + 2x = 4 and y - 3x = -1. First, we can isolate y in one of the equations by adding or subtracting the other equation. For example, if we add the two equations, we get: 2y = 3 Then, we can solve for y by dividing both sides by 2: y = 3/2 Next, we can substitute the value of y back into one of the original equations to solve for x. For example, using the equation y + 2x = 4: 3/2 + 2x = 4 Subtracting 3/2 from both sides: 2x = 7/2 And finally, dividing both sides by 2: x = 7/4 Now that we have found the values of x and y, we can add them to find (x + y): x + y = 7/4 + 3/2 Combining the fractional terms: x + y = 7/4 + 6/4 And finally, adding the whole numbers: x + y = 13/4 = 3.25 So, (x + y) is approximately equal to 3.
Ajụjụ 39 Ripọtì
A piece of thread of length 21.4cm is used to form a sector of a circle of radius 4.2cm on a piece of cloth. Calculate, correct to the nearest degree, the angle of the sector. [Take \(\pi = \frac{22}{7}\)]
Akọwa Nkọwa
Ajụjụ 41 Ripọtì
Given that Y is 20cm on a bearing of 300\(^o\) from x, how far south of y is x?
Ajụjụ 42 Ripọtì
The table shows the distribution of goals scored by 25 teams in a football competition. Calculate the probability that a team selected at randon scored either 4 or 7 goals.
Akọwa Nkọwa
To calculate the probability that a team selected at random scored either 4 or 7 goals, we need to add the frequency of teams that scored 4 goals to the frequency of teams that scored 7 goals and divide by the total number of teams (25). From the table, we can see that 6 teams scored 4 goals and 3 teams scored 7 goals. Therefore, the total number of teams that scored either 4 or 7 goals is: 6 + 3 = 9 The total number of teams in the competition is 25. Therefore, the probability that a team selected at random scored either 4 or 7 goals is: 9/25 This can be simplified as follows: 9/25 = 0.36 Therefore, the probability that a team selected at random scored either 4 or 7 goals is 0.36 or 36%. Hence, the correct option is: - \(\frac{9}{25}\)
Ajụjụ 44 Ripọtì
The angle of elevation of the top of a tree from a point 27m away and on the same horizontal ground as the foot of the tree is 30\(^o\). Find the height of the tree.
Akọwa Nkọwa
We can solve this problem using trigonometry, specifically the tangent function. Let's draw a diagram to visualize the situation. We have a right triangle with the height of the tree as one of the legs, the distance from the tree to the point on the ground as the other leg, and the angle of elevation (30 degrees) as the angle opposite the height of the tree.  We can use the tangent function to find the height of the tree: $$\tan(30^\circ) = \frac{\text{height of tree}}{27\text{ m}}$$ We know that the tangent of 30 degrees is equal to 1/\(\sqrt{3}\) (or approximately 0.577), so we can substitute that in and solve for the height of the tree: $$\frac{1}{\sqrt{3}} = \frac{\text{height of tree}}{27\text{ m}}$$ Multiplying both sides by 27 m gives: $$\text{height of tree} = \frac{27\text{ m}}{\sqrt{3}} = 9\sqrt{3}\text{ m}$$ Therefore, the height of the tree is 9\(\sqrt{3}\) meters. Answer option (D) is correct.
Ajụjụ 46 Ripọtì
Evaluate: \((64^{\frac{1}{2}} + 125^{\frac{1}{3}})^2\)
Ajụjụ 47 Ripọtì
Factorise completely the expression
\((x + 2)^2\) - \((2x + 1)^2\)
Ajụjụ 49 Ripọtì
Given that y varies inversely as the square of x. If x = 3 when y = 100, find the equation connecting x and y.
Akọwa Nkọwa
Ajụjụ 50 Ripọtì
The diagram shows a trapezium inscribed in a semi-circle. If O is the mid-point of WZ and |WX| = |XY| = |YZ|, calculate the value of m
Akọwa Nkọwa
Since O is the midpoint of WZ and the diameter of the semicircle, it follows that O lies on the circumference of the semicircle. Let angle WOZ be m. Then angle WXY is also m, since |WX| = |XY|. Similarly, angle YXZ is also m. Since the sum of angles in a triangle is 180 degrees, we have: angle WXY + angle YXZ + angle WYZ = 180 degrees Substituting m for angle WXY and angle YXZ, we get: 2m + angle WYZ = 180 degrees Since WXYZ is a trapezium, we have: angle WYZ + angle WXY = 180 degrees Substituting m for angle WXY, we get: angle WYZ + m = 180 degrees Combining this equation with the previous one, we get: 2m + angle WYZ = angle WYZ + m + 180 degrees Simplifying, we get: m = 60 degrees Therefore, the value of m is 60 degrees.
Ajụjụ 51 Ripọtì
(a) 
In the diagram
(i) The value of x; (ii)
(b) If \(2N4_{seven} = 15N_{nine}\), find the value of N.
Ajụjụ 52 Ripọtì
(a) Using ruler a pair of compasses only, construct:
(i) a trapezium PQRS such that |PQ| = 6.8 cm, < PQR = 120\(^o\), QR||PS, |PS| =10.6 cm, and IPRI = 93 cm;
(ii) locus \(l_1\) of points equidistant from P and R;
(iii) locus \(l_2\) of points equidistant from Q and R
(b) Measure: (i) |QR|;
ii. < PSR
ii. < PSR
iii. |QY|, where Y is the point of intersection h and h.
Ajụjụ 53 Ripọtì
If log\(_a\)(y + 2) = 1 + log\(_a\) x, find x in terms of y.
2. The table shows the distribution of timber production in five communities in a certain year
| Community |
Timber Production (tonnes) |
| Bibiani Amenfi Oda Wiawso Sankore |
600 900 1800 1500 2400 |
1. Draw a pie chart to represent the information.
2. What percentage of timber produced that year was from Amenfi?
3. If a tonne of timber is sold at $560.00, how much more revenue would Oda community receive than Bibiani?
Akọwa Nkọwa
None
Ajụjụ 54 Ripọtì
The sum of the first ten terms of an Arithmetic Progression (A.P.) is 130. If the fifth term is 3 times the first term, find the:
- common difference;
- first term;
- number of terms of the A.P. if the last term is 28.
Ajụjụ 55 Ripọtì
The table shows the distribution of sources obtained when a fair diwe was rolled 50 times.
| Score | 1 | 2 | 3 | 4 | 5 | 6 |
| Frequency | 2 | 5 | 13 | 11 | 9 | 10 |
1. Draw a bar chart for the distribution
2. Calculate the mean score of the distribution
Akọwa Nkọwa
None
Ajụjụ 56 Ripọtì
Using ruler and a pair of compasses only, construct:
(a) (i) quadrilateral PQRS with |PQ| = 6 cm, |PS| = 8 cm, < PSR = 90\(^o\), |SR| = 12 cm and |QR| = 11 cm;
(ii) perpendicular from Q to cut \(\over{SR}\) at K.
(b) Measure:
(i) IQRI;
(ii) < QRS.
Akọwa Nkọwa
None
Ajụjụ 57 Ripọtì
a. Find the range of values of x which satisfy the following inequalities simultaneously: 5 - x > 1 and 9 + x \(\geq\) 8
In the diagram, O is the centre of the circle, IPQI = IQRI and < PSR = 56°. Find < QRS.
Akọwa Nkọwa
None
Ajụjụ 58 Ripọtì
(a) 
In < PQS, |PQ| = 12 cm, |PS| = 5 cm, < SPQ = < PRQ = 90°, Find, correct to three significant figures, |PR|.
(b) The length of two ladders, L and M are 10m and 12m respectively. They are placed against a wall such that each ladder makes angle with the horizontal ground. If the foot of L is 8m from the foot of the wall.
(i) Draw a diagram to illustrate this information; (ii) Calculate the height at which M touches the wall.
Akọwa Nkọwa
None
Ajụjụ 59 Ripọtì
(a) The diagonals of a rhombus are 10.2 cm and 9.3 cm long. Calculate, correct to one decimal place, the perimeter of the rhombus.
(b) Given that \(\sin x = \frac{3}{5}, 0° < x < 90°\), find the value of \(5\cos x - 4\tan x\).
Ajụjụ 60 Ripọtì
a. A textbook company discovered that the profit made from selling its books is given by y = \(\frac{x^2}{8}\) + 5x, where x is the number of textbooks sold (in thousands) and y is the corresponding profit (in Ghana Cedis). If the company made a profit of GH₵ 20,000.00
i. form a quadratic equation in x;
ii. (using the quadratic formula, find, correct to the nearest whole number, the number of textbooks sold to make the profit.
b. The angle of elevation of the top T of a tree from a point P on the same ground level as the foot Q of a tree is 28\(^o\). A bird perched at a point R, halfway up the tree.
i. Represent the information in a diagram.
ii. Calculate, correct to the nearest degree, the angle of elevation of R from P.
Akọwa Nkọwa
None
Ajụjụ 61 Ripọtì
(a) Mr John paid N4,800.00 in N1.00 ordinary shares of a company which sold at N2.50 per share. If dividend was declared at 25k per share, how much dividend did he get?
(b) Using the method of completing the square, solve \(\frac{1 - x}{x} + \frac{x}{1 - x} = \frac{5}{2}\)
Akọwa Nkọwa
None
Ajụjụ 62 Ripọtì
In the diagram. PQR is an isosceles triangle. If the perimeter of the triangle is 28 cm, find the:
a. values of x and y;
b. lengths of the sides of the triangle.
Akọwa Nkọwa
None
Ajụjụ 63 Ripọtì
(a) Evaluate: \(\int \limits_1^2 (2x^3 - 4x + 3) dx\)
(b) Given that P\(^{-1}\) = \(\begin{pmatrix} -1 & 1 \\ 4 & -3 \end{pmatrix}\), find the matrix P.
Ajụjụ 64 Ripọtì
(a) If \(x = \begin{pmatrix} 2 \\ 3 \end{pmatrix}, y = \begin{pmatrix} 5 \\ -2 \end{pmatrix}\) and \(z = \begin{pmatrix} -4 \\ 13 \end{pmatrix}\), find the scalars p and q such that \(px + qy = z\).
(b)(i) Using the scale of 2cm to 2 units on both axis, draw on a graph paper two perpendicular axis x and y for \(-5 \leq x \leq 5, -5 \leq y \leq 5\) respectively.
(ii) Draw, on the graph paper, indicating clearly the vertices and their coordinates,
(1) the quadrilateral WXYZ with W(2, 3), X(4, -1), Y(-3, -4) and Z(-3, 2).
(2) the image \(W_{1}X_{1}Y_{1}Z_{1}\) of the quadrilateral WXYZ under an anti-clockwise rotation of 90° about the origin where \(W \to W_{1}, X \to X_{1}, Y \to Y_{1}\) and \(Z \to Z_{1}\).
Akọwa Nkọwa
None
Ajụjụ 65 Ripọtì
(a) Using a scale of 2cm to 2units on both axes, draw on a sheet of graph paper two perpendicular axes 0x and 0y for - 10 \(\leq\) x \(\leq\) 10 and -10 \(\leq\) y \(\leq\)10
(b) Given the points P(3, 2). Q(-1. 5). R(0. 8) and S(3, 7). draw on the same graph, indicating clearly the vertices and their coordinates, the:
(i) quadrilateral PQRS;
(ii) image P\(_1\)Q\(_1\)R\(_1\)S\(_1\) of PQRS under an anticlockwise rotation of 90\(^o\) about the origin where P \(\to\) P\(_1\), Q \(\to\) Q\(_{1}\), R \(\to\) R\(_{1}\) and S \(\to\) S\(_{1}\)
(iii) image P\(_2\)Q\(_2\)R\(_2\)S\(_2\) of P\(_1\)Q\(_1\)R\(_1\)S\(_1\) under a reflection in the line y - x = 0 where P\(_1\) \(\to\) P\(_2\), Q\(_1\) \(\to\) Q\(_{2}\), R\(_1\) \(\to\) R\(_{2}\) and S\(_1\) \(\to\) S\(_{2}\)
(c) Describe precisely the single transformation T for which T : PQRS \(\to\) P\(_2\)Q\(_2\)R\(_2\)S\(_2\)
(d) The side P\(_1\)Q\(_1\) of the quadrilateral P\(_1\)Q\(_1\)R\(_1\)S\(_1\) cuts the x-axis at the point W. What type of quadrilateral is P\(_1\)S\(_1\)R\(_1\)W?
None
Akọwa Nkọwa
None
Ajụjụ 66 Ripọtì
(a) If tan x = \(\frac{5}{12}\), \(0^o\). < x < 90°, evaluate, without using Mathematical tables or calculator, \(\frac{sin x}{(sin x)^2 + cosx}\)
(b) The diagram shows a rectangular lawn measuring 14m by 11m. A path of uniform width \(x\)m surrounds it. If the total area of the path is 186 m\(^2\), how wide is the path?
Akọwa Nkọwa
None
Ajụjụ 67 Ripọtì
(a) Find the equation of the line passing through the points (2, 5) and (-4, -7).
(b) Three ships P, Q and R are at sea. The bearing of Q from P is 030° and the bearing of P and R is 300°. If |PQ| = 5 km and |PR| = 8 km,
(i) Illustrate the information in a diagram.
(ii) Calculate, correct to three significant figures, the:
(1) distance between Q and R
(2) bearing of R from Q.
Ajụjụ 68 Ripọtì
In a road worthiness test on 240 cars, 60% passed. The number that failed had faults in Clutch, Brakes and Steering as follows: Clutch only - 28, Clutch and Steering - 14; Clutch, Steering and Brakes - 8; Clutch and Brakes - 20; Brakes and Steering only - 6. The number of cars with faults in Steering only is twice the number of cars with faults in Brakes only.
(a) Draw a Venn Diagram to illustrate this information.
(b) How many cars had : (i) Faulty Brakes? (ii) Only one fault?
None
Akọwa Nkọwa
None
Ajụjụ 69 Ripọtì
A container, in the form of a cone resting on its vertex, is full when 4.158 litres of water is poured into it.
(a) If the radius of its base is 21 cm,
(i) represent the information in a diagram;
(ii) calculate the height of the container.
(b) A certain amount of water is drawn out of the container such that the surface diameter of the water drops to 28 cm. Calculate the volume of the water drawn out. (Take \(\pi\) = \(\frac{22}{7}\))
Ajụjụ 70 Ripọtì
1. A donkey is tied with a rope to a post which is 15 m from a fence. If the length of the rope between the donkey and the post is 17m, calculate the length of the fence within the reach of the donkey.
2. The base of a right pyramid with vertex, V, is a square, PQRS, of side 15 cm. If the slant height is 32 cm long:
3. represent the information in a diagram;
4. calculate its:
5. height, correct to one decimal place;
6. volume, correct to the nearest \(cm^3\)
Ajụjụ 72 Ripọtì
| Marks | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 | 90 |
| Frequency | 1 | 1 | x | 5 | y | 1 | 4 | 3 | 1 |
The frequency distribution shows the marks distribution of a class of 30 students in an examination.
The mean mark of the distribution is 52.
(a) Find the values of x and y.
(b) Construct a group frequency distribution table starting with a lower class limit of 1 and class interval of 10.
(c) Draw a histogram for the distribution
(d) Use the histogram to estimate the mode.
Akọwa Nkọwa
None
Ajụjụ 73 Ripọtì
(a) If the mean of m, n, s, p and q is 12, calculate the mean of (m + 4), (n - 3), (s + 6), (p - 2) and (q + 8).
(b) In a community of 500 people, the 75th percentile age is 65 years while the 25th percentile age is 15 years. How many of the people are between 15 and 65 years?
Akọwa Nkọwa
None
Ajụjụ 74 Ripọtì
The diagram shows an athletics track with two parallel sides and two semi-circular ends Each of the parallel sides is 60 metres, long and the diameter of each semi-circular end is 120 metres long.
(a) Calculate the distance covered by an athlete who runs round the tack the two times. [Take \(\pi\) = \(\frac{22}{7}\)]
(b) If the athlete spends 200 seconds for the race, calculate the speed in km/h.
Ajụjụ 75 Ripọtì
In the diagram, |PT| = 4 cm, |TS| = 6 cm, |PQ| = 6 cm and < SPR = 30°. Calculate, correct to the nearest whole number:
(a) |SR| ;
(b) area of TQRS.
Akọwa Nkọwa
None
Ajụjụ 76 Ripọtì
A used car was purchased at N900,000.00. Its value depreciated by 30% in the first year. In each subsequent year, the depreciation was 22% of its value at the beginning of the year. If the car was bought on the 1st of March, 2011, calculate, correct to the nearest hundred naira, the value of the car on the 28th of February, 2015.
Ajụjụ 77 Ripọtì
(a)
(i) Copy and complete the addition \(\oplus\) and multiplication \(\otimes\) tables in modulo 5 on the set {2, 3, 4}.
| \(\oplus\) | 2 | 3 | 4 |
| 2 | |||
| 3 | |||
| 4 |
| \(\otimes\) | 2 | 3 | 4 |
| 2 | |||
| 3 | |||
| 4 |
(ii) Use the tables to:
(a) solve the equation 4 \(\otimes\)e\(\oplus\) 2 \(\equiv\) 1 (mod 5):
(b) find the value of n, if 4 \(\oplus\)n\(\otimes\) 2 \(\equiv\) (mod 5).
(b) Consider the following statements:
p: Landi has cholera,
q: Landi is in the hospital.
If p = q, state whether or not the following statements are valid:
(i) If Landi is in the hospital, then he has cholera.
(ii) If Landi is not in the hospital, then he does not have cholera.
(iii) If Landi does not have cholera, then he is not in the hospital.
Ajụjụ 78 Ripọtì
(a) The frequency distribution shows the range of prices of a brand of a car sold by a dealer and the corresponding quantity demanded.
| Price (N1,000,000.00 |
1.0 - 1.9 | 2.0 - 2.9 | 3.0 - 3.9 | 4.0 - 4.9 | 5.0 - 5.9 |
| Number of Vehicles | 23 | 48 | 107 | 90 | 32 |
(b) Represent the information in a histogram and use the histogram to determine the most preferred selling price for the brand of car.
Akọwa Nkọwa
None
Ajụjụ 79 Ripọtì
(a) In a right-angled triangle, sin X = \(\frac{3}{5}\). Evaluate, leaving the answer as a fraction, 5 (cosX)\(^2\) – 3.
(b) The base of a pyramid, 12 cm high, is a rectangle with dimensions 42 cm by 11 cm. if the pyramid is filled with water and emptied into a conical container of equal height and volume, calculate, leaving the answer in surd form (radicals), the base radius of the container. [Take π=\(\frac{22}{7}\)]
Ajụjụ 80 Ripọtì
A shop had two reduction sales during which prices of all items were reduced by 40% in the first sales and 30% in the second.
- If a shirt was sold at GH₵35.00 during the second reduction sales, find the price before the first sale.
- If the price of an article before the first reduction was GH₵180.00, find the total:
- reduction in the price due to the two sales;
- percentage reduction in the price of the article.
Akọwa Nkọwa
None
Ajụjụ 81 Ripọtì
(a) Copy and complete the table of values for y = 2x\(^{2}\) + x - 10 for -5 \(\leq\) x \(\leq\) 4.
| x | -5 | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
| y | 5 | -9 | -10 | 0 |
(b) Using scales of 2cm to 1 unit on the x- axis and 2cm to 5 units on the y- axis, Draw the graph of y = 2x\(^{2}\) + x - 10 for -5 \(\leq\) x \(\leq\) 4.
(c) Use the graph to find the solution of :
(i) 2x\(^{2}\) + x = 10
(ii) 2x\(^{2}\) + x - 10 = 2x
Ajụjụ 82 Ripọtì
A man starts from a point X and walk 285 m to Y on a bearing of 078\(^o\). He then walks due South to a point Z which is 307 m from X.
(a) Illustrate the information on a diagram.
(b) Find, correct to the nearest whole number, the:
(i) bearing of X from Z;
(ii) distance between Y and Z.
Ajụjụ 83 Ripọtì
(a) The graph of \(y = 2px^{2} - p^{2}x - 14\) passes through the point (3, 10). Find the values of p.
(b) Two lines, \(3y - 2x = 21\) and \(4y + 5x = 5\) intersect at the point Q. Find the coordinates of Q.
Akọwa Nkọwa
None
Ajụjụ 84 Ripọtì
(a) Lamin bought a book for N300.00 and sold it to Bola at a profit of x%. Bola then sold the same book at a profit of x%. If James paid \(N(6x + \frac{3}{4})\) more for the book than Lamin paid, find the value of x.
(b) Find the range of values of x which satisfies the inequality \(3x - 2 < 10 + x < 2 + 5x\).
Akọwa Nkọwa
None
Ajụjụ 85 Ripọtì
(a) Evaluate without using calculator, (\(\frac{1}{4} \times 9\frac{1}{7} + \frac{2}{5} (\frac{2}{2} + \frac{3}{4})) \div (\frac{2}{5} - \frac{1}{4})\)
(b) A hunter walked 250 m from point P to Q on a bearing 042\(^o\). Calculate, correct to the nearest meter the vertical distance he has moved.
Ajụjụ 86 Ripọtì
Musa is three years older than Manya. Seven years ago, Musa was twice as old as Manya. (1) How old are they now? (2) When will the sum of them be 45?
Akọwa Nkọwa
None
Ajụjụ 87 Ripọtì
The table shows the distribution of marks scored by students in a test.
| Mark (%) |
10 - 19 | 20 - 29 | 30 - 39 | 40 - 49 | 50 - 59 | 60 - 69 | 70 - 79 | 80 - 89 | 90 - 99 |
| Frequency | 4 | 7 | 12 | 18 | 20 | 14 | 9 | 4 | 2 |
(a) Construct a cumulative frequency table for the distribution.
(b) Draw a cumulative frequency curve for the distribution.
(c) Use the curve to estimate the:
(i) median;
(ii) probability that a student selected at random obtained distinction, if the lowest mark for distinction is 75%.
None
Akọwa Nkọwa
None
Ajụjụ 88 Ripọtì
a) Copy and complete the following table of values for y = 2 cos x – sin x ,\(0^o \leq x \leq 300^o\)
\[\begin{array}{c|c} x & 0^o & 30^o & 60^o & 90^o & 120^o & 150^o & 180^o & 210^o & 240^o & 270^o& 300^o \\ \hline Y & 2.00 & & 0.13 & & -1.87 & & -2.00 & & -0.13 & & \end{array}\]
(b) Using scales of 2 cm to 30\(^o\) on the x-axis and 2cm to 1 unit on the y-axis, draw the graph of y = 2 cos x – sin x for \(0^o \leq x \leq 300^o\)
(c) Use the graph to find the value(s) of x for which:
(i) 2 cos x – sin x = 1;
(ii) tan x = 2.
Akọwa Nkọwa
None
