WAEC SSCE - General Mathematics - 2021
Ajụjụ 1 Ripọtì
For what value of x is \(\frac{4 - 2x}{x + 1}\) undefined.
Akọwa Nkọwa
A fraction is undefined when its denominator is equal to zero. Therefore, we need to find the value of x that makes the denominator x + 1 equal to zero.
x + 1 = 0
x = -1
Therefore, the value of x that makes the fraction undefined is x = -1
Ajụjụ 2 Ripọtì
The distance d between two villages east more than 18 KM but not more than 23KM.
which of these inequalities represents the statements?
Akọwa Nkọwa
Ajụjụ 3 Ripọtì
A cone has a base radius of 8cm and height 11cm. calculate , correct to 2d.p, the curved surface area
Akọwa Nkọwa
Ajụjụ 4 Ripọtì
The sum of the interior angles of a regular polygon with k sides is (3k-10) right angles. Find the size of the exterior angle?
Akọwa Nkọwa
Ajụjụ 8 Ripọtì
In the diagram, ∠ABC and ∠BCD are right angles, ∠BAD = t and ∠EDF = 70°. Find the value of t.
Ajụjụ 9 Ripọtì
\(\overline{XY}\) is a line segments with the coordinates X (- 8,- 12) and Y(p,q). if the midpoint of \(\overline{XY}\) is (-4,-2) find the coordinates of Y.
Akọwa Nkọwa
The midpoint of a line segment is the point that is exactly halfway between the two endpoints of the segment. To find the midpoint, we take the average of the x-coordinates and the y-coordinates of the endpoints. So, for line segment XY with endpoints X (-8,-12) and Y (p,q), the midpoint is (-4,-2). Therefore, the average of the x-coordinates of X and Y is -4: (-8 + p)/2 = -4 Solving for p, we find: p = -8 + 2 * -4 = 0 Similarly, the average of the y-coordinates of X and Y is -2: (-12 + q)/2 = -2 Solving for q, we find: q = -12 + 2 * -2 = 8 Therefore, the coordinates of Y are (0,8).
Ajụjụ 11 Ripọtì
| Height(cm) | 160 | 161 | 162 | 163 | 164 | 165 |
| No. of players | 4 | 6 | 3 | 7 | 8 | 9 |
the table shows the height of 37 players of a basketball team calculates correct to one decimal place the mean height of the players.
Akọwa Nkọwa
Ajụjụ 13 Ripọtì
In △LMN, |LM| = 6cm, ∠LNM = x and sin x = sin x = \(\frac{3}{5}\).
Find the area of △LMN
Akọwa Nkọwa
Ajụjụ 15 Ripọtì
Find, correct to two decimal, the mean of 1\(\frac{1}{2}\), 2\(\frac{2}{3}\), 3\(\frac{3}{4}\), 4\(\frac{4}{5}\), and 5\(\frac{5}{6}\).
Ajụjụ 17 Ripọtì
The equation of a line is given as 3 x - 5y = 7. Find its gradient (slope)
Akọwa Nkọwa
The equation of a line in the slope-intercept form is y = mx + c, where m is the gradient (slope) of the line and c is the y-intercept. However, the given equation of the line is not in the slope-intercept form, but in the standard form. To find the slope of the line, we need to rewrite the equation in the slope-intercept form. We can do this by solving the equation for y: 3x - 5y = 7 -5y = -3x + 7 y = (3/5)x - 7/5 Comparing this with the slope-intercept form, we can see that the gradient (slope) of the line is 3/5. Therefore, the answer is 3/5.
Ajụjụ 18 Ripọtì
A cyclist moved at a speed of Xkm/h for 2 hours. He then increased his speed by 2 km/h for the next 3 hours.
If the total distance covered is 36 km, calculate his initials speed.
Ajụjụ 19 Ripọtì
In the diagram O is the centre of the circle PQRS, ?PQR = 72° and OR is parallel to PS. Find ∠ OPS
Ajụjụ 20 Ripọtì
500 tickets were sold for a concert tickets for adults and children were sold at $4.50 and $3.00 respectively if the total receipts for the concerts was $1987.50 how many tickets for adults were sold?
Akọwa Nkọwa
Let's assume that x is the number of tickets sold for adults, and y is the number of tickets sold for children. We know that the total number of tickets sold is 500, so we can write an equation: x + y = 500 We also know that the price of an adult ticket is $4.50 and the price of a child ticket is $3.00. So we can write another equation based on the total receipts: 4.5x + 3y = 1987.5 Now we have two equations with two unknowns, and we can solve for x, the number of adult tickets sold. One way to do this is to use the first equation to solve for y: y = 500 - x Then we can substitute this expression for y into the second equation: 4.5x + 3(500 - x) = 1987.5 Simplifying and solving for x: 4.5x + 1500 - 3x = 1987.5 1.5x = 487.5 x = 325 So the number of adult tickets sold is 325. Therefore, the correct answer is option (A): 325.
Ajụjụ 21 Ripọtì
Which of the following is not an exterior angle of a regular polygon?
Ajụjụ 23 Ripọtì
If 16 * 2\(^{(x + 1)}\) = 4\(^x\) * 8\(^{(1 - x)}\), find the value of x.
Akọwa Nkọwa
Ajụjụ 24 Ripọtì
If log\(_{10}\) 2 = m and log\(_{10}\) 3 = n, find log\(_{10}\) 24 in terms of m and n.
Akọwa Nkọwa
Ajụjụ 25 Ripọtì
The pie chart represents the distribution of fruits on display in the shop if there are 60 apples on display how many oranges are there?
Ajụjụ 27 Ripọtì
If \(\frac{2}{x-3}\) - \(\frac{3}{x-2}\) = \(\frac{p}{(x-3)(x -2)}\), find p.
Akọwa Nkọwa
Ajụjụ 28 Ripọtì
find the value of (x+y)
Ajụjụ 29 Ripọtì
From a point T, a man moves 12km due west and then moves 12km due south to another point Q. Calculate the bearing of T from Q.
Akọwa Nkọwa
Ajụjụ 30 Ripọtì
if p = {-3<x<1} and Q = {-1<x<3}, where x is a real number, find P n Q.
Ajụjụ 31 Ripọtì
In the diagram, ?XYZ is produced to T. if |XY| = |ZY| and ?XYT = 40°, find ?XZT
Ajụjụ 35 Ripọtì
A trader paid import duty of 38 kobo in the naira on the cost of an engine. If a total of #22,800.00 was paid as import duty, calculate the cost of the engine.
Ajụjụ 37 Ripọtì
A fair die is tossed twice what is the probability of get a sum of at least 10.
Akọwa Nkọwa
Ajụjụ 38 Ripọtì
In the diagram, \(\overline{MP}\) is a tangent to the circle NQR, ∠NQR, ∠PNQ = 64 and | \(\overline{RQ}\) | = | \(\overline{RN}\) |. Find the angle market t.
Akọwa Nkọwa
Ajụjụ 39 Ripọtì
correct 0.007985 to three significant figures.
Ajụjụ 42 Ripọtì
The height of an equilateral triangle of side is 10 3√ cm. calculate its perimeter.
Akọwa Nkọwa
Ajụjụ 43 Ripọtì
What number should be subtracted from the sum of 2 \(\frac{1}{6}\) and 2\(\frac{7}{12}\) to give 3\(\frac{1}{4}\)?
Akọwa Nkọwa
Ajụjụ 44 Ripọtì
The circumference of a circular track is 9km. A cyclist rides round it a number of times and stops after covering a distance of 302km. How far is the cyclist from the starting point?
Akọwa Nkọwa
Ajụjụ 46 Ripọtì
consider the statements:
P = All students offering Literature(L) also offer History(H);
Q = Students offering History(H) do not offer Geography(G).
Which of the Venn diagram correctly illustrate the two statements?
Ajụjụ 47 Ripọtì
In the diagram, line \(\overline{EC}\) is a diameter of the circle ABCDE.
If angle ABC equals 158°, find ?ADE
Akọwa Nkọwa
Ajụjụ 48 Ripọtì
A man will be (x+10)years old in 8years time. If 2years ago he was 63 years., find the value of x
Akọwa Nkọwa
A man will be (x+10) years old in 8years time.
As at today, he is x + 2 years of age.
The man was 63 years old 2 years ago, so he is 63+2=65 now.
8 years from now, he will be 65+8=73.
He will be (x+10) years old when he is 73. So
x+10=73
x=73-10=63
Ajụjụ 50 Ripọtì
.(a) In a class of 80 students,\(\frac{3}{4}\) study Biology and \(\frac{3}{5}\) study Physics.
If each student studies at least one of the subjects: (i) draw a Venn diagram to represent this information
(ii) how many students study both subjects
(iii) find the fraction of the class that study Biology but not Physics.
(b) Johnson and Jocatol Ltd. owned a business office with floor measuring 15m by 8 m which was to be carpeted.
The cost of carpeting was Gh¢ 890.00 per square metre. If a total of GH 216,120.00 was spent on painting and carpeting, how much was the cost of painting?
(a)
(i) Here is a Venn diagram to represent the information given:
B / \ / \ P/_____\ | | 80 ---
where P represents the students who study Physics and B represents those who study Biology. The region inside the circle P but outside the intersection represents students who only study Physics, while the region inside the circle B but outside the intersection represents students who only study Biology. The intersection represents students who study both subjects.
(ii) To find the number of students who study both subjects, we need to find the size of the intersection. Let x be the number of students who study both subjects. Then we can write two equations based on the given information:
- \(\frac{3}{4}\times80\) students study Biology.
- \(\frac{3}{5}\times80\) students study Physics.
Using these equations, we can solve for x:
- Biology students: \(\frac{3}{4}\times80 = 60\)
- Physics students: \(\frac{3}{5}\times80 = 48\)
- Total students: 80
- Students who study both subjects: x
From the diagram, we can see that the total number of students is 80, so we have:
60 + x + 48 = 80
Simplifying this equation, we get:
x = 12
Therefore, 12 students study both subjects.
(iii) To find the fraction of the class that study Biology but not Physics, we need to find the size of the region inside the circle B but outside the intersection, and divide by the total number of students. Let y be the number of students who study Biology but not Physics. Then we can write another equation based on the given information:
- \(\frac{3}{4}\times80\) students study Biology.
- x students study both subjects.
Using these equations, we can solve for y:
- Biology students: \(\frac{3}{4}\times80 = 60\)
- Students who study both subjects: x = 12
- Students who study Biology but not Physics: y
From the diagram, we can see that:
y + x = 60
Substituting x = 12, we get:
y + 12 = 60
Simplifying this equation, we get:
y = 48
Therefore, 48 students study Biology but not Physics. The fraction of the class that study Biology but not Physics is:
\(\frac{48}{80} = \frac{3}{5
Akọwa Nkọwa
(a)
(i) Here is a Venn diagram to represent the information given:
B / \ / \ P/_____\ | | 80 ---
where P represents the students who study Physics and B represents those who study Biology. The region inside the circle P but outside the intersection represents students who only study Physics, while the region inside the circle B but outside the intersection represents students who only study Biology. The intersection represents students who study both subjects.
(ii) To find the number of students who study both subjects, we need to find the size of the intersection. Let x be the number of students who study both subjects. Then we can write two equations based on the given information:
- \(\frac{3}{4}\times80\) students study Biology.
- \(\frac{3}{5}\times80\) students study Physics.
Using these equations, we can solve for x:
- Biology students: \(\frac{3}{4}\times80 = 60\)
- Physics students: \(\frac{3}{5}\times80 = 48\)
- Total students: 80
- Students who study both subjects: x
From the diagram, we can see that the total number of students is 80, so we have:
60 + x + 48 = 80
Simplifying this equation, we get:
x = 12
Therefore, 12 students study both subjects.
(iii) To find the fraction of the class that study Biology but not Physics, we need to find the size of the region inside the circle B but outside the intersection, and divide by the total number of students. Let y be the number of students who study Biology but not Physics. Then we can write another equation based on the given information:
- \(\frac{3}{4}\times80\) students study Biology.
- x students study both subjects.
Using these equations, we can solve for y:
- Biology students: \(\frac{3}{4}\times80 = 60\)
- Students who study both subjects: x = 12
- Students who study Biology but not Physics: y
From the diagram, we can see that:
y + x = 60
Substituting x = 12, we get:
y + 12 = 60
Simplifying this equation, we get:
y = 48
Therefore, 48 students study Biology but not Physics. The fraction of the class that study Biology but not Physics is:
\(\frac{48}{80} = \frac{3}{5
Ajụjụ 51 Ripọtì
(a) On Sam's first birthday celebration, his grandfather deposited an amount of S 1,000.00 in a bank compound at 4 % interest annually.
Find how much is in the account if Sam is 4 years old.
In the diagram, ABCD are points on the circle centre O. If |AB| = |BC| and ∠ADC= 50°, find ∠BAD.
(a) Amount at end of the 2nd year = \(\frac{104}{100}\) x $1000
= $1,040.00
Amount at end of 3rd year = \(\frac{104}{100}\) x $1040
= $1,081.60
Amount at end of 4th year = \(\frac{104}{100}\) x $1081.6
= $1,124.86
(b) ∠ADC + ∠DCA+ ∠CAD = 180°
50° + 90°+ ∠CAD = 180°
∠CAD = (180 - 140)°
= 40
∠ADC + ∠ABC = 180°
50° + ∠ABC = 180°
∠ABC =130°
∠BAC+ ∠BCA + ∠ABC = 180°
But ∠BAC = ∠BCA
2∠BAC+ ∠130° = 180°
2 ∠BAC=50°
∠BAC = 25°
∠BAD = ∠CAD + ∠BAC
= 40°+25° = 65°
(a) If Sam's grandfather deposited $1,000.00 in a bank at 4% interest annually, after 4 years, the total amount in the account would be $1,124.86.
(b) To find ∠BAD, we need to find ∠CAD and ∠BAC first. We know that ∠ADC = 50° and ∠ADC + ∠DCA + ∠CAD = 180°, so ∠CAD = 180° - 50° - 90° = 40°. Next, we know that ∠ADC + ∠ABC = 180°, so ∠ABC = 180° - 50° = 130°. Also, ∠BAC + ∠BCA + ∠ABC = 180° and ∠BAC = ∠BCA, so 2∠BAC + 130° = 180° and ∠BAC = (180° - 130°) / 2 = 25°. Finally, ∠BAD = ∠CAD + ∠BAC = 40° + 25° = 65°.
Akọwa Nkọwa
(a) If Sam's grandfather deposited $1,000.00 in a bank at 4% interest annually, after 4 years, the total amount in the account would be $1,124.86.
(b) To find ∠BAD, we need to find ∠CAD and ∠BAC first. We know that ∠ADC = 50° and ∠ADC + ∠DCA + ∠CAD = 180°, so ∠CAD = 180° - 50° - 90° = 40°. Next, we know that ∠ADC + ∠ABC = 180°, so ∠ABC = 180° - 50° = 130°. Also, ∠BAC + ∠BCA + ∠ABC = 180° and ∠BAC = ∠BCA, so 2∠BAC + 130° = 180° and ∠BAC = (180° - 130°) / 2 = 25°. Finally, ∠BAD = ∠CAD + ∠BAC = 40° + 25° = 65°.
Ajụjụ 52 Ripọtì
The table shows the distribution of the number of hours per day spent in studying by 50 students.
| Number of hours per day |
4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| Number of students |
5 | 7 | 5 | 9 | 12 | 4 | 3 | 5 |
Calculate, correct to two decimal places,
the: (a) mean; (b) standard deviation.
To find the mean and standard deviation, we can use the following formulas:
Mean (average) = (sum of all values) / (total number of values)
Standard deviation = sqrt((sum of (value - mean)^2) / (total number of values))
Let's first find the mean:
Multiply the number of hours by the number of students in that category and add up the results for all categories:
(4 x 5) + (5 x 7) + (6 x 5) + (7 x 9) + (8 x 12) + (9 x 4) + (10 x 3) + (11 x 5) = 411
The total number of students is 50, so:
Mean (average) = 411 / 50 = 8.22 (correct to two decimal places)
Therefore, the mean number of hours per day spent in studying by these 50 students is 8.22.
Next, let's find the standard deviation:
First, find the difference between each value and the mean, square each difference, and then add up the results:
[(4 - 8.22)^2 x 5] + [(5 - 8.22)^2 x 7] + [(6 - 8.22)^2 x 5] + [(7 - 8.22)^2 x 9] + [(8 - 8.22)^2 x 12] + [(9 - 8.22)^2 x 4] + [(10 - 8.22)^2 x 3] + [(11 - 8.22)^2 x 5] = 201.2
Divide the sum by the total number of values and take the square root:
Standard deviation = sqrt(201.2 / 50) = 1.99 (correct to two decimal places)
Therefore, the standard deviation is 1.99.
In conclusion, the mean number of hours per day spent in studying by these 50 students is 8.22, and the standard deviation is 1.99.
Akọwa Nkọwa
To find the mean and standard deviation, we can use the following formulas:
Mean (average) = (sum of all values) / (total number of values)
Standard deviation = sqrt((sum of (value - mean)^2) / (total number of values))
Let's first find the mean:
Multiply the number of hours by the number of students in that category and add up the results for all categories:
(4 x 5) + (5 x 7) + (6 x 5) + (7 x 9) + (8 x 12) + (9 x 4) + (10 x 3) + (11 x 5) = 411
The total number of students is 50, so:
Mean (average) = 411 / 50 = 8.22 (correct to two decimal places)
Therefore, the mean number of hours per day spent in studying by these 50 students is 8.22.
Next, let's find the standard deviation:
First, find the difference between each value and the mean, square each difference, and then add up the results:
[(4 - 8.22)^2 x 5] + [(5 - 8.22)^2 x 7] + [(6 - 8.22)^2 x 5] + [(7 - 8.22)^2 x 9] + [(8 - 8.22)^2 x 12] + [(9 - 8.22)^2 x 4] + [(10 - 8.22)^2 x 3] + [(11 - 8.22)^2 x 5] = 201.2
Divide the sum by the total number of values and take the square root:
Standard deviation = sqrt(201.2 / 50) = 1.99 (correct to two decimal places)
Therefore, the standard deviation is 1.99.
In conclusion, the mean number of hours per day spent in studying by these 50 students is 8.22, and the standard deviation is 1.99.
Ajụjụ 53 Ripọtì
.(a) In APQR, ∠PQR= 90°. If its area is 216cm\(^2\) and |PQ|:|QR| is 3:4, find |PR|.
(b) The present ages of a man and his son are 47 years and 17 years respectively. In how many years would the man's age be twice that of his son?
(a) We can use the formula for the area of a triangle, which is A = 1/2 * base * height, where the base is PQ and the height is PR. Since we know that PQ:QR is 3:4, we can let PQ be 3x and QR be 4x for some value of x.
Then, using the Pythagorean theorem, we can find PR: PR2 = PQ2 + QR2 = (3x)2 + (4x)2 = 25x2.
Substituting this into the formula for the area, we get:
216 = 1/2 * 3x * PR
PR = 144/3x = 48/x
So, we need to find x to determine PR. To do this, we can use the fact that the area of the triangle is also equal to 1/2 * PQ * PR. Substituting in the values we know, we get:
216 = 1/2 * 3x * PR
216 = 3/2 * x * (48/x)
216 = 72
x = 3
Therefore, PR = 48/3 = 16 cm.
(b) Let x be the number of years from now when the man's age will be twice that of his son. Then, in x years, the man's age will be 47 + x and the son's age will be 17 + x.
We can set up an equation using the fact that the man's age will be twice that of his son:
47 + x = 2(17 + x)
Solving for x, we get:
x = 13
Therefore, in 13 years, the man's age will be twice that of his son.
Akọwa Nkọwa
(a) We can use the formula for the area of a triangle, which is A = 1/2 * base * height, where the base is PQ and the height is PR. Since we know that PQ:QR is 3:4, we can let PQ be 3x and QR be 4x for some value of x.
Then, using the Pythagorean theorem, we can find PR: PR2 = PQ2 + QR2 = (3x)2 + (4x)2 = 25x2.
Substituting this into the formula for the area, we get:
216 = 1/2 * 3x * PR
PR = 144/3x = 48/x
So, we need to find x to determine PR. To do this, we can use the fact that the area of the triangle is also equal to 1/2 * PQ * PR. Substituting in the values we know, we get:
216 = 1/2 * 3x * PR
216 = 3/2 * x * (48/x)
216 = 72
x = 3
Therefore, PR = 48/3 = 16 cm.
(b) Let x be the number of years from now when the man's age will be twice that of his son. Then, in x years, the man's age will be 47 + x and the son's age will be 17 + x.
We can set up an equation using the fact that the man's age will be twice that of his son:
47 + x = 2(17 + x)
Solving for x, we get:
x = 13
Therefore, in 13 years, the man's age will be twice that of his son.
Ajụjụ 54 Ripọtì
In the diagram, \(\overline{PQ//RS}\) is a trapezium with QR//PS. U and T are points on \(\overline{PS}\) such that \(\overline{|PU|}\) = 5 cm,
\(\overline{|QU|}\) = 12 cm and ?PUQ= ?STR =90°. If the area of PQR = 20 cm\(^2\),
calculate, correct to the nearest whole number, the:
(a) perimeter; (b) area; of the trapezium
(a) Perimeter: We can calculate the perimeter of a trapezium by adding up the lengths of all four sides. If we call the length of side PR "x", then the length of side QR is 5 cm + x. The length of side PS is 12 cm + 5 cm = 17 cm, and the length of side QS is 12 cm.
So the perimeter is: x + 5 + 17 + 12 + x = 2x + 34 cm
(b) Area: To find the area of the trapezium, we first need to find the height. We can use the Pythagorean theorem to find the height of the trapezium.
Let h be the height of the trapezium. Then, h2 = PR2 - PU2 = x2 - 25
And, h2 = QR2 - QU2 = (5 + x)2 - 144
So, x2 - 25 = (5 + x)2 - 144
Expanding both sides: x2 - 25 = x2 + 10x + 25 - 144
10x = -144 + 25 - 25 = -144
x = -14.4 cm
Since x has to be positive, the height of the trapezium is not possible. So, the problem has no solution.
Explanation: In this problem, we are given a trapezium PQRS with parallel sides PQ and RS, and we are asked to find the perimeter and area of the trapezium. To find the perimeter, we simply add up the lengths of all four sides. To find the area, we need to find the height of the trapezium, which we can do using the Pythagorean theorem. However, in this case, the height of the trapezium is not possible, so the problem has no solution.
Akọwa Nkọwa
(a) Perimeter: We can calculate the perimeter of a trapezium by adding up the lengths of all four sides. If we call the length of side PR "x", then the length of side QR is 5 cm + x. The length of side PS is 12 cm + 5 cm = 17 cm, and the length of side QS is 12 cm.
So the perimeter is: x + 5 + 17 + 12 + x = 2x + 34 cm
(b) Area: To find the area of the trapezium, we first need to find the height. We can use the Pythagorean theorem to find the height of the trapezium.
Let h be the height of the trapezium. Then, h2 = PR2 - PU2 = x2 - 25
And, h2 = QR2 - QU2 = (5 + x)2 - 144
So, x2 - 25 = (5 + x)2 - 144
Expanding both sides: x2 - 25 = x2 + 10x + 25 - 144
10x = -144 + 25 - 25 = -144
x = -14.4 cm
Since x has to be positive, the height of the trapezium is not possible. So, the problem has no solution.
Explanation: In this problem, we are given a trapezium PQRS with parallel sides PQ and RS, and we are asked to find the perimeter and area of the trapezium. To find the perimeter, we simply add up the lengths of all four sides. To find the area, we need to find the height of the trapezium, which we can do using the Pythagorean theorem. However, in this case, the height of the trapezium is not possible, so the problem has no solution.
Ajụjụ 55 Ripọtì
(a) Copy and complete the table of values for the relation y=2x\(^2\) - x - 2 for 4 ≤ x ≤ 4.
| x | -4 | -3 | -2 | -2 | 0 | 1 | 2 | 3 | 4 |
| y | 19 | -2 | 26 |
(b) Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw the graph of y = 2x\(^2\) - x - 2 for 4 ≤ x ≤ 4.
(c) On the same axes, draw the graph of y = 2x + 3.
(d) Use the graph to find the: (i) roots of the equation 2x-3r-5 0; (i) range of values of x for which 2x\(^2\) -x - 2<0.
(a)
To complete the table of values for the relation y=2x\(^2\) - x - 2 for 4 ≤ x ≤ 4, we need to substitute each value of x into the equation and calculate the corresponding value of y.
| x | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
| y | 30 | 19 | 8 | -1 | -2 | -1 | 4 | 13 | 26 |
(b)
Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, we can plot the points from the completed table of values to draw the graph of y = 2x\(^2\) - x - 2 for 4 ≤ x ≤ 4. The graph should be a smooth curve passing through the plotted points.
(c)
On the same axes, we can draw the graph of y = 2x + 3 by plotting some points and drawing a straight line passing through them. For example, when x = -2, y = -1, and when x = 2, y = 7. This gives us two points (-2,-1) and (2,7), which we can use to draw the line.
(d)
(i)
To find the roots of the equation 2x - 3r - 5 = 0 using the graph, we need to look for the points where the graph intersects the line y = 2x - 5. These points correspond to the values of x that satisfy the equation. From the graph, it appears that the intersection point is around x = 3. Therefore, we can estimate that the root of the equation is x ≈ 3.
(ii)
To find the range of values of x for which 2x\(^2\) -x - 2 < 0 using the graph, we need to look for the points where the graph is below the x-axis. These points correspond to the values of x that satisfy the inequality. From the graph, it appears that the graph is below the x-axis for values of x between approximately -1.5 and 1.5. Therefore, the range of values of x for which 2x\(^2\) -x - 2 < 0 is -1.5 < x < 1.5.
Akọwa Nkọwa
(a)
To complete the table of values for the relation y=2x\(^2\) - x - 2 for 4 ≤ x ≤ 4, we need to substitute each value of x into the equation and calculate the corresponding value of y.
| x | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
| y | 30 | 19 | 8 | -1 | -2 | -1 | 4 | 13 | 26 |
(b)
Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, we can plot the points from the completed table of values to draw the graph of y = 2x\(^2\) - x - 2 for 4 ≤ x ≤ 4. The graph should be a smooth curve passing through the plotted points.
(c)
On the same axes, we can draw the graph of y = 2x + 3 by plotting some points and drawing a straight line passing through them. For example, when x = -2, y = -1, and when x = 2, y = 7. This gives us two points (-2,-1) and (2,7), which we can use to draw the line.
(d)
(i)
To find the roots of the equation 2x - 3r - 5 = 0 using the graph, we need to look for the points where the graph intersects the line y = 2x - 5. These points correspond to the values of x that satisfy the equation. From the graph, it appears that the intersection point is around x = 3. Therefore, we can estimate that the root of the equation is x ≈ 3.
(ii)
To find the range of values of x for which 2x\(^2\) -x - 2 < 0 using the graph, we need to look for the points where the graph is below the x-axis. These points correspond to the values of x that satisfy the inequality. From the graph, it appears that the graph is below the x-axis for values of x between approximately -1.5 and 1.5. Therefore, the range of values of x for which 2x\(^2\) -x - 2 < 0 is -1.5 < x < 1.5.
Ajụjụ 56 Ripọtì
The points X, Y and Z are located such that Y is 15 km south of X, Z is 20 km from X on a bearing of 270".
Calculate, correct: (a) two significant figures, |YZ|
(b) The nearest degree, the bearing of Y from Z
(a)
To find |YZ|, we need to first draw a diagram of the points X, Y, and Z. Since Y is 15 km south of X, we can draw a line segment from X going south with a length of 15 km, and label the endpoint as Y. Next, we know that Z is 20 km from X on a bearing of 270°, which means Z is directly west of X. Therefore, we can draw a line segment from X going west with a length of 20 km, and label the endpoint as Z. Finally, we can draw a line segment from Y to Z.
To find the length of |YZ|, we can use the Pythagorean theorem, which states that for a right triangle with legs of lengths a and b and hypotenuse of length c, \(c^2 = a^2 + b^2\). In this case, YZ is the hypotenuse, and its length can be found by:
\(|YZ|^2 = |XY|^2 + |XZ|^2\)
where \(|XY|\) is the length of the segment from X to Y, and \(|XZ|\) is the length of the segment from X to Z.
We know that \(|XY|\) is 15 km, and \(|XZ|\) is 20 km, so:
\(|YZ|^2 = 15^2 + 20^2\) \(|YZ|^2 = 225 + 400\) \(|YZ|^2 = 625\) \(|YZ| = \sqrt{625}\) \(|YZ| = 25\) km (rounded to two significant figures)
Therefore, the length of \(|YZ|\) is 25 km, correct to two significant figures.
(b)
To find the bearing of Y from Z, we can use trigonometry. Specifically, we can use the tangent function, which is defined as the ratio of the opposite side to the adjacent side in a right triangle. In this case, we can use the tangent of the angle formed by the line segment YZ and the horizontal axis to find the bearing of Y from Z.
Let \(\theta\) be the angle formed by the line segment YZ and the horizontal axis. Then:
\(\tan(\theta) = \frac{{\text{{opposite}}}}{{\text{{adjacent}}}} = \frac{{|XY|}}{{|XZ|}}\)
We know that \(|XY|\) is 15 km and \(|XZ|\) is 20 km, so:
\(\tan(\theta) = \frac{{15}}{{20}}\) \(\tan(\theta) = 0.75\)
To find \(\theta\), we can take the inverse tangent (also called arctangent) of both sides:
\(\theta = \tan^{-1}(0.75)\) \(\theta = 36.87^\circ\)
Therefore, the bearing of Y from Z is 37°, rounded to the nearest degree. This means that if you were standing at point Z and facing due north, you would need to turn
Akọwa Nkọwa
(a)
To find |YZ|, we need to first draw a diagram of the points X, Y, and Z. Since Y is 15 km south of X, we can draw a line segment from X going south with a length of 15 km, and label the endpoint as Y. Next, we know that Z is 20 km from X on a bearing of 270°, which means Z is directly west of X. Therefore, we can draw a line segment from X going west with a length of 20 km, and label the endpoint as Z. Finally, we can draw a line segment from Y to Z.
To find the length of |YZ|, we can use the Pythagorean theorem, which states that for a right triangle with legs of lengths a and b and hypotenuse of length c, \(c^2 = a^2 + b^2\). In this case, YZ is the hypotenuse, and its length can be found by:
\(|YZ|^2 = |XY|^2 + |XZ|^2\)
where \(|XY|\) is the length of the segment from X to Y, and \(|XZ|\) is the length of the segment from X to Z.
We know that \(|XY|\) is 15 km, and \(|XZ|\) is 20 km, so:
\(|YZ|^2 = 15^2 + 20^2\) \(|YZ|^2 = 225 + 400\) \(|YZ|^2 = 625\) \(|YZ| = \sqrt{625}\) \(|YZ| = 25\) km (rounded to two significant figures)
Therefore, the length of \(|YZ|\) is 25 km, correct to two significant figures.
(b)
To find the bearing of Y from Z, we can use trigonometry. Specifically, we can use the tangent function, which is defined as the ratio of the opposite side to the adjacent side in a right triangle. In this case, we can use the tangent of the angle formed by the line segment YZ and the horizontal axis to find the bearing of Y from Z.
Let \(\theta\) be the angle formed by the line segment YZ and the horizontal axis. Then:
\(\tan(\theta) = \frac{{\text{{opposite}}}}{{\text{{adjacent}}}} = \frac{{|XY|}}{{|XZ|}}\)
We know that \(|XY|\) is 15 km and \(|XZ|\) is 20 km, so:
\(\tan(\theta) = \frac{{15}}{{20}}\) \(\tan(\theta) = 0.75\)
To find \(\theta\), we can take the inverse tangent (also called arctangent) of both sides:
\(\theta = \tan^{-1}(0.75)\) \(\theta = 36.87^\circ\)
Therefore, the bearing of Y from Z is 37°, rounded to the nearest degree. This means that if you were standing at point Z and facing due north, you would need to turn
Ajụjụ 57 Ripọtì
(a) A man shared his property among his children as follows:
| Child's name |
Ann | Afia | Kojo | Nuno | Akom |
| Percentage share |
5 | 15 | 10 | 45 | 25 |
Represent the information on a pie chart
(b) A box contains 5 red, 3 green and 4 blue identical beads. Calculate the probability th a girl takes away two red beads, one after the other, from the box.
(a) To represent the given information on a pie chart, we need to find the angles of the sectors corresponding to each child's percentage share.
Let the total value of the property be 100 units. Then, Ann's share is 5 units, Afia's share is 15 units, Kojo's share is 10 units, Nuno's share is 45 units, and Akom's share is 25 units.
To find the angle of the sector corresponding to each share, we use the formula:
Angle of sector = (Percentage share/100) x 360
Using this formula, we get:
Angle of Ann's sector = (5/100) x 360 = 18 degrees
Angle of Afia's sector = (15/100) x 360 = 54 degrees
Angle of Kojo's sector = (10/100) x 360 = 36 degrees
Angle of Nuno's sector = (45/100) x 360 = 162 degrees
Angle of Akom's sector = (25/100) x 360 = 90 degrees
We can now draw a circle representing the total value of the property, and divide it into sectors of the calculated angles for each child. Label each sector with the corresponding child's name.
(b) The probability of taking two red beads, one after the other, from the box can be calculated as follows:
The probability of taking a red bead on the first draw is 5/12 (since there are 5 red beads out of 12 beads in total).
After taking out one red bead, there are now 4 red beads and 11 beads in total. So the probability of taking out another red bead on the second draw is 4/11.
To find the probability of both events happening, we multiply the probabilities of each event. Therefore, the probability of taking two red beads, one after the other, is:
(5/12) x (4/11) = 5/33
So the probability of a girl taking two red beads, one after the other, from the box is 5/33.
Akọwa Nkọwa
(a) To represent the given information on a pie chart, we need to find the angles of the sectors corresponding to each child's percentage share.
Let the total value of the property be 100 units. Then, Ann's share is 5 units, Afia's share is 15 units, Kojo's share is 10 units, Nuno's share is 45 units, and Akom's share is 25 units.
To find the angle of the sector corresponding to each share, we use the formula:
Angle of sector = (Percentage share/100) x 360
Using this formula, we get:
Angle of Ann's sector = (5/100) x 360 = 18 degrees
Angle of Afia's sector = (15/100) x 360 = 54 degrees
Angle of Kojo's sector = (10/100) x 360 = 36 degrees
Angle of Nuno's sector = (45/100) x 360 = 162 degrees
Angle of Akom's sector = (25/100) x 360 = 90 degrees
We can now draw a circle representing the total value of the property, and divide it into sectors of the calculated angles for each child. Label each sector with the corresponding child's name.
(b) The probability of taking two red beads, one after the other, from the box can be calculated as follows:
The probability of taking a red bead on the first draw is 5/12 (since there are 5 red beads out of 12 beads in total).
After taking out one red bead, there are now 4 red beads and 11 beads in total. So the probability of taking out another red bead on the second draw is 4/11.
To find the probability of both events happening, we multiply the probabilities of each event. Therefore, the probability of taking two red beads, one after the other, is:
(5/12) x (4/11) = 5/33
So the probability of a girl taking two red beads, one after the other, from the box is 5/33.
Ajụjụ 58 Ripọtì
In the diagram, PQRS is a circle. = . ?SPR = 26° and the interior angles of PQS are in the ratio 2:3 :3.
Calculate: (i) PQR; (ii) RPQ; (iii) PRQ
(b) The coordinates of two points P and Q in a plane are (7, 3) and (5, x) respectively, where X is a real number.
If |PQ| = 29 units, find the value of x.
(a)
(i) Since PQRS is a circle and |PQ| = |QS|, we have ?PQR = ?QRS (angles subtended by equal chords are equal).
Let ?PQS = 2x. Then, ?PQR = 3x and ?QRS = 3x.
We know that the sum of interior angles of a triangle is 180°. So, in ?PQS, we have:
2x + 3x + 3x = 180°
8x = 180°
x = 22.5°
Now, in ?PQR, we have:
?PQR + ?QPR + ?RPQ = 180° (sum of interior angles of a triangle)
3x + 90° + 26° = 180° (since ?QPR is a right angle)
3x = 64°
x = 21.33°
Therefore, PQR = 3x = 64°.
(ii) RPQ = 180° - ?PQR - ?QPR = 180° - 64° - 90° = 26°.
(iii) PRQ = 180° - ?PQR - ?RPQ = 180° - 64° - 26° = 90°.
(b) Using the distance formula, we can find the distance between P and Q:
|PQ|² = (5-7)² + (x-3)²
|PQ|² = 4 + (x-3)²
Since |PQ| = 29 units, we have:
29² = 4 + (x-3)²
841 = (x-3)² + 4
837 = (x-3)²
Taking the square root of both sides, we get:
x-3 = ±?837
x = 3 ± ?837
Since x is a real number, we take the positive square root:
x = 3 + ?837
Therefore, the value of x is 3 + ?837 units.
Akọwa Nkọwa
(a)
(i) Since PQRS is a circle and |PQ| = |QS|, we have ?PQR = ?QRS (angles subtended by equal chords are equal).
Let ?PQS = 2x. Then, ?PQR = 3x and ?QRS = 3x.
We know that the sum of interior angles of a triangle is 180°. So, in ?PQS, we have:
2x + 3x + 3x = 180°
8x = 180°
x = 22.5°
Now, in ?PQR, we have:
?PQR + ?QPR + ?RPQ = 180° (sum of interior angles of a triangle)
3x + 90° + 26° = 180° (since ?QPR is a right angle)
3x = 64°
x = 21.33°
Therefore, PQR = 3x = 64°.
(ii) RPQ = 180° - ?PQR - ?QPR = 180° - 64° - 90° = 26°.
(iii) PRQ = 180° - ?PQR - ?RPQ = 180° - 64° - 26° = 90°.
(b) Using the distance formula, we can find the distance between P and Q:
|PQ|² = (5-7)² + (x-3)²
|PQ|² = 4 + (x-3)²
Since |PQ| = 29 units, we have:
29² = 4 + (x-3)²
841 = (x-3)² + 4
837 = (x-3)²
Taking the square root of both sides, we get:
x-3 = ±?837
x = 3 ± ?837
Since x is a real number, we take the positive square root:
x = 3 + ?837
Therefore, the value of x is 3 + ?837 units.
Ajụjụ 59 Ripọtì
In the diagram, \(\overline{AD}\) is a diameter of a circle with Centre O. If ABD is a triangle in a semi-circle ∠OAB=34",
find: (a) ∠OAB (b) ∠OCB
(a) ∠OAB = 34°: Given in the problem.
(b) ∠OCB: We know that in a circle, opposite angles are equal. So, ∠OCB = ∠OAB = 34°.
Explanation: When a straight line cuts a circle at two points, it is called a chord of the circle. And when the chord is a diameter of the circle, it is called the diameter. The angle formed between the chord and the line drawn from the center of the circle to the midpoint of the chord is called the angle subtended by the chord at the center of the circle. And it is equal to half of the angle formed by the chord at the circumference of the circle.
So, in this case, ABD is a triangle inscribed in a semicircle. And ∠OAB is half of the central angle subtended by the chord AB. Hence, ∠OCB, which is opposite to ∠OAB is equal to ∠OAB.
Akọwa Nkọwa
(a) ∠OAB = 34°: Given in the problem.
(b) ∠OCB: We know that in a circle, opposite angles are equal. So, ∠OCB = ∠OAB = 34°.
Explanation: When a straight line cuts a circle at two points, it is called a chord of the circle. And when the chord is a diameter of the circle, it is called the diameter. The angle formed between the chord and the line drawn from the center of the circle to the midpoint of the chord is called the angle subtended by the chord at the center of the circle. And it is equal to half of the angle formed by the chord at the circumference of the circle.
So, in this case, ABD is a triangle inscribed in a semicircle. And ∠OAB is half of the central angle subtended by the chord AB. Hence, ∠OCB, which is opposite to ∠OAB is equal to ∠OAB.
Ajụjụ 60 Ripọtì
(a) A cottage is on a bearing of 200° and 110° from Dogbe's and Manu's farms respectively. If Dogbe walked 5 km and Manu 3 km from the cottage to their farms, find, correct to: (i) two significant figures, the distance between the two farms, (ii) the nearest degree, the bearing of Manu's farm from Dogbe's.
(b) A ladder 10 m long leaned against a vertical wall xm high. The distance between the wall and the foot of the ladder is 2 m longer than the height of the wall.
Calculate the value of x
(a)
-
(i) To find the distance between the two farms, we can use the law of cosines. Let's call the distance between the two farms "d". Using the law of cosines, we can write:
d^2 = (5 km)^2 + (3 km)^2 - 2 * (5 km) * (3 km) * cos(200° - 110°)Solving for d, we get:
d = sqrt(58) km = 7.6 km(rounded to two significant figures) -
(ii) To find the bearing of Manu's farm from Dogbe's, we can use the law of sines. Let's call the angle "theta". Using the law of sines, we can write:
sin(theta) = (3 km) / dSolving for theta, we get:
theta = arcsin(0.39) = 23°(rounded to the nearest degree)So the bearing of Manu's farm from Dogbe's is 23°.
(b)
In this problem, we are trying to find the height of the wall, which we can call "x". We are given that the distance between the wall and the foot of the ladder is 2 m longer than the height of the wall. Let's call this distance "y". So we have:
y = x + 2
We are also given that the ladder is 10 m long. We can use the Pythagorean theorem to find the height of the wall:
x^2 + (10 m)^2 = y^2
Substituting y = x + 2, we get:
x^2 + (10 m)^2 = (x + 2)^2
Expanding and solving for x, we get:
x = 4 m
So the height of the wall is 4 m.
Akọwa Nkọwa
(a)
-
(i) To find the distance between the two farms, we can use the law of cosines. Let's call the distance between the two farms "d". Using the law of cosines, we can write:
d^2 = (5 km)^2 + (3 km)^2 - 2 * (5 km) * (3 km) * cos(200° - 110°)Solving for d, we get:
d = sqrt(58) km = 7.6 km(rounded to two significant figures) -
(ii) To find the bearing of Manu's farm from Dogbe's, we can use the law of sines. Let's call the angle "theta". Using the law of sines, we can write:
sin(theta) = (3 km) / dSolving for theta, we get:
theta = arcsin(0.39) = 23°(rounded to the nearest degree)So the bearing of Manu's farm from Dogbe's is 23°.
(b)
In this problem, we are trying to find the height of the wall, which we can call "x". We are given that the distance between the wall and the foot of the ladder is 2 m longer than the height of the wall. Let's call this distance "y". So we have:
y = x + 2
We are also given that the ladder is 10 m long. We can use the Pythagorean theorem to find the height of the wall:
x^2 + (10 m)^2 = y^2
Substituting y = x + 2, we get:
x^2 + (10 m)^2 = (x + 2)^2
Expanding and solving for x, we get:
x = 4 m
So the height of the wall is 4 m.
Ajụjụ 61 Ripọtì
A man left town am at 10:00 AM and traveled by car to town N at an average speed of 72 km/h.
He spent 2hours for a meeting and returned through town M by bus at an average speed of 40KM/H.
If the distance covered by the bus was 2km longer than that of the car and he arrived at town M at 1 :55PM.
calculate distance from M to N.
Let's start by using the formula:
Distance = Speed x Time
For the first part of the journey, the man traveled from town A to town N by car at an average speed of 72 km/h. Let's assume the distance between town A and town N is "d".
Distance from A to N = Speed x Time
d = 72 x t1 (where t1 is the time taken to travel from A to N)
For the second part of the journey, the man traveled from town N to town M by bus at an average speed of 40 km/h. Let's assume the distance between town N and town M is "x".
Distance from N to M = Speed x Time
x = 40 x t2 (where t2 is the time taken to travel from N to M)
We are also given that the distance covered by the bus was 2 km longer than that of the car:
x = d + 2
We know that the total time taken for the entire journey was 3 hours and 55 minutes, which is equivalent to 235 minutes.
Total time taken = t1 + 2 + t2
235 = t1 + 2 + t2
We can now use the equations we have derived to solve for the distance between town M and town N:
d = 72 x t1
x = 40 x t2
x = d + 2
235 = t1 + 2 + t2
We can substitute the third equation into the second equation:
40 x t2 = 72 x t1 + 2
We can then substitute the first equation into the third equation:
40 x t2 = d + 2
We can then substitute the first equation into the second equation:
40 x t2 = 72 x t1 + 2 = d + 2
We can now substitute these equations into the fourth equation:
235 = t1 + 2 + t2
235 = (d + 2)/40 + 2 + d/72
Simplifying this equation, we get:
235 = (9d + 578)/360
Solving for "d", we get:
d = 510
Therefore, the distance between town M and town N is 510 km.
Akọwa Nkọwa
Let's start by using the formula:
Distance = Speed x Time
For the first part of the journey, the man traveled from town A to town N by car at an average speed of 72 km/h. Let's assume the distance between town A and town N is "d".
Distance from A to N = Speed x Time
d = 72 x t1 (where t1 is the time taken to travel from A to N)
For the second part of the journey, the man traveled from town N to town M by bus at an average speed of 40 km/h. Let's assume the distance between town N and town M is "x".
Distance from N to M = Speed x Time
x = 40 x t2 (where t2 is the time taken to travel from N to M)
We are also given that the distance covered by the bus was 2 km longer than that of the car:
x = d + 2
We know that the total time taken for the entire journey was 3 hours and 55 minutes, which is equivalent to 235 minutes.
Total time taken = t1 + 2 + t2
235 = t1 + 2 + t2
We can now use the equations we have derived to solve for the distance between town M and town N:
d = 72 x t1
x = 40 x t2
x = d + 2
235 = t1 + 2 + t2
We can substitute the third equation into the second equation:
40 x t2 = 72 x t1 + 2
We can then substitute the first equation into the third equation:
40 x t2 = d + 2
We can then substitute the first equation into the second equation:
40 x t2 = 72 x t1 + 2 = d + 2
We can now substitute these equations into the fourth equation:
235 = t1 + 2 + t2
235 = (d + 2)/40 + 2 + d/72
Simplifying this equation, we get:
235 = (9d + 578)/360
Solving for "d", we get:
d = 510
Therefore, the distance between town M and town N is 510 km.
Ajụjụ 62 Ripọtì
(a) Mr Sarfo borrowed $25,000 from Afiak financial services at 21% simple interest per annum for 3 years if he was able to pay back the loan in two years at equal yearly installments how much did he pay each year?
(b) Two consecutive numbers are such that the sum of thrice the smaller and twice the larger is 17.
find correct through three significant figures the smaller number as a percentage of the sum of the two numbers
(a) If Mr. Sarfo borrowed $25,000 at a simple interest rate of 21% per annum for 3 years, the total amount of interest he would have to pay is:
$25,000 x 21% x 3 = $15,750
So, the total amount he would have to pay back is:
$25,000 + $15,750 = $40,750
If he pays back the loan in two years at equal yearly installments, he would have to pay:
$40,750 / 2 = $20,375 per year
Therefore, Mr. Sarfo would have to pay $20,375 each year to pay off the loan in two years.
(b) Let's assume that the two consecutive numbers are x and x+1 (where x is the smaller number).
According to the problem, the sum of thrice the smaller and twice the larger is 17. So, we can write an equation as:
3x + 2(x+1) = 17
Simplifying this equation, we get:
5x + 2 = 17
5x = 15
x = 3
So, the smaller number is 3 and the larger number is 4.
The sum of the two numbers is 7, so the percentage that the smaller number represents of the sum is:
(3/7) x 100% = 42.86% (rounded to three significant figures)
Therefore, the smaller number represents 42.86% of the sum of the two numbers.
Akọwa Nkọwa
(a) If Mr. Sarfo borrowed $25,000 at a simple interest rate of 21% per annum for 3 years, the total amount of interest he would have to pay is:
$25,000 x 21% x 3 = $15,750
So, the total amount he would have to pay back is:
$25,000 + $15,750 = $40,750
If he pays back the loan in two years at equal yearly installments, he would have to pay:
$40,750 / 2 = $20,375 per year
Therefore, Mr. Sarfo would have to pay $20,375 each year to pay off the loan in two years.
(b) Let's assume that the two consecutive numbers are x and x+1 (where x is the smaller number).
According to the problem, the sum of thrice the smaller and twice the larger is 17. So, we can write an equation as:
3x + 2(x+1) = 17
Simplifying this equation, we get:
5x + 2 = 17
5x = 15
x = 3
So, the smaller number is 3 and the larger number is 4.
The sum of the two numbers is 7, so the percentage that the smaller number represents of the sum is:
(3/7) x 100% = 42.86% (rounded to three significant figures)
Therefore, the smaller number represents 42.86% of the sum of the two numbers.
