WAEC SSCE - General Mathematics - 2021
Ajụjụ 2 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 5 Ripọtì
In the diagram, line \(\overline{EC}\) is a diameter of the circle ABCDE.
If angle ABC equals 158°, find ?ADE
Akọwa Nkọwa
Ajụjụ 7 Ripọtì
A fair die is tossed twice what is the probability of get a sum of at least 10.
Akọwa Nkọwa
Ajụjụ 9 Ripọtì
A cone has a base radius of 8cm and height 11cm. calculate , correct to 2d.p, the curved surface area
Akọwa Nkọwa
Ajụjụ 11 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 12 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 14 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 15 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 17 Ripọtì
find the value of (x+y)
Ajụjụ 18 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 22 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 23 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 24 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 25 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 26 Ripọtì
If \(\frac{2}{x-3}\) - \(\frac{3}{x-2}\) = \(\frac{p}{(x-3)(x -2)}\), find p.
Akọwa Nkọwa
Ajụjụ 28 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 29 Ripọtì
In the diagram O is the centre of the circle PQRS, ?PQR = 72° and OR is parallel to PS. Find ∠ OPS
Ajụjụ 30 Ripọtì
If log\(_{10}\) 2 = m and log\(_{10}\) 3 = n, find log\(_{10}\) 24 in terms of m and n.
Akọwa Nkọwa
Ajụjụ 32 Ripọtì
Which of the following is not an exterior angle of a regular polygon?
Ajụjụ 33 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 34 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 36 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 37 Ripọtì
if p = {-3<x<1} and Q = {-1<x<3}, where x is a real number, find P n Q.
Ajụjụ 38 Ripọtì
In the diagram, ∠ABC and ∠BCD are right angles, ∠BAD = t and ∠EDF = 70°. Find the value of t.
Ajụjụ 39 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 40 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
If 16 * 2\(^{(x + 1)}\) = 4\(^x\) * 8\(^{(1 - x)}\), find the value of x.
Akọwa Nkọwa
Ajụjụ 44 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 45 Ripọtì
In △LMN, |LM| = 6cm, ∠LNM = x and sin x = sin x = \(\frac{3}{5}\).
Find the area of △LMN
Akọwa Nkọwa
Ajụjụ 46 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 47 Ripọtì
In the diagram, ?XYZ is produced to T. if |XY| = |ZY| and ?XYT = 40°, find ?XZT
Ajụjụ 50 Ripọtì
(a) Copy and complete the table of values for the relation \(y=2x^2-x-2\) for \(4 \le x \le 4\).
| 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 \(4 \le x \le 4\).
(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) Completed table of values for \( y = 2x^2 - x - 2 \)
Substituting each value of \(x\) into \( y = 2x^2 - x - 2 \):
At \(x=-4:\; y = 2(-4)^2-(-4)-2 = 32+4-2 = 34\)
At \(x=-2:\; y = 2(4)-(-2)-2 = 8+2-2 = 8\)
At \(x=-1:\; y = 2(1)-(-1)-2 = 2+1-2 = 1\)
At \(x=1:\; y = 2(1)-1-2 = 2-1-2 = -1\)
At \(x=2:\; y = 2(4)-2-2 = 8-2-2 = 4\)
At \(x=3:\; y = 2(9)-3-2 = 18-3-2 = 13\)
| x | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
| y | 34 | 19 | 8 | 1 | -2 | -1 | 4 | 13 | 26 |
(b) and (c) The graphs of \( y = 2x^2 - x - 2 \) and \( y = 2x + 3 \)
Using a scale of 2 cm to 1 unit on the \(x\)-axis and 2 cm to 5 units on the \(y\)-axis, the points from the table are plotted and joined with a smooth curve. The straight line \( y = 2x + 3 \) is drawn on the same axes (it passes through \((-4,-5)\) and \((4,11)\), with \(y\)-intercept \(3\)).
(d) Using the graph
(i) Roots of \( 2x^2 - 3x - 5 = 0 \).
Subtracting this equation from the plotted curve:
\[ (2x^2 - x - 2) - (2x^2 - 3x - 5) = 2x + 3 \]So the solutions of \( 2x^2 - 3x - 5 = 0 \) are given by the values of \(x\) where the curve \( y = 2x^2 - x - 2 \) meets the line \( y = 2x + 3 \). Reading off the two points of intersection on the graph:
\[ x = -1 \quad \text{and} \quad x = 2.5 \](each to within \(\pm 0.1\)).
(ii) Range of values of \(x\) for which \( 2x^2 - x - 2 < 0 \).
This is the range where the curve lies below the \(x\)-axis, i.e. between the two points where the curve cuts the \(x\)-axis. Reading these cutting points from the graph:
\[ -0.8 < x < 1.3 \](each end to within \(\pm 0.1\)).
Akọwa Nkọwa
(a) Completed table of values for \( y = 2x^2 - x - 2 \)
Substituting each value of \(x\) into \( y = 2x^2 - x - 2 \):
At \(x=-4:\; y = 2(-4)^2-(-4)-2 = 32+4-2 = 34\)
At \(x=-2:\; y = 2(4)-(-2)-2 = 8+2-2 = 8\)
At \(x=-1:\; y = 2(1)-(-1)-2 = 2+1-2 = 1\)
At \(x=1:\; y = 2(1)-1-2 = 2-1-2 = -1\)
At \(x=2:\; y = 2(4)-2-2 = 8-2-2 = 4\)
At \(x=3:\; y = 2(9)-3-2 = 18-3-2 = 13\)
| x | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
| y | 34 | 19 | 8 | 1 | -2 | -1 | 4 | 13 | 26 |
(b) and (c) The graphs of \( y = 2x^2 - x - 2 \) and \( y = 2x + 3 \)
Using a scale of 2 cm to 1 unit on the \(x\)-axis and 2 cm to 5 units on the \(y\)-axis, the points from the table are plotted and joined with a smooth curve. The straight line \( y = 2x + 3 \) is drawn on the same axes (it passes through \((-4,-5)\) and \((4,11)\), with \(y\)-intercept \(3\)).
(d) Using the graph
(i) Roots of \( 2x^2 - 3x - 5 = 0 \).
Subtracting this equation from the plotted curve:
\[ (2x^2 - x - 2) - (2x^2 - 3x - 5) = 2x + 3 \]So the solutions of \( 2x^2 - 3x - 5 = 0 \) are given by the values of \(x\) where the curve \( y = 2x^2 - x - 2 \) meets the line \( y = 2x + 3 \). Reading off the two points of intersection on the graph:
\[ x = -1 \quad \text{and} \quad x = 2.5 \](each to within \(\pm 0.1\)).
(ii) Range of values of \(x\) for which \( 2x^2 - x - 2 < 0 \).
This is the range where the curve lies below the \(x\)-axis, i.e. between the two points where the curve cuts the \(x\)-axis. Reading these cutting points from the graph:
\[ -0.8 < x < 1.3 \](each end to within \(\pm 0.1\)).
Ajụjụ 51 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) The cottage \(C\) is at bearing \(200^{\circ}\) from Dogbe's farm \(D\) and \(110^{\circ}\) from Manu's farm \(M\). Reversing bearings, from the cottage:
- Bearing of \(D\) from \(C=200^{\circ}-180^{\circ}=020^{\circ}\), with \(|CD|=5\) km.
- Bearing of \(M\) from \(C=110^{\circ}+180^{\circ}=290^{\circ}\), with \(|CM|=3\) km.
Angle at the cottage \(=290^{\circ}-020^{\circ}=270^{\circ}\Rightarrow\) reflex; the angle between the two directions is \(360^{\circ}-270^{\circ}=90^{\circ}\).
(i) With \(\angle DCM=90^{\circ}\):
\[|DM|=\sqrt{5^{2}+3^{2}}=\sqrt{34}=5.83\approx\mathbf{5.8\text{ km}}\ (2\text{ s.f.}).\]
(ii) Using coordinates \(D=(1.71,4.70),\ M=(-2.82,1.03)\): \(D\!\to\!M=(-4.53,-3.67)\) points south-west.
\[\tan\alpha=\frac{4.53}{3.67}=1.234\Rightarrow\alpha=51^{\circ};\quad\text{bearing}=180^{\circ}+51^{\circ}=\mathbf{231^{\circ}}.\]
(b) Ladder 10 m, wall \(x\) m, foot distance \((x+2)\) m:
\[x^{2}+(x+2)^{2}=10^{2}\Rightarrow 2x^{2}+4x+4=100\Rightarrow x^{2}+2x-48=0\Rightarrow (x+8)(x-6)=0.\]
Since \(x>0,\ \mathbf{x=6\text{ m}}.\)
Akọwa Nkọwa
(a) The cottage \(C\) is at bearing \(200^{\circ}\) from Dogbe's farm \(D\) and \(110^{\circ}\) from Manu's farm \(M\). Reversing bearings, from the cottage:
- Bearing of \(D\) from \(C=200^{\circ}-180^{\circ}=020^{\circ}\), with \(|CD|=5\) km.
- Bearing of \(M\) from \(C=110^{\circ}+180^{\circ}=290^{\circ}\), with \(|CM|=3\) km.
Angle at the cottage \(=290^{\circ}-020^{\circ}=270^{\circ}\Rightarrow\) reflex; the angle between the two directions is \(360^{\circ}-270^{\circ}=90^{\circ}\).
(i) With \(\angle DCM=90^{\circ}\):
\[|DM|=\sqrt{5^{2}+3^{2}}=\sqrt{34}=5.83\approx\mathbf{5.8\text{ km}}\ (2\text{ s.f.}).\]
(ii) Using coordinates \(D=(1.71,4.70),\ M=(-2.82,1.03)\): \(D\!\to\!M=(-4.53,-3.67)\) points south-west.
\[\tan\alpha=\frac{4.53}{3.67}=1.234\Rightarrow\alpha=51^{\circ};\quad\text{bearing}=180^{\circ}+51^{\circ}=\mathbf{231^{\circ}}.\]
(b) Ladder 10 m, wall \(x\) m, foot distance \((x+2)\) m:
\[x^{2}+(x+2)^{2}=10^{2}\Rightarrow 2x^{2}+4x+4=100\Rightarrow x^{2}+2x-48=0\Rightarrow (x+8)(x-6)=0.\]
Since \(x>0,\ \mathbf{x=6\text{ m}}.\)
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.
Given distribution.
| Hours per day (x) | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
|---|---|---|---|---|---|---|---|---|
| Students (f) | 5 | 7 | 5 | 9 | 12 | 4 | 3 | 5 |
Total number of students: \(\sum f = 50\).
(a) Mean. Compute \(\sum fx\):
| x | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
|---|---|---|---|---|---|---|---|---|
| f | 5 | 7 | 5 | 9 | 12 | 4 | 3 | 5 |
| fx | 20 | 35 | 30 | 63 | 96 | 36 | 30 | 55 |
\[\sum fx = 20+35+30+63+96+36+30+55 = 365.\]
\[\bar{x} = \frac{\sum fx}{\sum f} = \frac{365}{50} = 7.30.\]
(b) Standard deviation. Compute \(\sum fx^{2}\):
| x | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
|---|---|---|---|---|---|---|---|---|
| fx\(^2\) | 80 | 175 | 180 | 441 | 768 | 324 | 300 | 605 |
\[\sum fx^{2} = 80+175+180+441+768+324+300+605 = 2873.\]
\[\text{Variance} = \frac{\sum fx^{2}}{\sum f} - \bar{x}^{2} = \frac{2873}{50} - (7.3)^{2} = 57.46 - 53.29 = 4.17.\]
\[\text{S.D.} = \sqrt{4.17} = 2.04 \text{ (2 d.p.).}\]
The mean study time is 7.30 hours and the standard deviation is 2.04 hours.
Akọwa Nkọwa
Given distribution.
| Hours per day (x) | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
|---|---|---|---|---|---|---|---|---|
| Students (f) | 5 | 7 | 5 | 9 | 12 | 4 | 3 | 5 |
Total number of students: \(\sum f = 50\).
(a) Mean. Compute \(\sum fx\):
| x | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
|---|---|---|---|---|---|---|---|---|
| f | 5 | 7 | 5 | 9 | 12 | 4 | 3 | 5 |
| fx | 20 | 35 | 30 | 63 | 96 | 36 | 30 | 55 |
\[\sum fx = 20+35+30+63+96+36+30+55 = 365.\]
\[\bar{x} = \frac{\sum fx}{\sum f} = \frac{365}{50} = 7.30.\]
(b) Standard deviation. Compute \(\sum fx^{2}\):
| x | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
|---|---|---|---|---|---|---|---|---|
| fx\(^2\) | 80 | 175 | 180 | 441 | 768 | 324 | 300 | 605 |
\[\sum fx^{2} = 80+175+180+441+768+324+300+605 = 2873.\]
\[\text{Variance} = \frac{\sum fx^{2}}{\sum f} - \bar{x}^{2} = \frac{2873}{50} - (7.3)^{2} = 57.46 - 53.29 = 4.17.\]
\[\text{S.D.} = \sqrt{4.17} = 2.04 \text{ (2 d.p.).}\]
The mean study time is 7.30 hours and the standard deviation is 2.04 hours.
Ajụjụ 53 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
Reading the diagram. The trapezium has \(QR \parallel PS\). \(QU\) and \(RT\) are drawn perpendicular to \(PS\), so \(QU = RT = 12\text{ cm}\) is the vertical height of the trapezium. From the figure: \(|PU| = 5\text{ cm}\), \(|QU| = 12\text{ cm}\), \(\angle PUQ = \angle STR = 90^\circ\), the angle at \(S\) is \(50^\circ\), and the area of \(\triangle PQR = 20\text{ cm}^2\).
Step 1: The slant side \(PQ\). Triangle \(PUQ\) is right-angled at \(U\):
\[|PQ| = \sqrt{5^2 + 12^2} = \sqrt{25+144} = \sqrt{169} = 13\text{ cm}.\]Step 2: The top side \(QR\). Since \(QU \perp PS\), \(RT \perp PS\) and \(QR \parallel PS\), the figure \(QURT\) is a rectangle, so \(QR = UT\). Vertices \(Q\) and \(R\) both lie \(12\text{ cm}\) above line \(PS\), while \(P\) lies on \(PS\); hence the height of \(\triangle PQR\) on base \(QR\) is \(12\text{ cm}\):
\[\text{Area }\triangle PQR = \tfrac{1}{2}\times QR \times 12 = 6\,QR.\]\[6\,QR = 20 \;\Rightarrow\; QR = \tfrac{20}{6} = 3.33\text{ cm}.\]Step 3: The right-hand side using \(\angle S = 50^\circ\). Triangle \(STR\) is right-angled at \(T\) with \(RT = 12\):
\[|TS| = \frac{RT}{\tan 50^\circ} = \frac{12}{1.1918} = 10.07\text{ cm},\qquad |RS| = \frac{RT}{\sin 50^\circ} = \frac{12}{0.7660} = 15.67\text{ cm}.\]Step 4: The base \(PS\).
\[|PS| = |PU| + |UT| + |TS| = 5 + 3.33 + 10.07 = 18.40\text{ cm}.\](a) Perimeter of the trapezium.
\[P = |PQ| + |QR| + |RS| + |SP| = 13 + 3.33 + 15.67 + 18.40 = 50.40\text{ cm}.\]Perimeter \(\approx \mathbf{50\text{ cm}}\) (nearest whole number).
(b) Area of the trapezium. Parallel sides \(QR = 3.33\) and \(PS = 18.40\), height \(12\):
\[A = \tfrac{1}{2}(QR + PS)\times h = \tfrac{1}{2}(3.33 + 18.40)\times 12 = \tfrac{1}{2}(21.73)(12) = 130.4\text{ cm}^2.\]Area \(\approx \mathbf{130\text{ cm}^2}\) (nearest whole number).
Akọwa Nkọwa
Reading the diagram. The trapezium has \(QR \parallel PS\). \(QU\) and \(RT\) are drawn perpendicular to \(PS\), so \(QU = RT = 12\text{ cm}\) is the vertical height of the trapezium. From the figure: \(|PU| = 5\text{ cm}\), \(|QU| = 12\text{ cm}\), \(\angle PUQ = \angle STR = 90^\circ\), the angle at \(S\) is \(50^\circ\), and the area of \(\triangle PQR = 20\text{ cm}^2\).
Step 1: The slant side \(PQ\). Triangle \(PUQ\) is right-angled at \(U\):
\[|PQ| = \sqrt{5^2 + 12^2} = \sqrt{25+144} = \sqrt{169} = 13\text{ cm}.\]Step 2: The top side \(QR\). Since \(QU \perp PS\), \(RT \perp PS\) and \(QR \parallel PS\), the figure \(QURT\) is a rectangle, so \(QR = UT\). Vertices \(Q\) and \(R\) both lie \(12\text{ cm}\) above line \(PS\), while \(P\) lies on \(PS\); hence the height of \(\triangle PQR\) on base \(QR\) is \(12\text{ cm}\):
\[\text{Area }\triangle PQR = \tfrac{1}{2}\times QR \times 12 = 6\,QR.\]\[6\,QR = 20 \;\Rightarrow\; QR = \tfrac{20}{6} = 3.33\text{ cm}.\]Step 3: The right-hand side using \(\angle S = 50^\circ\). Triangle \(STR\) is right-angled at \(T\) with \(RT = 12\):
\[|TS| = \frac{RT}{\tan 50^\circ} = \frac{12}{1.1918} = 10.07\text{ cm},\qquad |RS| = \frac{RT}{\sin 50^\circ} = \frac{12}{0.7660} = 15.67\text{ cm}.\]Step 4: The base \(PS\).
\[|PS| = |PU| + |UT| + |TS| = 5 + 3.33 + 10.07 = 18.40\text{ cm}.\](a) Perimeter of the trapezium.
\[P = |PQ| + |QR| + |RS| + |SP| = 13 + 3.33 + 15.67 + 18.40 = 50.40\text{ cm}.\]Perimeter \(\approx \mathbf{50\text{ cm}}\) (nearest whole number).
(b) Area of the trapezium. Parallel sides \(QR = 3.33\) and \(PS = 18.40\), height \(12\):
\[A = \tfrac{1}{2}(QR + PS)\times h = \tfrac{1}{2}(3.33 + 18.40)\times 12 = \tfrac{1}{2}(21.73)(12) = 130.4\text{ cm}^2.\]Area \(\approx \mathbf{130\text{ cm}^2}\) (nearest whole number).
Ajụjụ 54 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
Take \(X\) as origin. \(Y\) is 15 km due south of \(X\); \(Z\) is 20 km on bearing \(270^{\circ}\), i.e. due west of \(X\). So \(\angle YXZ=90^{\circ}\) (south and west are perpendicular).
(a) \(|YZ|\) by Pythagoras:
\[|YZ|=\sqrt{15^{2}+20^{2}}=\sqrt{225+400}=\sqrt{625}=25\text{ km (2 s.f.)}.\]
(b) Bearing of \(Y\) from \(Z\)
Place \(Z=(-20,0)\), \(Y=(0,-15)\). The vector \(Z\!\to\!Y=(20,-15)\): from \(Z\), \(Y\) lies east and south. The angle east of south:
\[\tan\theta=\frac{20}{15}=1.333\Rightarrow\theta=53.13^{\circ}\ \text{(east of due south)}.\]
Bearing \(=180^{\circ}-53.13^{\circ}=126.87^{\circ}\approx \mathbf{127^{\circ}}\) (nearest degree).
Akọwa Nkọwa
Take \(X\) as origin. \(Y\) is 15 km due south of \(X\); \(Z\) is 20 km on bearing \(270^{\circ}\), i.e. due west of \(X\). So \(\angle YXZ=90^{\circ}\) (south and west are perpendicular).
(a) \(|YZ|\) by Pythagoras:
\[|YZ|=\sqrt{15^{2}+20^{2}}=\sqrt{225+400}=\sqrt{625}=25\text{ km (2 s.f.)}.\]
(b) Bearing of \(Y\) from \(Z\)
Place \(Z=(-20,0)\), \(Y=(0,-15)\). The vector \(Z\!\to\!Y=(20,-15)\): from \(Z\), \(Y\) lies east and south. The angle east of south:
\[\tan\theta=\frac{20}{15}=1.333\Rightarrow\theta=53.13^{\circ}\ \text{(east of due south)}.\]
Bearing \(=180^{\circ}-53.13^{\circ}=126.87^{\circ}\approx \mathbf{127^{\circ}}\) (nearest degree).
Ajụjụ 55 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) In a class of \(n=80\) students, let \(B\) be the set who study Biology and \(P\) the set who study Physics.
Number studying Biology \(=\tfrac{3}{4}\times80=60\).
Number studying Physics \(=\tfrac{3}{5}\times80=48\).
Since every student studies at least one subject, no one lies outside the two sets, i.e. \(n(B\cup P)=80\).
(i) Venn diagram. Let \(x\) be the number who study both subjects. Then Biology-only \(=60-x\) and Physics-only \(=48-x\):
(ii) How many study both subjects. Using the total,
\[ (60-x)+x+(48-x)=80 \] \[ 108-x=80\;\Rightarrow\;x=108-80=28. \]So 28 students study both Biology and Physics. Hence Biology-only \(=60-28=32\) and Physics-only \(=48-28=20\). Check: \(32+28+20=80\). \(\checkmark\)
(iii) Fraction that study Biology but not Physics. These are the Biology-only students:
\[ \frac{60-28}{80}=\frac{32}{80}=\frac{2}{5}. \]The fraction of the class that study Biology but not Physics is \(\dfrac{2}{5}\).
(b) Area of the office floor \(=15\text{ m}\times8\text{ m}=120\text{ m}^{2}\).
Cost of carpeting \(=120\text{ m}^{2}\times\text{GH}\cent890.00=\text{GH}\cent106{,}800.00\).
Total spent on painting and carpeting \(=\text{GH}\cent216{,}120.00\). Therefore
\[ \text{Cost of painting}=216{,}120.00-106{,}800.00=\text{GH}\cent109{,}320.00. \]The cost of painting was GH\(\cent\)109,320.00.
Akọwa Nkọwa
(a) In a class of \(n=80\) students, let \(B\) be the set who study Biology and \(P\) the set who study Physics.
Number studying Biology \(=\tfrac{3}{4}\times80=60\).
Number studying Physics \(=\tfrac{3}{5}\times80=48\).
Since every student studies at least one subject, no one lies outside the two sets, i.e. \(n(B\cup P)=80\).
(i) Venn diagram. Let \(x\) be the number who study both subjects. Then Biology-only \(=60-x\) and Physics-only \(=48-x\):
(ii) How many study both subjects. Using the total,
\[ (60-x)+x+(48-x)=80 \] \[ 108-x=80\;\Rightarrow\;x=108-80=28. \]So 28 students study both Biology and Physics. Hence Biology-only \(=60-28=32\) and Physics-only \(=48-28=20\). Check: \(32+28+20=80\). \(\checkmark\)
(iii) Fraction that study Biology but not Physics. These are the Biology-only students:
\[ \frac{60-28}{80}=\frac{32}{80}=\frac{2}{5}. \]The fraction of the class that study Biology but not Physics is \(\dfrac{2}{5}\).
(b) Area of the office floor \(=15\text{ m}\times8\text{ m}=120\text{ m}^{2}\).
Cost of carpeting \(=120\text{ m}^{2}\times\text{GH}\cent890.00=\text{GH}\cent106{,}800.00\).
Total spent on painting and carpeting \(=\text{GH}\cent216{,}120.00\). Therefore
\[ \text{Cost of painting}=216{,}120.00-106{,}800.00=\text{GH}\cent109{,}320.00. \]The cost of painting was GH\(\cent\)109,320.00.
Ajụjụ 56 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) In \(\triangle PQR\), \(\angle PQR=90^{\circ}\), so \(PQ\) and \(QR\) are the perpendicular sides. Let \(PQ=3k,\ QR=4k\).
\[\text{Area}=\tfrac12\,PQ\cdot QR=216\Rightarrow\tfrac12(3k)(4k)=216\Rightarrow 6k^{2}=216\Rightarrow k^{2}=36\Rightarrow k=6.\]
So \(PQ=18\text{ cm},\ QR=24\text{ cm}\). By Pythagoras:
\[|PR|=\sqrt{18^{2}+24^{2}}=\sqrt{324+576}=\sqrt{900}=30\text{ cm}.\]
(b) Let the man's age be twice his son's after \(x\) years:
\[47+x=2(17+x)\Rightarrow 47+x=34+2x\Rightarrow x=13.\]
In 13 years' time the man will be twice as old as his son (60 and 30).
Akọwa Nkọwa
(a) In \(\triangle PQR\), \(\angle PQR=90^{\circ}\), so \(PQ\) and \(QR\) are the perpendicular sides. Let \(PQ=3k,\ QR=4k\).
\[\text{Area}=\tfrac12\,PQ\cdot QR=216\Rightarrow\tfrac12(3k)(4k)=216\Rightarrow 6k^{2}=216\Rightarrow k^{2}=36\Rightarrow k=6.\]
So \(PQ=18\text{ cm},\ QR=24\text{ cm}\). By Pythagoras:
\[|PR|=\sqrt{18^{2}+24^{2}}=\sqrt{324+576}=\sqrt{900}=30\text{ cm}.\]
(b) Let the man's age be twice his son's after \(x\) years:
\[47+x=2(17+x)\Rightarrow 47+x=34+2x\Rightarrow x=13.\]
In 13 years' time the man will be twice as old as his son (60 and 30).
Ajụjụ 57 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
In the diagram, \(\overline{AD}\) is a diameter of the circle with centre \(O\), \(B\) lies on the circle, and \(BC\) is a tangent to the circle at \(B\). It is given that \(\angle OAB = 34^\circ\).
(a) \(\angle OAB\)
\(OA\) and \(OB\) are radii, so triangle \(OAB\) is isosceles and \(\angle OBA = \angle OAB\):
\[ \angle OAB = 34^\circ \](b) \(\angle OCB\)
Since \(AD\) is a diameter, the angle in the semicircle is a right angle:
\[ \angle ABD = 90^\circ \]The angle subtended at the centre is twice the angle at the circumference on the same arc \(BD\), so
\[ \angle BOC = \angle BOD = 2 \times \angle OAB = 2 \times 34^\circ = 68^\circ \]A tangent is perpendicular to the radius at the point of contact, so at \(B\):
\[ \angle OBC = 90^\circ \]In triangle \(OBC\), the angles sum to \(180^\circ\):
\[ \angle BOC + \angle OBC + \angle OCB = 180^\circ \] \[ 68^\circ + 90^\circ + \angle OCB = 180^\circ \] \[ \angle OCB = 180^\circ - 158^\circ = 22^\circ \]Akọwa Nkọwa
In the diagram, \(\overline{AD}\) is a diameter of the circle with centre \(O\), \(B\) lies on the circle, and \(BC\) is a tangent to the circle at \(B\). It is given that \(\angle OAB = 34^\circ\).
(a) \(\angle OAB\)
\(OA\) and \(OB\) are radii, so triangle \(OAB\) is isosceles and \(\angle OBA = \angle OAB\):
\[ \angle OAB = 34^\circ \](b) \(\angle OCB\)
Since \(AD\) is a diameter, the angle in the semicircle is a right angle:
\[ \angle ABD = 90^\circ \]The angle subtended at the centre is twice the angle at the circumference on the same arc \(BD\), so
\[ \angle BOC = \angle BOD = 2 \times \angle OAB = 2 \times 34^\circ = 68^\circ \]A tangent is perpendicular to the radius at the point of contact, so at \(B\):
\[ \angle OBC = 90^\circ \]In triangle \(OBC\), the angles sum to \(180^\circ\):
\[ \angle BOC + \angle OBC + \angle OCB = 180^\circ \] \[ 68^\circ + 90^\circ + \angle OCB = 180^\circ \] \[ \angle OCB = 180^\circ - 158^\circ = 22^\circ \]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
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.
Setting up the time equation
Total time from 10:00 a.m. to 1:55 p.m. \(=3\text{ h }55\text{ min}=3\tfrac{55}{60}\text{ h}=3\tfrac{11}{12}\text{ h}\).
Removing the 2-hour meeting, the driving+bus time is
\[3\tfrac{11}{12}-2=1\tfrac{11}{12}\text{ h}=\tfrac{23}{12}\text{ h}.\]
Let the car distance \(M\!\to\!N\) be \(d\) km (at 72 km/h). The bus distance is \((d+2)\) km (at 40 km/h). Then
\[\frac{d}{72}+\frac{d+2}{40}=\frac{23}{12}.\]
Multiply through by 360:
\[5d+9(d+2)=690\Rightarrow 14d+18=690\Rightarrow 14d=672\Rightarrow d=48.\]
Check: car \(=48/72=40\) min, bus \(=50/40=1\text{ h }15\text{ min}\); driving \(=1\text{ h }55\text{ min}\); with the 2 h meeting the total is 3 h 55 min, arriving 1:55 p.m. \(\checkmark\)
Distance from \(M\) to \(N\) \(= 48\) km.
Akọwa Nkọwa
Setting up the time equation
Total time from 10:00 a.m. to 1:55 p.m. \(=3\text{ h }55\text{ min}=3\tfrac{55}{60}\text{ h}=3\tfrac{11}{12}\text{ h}\).
Removing the 2-hour meeting, the driving+bus time is
\[3\tfrac{11}{12}-2=1\tfrac{11}{12}\text{ h}=\tfrac{23}{12}\text{ h}.\]
Let the car distance \(M\!\to\!N\) be \(d\) km (at 72 km/h). The bus distance is \((d+2)\) km (at 40 km/h). Then
\[\frac{d}{72}+\frac{d+2}{40}=\frac{23}{12}.\]
Multiply through by 360:
\[5d+9(d+2)=690\Rightarrow 14d+18=690\Rightarrow 14d=672\Rightarrow d=48.\]
Check: car \(=48/72=40\) min, bus \(=50/40=1\text{ h }15\text{ min}\); driving \(=1\text{ h }55\text{ min}\); with the 2 h meeting the total is 3 h 55 min, arriving 1:55 p.m. \(\checkmark\)
Distance from \(M\) to \(N\) \(= 48\) km.
Ajụjụ 60 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) Pie chart of the property share
The shares sum to \(5+15+10+45+25 = 100\). A full circle is \(360^\circ\), so each child's sector angle is their percentage of \(360^\circ\), i.e. percentage \(\times \frac{360}{100} = \text{percentage} \times 3.6^\circ\):
| Child | Percentage share (%) | Sector angle |
|---|---|---|
| Ann | 5 | \(\frac{5}{100}\times 360^\circ = 18^\circ\) |
| Afia | 15 | \(\frac{15}{100}\times 360^\circ = 54^\circ\) |
| Kojo | 10 | \(\frac{10}{100}\times 360^\circ = 36^\circ\) |
| Nuno | 45 | \(\frac{45}{100}\times 360^\circ = 162^\circ\) |
| Akom | 25 | \(\frac{25}{100}\times 360^\circ = 90^\circ\) |
| Total | 100 | \(360^\circ\) |
Check: \(18^\circ+54^\circ+36^\circ+162^\circ+90^\circ = 360^\circ\). Drawing a circle and marking each sector with a protractor gives:
(b) Probability of taking two red beads, one after the other
The box holds \(5 + 3 + 4 = 12\) identical beads, of which 5 are red. The two beads are taken one after the other without replacement.
For the first draw, 5 of the 12 beads are red:
\[P(\text{1st red}) = \frac{5}{12}.\]After one red bead is removed, 4 red beads remain out of 11 beads in total:
\[P(\text{2nd red}\mid\text{1st red}) = \frac{4}{11}.\]Since both events must happen, multiply the probabilities:
\[P(\text{two red}) = \frac{5}{12}\times\frac{4}{11} = \frac{20}{132} = \frac{5}{33}\approx 0.1515.\]Therefore the probability of taking two red beads one after the other is \(\dfrac{5}{33}\) (about \(0.152\)).
Akọwa Nkọwa
(a) Pie chart of the property share
The shares sum to \(5+15+10+45+25 = 100\). A full circle is \(360^\circ\), so each child's sector angle is their percentage of \(360^\circ\), i.e. percentage \(\times \frac{360}{100} = \text{percentage} \times 3.6^\circ\):
| Child | Percentage share (%) | Sector angle |
|---|---|---|
| Ann | 5 | \(\frac{5}{100}\times 360^\circ = 18^\circ\) |
| Afia | 15 | \(\frac{15}{100}\times 360^\circ = 54^\circ\) |
| Kojo | 10 | \(\frac{10}{100}\times 360^\circ = 36^\circ\) |
| Nuno | 45 | \(\frac{45}{100}\times 360^\circ = 162^\circ\) |
| Akom | 25 | \(\frac{25}{100}\times 360^\circ = 90^\circ\) |
| Total | 100 | \(360^\circ\) |
Check: \(18^\circ+54^\circ+36^\circ+162^\circ+90^\circ = 360^\circ\). Drawing a circle and marking each sector with a protractor gives:
(b) Probability of taking two red beads, one after the other
The box holds \(5 + 3 + 4 = 12\) identical beads, of which 5 are red. The two beads are taken one after the other without replacement.
For the first draw, 5 of the 12 beads are red:
\[P(\text{1st red}) = \frac{5}{12}.\]After one red bead is removed, 4 red beads remain out of 11 beads in total:
\[P(\text{2nd red}\mid\text{1st red}) = \frac{4}{11}.\]Since both events must happen, multiply the probabilities:
\[P(\text{two red}) = \frac{5}{12}\times\frac{4}{11} = \frac{20}{132} = \frac{5}{33}\approx 0.1515.\]Therefore the probability of taking two red beads one after the other is \(\dfrac{5}{33}\) (about \(0.152\)).
Ajụjụ 61 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.
Ajụjụ 62 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.
%IMG%
In the diagram, ABCD are points on the circle centre O. If \(|AB| = |BC|\) and \(\angle ADC = 50^\circ\), find \(\angle BAD\).
(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) \(\angle ADC + \angle DCA + \angle CAD = 180^\circ\)
\(50^\circ + 90^\circ + \angle CAD = 180^\circ\)
\(\angle CAD = (180 - 140)^\circ\)
= 40
\(\angle ADC + \angle ABC = 180^\circ\)
\(50^\circ + \angle ABC = 180^\circ\)
\(\angle ABC = 130^\circ\)
\(\angle BAC + \angle BCA + \angle ABC = 180^\circ\)
But \(\angle BAC = \angle BCA\)
\(2\angle BAC + \angle 130^\circ = 180^\circ\)
\(2\angle BAC = 50^\circ\)
\(\angle BAC = 25^\circ\)
\(\angle BAD = \angle CAD + \angle BAC\)
= \(40^\circ + 25^\circ = 65^\circ\)
(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°.
