WAEC SSCE - General Mathematics - 2020 (Essay)
Question 1 Report
(a) Given that P = (\(\frac{rk}{Q} - ms\))\(^{\frac{2}{3}}\)
(i) Make Q the subject of the relation;
(ii) find, correct to two decimal places, the value of Q when P = 3, m = 15, s = 0.2, k = 4 and r = 10.
(b) Given that \(\frac{x + 2y}{5}\) = x - 2y, find x : y
(a)(i) Make Q the subject of \(P = \left(\dfrac{rk}{Q} - ms\right)^{\frac{2}{3}}\).
Raise both sides to the power \(\tfrac{3}{2}\):
\[P^{\frac{3}{2}} = \frac{rk}{Q} - ms.\]
\[\frac{rk}{Q} = P^{\frac{3}{2}} + ms \Rightarrow Q = \frac{rk}{P^{\frac{3}{2}} + ms}.\]
(a)(ii) With \(P = 3, m = 15, s = 0.2, k = 4, r = 10\):
- \(ms = 15 \times 0.2 = 3\)
- \(rk = 10 \times 4 = 40\)
- \(P^{\frac{3}{2}} = 3^{1.5} = \sqrt{27} = 5.196\)
\[Q = \frac{40}{5.196 + 3} = \frac{40}{8.196} = 4.88\ (\text{2 d.p.}).\]
(b) Given \(\dfrac{x + 2y}{5} = x - 2y\). Cross-multiplying:
\[x + 2y = 5(x - 2y) = 5x - 10y.\]
\[12y = 4x \Rightarrow \frac{x}{y} = \frac{12}{4} = 3.\]
Therefore \(x : y = 3 : 1\).
Answer Details
(a)(i) Make Q the subject of \(P = \left(\dfrac{rk}{Q} - ms\right)^{\frac{2}{3}}\).
Raise both sides to the power \(\tfrac{3}{2}\):
\[P^{\frac{3}{2}} = \frac{rk}{Q} - ms.\]
\[\frac{rk}{Q} = P^{\frac{3}{2}} + ms \Rightarrow Q = \frac{rk}{P^{\frac{3}{2}} + ms}.\]
(a)(ii) With \(P = 3, m = 15, s = 0.2, k = 4, r = 10\):
- \(ms = 15 \times 0.2 = 3\)
- \(rk = 10 \times 4 = 40\)
- \(P^{\frac{3}{2}} = 3^{1.5} = \sqrt{27} = 5.196\)
\[Q = \frac{40}{5.196 + 3} = \frac{40}{8.196} = 4.88\ (\text{2 d.p.}).\]
(b) Given \(\dfrac{x + 2y}{5} = x - 2y\). Cross-multiplying:
\[x + 2y = 5(x - 2y) = 5x - 10y.\]
\[12y = 4x \Rightarrow \frac{x}{y} = \frac{12}{4} = 3.\]
Therefore \(x : y = 3 : 1\).
Question 2 Report
(a) If A = {multiples of 2}, B = {multiples of 3} and C = {factors of 6} are subsets of \(\mu\) = {x: \(1 \leq x \leq 10\)} find A′ \(\cap\) B′ \(\cap\) C′
(b) Tickets for a movie premiere cost $18.50 each while the bulk purchase price for 5 tickets is $80.00. If 4 gentlemen decide to get a fifth person to join them so that they can share the bulk purchase price equally, how much would each person save?
(a) The universal set is \(\mu = \{1,2,3,4,5,6,7,8,9,10\}\).
- \(A = \{\text{multiples of }2\} = \{2,4,6,8,10\}\)
- \(B = \{\text{multiples of }3\} = \{3,6,9\}\)
- \(C = \{\text{factors of }6\} = \{1,2,3,6\}\)
By De Morgan's law, \(A' \cap B' \cap C' = (A \cup B \cup C)'\).
\[A \cup B \cup C = \{1,2,3,4,6,8,9,10\}.\]
Therefore the elements of \(\mu\) not in this union are:
\[A' \cap B' \cap C' = \{5, 7\}.\]
(b) Buying singly, one ticket costs \(\$18.50\). The bulk price for 5 tickets is \(\$80.00\), shared equally among the 5 people:
\[\text{Cost per person} = \frac{80.00}{5} = \$16.00.\]
Each person's saving compared with buying a single ticket:
\[18.50 - 16.00 = \$2.50.\]
Each person would save \(\mathbf{\$2.50}\).
Answer Details
(a) The universal set is \(\mu = \{1,2,3,4,5,6,7,8,9,10\}\).
- \(A = \{\text{multiples of }2\} = \{2,4,6,8,10\}\)
- \(B = \{\text{multiples of }3\} = \{3,6,9\}\)
- \(C = \{\text{factors of }6\} = \{1,2,3,6\}\)
By De Morgan's law, \(A' \cap B' \cap C' = (A \cup B \cup C)'\).
\[A \cup B \cup C = \{1,2,3,4,6,8,9,10\}.\]
Therefore the elements of \(\mu\) not in this union are:
\[A' \cap B' \cap C' = \{5, 7\}.\]
(b) Buying singly, one ticket costs \(\$18.50\). The bulk price for 5 tickets is \(\$80.00\), shared equally among the 5 people:
\[\text{Cost per person} = \frac{80.00}{5} = \$16.00.\]
Each person's saving compared with buying a single ticket:
\[18.50 - 16.00 = \$2.50.\]
Each person would save \(\mathbf{\$2.50}\).
Question 3 Report
A die was rolled a number of times. The outcomes are as shown in the table
| Number | 1 | 2 | 3 | 4 | 5 | 6 |
| Outcomes | 32 | m | 25 | 40 | 28 | 45 |
If the probability of obtaining 2 is 0.15, find the:
(a) value of m;
(b) number of times the die was rolled;
(c) probability of obtaining an even number.
| Number | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Outcomes | 32 | m | 25 | 40 | 28 | 45 |
(a) Value of m. The total number of rolls is \(N=32+m+25+40+28+45=170+m\). Since \(P(2)=0.15\):
\[ \frac{m}{170+m}=0.15 \;\Rightarrow\; m=0.15(170+m)=25.5+0.15m \] \[ 0.85m=25.5 \;\Rightarrow\; m=30 \](b) Number of times the die was rolled.
\[ N=170+30=200 \](c) Probability of an even number. Even outcomes are 2, 4 and 6:
\[ 30+40+45=115 \] \[ P(\text{even})=\frac{115}{200}=\frac{23}{40}=0.575 \]Answer Details
| Number | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Outcomes | 32 | m | 25 | 40 | 28 | 45 |
(a) Value of m. The total number of rolls is \(N=32+m+25+40+28+45=170+m\). Since \(P(2)=0.15\):
\[ \frac{m}{170+m}=0.15 \;\Rightarrow\; m=0.15(170+m)=25.5+0.15m \] \[ 0.85m=25.5 \;\Rightarrow\; m=30 \](b) Number of times the die was rolled.
\[ N=170+30=200 \](c) Probability of an even number. Even outcomes are 2, 4 and 6:
\[ 30+40+45=115 \] \[ P(\text{even})=\frac{115}{200}=\frac{23}{40}=0.575 \]Question 4 Report
The total surface area of a cone of slant height 1cm and base radius rcm is 224\(\pi\) cm\(^2\). If r : 1 = 2.5, find:
(a) correct to one decimal place, the value of r
(b) correct to the nearest whole number, the volume of the cone [Take \(\pi\) = \(\frac{22}{7}\)]
Setting up. The total surface area of a cone is \(\pi r(r + l) = 224\pi\), so
\[r(r + l) = 224.\]
The given ratio of slant height to base radius is \(l : r = 2.5\), i.e. \(l = 2.5r\) (the slant height must exceed the radius). Substituting,
\[r(r + 2.5r) = 224 \Rightarrow 3.5r^2 = 224 \Rightarrow r^2 = 64.\]
(a) \(r = 8.0\text{ cm}\), and \(l = 2.5\times8 = 20\text{ cm}\).
(b) Height \(h = \sqrt{l^2 - r^2} = \sqrt{20^2 - 8^2} = \sqrt{336} = 18.33\text{ cm}.\)
\[V = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\times\tfrac{22}{7}\times64\times18.33 = 1229\text{ cm}^3\text{ (to the nearest whole number)}.\]
Answer Details
Setting up. The total surface area of a cone is \(\pi r(r + l) = 224\pi\), so
\[r(r + l) = 224.\]
The given ratio of slant height to base radius is \(l : r = 2.5\), i.e. \(l = 2.5r\) (the slant height must exceed the radius). Substituting,
\[r(r + 2.5r) = 224 \Rightarrow 3.5r^2 = 224 \Rightarrow r^2 = 64.\]
(a) \(r = 8.0\text{ cm}\), and \(l = 2.5\times8 = 20\text{ cm}\).
(b) Height \(h = \sqrt{l^2 - r^2} = \sqrt{20^2 - 8^2} = \sqrt{336} = 18.33\text{ cm}.\)
\[V = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\times\tfrac{22}{7}\times64\times18.33 = 1229\text{ cm}^3\text{ (to the nearest whole number)}.\]
Question 5 Report
(a) In the diagram, AB is a tangent to the circle with centre O, and COB is a straight line. If CD//AB and < ABE = 40°, find: < ODE.
(b) ABCD is a parallelogram in which |\(\overline{CD}\)| = 7 cm, I\(\overline{AD}\)I = 5 cm and < ADC= 125°.
(i) Illustrate the information in a diagram.
(ii) Find, correct to one decimal place, the area of the parallelogram.
(c) If x = \(\frac{1}{2}\)(1 - \(\sqrt{2}\)). Evaluate (2x\(^2\) - 2x).
(a)
<ABE = <OBC (tangent and radius form a right angle)
<ABE = 40° (given)
Therefore, <OBC = 40°.
Since CD//AB, then <CDE = <ABE = 40° (alternate angles)
<ODE = <OBC + <CDE = 40° + 40° = 80°.
(b)
(i)
A ----------- B
\ /
\ /
\ /
\ /
\ /
\ /
D-------C
(125°)
(ii)
The area of the parallelogram is given by A = base x height. Since AD || BC, then the height of the parallelogram is given by the perpendicular distance between AD and BC, which is 7 cm. To find the length of the base, we use the cosine rule:
|\(\overline{AD}\)|² = |\(\overline{AB}\)|² + |\(\overline{BD}\)|² - 2|\(\overline{AB}\)||\(\overline{BD}\)| cos(ADC)
5² = x² + (x+7)² - 2x(x+7)cos(125°)
25 = 2x² + 14x + 49 + 2x(x+7)(-0.5736)
25 = 2x² + 14x + 49 - 1.1472x² - 10.0317x - 20.7032
0 = 0.8528x² + 4.0317x + 4.7032
Using the quadratic formula: x = (-b ± ?(b² - 4ac))/2a, we have:
x = (-4.0317 ± ?(4.0317² - 4(0.8528)(4.7032)))/(2(0.8528))
x = (-4.0317 ± 4.9988)/1.7056
x = 0.6559 or x = -3.6582
Since x represents a length, we discard the negative solution. Therefore, x = 0.6559 cm.
So, the area of the parallelogram is A = base x height = 0.6559 cm x 7 cm = 4.5913 cm² (to one decimal place).
(c)
x = \(\frac{1}{2}\)(1 - \(\sqrt{2}\))
2x² - 2x = 2\(\left(\frac{1}{2}\)(1 - \(\sqrt{2}\))²\) - 2(1 - \(\sqrt{2}\))
= (1 - 2\(\sqrt{2}\) + 2) - 2 + 2\(\sqrt{2}\)
= 1 + 2\(\sqrt{2}\).
Therefore, 2x² - 2x = 1 + 2\(\sqrt{2}\).
Answer Details
(a)
<ABE = <OBC (tangent and radius form a right angle)
<ABE = 40° (given)
Therefore, <OBC = 40°.
Since CD//AB, then <CDE = <ABE = 40° (alternate angles)
<ODE = <OBC + <CDE = 40° + 40° = 80°.
(b)
(i)
A ----------- B
\ /
\ /
\ /
\ /
\ /
\ /
D-------C
(125°)
(ii)
The area of the parallelogram is given by A = base x height. Since AD || BC, then the height of the parallelogram is given by the perpendicular distance between AD and BC, which is 7 cm. To find the length of the base, we use the cosine rule:
|\(\overline{AD}\)|² = |\(\overline{AB}\)|² + |\(\overline{BD}\)|² - 2|\(\overline{AB}\)||\(\overline{BD}\)| cos(ADC)
5² = x² + (x+7)² - 2x(x+7)cos(125°)
25 = 2x² + 14x + 49 + 2x(x+7)(-0.5736)
25 = 2x² + 14x + 49 - 1.1472x² - 10.0317x - 20.7032
0 = 0.8528x² + 4.0317x + 4.7032
Using the quadratic formula: x = (-b ± ?(b² - 4ac))/2a, we have:
x = (-4.0317 ± ?(4.0317² - 4(0.8528)(4.7032)))/(2(0.8528))
x = (-4.0317 ± 4.9988)/1.7056
x = 0.6559 or x = -3.6582
Since x represents a length, we discard the negative solution. Therefore, x = 0.6559 cm.
So, the area of the parallelogram is A = base x height = 0.6559 cm x 7 cm = 4.5913 cm² (to one decimal place).
(c)
x = \(\frac{1}{2}\)(1 - \(\sqrt{2}\))
2x² - 2x = 2\(\left(\frac{1}{2}\)(1 - \(\sqrt{2}\))²\) - 2(1 - \(\sqrt{2}\))
= (1 - 2\(\sqrt{2}\) + 2) - 2 + 2\(\sqrt{2}\)
= 1 + 2\(\sqrt{2}\).
Therefore, 2x² - 2x = 1 + 2\(\sqrt{2}\).
Question 6 Report
The table shows the distribution of marks obtained by students in an examination.
| Marks (%) | 0 - 9 | 10 - 19 | 20 - 29 | 30 - 39 | 40 - 49 | 50 - 59 | 60 - 69 | 70 - 79 | 80 - 89 | 90 - 99 |
| Frequency | 7 | 11 | 17 | 20 | 29 | 34 | 30 | 25 | 21 | 6 |
(a) Construct a cumulative frequency table for the distribution.
(b) Draw the cumulative frequency curve for the distribution.
(c) Using the curve, find correct to one decimal place, the:
(i) median mark;
(ii) lowest mark for the distinction if 5% of the students passed with distinction
(a) Cumulative Frequency Table
| Marks (%) | Upper Class Boundary | Frequency | Cumulative Frequency |
|---|---|---|---|
| 0 – 9 | 9.5 | 7 | 7 |
| 10 – 19 | 19.5 | 11 | 18 |
| 20 – 29 | 29.5 | 17 | 35 |
| 30 – 39 | 39.5 | 20 | 55 |
| 40 – 49 | 49.5 | 29 | 84 |
| 50 – 59 | 59.5 | 34 | 118 |
| 60 – 69 | 69.5 | 30 | 148 |
| 70 – 79 | 79.5 | 25 | 173 |
| 80 – 89 | 89.5 | 21 | 194 |
| 90 – 99 | 99.5 | 6 | 200 |
(b) Cumulative Frequency Curve
Using the upper class boundaries, plot the points:
(9.5, 7), (19.5, 18), (29.5, 35), (39.5, 55), (49.5, 84),
(59.5, 118), (69.5, 148), (79.5, 173), (89.5, 194) and (99.5, 200).
Join the points with a smooth cumulative frequency curve (ogive).
(c)
(i) Median mark
Total frequency, N = 200.
Median position = N⁄2 = 200⁄2 = 100th value.
The 100th value lies in the class 50
Answer Details
(a) Cumulative Frequency Table
| Marks (%) | Upper Class Boundary | Frequency | Cumulative Frequency |
|---|---|---|---|
| 0 – 9 | 9.5 | 7 | 7 |
| 10 – 19 | 19.5 | 11 | 18 |
| 20 – 29 | 29.5 | 17 | 35 |
| 30 – 39 | 39.5 | 20 | 55 |
| 40 – 49 | 49.5 | 29 | 84 |
| 50 – 59 | 59.5 | 34 | 118 |
| 60 – 69 | 69.5 | 30 | 148 |
| 70 – 79 | 79.5 | 25 | 173 |
| 80 – 89 | 89.5 | 21 | 194 |
| 90 – 99 | 99.5 | 6 | 200 |
(b) Cumulative Frequency Curve
Using the upper class boundaries, plot the points:
(9.5, 7), (19.5, 18), (29.5, 35), (39.5, 55), (49.5, 84),
(59.5, 118), (69.5, 148), (79.5, 173), (89.5, 194) and (99.5, 200).
Join the points with a smooth cumulative frequency curve (ogive).
(c)
(i) Median mark
Total frequency, N = 200.
Median position = N⁄2 = 200⁄2 = 100th value.
The 100th value lies in the class 50
Question 7 Report
(a) Copy and complete the table of values for the relation \(y = 3 \sin 2x\).
| x | \(o^o\) | \(15^o\) | \(30^o\) | \(45^o\) | \(60^o\) | \(75^o\) | \(90^o\) | \(105^o\) | \(120^o\) | \(135^o\) | \(^o\) |
| y | 0.0 | 1.5 | -2.6 |
(b) Using a scale of 2 cm to 15° on the x-axis and 2cm to I unit on the y-axis, draw the graph of \(y = 3 \sin 2x\) for \(0° \geq x \geq 150°\).
(c) Use the graph to find the truth set of;
(i) \(3 \sin 2x + 2 = 0\);
(ii ) \(\frac{3}{2} \sin 2x = 0.25\).
(a) For each value of \(x\), calculate \(y=3\sin 2x\), correct to 1 decimal place where necessary.
| \(x\) (degrees) | 0 | 15 | 30 | 45 | 60 | 75 | 90 | 105 | 120 | 135 | 150 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| \(y=3\sin2x\) | 0.0 | 1.5 | 2.6 | 3.0 | 2.6 | 1.5 | 0.0 | -1.5 | -2.6 | -3.0 | -2.6 |
For example, \(3\sin 2(30^\circ)=3\sin60^\circ=2.598\ldots\approx2.6\).
(b) The plotted graph of \(y=3\sin2x\) is shown below. The curve is drawn through the tabulated points, using the stated scales.
(c)(i) \(3\sin2x+2=0\) gives \(y=-2\). Reading the intersection of the curve with \(y=-2\) gives
\[x\approx111^\circ.\]
Hence the truth set is \(\{111^\circ\}\), approximately.
(c)(ii) \(\frac32\sin2x=0.25\). Since \(y=3\sin2x\), this is equivalent to \(y=0.5\). Reading the intersections of the curve with \(y=0.5\) gives
\[x\approx5^\circ\quad\text{or}\quad x\approx85^\circ.\]
Therefore, the truth set is \(\{5^\circ,\ 85^\circ\}\), approximately.
Answer Details
(a) For each value of \(x\), calculate \(y=3\sin 2x\), correct to 1 decimal place where necessary.
| \(x\) (degrees) | 0 | 15 | 30 | 45 | 60 | 75 | 90 | 105 | 120 | 135 | 150 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| \(y=3\sin2x\) | 0.0 | 1.5 | 2.6 | 3.0 | 2.6 | 1.5 | 0.0 | -1.5 | -2.6 | -3.0 | -2.6 |
For example, \(3\sin 2(30^\circ)=3\sin60^\circ=2.598\ldots\approx2.6\).
(b) The plotted graph of \(y=3\sin2x\) is shown below. The curve is drawn through the tabulated points, using the stated scales.
(c)(i) \(3\sin2x+2=0\) gives \(y=-2\). Reading the intersection of the curve with \(y=-2\) gives
\[x\approx111^\circ.\]
Hence the truth set is \(\{111^\circ\}\), approximately.
(c)(ii) \(\frac32\sin2x=0.25\). Since \(y=3\sin2x\), this is equivalent to \(y=0.5\). Reading the intersections of the curve with \(y=0.5\) gives
\[x\approx5^\circ\quad\text{or}\quad x\approx85^\circ.\]
Therefore, the truth set is \(\{5^\circ,\ 85^\circ\}\), approximately.
Question 8 Report
(a) In the diagram, MNPQ is a circle with centre O, |MN| = |NP| and < OMN = 50°. Find:
(I) < MNP
(ii) < POQ
(b) Find the equation of the line which has the same gradient as 8y + 4xy = 24 and passes through the point (-8, 12)
(a) Circle \(MNPQ\) with centre \(O\), \(|MN|=|NP|\) and \(\angle OMN=50^{\circ}\).
From the diagram \(M\) and \(Q\) are the ends of a diameter through \(O\), and the points lie in the order \(M, N, P, Q\) round the circle.
(i) Find \(\angle MNP\).
\(OM\) and \(ON\) are radii, so triangle \(OMN\) is isosceles with \(OM=ON\). Hence the base angles are equal:
\[\angle ONM=\angle OMN=50^{\circ}\]Angle at the centre of triangle \(OMN\):
\[\angle MON=180^{\circ}-50^{\circ}-50^{\circ}=80^{\circ}\]Equal chords subtend equal angles at the centre, and \(|MN|=|NP|\), so:
\[\angle NOP=\angle MON=80^{\circ}\]The central angle standing on the minor arc \(MNP\) is therefore
\[\angle MON+\angle NOP=80^{\circ}+80^{\circ}=160^{\circ}.\]The reflex central angle on the major arc \(MP\) (the arc not containing \(N\)) is
\[360^{\circ}-160^{\circ}=200^{\circ}.\]The inscribed angle \(\angle MNP\) stands on this major arc \(MP\), so it is half of it:
\[\angle MNP=\frac{1}{2}\times 200^{\circ}=100^{\circ}\](ii) Find \(\angle POQ\).
\(M\), \(O\), \(Q\) are collinear (diameter), so the central angles along the arc \(M\to N\to P\to Q\) add up to a straight angle:
\[\angle MON+\angle NOP+\angle POQ=180^{\circ}\]\[80^{\circ}+80^{\circ}+\angle POQ=180^{\circ}\]\[\angle POQ=20^{\circ}\](b) Line with the same gradient as \(8y+4x=24\) through \((-8,\,12)\).
Make \(y\) the subject to read off the gradient:
\[8y=24-4x\]\[y=3-\tfrac{1}{2}x\]The gradient is \(m=-\dfrac{1}{2}\). Using \(y-y_1=m(x-x_1)\) with \((x_1,y_1)=(-8,12)\):
\[y-12=-\tfrac{1}{2}(x+8)\]\[y-12=-\tfrac{1}{2}x-4\]\[y=-\tfrac{1}{2}x+8\]The required line is \(y=-\dfrac{1}{2}x+8\), i.e. \(x+2y=16\).
Answer Details
(a) Circle \(MNPQ\) with centre \(O\), \(|MN|=|NP|\) and \(\angle OMN=50^{\circ}\).
From the diagram \(M\) and \(Q\) are the ends of a diameter through \(O\), and the points lie in the order \(M, N, P, Q\) round the circle.
(i) Find \(\angle MNP\).
\(OM\) and \(ON\) are radii, so triangle \(OMN\) is isosceles with \(OM=ON\). Hence the base angles are equal:
\[\angle ONM=\angle OMN=50^{\circ}\]Angle at the centre of triangle \(OMN\):
\[\angle MON=180^{\circ}-50^{\circ}-50^{\circ}=80^{\circ}\]Equal chords subtend equal angles at the centre, and \(|MN|=|NP|\), so:
\[\angle NOP=\angle MON=80^{\circ}\]The central angle standing on the minor arc \(MNP\) is therefore
\[\angle MON+\angle NOP=80^{\circ}+80^{\circ}=160^{\circ}.\]The reflex central angle on the major arc \(MP\) (the arc not containing \(N\)) is
\[360^{\circ}-160^{\circ}=200^{\circ}.\]The inscribed angle \(\angle MNP\) stands on this major arc \(MP\), so it is half of it:
\[\angle MNP=\frac{1}{2}\times 200^{\circ}=100^{\circ}\](ii) Find \(\angle POQ\).
\(M\), \(O\), \(Q\) are collinear (diameter), so the central angles along the arc \(M\to N\to P\to Q\) add up to a straight angle:
\[\angle MON+\angle NOP+\angle POQ=180^{\circ}\]\[80^{\circ}+80^{\circ}+\angle POQ=180^{\circ}\]\[\angle POQ=20^{\circ}\](b) Line with the same gradient as \(8y+4x=24\) through \((-8,\,12)\).
Make \(y\) the subject to read off the gradient:
\[8y=24-4x\]\[y=3-\tfrac{1}{2}x\]The gradient is \(m=-\dfrac{1}{2}\). Using \(y-y_1=m(x-x_1)\) with \((x_1,y_1)=(-8,12)\):
\[y-12=-\tfrac{1}{2}(x+8)\]\[y-12=-\tfrac{1}{2}x-4\]\[y=-\tfrac{1}{2}x+8\]The required line is \(y=-\dfrac{1}{2}x+8\), i.e. \(x+2y=16\).
Question 9 Report
(a) Two cyclists X and Y leave town Q at the same time. Cyclist X travels at the rate of 5 km/h on a bearing of 049° and cyclist Y travels at the rate of 9 km/h on a bearing of 319°.
(a) Illustrate the information on a diagram.
(b) After travelling for two hours, calculate. correct to the nearest whole number, the:
(i) distance between cyclist X and Y;
(ii) bearing of cyclist X from Y.
(c) Find the average speed at which cyclist X will get to Y in 4 hours.
(a) Diagram. From Q draw QX on bearing 049° and QY on bearing 319°. The angle between the two paths is \(\angle XQY = 360^\circ - (319^\circ - 49^\circ) = 360^\circ - 270^\circ = 90^\circ\), a right angle at Q.
(b) After two hours: \(QX = 5\times2 = 10\) km and \(QY = 9\times2 = 18\) km.
(i) Distance XY (right angle at Q, so use Pythagoras):
\[XY = \sqrt{10^2 + 18^2} = \sqrt{424} = 20.6 \approx 21\text{ km}.\]
(ii) Bearing of X from Y. Taking Q as origin, \(X = (10\sin49^\circ, 10\cos49^\circ) = (7.55, 6.56)\) and \(Y = (18\sin319^\circ, 18\cos319^\circ) = (-11.81, 13.59)\). Then \(\vec{YX} = (19.36, -7.02)\), which points into the south-east region:
\[\text{bearing} = 180^\circ - \tan^{-1}\!\frac{19.36}{7.02} = 180^\circ - 70^\circ = 110^\circ.\]
(c) Average speed to cover XY in 4 hours:
\[\text{speed} = \frac{20.6}{4} = 5.1 \approx 5\text{ km/h}.\]
Answer Details
(a) Diagram. From Q draw QX on bearing 049° and QY on bearing 319°. The angle between the two paths is \(\angle XQY = 360^\circ - (319^\circ - 49^\circ) = 360^\circ - 270^\circ = 90^\circ\), a right angle at Q.
(b) After two hours: \(QX = 5\times2 = 10\) km and \(QY = 9\times2 = 18\) km.
(i) Distance XY (right angle at Q, so use Pythagoras):
\[XY = \sqrt{10^2 + 18^2} = \sqrt{424} = 20.6 \approx 21\text{ km}.\]
(ii) Bearing of X from Y. Taking Q as origin, \(X = (10\sin49^\circ, 10\cos49^\circ) = (7.55, 6.56)\) and \(Y = (18\sin319^\circ, 18\cos319^\circ) = (-11.81, 13.59)\). Then \(\vec{YX} = (19.36, -7.02)\), which points into the south-east region:
\[\text{bearing} = 180^\circ - \tan^{-1}\!\frac{19.36}{7.02} = 180^\circ - 70^\circ = 110^\circ.\]
(c) Average speed to cover XY in 4 hours:
\[\text{speed} = \frac{20.6}{4} = 5.1 \approx 5\text{ km/h}.\]
Question 10 Report
(a) Ms. Maureen spent \(\frac{1}{4}\) of her monthly income at a shopping mall, \(\frac{1}{3}\) at an open market and \(\frac{2}{5}\) of the remaining amount at a Mechanic workshop. If she had N222,000.00 left, find:
(i) her monthly income.
(ii) the amount spent at the open market.
(b) The third term of an Arithmetic Progression (A. P.) is 4m - 2n. If the ninth term of the progression is 2m - 8n. find the common difference in terms of m and n.
a) (i) To find Ms. Maureen's monthly income:
To find Ms. Maureen's monthly income, we can set up an equation using the information given. Let's call her monthly income "M".
Ms. Maureen spent 1/4 of her monthly income at the shopping mall, so: M * 1/4 = M/4
She spent 1/3 of her monthly income at the open market, so: M * 1/3 = M/3
The remaining amount after spending at the shopping mall and open market is M - M/4 - M/3 = M * (3/4) - M/3.
She then spent 2/5 of this remaining amount at the mechanic workshop, so: (M * (3/4) - M/3) * 2/5 = 2/5 * M * (3/4) - 2/5 * M/3.
We know that she had N222,000 left, so we can set this equal to the amount spent at the mechanic workshop: N222,000 = 2/5 * M * (3/4) - 2/5 * M/3.
Solving for M, we can find her monthly income:
M = N222,000 * 5/2 * 4/3 / (3/4 - 2/5) = N222,000 * 5/2 * 4/3 / (5/4 - 6/5) = N222,000 * 20/3 / (25/20 - 24/20) = N222,000 * 20/3 / (5/20) = N222,000 * 20/3 * 20/5 = N222,000 * 4 = N888,000.
So Ms. Maureen's monthly income is N888,000.
(ii) To find the amount spent at the open market:
To find the amount spent at the open market, we use the equation M * 1/3 = M/3 and substitute M = N888,000:
N888,000 * 1/3 = N888,000/3
N888,000/3 = N296,000
So the amount spent at the open market is N296,000.
b) The third term of an arithmetic progression (A.P.) is 4m - 2n, and the ninth term is 2m - 8n. To find the common difference:
To find the common difference, we can subtract the third term from the ninth term and divide by 6, since there are 6 terms between the third and ninth terms:
(2m - 8n) - (4m - 2n) = -2m + 6n
The common difference is -2m + 6n / 6 = -m + n.
So the common difference in terms of m and n is -m + n.
Answer Details
a) (i) To find Ms. Maureen's monthly income:
To find Ms. Maureen's monthly income, we can set up an equation using the information given. Let's call her monthly income "M".
Ms. Maureen spent 1/4 of her monthly income at the shopping mall, so: M * 1/4 = M/4
She spent 1/3 of her monthly income at the open market, so: M * 1/3 = M/3
The remaining amount after spending at the shopping mall and open market is M - M/4 - M/3 = M * (3/4) - M/3.
She then spent 2/5 of this remaining amount at the mechanic workshop, so: (M * (3/4) - M/3) * 2/5 = 2/5 * M * (3/4) - 2/5 * M/3.
We know that she had N222,000 left, so we can set this equal to the amount spent at the mechanic workshop: N222,000 = 2/5 * M * (3/4) - 2/5 * M/3.
Solving for M, we can find her monthly income:
M = N222,000 * 5/2 * 4/3 / (3/4 - 2/5) = N222,000 * 5/2 * 4/3 / (5/4 - 6/5) = N222,000 * 20/3 / (25/20 - 24/20) = N222,000 * 20/3 / (5/20) = N222,000 * 20/3 * 20/5 = N222,000 * 4 = N888,000.
So Ms. Maureen's monthly income is N888,000.
(ii) To find the amount spent at the open market:
To find the amount spent at the open market, we use the equation M * 1/3 = M/3 and substitute M = N888,000:
N888,000 * 1/3 = N888,000/3
N888,000/3 = N296,000
So the amount spent at the open market is N296,000.
b) The third term of an arithmetic progression (A.P.) is 4m - 2n, and the ninth term is 2m - 8n. To find the common difference:
To find the common difference, we can subtract the third term from the ninth term and divide by 6, since there are 6 terms between the third and ninth terms:
(2m - 8n) - (4m - 2n) = -2m + 6n
The common difference is -2m + 6n / 6 = -m + n.
So the common difference in terms of m and n is -m + n.
Question 11 Report
(a} In the diagram, O is the centre of the circle ABCDE, = I\(\overline{BC}\)I = |\(\overline{CD}\)| and < BCD = 108°. Find < CDE.
(b) Given that tan x = \(\sqrt{3}\), 0\(^o\) \(\geq\) x \(\geq\) 90\(^o\), evaluate
\(\frac{(cos x)^2 - sin x}{(sin x)^2 + cos x}\)
(a) To find < CDE, we need to first find < BOC.
Since O is the center of the circle, < BOC is twice the angle < BAC. Therefore,
< BAC = 1/2 * < BOC
= 1/2 * (360° - < BCD - < ABC)
= 1/2 * (360° - 108° - 36°)
= 1/2 * 216°
= 108°
Similarly, we can find that < BDC = 1/2 * (360° - < BCD - < CDE)
= 1/2 * (360° - 108° - < CDE)
= 126° - 1/2 * < CDE
Since < BDC and < BCD are equal, we have:
126° - 1/2 * < CDE = 108°
1/2 * < CDE = 18°
< CDE = 36°
Therefore, < CDE is 36°.
(b) We are given that tan x = √3, and we need to find the value of:
(cos x)^2 - sin x
---------------------
(sin x)^2 + cos x
Using the identity (cos x)^2 + (sin x)^2 = 1, we can write:
(cos x)^2 = 1 - (sin x)^2
Substituting this into the expression, we get:
[(1 - (sin x)^2) - sin x] / [(sin x)^2 + cos x]
Simplifying, we get:
(1 - 2(sin x)^2 - sin x) / [(sin x)^2 + cos x]
Substituting tan x = √3, we have:
sin x / cos x = √3
sin x = √3 cos x
Substituting this into the expression, we get:
(1 - 2(3cos^2 x) - √3 cos x) / (3cos^2 x + cos x)
Multiplying both the numerator and denominator by -1, we get:
(2(3cos^2 x) + √3 cos x - 1) / (3cos^2 x + cos x)
Simplifying, we get:
(6cos^2 x + √3 cos x - 1) / (3cos^2 x + cos x)
Substituting √3 for sin x / cos x, we have:
(6/4 + √3/4 - 1) / (3/4 + 1/4)
= (9/4 + √3/4) / 1
= 9 + √3
Therefore, the value of the expression is
Answer Details
(a) To find < CDE, we need to first find < BOC.
Since O is the center of the circle, < BOC is twice the angle < BAC. Therefore,
< BAC = 1/2 * < BOC
= 1/2 * (360° - < BCD - < ABC)
= 1/2 * (360° - 108° - 36°)
= 1/2 * 216°
= 108°
Similarly, we can find that < BDC = 1/2 * (360° - < BCD - < CDE)
= 1/2 * (360° - 108° - < CDE)
= 126° - 1/2 * < CDE
Since < BDC and < BCD are equal, we have:
126° - 1/2 * < CDE = 108°
1/2 * < CDE = 18°
< CDE = 36°
Therefore, < CDE is 36°.
(b) We are given that tan x = √3, and we need to find the value of:
(cos x)^2 - sin x
---------------------
(sin x)^2 + cos x
Using the identity (cos x)^2 + (sin x)^2 = 1, we can write:
(cos x)^2 = 1 - (sin x)^2
Substituting this into the expression, we get:
[(1 - (sin x)^2) - sin x] / [(sin x)^2 + cos x]
Simplifying, we get:
(1 - 2(sin x)^2 - sin x) / [(sin x)^2 + cos x]
Substituting tan x = √3, we have:
sin x / cos x = √3
sin x = √3 cos x
Substituting this into the expression, we get:
(1 - 2(3cos^2 x) - √3 cos x) / (3cos^2 x + cos x)
Multiplying both the numerator and denominator by -1, we get:
(2(3cos^2 x) + √3 cos x - 1) / (3cos^2 x + cos x)
Simplifying, we get:
(6cos^2 x + √3 cos x - 1) / (3cos^2 x + cos x)
Substituting √3 for sin x / cos x, we have:
(6/4 + √3/4 - 1) / (3/4 + 1/4)
= (9/4 + √3/4) / 1
= 9 + √3
Therefore, the value of the expression is
Question 12 Report
(a) The diagram shows a wooden structure in the form of a cone, mounted on a hemispherical base. The vertical height of the cone is 48 m and the base radius is 14. Calculate, correct to three significant figures, the surface area of the structure, [Take \(\pi = \frac{22}{7}\)]
(b) Five years ago, Musah was twice as old as Sesay. If the sum of their ages is 100, find Sesay's present age.
(a) Surface Area of the Wooden Structure
The structure consists of a cone mounted on a hemispherical base. We need to find the total surface area of this structure.
Given:
- Radius of both the cone and the hemisphere, r=14 m
- Height of the cone, h=48 m
- π=722
Step 1: Calculate the Slant Height of the Cone
The slant height
l of the cone can be found using the Pythagorean theorem:
l=h2+r2=482+142
l=2304+196=2500=50 m
Step 2: Calculate the Surface Area of the Cone
The surface area of the cone (excluding the base) is:
Acone=πrl=722×14×50 Acone=22×50=1100 m2
Step 3: Calculate the Surface Area of the Hemisphere
The surface area of a hemisphere (excluding the base) is:
Ahemisphere=2πr2=2×722×142
Ahemisphere=2×722×196
Ahemisphere=2×22×28=1232 m2
Step 4: Calculate the Total Surface Area
The total surface area of the structure is the sum of the surface area of the cone and the surface area of the hemisphere:
Atotal=Acone+Ahemisphere=1100+1232=2332 m2
Thus, the surface area of the structure, correct to three significant figures, is: 2330 m2
(b) Finding Sesay's Present Age
Given:
- Five years ago, Musah was twice as old as Sesay.
- The sum of their current ages is 100.
Let Musah's present age be M and Sesay's present age be S.
Five years ago:
- Musah's age was M−5
- Sesay's age was S−5
According to the problem, five years ago, Musah was twice as old as Sesay:
M−5=2(S−5)
M−5=2S−10
M=2S−5
We are also given that the sum of their current ages is 100: M+S=100
Substitute:
M=2S−5 into
M+S=100:
(2S−5)+S=100
3S−5=100
3S=105
S=35
So, Sesay's present age is:
S=35
To verify, calculate Musah's present age:
M=2S−5=2(35)−5=70−5=65
Check the sum: M+S=65+35=100
Thus, Sesay's present age is 35 years.
Answer Details
(a) Surface Area of the Wooden Structure
The structure consists of a cone mounted on a hemispherical base. We need to find the total surface area of this structure.
Given:
- Radius of both the cone and the hemisphere, r=14 m
- Height of the cone, h=48 m
- π=722
Step 1: Calculate the Slant Height of the Cone
The slant height
l of the cone can be found using the Pythagorean theorem:
l=h2+r2=482+142
l=2304+196=2500=50 m
Step 2: Calculate the Surface Area of the Cone
The surface area of the cone (excluding the base) is:
Acone=πrl=722×14×50 Acone=22×50=1100 m2
Step 3: Calculate the Surface Area of the Hemisphere
The surface area of a hemisphere (excluding the base) is:
Ahemisphere=2πr2=2×722×142
Ahemisphere=2×722×196
Ahemisphere=2×22×28=1232 m2
Step 4: Calculate the Total Surface Area
The total surface area of the structure is the sum of the surface area of the cone and the surface area of the hemisphere:
Atotal=Acone+Ahemisphere=1100+1232=2332 m2
Thus, the surface area of the structure, correct to three significant figures, is: 2330 m2
(b) Finding Sesay's Present Age
Given:
- Five years ago, Musah was twice as old as Sesay.
- The sum of their current ages is 100.
Let Musah's present age be M and Sesay's present age be S.
Five years ago:
- Musah's age was M−5
- Sesay's age was S−5
According to the problem, five years ago, Musah was twice as old as Sesay:
M−5=2(S−5)
M−5=2S−10
M=2S−5
We are also given that the sum of their current ages is 100: M+S=100
Substitute:
M=2S−5 into
M+S=100:
(2S−5)+S=100
3S−5=100
3S=105
S=35
So, Sesay's present age is:
S=35
To verify, calculate Musah's present age:
M=2S−5=2(35)−5=70−5=65
Check the sum: M+S=65+35=100
Thus, Sesay's present age is 35 years.
