WAEC SSCE - General Mathematics - 1994 (Essay)
Question 1 Report
(a) Evaluate, without using mathematical tables, \(17.57^{2} - 12.43^{2}\).
(b) Prove that angles in the same segment of a circle are equal.
(a) Use the difference of two squares, \(a^2 - b^2 = (a-b)(a+b)\):
\(17.57^2 - 12.43^2 = (17.57 - 12.43)(17.57 + 12.43)\)
\(= (5.14)(30.00) = \mathbf{154.2}\)
(b) Theorem: Angles in the same segment of a circle are equal.
Given: A circle with centre \(O\); a chord \(AB\); points \(P\) and \(Q\) on the major arc (the same segment), so \(\angle APB\) and \(\angle AQB\) both stand on chord \(AB\).
To prove: \(\angle APB = \angle AQB\).
Construction: Join \(OA\) and \(OB\).
Proof: By the theorem that the angle subtended by an arc at the centre is twice the angle it subtends at the circumference (on the same arc):
\(\angle AOB = 2\,\angle APB \quad\text{and}\quad \angle AOB = 2\,\angle AQB\)
Therefore \(2\,\angle APB = 2\,\angle AQB\), giving \(\angle APB = \angle AQB\). Hence angles in the same segment are equal. \(\blacksquare\)
Answer Details
(a) Use the difference of two squares, \(a^2 - b^2 = (a-b)(a+b)\):
\(17.57^2 - 12.43^2 = (17.57 - 12.43)(17.57 + 12.43)\)
\(= (5.14)(30.00) = \mathbf{154.2}\)
(b) Theorem: Angles in the same segment of a circle are equal.
Given: A circle with centre \(O\); a chord \(AB\); points \(P\) and \(Q\) on the major arc (the same segment), so \(\angle APB\) and \(\angle AQB\) both stand on chord \(AB\).
To prove: \(\angle APB = \angle AQB\).
Construction: Join \(OA\) and \(OB\).
Proof: By the theorem that the angle subtended by an arc at the centre is twice the angle it subtends at the circumference (on the same arc):
\(\angle AOB = 2\,\angle APB \quad\text{and}\quad \angle AOB = 2\,\angle AQB\)
Therefore \(2\,\angle APB = 2\,\angle AQB\), giving \(\angle APB = \angle AQB\). Hence angles in the same segment are equal. \(\blacksquare\)
Question 2 Report
A bag contains 12 white balls and 8 black balls, another contains 10 white balls and 15 black balls. If two balls are drawn, without replacement from each bag, find the probability that :
(a) all four balls are black ;
(b) exactly one of the four balls is white.
Setup. Bag A: 12 white, 8 black (20 balls). Bag B: 10 white, 15 black (25 balls). Two balls are drawn from each bag without replacement.
Useful probabilities per bag
Bag A, two black: \(\dfrac{8}{20}\times\dfrac{7}{19} = \dfrac{14}{95}\)
Bag A, exactly one white: \(2\times\dfrac{12}{20}\times\dfrac{8}{19} = \dfrac{48}{95}\)
Bag B, two black: \(\dfrac{15}{25}\times\dfrac{14}{24} = \dfrac{7}{20}\)
Bag B, exactly one white: \(2\times\dfrac{10}{25}\times\dfrac{15}{24} = \dfrac{1}{2}\)
(a) All four balls black
Both bags must give two blacks:
\[P = \frac{14}{95}\times\frac{7}{20} = \frac{98}{1900} = \frac{49}{950} \approx 0.052\]
(b) Exactly one of the four balls is white
The single white must come from one bag while the other bag yields two blacks.
White from A, two black from B: \(\dfrac{48}{95}\times\dfrac{7}{20} = \dfrac{84}{475}\)
Two black from A, white from B: \(\dfrac{14}{95}\times\dfrac{1}{2} = \dfrac{7}{95} = \dfrac{35}{475}\)
Add the mutually exclusive cases:
\[P = \frac{84}{475} + \frac{35}{475} = \frac{119}{475} \approx 0.251\]
Answer Details
Setup. Bag A: 12 white, 8 black (20 balls). Bag B: 10 white, 15 black (25 balls). Two balls are drawn from each bag without replacement.
Useful probabilities per bag
Bag A, two black: \(\dfrac{8}{20}\times\dfrac{7}{19} = \dfrac{14}{95}\)
Bag A, exactly one white: \(2\times\dfrac{12}{20}\times\dfrac{8}{19} = \dfrac{48}{95}\)
Bag B, two black: \(\dfrac{15}{25}\times\dfrac{14}{24} = \dfrac{7}{20}\)
Bag B, exactly one white: \(2\times\dfrac{10}{25}\times\dfrac{15}{24} = \dfrac{1}{2}\)
(a) All four balls black
Both bags must give two blacks:
\[P = \frac{14}{95}\times\frac{7}{20} = \frac{98}{1900} = \frac{49}{950} \approx 0.052\]
(b) Exactly one of the four balls is white
The single white must come from one bag while the other bag yields two blacks.
White from A, two black from B: \(\dfrac{48}{95}\times\dfrac{7}{20} = \dfrac{84}{475}\)
Two black from A, white from B: \(\dfrac{14}{95}\times\dfrac{1}{2} = \dfrac{7}{95} = \dfrac{35}{475}\)
Add the mutually exclusive cases:
\[P = \frac{84}{475} + \frac{35}{475} = \frac{119}{475} \approx 0.251\]
Question 3 Report
(a) Using a ruler and a pair of compasses only, construct (i) a triangle XYZ in which /YZ/ = 8cm, < XYZ = 60° and < XZY = 75°. Measure /XY/; (ii) the locus \(l_{1}\) of points equidistant from Y and Z ; (iii) the locus \(l_{2}\) of points equidistant from XY and YZ.
(b) Measure QY where Q is the point of intersection of \(l_{1}\) and \(l_{2}\).
(a)(i) Construction of triangle \(XYZ\)
- Draw \(YZ=8\text{ cm}\).
- At \(Y\), construct an angle of \(60^\circ\). At \(Z\), construct an angle of \(75^\circ\), using ruler and compasses only.
- Let the two constructed rays meet at \(X\). This gives triangle \(XYZ\).
From the completed construction,
\[|XY|\approx 10.9\text{ cm}.\]
(a)(ii) The locus \(l_1\) of points equidistant from \(Y\) and \(Z\) is the perpendicular bisector of \(YZ\).
(a)(iii) The locus \(l_2\) of points equidistant from \(XY\) and \(YZ\) is the internal bisector of \(\angle XYZ\).
(b) Let \(Q\) be the intersection of \(l_1\) and \(l_2\). Measuring from the construction gives
\[\boxed{|QY|\approx 4.8\text{ cm}}\]
Answer Details
(a)(i) Construction of triangle \(XYZ\)
- Draw \(YZ=8\text{ cm}\).
- At \(Y\), construct an angle of \(60^\circ\). At \(Z\), construct an angle of \(75^\circ\), using ruler and compasses only.
- Let the two constructed rays meet at \(X\). This gives triangle \(XYZ\).
From the completed construction,
\[|XY|\approx 10.9\text{ cm}.\]
(a)(ii) The locus \(l_1\) of points equidistant from \(Y\) and \(Z\) is the perpendicular bisector of \(YZ\).
(a)(iii) The locus \(l_2\) of points equidistant from \(XY\) and \(YZ\) is the internal bisector of \(\angle XYZ\).
(b) Let \(Q\) be the intersection of \(l_1\) and \(l_2\). Measuring from the construction gives
\[\boxed{|QY|\approx 4.8\text{ cm}}\]
Question 4 Report
The table below gives the ages, to the nearest 5 years of 50 people.
| Age in years | 10 | 15 | 20 | 25 | 30 |
| No of people | 8 | 19 | 10 | 7 | 6 |
(a) Construct a cumulative frequency table for the distribution.
(b) Draw a cumulative frequency curve (Ogive)
(c) From your Ogive, find the : (i) median age ; (ii) number of people who are at most 15 years of age ; (iii) number of people who are between 20 and 25 years of age.
(a) Cumulative frequency table
| Age (years) | Number of people | Cumulative frequency |
|---|---|---|
| 10 | 8 | 8 |
| 15 | 19 | 27 |
| 20 | 10 | 37 |
| 25 | 7 | 44 |
| 30 | 6 | 50 |
(b) Cumulative frequency curve (ogive)
Plot the points \((10,8), (15,27), (20,37), (25,44)\) and \((30,50)\), then join them with a smooth increasing curve.
(c)
(i) Median age
\(N=50\), so the median corresponds to cumulative frequency \(\frac{50}{2}=25\). Reading across from 25 to the ogive and down to the age axis gives:
\[\boxed{\text{Median age} \approx 14.5\text{ years}}\]
(ii) Number of people at most 15 years old
At age 15 years, the cumulative frequency is \(27\).
\[\boxed{27\text{ people}}\]
(iii) Number of people between 20 and 25 years old
\[44-37=7\]
\[\boxed{7\text{ people}}\]
Answer Details
(a) Cumulative frequency table
| Age (years) | Number of people | Cumulative frequency |
|---|---|---|
| 10 | 8 | 8 |
| 15 | 19 | 27 |
| 20 | 10 | 37 |
| 25 | 7 | 44 |
| 30 | 6 | 50 |
(b) Cumulative frequency curve (ogive)
Plot the points \((10,8), (15,27), (20,37), (25,44)\) and \((30,50)\), then join them with a smooth increasing curve.
(c)
(i) Median age
\(N=50\), so the median corresponds to cumulative frequency \(\frac{50}{2}=25\). Reading across from 25 to the ogive and down to the age axis gives:
\[\boxed{\text{Median age} \approx 14.5\text{ years}}\]
(ii) Number of people at most 15 years old
At age 15 years, the cumulative frequency is \(27\).
\[\boxed{27\text{ people}}\]
(iii) Number of people between 20 and 25 years old
\[44-37=7\]
\[\boxed{7\text{ people}}\]
Question 5 Report
(a) Divide \(11111111_{two}\) by \(101_{two}\)
(b) A sector of radius 6 cm has an angle of 105° at the centre. Calculate its:
(i) perimeter ; (ii) area . [Take \(\pi = \frac{22}{7}\)]
(a) Divide \(11111111_{two}\) by \(101_{two}\)
We use binary long division. Divisor is \(101_{two}\).
1 1 0 0 1 1
___________________
1 0 1 ) 1 1 1 1 1 1 1 1
1 0 1
-----
1 0 1
1 0 1
-----
0 0 1 1
1 1 1
1 0 1
-----
1 0 1
1 0 1
-----
0 0The quotient is \(11111111_{two} \div 101_{two} = 110011_{two}\), remainder \(0\).
Check in base ten: \(11111111_{two}=255\), \(101_{two}=5\), and \(255 \div 5 = 51 = 110011_{two}\). Correct.
(b) Sector: radius 6 cm, angle 105°, \(\pi=\tfrac{22}{7}\)
(i) Perimeter. First find the arc length.
\[\text{Arc} = \frac{\theta}{360}\times 2\pi r = \frac{105}{360}\times 2\times\frac{22}{7}\times 6 = \frac{7}{24}\times\frac{264}{7} = 11\text{ cm}\]
Perimeter = arc + the two straight radii:
\[P = 11 + 2(6) = 23\text{ cm}\]
(ii) Area.
\[A = \frac{\theta}{360}\times \pi r^{2} = \frac{105}{360}\times\frac{22}{7}\times 6^{2} = \frac{7}{24}\times\frac{792}{7} = 33\text{ cm}^{2}\]
Answer Details
(a) Divide \(11111111_{two}\) by \(101_{two}\)
We use binary long division. Divisor is \(101_{two}\).
1 1 0 0 1 1
___________________
1 0 1 ) 1 1 1 1 1 1 1 1
1 0 1
-----
1 0 1
1 0 1
-----
0 0 1 1
1 1 1
1 0 1
-----
1 0 1
1 0 1
-----
0 0The quotient is \(11111111_{two} \div 101_{two} = 110011_{two}\), remainder \(0\).
Check in base ten: \(11111111_{two}=255\), \(101_{two}=5\), and \(255 \div 5 = 51 = 110011_{two}\). Correct.
(b) Sector: radius 6 cm, angle 105°, \(\pi=\tfrac{22}{7}\)
(i) Perimeter. First find the arc length.
\[\text{Arc} = \frac{\theta}{360}\times 2\pi r = \frac{105}{360}\times 2\times\frac{22}{7}\times 6 = \frac{7}{24}\times\frac{264}{7} = 11\text{ cm}\]
Perimeter = arc + the two straight radii:
\[P = 11 + 2(6) = 23\text{ cm}\]
(ii) Area.
\[A = \frac{\theta}{360}\times \pi r^{2} = \frac{105}{360}\times\frac{22}{7}\times 6^{2} = \frac{7}{24}\times\frac{792}{7} = 33\text{ cm}^{2}\]
Question 6 Report
(a) Copy and complete the following table of values for \(y = 2x^{2} - 9x - 1\).
| x | -1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| y | -1 | -8 | -11 | 17 |
(b) Using a scale of 2cm to represent 1 unit on the x- axis and 2cm to represent 5 units on the y- axis, draw the graph of \(y = 2x^{2} - 9x - 1\).
(c) Use your graph to find the : (i) roots of the equation \(2x^{2} - 9x = 4\), correct to one decimal place ; (ii) gradient of the curve \(y = 2x^{2} - 9x - 1\) at x = 3.
(a) For each value of \(x\), calculate \(y=2x^2-9x-1\).
| \(x\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) |
|---|---|---|---|---|---|---|---|---|
| \(y\) | \(10\) | \(-1\) | \(-8\) | \(-11\) | \(-10\) | \(-5\) | \(4\) | \(17\) |
(b) The plotted curve, together with the construction line \(y=3\) and the tangent at \(x=3\), is shown below.
(c)(i) Since \(2x^2-9x=4\), then
\[2x^2-9x-1=3.\]
Thus, draw or consider the horizontal line \(y=3\). From its intersections with the curve,
\[\boxed{x=-0.4\text{ or }x=4.9}\]
correct to one decimal place.
(ii) At \(x=3\), the point on the curve is \((3,-10)\). The tangent shown passes through \((3,-10)\) and \((5,-4)\). Hence,
\[\text{Gradient}=\frac{-4-(-10)}{5-3}=\frac{6}{2}=\boxed{3}.\]
Answer Details
(a) For each value of \(x\), calculate \(y=2x^2-9x-1\).
| \(x\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) |
|---|---|---|---|---|---|---|---|---|
| \(y\) | \(10\) | \(-1\) | \(-8\) | \(-11\) | \(-10\) | \(-5\) | \(4\) | \(17\) |
(b) The plotted curve, together with the construction line \(y=3\) and the tangent at \(x=3\), is shown below.
(c)(i) Since \(2x^2-9x=4\), then
\[2x^2-9x-1=3.\]
Thus, draw or consider the horizontal line \(y=3\). From its intersections with the curve,
\[\boxed{x=-0.4\text{ or }x=4.9}\]
correct to one decimal place.
(ii) At \(x=3\), the point on the curve is \((3,-10)\). The tangent shown passes through \((3,-10)\) and \((5,-4)\). Hence,
\[\text{Gradient}=\frac{-4-(-10)}{5-3}=\frac{6}{2}=\boxed{3}.\]
Question 7 Report
(a) Given that \(3 \times 9^{1 + x} = 27^{-x}\), find x.
(b) Evaluate \(\log_{10} \sqrt{35} + \log_{10} \sqrt{2} - \log_{10} \sqrt{7}\)
(a) Express everything as powers of 3: \(9 = 3^2\), \(27 = 3^3\).
\(3\times 9^{1+x} = 3^1 \times 3^{2(1+x)} = 3^{1 + 2 + 2x} = 3^{3 + 2x}\)
\(27^{-x} = 3^{-3x}\)
Equating indices: \(3 + 2x = -3x \ \Rightarrow\ 5x = -3 \ \Rightarrow\ \mathbf{x = -\dfrac{3}{5}}\).
(b) Use \(\log\sqrt{a} = \tfrac{1}{2}\log a\) and combine:
\(\log_{10}\sqrt{35} + \log_{10}\sqrt{2} - \log_{10}\sqrt{7} = \log_{10}\!\left(\dfrac{\sqrt{35}\times\sqrt{2}}{\sqrt{7}}\right)\)
\(= \log_{10}\sqrt{\dfrac{35\times 2}{7}} = \log_{10}\sqrt{10} = \log_{10} 10^{1/2} = \dfrac{1}{2}\)
Value \(= \mathbf{\dfrac{1}{2}}\).
Answer Details
(a) Express everything as powers of 3: \(9 = 3^2\), \(27 = 3^3\).
\(3\times 9^{1+x} = 3^1 \times 3^{2(1+x)} = 3^{1 + 2 + 2x} = 3^{3 + 2x}\)
\(27^{-x} = 3^{-3x}\)
Equating indices: \(3 + 2x = -3x \ \Rightarrow\ 5x = -3 \ \Rightarrow\ \mathbf{x = -\dfrac{3}{5}}\).
(b) Use \(\log\sqrt{a} = \tfrac{1}{2}\log a\) and combine:
\(\log_{10}\sqrt{35} + \log_{10}\sqrt{2} - \log_{10}\sqrt{7} = \log_{10}\!\left(\dfrac{\sqrt{35}\times\sqrt{2}}{\sqrt{7}}\right)\)
\(= \log_{10}\sqrt{\dfrac{35\times 2}{7}} = \log_{10}\sqrt{10} = \log_{10} 10^{1/2} = \dfrac{1}{2}\)
Value \(= \mathbf{\dfrac{1}{2}}\).
Question 8 Report
(a) If \(\log_{10} (3x - 1) - \log_{10} 2 = 3\), find the value of x.
(b) Use logarithm tables to evaluate \(\sqrt{\frac{0.897 \times 3.536}{0.00249}}\), correct to 3 significant figures.
(a) Solve \(\log_{10}(3x-1) - \log_{10}2 = 3\)
Combine the logs using \(\log a - \log b = \log\frac{a}{b}\):
\[\log_{10}\left(\frac{3x-1}{2}\right) = 3\]
Rewrite in index form (\(\log_{10}N = 3 \Rightarrow N = 10^{3}\)):
\[\frac{3x-1}{2} = 1000\]
\[3x-1 = 2000 \;\Rightarrow\; 3x = 2001 \;\Rightarrow\; x = 667\]
(b) Evaluate \(\sqrt{\dfrac{0.897\times 3.536}{0.00249}}\) to 3 s.f. using logarithm tables
| Number | Logarithm |
|---|---|
| 0.897 | \(\bar{1}.9528\) |
| 3.536 | \(0.5485\) |
| Numerator sum | \(0.5013\) |
| 0.00249 | \(\bar{3}.3963\) |
Divide (subtract the log of the denominator):
\[0.5013 - \bar{3}.3963 = 0.5013 - (-2.6037) = 3.1050\]
Take the square root (divide the log by 2):
\[\tfrac{1}{2}\times 3.1050 = 1.5525\]
Antilog of \(1.5525\) gives \(3.570\times 10^{1}\).
\[\sqrt{\frac{0.897\times 3.536}{0.00249}} \approx 35.7\]
Answer Details
(a) Solve \(\log_{10}(3x-1) - \log_{10}2 = 3\)
Combine the logs using \(\log a - \log b = \log\frac{a}{b}\):
\[\log_{10}\left(\frac{3x-1}{2}\right) = 3\]
Rewrite in index form (\(\log_{10}N = 3 \Rightarrow N = 10^{3}\)):
\[\frac{3x-1}{2} = 1000\]
\[3x-1 = 2000 \;\Rightarrow\; 3x = 2001 \;\Rightarrow\; x = 667\]
(b) Evaluate \(\sqrt{\dfrac{0.897\times 3.536}{0.00249}}\) to 3 s.f. using logarithm tables
| Number | Logarithm |
|---|---|
| 0.897 | \(\bar{1}.9528\) |
| 3.536 | \(0.5485\) |
| Numerator sum | \(0.5013\) |
| 0.00249 | \(\bar{3}.3963\) |
Divide (subtract the log of the denominator):
\[0.5013 - \bar{3}.3963 = 0.5013 - (-2.6037) = 3.1050\]
Take the square root (divide the log by 2):
\[\tfrac{1}{2}\times 3.1050 = 1.5525\]
Antilog of \(1.5525\) gives \(3.570\times 10^{1}\).
\[\sqrt{\frac{0.897\times 3.536}{0.00249}} \approx 35.7\]
Question 9 Report
(a) Two points X(32°N, 47°W) and Y(32°N, 25°E) are on the earth's surface. If it takes an aeroplane 11 hours to fly from X to Y along the parallel of latitude, calculate its speed, correct to the nearest kilometre per hour. [Radius of the earth = 6400km; \(\pi = \frac{22}{7}\)]
(b) Two observers P and Q, 15metres apart observe a kite (K) in the same vertical plane and from the same side of the kite. The angles of elevation of the kite from P and Q are 35° and 45° respectively. Find the height of the kite to the nearest metre.
(a) Speed along the parallel of latitude
Both points are on latitude \(32^\circ\)N. The difference in longitude is
\[47^\circ + 25^\circ = 72^\circ\]
Distance along a parallel of latitude \(\phi\):
\[d = \frac{\theta}{360}\times 2\pi R\cos\phi\]
\[d = \frac{72}{360}\times 2\times\frac{22}{7}\times 6400\times\cos32^\circ\]
Using \(\cos32^\circ = 0.8480\):
\[d = 0.2\times \frac{44}{7}\times 6400\times 0.8480 \approx 6823\text{ km}\]
Speed \(= \dfrac{\text{distance}}{\text{time}} = \dfrac{6823}{11} \approx 620\) km/h.
Speed \(\approx 620\) km/h.
(b) Height of the kite
Let the foot of the vertical from the kite \(K\) meet the ground at \(O\). \(Q\) is nearer (elevation \(45^\circ\)) and \(P\) is \(15\) m further back (elevation \(35^\circ\)). Let \(OQ = d\) and height \(= h\).
From \(Q\): \(\tan45^\circ = \dfrac{h}{d} = 1 \Rightarrow d = h\).
From \(P\): \(\tan35^\circ = \dfrac{h}{d+15}\).
Substitute \(d = h\):
\[h = (h+15)\tan35^\circ = 0.7002(h+15)\]
\[h - 0.7002h = 10.503 \;\Rightarrow\; 0.2998h = 10.503 \;\Rightarrow\; h \approx 35\text{ m}\]
Height \(\approx 35\) m.
Answer Details
(a) Speed along the parallel of latitude
Both points are on latitude \(32^\circ\)N. The difference in longitude is
\[47^\circ + 25^\circ = 72^\circ\]
Distance along a parallel of latitude \(\phi\):
\[d = \frac{\theta}{360}\times 2\pi R\cos\phi\]
\[d = \frac{72}{360}\times 2\times\frac{22}{7}\times 6400\times\cos32^\circ\]
Using \(\cos32^\circ = 0.8480\):
\[d = 0.2\times \frac{44}{7}\times 6400\times 0.8480 \approx 6823\text{ km}\]
Speed \(= \dfrac{\text{distance}}{\text{time}} = \dfrac{6823}{11} \approx 620\) km/h.
Speed \(\approx 620\) km/h.
(b) Height of the kite
Let the foot of the vertical from the kite \(K\) meet the ground at \(O\). \(Q\) is nearer (elevation \(45^\circ\)) and \(P\) is \(15\) m further back (elevation \(35^\circ\)). Let \(OQ = d\) and height \(= h\).
From \(Q\): \(\tan45^\circ = \dfrac{h}{d} = 1 \Rightarrow d = h\).
From \(P\): \(\tan35^\circ = \dfrac{h}{d+15}\).
Substitute \(d = h\):
\[h = (h+15)\tan35^\circ = 0.7002(h+15)\]
\[h - 0.7002h = 10.503 \;\Rightarrow\; 0.2998h = 10.503 \;\Rightarrow\; h \approx 35\text{ m}\]
Height \(\approx 35\) m.
Question 10 Report
(a) The fourth term of an A.P is 37 and 6th term is 12 more than the fourth term . Find the first and seventh terms.
(b) If \(P = {1, 2, 3, 4}\) and \(Q = {3, 5, 6}\), find (i) \(P \cap Q\) ; (ii) \(P \cup Q\) ; (iii) \((P \cap Q) \cup Q\) ; (iv) \((P \cap Q) \cup P\).
(a) Arithmetic Progression
Let the first term be \(a\) and common difference \(d\). The \(n\)th term is \(a+(n-1)d\).
Fourth term: \(a + 3d = 37\).
The 6th term is 12 more than the 4th term: \((a+5d) = 37 + 12 = 49\), so \(a + 5d = 49\).
Subtract the equations:
\[(a+5d)-(a+3d) = 49-37 \;\Rightarrow\; 2d = 12 \;\Rightarrow\; d = 6\]
Then \(a = 37 - 3(6) = 19\).
First term \(= 19\). Seventh term:
\[a + 6d = 19 + 6(6) = 55\]
Seventh term \(= 55\).
(b) Sets with \(P=\{1,2,3,4\}\) and \(Q=\{3,5,6\}\).
(i) \(P\cap Q = \{3\}\)
(ii) \(P\cup Q = \{1,2,3,4,5,6\}\)
(iii) \((P\cap Q)\cup Q = \{3\}\cup\{3,5,6\} = \{3,5,6\}\)
(iv) \((P\cap Q)\cup P = \{3\}\cup\{1,2,3,4\} = \{1,2,3,4\}\)
Answer Details
(a) Arithmetic Progression
Let the first term be \(a\) and common difference \(d\). The \(n\)th term is \(a+(n-1)d\).
Fourth term: \(a + 3d = 37\).
The 6th term is 12 more than the 4th term: \((a+5d) = 37 + 12 = 49\), so \(a + 5d = 49\).
Subtract the equations:
\[(a+5d)-(a+3d) = 49-37 \;\Rightarrow\; 2d = 12 \;\Rightarrow\; d = 6\]
Then \(a = 37 - 3(6) = 19\).
First term \(= 19\). Seventh term:
\[a + 6d = 19 + 6(6) = 55\]
Seventh term \(= 55\).
(b) Sets with \(P=\{1,2,3,4\}\) and \(Q=\{3,5,6\}\).
(i) \(P\cap Q = \{3\}\)
(ii) \(P\cup Q = \{1,2,3,4,5,6\}\)
(iii) \((P\cap Q)\cup Q = \{3\}\cup\{3,5,6\} = \{3,5,6\}\)
(iv) \((P\cap Q)\cup P = \{3\}\cup\{1,2,3,4\} = \{1,2,3,4\}\)
Question 11 Report
(a) A tower and a building stand on the same horizontal level. From the point P at the bottom of the building, the angle of elevation of the top, T of the tower is 65°. From the top Q of the building, the angle of elevation of the point T is 25°. If the building is 20m high, calculate the distance PT.
(b) Hence or otherwise, calculate the height of the tower. [Give your answers correct to 3 significant figures].
Let the horizontal distance between the building and the foot \(F\) of the tower be \(x\), and the tower height be \(H\). \(P\) is at the foot of the building, \(Q\) is its top, \(20\text{ m}\) directly above \(P\).
From \(P\): \(\tan 65° = \dfrac{H}{x}\), so \(H = x\tan 65°\).
From \(Q\) (height 20 m, same horizontal distance \(x\)): \(\tan 25° = \dfrac{H - 20}{x}\), so \(H - 20 = x\tan 25°\).
Subtracting: \(20 = x(\tan 65° - \tan 25°) = x(2.1445 - 0.4663) = 1.6782x\).
\(x = \dfrac{20}{1.6782} = 11.92\text{ m}\).
(a) Distance PT: \(PT\) is the line of sight from \(P\) to \(T\), at elevation \(65°\):
\(PT = \dfrac{x}{\cos 65°} = \dfrac{11.92}{0.4226} = 28.20 \approx \mathbf{28.2\text{ m}}\) (3 s.f.).
(b) Height of tower: \(H = x\tan 65° = 11.92 \times 2.1445 = 25.56 \approx \mathbf{25.6\text{ m}}\) (3 s.f.).
Answer Details
Let the horizontal distance between the building and the foot \(F\) of the tower be \(x\), and the tower height be \(H\). \(P\) is at the foot of the building, \(Q\) is its top, \(20\text{ m}\) directly above \(P\).
From \(P\): \(\tan 65° = \dfrac{H}{x}\), so \(H = x\tan 65°\).
From \(Q\) (height 20 m, same horizontal distance \(x\)): \(\tan 25° = \dfrac{H - 20}{x}\), so \(H - 20 = x\tan 25°\).
Subtracting: \(20 = x(\tan 65° - \tan 25°) = x(2.1445 - 0.4663) = 1.6782x\).
\(x = \dfrac{20}{1.6782} = 11.92\text{ m}\).
(a) Distance PT: \(PT\) is the line of sight from \(P\) to \(T\), at elevation \(65°\):
\(PT = \dfrac{x}{\cos 65°} = \dfrac{11.92}{0.4226} = 28.20 \approx \mathbf{28.2\text{ m}}\) (3 s.f.).
(b) Height of tower: \(H = x\tan 65° = 11.92 \times 2.1445 = 25.56 \approx \mathbf{25.6\text{ m}}\) (3 s.f.).
Question 12 Report
The table below gives the frequency distribution of the marks obtained by some students in a scholarship examination.
| Scores (x) | 15 | 25 | 35 | 45 | 55 | 65 | 75 |
| Freq (f) | 1 | 4 | 12 | 24 | 18 | 8 | 3 |
(a) Calculate, correct to 3 significant figures, the mean mark.
(b) Find the : (i) mode ; (ii) range of the distribution.
This is a discrete frequency distribution, so each score \(x\) already has its frequency \(f\). Build an \(fx\) column, add the columns, then apply the mean formula.
| Scores \(x\) | 15 | 25 | 35 | 45 | 55 | 65 | 75 | Total |
|---|---|---|---|---|---|---|---|---|
| Freq \(f\) | 1 | 4 | 12 | 24 | 18 | 8 | 3 | 70 |
| \(fx\) | 15 | 100 | 420 | 1080 | 990 | 520 | 225 | 3350 |
(a) Mean mark.
\[\bar{x} = \frac{\sum fx}{\sum f} = \frac{3350}{70} = 47.857\ldots\]Correct to 3 significant figures, the mean mark is \(\mathbf{47.9}\).
(b)(i) Mode. The mode is the score with the highest frequency. The largest frequency is \(24\), which belongs to the score \(45\), so the mode is \(\mathbf{45}\).
(b)(ii) Range.
\[\text{Range} = \text{highest score} - \text{lowest score} = 75 - 15 = \mathbf{60}.\]Examination note: multiply every score by its own frequency before adding, and divide by \(\sum f = 70\), not by the number of different scores (7). Dividing by 7 is the usual slip that gives a wrong mean near \(46\text{-}48\) instead of the correct \(47.9\).
Answer Details
This is a discrete frequency distribution, so each score \(x\) already has its frequency \(f\). Build an \(fx\) column, add the columns, then apply the mean formula.
| Scores \(x\) | 15 | 25 | 35 | 45 | 55 | 65 | 75 | Total |
|---|---|---|---|---|---|---|---|---|
| Freq \(f\) | 1 | 4 | 12 | 24 | 18 | 8 | 3 | 70 |
| \(fx\) | 15 | 100 | 420 | 1080 | 990 | 520 | 225 | 3350 |
(a) Mean mark.
\[\bar{x} = \frac{\sum fx}{\sum f} = \frac{3350}{70} = 47.857\ldots\]Correct to 3 significant figures, the mean mark is \(\mathbf{47.9}\).
(b)(i) Mode. The mode is the score with the highest frequency. The largest frequency is \(24\), which belongs to the score \(45\), so the mode is \(\mathbf{45}\).
(b)(ii) Range.
\[\text{Range} = \text{highest score} - \text{lowest score} = 75 - 15 = \mathbf{60}.\]Examination note: multiply every score by its own frequency before adding, and divide by \(\sum f = 70\), not by the number of different scores (7). Dividing by 7 is the usual slip that gives a wrong mean near \(46\text{-}48\) instead of the correct \(47.9\).
