WAEC SSCE - General Mathematics - 1995 (Essay)
Question 1 Report
(a) A pack of 52 playing cards is shuffled and a card is drawn at random. Calculate the probability that it is either a five or a red nine.
[Hint : There are 4 fives and 2 red nines in a pack of 52 cards]
(b) P, Q and R are points in the same horizontal plane. The bearing of Q from P is 150° and the bearing of R from Q is 060°. If /PQ/ = 5m and /QR/ = 3m, find the bearing of R from P, correct to the nearest degree.
(a) A pack has 4 fives and 2 red nines. Drawing a five and drawing a red nine are mutually exclusive events (a five cannot also be a red nine), so the probabilities add.
\(P(\text{five or red nine}) = \frac{4}{52} + \frac{2}{52} = \frac{6}{52} = \dfrac{3}{26}\)
(b) Take P as origin and measure position as (East, North). A bearing \(\theta\) gives the direction \((\sin\theta, \cos\theta)\).
- \(Q = 5(\sin150°,\ \cos150°) = (2.500,\ -4.330)\)
- \(\vec{QR} = 3(\sin060°,\ \cos060°) = (2.598,\ 1.500)\)
- \(R = (2.500+2.598,\ -4.330+1.500) = (5.098,\ -2.830)\)
Relative to P, R lies to the East and to the South, so the bearing is between 090° and 180°.
\(\text{Bearing} = 180° - \tan^{-1}\!\left(\dfrac{5.098}{2.830}\right) = 180° - 60.96° = 119.04°\)
Bearing of R from P \(\approx \mathbf{119°}\) (to the nearest degree).
Answer Details
(a) A pack has 4 fives and 2 red nines. Drawing a five and drawing a red nine are mutually exclusive events (a five cannot also be a red nine), so the probabilities add.
\(P(\text{five or red nine}) = \frac{4}{52} + \frac{2}{52} = \frac{6}{52} = \dfrac{3}{26}\)
(b) Take P as origin and measure position as (East, North). A bearing \(\theta\) gives the direction \((\sin\theta, \cos\theta)\).
- \(Q = 5(\sin150°,\ \cos150°) = (2.500,\ -4.330)\)
- \(\vec{QR} = 3(\sin060°,\ \cos060°) = (2.598,\ 1.500)\)
- \(R = (2.500+2.598,\ -4.330+1.500) = (5.098,\ -2.830)\)
Relative to P, R lies to the East and to the South, so the bearing is between 090° and 180°.
\(\text{Bearing} = 180° - \tan^{-1}\!\left(\dfrac{5.098}{2.830}\right) = 180° - 60.96° = 119.04°\)
Bearing of R from P \(\approx \mathbf{119°}\) (to the nearest degree).
Question 2 Report
(a) Factorise : \(px - 2px - 4qy + 2py\)
(b) Given that the universal set U = {1, 2, 3, 4,5, 6, 7, 8, 9, 10}, P = {1, 2, 4, 6, 10} and Q = {2, 3, 6, 9}; show that \((P \cup Q)' = P' \cap Q'\)
(a) Grouping the four terms \(px - 2qx - 4qy + 2py\) in pairs: \[(px - 2qx) + (2py - 4qy) = x(p - 2q) + 2y(p - 2q) = (p - 2q)(x + 2y).\] So the expression factorises as \(\mathbf{(p - 2q)(x + 2y)}.\)
(b) \(U = \{1,2,3,4,5,6,7,8,9,10\},\ P = \{1,2,4,6,10\},\ Q = \{2,3,6,9\}.\)
Left side: \(P \cup Q = \{1,2,3,4,6,9,10\}\), so \[(P \cup Q)' = \{5, 7, 8\}.\]
Right side: \(P' = \{3,5,7,8,9\}\) and \(Q' = \{1,4,5,7,8,10\}\), so \[P' \cap Q' = \{5, 7, 8\}.\]
Since \((P \cup Q)' = \{5,7,8\} = P' \cap Q'\), the identity is verified (De Morgan's law).
Answer Details
(a) Grouping the four terms \(px - 2qx - 4qy + 2py\) in pairs: \[(px - 2qx) + (2py - 4qy) = x(p - 2q) + 2y(p - 2q) = (p - 2q)(x + 2y).\] So the expression factorises as \(\mathbf{(p - 2q)(x + 2y)}.\)
(b) \(U = \{1,2,3,4,5,6,7,8,9,10\},\ P = \{1,2,4,6,10\},\ Q = \{2,3,6,9\}.\)
Left side: \(P \cup Q = \{1,2,3,4,6,9,10\}\), so \[(P \cup Q)' = \{5, 7, 8\}.\]
Right side: \(P' = \{3,5,7,8,9\}\) and \(Q' = \{1,4,5,7,8,10\}\), so \[P' \cap Q' = \{5, 7, 8\}.\]
Since \((P \cup Q)' = \{5,7,8\} = P' \cap Q'\), the identity is verified (De Morgan's law).
Question 3 Report
(a) Using mathematical tables, find ; (i) \(2 \sin 63.35°\) ; (ii) \(\log \cos 44.74°\);
(b) Find the value of K given that \(\log K - \log (K - 2) = \log 5\);
(c) Use logarithm tables to evaluate \(\frac{(3.68)^{2} \times 6.705}{\sqrt{0.3581}}\)
(a)(i) From tables, \(\sin 63.35° = 0.8937\).
\(2\sin 63.35° = 2 \times 0.8937 = \mathbf{1.787}\)
(ii) \(\cos 44.74° = 0.7103\).
\(\log \cos 44.74° = \log 0.7103 = \bar{1}.8515 \;(= -0.1485)\)
(b) \(\log K - \log(K-2) = \log 5\)
\(\log\!\left(\dfrac{K}{K-2}\right) = \log 5 \ \Rightarrow\ \dfrac{K}{K-2} = 5\)
\(K = 5(K-2) = 5K - 10 \ \Rightarrow\ 4K = 10 \ \Rightarrow\ \mathbf{K = 2.5}\)
(c) Evaluate \(\dfrac{(3.68)^2 \times 6.705}{\sqrt{0.3581}}\) using logarithms.
| Expression | Log |
|---|---|
| \((3.68)^2\) | \(2 \times 0.5658 = 1.1316\) |
| \(6.705\) | \(0.8264\) |
| Numerator total | \(1.9580\) |
| \(\sqrt{0.3581}\) | \(\tfrac{1}{2}(\bar{1}.5540) = \bar{1}.7770\) |
\(\text{Log of answer} = 1.9580 - \bar{1}.7770 = 1.9580 + 0.2230 = 2.1810\)
Antilog \(2.1810 = \mathbf{151.7}\) (4 s.f.).
Answer Details
(a)(i) From tables, \(\sin 63.35° = 0.8937\).
\(2\sin 63.35° = 2 \times 0.8937 = \mathbf{1.787}\)
(ii) \(\cos 44.74° = 0.7103\).
\(\log \cos 44.74° = \log 0.7103 = \bar{1}.8515 \;(= -0.1485)\)
(b) \(\log K - \log(K-2) = \log 5\)
\(\log\!\left(\dfrac{K}{K-2}\right) = \log 5 \ \Rightarrow\ \dfrac{K}{K-2} = 5\)
\(K = 5(K-2) = 5K - 10 \ \Rightarrow\ 4K = 10 \ \Rightarrow\ \mathbf{K = 2.5}\)
(c) Evaluate \(\dfrac{(3.68)^2 \times 6.705}{\sqrt{0.3581}}\) using logarithms.
| Expression | Log |
|---|---|
| \((3.68)^2\) | \(2 \times 0.5658 = 1.1316\) |
| \(6.705\) | \(0.8264\) |
| Numerator total | \(1.9580\) |
| \(\sqrt{0.3581}\) | \(\tfrac{1}{2}(\bar{1}.5540) = \bar{1}.7770\) |
\(\text{Log of answer} = 1.9580 - \bar{1}.7770 = 1.9580 + 0.2230 = 2.1810\)
Antilog \(2.1810 = \mathbf{151.7}\) (4 s.f.).
Question 4 Report
(a) A man travels from a village X on a bearing of 060° to a village Y which is 20km away. From Y, he travels to a village Z, on a bearing of 195°. If Z is directly east of X, calculate, correct to three significant figures, the distance of :
(i) Y from Z ; (ii) Z from X .
(b) An aircraft flies due South from an airfield on latitude 36°N, longitude 138°E to an airfield on latitude 36°S, longitude 138°E.
(i) Calculate the distance travelled, correct to three significant figures ; (ii) if the speed of the aircraft is 800km per hour, calculate the time taken, correct to the nearest hour.
[Take \(\pi = \frac{22}{7}\), R = 6400km].
(a) Take X as origin with (East, North) components; a bearing \(\theta\) gives direction \((\sin\theta,\cos\theta)\).
\(Y = 20(\sin060°,\cos060°) = (17.32,\ 10.00)\)
From Y on bearing \(195°\), distance \(YZ = d\): \(Z = Y + d(\sin195°,\cos195°) = (17.32 - 0.2588d,\ 10.00 - 0.9659d)\).
Z is due East of X, so its North component is \(0\):
\(10.00 - 0.9659d = 0 \ \Rightarrow\ d = \dfrac{10.00}{0.9659} = 10.35\)
(i) \(|YZ| \approx \mathbf{10.4\text{ km}}\).
East component of Z: \(17.32 - 0.2588(10.35) = 17.32 - 2.68 = 14.64\).
(ii) \(|ZX| \approx \mathbf{14.6\text{ km}}\).
(b) The route is along the same meridian (138°E) from 36°N to 36°S, an angular change of \(36° + 36° = 72°\).
(i) \(\text{Distance} = \dfrac{72}{360}\times 2\pi R = \dfrac{72}{360}\times 2\times\dfrac{22}{7}\times 6400\)
\(= 0.2 \times 40228.57 = 8045.7 \approx \mathbf{8050\text{ km}}\) (3 s.f.).
(ii) \(\text{Time} = \dfrac{8045.7}{800} = 10.06 \approx \mathbf{10\text{ hours}}\).
Answer Details
(a) Take X as origin with (East, North) components; a bearing \(\theta\) gives direction \((\sin\theta,\cos\theta)\).
\(Y = 20(\sin060°,\cos060°) = (17.32,\ 10.00)\)
From Y on bearing \(195°\), distance \(YZ = d\): \(Z = Y + d(\sin195°,\cos195°) = (17.32 - 0.2588d,\ 10.00 - 0.9659d)\).
Z is due East of X, so its North component is \(0\):
\(10.00 - 0.9659d = 0 \ \Rightarrow\ d = \dfrac{10.00}{0.9659} = 10.35\)
(i) \(|YZ| \approx \mathbf{10.4\text{ km}}\).
East component of Z: \(17.32 - 0.2588(10.35) = 17.32 - 2.68 = 14.64\).
(ii) \(|ZX| \approx \mathbf{14.6\text{ km}}\).
(b) The route is along the same meridian (138°E) from 36°N to 36°S, an angular change of \(36° + 36° = 72°\).
(i) \(\text{Distance} = \dfrac{72}{360}\times 2\pi R = \dfrac{72}{360}\times 2\times\dfrac{22}{7}\times 6400\)
\(= 0.2 \times 40228.57 = 8045.7 \approx \mathbf{8050\text{ km}}\) (3 s.f.).
(ii) \(\text{Time} = \dfrac{8045.7}{800} = 10.06 \approx \mathbf{10\text{ hours}}\).
Question 5 Report
The quantity y is partly constant and partly varies inversely as the square of x.
(a) Write down the relationship between x and y.
(b) When x = 1, y = 11 and when x = 2, y = 5, find the value of y when x = 4.
(a) "Partly constant and partly varies inversely as the square of \(x\)" means \[y = a + \frac{b}{x^2},\] where \(a\) and \(b\) are constants.
(b) Using the given values:
When \(x = 1, y = 11\): \(a + b = 11.\quad(1)\)
When \(x = 2, y = 5\): \(a + \dfrac{b}{4} = 5.\quad(2)\)
Subtract \((2)\) from \((1)\): \(b - \dfrac{b}{4} = 6 \Rightarrow \dfrac{3b}{4} = 6 \Rightarrow b = 8.\) Then \(a = 11 - 8 = 3.\)
So \(y = 3 + \dfrac{8}{x^2}.\) When \(x = 4\): \[y = 3 + \frac{8}{16} = 3 + 0.5 = \mathbf{3.5}.\]
Answer Details
(a) "Partly constant and partly varies inversely as the square of \(x\)" means \[y = a + \frac{b}{x^2},\] where \(a\) and \(b\) are constants.
(b) Using the given values:
When \(x = 1, y = 11\): \(a + b = 11.\quad(1)\)
When \(x = 2, y = 5\): \(a + \dfrac{b}{4} = 5.\quad(2)\)
Subtract \((2)\) from \((1)\): \(b - \dfrac{b}{4} = 6 \Rightarrow \dfrac{3b}{4} = 6 \Rightarrow b = 8.\) Then \(a = 11 - 8 = 3.\)
So \(y = 3 + \dfrac{8}{x^2}.\) When \(x = 4\): \[y = 3 + \frac{8}{16} = 3 + 0.5 = \mathbf{3.5}.\]
Question 6 Report
(a) Copy and complete the table for the relation \(y = 2 \cos 2x - 1\).
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° |
| \(y = 2\cos 2x - 1\) | 1.0 | 0 | 1.0 |
(b) Using a scale of 2cm = 30° on the x- axis and 2cm = 1 unit on the y- axis, draw the graph of \(y = 2 \cos 2x - 1\) for \(0° \leq x \leq 180°\).
(c) On the same axis, draw the graph of \(y = \frac{1}{180} (x - 360)\)
(d) Use your graphs to find the : (i) values of x for which \(2 \cos 2x + \frac{1}{2} = 0\); (ii) roots of the equation \(2 \cos 2x - \frac{x}{180} + 1 = 0\).
(a) For each value of x, evaluate \(y=2\cos 2x-1\).
| \(x\) | \(0^\circ\) | \(30^\circ\) | \(60^\circ\) | \(90^\circ\) | \(120^\circ\) | \(150^\circ\) | \(180^\circ\) |
|---|---|---|---|---|---|---|---|
| \(y=2\cos2x-1\) | 1.0 | 0 | -2.0 | -3.0 | -2.0 | 0 | 1.0 |
(b) and (c) The graphs of \(y=2\cos2x-1\) and \(y=\dfrac{1}{180}(x-360)\), drawn on the same axes using the stated scales, are shown below.
For the straight line, the plotting values are:
| \(x\) | \(0^\circ\) | \(30^\circ\) | \(60^\circ\) | \(90^\circ\) | \(120^\circ\) | \(150^\circ\) | \(180^\circ\) |
|---|---|---|---|---|---|---|---|
| \(y=\dfrac{x-360}{180}\) | -2.00 | -1.83 | -1.67 | -1.50 | -1.33 | -1.17 | -1.00 |
(d)(i) \[2\cos2x+\frac12=0\] \[2\cos2x-1=-\frac32=-1.5\] Hence, reading the points where the cosine curve has \(y=-1.5\), \[\boxed{x\approx51^\circ\text{ or }129^\circ.}\]
(d)(ii) \[2\cos2x-\frac{x}{180}+1=0\] \[2\cos2x-1=\frac{x}{180}-2=\frac{1}{180}(x-360).\] Thus, the required roots are the \(x\)-coordinates of the intersections of the curve and the straight line: \[\boxed{x\approx54^\circ\text{ or }132^\circ.}\]
Answer Details
(a) For each value of x, evaluate \(y=2\cos 2x-1\).
| \(x\) | \(0^\circ\) | \(30^\circ\) | \(60^\circ\) | \(90^\circ\) | \(120^\circ\) | \(150^\circ\) | \(180^\circ\) |
|---|---|---|---|---|---|---|---|
| \(y=2\cos2x-1\) | 1.0 | 0 | -2.0 | -3.0 | -2.0 | 0 | 1.0 |
(b) and (c) The graphs of \(y=2\cos2x-1\) and \(y=\dfrac{1}{180}(x-360)\), drawn on the same axes using the stated scales, are shown below.
For the straight line, the plotting values are:
| \(x\) | \(0^\circ\) | \(30^\circ\) | \(60^\circ\) | \(90^\circ\) | \(120^\circ\) | \(150^\circ\) | \(180^\circ\) |
|---|---|---|---|---|---|---|---|
| \(y=\dfrac{x-360}{180}\) | -2.00 | -1.83 | -1.67 | -1.50 | -1.33 | -1.17 | -1.00 |
(d)(i) \[2\cos2x+\frac12=0\] \[2\cos2x-1=-\frac32=-1.5\] Hence, reading the points where the cosine curve has \(y=-1.5\), \[\boxed{x\approx51^\circ\text{ or }129^\circ.}\]
(d)(ii) \[2\cos2x-\frac{x}{180}+1=0\] \[2\cos2x-1=\frac{x}{180}-2=\frac{1}{180}(x-360).\] Thus, the required roots are the \(x\)-coordinates of the intersections of the curve and the straight line: \[\boxed{x\approx54^\circ\text{ or }132^\circ.}\]
Question 7 Report
(a) Using a ruler and a pair of compasses only, construct triangle ABC with /AB/ = 7.5cm, /BC/ = 8.1cm and < ABC = 105°.
(b) Locate a point D on BC such that /BD/ : /DC/ is 3 : 2.
(c) Through D, construct a line I perpendicular to BC.
(d) If the line I meets AC at P, measure /BP/.
Construction and measurement
- Draw a straight line and mark a point B on it. With centre B and radius 8.1 cm, mark C on the line.
- At B, construct angles of 90° and 120° with BC. Bisect the angle between them to obtain a ray making 105° with BC.
- With centre B and radius 7.5 cm, cut this ray at A. Join A to C to complete triangle ABC.
- To divide BC in the ratio \(3:2\), draw any auxiliary ray from B. Mark five equal intervals on it, join the fifth mark to C, and draw through the third mark a parallel to this joining line. Its intersection with BC is D. Thus \(BD=4.86\text{ cm}\) and \(DC=3.24\text{ cm}\).
- Construct the perpendicular to BC through D. Let it meet AC at P.
The completed construction is shown below.
Measuring the constructed length gives:
\[\boxed{BP \approx 5.4\text{ cm}}\]
More accurately, \(BP\approx 5.39\text{ cm}\).
Answer Details
Construction and measurement
- Draw a straight line and mark a point B on it. With centre B and radius 8.1 cm, mark C on the line.
- At B, construct angles of 90° and 120° with BC. Bisect the angle between them to obtain a ray making 105° with BC.
- With centre B and radius 7.5 cm, cut this ray at A. Join A to C to complete triangle ABC.
- To divide BC in the ratio \(3:2\), draw any auxiliary ray from B. Mark five equal intervals on it, join the fifth mark to C, and draw through the third mark a parallel to this joining line. Its intersection with BC is D. Thus \(BD=4.86\text{ cm}\) and \(DC=3.24\text{ cm}\).
- Construct the perpendicular to BC through D. Let it meet AC at P.
The completed construction is shown below.
Measuring the constructed length gives:
\[\boxed{BP \approx 5.4\text{ cm}}\]
More accurately, \(BP\approx 5.39\text{ cm}\).
Question 8 Report
The frequency table shows the marks scored by 32 students in a test.
| Marks scored | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| No of students | 2 | 3 | 4 | 4 | 4 | 4 | 5 | 3 | 2 | 1 |
Find the :
(a)(i) mean ; (ii) median ; (iii) mode of the marks;
(b) percentage of the students who scored at least 8 marks.
Total number of students: \(2+3+4+4+4+4+5+3+2+1 = 32\).
| Mark (x) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| f | 2 | 3 | 4 | 4 | 4 | 4 | 5 | 3 | 2 | 1 |
| fx | 2 | 6 | 12 | 16 | 20 | 24 | 35 | 24 | 18 | 10 |
| Cum. f | 2 | 5 | 9 | 13 | 17 | 21 | 26 | 29 | 31 | 32 |
(a)(i) Mean. \(\sum fx = 167\).
\[\bar{x}=\frac{167}{32}\approx 5.22.\]
(a)(ii) Median. With \(N=32\), the median is the mean of the 16th and 17th values. The cumulative frequency reaches 13 at mark 4 and 17 at mark 5, so both the 16th and 17th values are 5.
\[\text{Median}=\frac{5+5}{2}=5.\]
(a)(iii) Mode. The highest frequency (5) is at mark 7, so the mode = 7.
(b) Percentage scoring at least 8 marks. Marks 8, 9, 10 have \(3+2+1=6\) students.
\[\frac{6}{32}\times100\% = 18.75\%.\]
Answer Details
Total number of students: \(2+3+4+4+4+4+5+3+2+1 = 32\).
| Mark (x) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| f | 2 | 3 | 4 | 4 | 4 | 4 | 5 | 3 | 2 | 1 |
| fx | 2 | 6 | 12 | 16 | 20 | 24 | 35 | 24 | 18 | 10 |
| Cum. f | 2 | 5 | 9 | 13 | 17 | 21 | 26 | 29 | 31 | 32 |
(a)(i) Mean. \(\sum fx = 167\).
\[\bar{x}=\frac{167}{32}\approx 5.22.\]
(a)(ii) Median. With \(N=32\), the median is the mean of the 16th and 17th values. The cumulative frequency reaches 13 at mark 4 and 17 at mark 5, so both the 16th and 17th values are 5.
\[\text{Median}=\frac{5+5}{2}=5.\]
(a)(iii) Mode. The highest frequency (5) is at mark 7, so the mode = 7.
(b) Percentage scoring at least 8 marks. Marks 8, 9, 10 have \(3+2+1=6\) students.
\[\frac{6}{32}\times100\% = 18.75\%.\]
Question 9 Report
(a) In the diagram, PQSR and SRYZ are parallelograms and PQYZ is a straight line. If /QY/ = 2cm and /RS/ = 3cm, find /PZ/.
(b) P and Q are two towns on the earth's surface on latitude 56°N. Thei longitudes are 25°E and 95°E respectively. Find the distance PQ along their parallel of latitude, correct to the nearest km. [Take radius of the earth as 6400km and \(\pi = \frac{22}{7}\)]
(a) Finding /PZ/
The diagram shows the parallelograms PQSR and SRYZ sharing the common side SR, with the points P, Q, Y, Z lying on one straight line in that order.
In a parallelogram opposite sides are equal, so:
- In parallelogram PQSR, side \(PQ\) is opposite side \(SR\), hence \(PQ = SR = 3\text{ cm}\).
- In parallelogram SRYZ, side \(ZY\) is opposite side \(SR\), hence \(ZY = SR = 3\text{ cm}\).
Since \(P, Q, Y, Z\) are collinear in that order:
\[ PZ = PQ + QY + YZ \]
\[ PZ = 3 + 2 + 3 = 8\text{ cm} \]
/PZ/ = 8 cm.
(b) Distance PQ along the parallel of latitude 56\(^{\circ}\)N
Both towns lie on latitude \(56^{\circ}\)N. Their longitudes are \(25^{\circ}\)E and \(95^{\circ}\)E, so the difference in longitude is:
\[ \theta = 95^{\circ} - 25^{\circ} = 70^{\circ} \]
The radius of the parallel of latitude is \(r = R\cos 56^{\circ}\), and the required distance is an arc of that parallel:
\[ PQ = \frac{\theta}{360^{\circ}} \times 2\pi R\cos 56^{\circ} \]
Substituting \(R = 6400\text{ km}\), \(\pi = \frac{22}{7}\) and \(\cos 56^{\circ} = 0.5592\):
\[ PQ = \frac{70}{360} \times 2 \times \frac{22}{7} \times 6400 \times 0.5592 \]
\[ PQ = \frac{70}{360} \times 40228.57 \times 0.5592 \]
\[ PQ = 0.19444 \times 40228.57 \times 0.5592 \approx 4374\text{ km} \]
Distance PQ \(\approx\) 4374 km (to the nearest km).
Answer Details
(a) Finding /PZ/
The diagram shows the parallelograms PQSR and SRYZ sharing the common side SR, with the points P, Q, Y, Z lying on one straight line in that order.
In a parallelogram opposite sides are equal, so:
- In parallelogram PQSR, side \(PQ\) is opposite side \(SR\), hence \(PQ = SR = 3\text{ cm}\).
- In parallelogram SRYZ, side \(ZY\) is opposite side \(SR\), hence \(ZY = SR = 3\text{ cm}\).
Since \(P, Q, Y, Z\) are collinear in that order:
\[ PZ = PQ + QY + YZ \]
\[ PZ = 3 + 2 + 3 = 8\text{ cm} \]
/PZ/ = 8 cm.
(b) Distance PQ along the parallel of latitude 56\(^{\circ}\)N
Both towns lie on latitude \(56^{\circ}\)N. Their longitudes are \(25^{\circ}\)E and \(95^{\circ}\)E, so the difference in longitude is:
\[ \theta = 95^{\circ} - 25^{\circ} = 70^{\circ} \]
The radius of the parallel of latitude is \(r = R\cos 56^{\circ}\), and the required distance is an arc of that parallel:
\[ PQ = \frac{\theta}{360^{\circ}} \times 2\pi R\cos 56^{\circ} \]
Substituting \(R = 6400\text{ km}\), \(\pi = \frac{22}{7}\) and \(\cos 56^{\circ} = 0.5592\):
\[ PQ = \frac{70}{360} \times 2 \times \frac{22}{7} \times 6400 \times 0.5592 \]
\[ PQ = \frac{70}{360} \times 40228.57 \times 0.5592 \]
\[ PQ = 0.19444 \times 40228.57 \times 0.5592 \approx 4374\text{ km} \]
Distance PQ \(\approx\) 4374 km (to the nearest km).
Question 10 Report
A box contains 5 blue balls, 3 black balls and 2 red balls of the same size. A ball is selected at random from the box and then replaced. A second ball is then selected. Find the probability of obtaining
(a) two red balls ;
(b) two blue balls or two black balls ;
(c) one black and one red ball in any order.
Total balls \(= 5 + 3 + 2 = 10\). Since each ball is replaced, the two draws are independent and the probabilities on each draw stay the same.
\(P(\text{blue}) = \tfrac{5}{10} = \tfrac{1}{2},\quad P(\text{black}) = \tfrac{3}{10},\quad P(\text{red}) = \tfrac{2}{10} = \tfrac{1}{5}\)
(a) Two red balls:
\(P = \tfrac{1}{5}\times\tfrac{1}{5} = \dfrac{1}{25}\)
(b) Two blue OR two black (mutually exclusive, so add):
\(P = \left(\tfrac{1}{2}\right)^2 + \left(\tfrac{3}{10}\right)^2 = \tfrac{1}{4} + \tfrac{9}{100} = \tfrac{25}{100} + \tfrac{9}{100} = \dfrac{34}{100} = \dfrac{17}{50}\)
(c) One black and one red in any order (black-then-red or red-then-black):
\(P = 2\times\tfrac{3}{10}\times\tfrac{2}{10} = 2\times\tfrac{6}{100} = \dfrac{12}{100} = \dfrac{3}{25}\)
Answer Details
Total balls \(= 5 + 3 + 2 = 10\). Since each ball is replaced, the two draws are independent and the probabilities on each draw stay the same.
\(P(\text{blue}) = \tfrac{5}{10} = \tfrac{1}{2},\quad P(\text{black}) = \tfrac{3}{10},\quad P(\text{red}) = \tfrac{2}{10} = \tfrac{1}{5}\)
(a) Two red balls:
\(P = \tfrac{1}{5}\times\tfrac{1}{5} = \dfrac{1}{25}\)
(b) Two blue OR two black (mutually exclusive, so add):
\(P = \left(\tfrac{1}{2}\right)^2 + \left(\tfrac{3}{10}\right)^2 = \tfrac{1}{4} + \tfrac{9}{100} = \tfrac{25}{100} + \tfrac{9}{100} = \dfrac{34}{100} = \dfrac{17}{50}\)
(c) One black and one red in any order (black-then-red or red-then-black):
\(P = 2\times\tfrac{3}{10}\times\tfrac{2}{10} = 2\times\tfrac{6}{100} = \dfrac{12}{100} = \dfrac{3}{25}\)
Question 11 Report
(a) Given that \(p = x + ym^{3}\), find m in terms of p, x and y.
(b) Using the method of completing the square, find the roots of the equation \(x^{2} - 6x + 7 = 0\), correct to 1 decimal place.
(c) The product of two consecutive positive odd numbers is 195. By constructing a quadratic equation and solving it, find the two numbers.
(a) \(p = x + ym^3\). Make \(m\) the subject.
\(ym^3 = p - x \ \Rightarrow\ m^3 = \dfrac{p-x}{y} \ \Rightarrow\ \mathbf{m = \sqrt[3]{\dfrac{p-x}{y}}}\)
(b) \(x^2 - 6x + 7 = 0\) by completing the square.
\(x^2 - 6x = -7\)
Add \(\left(\tfrac{6}{2}\right)^2 = 9\) to both sides:
\((x-3)^2 = -7 + 9 = 2\)
\(x - 3 = \pm\sqrt{2} = \pm 1.414\)
\(x = 3 + 1.414 = 4.4\quad\text{or}\quad x = 3 - 1.414 = 1.6\) (1 d.p.)
(c) Let the consecutive odd numbers be \(n\) and \(n+2\).
\(n(n+2) = 195 \ \Rightarrow\ n^2 + 2n - 195 = 0\)
\((n+15)(n-13) = 0 \ \Rightarrow\ n = -15\ \text{or}\ n = 13\)
Since the numbers are positive, \(n = 13\). The numbers are \(\mathbf{13}\) and \(\mathbf{15}\).
Answer Details
(a) \(p = x + ym^3\). Make \(m\) the subject.
\(ym^3 = p - x \ \Rightarrow\ m^3 = \dfrac{p-x}{y} \ \Rightarrow\ \mathbf{m = \sqrt[3]{\dfrac{p-x}{y}}}\)
(b) \(x^2 - 6x + 7 = 0\) by completing the square.
\(x^2 - 6x = -7\)
Add \(\left(\tfrac{6}{2}\right)^2 = 9\) to both sides:
\((x-3)^2 = -7 + 9 = 2\)
\(x - 3 = \pm\sqrt{2} = \pm 1.414\)
\(x = 3 + 1.414 = 4.4\quad\text{or}\quad x = 3 - 1.414 = 1.6\) (1 d.p.)
(c) Let the consecutive odd numbers be \(n\) and \(n+2\).
\(n(n+2) = 195 \ \Rightarrow\ n^2 + 2n - 195 = 0\)
\((n+15)(n-13) = 0 \ \Rightarrow\ n = -15\ \text{or}\ n = 13\)
Since the numbers are positive, \(n = 13\). The numbers are \(\mathbf{13}\) and \(\mathbf{15}\).
Question 12 Report
The table shows the scores of 2000 candidates in an entrance examination into a private secondary school.
| % Mark | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 | 61-70 | 71-80 | 81-90 |
| No of pupils | 68 | 184 | 294 | 402 | 480 | 310 | 164 | 98 |
(a) Prepare a cumulative frequency table and draw the cumulative frequency curve for the distribution.
(b) Use your curve to estimate the : (i) cut off mark, if 300 candidates are to be offered admission ; (ii) probability that a candidate picked at random, scored at least 45%.
(a) Cumulative frequency table
| Marks (%) | Number of pupils, \(f\) | Cumulative frequency |
|---|---|---|
| 11 - 20 | 68 | 68 |
| 21 - 30 | 184 | 252 |
| 31 - 40 | 294 | 546 |
| 41 - 50 | 402 | 948 |
| 51 - 60 | 480 | 1428 |
| 61 - 70 | 310 | 1738 |
| 71 - 80 | 164 | 1902 |
| 81 - 90 | 98 | 2000 |
Using upper class boundaries, plot the cumulative frequencies and join the points with a smooth increasing curve.
(b)(i) Cut-off mark
The 300 candidates admitted are the highest scorers. Hence the number below the cut-off mark is
\[2000-300=1700.\]
From the ogive, at cumulative frequency \(1700\), the corresponding mark is approximately \(68\).
Therefore, the cut-off mark is 68 marks (68%).
(b)(ii) Probability of scoring at least \(45\%\)
From the ogive, the cumulative frequency at \(45\%\) is approximately \(720\).
\[\text{Number scoring at least }45\%=2000-720=1280.\]
\[P(\text{score at least }45\%)=\frac{1280}{2000}=0.64.\]
Answer Details
(a) Cumulative frequency table
| Marks (%) | Number of pupils, \(f\) | Cumulative frequency |
|---|---|---|
| 11 - 20 | 68 | 68 |
| 21 - 30 | 184 | 252 |
| 31 - 40 | 294 | 546 |
| 41 - 50 | 402 | 948 |
| 51 - 60 | 480 | 1428 |
| 61 - 70 | 310 | 1738 |
| 71 - 80 | 164 | 1902 |
| 81 - 90 | 98 | 2000 |
Using upper class boundaries, plot the cumulative frequencies and join the points with a smooth increasing curve.
(b)(i) Cut-off mark
The 300 candidates admitted are the highest scorers. Hence the number below the cut-off mark is
\[2000-300=1700.\]
From the ogive, at cumulative frequency \(1700\), the corresponding mark is approximately \(68\).
Therefore, the cut-off mark is 68 marks (68%).
(b)(ii) Probability of scoring at least \(45\%\)
From the ogive, the cumulative frequency at \(45\%\) is approximately \(720\).
\[\text{Number scoring at least }45\%=2000-720=1280.\]
\[P(\text{score at least }45\%)=\frac{1280}{2000}=0.64.\]
