WAEC SSCE - General Mathematics - 2018 (Essay)
Tambaya 1 Rahoto
The diagram shows an athletics track with two parallel sides and two semi-circular ends Each of the parallel sides is 60 metres, long and the diameter of each semi-circular end is 120 metres long.
(a) Calculate the distance covered by an athlete who runs round the tack the two times. [Take \(\pi\) = \(\frac{22}{7}\)]
(b) If the athlete spends 200 seconds for the race, calculate the speed in km/h.
From the diagram the track is a rectangle capped by two semicircular ends. The two straight (parallel) sides are each 60 m long, and each semicircular end has diameter 120 m (the marked vertical distance between the two straights).
(a) Distance for one lap, then two laps.
The two semicircular ends together form one complete circle of diameter \(d = 120\) m. So one lap consists of the two straights plus one full circle:
\[ \text{One lap} = 2(60) + \pi d = 120 + \frac{22}{7}\times 120. \]
\[ \frac{22}{7}\times 120 = \frac{2640}{7} = 377.14\text{ m}. \]
\[ \text{One lap} = 120 + \frac{2640}{7} = \frac{840 + 2640}{7} = \frac{3480}{7} = 497.14\text{ m}. \]
Running round the track two times:
\[ D = 2 \times \frac{3480}{7} = \frac{6960}{7} = 994.29\text{ m} \;(\text{to 2 d.p.}). \]
Distance covered \(\approx 994.29\) m.
(b) Speed in km/h.
Time \(= 200\) s. First find the speed in metres per second:
\[ \text{Speed} = \frac{D}{t} = \frac{6960/7}{200} = \frac{6960}{1400} = 4.971\text{ m/s}. \]
Convert to km/h (multiply by \(\tfrac{3600}{1000} = 3.6\)):
\[ \text{Speed} = 4.971 \times 3.6 = 17.90\text{ km/h}. \]
Alternatively: \(D = \dfrac{6960}{7000}\) km \(= 0.99429\) km, \(t = \dfrac{200}{3600}\) h \(= \dfrac{1}{18}\) h, so speed \(= 0.99429 \times 18 = 17.90\) km/h.
Speed \(\approx 17.9\) km/h.
Bayanin Amsa
From the diagram the track is a rectangle capped by two semicircular ends. The two straight (parallel) sides are each 60 m long, and each semicircular end has diameter 120 m (the marked vertical distance between the two straights).
(a) Distance for one lap, then two laps.
The two semicircular ends together form one complete circle of diameter \(d = 120\) m. So one lap consists of the two straights plus one full circle:
\[ \text{One lap} = 2(60) + \pi d = 120 + \frac{22}{7}\times 120. \]
\[ \frac{22}{7}\times 120 = \frac{2640}{7} = 377.14\text{ m}. \]
\[ \text{One lap} = 120 + \frac{2640}{7} = \frac{840 + 2640}{7} = \frac{3480}{7} = 497.14\text{ m}. \]
Running round the track two times:
\[ D = 2 \times \frac{3480}{7} = \frac{6960}{7} = 994.29\text{ m} \;(\text{to 2 d.p.}). \]
Distance covered \(\approx 994.29\) m.
(b) Speed in km/h.
Time \(= 200\) s. First find the speed in metres per second:
\[ \text{Speed} = \frac{D}{t} = \frac{6960/7}{200} = \frac{6960}{1400} = 4.971\text{ m/s}. \]
Convert to km/h (multiply by \(\tfrac{3600}{1000} = 3.6\)):
\[ \text{Speed} = 4.971 \times 3.6 = 17.90\text{ km/h}. \]
Alternatively: \(D = \dfrac{6960}{7000}\) km \(= 0.99429\) km, \(t = \dfrac{200}{3600}\) h \(= \dfrac{1}{18}\) h, so speed \(= 0.99429 \times 18 = 17.90\) km/h.
Speed \(\approx 17.9\) km/h.
Tambaya 2 Rahoto
1. A donkey is tied with a rope to a post which is 15 m from a fence. If the length of the rope between the donkey and the post is 17m, calculate the length of the fence within the reach of the donkey.
2. The base of a right pyramid with vertex, V, is a square, PQRS, of side 15 cm. If the slant height is 32 cm long:
3. represent the information in a diagram;
4. calculate its:
5. height, correct to one decimal place;
6. volume, correct to the nearest \(cm^3\)
1. Length of fence within reach
The donkey can reach every point 17 m from the post, so its path forms a circle of radius 17 m. The fence is 15 m from the post. This makes a right-angled triangle from the post to the nearest point on the fence and to one end of the reachable section.
Using Pythagoras’ theorem on one half of the reachable fence length:
\[ x^2=17^2-15^2=289-225=64 \] \[ x=8\text{ m} \]The total length of fence is twice this value:
\[ 2x=2(8)=\boxed{16\text{ m}} \]2. Right square pyramid
The reference answer treats the given 32 cm as the sloping edge from the vertex \(V\) to a corner of the square base, for example \(VP\). In a square, the centre \(O\) is halfway along a diagonal, so \(OP\) is half the diagonal. The vertical height is \(VO\).
Height
First find the diagonal \(PR\) of the square base:
\[ PR=\sqrt{15^2+15^2}=15\sqrt2\text{ cm} \]Since \(O\) is the centre of the square,
\[ OP=\frac{PR}{2}=\frac{15\sqrt2}{2}\approx10.6066\text{ cm} \]Triangle \(VOP\) is right-angled at \(O\), so:
\[ 32^2=h^2+10.6066^2 \] \[ h^2=1024-112.5=911.5 \] \[ h=\sqrt{911.5}=30.191\ldots \] \[ \boxed{h=30.2\text{ cm}} \]Volume
\[ \text{Volume}=\frac13\times\text{base area}\times\text{perpendicular height} \] \[ =\frac13\times15^2\times30.191\ldots \] \[ =2264.33\ldots\text{ cm}^3 \] \[ \boxed{2264\text{ cm}^3} \]The unit for volume must be \(\text{cm}^3\), not \(\text{cm}^2\). Also, distinguish between half a base edge, \(7.5\) cm, and the distance from the centre to a corner, \(OP=\frac{15\sqrt2}{2}\) cm. The latter is required here because the 32 cm line is \(VP\), ending at a corner of the base.
Bayanin Amsa
1. Length of fence within reach
The donkey can reach every point 17 m from the post, so its path forms a circle of radius 17 m. The fence is 15 m from the post. This makes a right-angled triangle from the post to the nearest point on the fence and to one end of the reachable section.
Using Pythagoras’ theorem on one half of the reachable fence length:
\[ x^2=17^2-15^2=289-225=64 \] \[ x=8\text{ m} \]The total length of fence is twice this value:
\[ 2x=2(8)=\boxed{16\text{ m}} \]2. Right square pyramid
The reference answer treats the given 32 cm as the sloping edge from the vertex \(V\) to a corner of the square base, for example \(VP\). In a square, the centre \(O\) is halfway along a diagonal, so \(OP\) is half the diagonal. The vertical height is \(VO\).
Height
First find the diagonal \(PR\) of the square base:
\[ PR=\sqrt{15^2+15^2}=15\sqrt2\text{ cm} \]Since \(O\) is the centre of the square,
\[ OP=\frac{PR}{2}=\frac{15\sqrt2}{2}\approx10.6066\text{ cm} \]Triangle \(VOP\) is right-angled at \(O\), so:
\[ 32^2=h^2+10.6066^2 \] \[ h^2=1024-112.5=911.5 \] \[ h=\sqrt{911.5}=30.191\ldots \] \[ \boxed{h=30.2\text{ cm}} \]Volume
\[ \text{Volume}=\frac13\times\text{base area}\times\text{perpendicular height} \] \[ =\frac13\times15^2\times30.191\ldots \] \[ =2264.33\ldots\text{ cm}^3 \] \[ \boxed{2264\text{ cm}^3} \]The unit for volume must be \(\text{cm}^3\), not \(\text{cm}^2\). Also, distinguish between half a base edge, \(7.5\) cm, and the distance from the centre to a corner, \(OP=\frac{15\sqrt2}{2}\) cm. The latter is required here because the 32 cm line is \(VP\), ending at a corner of the base.
Tambaya 3 Rahoto
(a) Evaluate without using calculator, (\(\frac{1}{4} \times 9\frac{1}{7} + \frac{2}{5} (\frac{2}{2} + \frac{3}{4})) \div (\frac{2}{5} - \frac{1}{4})\)
(b) A hunter walked 250 m from point P to Q on a bearing 042\(^o\). Calculate, correct to the nearest meter the vertical distance he has moved.
(a) Work part by part (reading \(\frac{2}{2} = 1\) as written).
\[\tfrac{1}{4} \times 9\tfrac{1}{7} = \tfrac{1}{4} \times \tfrac{64}{7} = \tfrac{16}{7}\] \[\tfrac{2}{5}\left(1 + \tfrac{3}{4}\right) = \tfrac{2}{5} \times \tfrac{7}{4} = \tfrac{7}{10}\] \[\tfrac{16}{7} + \tfrac{7}{10} = \tfrac{160 + 49}{70} = \tfrac{209}{70}\] \[\tfrac{2}{5} - \tfrac{1}{4} = \tfrac{3}{20}\] \[\tfrac{209}{70} \div \tfrac{3}{20} = \tfrac{209}{70} \times \tfrac{20}{3} = \tfrac{418}{21} = 19\tfrac{19}{21}\](b) The vertical (north-south) distance is the northward component of the walk.
\[\text{Vertical distance} = 250\cos 42^{\circ} = 250 \times 0.7431 = 185.8 \approx 186 \text{ m}\]Bayanin Amsa
(a) Work part by part (reading \(\frac{2}{2} = 1\) as written).
\[\tfrac{1}{4} \times 9\tfrac{1}{7} = \tfrac{1}{4} \times \tfrac{64}{7} = \tfrac{16}{7}\] \[\tfrac{2}{5}\left(1 + \tfrac{3}{4}\right) = \tfrac{2}{5} \times \tfrac{7}{4} = \tfrac{7}{10}\] \[\tfrac{16}{7} + \tfrac{7}{10} = \tfrac{160 + 49}{70} = \tfrac{209}{70}\] \[\tfrac{2}{5} - \tfrac{1}{4} = \tfrac{3}{20}\] \[\tfrac{209}{70} \div \tfrac{3}{20} = \tfrac{209}{70} \times \tfrac{20}{3} = \tfrac{418}{21} = 19\tfrac{19}{21}\](b) The vertical (north-south) distance is the northward component of the walk.
\[\text{Vertical distance} = 250\cos 42^{\circ} = 250 \times 0.7431 = 185.8 \approx 186 \text{ m}\]Tambaya 4 Rahoto
(a) Find the equation of the line passing through the points (2, 5) and (-4, -7).
(b) Three ships P, Q and R are at sea. The bearing of Q from P is 030° and the bearing of P and R is 300°. If |PQ| = 5 km and |PR| = 8 km,
(i) Illustrate the information in a diagram.
(ii) Calculate, correct to three significant figures, the:
(1) distance between Q and R
(2) bearing of R from Q.
(a) Equation of the line
The gradient of the line through \((2,5)\) and \((-4,-7)\) is
\[ m=\frac{-7-5}{-4-2}=\frac{-12}{-6}=2. \]
Using the point \((2,5)\),
\[ y-5=2(x-2). \]
\[ y-5=2x-4 \]
\[ \boxed{y=2x+1}. \]
(b)(i) Illustration
As shown in the diagram above, \(Q\) is on a bearing of \(030^\circ\) from \(P\), while \(R\) is on a bearing of \(300^\circ\) from \(P\). Therefore, the angle between \(PQ\) and \(PR\) is
\[ \angle QPR=30^\circ+60^\circ=90^\circ. \]
Thus, \(\triangle PQR\) is right-angled at \(P\), with \(PQ=5\text{ km}\) and \(PR=8\text{ km}\).
(b)(ii)(1) Distance between \(Q\) and \(R\)
By Pythagoras' theorem,
\[ QR^2=PQ^2+PR^2 \]
\[ QR^2=5^2+8^2=25+64=89. \]
\[ QR=\sqrt{89}=9.433\ldots \]
\[ \boxed{QR=9.43\text{ km}} \]
correct to three significant figures.
(b)(ii)(2) Bearing of \(R\) from \(Q\)
First, find the angle at \(Q\):
\[ \tan \angle PQR=\frac{PR}{PQ}=\frac{8}{5}. \]
\[ \angle PQR=\tan^{-1}\left(\frac{8}{5}\right)=58.0^\circ. \]
The bearing of \(P\) from \(Q\) is the reverse bearing of \(030^\circ\):
\[ 030^\circ+180^\circ=210^\circ. \]
From the diagram, \(R\) lies clockwise from the direction \(QP\). Hence,
\[ \text{Bearing of }R\text{ from }Q=210^\circ+58^\circ=268^\circ. \]
\[ \boxed{268^\circ} \]
Bayanin Amsa
(a) Equation of the line
The gradient of the line through \((2,5)\) and \((-4,-7)\) is
\[ m=\frac{-7-5}{-4-2}=\frac{-12}{-6}=2. \]
Using the point \((2,5)\),
\[ y-5=2(x-2). \]
\[ y-5=2x-4 \]
\[ \boxed{y=2x+1}. \]
(b)(i) Illustration
As shown in the diagram above, \(Q\) is on a bearing of \(030^\circ\) from \(P\), while \(R\) is on a bearing of \(300^\circ\) from \(P\). Therefore, the angle between \(PQ\) and \(PR\) is
\[ \angle QPR=30^\circ+60^\circ=90^\circ. \]
Thus, \(\triangle PQR\) is right-angled at \(P\), with \(PQ=5\text{ km}\) and \(PR=8\text{ km}\).
(b)(ii)(1) Distance between \(Q\) and \(R\)
By Pythagoras' theorem,
\[ QR^2=PQ^2+PR^2 \]
\[ QR^2=5^2+8^2=25+64=89. \]
\[ QR=\sqrt{89}=9.433\ldots \]
\[ \boxed{QR=9.43\text{ km}} \]
correct to three significant figures.
(b)(ii)(2) Bearing of \(R\) from \(Q\)
First, find the angle at \(Q\):
\[ \tan \angle PQR=\frac{PR}{PQ}=\frac{8}{5}. \]
\[ \angle PQR=\tan^{-1}\left(\frac{8}{5}\right)=58.0^\circ. \]
The bearing of \(P\) from \(Q\) is the reverse bearing of \(030^\circ\):
\[ 030^\circ+180^\circ=210^\circ. \]
From the diagram, \(R\) lies clockwise from the direction \(QP\). Hence,
\[ \text{Bearing of }R\text{ from }Q=210^\circ+58^\circ=268^\circ. \]
\[ \boxed{268^\circ} \]
Tambaya 5 Rahoto
A container, in the form of a cone resting on its vertex, is full when 4.158 litres of water is poured into it.
(a) If the radius of its base is 21 cm,
(i) represent the information in a diagram;
(ii) calculate the height of the container.
(b) A certain amount of water is drawn out of the container such that the surface diameter of the water drops to 28 cm. Calculate the volume of the water drawn out. (Take \(\pi\) = \(\frac{22}{7}\))
(a)(i) Diagram: a cone standing on its vertex (point downwards), base radius 21 cm at the top, water filling it to height \(h\).
(a)(ii) Height. \(4.158\) litres \(= 4158 \text{ cm}^3\).
\[\tfrac{1}{3} \times \tfrac{22}{7} \times 21^2 \times h = 4158\] \[\tfrac{1}{3} \times \tfrac{22}{7} \times 441 \times h = 462h = 4158 \Rightarrow h = 9 \text{ cm}\](b) When the surface diameter drops to 28 cm, the surface radius is 14 cm. By similar cones the remaining water height is
\[h' = 9 \times \frac{14}{21} = 6 \text{ cm}\]Volume remaining:
\[\tfrac{1}{3} \times \tfrac{22}{7} \times 14^2 \times 6 = \tfrac{1}{3} \times 616 \times 6 = 1232 \text{ cm}^3\]Volume drawn out \(= 4158 - 1232 = 2926 \text{ cm}^3\).
Bayanin Amsa
(a)(i) Diagram: a cone standing on its vertex (point downwards), base radius 21 cm at the top, water filling it to height \(h\).
(a)(ii) Height. \(4.158\) litres \(= 4158 \text{ cm}^3\).
\[\tfrac{1}{3} \times \tfrac{22}{7} \times 21^2 \times h = 4158\] \[\tfrac{1}{3} \times \tfrac{22}{7} \times 441 \times h = 462h = 4158 \Rightarrow h = 9 \text{ cm}\](b) When the surface diameter drops to 28 cm, the surface radius is 14 cm. By similar cones the remaining water height is
\[h' = 9 \times \frac{14}{21} = 6 \text{ cm}\]Volume remaining:
\[\tfrac{1}{3} \times \tfrac{22}{7} \times 14^2 \times 6 = \tfrac{1}{3} \times 616 \times 6 = 1232 \text{ cm}^3\]Volume drawn out \(= 4158 - 1232 = 2926 \text{ cm}^3\).
Tambaya 6 Rahoto
In a road worthiness test on 240 cars, 60% passed. The number that failed had faults in Clutch, Brakes and Steering as follows: Clutch only - 28, Clutch and Steering - 14; Clutch, Steering and Brakes - 8; Clutch and Brakes - 20; Brakes and Steering only - 6. The number of cars with faults in Steering only is twice the number of cars with faults in Brakes only.
(a) Draw a Venn Diagram to illustrate this information.
(b) How many cars had : (i) Faulty Brakes? (ii) Only one fault?
(a) Venn diagram
Let the three sets be:
- C = cars with faulty clutch
- B = cars with faulty brakes
- S = cars with faulty steering
(b) Calculation
Number that passed:
\[60\%\text{ of }240=\frac{60}{100}\times240=144\]
Therefore, number that failed:
\[240-144=96\]
The given numbers for the two-set intersections include the cars with all three faults. Hence:
\[C\cap S\text{ only}=14-8=6\]
\[C\cap B\text{ only}=20-8=12\]
Let the number with brakes only be \(x\). Then the number with steering only is \(2x\).
From the Venn diagram:
\[28+6+8+12+6+x+2x=96\]
\[60+3x=96\]
\[3x=36\]
\[x=12\]
Thus, brakes only \(=12\), while steering only \(=2(12)=24\).
(i) Cars with faulty brakes
\[12+12+6+8=38\]
\[\boxed{38\text{ cars}}\]
(ii) Cars with only one fault
\[28+12+24=64\]
\[\boxed{64\text{ cars}}\]
Bayanin Amsa
(a) Venn diagram
Let the three sets be:
- C = cars with faulty clutch
- B = cars with faulty brakes
- S = cars with faulty steering
(b) Calculation
Number that passed:
\[60\%\text{ of }240=\frac{60}{100}\times240=144\]
Therefore, number that failed:
\[240-144=96\]
The given numbers for the two-set intersections include the cars with all three faults. Hence:
\[C\cap S\text{ only}=14-8=6\]
\[C\cap B\text{ only}=20-8=12\]
Let the number with brakes only be \(x\). Then the number with steering only is \(2x\).
From the Venn diagram:
\[28+6+8+12+6+x+2x=96\]
\[60+3x=96\]
\[3x=36\]
\[x=12\]
Thus, brakes only \(=12\), while steering only \(=2(12)=24\).
(i) Cars with faulty brakes
\[12+12+6+8=38\]
\[\boxed{38\text{ cars}}\]
(ii) Cars with only one fault
\[28+12+24=64\]
\[\boxed{64\text{ cars}}\]
Tambaya 7 Rahoto
a. A textbook company discovered that the profit made from selling its books is given by y = \(\frac{x^2}{8}\) + 5x, where x is the number of textbooks sold (in thousands) and y is the corresponding profit (in Ghana Cedis). If the company made a profit of GH₵ 20,000.00
i. form a quadratic equation in x;
ii. (using the quadratic formula, find, correct to the nearest whole number, the number of textbooks sold to make the profit.
b. The angle of elevation of the top T of a tree from a point P on the same ground level as the foot Q of a tree is 28\(^o\). A bird perched at a point R, halfway up the tree.
i. Represent the information in a diagram.
ii. Calculate, correct to the nearest degree, the angle of elevation of R from P.
(a)(i) With profit \(y = 20000\):
\[\frac{x^2}{8} + 5x = 20000 \;\;(\times 8) \Rightarrow x^2 + 40x - 160000 = 0\](a)(ii) Quadratic formula:
\[x = \frac{-40 \pm \sqrt{40^2 + 4(160000)}}{2} = \frac{-40 \pm \sqrt{641600}}{2} = \frac{-40 \pm 801.0}{2}\]Taking the positive root: \(x = \dfrac{761.0}{2} \approx 380\) (in thousands), i.e. about \(380{,}000\) textbooks.
(b)(i) Diagram: vertical tree \(TQ\) with P on the ground so that \(\angle TPQ = 28^{\circ}\); R is the midpoint of \(TQ\).
(b)(ii) Let \(PQ = d\). Then \(TQ = d\tan 28^{\circ}\) and \(RQ = \tfrac{1}{2}TQ = \tfrac{1}{2}d\tan 28^{\circ}\).
\[\tan(\angle RPQ) = \frac{RQ}{PQ} = \tfrac{1}{2}\tan 28^{\circ} = \tfrac{1}{2}(0.5317) = 0.2659\] \[\angle RPQ = \tan^{-1}(0.2659) \approx 15^{\circ}\]Bayanin Amsa
(a)(i) With profit \(y = 20000\):
\[\frac{x^2}{8} + 5x = 20000 \;\;(\times 8) \Rightarrow x^2 + 40x - 160000 = 0\](a)(ii) Quadratic formula:
\[x = \frac{-40 \pm \sqrt{40^2 + 4(160000)}}{2} = \frac{-40 \pm \sqrt{641600}}{2} = \frac{-40 \pm 801.0}{2}\]Taking the positive root: \(x = \dfrac{761.0}{2} \approx 380\) (in thousands), i.e. about \(380{,}000\) textbooks.
(b)(i) Diagram: vertical tree \(TQ\) with P on the ground so that \(\angle TPQ = 28^{\circ}\); R is the midpoint of \(TQ\).
(b)(ii) Let \(PQ = d\). Then \(TQ = d\tan 28^{\circ}\) and \(RQ = \tfrac{1}{2}TQ = \tfrac{1}{2}d\tan 28^{\circ}\).
\[\tan(\angle RPQ) = \frac{RQ}{PQ} = \tfrac{1}{2}\tan 28^{\circ} = \tfrac{1}{2}(0.5317) = 0.2659\] \[\angle RPQ = \tan^{-1}(0.2659) \approx 15^{\circ}\]Tambaya 8 Rahoto
The table shows the distribution of marks scored by students in a test.
| Mark (%) |
10 - 19 | 20 - 29 | 30 - 39 | 40 - 49 | 50 - 59 | 60 - 69 | 70 - 79 | 80 - 89 | 90 - 99 |
| Frequency | 4 | 7 | 12 | 18 | 20 | 14 | 9 | 4 | 2 |
(a) Construct a cumulative frequency table for the distribution.
(b) Draw a cumulative frequency curve for the distribution.
(c) Use the curve to estimate the:
(i) median;
(ii) probability that a student selected at random obtained distinction, if the lowest mark for distinction is 75%.
(a) Cumulative frequency table
| Marks (%) | Class boundaries | Frequency | Cumulative frequency |
|---|---|---|---|
| 10–19 | 9.5–19.5 | 4 | 4 |
| 20–29 | 19.5–29.5 | 7 | 11 |
| 30–39 | 29.5–39.5 | 12 | 23 |
| 40–49 | 39.5–49.5 | 18 | 41 |
| 50–59 | 49.5–59.5 | 20 | 61 |
| 60–69 | 59.5–69.5 | 14 | 75 |
| 70–79 | 69.5–79.5 | 9 | 84 |
| 80–89 | 79.5–89.5 | 4 | 88 |
| 90–99 | 89.5–99.5 | 2 | 90 |
Total frequency, \(N=90\).
(b) Cumulative frequency curve
The less-than ogive is plotted using the upper class boundaries and their cumulative frequencies.
(c)(i) Median
\[\frac{N}{2}=\frac{90}{2}=45.\]
From the curve, the mark corresponding to cumulative frequency 45 is approximately \(51.5\%\).
\[\boxed{\text{Median}=51.5\%}\]
(c)(ii) Probability of obtaining distinction
At \(75\%\), the cumulative frequency read from the curve is approximately \(80\). Hence, the number scoring at least \(75\%\) is
\[90-80=10.\]
Therefore,
\[P(\text{distinction})=\frac{10}{90}=\frac{1}{9}\approx0.111.\]
Bayanin Amsa
(a) Cumulative frequency table
| Marks (%) | Class boundaries | Frequency | Cumulative frequency |
|---|---|---|---|
| 10–19 | 9.5–19.5 | 4 | 4 |
| 20–29 | 19.5–29.5 | 7 | 11 |
| 30–39 | 29.5–39.5 | 12 | 23 |
| 40–49 | 39.5–49.5 | 18 | 41 |
| 50–59 | 49.5–59.5 | 20 | 61 |
| 60–69 | 59.5–69.5 | 14 | 75 |
| 70–79 | 69.5–79.5 | 9 | 84 |
| 80–89 | 79.5–89.5 | 4 | 88 |
| 90–99 | 89.5–99.5 | 2 | 90 |
Total frequency, \(N=90\).
(b) Cumulative frequency curve
The less-than ogive is plotted using the upper class boundaries and their cumulative frequencies.
(c)(i) Median
\[\frac{N}{2}=\frac{90}{2}=45.\]
From the curve, the mark corresponding to cumulative frequency 45 is approximately \(51.5\%\).
\[\boxed{\text{Median}=51.5\%}\]
(c)(ii) Probability of obtaining distinction
At \(75\%\), the cumulative frequency read from the curve is approximately \(80\). Hence, the number scoring at least \(75\%\) is
\[90-80=10.\]
Therefore,
\[P(\text{distinction})=\frac{10}{90}=\frac{1}{9}\approx0.111.\]
Tambaya 9 Rahoto
(a) Evaluate: \(\int \limits_1^2 (2x^3 - 4x + 3) dx\)
(b) Given that P\(^{-1}\) = \(\begin{pmatrix} -1 & 1 \\ 4 & -3 \end{pmatrix}\), find the matrix P.
(a)
\[\int_1^2 (2x^3 - 4x + 3)\,dx = \left[\frac{x^4}{2} - 2x^2 + 3x\right]_1^2\]At \(x = 2\): \(8 - 8 + 6 = 6\). At \(x = 1\): \(\tfrac{1}{2} - 2 + 3 = \tfrac{3}{2}\).
\[= 6 - \tfrac{3}{2} = \tfrac{9}{2} = 4.5\](b) \(P\) is the inverse of \(P^{-1}\). With \(P^{-1} = \begin{pmatrix} -1 & 1 \\ 4 & -3 \end{pmatrix}\), \(\det = (-1)(-3) - (1)(4) = -1\).
\[P = \frac{1}{-1}\begin{pmatrix} -3 & -1 \\ -4 & -1 \end{pmatrix} = \begin{pmatrix} 3 & 1 \\ 4 & 1 \end{pmatrix}\]Check: \(P\,P^{-1} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\).
Bayanin Amsa
(a)
\[\int_1^2 (2x^3 - 4x + 3)\,dx = \left[\frac{x^4}{2} - 2x^2 + 3x\right]_1^2\]At \(x = 2\): \(8 - 8 + 6 = 6\). At \(x = 1\): \(\tfrac{1}{2} - 2 + 3 = \tfrac{3}{2}\).
\[= 6 - \tfrac{3}{2} = \tfrac{9}{2} = 4.5\](b) \(P\) is the inverse of \(P^{-1}\). With \(P^{-1} = \begin{pmatrix} -1 & 1 \\ 4 & -3 \end{pmatrix}\), \(\det = (-1)(-3) - (1)(4) = -1\).
\[P = \frac{1}{-1}\begin{pmatrix} -3 & -1 \\ -4 & -1 \end{pmatrix} = \begin{pmatrix} 3 & 1 \\ 4 & 1 \end{pmatrix}\]Check: \(P\,P^{-1} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\).
Tambaya 10 Rahoto
Musa is three years older than Manya. Seven years ago, Musa was twice as old as Manya. (1) How old are they now? (2) When will the sum of them be 45?
Let Manya's present age be \(m\). Then Musa \(= m + 3\).
Seven years ago Musa was twice Manya:
\[(m + 3) - 7 = 2(m - 7)\] \[m - 4 = 2m - 14 \Rightarrow m = 10\](1) Manya is \(10\) years and Musa is \(13\) years now.
(2) The present sum of ages is \(23\), and it rises by 2 each year. Let it take \(t\) years:
\[23 + 2t = 45 \Rightarrow 2t = 22 \Rightarrow t = 11\]The sum of their ages will be 45 in 11 years.
Bayanin Amsa
Let Manya's present age be \(m\). Then Musa \(= m + 3\).
Seven years ago Musa was twice Manya:
\[(m + 3) - 7 = 2(m - 7)\] \[m - 4 = 2m - 14 \Rightarrow m = 10\](1) Manya is \(10\) years and Musa is \(13\) years now.
(2) The present sum of ages is \(23\), and it rises by 2 each year. Let it take \(t\) years:
\[23 + 2t = 45 \Rightarrow 2t = 22 \Rightarrow t = 11\]The sum of their ages will be 45 in 11 years.
Tambaya 11 Rahoto
(a) The diagonals of a rhombus are 10.2 cm and 9.3 cm long. Calculate, correct to one decimal place, the perimeter of the rhombus.
(b) Given that \(\sin x = \frac{3}{5}, 0° < x < 90°\), find the value of \(5\cos x - 4\tan x\).
(a) The diagonals of a rhombus bisect each other at right angles. Half-diagonals are
\[ \frac{10.2}{2} = 5.1\ \text{cm}, \qquad \frac{9.3}{2} = 4.65\ \text{cm}. \]A side of the rhombus is the hypotenuse of a right triangle with these legs:
\[ s = \sqrt{5.1^2 + 4.65^2} = \sqrt{26.01 + 21.6225} = \sqrt{47.6325} = 6.9016\ \text{cm}. \]Perimeter \(= 4s = 4 \times 6.9016 = 27.6\ \text{cm (to 1 d.p.).}\)
(b) Given \(\sin x = \tfrac{3}{5}\) with \(0^\circ < x < 90^\circ\), this is a 3-4-5 right triangle, so
\[ \cos x = \frac{4}{5}, \qquad \tan x = \frac{3}{4}. \]Then
\[ 5\cos x - 4\tan x = 5\left(\frac{4}{5}\right) - 4\left(\frac{3}{4}\right) = 4 - 3 = 1. \]Bayanin Amsa
(a) The diagonals of a rhombus bisect each other at right angles. Half-diagonals are
\[ \frac{10.2}{2} = 5.1\ \text{cm}, \qquad \frac{9.3}{2} = 4.65\ \text{cm}. \]A side of the rhombus is the hypotenuse of a right triangle with these legs:
\[ s = \sqrt{5.1^2 + 4.65^2} = \sqrt{26.01 + 21.6225} = \sqrt{47.6325} = 6.9016\ \text{cm}. \]Perimeter \(= 4s = 4 \times 6.9016 = 27.6\ \text{cm (to 1 d.p.).}\)
(b) Given \(\sin x = \tfrac{3}{5}\) with \(0^\circ < x < 90^\circ\), this is a 3-4-5 right triangle, so
\[ \cos x = \frac{4}{5}, \qquad \tan x = \frac{3}{4}. \]Then
\[ 5\cos x - 4\tan x = 5\left(\frac{4}{5}\right) - 4\left(\frac{3}{4}\right) = 4 - 3 = 1. \]Tambaya 12 Rahoto
(a) Copy and complete the table of values for \(y = 2x^{2} + x - 10\) for \(-5 \leq x \leq 4\).
| x | -5 | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
| y | 5 | -9 | -10 | 0 |
(b) Using scales of 2cm to 1 unit on the x- axis and 2cm to 5 units on the y- axis, Draw the graph of \(y = 2x^{2} + x - 10\) for \(-5 \leq x \leq 4\).
(c) Use the graph to find the solution of :
(i) \(2x^{2} + x = 10\)
(ii) \(2x^{2} + x - 10 = 2x\)
(a) For \(y=2x^{2}+x-10\), the completed table is:
| \(x\) | \(-5\) | \(-4\) | \(-3\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) |
|---|---|---|---|---|---|---|---|---|---|---|
| \(y\) | \(35\) | \(18\) | \(5\) | \(-4\) | \(-9\) | \(-10\) | \(-7\) | \(0\) | \(11\) | \(26\) |
(b) The graph of \(y=2x^{2}+x-10\) is shown below. The straight line \(y=2x\), used in part (c)(ii), is also shown.
(c)(i) \[2x^{2}+x=10\iff 2x^{2}+x-10=0.\] The solutions are the \(x\)-coordinates where the parabola cuts the \(x\)-axis. Hence, \[\boxed{x=-2.5\text{ or }x=2}.\]
(c)(ii) The solutions of \(2x^{2}+x-10=2x\) are the \(x\)-coordinates of the points where the parabola intersects the line \(y=2x\). From the graph, \[\boxed{x=-2\text{ or }x=2.5}.\]
Bayanin Amsa
(a) For \(y=2x^{2}+x-10\), the completed table is:
| \(x\) | \(-5\) | \(-4\) | \(-3\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) |
|---|---|---|---|---|---|---|---|---|---|---|
| \(y\) | \(35\) | \(18\) | \(5\) | \(-4\) | \(-9\) | \(-10\) | \(-7\) | \(0\) | \(11\) | \(26\) |
(b) The graph of \(y=2x^{2}+x-10\) is shown below. The straight line \(y=2x\), used in part (c)(ii), is also shown.
(c)(i) \[2x^{2}+x=10\iff 2x^{2}+x-10=0.\] The solutions are the \(x\)-coordinates where the parabola cuts the \(x\)-axis. Hence, \[\boxed{x=-2.5\text{ or }x=2}.\]
(c)(ii) The solutions of \(2x^{2}+x-10=2x\) are the \(x\)-coordinates of the points where the parabola intersects the line \(y=2x\). From the graph, \[\boxed{x=-2\text{ or }x=2.5}.\]
Tambaya 13 Rahoto
A man starts from a point X and walk 285 m to Y on a bearing of 078\(^o\). He then walks due South to a point Z which is 307 m from X.
(a) Illustrate the information on a diagram.
(b) Find, correct to the nearest whole number, the:
(i) bearing of X from Z;
(ii) distance between Y and Z.
(a) Diagram: from X, line XY runs at bearing \(078^{\circ}\) (285 m); from Y the man walks due south to Z; XZ \(= 307\) m closes the triangle.
Place X at the origin:
\[Y = (285\sin 78^{\circ},\ 285\cos 78^{\circ}) = (278.8,\ 59.3)\]Z has the same easting as Y (a due-south move): \(Z = (278.8,\ z)\).
\[|XZ| = 307:\ 278.8^2 + z^2 = 307^2 \Rightarrow z^2 = 16520 \Rightarrow z = -128.5\](b)(ii) Distance YZ \(= 59.3 - (-128.5) = 187.8 \approx 188\) m.
(b)(i) Bearing of X from Z. \(\vec{ZX} = (-278.8,\ 128.5)\), pointing north-west.
\[\text{angle west of north} = \tan^{-1}\frac{278.8}{128.5} = 65^{\circ} \Rightarrow \text{bearing} = 360^{\circ} - 65^{\circ} = 295^{\circ}\]Bayanin Amsa
(a) Diagram: from X, line XY runs at bearing \(078^{\circ}\) (285 m); from Y the man walks due south to Z; XZ \(= 307\) m closes the triangle.
Place X at the origin:
\[Y = (285\sin 78^{\circ},\ 285\cos 78^{\circ}) = (278.8,\ 59.3)\]Z has the same easting as Y (a due-south move): \(Z = (278.8,\ z)\).
\[|XZ| = 307:\ 278.8^2 + z^2 = 307^2 \Rightarrow z^2 = 16520 \Rightarrow z = -128.5\](b)(ii) Distance YZ \(= 59.3 - (-128.5) = 187.8 \approx 188\) m.
(b)(i) Bearing of X from Z. \(\vec{ZX} = (-278.8,\ 128.5)\), pointing north-west.
\[\text{angle west of north} = \tan^{-1}\frac{278.8}{128.5} = 65^{\circ} \Rightarrow \text{bearing} = 360^{\circ} - 65^{\circ} = 295^{\circ}\]Tambaya 14 Rahoto
A used car was purchased at N900,000.00. Its value depreciated by 30% in the first year. In each subsequent year, the depreciation was 22% of its value at the beginning of the year. If the car was bought on the 1st of March, 2011, calculate, correct to the nearest hundred naira, the value of the car on the 28th of February, 2015.
Purchase price \(= \text{N}900{,}000.00\) on 1 March 2011. The value on 28 February 2015 is after 4 full years.
Year 1 (to Feb 2012): depreciation 30%, so the value is multiplied by \(0.70\).
Years 2, 3, 4: depreciation 22% each year, so each year the value is multiplied by \(0.78\).
\[ V = 900000 \times 0.70 \times (0.78)^3. \]Now \((0.78)^3 = 0.474552\), and \(900000 \times 0.70 = 630000\):
\[ V = 630000 \times 0.474552 = 298967.76. \]Correct to the nearest hundred naira, the value of the car on 28 February 2015 is N299,000.00.
Bayanin Amsa
Purchase price \(= \text{N}900{,}000.00\) on 1 March 2011. The value on 28 February 2015 is after 4 full years.
Year 1 (to Feb 2012): depreciation 30%, so the value is multiplied by \(0.70\).
Years 2, 3, 4: depreciation 22% each year, so each year the value is multiplied by \(0.78\).
\[ V = 900000 \times 0.70 \times (0.78)^3. \]Now \((0.78)^3 = 0.474552\), and \(900000 \times 0.70 = 630000\):
\[ V = 630000 \times 0.474552 = 298967.76. \]Correct to the nearest hundred naira, the value of the car on 28 February 2015 is N299,000.00.
Tambaya 15 Rahoto
In the diagram. PQR is an isosceles triangle. If the perimeter of the triangle is 28 cm, find the:
a. values of x and y;
b. lengths of the sides of the triangle.
In triangle PQR the three sides are marked as follows: \(|PQ| = 2y + x\), \(|QR| = 4y\) and \(|PR| = 6y - 2x + 1\). Since PQR is isosceles the two slant sides are equal, \(|QR| = |PR|\).
(a) Values of x and y
Perimeter equation:
\[ (2y + x) + 4y + (6y - 2x + 1) = 28 \]
\[ 12y - x + 1 = 28 \quad\Rightarrow\quad 12y - x = 27 \quad\text{...(1)} \]
Isosceles equation \((|QR| = |PR|)\):
\[ 4y = 6y - 2x + 1 \quad\Rightarrow\quad 2x - 2y = 1 \quad\text{...(2)} \]
From (1), \(x = 12y - 27\). Substituting into (2):
\[ 2(12y - 27) - 2y = 1 \Rightarrow 24y - 54 - 2y = 1 \Rightarrow 22y = 55 \]
\[ y = \frac{55}{22} = \frac{5}{2} = 2\tfrac{1}{2} \]
\[ x = 12\left(\tfrac{5}{2}\right) - 27 = 30 - 27 = 3 \]
Hence x = 3 and y = 2\(\tfrac{1}{2}\) (i.e. 2.5).
(b) Lengths of the sides
- \(|PQ| = 2y + x = 2\left(\tfrac{5}{2}\right) + 3 = 5 + 3 = \textbf{8 cm}\)
- \(|QR| = 4y = 4\left(\tfrac{5}{2}\right) = \textbf{10 cm}\)
- \(|PR| = 6y - 2x + 1 = 6\left(\tfrac{5}{2}\right) - 2(3) + 1 = 15 - 6 + 1 = \textbf{10 cm}\)
Check: \(8 + 10 + 10 = 28\) cm, which agrees with the given perimeter, and \(|QR| = |PR| = 10\) cm confirms the triangle is isosceles.
Bayanin Amsa
In triangle PQR the three sides are marked as follows: \(|PQ| = 2y + x\), \(|QR| = 4y\) and \(|PR| = 6y - 2x + 1\). Since PQR is isosceles the two slant sides are equal, \(|QR| = |PR|\).
(a) Values of x and y
Perimeter equation:
\[ (2y + x) + 4y + (6y - 2x + 1) = 28 \]
\[ 12y - x + 1 = 28 \quad\Rightarrow\quad 12y - x = 27 \quad\text{...(1)} \]
Isosceles equation \((|QR| = |PR|)\):
\[ 4y = 6y - 2x + 1 \quad\Rightarrow\quad 2x - 2y = 1 \quad\text{...(2)} \]
From (1), \(x = 12y - 27\). Substituting into (2):
\[ 2(12y - 27) - 2y = 1 \Rightarrow 24y - 54 - 2y = 1 \Rightarrow 22y = 55 \]
\[ y = \frac{55}{22} = \frac{5}{2} = 2\tfrac{1}{2} \]
\[ x = 12\left(\tfrac{5}{2}\right) - 27 = 30 - 27 = 3 \]
Hence x = 3 and y = 2\(\tfrac{1}{2}\) (i.e. 2.5).
(b) Lengths of the sides
- \(|PQ| = 2y + x = 2\left(\tfrac{5}{2}\right) + 3 = 5 + 3 = \textbf{8 cm}\)
- \(|QR| = 4y = 4\left(\tfrac{5}{2}\right) = \textbf{10 cm}\)
- \(|PR| = 6y - 2x + 1 = 6\left(\tfrac{5}{2}\right) - 2(3) + 1 = 15 - 6 + 1 = \textbf{10 cm}\)
Check: \(8 + 10 + 10 = 28\) cm, which agrees with the given perimeter, and \(|QR| = |PR| = 10\) cm confirms the triangle is isosceles.
Tambaya 16 Rahoto
(a) In a right-angled triangle, sin X = \(\frac{3}{5}\). Evaluate, leaving the answer as a fraction, 5 (cosX)\(^2\) – 3.
(b) The base of a pyramid, 12 cm high, is a rectangle with dimensions 42 cm by 11 cm. if the pyramid is filled with water and emptied into a conical container of equal height and volume, calculate, leaving the answer in surd form (radicals), the base radius of the container. [Take π=\(\frac{22}{7}\)]
(a) \(\sin X = \tfrac{3}{5} \Rightarrow \cos X = \tfrac{4}{5}\) (a 3-4-5 triangle).
\[5\cos^2 X - 3 = 5\left(\tfrac{16}{25}\right) - 3 = \tfrac{16}{5} - 3 = \tfrac{1}{5}\](b) Volume of the pyramid \(= \tfrac{1}{3} \times (42 \times 11) \times 12 = \tfrac{1}{3} \times 462 \times 12 = 1848 \text{ cm}^3\).
The cone has the same height (12 cm) and the same volume (1848):
\[\tfrac{1}{3} \times \tfrac{22}{7} \times r^2 \times 12 = 1848\] \[\tfrac{88}{7}r^2 = 1848 \Rightarrow r^2 = \frac{1848 \times 7}{88} = 147\] \[r = \sqrt{147} = 7\sqrt{3} \text{ cm}\]Bayanin Amsa
(a) \(\sin X = \tfrac{3}{5} \Rightarrow \cos X = \tfrac{4}{5}\) (a 3-4-5 triangle).
\[5\cos^2 X - 3 = 5\left(\tfrac{16}{25}\right) - 3 = \tfrac{16}{5} - 3 = \tfrac{1}{5}\](b) Volume of the pyramid \(= \tfrac{1}{3} \times (42 \times 11) \times 12 = \tfrac{1}{3} \times 462 \times 12 = 1848 \text{ cm}^3\).
The cone has the same height (12 cm) and the same volume (1848):
\[\tfrac{1}{3} \times \tfrac{22}{7} \times r^2 \times 12 = 1848\] \[\tfrac{88}{7}r^2 = 1848 \Rightarrow r^2 = \frac{1848 \times 7}{88} = 147\] \[r = \sqrt{147} = 7\sqrt{3} \text{ cm}\]Tambaya 17 Rahoto
A shop had two reduction sales during which prices of all items were reduced by 40% in the first sales and 30% in the second.
- If a shirt was sold at GH₵35.00 during the second reduction sales, find the price before the first sale.
- If the price of an article before the first reduction was GH₵180.00, find the total:
- reduction in the price due to the two sales;
- percentage reduction in the price of the article.
Price before the first sale. After a 40% cut the price is \(0.6\) of the original; after a further 30% cut it is \(0.7 \times 0.6 = 0.42\) of the original.
\[0.42 \times P = 35 \Rightarrow P = \frac{35}{0.42} = \text{GH}\unicode{x20B5}83.33\]Article costing GH\unicode{x20B5}180.00. Final price \(= 0.42 \times 180 = \text{GH}\unicode{x20B5}75.60\).
Total reduction \(= 180 - 75.60 = \text{GH}\unicode{x20B5}104.40\).
Percentage reduction \(= \dfrac{104.40}{180} \times 100\% = 58\%\).
Bayanin Amsa
Price before the first sale. After a 40% cut the price is \(0.6\) of the original; after a further 30% cut it is \(0.7 \times 0.6 = 0.42\) of the original.
\[0.42 \times P = 35 \Rightarrow P = \frac{35}{0.42} = \text{GH}\unicode{x20B5}83.33\]Article costing GH\unicode{x20B5}180.00. Final price \(= 0.42 \times 180 = \text{GH}\unicode{x20B5}75.60\).
Total reduction \(= 180 - 75.60 = \text{GH}\unicode{x20B5}104.40\).
Percentage reduction \(= \dfrac{104.40}{180} \times 100\% = 58\%\).
Tambaya 18 Rahoto
(a) Using a scale of 2cm to 2units on both axes, draw on a sheet of graph paper two perpendicular axes 0x and 0y for \(-10 \leq x \leq 10\) and \(-10 \leq y \leq 10\)
(b) Given the points P(3, 2). Q(-1. 5). R(0. 8) and S(3, 7). draw on the same graph, indicating clearly the vertices and their coordinates, the:
(i) quadrilateral PQRS;
(ii) image \(P_1Q_1R_1S_1\) of PQRS under an anticlockwise rotation of \(90^o\) about the origin where \(P \to P_1\), \(Q \to Q_1\), \(R \to R_1\) and \(S \to S_1\)
(iii) image \(P_2Q_2R_2S_2\) of \(P_1Q_1R_1S_1\) under a reflection in the line \(y - x = 0\) where \(P_1 \to P_2\), \(Q_1 \to Q_2\), \(R_1 \to R_2\) and \(S_1 \to S_2\)
(c) Describe precisely the single transformation T for which \(T : PQRS \to P_2Q_2R_2S_2\)
(d) The side \(P_1Q_1\) of the quadrilateral \(P_1Q_1R_1S_1\) cuts the x-axis at the point W. What type of quadrilateral is \(P_1S_1R_1W\)?
(a) and (b) The required coordinate graph, drawn to equal scales on both axes, is shown below. The vertices are joined in the order in which they are named.
The coordinates plotted are:
| Object | Vertices |
|---|---|
| PQRS | P(3,2), Q(-1,5), R(0,8), S(3,7) |
| P1Q1R1S1 | P1(-2,3), Q1(-5,-1), R1(-8,0), S1(-7,3) |
| P2Q2R2S2 | P2(3,-2), Q2(-1,-5), R2(0,-8), S2(3,-7) |
For a rotation of \(90^\circ\) anticlockwise about the origin, \((x,y)\mapsto(-y,x)\). Thus, for example, \(P(3,2)\mapsto P_1(-2,3)\).
Reflection in \(y=x\) maps \((x,y)\mapsto(y,x)\). Thus \(P_1(-2,3)\mapsto P_2(3,-2)\).
(c) Combining the two transformations gives
\[(x,y)\mapsto(-y,x)\mapsto(x,-y).\]
Hence \(T\) is a reflection in the x-axis.
(d) The line \(P_1Q_1\) has gradient
\[m=\frac{-1-3}{-5-(-2)}=\frac{-4}{-3}=\frac{4}{3}.\]
Its equation is
\[y-3=\frac{4}{3}(x+2).\]
At the x-axis, \(y=0\), so
\[-3=\frac{4}{3}(x+2),\qquad x=-\frac{17}{4}.\]
Therefore \(W\left(-\frac{17}{4},0\right)\). Since \(P_1S_1\) and \(R_1W\) are both horizontal, they are parallel. Thus \(P_1S_1R_1W\) is a trapezium.
Bayanin Amsa
(a) and (b) The required coordinate graph, drawn to equal scales on both axes, is shown below. The vertices are joined in the order in which they are named.
The coordinates plotted are:
| Object | Vertices |
|---|---|
| PQRS | P(3,2), Q(-1,5), R(0,8), S(3,7) |
| P1Q1R1S1 | P1(-2,3), Q1(-5,-1), R1(-8,0), S1(-7,3) |
| P2Q2R2S2 | P2(3,-2), Q2(-1,-5), R2(0,-8), S2(3,-7) |
For a rotation of \(90^\circ\) anticlockwise about the origin, \((x,y)\mapsto(-y,x)\). Thus, for example, \(P(3,2)\mapsto P_1(-2,3)\).
Reflection in \(y=x\) maps \((x,y)\mapsto(y,x)\). Thus \(P_1(-2,3)\mapsto P_2(3,-2)\).
(c) Combining the two transformations gives
\[(x,y)\mapsto(-y,x)\mapsto(x,-y).\]
Hence \(T\) is a reflection in the x-axis.
(d) The line \(P_1Q_1\) has gradient
\[m=\frac{-1-3}{-5-(-2)}=\frac{-4}{-3}=\frac{4}{3}.\]
Its equation is
\[y-3=\frac{4}{3}(x+2).\]
At the x-axis, \(y=0\), so
\[-3=\frac{4}{3}(x+2),\qquad x=-\frac{17}{4}.\]
Therefore \(W\left(-\frac{17}{4},0\right)\). Since \(P_1S_1\) and \(R_1W\) are both horizontal, they are parallel. Thus \(P_1S_1R_1W\) is a trapezium.
Tambaya 19 Rahoto
(a) The diagram, = IWYI = IXZI and < WXY = 80\(^o\). What is the size of < XWZ?
(b) A man was charged 2 kobo per month for every N1.00 he borrowed from a bank. At what rate per annum was the interest charged?
(a) Finding \(\angle XWZ\)
\(WXYZ\) is a quadrilateral inscribed in the circle (a cyclic quadrilateral) with vertices in order \(W, X, Y, Z\). Its diagonals are equal: \(|WY| = |XZ|\), and \(\angle WXY = 80^{\circ}\).
Step 1: Interpret the equal diagonals. A cyclic quadrilateral whose diagonals are equal is an isosceles trapezium. From the diagram, side \(WX\) is parallel to side \(ZY\) (\(WX \parallel ZY\)), and the legs \(WZ\) and \(XY\) are the equal sides.
Step 2: Apply the base-angle property. In an isosceles trapezium the two angles standing on the same parallel side are equal. Here \(\angle XWZ\) (at \(W\)) and \(\angle WXY\) (at \(X\)) both stand on the base \(WX\), so they are equal:
\[ \angle XWZ = \angle WXY = 80^{\circ} \]
Check (cyclic quadrilateral): The opposite angles are then \(\angle WZY = \angle XYZ = 180^{\circ} - 80^{\circ} = 100^{\circ}\), and \(80^{\circ} + 100^{\circ} = 180^{\circ}\), confirming that opposite angles are supplementary.
\(\angle XWZ = 80^{\circ}\).
(b) Rate of interest per annum
The man is charged \(2\) kobo per month for every \(N1.00\) borrowed. Since \(N1.00 = 100\) kobo, the monthly interest on \(N1.00\) is:
\[ \text{monthly rate} = \frac{2\text{ kobo}}{100\text{ kobo}} \times 100\% = 2\%\text{ per month} \]
There are \(12\) months in a year, so:
\[ \text{rate per annum} = 2\% \times 12 = 24\% \]
The interest was charged at 24% per annum.
Bayanin Amsa
(a) Finding \(\angle XWZ\)
\(WXYZ\) is a quadrilateral inscribed in the circle (a cyclic quadrilateral) with vertices in order \(W, X, Y, Z\). Its diagonals are equal: \(|WY| = |XZ|\), and \(\angle WXY = 80^{\circ}\).
Step 1: Interpret the equal diagonals. A cyclic quadrilateral whose diagonals are equal is an isosceles trapezium. From the diagram, side \(WX\) is parallel to side \(ZY\) (\(WX \parallel ZY\)), and the legs \(WZ\) and \(XY\) are the equal sides.
Step 2: Apply the base-angle property. In an isosceles trapezium the two angles standing on the same parallel side are equal. Here \(\angle XWZ\) (at \(W\)) and \(\angle WXY\) (at \(X\)) both stand on the base \(WX\), so they are equal:
\[ \angle XWZ = \angle WXY = 80^{\circ} \]
Check (cyclic quadrilateral): The opposite angles are then \(\angle WZY = \angle XYZ = 180^{\circ} - 80^{\circ} = 100^{\circ}\), and \(80^{\circ} + 100^{\circ} = 180^{\circ}\), confirming that opposite angles are supplementary.
\(\angle XWZ = 80^{\circ}\).
(b) Rate of interest per annum
The man is charged \(2\) kobo per month for every \(N1.00\) borrowed. Since \(N1.00 = 100\) kobo, the monthly interest on \(N1.00\) is:
\[ \text{monthly rate} = \frac{2\text{ kobo}}{100\text{ kobo}} \times 100\% = 2\%\text{ per month} \]
There are \(12\) months in a year, so:
\[ \text{rate per annum} = 2\% \times 12 = 24\% \]
The interest was charged at 24% per annum.
Tambaya 20 Rahoto
The table shows the distribution of sources obtained when a fair diwe was rolled 50 times.
| Score | 1 | 2 | 3 | 4 | 5 | 6 |
| Frequency | 2 | 5 | 13 | 11 | 9 | 10 |
1. Draw a bar chart for the distribution
2. Calculate the mean score of the distribution
1. Bar chart
2. Mean score
| Score, \(x\) | 1 | 2 | 3 | 4 | 5 | 6 | Total |
|---|---|---|---|---|---|---|---|
| Frequency, \(f\) | 2 | 5 | 13 | 11 | 9 | 10 | \(50\) |
| \(fx\) | 2 | 10 | 39 | 44 | 45 | 60 | \(200\) |
\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{200}{50}=4\]
Therefore, the mean score is \(4\).
Bayanin Amsa
1. Bar chart
2. Mean score
| Score, \(x\) | 1 | 2 | 3 | 4 | 5 | 6 | Total |
|---|---|---|---|---|---|---|---|
| Frequency, \(f\) | 2 | 5 | 13 | 11 | 9 | 10 | \(50\) |
| \(fx\) | 2 | 10 | 39 | 44 | 45 | 60 | \(200\) |
\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{200}{50}=4\]
Therefore, the mean score is \(4\).
Tambaya 21 Rahoto
a) Copy and complete the following table of values for y = 2 cos x – sin x ,\(0^o \leq x \leq 300^o\)
\[\begin{array}{c|c} x & 0^o & 30^o & 60^o & 90^o & 120^o & 150^o & 180^o & 210^o & 240^o & 270^o& 300^o \\ \hline Y & 2.00 & & 0.13 & & -1.87 & & -2.00 & & -0.13 & & \end{array}\]
(b) Using scales of 2 cm to 30\(^o\) on the x-axis and 2cm to 1 unit on the y-axis, draw the graph of y = 2 cos x – sin x for \(0^o \leq x \leq 300^o\)
(c) Use the graph to find the value(s) of x for which:
(i) 2 cos x – sin x = 1;
(ii) tan x = 2.
(a) Completed table for \(y=2\cos x-\sin x\):
| \(x\) | \(0^\circ\) | \(30^\circ\) | \(60^\circ\) | \(90^\circ\) | \(120^\circ\) | \(150^\circ\) | \(180^\circ\) | \(210^\circ\) | \(240^\circ\) | \(270^\circ\) | \(300^\circ\) |
|---|---|---|---|---|---|---|---|---|---|---|---|
| \(y\) | 2.00 | 1.23 | 0.13 | -1.00 | -1.87 | -2.23 | -2.00 | -1.23 | -0.13 | 1.00 | 1.87 |
For example, at \(x=30^\circ\):
\[y=2\cos30^\circ-\sin30^\circ=2(0.866)-0.5=1.232\ldots\approx1.23.\]
(b) Graph of \(y=2\cos x-\sin x\)
(c)(i) To solve \(2\cos x-\sin x=1\), draw or imagine the horizontal line \(y=1\). The required values are the \(x\)-coordinates where this line meets the curve:
\[x\approx36^\circ\quad\text{or}\quad x\approx270^\circ.\]
(c)(ii) Since
\[\tan x=2 \quad\Rightarrow\quad \frac{\sin x}{\cos x}=2 \quad\Rightarrow\quad \sin x=2\cos x,\]
this is equivalent to
\[2\cos x-\sin x=0.\]
Therefore, read the \(x\)-coordinates where the curve crosses the \(x\)-axis:
\[x\approx63^\circ\quad\text{or}\quad x\approx243^\circ.\]
Examination reminder: For an equation involving \(2\cos x-\sin x\), use the graph’s \(y\)-value. In particular, \(\tan x=2\) rearranges to \(2\cos x-\sin x=0\), so look for the \(x\)-intercepts of the curve.
Bayanin Amsa
(a) Completed table for \(y=2\cos x-\sin x\):
| \(x\) | \(0^\circ\) | \(30^\circ\) | \(60^\circ\) | \(90^\circ\) | \(120^\circ\) | \(150^\circ\) | \(180^\circ\) | \(210^\circ\) | \(240^\circ\) | \(270^\circ\) | \(300^\circ\) |
|---|---|---|---|---|---|---|---|---|---|---|---|
| \(y\) | 2.00 | 1.23 | 0.13 | -1.00 | -1.87 | -2.23 | -2.00 | -1.23 | -0.13 | 1.00 | 1.87 |
For example, at \(x=30^\circ\):
\[y=2\cos30^\circ-\sin30^\circ=2(0.866)-0.5=1.232\ldots\approx1.23.\]
(b) Graph of \(y=2\cos x-\sin x\)
(c)(i) To solve \(2\cos x-\sin x=1\), draw or imagine the horizontal line \(y=1\). The required values are the \(x\)-coordinates where this line meets the curve:
\[x\approx36^\circ\quad\text{or}\quad x\approx270^\circ.\]
(c)(ii) Since
\[\tan x=2 \quad\Rightarrow\quad \frac{\sin x}{\cos x}=2 \quad\Rightarrow\quad \sin x=2\cos x,\]
this is equivalent to
\[2\cos x-\sin x=0.\]
Therefore, read the \(x\)-coordinates where the curve crosses the \(x\)-axis:
\[x\approx63^\circ\quad\text{or}\quad x\approx243^\circ.\]
Examination reminder: For an equation involving \(2\cos x-\sin x\), use the graph’s \(y\)-value. In particular, \(\tan x=2\) rearranges to \(2\cos x-\sin x=0\), so look for the \(x\)-intercepts of the curve.
Tambaya 22 Rahoto
| Marks | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 | 90 |
| Frequency | 1 | 1 | x | 5 | y | 1 | 4 | 3 | 1 |
The frequency distribution shows the marks distribution of a class of 30 students in an examination.
The mean mark of the distribution is 52.
(a) Find the values of x and y.
(b) Construct a group frequency distribution table starting with a lower class limit of 1 and class interval of 10.
(c) Draw a histogram for the distribution
(d) Use the histogram to estimate the mode.
(a) Determination of \(x\) and \(y\)
| Mark, \(m\) | Frequency, \(f\) | \(fm\) |
|---|---|---|
| 10 | 1 | 10 |
| 20 | 1 | 20 |
| 30 | \(x\) | \(30x\) |
| 40 | 5 | 200 |
| 50 | \(y\) | \(50y\) |
| 60 | 1 | 60 |
| 70 | 4 | 280 |
| 80 | 3 | 240 |
| 90 | 1 | 90 |
| Total | \(16+x+y\) | \(900+30x+50y\) |
Since there are 30 students,
\[16+x+y=30 \quad\Rightarrow\quad x+y=14\tag{1}\]
Also,
\[52=\frac{900+30x+50y}{30}\]
\[1560=900+30x+50y\]
\[3x+5y=66.\tag{2}\]
From (1), \(x=14-y\). Hence,
\[3(14-y)+5y=66\]
\[42+2y=66\Rightarrow y=12.\]
Therefore, \(x=14-12=2\).
\(\boxed{x=2,\ y=12}\)
(b) Grouped frequency distribution
| Class interval | Class boundaries | Frequency |
|---|---|---|
| 1 - 10 | 0.5 - 10.5 | 1 |
| 11 - 20 | 10.5 - 20.5 | 1 |
| 21 - 30 | 20.5 - 30.5 | 2 |
| 31 - 40 | 30.5 - 40.5 | 5 |
| 41 - 50 | 40.5 - 50.5 | 12 |
| 51 - 60 | 50.5 - 60.5 | 1 |
| 61 - 70 | 60.5 - 70.5 | 4 |
| 71 - 80 | 70.5 - 80.5 | 3 |
| 81 - 90 | 80.5 - 90.5 | 1 |
(c) Histogram
(d) Mode
The modal class is \(41-50\). Reading the peak position from the histogram by the usual intersecting-lines method gives a mode of approximately
\[\boxed{44\text{ marks}}\]
Bayanin Amsa
(a) Determination of \(x\) and \(y\)
| Mark, \(m\) | Frequency, \(f\) | \(fm\) |
|---|---|---|
| 10 | 1 | 10 |
| 20 | 1 | 20 |
| 30 | \(x\) | \(30x\) |
| 40 | 5 | 200 |
| 50 | \(y\) | \(50y\) |
| 60 | 1 | 60 |
| 70 | 4 | 280 |
| 80 | 3 | 240 |
| 90 | 1 | 90 |
| Total | \(16+x+y\) | \(900+30x+50y\) |
Since there are 30 students,
\[16+x+y=30 \quad\Rightarrow\quad x+y=14\tag{1}\]
Also,
\[52=\frac{900+30x+50y}{30}\]
\[1560=900+30x+50y\]
\[3x+5y=66.\tag{2}\]
From (1), \(x=14-y\). Hence,
\[3(14-y)+5y=66\]
\[42+2y=66\Rightarrow y=12.\]
Therefore, \(x=14-12=2\).
\(\boxed{x=2,\ y=12}\)
(b) Grouped frequency distribution
| Class interval | Class boundaries | Frequency |
|---|---|---|
| 1 - 10 | 0.5 - 10.5 | 1 |
| 11 - 20 | 10.5 - 20.5 | 1 |
| 21 - 30 | 20.5 - 30.5 | 2 |
| 31 - 40 | 30.5 - 40.5 | 5 |
| 41 - 50 | 40.5 - 50.5 | 12 |
| 51 - 60 | 50.5 - 60.5 | 1 |
| 61 - 70 | 60.5 - 70.5 | 4 |
| 71 - 80 | 70.5 - 80.5 | 3 |
| 81 - 90 | 80.5 - 90.5 | 1 |
(c) Histogram
(d) Mode
The modal class is \(41-50\). Reading the peak position from the histogram by the usual intersecting-lines method gives a mode of approximately
\[\boxed{44\text{ marks}}\]
Tambaya 23 Rahoto
(a) 
In the diagram
(i) The value of x; (ii)
(b) If \(2N4_{seven} = 15N_{nine}\), find the value of N.
(a) Finding \(x\).
Reading the diagram. \(Q\), \(R\) and \(S\) lie on a circle with centre \(O\). The straight line \(Q\,O\,S\) passes through the centre, so \(QS\) is a diameter, and it is produced beyond \(S\) to the external point \(T\). The angle at \(Q\) is \(x\), the right-angle mark is at \(R\), and the exterior angle at \(S\) (between \(SR\) and \(ST\)) is \((3x + 15)^\circ\).
Angle in a semicircle. Since \(QS\) is a diameter, the angle it subtends at \(R\) is a right angle:
\[\angle QRS = 90^\circ,\] which agrees with the mark at \(R\).Exterior-angle theorem in \(\triangle QRS\). The line \(Q\,S\,T\) is straight, so \((3x+15)^\circ\) is the exterior angle at \(S\); it equals the sum of the two remote interior angles \(\angle Q\) and \(\angle R\):
\[3x + 15 = x + 90.\]\[3x - x = 90 - 15 \;\Rightarrow\; 2x = 75 \;\Rightarrow\; x = \mathbf{37.5^\circ}.\](The follow-up part (ii) is not legible in the source text, so only \(x\) is evaluated here.)
(b) Number-base equation: \(2N4_{\text{seven}} = 15N_{\text{nine}}\).
Expand each numeral in powers of its base, treating \(N\) as an unknown digit.
Left side (base 7):
\[2N4_{\text{seven}} = 2(7^2) + N(7) + 4 = 98 + 7N + 4 = 102 + 7N.\]Right side (base 9):
\[15N_{\text{nine}} = 1(9^2) + 5(9) + N = 81 + 45 + N = 126 + N.\]Equate:
\[102 + 7N = 126 + N \;\Rightarrow\; 6N = 24 \;\Rightarrow\; N = \mathbf{4}.\]Check: \(N = 4\) is a valid digit in both base 7 and base 9. Left \(= 102 + 28 = 130\); right \(= 126 + 4 = 130\). Both equal \(130_{\text{ten}}\). Correct.
Bayanin Amsa
(a) Finding \(x\).
Reading the diagram. \(Q\), \(R\) and \(S\) lie on a circle with centre \(O\). The straight line \(Q\,O\,S\) passes through the centre, so \(QS\) is a diameter, and it is produced beyond \(S\) to the external point \(T\). The angle at \(Q\) is \(x\), the right-angle mark is at \(R\), and the exterior angle at \(S\) (between \(SR\) and \(ST\)) is \((3x + 15)^\circ\).
Angle in a semicircle. Since \(QS\) is a diameter, the angle it subtends at \(R\) is a right angle:
\[\angle QRS = 90^\circ,\] which agrees with the mark at \(R\).Exterior-angle theorem in \(\triangle QRS\). The line \(Q\,S\,T\) is straight, so \((3x+15)^\circ\) is the exterior angle at \(S\); it equals the sum of the two remote interior angles \(\angle Q\) and \(\angle R\):
\[3x + 15 = x + 90.\]\[3x - x = 90 - 15 \;\Rightarrow\; 2x = 75 \;\Rightarrow\; x = \mathbf{37.5^\circ}.\](The follow-up part (ii) is not legible in the source text, so only \(x\) is evaluated here.)
(b) Number-base equation: \(2N4_{\text{seven}} = 15N_{\text{nine}}\).
Expand each numeral in powers of its base, treating \(N\) as an unknown digit.
Left side (base 7):
\[2N4_{\text{seven}} = 2(7^2) + N(7) + 4 = 98 + 7N + 4 = 102 + 7N.\]Right side (base 9):
\[15N_{\text{nine}} = 1(9^2) + 5(9) + N = 81 + 45 + N = 126 + N.\]Equate:
\[102 + 7N = 126 + N \;\Rightarrow\; 6N = 24 \;\Rightarrow\; N = \mathbf{4}.\]Check: \(N = 4\) is a valid digit in both base 7 and base 9. Left \(= 102 + 28 = 130\); right \(= 126 + 4 = 130\). Both equal \(130_{\text{ten}}\). Correct.
Tambaya 24 Rahoto
In the diagram, |PT| = 4 cm, |TS| = 6 cm, |PQ| = 6 cm and < SPR = 30°. Calculate, correct to the nearest whole number:
(a) |SR| ;
(b) area of TQRS.
From the diagram, in triangle PSR the point T lies on PS and Q lies on PR, with TQ parallel to SR (shown by the arrows). Given \(|PT| = 4\text{ cm}\), \(|TS| = 6\text{ cm}\), \(|PQ| = 6\text{ cm}\) and \(\angle SPR = 30^\circ\).
Preliminary - use the parallel lines. Because \(TQ \parallel SR\), triangles PTQ and PSR are similar (equiangular). The full side PS is
\[|PS| = |PT| + |TS| = 4 + 6 = 10\text{ cm}.\]
The ratio of similarity is \(\dfrac{PT}{PS} = \dfrac{4}{10} = 0.4\), so
\[\frac{PQ}{PR} = 0.4 \;\Rightarrow\; |PR| = \frac{PQ}{0.4} = \frac{6}{0.4} = 15\text{ cm}.\]
(a) Finding |SR|. In triangle PSR use the Cosine Rule with \(|PS| = 10\), \(|PR| = 15\), \(\angle P = 30^\circ\):
\[|SR|^2 = |PS|^2 + |PR|^2 - 2|PS||PR|\cos 30^\circ.\]
\[|SR|^2 = 10^2 + 15^2 - 2(10)(15)\cos 30^\circ = 100 + 225 - 300(0.8660).\]
\[|SR|^2 = 325 - 259.8 = 65.2.\]
\[|SR| = \sqrt{65.2} = 8.07\ldots \approx 8\text{ cm}.\]
(b) Area of TQRS. TQRS is the trapezium left when the small triangle PTQ is removed from triangle PSR:
\[\text{Area of } TQRS = \text{Area of } \triangle PSR - \text{Area of } \triangle PTQ.\]
Using \(\text{Area} = \tfrac{1}{2}ab\sin C\) with the common angle \(30^\circ\):
\[\text{Area of } \triangle PSR = \tfrac{1}{2}(10)(15)\sin 30^\circ = \tfrac{1}{2}(150)(0.5) = 37.5\text{ cm}^2.\]
\[\text{Area of } \triangle PTQ = \tfrac{1}{2}(4)(6)\sin 30^\circ = \tfrac{1}{2}(24)(0.5) = 6\text{ cm}^2.\]
\[\text{Area of } TQRS = 37.5 - 6 = 31.5 \approx 32\text{ cm}^2.\]
\[\boxed{|SR| \approx 8\text{ cm},\qquad \text{Area of } TQRS \approx 32\text{ cm}^2.}\]
Bayanin Amsa
From the diagram, in triangle PSR the point T lies on PS and Q lies on PR, with TQ parallel to SR (shown by the arrows). Given \(|PT| = 4\text{ cm}\), \(|TS| = 6\text{ cm}\), \(|PQ| = 6\text{ cm}\) and \(\angle SPR = 30^\circ\).
Preliminary - use the parallel lines. Because \(TQ \parallel SR\), triangles PTQ and PSR are similar (equiangular). The full side PS is
\[|PS| = |PT| + |TS| = 4 + 6 = 10\text{ cm}.\]
The ratio of similarity is \(\dfrac{PT}{PS} = \dfrac{4}{10} = 0.4\), so
\[\frac{PQ}{PR} = 0.4 \;\Rightarrow\; |PR| = \frac{PQ}{0.4} = \frac{6}{0.4} = 15\text{ cm}.\]
(a) Finding |SR|. In triangle PSR use the Cosine Rule with \(|PS| = 10\), \(|PR| = 15\), \(\angle P = 30^\circ\):
\[|SR|^2 = |PS|^2 + |PR|^2 - 2|PS||PR|\cos 30^\circ.\]
\[|SR|^2 = 10^2 + 15^2 - 2(10)(15)\cos 30^\circ = 100 + 225 - 300(0.8660).\]
\[|SR|^2 = 325 - 259.8 = 65.2.\]
\[|SR| = \sqrt{65.2} = 8.07\ldots \approx 8\text{ cm}.\]
(b) Area of TQRS. TQRS is the trapezium left when the small triangle PTQ is removed from triangle PSR:
\[\text{Area of } TQRS = \text{Area of } \triangle PSR - \text{Area of } \triangle PTQ.\]
Using \(\text{Area} = \tfrac{1}{2}ab\sin C\) with the common angle \(30^\circ\):
\[\text{Area of } \triangle PSR = \tfrac{1}{2}(10)(15)\sin 30^\circ = \tfrac{1}{2}(150)(0.5) = 37.5\text{ cm}^2.\]
\[\text{Area of } \triangle PTQ = \tfrac{1}{2}(4)(6)\sin 30^\circ = \tfrac{1}{2}(24)(0.5) = 6\text{ cm}^2.\]
\[\text{Area of } TQRS = 37.5 - 6 = 31.5 \approx 32\text{ cm}^2.\]
\[\boxed{|SR| \approx 8\text{ cm},\qquad \text{Area of } TQRS \approx 32\text{ cm}^2.}\]
Tambaya 25 Rahoto
(a) The graph of \(y = 2px^{2} - p^{2}x - 14\) passes through the point (3, 10). Find the values of p.
(b) Two lines, \(3y - 2x = 21\) and \(4y + 5x = 5\) intersect at the point Q. Find the coordinates of Q.
(a) Since the graph passes through \((3,10)\), substitute \(x=3\) and \(y=10\) into \(y=2px^2-p^2x-14\):
\[10=2p(3)^2-p^2(3)-14.\]
\[10=18p-3p^2-14\]
\[3p^2-18p+24=0\]
\[p^2-6p+8=0\]
\[(p-2)(p-4)=0.\]
Therefore,
\[\boxed{p=2\text{ or }p=4}.\]
For these two values, the corresponding curves are \(y=4x^2-4x-14\) and \(y=8x^2-16x-14\). Both pass through \((3,10)\), as shown below.
(b) The equations of the two lines are
\[3y-2x=21 \quad \text{and} \quad 4y+5x=5.\]
Multiply the first equation by 5 and the second equation by 2:
\[15y-10x=105\]
\[8y+10x=10.\]
Adding the equations gives
\[23y=115\]
\[y=5.\]
Substitute \(y=5\) into \(3y-2x=21\):
\[3(5)-2x=21\]
\[15-2x=21\]
\[-2x=6\]
\[x=-3.\]
Hence, the coordinates of \(Q\) are
\[\boxed{Q=(-3,5)}.\]
Bayanin Amsa
(a) Since the graph passes through \((3,10)\), substitute \(x=3\) and \(y=10\) into \(y=2px^2-p^2x-14\):
\[10=2p(3)^2-p^2(3)-14.\]
\[10=18p-3p^2-14\]
\[3p^2-18p+24=0\]
\[p^2-6p+8=0\]
\[(p-2)(p-4)=0.\]
Therefore,
\[\boxed{p=2\text{ or }p=4}.\]
For these two values, the corresponding curves are \(y=4x^2-4x-14\) and \(y=8x^2-16x-14\). Both pass through \((3,10)\), as shown below.
(b) The equations of the two lines are
\[3y-2x=21 \quad \text{and} \quad 4y+5x=5.\]
Multiply the first equation by 5 and the second equation by 2:
\[15y-10x=105\]
\[8y+10x=10.\]
Adding the equations gives
\[23y=115\]
\[y=5.\]
Substitute \(y=5\) into \(3y-2x=21\):
\[3(5)-2x=21\]
\[15-2x=21\]
\[-2x=6\]
\[x=-3.\]
Hence, the coordinates of \(Q\) are
\[\boxed{Q=(-3,5)}.\]
Tambaya 26 Rahoto
(a) 
In < PQS, |PQ| = 12 cm, |PS| = 5 cm, < SPQ = < PRQ = 90°, Find, correct to three significant figures, |PR|.
(b) The length of two ladders, L and M are 10m and 12m respectively. They are placed against a wall such that each ladder makes angle with the horizontal ground. If the foot of L is 8m from the foot of the wall.
(i) Draw a diagram to illustrate this information; (ii) Calculate the height at which M touches the wall.
(a)
In right-angled triangle \(SPQ\),
\[SQ^2=PS^2+PQ^2=5^2+12^2=169.\]
Hence, \(SQ=13\text{ cm}\).
\(PR\) is the perpendicular height from \(P\) to the hypotenuse \(SQ\). Equating the two expressions for the area of \(\triangle SPQ\):
\[\frac12(5)(12)=\frac12(13)(PR).\]
\[PR=\frac{5\times12}{13}=4.615\ldots\text{ cm}.\]
\[\boxed{PR=4.62\text{ cm}}\qquad\text{(correct to 3 significant figures).}\]
(b)(i) The required diagram is:
(b)(ii)
For ladder \(L\), let \(\theta\) be its angle with the horizontal ground. Then
\[\cos\theta=\frac{8}{10}=0.8.\]
Therefore,
\[\sin\theta=\sqrt{1-0.8^2}=\sqrt{0.36}=0.6.\]
Since ladder \(M\) makes the same angle \(\theta\) with the ground, if \(h\) is the height at which it touches the wall,
\[\sin\theta=\frac{h}{12}.\]
\[h=12(0.6)=7.2\text{ m}.\]
\[\boxed{\text{Ladder }M\text{ touches the wall }7.2\text{ m above the ground.}}\]
Bayanin Amsa
(a)
In right-angled triangle \(SPQ\),
\[SQ^2=PS^2+PQ^2=5^2+12^2=169.\]
Hence, \(SQ=13\text{ cm}\).
\(PR\) is the perpendicular height from \(P\) to the hypotenuse \(SQ\). Equating the two expressions for the area of \(\triangle SPQ\):
\[\frac12(5)(12)=\frac12(13)(PR).\]
\[PR=\frac{5\times12}{13}=4.615\ldots\text{ cm}.\]
\[\boxed{PR=4.62\text{ cm}}\qquad\text{(correct to 3 significant figures).}\]
(b)(i) The required diagram is:
(b)(ii)
For ladder \(L\), let \(\theta\) be its angle with the horizontal ground. Then
\[\cos\theta=\frac{8}{10}=0.8.\]
Therefore,
\[\sin\theta=\sqrt{1-0.8^2}=\sqrt{0.36}=0.6.\]
Since ladder \(M\) makes the same angle \(\theta\) with the ground, if \(h\) is the height at which it touches the wall,
\[\sin\theta=\frac{h}{12}.\]
\[h=12(0.6)=7.2\text{ m}.\]
\[\boxed{\text{Ladder }M\text{ touches the wall }7.2\text{ m above the ground.}}\]
Tambaya 27 Rahoto
(a) Using ruler a pair of compasses only, construct:
(i) a trapezium PQRS such that |PQ| = 6.8 cm, < PQR = 120\(^o\), QR||PS, |PS| =10.6 cm, and IPRI = 93 cm;
(ii) locus \(l_1\) of points equidistant from P and R;
(iii) locus \(l_2\) of points equidistant from Q and R
(b) Measure: (i) |QR|;
ii. < PSR
ii. < PSR
iii. |QY|, where Y is the point of intersection h and h.
(a) Construction (ruler and compasses only).
- Draw \(|PQ| = 6.8\) cm. At Q construct \(\angle PQR = 120^{\circ}\) and draw the ray along which R lies.
- Through P draw a line parallel to QR (since \(QR \parallel PS\)) and mark \(|PS| = 10.6\) cm. The diagonal \(|PR| = 9.3\) cm (arc from P) fixes R on its ray. Join R to S to complete trapezium PQRS.
- \(l_1\): construct the perpendicular bisector of \(PR\) (points equidistant from P and R).
- \(l_2\): construct the perpendicular bisector of \(QR\) (points equidistant from Q and R). Let \(l_1\) and \(l_2\) meet at Y.
(b) Expected measurements from an accurate drawing:
- \(|QR| \approx 3.8\) cm
- \(\angle PSR \approx 60^{\circ}\)
- \(|QY| \approx 5.4\) cm
(Accept small tolerances, since these are read off the construction.)
Bayanin Amsa
(a) Construction (ruler and compasses only).
- Draw \(|PQ| = 6.8\) cm. At Q construct \(\angle PQR = 120^{\circ}\) and draw the ray along which R lies.
- Through P draw a line parallel to QR (since \(QR \parallel PS\)) and mark \(|PS| = 10.6\) cm. The diagonal \(|PR| = 9.3\) cm (arc from P) fixes R on its ray. Join R to S to complete trapezium PQRS.
- \(l_1\): construct the perpendicular bisector of \(PR\) (points equidistant from P and R).
- \(l_2\): construct the perpendicular bisector of \(QR\) (points equidistant from Q and R). Let \(l_1\) and \(l_2\) meet at Y.
(b) Expected measurements from an accurate drawing:
- \(|QR| \approx 3.8\) cm
- \(\angle PSR \approx 60^{\circ}\)
- \(|QY| \approx 5.4\) cm
(Accept small tolerances, since these are read off the construction.)
Tambaya 28 Rahoto
(a) Mr John paid N4,800.00 in N1.00 ordinary shares of a company which sold at N2.50 per share. If dividend was declared at 25k per share, how much dividend did he get?
(b) Using the method of completing the square, solve \(\frac{1 - x}{x} + \frac{x}{1 - x} = \frac{5}{2}\)
(a) Number of shares bought \(= \dfrac{4800}{2.50} = 1920\) shares.
Dividend of 25k \(= N0.25\) per share:
\[1920 \times N0.25 = N480.00\](b) Let \(t = \dfrac{1 - x}{x}\); then \(\dfrac{x}{1 - x} = \dfrac{1}{t}\), and the equation becomes \(t + \dfrac{1}{t} = \dfrac{5}{2}\), i.e.
\[2t^2 - 5t + 2 = 0 \Rightarrow t^2 - \tfrac{5}{2}t = -1\]Completing the square:
\[\left(t - \tfrac{5}{4}\right)^2 = -1 + \tfrac{25}{16} = \tfrac{9}{16} \Rightarrow t - \tfrac{5}{4} = \pm\tfrac{3}{4}\] \[t = 2 \quad \text{or} \quad t = \tfrac{1}{2}\]\(\dfrac{1 - x}{x} = 2 \Rightarrow 1 - x = 2x \Rightarrow x = \tfrac{1}{3}\).
\(\dfrac{1 - x}{x} = \tfrac{1}{2} \Rightarrow 2 - 2x = x \Rightarrow x = \tfrac{2}{3}\).
So \(x = \dfrac{1}{3}\) or \(x = \dfrac{2}{3}\).
Bayanin Amsa
(a) Number of shares bought \(= \dfrac{4800}{2.50} = 1920\) shares.
Dividend of 25k \(= N0.25\) per share:
\[1920 \times N0.25 = N480.00\](b) Let \(t = \dfrac{1 - x}{x}\); then \(\dfrac{x}{1 - x} = \dfrac{1}{t}\), and the equation becomes \(t + \dfrac{1}{t} = \dfrac{5}{2}\), i.e.
\[2t^2 - 5t + 2 = 0 \Rightarrow t^2 - \tfrac{5}{2}t = -1\]Completing the square:
\[\left(t - \tfrac{5}{4}\right)^2 = -1 + \tfrac{25}{16} = \tfrac{9}{16} \Rightarrow t - \tfrac{5}{4} = \pm\tfrac{3}{4}\] \[t = 2 \quad \text{or} \quad t = \tfrac{1}{2}\]\(\dfrac{1 - x}{x} = 2 \Rightarrow 1 - x = 2x \Rightarrow x = \tfrac{1}{3}\).
\(\dfrac{1 - x}{x} = \tfrac{1}{2} \Rightarrow 2 - 2x = x \Rightarrow x = \tfrac{2}{3}\).
So \(x = \dfrac{1}{3}\) or \(x = \dfrac{2}{3}\).
Tambaya 29 Rahoto
(a) Lamin bought a book for N300.00 and sold it to Bola at a profit of x%. Bola then sold the same book at a profit of x%. If James paid \(N(6x + \frac{3}{4})\) more for the book than Lamin paid, find the value of x.
(b) Find the range of values of x which satisfies the inequality \(3x - 2 < 10 + x < 2 + 5x\).
(a) Lamin's cost is \(N300\). Selling to Bola at \(x\%\) profit means Bola pays \(300\left(1 + \frac{x}{100}\right)\).
Bola resells at another \(x\%\) profit, so James pays \(300\left(1 + \frac{x}{100}\right)^2\).
James pays \(N\left(6x + \frac{3}{4}\right)\) more than Lamin's \(N300\):
\[300\left(1 + \tfrac{x}{100}\right)^2 - 300 = 6x + \tfrac{3}{4}\] \[300\left(\tfrac{2x}{100} + \tfrac{x^2}{10000}\right) = 6x + \tfrac{3}{4} \Rightarrow 6x + 0.03x^2 = 6x + 0.75\] \[0.03x^2 = 0.75 \Rightarrow x^2 = 25 \Rightarrow x = 5\]So \(x = 5\) (taking the positive value).
(b) Split \(3x - 2 < 10 + x < 2 + 5x\) into two inequalities.
Left: \(3x - 2 < 10 + x \Rightarrow 2x < 12 \Rightarrow x < 6\).
Right: \(10 + x < 2 + 5x \Rightarrow 8 < 4x \Rightarrow x > 2\).
Therefore \(2 < x < 6\).
Bayanin Amsa
(a) Lamin's cost is \(N300\). Selling to Bola at \(x\%\) profit means Bola pays \(300\left(1 + \frac{x}{100}\right)\).
Bola resells at another \(x\%\) profit, so James pays \(300\left(1 + \frac{x}{100}\right)^2\).
James pays \(N\left(6x + \frac{3}{4}\right)\) more than Lamin's \(N300\):
\[300\left(1 + \tfrac{x}{100}\right)^2 - 300 = 6x + \tfrac{3}{4}\] \[300\left(\tfrac{2x}{100} + \tfrac{x^2}{10000}\right) = 6x + \tfrac{3}{4} \Rightarrow 6x + 0.03x^2 = 6x + 0.75\] \[0.03x^2 = 0.75 \Rightarrow x^2 = 25 \Rightarrow x = 5\]So \(x = 5\) (taking the positive value).
(b) Split \(3x - 2 < 10 + x < 2 + 5x\) into two inequalities.
Left: \(3x - 2 < 10 + x \Rightarrow 2x < 12 \Rightarrow x < 6\).
Right: \(10 + x < 2 + 5x \Rightarrow 8 < 4x \Rightarrow x > 2\).
Therefore \(2 < x < 6\).
Tambaya 30 Rahoto
(a)
(i) Copy and complete the addition \(\oplus\) and multiplication \(\otimes\) tables in modulo 5 on the set {2, 3, 4}.
| \(\oplus\) | 2 | 3 | 4 |
| 2 | |||
| 3 | |||
| 4 |
| \(\otimes\) | 2 | 3 | 4 |
| 2 | |||
| 3 | |||
| 4 |
(ii) Use the tables to:
(a) solve the equation \(4 \otimes e \oplus 2 \equiv 1 \pmod{5}\):
(b) find the value of n, if \(4 \oplus n \otimes 2 \equiv\) (mod 5).
(b) Consider the following statements:
p: Landi has cholera,
q: Landi is in the hospital.
If p = q, state whether or not the following statements are valid:
(i) If Landi is in the hospital, then he has cholera.
(ii) If Landi is not in the hospital, then he does not have cholera.
(iii) If Landi does not have cholera, then he is not in the hospital.
(a)(i) Modulo 5 tables on {2, 3, 4}
Addition \(\oplus\) (add, then take the remainder on division by 5):
| \(\oplus\) | 2 | 3 | 4 |
|---|---|---|---|
| 2 | 4 | 0 | 1 |
| 3 | 0 | 1 | 2 |
| 4 | 1 | 2 | 3 |
Multiplication \(\otimes\) (multiply, then take the remainder on division by 5):
| \(\otimes\) | 2 | 3 | 4 |
|---|---|---|---|
| 2 | 4 | 1 | 3 |
| 3 | 1 | 4 | 2 |
| 4 | 3 | 2 | 1 |
(For example \(3\otimes 4 = 12 = 2\times 5 + 2 \equiv 2\), and \(4\oplus 3 = 7 = 5 + 2 \equiv 2\).)
(a)(ii)(a) Solve \(4\otimes e \oplus 2 \equiv 1 \ (\text{mod }5)\)
\(4\otimes e \equiv 1 \ominus 2 \equiv -1 \equiv 4 \ (\text{mod }5)\). So \(4e \equiv 4 \ (\text{mod }5)\). Multiplying both sides by the inverse of 4 (which is 4, since \(4\times 4 = 16 \equiv 1\)) gives \(e \equiv 16 \equiv 1 \ (\text{mod }5)\). Hence \(e = 1\).
(a)(ii)(b) Find n if \(4 \oplus n\otimes 2 \equiv 1 \ (\text{mod }5)\)
Multiplication first: \(n\otimes 2 = 2n\). Then \(4 \oplus 2n \equiv 1\), so \(2n \equiv 1 - 4 \equiv -3 \equiv 2 \ (\text{mod }5)\). Multiplying by the inverse of 2 (which is 3, since \(2\times 3 = 6 \equiv 1\)) gives \(n \equiv 6 \equiv 1 \ (\text{mod }5)\). Hence \(n = 1\).
(b) Validity of the statements (given \(p \Rightarrow q\))
Here p: Landi has cholera, q: Landi is in the hospital, and we are told \(p \Rightarrow q\) is true.
- (i) "If in hospital, then has cholera" is \(q \Rightarrow p\), the converse. It does not follow from \(p\Rightarrow q\), so it is not valid.
- (ii) "If not in hospital, then no cholera" is \(\lnot q \Rightarrow \lnot p\), the contrapositive of \(p\Rightarrow q\). It is logically equivalent, so it is valid.
- (iii) "If no cholera, then not in hospital" is \(\lnot p \Rightarrow \lnot q\), the inverse. It does not follow, so it is not valid.
Bayanin Amsa
(a)(i) Modulo 5 tables on {2, 3, 4}
Addition \(\oplus\) (add, then take the remainder on division by 5):
| \(\oplus\) | 2 | 3 | 4 |
|---|---|---|---|
| 2 | 4 | 0 | 1 |
| 3 | 0 | 1 | 2 |
| 4 | 1 | 2 | 3 |
Multiplication \(\otimes\) (multiply, then take the remainder on division by 5):
| \(\otimes\) | 2 | 3 | 4 |
|---|---|---|---|
| 2 | 4 | 1 | 3 |
| 3 | 1 | 4 | 2 |
| 4 | 3 | 2 | 1 |
(For example \(3\otimes 4 = 12 = 2\times 5 + 2 \equiv 2\), and \(4\oplus 3 = 7 = 5 + 2 \equiv 2\).)
(a)(ii)(a) Solve \(4\otimes e \oplus 2 \equiv 1 \ (\text{mod }5)\)
\(4\otimes e \equiv 1 \ominus 2 \equiv -1 \equiv 4 \ (\text{mod }5)\). So \(4e \equiv 4 \ (\text{mod }5)\). Multiplying both sides by the inverse of 4 (which is 4, since \(4\times 4 = 16 \equiv 1\)) gives \(e \equiv 16 \equiv 1 \ (\text{mod }5)\). Hence \(e = 1\).
(a)(ii)(b) Find n if \(4 \oplus n\otimes 2 \equiv 1 \ (\text{mod }5)\)
Multiplication first: \(n\otimes 2 = 2n\). Then \(4 \oplus 2n \equiv 1\), so \(2n \equiv 1 - 4 \equiv -3 \equiv 2 \ (\text{mod }5)\). Multiplying by the inverse of 2 (which is 3, since \(2\times 3 = 6 \equiv 1\)) gives \(n \equiv 6 \equiv 1 \ (\text{mod }5)\). Hence \(n = 1\).
(b) Validity of the statements (given \(p \Rightarrow q\))
Here p: Landi has cholera, q: Landi is in the hospital, and we are told \(p \Rightarrow q\) is true.
- (i) "If in hospital, then has cholera" is \(q \Rightarrow p\), the converse. It does not follow from \(p\Rightarrow q\), so it is not valid.
- (ii) "If not in hospital, then no cholera" is \(\lnot q \Rightarrow \lnot p\), the contrapositive of \(p\Rightarrow q\). It is logically equivalent, so it is valid.
- (iii) "If no cholera, then not in hospital" is \(\lnot p \Rightarrow \lnot q\), the inverse. It does not follow, so it is not valid.
Tambaya 31 Rahoto
Using ruler and a pair of compasses only, construct:
(a) (i) quadrilateral PQRS with |PQ| = 6 cm, |PS| = 8 cm, < PSR = 90\(^o\), |SR| = 12 cm and |QR| = 11 cm;
(ii) perpendicular from Q to cut \(\over{SR}\) at K.
(b) Measure:
(i) IQRI;
(ii) < QRS.
(a) Construction
- Draw the line segment \(SR=12\text{ cm}\).
- At \(S\), construct a perpendicular to \(SR\). Mark \(P\) on it so that \(PS=8\text{ cm}\).
- With centre \(P\) and radius \(6\text{ cm}\), draw an arc. With centre \(R\) and radius \(11\text{ cm}\), draw another arc cutting the first at \(Q\).
- Join \(PQ\) and \(QR\) to complete quadrilateral \(PQRS\).
- Construct the perpendicular from \(Q\) to \(SR\), meeting \(SR\) at \(K\).
(b) Measurements from the construction
(i) \(\lvert RK\rvert \approx 9.3\text{ cm}\).
(ii) \(\angle QRS \approx 55^\circ\).
Bayanin Amsa
(a) Construction
- Draw the line segment \(SR=12\text{ cm}\).
- At \(S\), construct a perpendicular to \(SR\). Mark \(P\) on it so that \(PS=8\text{ cm}\).
- With centre \(P\) and radius \(6\text{ cm}\), draw an arc. With centre \(R\) and radius \(11\text{ cm}\), draw another arc cutting the first at \(Q\).
- Join \(PQ\) and \(QR\) to complete quadrilateral \(PQRS\).
- Construct the perpendicular from \(Q\) to \(SR\), meeting \(SR\) at \(K\).
(b) Measurements from the construction
(i) \(\lvert RK\rvert \approx 9.3\text{ cm}\).
(ii) \(\angle QRS \approx 55^\circ\).
Tambaya 32 Rahoto
(a) If \(x = \begin{pmatrix} 2 \\ 3 \end{pmatrix}, y = \begin{pmatrix} 5 \\ -2 \end{pmatrix}\) and \(z = \begin{pmatrix} -4 \\ 13 \end{pmatrix}\), find the scalars p and q such that \(px + qy = z\).
(b)(i) Using the scale of 2cm to 2 units on both axis, draw on a graph paper two perpendicular axis x and y for \(-5 \leq x \leq 5, -5 \leq y \leq 5\) respectively.
(ii) Draw, on the graph paper, indicating clearly the vertices and their coordinates,
(1) the quadrilateral WXYZ with W(2, 3), X(4, -1), Y(-3, -4) and Z(-3, 2).
(2) the image \(W_{1}X_{1}Y_{1}Z_{1}\) of the quadrilateral WXYZ under an anti-clockwise rotation of 90° about the origin where \(W \to W_{1}, X \to X_{1}, Y \to Y_{1}\) and \(Z \to Z_{1}\).
(a)
Given
\[p\begin{pmatrix}2\\3\end{pmatrix}+q\begin{pmatrix}5\\-2\end{pmatrix}=\begin{pmatrix}-4\\13\end{pmatrix}.\]
Equating corresponding components gives
\[2p+5q=-4 \qquad (1)\]
\[3p-2q=13 \qquad (2)\]
Multiply (1) by 2 and (2) by 5:
\[4p+10q=-8\]
\[15p-10q=65\]
Adding,
\[19p=57\]
\[p=3.\]
Substituting into (1),
\[2(3)+5q=-4\]
\[5q=-10\]
\[q=-2.\]
Therefore, \(\boxed{p=3,\ q=-2}\).
(b)(i) and (ii)
Using the scale \(2\text{ cm}\) to \(2\) units on both axes, the quadrilateral and its image are plotted below. The vertices are joined in the order \(W\to X\to Y\to Z\to W\), and similarly for the image.
A rotation of \(90^\circ\) anticlockwise about the origin maps \((x,y)\) onto \((-y,x)\). Hence:
\[\begin{aligned}W(2,3)&\longmapsto W_1(-3,2),\\X(4,-1)&\longmapsto X_1(1,4),\\Y(-3,-4)&\longmapsto Y_1(4,-3),\\Z(-3,2)&\longmapsto Z_1(-2,-3).\end{aligned}\]
Bayanin Amsa
(a)
Given
\[p\begin{pmatrix}2\\3\end{pmatrix}+q\begin{pmatrix}5\\-2\end{pmatrix}=\begin{pmatrix}-4\\13\end{pmatrix}.\]
Equating corresponding components gives
\[2p+5q=-4 \qquad (1)\]
\[3p-2q=13 \qquad (2)\]
Multiply (1) by 2 and (2) by 5:
\[4p+10q=-8\]
\[15p-10q=65\]
Adding,
\[19p=57\]
\[p=3.\]
Substituting into (1),
\[2(3)+5q=-4\]
\[5q=-10\]
\[q=-2.\]
Therefore, \(\boxed{p=3,\ q=-2}\).
(b)(i) and (ii)
Using the scale \(2\text{ cm}\) to \(2\) units on both axes, the quadrilateral and its image are plotted below. The vertices are joined in the order \(W\to X\to Y\to Z\to W\), and similarly for the image.
A rotation of \(90^\circ\) anticlockwise about the origin maps \((x,y)\) onto \((-y,x)\). Hence:
\[\begin{aligned}W(2,3)&\longmapsto W_1(-3,2),\\X(4,-1)&\longmapsto X_1(1,4),\\Y(-3,-4)&\longmapsto Y_1(4,-3),\\Z(-3,2)&\longmapsto Z_1(-2,-3).\end{aligned}\]
Tambaya 33 Rahoto
a. Find the range of values of x which satisfy the following inequalities simultaneously: 5 - x > 1 and 9 + x \(\geq\) 8
In the diagram, O is the centre of the circle, IPQI = IQRI and < PSR = 56°. Find < QRS.
(a) Solving the inequalities simultaneously
First inequality:
\[ 5 - x > 1 \;\Rightarrow\; -x > 1 - 5 \;\Rightarrow\; -x > -4 \;\Rightarrow\; x < 4 \]
Second inequality:
\[ 9 + x \geq 8 \;\Rightarrow\; x \geq 8 - 9 \;\Rightarrow\; x \geq -1 \]
Combining both conditions:
\[ -1 \leq x < 4 \]
(b) Finding \(\angle QRS\)
From the diagram, \(O\) is the centre and \(PS\) passes through \(O\), so \(PS\) is a diameter. Also \(|PQ| = |QR|\) (equal chords, shown by the tick marks) and \(\angle PSR = 56^{\circ}\). The points \(P, Q, R\) lie on the arc on the same side of the diameter.
Step 1: Use equal chords. Equal chords subtend equal arcs, so \(\text{arc } PQ = \text{arc } QR\).
Step 2: Use the inscribed angle at S. \(\angle PSR = 56^{\circ}\) is the angle subtended at the circumference by the arc \(PQR\) (arc \(PR\) through \(Q\)). Hence
\[ \text{arc } PQR = 2 \times 56^{\circ} = 112^{\circ} \]
and since arc \(PQ = \) arc \(QR\), each equals \(56^{\circ}\).
Step 3: Find the remaining arc. Because \(PS\) is a diameter, the semicircle \(P\)-\(Q\)-\(R\)-\(S\) totals \(180^{\circ}\):
\[ \text{arc } RS = 180^{\circ} - \text{arc } PQ - \text{arc } QR = 180^{\circ} - 56^{\circ} - 56^{\circ} = 68^{\circ} \]
Step 4: Use the cyclic quadrilateral PQRS. Opposite angles of a cyclic quadrilateral are supplementary, so \(\angle QPS + \angle QRS = 180^{\circ}\).
\(\angle QPS\) subtends arc \(QRS = \text{arc } QR + \text{arc } RS = 56^{\circ} + 68^{\circ} = 124^{\circ}\), so
\[ \angle QPS = \tfrac{1}{2}\times 124^{\circ} = 62^{\circ} \]
Therefore
\[ \angle QRS = 180^{\circ} - 62^{\circ} = 118^{\circ} \]
\(\angle QRS = 118^{\circ}\).
Bayanin Amsa
(a) Solving the inequalities simultaneously
First inequality:
\[ 5 - x > 1 \;\Rightarrow\; -x > 1 - 5 \;\Rightarrow\; -x > -4 \;\Rightarrow\; x < 4 \]
Second inequality:
\[ 9 + x \geq 8 \;\Rightarrow\; x \geq 8 - 9 \;\Rightarrow\; x \geq -1 \]
Combining both conditions:
\[ -1 \leq x < 4 \]
(b) Finding \(\angle QRS\)
From the diagram, \(O\) is the centre and \(PS\) passes through \(O\), so \(PS\) is a diameter. Also \(|PQ| = |QR|\) (equal chords, shown by the tick marks) and \(\angle PSR = 56^{\circ}\). The points \(P, Q, R\) lie on the arc on the same side of the diameter.
Step 1: Use equal chords. Equal chords subtend equal arcs, so \(\text{arc } PQ = \text{arc } QR\).
Step 2: Use the inscribed angle at S. \(\angle PSR = 56^{\circ}\) is the angle subtended at the circumference by the arc \(PQR\) (arc \(PR\) through \(Q\)). Hence
\[ \text{arc } PQR = 2 \times 56^{\circ} = 112^{\circ} \]
and since arc \(PQ = \) arc \(QR\), each equals \(56^{\circ}\).
Step 3: Find the remaining arc. Because \(PS\) is a diameter, the semicircle \(P\)-\(Q\)-\(R\)-\(S\) totals \(180^{\circ}\):
\[ \text{arc } RS = 180^{\circ} - \text{arc } PQ - \text{arc } QR = 180^{\circ} - 56^{\circ} - 56^{\circ} = 68^{\circ} \]
Step 4: Use the cyclic quadrilateral PQRS. Opposite angles of a cyclic quadrilateral are supplementary, so \(\angle QPS + \angle QRS = 180^{\circ}\).
\(\angle QPS\) subtends arc \(QRS = \text{arc } QR + \text{arc } RS = 56^{\circ} + 68^{\circ} = 124^{\circ}\), so
\[ \angle QPS = \tfrac{1}{2}\times 124^{\circ} = 62^{\circ} \]
Therefore
\[ \angle QRS = 180^{\circ} - 62^{\circ} = 118^{\circ} \]
\(\angle QRS = 118^{\circ}\).
Tambaya 34 Rahoto
(a) If tan x = \(\frac{5}{12}\), \(0^o\). < x < 90°, evaluate, without using Mathematical tables or calculator, \(\frac{sin x}{(sin x)^2 + cosx}\)
(b) The diagram shows a rectangular lawn measuring 14m by 11m. A path of uniform width \(x\)m surrounds it. If the total area of the path is 186 m\(^2\), how wide is the path?
(a) \(\tan x = \frac{5}{12}\) gives opposite 5, adjacent 12, hypotenuse \(\sqrt{5^2 + 12^2} = 13\).
So \(\sin x = \frac{5}{13}\) and \(\cos x = \frac{12}{13}\).
\[\frac{\sin x}{\sin^2 x + \cos x} = \frac{\tfrac{5}{13}}{\tfrac{25}{169} + \tfrac{12}{13}} = \frac{\tfrac{5}{13}}{\tfrac{25 + 156}{169}} = \frac{5}{13} \times \frac{169}{181} = \frac{65}{181}\](b) Lawn is \(14 \times 11\); a path of width \(x\) all round gives an outer rectangle \((14 + 2x)\) by \((11 + 2x)\).
\[(14 + 2x)(11 + 2x) - 14 \times 11 = 186\] \[154 + 50x + 4x^2 - 154 = 186 \Rightarrow 4x^2 + 50x - 186 = 0\] \[2x^2 + 25x - 93 = 0\]Discriminant \(= 25^2 + 4(2)(93) = 625 + 744 = 1369 = 37^2\).
\[x = \frac{-25 + 37}{4} = 3\]The path is \(3\) m wide (the negative root is rejected).
Bayanin Amsa
(a) \(\tan x = \frac{5}{12}\) gives opposite 5, adjacent 12, hypotenuse \(\sqrt{5^2 + 12^2} = 13\).
So \(\sin x = \frac{5}{13}\) and \(\cos x = \frac{12}{13}\).
\[\frac{\sin x}{\sin^2 x + \cos x} = \frac{\tfrac{5}{13}}{\tfrac{25}{169} + \tfrac{12}{13}} = \frac{\tfrac{5}{13}}{\tfrac{25 + 156}{169}} = \frac{5}{13} \times \frac{169}{181} = \frac{65}{181}\](b) Lawn is \(14 \times 11\); a path of width \(x\) all round gives an outer rectangle \((14 + 2x)\) by \((11 + 2x)\).
\[(14 + 2x)(11 + 2x) - 14 \times 11 = 186\] \[154 + 50x + 4x^2 - 154 = 186 \Rightarrow 4x^2 + 50x - 186 = 0\] \[2x^2 + 25x - 93 = 0\]Discriminant \(= 25^2 + 4(2)(93) = 625 + 744 = 1369 = 37^2\).
\[x = \frac{-25 + 37}{4} = 3\]The path is \(3\) m wide (the negative root is rejected).
Tambaya 35 Rahoto
The sum of the first ten terms of an Arithmetic Progression (A.P.) is 130. If the fifth term is 3 times the first term, find the:
- common difference;
- first term;
- number of terms of the A.P. if the last term is 28.
Let the first term be \(a\) and the common difference \(d\).
Sum of 10 terms: \(S_{10} = \tfrac{10}{2}(2a + 9d) = 130 \Rightarrow 2a + 9d = 26\).
Fifth term is three times the first: \(a + 4d = 3a \Rightarrow 4d = 2a \Rightarrow a = 2d\).
Substitute: \(2(2d) + 9d = 26 \Rightarrow 13d = 26 \Rightarrow d = 2\), so \(a = 4\).
Common difference \(= 2\); first term \(= 4\).
Number of terms when the last term is 28:
\[a + (n - 1)d = 28 \Rightarrow 4 + 2(n - 1) = 28 \Rightarrow n - 1 = 12 \Rightarrow n = 13\]Bayanin Amsa
Let the first term be \(a\) and the common difference \(d\).
Sum of 10 terms: \(S_{10} = \tfrac{10}{2}(2a + 9d) = 130 \Rightarrow 2a + 9d = 26\).
Fifth term is three times the first: \(a + 4d = 3a \Rightarrow 4d = 2a \Rightarrow a = 2d\).
Substitute: \(2(2d) + 9d = 26 \Rightarrow 13d = 26 \Rightarrow d = 2\), so \(a = 4\).
Common difference \(= 2\); first term \(= 4\).
Number of terms when the last term is 28:
\[a + (n - 1)d = 28 \Rightarrow 4 + 2(n - 1) = 28 \Rightarrow n - 1 = 12 \Rightarrow n = 13\]Tambaya 36 Rahoto
If \(\log_a(y + 2) = 1 + \log_a x\), find x in terms of y.
2. The table shows the distribution of timber production in five communities in a certain year
| Community | Timber Production (tonnes) |
| Bibiani | 600 |
| Amenfi | 900 |
| Oda | 1800 |
| Wiawso | 1500 |
| Sankore | 2400 |
1. Draw a pie chart to represent the information.
2. What percentage of timber produced that year was from Amenfi?
3. If a tonne of timber is sold at $560.00, how much more revenue would Oda community receive than Bibiani?
1. Given
\[\log_a(y+2)=1+\log_a x\]
Since \(1=\log_a a\),
\[\log_a(y+2)=\log_a a+\log_a x=\log_a(ax).\]
Therefore,
\[y+2=ax\]
\[\boxed{x=\frac{y+2}{a}}\]
2(a)(i) Pie chart
Total timber production:
\[600+900+1800+1500+2400=7200\text{ tonnes}.\]
| Community | Production (tonnes) | Sector angle |
|---|---|---|
| Bibiani | 600 | \(\frac{600}{7200}\times360^\circ=30^\circ\) |
| Amenfi | 900 | \(\frac{900}{7200}\times360^\circ=45^\circ\) |
| Oda | 1800 | \(\frac{1800}{7200}\times360^\circ=90^\circ\) |
| Wiawso | 1500 | \(\frac{1500}{7200}\times360^\circ=75^\circ\) |
| Sankore | 2400 | \(\frac{2400}{7200}\times360^\circ=120^\circ\) |
2(a)(ii)
\[\frac{900}{7200}\times100\%=\boxed{12.5\%}.\]
2(a)(iii)
Difference in production:
\[1800-600=1200\text{ tonnes}.\]
Hence, the extra revenue received by Oda is
\[1200\times\$560.00=\boxed{\$672,000.00}.\]
Bayanin Amsa
1. Given
\[\log_a(y+2)=1+\log_a x\]
Since \(1=\log_a a\),
\[\log_a(y+2)=\log_a a+\log_a x=\log_a(ax).\]
Therefore,
\[y+2=ax\]
\[\boxed{x=\frac{y+2}{a}}\]
2(a)(i) Pie chart
Total timber production:
\[600+900+1800+1500+2400=7200\text{ tonnes}.\]
| Community | Production (tonnes) | Sector angle |
|---|---|---|
| Bibiani | 600 | \(\frac{600}{7200}\times360^\circ=30^\circ\) |
| Amenfi | 900 | \(\frac{900}{7200}\times360^\circ=45^\circ\) |
| Oda | 1800 | \(\frac{1800}{7200}\times360^\circ=90^\circ\) |
| Wiawso | 1500 | \(\frac{1500}{7200}\times360^\circ=75^\circ\) |
| Sankore | 2400 | \(\frac{2400}{7200}\times360^\circ=120^\circ\) |
2(a)(ii)
\[\frac{900}{7200}\times100\%=\boxed{12.5\%}.\]
2(a)(iii)
Difference in production:
\[1800-600=1200\text{ tonnes}.\]
Hence, the extra revenue received by Oda is
\[1200\times\$560.00=\boxed{\$672,000.00}.\]
Tambaya 37 Rahoto
(a) The frequency distribution shows the range of prices of a brand of a car sold by a dealer and the corresponding quantity demanded.
| Price (N1,000,000.00 |
1.0 - 1.9 | 2.0 - 2.9 | 3.0 - 3.9 | 4.0 - 4.9 | 5.0 - 5.9 |
| Number of Vehicles | 23 | 48 | 107 | 90 | 32 |
(b) Represent the information in a histogram and use the histogram to determine the most preferred selling price for the brand of car.
(a) Class boundaries
| Price (₦ million) | Class boundaries (₦ million) | Number of vehicles |
|---|---|---|
| 1.0 – 1.9 | 0.95 – 1.95 | 23 |
| 2.0 – 2.9 | 1.95 – 2.95 | 48 |
| 3.0 – 3.9 | 2.95 – 3.95 | 107 |
| 4.0 – 4.9 | 3.95 – 4.95 | 90 |
| 5.0 – 5.9 | 4.95 – 5.95 | 32 |
(b) Histogram
The bars have equal width of ₦1 million and are drawn contiguously, using the class boundaries on the horizontal axis.
The tallest bar is for the class ₦3.0 million to ₦3.9 million. Drawing the two modal lines across this bar and reading their point of intersection on the price axis gives approximately
\[2.95+0.75=3.70.\]
Therefore, the most preferred selling price is
\[\boxed{₦3,700,000.00}\]
Bayanin Amsa
(a) Class boundaries
| Price (₦ million) | Class boundaries (₦ million) | Number of vehicles |
|---|---|---|
| 1.0 – 1.9 | 0.95 – 1.95 | 23 |
| 2.0 – 2.9 | 1.95 – 2.95 | 48 |
| 3.0 – 3.9 | 2.95 – 3.95 | 107 |
| 4.0 – 4.9 | 3.95 – 4.95 | 90 |
| 5.0 – 5.9 | 4.95 – 5.95 | 32 |
(b) Histogram
The bars have equal width of ₦1 million and are drawn contiguously, using the class boundaries on the horizontal axis.
The tallest bar is for the class ₦3.0 million to ₦3.9 million. Drawing the two modal lines across this bar and reading their point of intersection on the price axis gives approximately
\[2.95+0.75=3.70.\]
Therefore, the most preferred selling price is
\[\boxed{₦3,700,000.00}\]
Tambaya 38 Rahoto
(a) If the mean of m, n, s, p and q is 12, calculate the mean of (m + 4), (n - 3), (s + 6), (p - 2) and (q + 8).
(b) In a community of 500 people, the 75th percentile age is 65 years while the 25th percentile age is 15 years. How many of the people are between 15 and 65 years?
(a) The mean of five numbers is their sum divided by 5.
Since the mean of \(m, n, s, p, q\) is 12, the sum is \(5 \times 12 = 60\).
Adding the constants changes the sum by \(+4 - 3 + 6 - 2 + 8 = +13\).
New sum \(= 60 + 13 = 73\), so new mean \(= \dfrac{73}{5} = 14.6\).
(b) The 25th percentile (15 years) and the 75th percentile (65 years) enclose the middle 50% of the data.
Number of people between 15 and 65 years \(= 50\% \times 500 = 250\).
Bayanin Amsa
(a) The mean of five numbers is their sum divided by 5.
Since the mean of \(m, n, s, p, q\) is 12, the sum is \(5 \times 12 = 60\).
Adding the constants changes the sum by \(+4 - 3 + 6 - 2 + 8 = +13\).
New sum \(= 60 + 13 = 73\), so new mean \(= \dfrac{73}{5} = 14.6\).
(b) The 25th percentile (15 years) and the 75th percentile (65 years) enclose the middle 50% of the data.
Number of people between 15 and 65 years \(= 50\% \times 500 = 250\).
