WAEC SSCE - General Mathematics - 2006 (Essay)
Tambaya 1 Rahoto
Two fair dice are tossed together once.
(a) Draw a sample space for the possible outcomes ;
(b) Find the probability of getting a total : (i) of 7 or 8 ; (ii) less than 4.
(a) Sample space
Let the first number in each ordered pair be the score on the first die and the second number be the score on the second die.
Hence, there are \(6\times6=36\) equally likely outcomes.
(b)(i) Probability of a total of 7 or 8
A total of \(7\) occurs in 6 outcomes:
\((1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\).
A total of \(8\) occurs in 5 outcomes:
\((2,6),(3,5),(4,4),(5,3),(6,2)\).
Therefore,
\[P(7\text{ or }8)=\frac{6+5}{36}=\frac{11}{36}.\]
(b)(ii) Probability of a total less than 4
The possible totals are \(2\) and \(3\), obtained from:
\((1,1),(1,2),(2,1)\).
Thus,
\[P(\text{total less than }4)=\frac{3}{36}=\frac{1}{12}.\]
Bayanin Amsa
(a) Sample space
Let the first number in each ordered pair be the score on the first die and the second number be the score on the second die.
Hence, there are \(6\times6=36\) equally likely outcomes.
(b)(i) Probability of a total of 7 or 8
A total of \(7\) occurs in 6 outcomes:
\((1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\).
A total of \(8\) occurs in 5 outcomes:
\((2,6),(3,5),(4,4),(5,3),(6,2)\).
Therefore,
\[P(7\text{ or }8)=\frac{6+5}{36}=\frac{11}{36}.\]
(b)(ii) Probability of a total less than 4
The possible totals are \(2\) and \(3\), obtained from:
\((1,1),(1,2),(2,1)\).
Thus,
\[P(\text{total less than }4)=\frac{3}{36}=\frac{1}{12}.\]
Tambaya 2 Rahoto
(a) 
The diagram shows a pyramid standing on a cuboid. The dimensions of the cuboid are 4m by 3m by 2m and the slant edge of the pyramid is 5m. Calculet the volume of the shape.
(b) The 2nd, 3rd and 4th terms of an A.P are x - 2, 5 and x + 2 respectively. Calculate the value of x.
(a) Volume of the solid (pyramid on a cuboid).
From the diagram the cuboid has base \(4\text{ m} \times 3\text{ m}\) and height \(2\text{ m}\); the pyramid stands on the top \(4\text{ m} \times 3\text{ m}\) face and has a slant edge (apex to base corner) of \(5\text{ m}\).
Volume of the cuboid:
\[V_{\text{cuboid}} = 4 \times 3 \times 2 = 24 \text{ m}^3\]Height of the pyramid. The apex is above the centre of the rectangular base. The distance from the centre to a base corner is half the diagonal of the \(4 \times 3\) rectangle:
\[\text{diagonal} = \sqrt{4^2 + 3^2} = \sqrt{25} = 5 \text{ m}, \qquad \text{half-diagonal} = \tfrac{5}{2} = 2.5 \text{ m}\]Using the right triangle (height, half-diagonal, slant edge):
\[h = \sqrt{5^2 - 2.5^2} = \sqrt{25 - 6.25} = \sqrt{18.75} \approx 4.330 \text{ m}\]Volume of the pyramid:
\[V_{\text{pyramid}} = \tfrac{1}{3} \times (4 \times 3) \times h = \tfrac{1}{3} \times 12 \times 4.330 = 17.32 \text{ m}^3\]Total volume of the shape:
\[V = 24 + 17.32 = 41.32 \text{ m}^3\]Total volume \(\approx 41.32\text{ m}^3\) (to 2 d.p.).
(b) Value of x in the A.P.
The 2nd, 3rd and 4th terms are \(x-2\), \(5\) and \(x+2\). In an A.P. the difference between successive terms is constant, so the middle term equals the average of its neighbours (equivalently, consecutive differences are equal):
\[5 - (x-2) = (x+2) - 5\]\[7 - x = x - 3\]\[10 = 2x \quad\Rightarrow\quad x = 5\]x = 5. (Check: terms are \(3, 5, 7\), common difference \(2\).)
Bayanin Amsa
(a) Volume of the solid (pyramid on a cuboid).
From the diagram the cuboid has base \(4\text{ m} \times 3\text{ m}\) and height \(2\text{ m}\); the pyramid stands on the top \(4\text{ m} \times 3\text{ m}\) face and has a slant edge (apex to base corner) of \(5\text{ m}\).
Volume of the cuboid:
\[V_{\text{cuboid}} = 4 \times 3 \times 2 = 24 \text{ m}^3\]Height of the pyramid. The apex is above the centre of the rectangular base. The distance from the centre to a base corner is half the diagonal of the \(4 \times 3\) rectangle:
\[\text{diagonal} = \sqrt{4^2 + 3^2} = \sqrt{25} = 5 \text{ m}, \qquad \text{half-diagonal} = \tfrac{5}{2} = 2.5 \text{ m}\]Using the right triangle (height, half-diagonal, slant edge):
\[h = \sqrt{5^2 - 2.5^2} = \sqrt{25 - 6.25} = \sqrt{18.75} \approx 4.330 \text{ m}\]Volume of the pyramid:
\[V_{\text{pyramid}} = \tfrac{1}{3} \times (4 \times 3) \times h = \tfrac{1}{3} \times 12 \times 4.330 = 17.32 \text{ m}^3\]Total volume of the shape:
\[V = 24 + 17.32 = 41.32 \text{ m}^3\]Total volume \(\approx 41.32\text{ m}^3\) (to 2 d.p.).
(b) Value of x in the A.P.
The 2nd, 3rd and 4th terms are \(x-2\), \(5\) and \(x+2\). In an A.P. the difference between successive terms is constant, so the middle term equals the average of its neighbours (equivalently, consecutive differences are equal):
\[5 - (x-2) = (x+2) - 5\]\[7 - x = x - 3\]\[10 = 2x \quad\Rightarrow\quad x = 5\]x = 5. (Check: terms are \(3, 5, 7\), common difference \(2\).)
Tambaya 3 Rahoto
The following table shows the distribution of test scores in a class.
| Scores | 1 | 2 | 3 | 4 | 5 | 7 | 8 | 9 | 10 |
| No of pupils | 1 | 1 | 5 | 3 | \(k^{2} + 1\) | 6 | 2 | 3 | 4 |
(a) If the mean score of the class is 6, find the : (i) value of k (ii) median score.
(b) Draw a bar chart for the distribution.
(c) If a pupil is picked at random, what is the probability that he/ she will score less than 6?
(a)(i) Finding the value of \(k\)
| Score, \(x\) | 1 | 2 | 3 | 4 | 5 | 7 | 8 | 9 | 10 | Total |
|---|---|---|---|---|---|---|---|---|---|---|
| Frequency, \(f\) | 1 | 1 | 5 | 3 | \(k^2+1\) | 6 | 2 | 3 | 4 | \(k^2+26\) |
| \(fx\) | 1 | 2 | 15 | 12 | \(5k^2+5\) | 42 | 16 | 27 | 40 | \(5k^2+160\) |
Since the mean score is 6,
\[6=\frac{5k^2+160}{k^2+26}\]
\[6k^2+156=5k^2+160\]
\[k^2=4\]
\[\therefore k=2\]
Thus, the frequency for score 5 is \(k^2+1=5\), and the total number of pupils is \(4+26=30\).
(a)(ii) Median score
| Score | 1 | 2 | 3 | 4 | 5 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|
| Cumulative frequency | 1 | 2 | 7 | 10 | 15 | 21 | 23 | 26 | 30 |
There are 30 scores. The 15th score is 5 and the 16th score is 7.
\[\text{Median}=\frac{5+7}{2}=6\]
(b) Bar chart of the distribution
(c) Probability of scoring less than 6
The scores less than 6 are 1, 2, 3, 4 and 5.
\[\text{Number of such pupils}=1+1+5+3+5=15\]
\[P(\text{score less than }6)=\frac{15}{30}=\boxed{\frac{1}{2}}\]
Bayanin Amsa
(a)(i) Finding the value of \(k\)
| Score, \(x\) | 1 | 2 | 3 | 4 | 5 | 7 | 8 | 9 | 10 | Total |
|---|---|---|---|---|---|---|---|---|---|---|
| Frequency, \(f\) | 1 | 1 | 5 | 3 | \(k^2+1\) | 6 | 2 | 3 | 4 | \(k^2+26\) |
| \(fx\) | 1 | 2 | 15 | 12 | \(5k^2+5\) | 42 | 16 | 27 | 40 | \(5k^2+160\) |
Since the mean score is 6,
\[6=\frac{5k^2+160}{k^2+26}\]
\[6k^2+156=5k^2+160\]
\[k^2=4\]
\[\therefore k=2\]
Thus, the frequency for score 5 is \(k^2+1=5\), and the total number of pupils is \(4+26=30\).
(a)(ii) Median score
| Score | 1 | 2 | 3 | 4 | 5 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|
| Cumulative frequency | 1 | 2 | 7 | 10 | 15 | 21 | 23 | 26 | 30 |
There are 30 scores. The 15th score is 5 and the 16th score is 7.
\[\text{Median}=\frac{5+7}{2}=6\]
(b) Bar chart of the distribution
(c) Probability of scoring less than 6
The scores less than 6 are 1, 2, 3, 4 and 5.
\[\text{Number of such pupils}=1+1+5+3+5=15\]
\[P(\text{score less than }6)=\frac{15}{30}=\boxed{\frac{1}{2}}\]
Tambaya 4 Rahoto
(a) Using a ruler and a pair of compasses only, (i) construct \(\Delta\) XYZ such that |XY| = 8 cm and < YXZ = < ZYX = 45°. (ii) locate a point P inside the triangle equidistant from XY and XZ and also equidistance from YX and YZ. (iii) construct a circle touching the three sides of the triangle (iv) measure the radius of the circle.
(b) The length of the sides of a hexagon are x - 5, 2x, 2x, 2x + 7, 2x and 2x - 1. If the perimeter is 144 cm, find the value of x.
(a) Construction of the triangle and its incircle
- Draw the line segment \(XY=8\text{ cm}\).
- At \(X\), construct a perpendicular to \(XY\), then bisect the \(90^\circ\) angle to make a ray at \(45^\circ\) to \(XY\).
- At \(Y\), construct a perpendicular to \(YX\), then bisect the \(90^\circ\) angle to make a ray at \(45^\circ\) to \(YX\). The two \(45^\circ\) rays meet at \(Z\).
- Bisect \(\angle YXZ\) and \(\angle XYZ\). Their intersection is \(P\).
A point on an angle bisector is equidistant from the two sides of that angle. Therefore, \(P\) is equidistant from \(XY\), \(XZ\), and \(YZ\): it is the incentre of the triangle.
To construct the circle, draw a perpendicular from \(P\) to any side, for example \(XY\). Use the perpendicular distance as the radius and draw a circle with centre \(P\). Because \(P\) is equidistant from all three sides, this circle touches all three sides.
The triangle is right-angled at \(Z\), with hypotenuse \(XY=8\text{ cm}\). Its two equal shorter sides are
\[ 8\cos45^\circ=4\sqrt{2}\text{ cm}. \]
For a right-angled triangle, the inradius is
\[ r=\frac{a+b-c}{2}. \]
Hence,
\[ r=\frac{4\sqrt2+4\sqrt2-8}{2} =4\sqrt2-4 =4(\sqrt2-1) \approx1.66\text{ cm}. \]
Therefore, the measured radius is approximately \(1.7\text{ cm}\).
(b) Add all six side lengths to make the perimeter:
\[ (x-5)+2x+2x+(2x+7)+2x+(2x-1)=144. \]
Collecting like terms gives
\[ 11x+1=144 \]
\[ 11x=143 \]
\[ x=\frac{143}{11}=13. \]
Therefore, \(x=13\).
Examination reminder: When finding a perimeter, include every side exactly once and keep brackets around expressions such as \(x-5\) and \(2x+7\) before simplifying.
Bayanin Amsa
(a) Construction of the triangle and its incircle
- Draw the line segment \(XY=8\text{ cm}\).
- At \(X\), construct a perpendicular to \(XY\), then bisect the \(90^\circ\) angle to make a ray at \(45^\circ\) to \(XY\).
- At \(Y\), construct a perpendicular to \(YX\), then bisect the \(90^\circ\) angle to make a ray at \(45^\circ\) to \(YX\). The two \(45^\circ\) rays meet at \(Z\).
- Bisect \(\angle YXZ\) and \(\angle XYZ\). Their intersection is \(P\).
A point on an angle bisector is equidistant from the two sides of that angle. Therefore, \(P\) is equidistant from \(XY\), \(XZ\), and \(YZ\): it is the incentre of the triangle.
To construct the circle, draw a perpendicular from \(P\) to any side, for example \(XY\). Use the perpendicular distance as the radius and draw a circle with centre \(P\). Because \(P\) is equidistant from all three sides, this circle touches all three sides.
The triangle is right-angled at \(Z\), with hypotenuse \(XY=8\text{ cm}\). Its two equal shorter sides are
\[ 8\cos45^\circ=4\sqrt{2}\text{ cm}. \]
For a right-angled triangle, the inradius is
\[ r=\frac{a+b-c}{2}. \]
Hence,
\[ r=\frac{4\sqrt2+4\sqrt2-8}{2} =4\sqrt2-4 =4(\sqrt2-1) \approx1.66\text{ cm}. \]
Therefore, the measured radius is approximately \(1.7\text{ cm}\).
(b) Add all six side lengths to make the perimeter:
\[ (x-5)+2x+2x+(2x+7)+2x+(2x-1)=144. \]
Collecting like terms gives
\[ 11x+1=144 \]
\[ 11x=143 \]
\[ x=\frac{143}{11}=13. \]
Therefore, \(x=13\).
Examination reminder: When finding a perimeter, include every side exactly once and keep brackets around expressions such as \(x-5\) and \(2x+7\) before simplifying.
Tambaya 5 Rahoto
(a) Copy and complete the table of the relation \(y = 2\sin x - \cos 2x\).
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° |
| y | 0.5 | -1.0 |
Using a scale of 2 cm to 30° on the x- axis and 2 cm to 0.5 unit on the y- axis, draw the graph of \(y = 2\sin x - \cos 2x\), for \(0° \leq x \leq 180°\).
(b) Using the same axes, draw the graph of \(y = 1.25\).
(c) Use your graphs to find the : (i) values of x for which \(2\sin x - \cos 2x = 0\) ; (ii) the roots of the equation \(2\sin x - \cos 2x = 1.25\).
(a) For \(y=2\sin x-\cos 2x\), the completed table is:
| \(x\) | \(0^\circ\) | \(30^\circ\) | \(60^\circ\) | \(90^\circ\) | \(120^\circ\) | \(150^\circ\) | \(180^\circ\) |
|---|---|---|---|---|---|---|---|
| \(y\) | \(-1.0\) | \(0.5\) | \(2.2\) | \(3.0\) | \(2.2\) | \(0.5\) | \(-1.0\) |
For example,
\[y(60^\circ)=2\sin60^\circ-\cos120^\circ=2\left(\frac{\sqrt3}{2}\right)-\left(-\frac12\right)=2.232\ldots\approx2.2.\]
(b) The graph of \(y=1.25\) is the horizontal line shown on the same axes.
(c)
(i) The curve cuts the \(x\)-axis at approximately
\[\boxed{x=21^\circ\text{ and }159^\circ}.\]
(ii) The intersections of the curve with \(y=1.25\) give
\[\boxed{x=42^\circ\text{ and }138^\circ}.\]
Bayanin Amsa
(a) For \(y=2\sin x-\cos 2x\), the completed table is:
| \(x\) | \(0^\circ\) | \(30^\circ\) | \(60^\circ\) | \(90^\circ\) | \(120^\circ\) | \(150^\circ\) | \(180^\circ\) |
|---|---|---|---|---|---|---|---|
| \(y\) | \(-1.0\) | \(0.5\) | \(2.2\) | \(3.0\) | \(2.2\) | \(0.5\) | \(-1.0\) |
For example,
\[y(60^\circ)=2\sin60^\circ-\cos120^\circ=2\left(\frac{\sqrt3}{2}\right)-\left(-\frac12\right)=2.232\ldots\approx2.2.\]
(b) The graph of \(y=1.25\) is the horizontal line shown on the same axes.
(c)
(i) The curve cuts the \(x\)-axis at approximately
\[\boxed{x=21^\circ\text{ and }159^\circ}.\]
(ii) The intersections of the curve with \(y=1.25\) give
\[\boxed{x=42^\circ\text{ and }138^\circ}.\]
Tambaya 6 Rahoto
(a) A dealer sold a car to a man and made a profit of 15%. The man then sold it to a woman for N120,175.00 at a loss of 5%. How much did the dealer buy the car?
(b) The diameter of the wheel of a car is 36cm. How many revolutions, correct to three significant figures, will it make to cover a distance of 1.05 km? [Take \(\pi = \frac{22}{7}\)].
(a) Work backwards. The man sold to the woman at a 5% loss for \(N120{,}175\), so this is \(95\%\) of the man's cost price (which is the dealer's selling price).
Dealer's selling price \(= \dfrac{120{,}175}{0.95} = N126{,}500\).
The dealer made a 15% profit, so \(N126{,}500 = 115\%\) of the dealer's cost price.
Dealer's cost price \(= \dfrac{126{,}500}{1.15} = \mathbf{N110{,}000.00}\).
(b) Wheel diameter \(= 36\text{ cm}\), so circumference \(= \pi d = \dfrac{22}{7}\times 36 = \dfrac{792}{7} = 113.14\text{ cm}\).
Distance \(= 1.05\text{ km} = 105{,}000\text{ cm}\).
Number of revolutions \(= \dfrac{105{,}000}{113.14} = 928.03 \approx \mathbf{928}\) (3 s.f.).
Bayanin Amsa
(a) Work backwards. The man sold to the woman at a 5% loss for \(N120{,}175\), so this is \(95\%\) of the man's cost price (which is the dealer's selling price).
Dealer's selling price \(= \dfrac{120{,}175}{0.95} = N126{,}500\).
The dealer made a 15% profit, so \(N126{,}500 = 115\%\) of the dealer's cost price.
Dealer's cost price \(= \dfrac{126{,}500}{1.15} = \mathbf{N110{,}000.00}\).
(b) Wheel diameter \(= 36\text{ cm}\), so circumference \(= \pi d = \dfrac{22}{7}\times 36 = \dfrac{792}{7} = 113.14\text{ cm}\).
Distance \(= 1.05\text{ km} = 105{,}000\text{ cm}\).
Number of revolutions \(= \dfrac{105{,}000}{113.14} = 928.03 \approx \mathbf{928}\) (3 s.f.).
Tambaya 7 Rahoto
(a) Simplify : \(\frac{4\frac{2}{9} - 1\frac{13}{15}}{2\frac{1}{5} + \frac{4}{7} \times 2\frac{1}{3}}\)
(b) By rationalising the denominator, simplify : \(\frac{7\sqrt{5}}{\sqrt{7}}\), leaving your answer in surd form.
(a) Convert mixed numbers to improper fractions.
Numerator: \(4\tfrac{2}{9} - 1\tfrac{13}{15} = \tfrac{38}{9} - \tfrac{28}{15} = \tfrac{190}{45} - \tfrac{84}{45} = \tfrac{106}{45}\).
Denominator: First \(\tfrac{4}{7}\times 2\tfrac{1}{3} = \tfrac{4}{7}\times\tfrac{7}{3} = \tfrac{4}{3}\). Then \(2\tfrac{1}{5} + \tfrac{4}{3} = \tfrac{11}{5} + \tfrac{4}{3} = \tfrac{33}{15} + \tfrac{20}{15} = \tfrac{53}{15}\).
\(\dfrac{106/45}{53/15} = \dfrac{106}{45}\times\dfrac{15}{53} = \dfrac{106\times 15}{45\times 53} = \dfrac{2\times 15}{45} = \dfrac{30}{45} = \mathbf{\dfrac{2}{3}}\)
(b) Rationalise \(\dfrac{7\sqrt{5}}{\sqrt{7}}\) by multiplying by \(\dfrac{\sqrt{7}}{\sqrt{7}}\):
\(\dfrac{7\sqrt{5}}{\sqrt{7}}\times\dfrac{\sqrt{7}}{\sqrt{7}} = \dfrac{7\sqrt{35}}{7} = \mathbf{\sqrt{35}}\)
Bayanin Amsa
(a) Convert mixed numbers to improper fractions.
Numerator: \(4\tfrac{2}{9} - 1\tfrac{13}{15} = \tfrac{38}{9} - \tfrac{28}{15} = \tfrac{190}{45} - \tfrac{84}{45} = \tfrac{106}{45}\).
Denominator: First \(\tfrac{4}{7}\times 2\tfrac{1}{3} = \tfrac{4}{7}\times\tfrac{7}{3} = \tfrac{4}{3}\). Then \(2\tfrac{1}{5} + \tfrac{4}{3} = \tfrac{11}{5} + \tfrac{4}{3} = \tfrac{33}{15} + \tfrac{20}{15} = \tfrac{53}{15}\).
\(\dfrac{106/45}{53/15} = \dfrac{106}{45}\times\dfrac{15}{53} = \dfrac{106\times 15}{45\times 53} = \dfrac{2\times 15}{45} = \dfrac{30}{45} = \mathbf{\dfrac{2}{3}}\)
(b) Rationalise \(\dfrac{7\sqrt{5}}{\sqrt{7}}\) by multiplying by \(\dfrac{\sqrt{7}}{\sqrt{7}}\):
\(\dfrac{7\sqrt{5}}{\sqrt{7}}\times\dfrac{\sqrt{7}}{\sqrt{7}} = \dfrac{7\sqrt{35}}{7} = \mathbf{\sqrt{35}}\)
Tambaya 8 Rahoto
(a) Two lines AB and CD intersect at x such that \(\stackrel\frown{CAX}\) is equal to \(\stackrel\frown{BDX}\). If |AX| = 6 cm, |XB| = 4 cm and |CX| = 3 cm, find |XD|.
(b) 
The diagram shows the positions of three points X, Y and Z on a horizontal plane. The bearing of Y from X is 312° and that of Y from Z is 022°. If |XY| = 32 km and |ZY| = 50 km, calculate, correct to one decimal place : (i) |XZ| ; (ii) the bearing of Z from X.
(a) Finding |XD|. Since \(\angle CAX=\angle BDX\) (given) and \(\angle AXC=\angle DXB\) (vertically opposite), triangles \(AXC\) and \(DXB\) are similar (AA). Matching the equal angles gives
\[\frac{AX}{DX}=\frac{CX}{BX}\Rightarrow \frac{6}{XD}=\frac{3}{4}\]
\[XD=\frac{6\times4}{3}=8\text{ cm}.\]
(b) The angle at Y. The bearing of Y from X is \(312°\), so the bearing of X from Y is \(312°-180°=132°\). The bearing of Y from Z is \(022°\), so the bearing of Z from Y is \(022°+180°=202°\). Hence
\[\angle XYZ=202°-132°=70°.\]
(i) |XZ|. By the cosine rule,
\[XZ^2=XY^2+ZY^2-2\,XY\cdot ZY\cos70°=32^2+50^2-2(32)(50)\cos70°\]
\[XZ^2=3524-3200(0.3420)=2429.5\Rightarrow XZ=\sqrt{2429.5}=49.3\text{ km}.\]
(ii) Bearing of Z from X. By the sine rule,
\[\frac{\sin\angle YXZ}{ZY}=\frac{\sin70°}{XZ}\Rightarrow \sin\angle YXZ=\frac{50\sin70°}{49.3}=0.9532\]
\[\angle YXZ=72.4°.\]
From the diagram Z lies to the west of Y as seen from X, so the bearing of Z from X is measured anticlockwise from that of Y:
\[312°-72.4°=239.6°.\]
Bayanin Amsa
(a) Finding |XD|. Since \(\angle CAX=\angle BDX\) (given) and \(\angle AXC=\angle DXB\) (vertically opposite), triangles \(AXC\) and \(DXB\) are similar (AA). Matching the equal angles gives
\[\frac{AX}{DX}=\frac{CX}{BX}\Rightarrow \frac{6}{XD}=\frac{3}{4}\]
\[XD=\frac{6\times4}{3}=8\text{ cm}.\]
(b) The angle at Y. The bearing of Y from X is \(312°\), so the bearing of X from Y is \(312°-180°=132°\). The bearing of Y from Z is \(022°\), so the bearing of Z from Y is \(022°+180°=202°\). Hence
\[\angle XYZ=202°-132°=70°.\]
(i) |XZ|. By the cosine rule,
\[XZ^2=XY^2+ZY^2-2\,XY\cdot ZY\cos70°=32^2+50^2-2(32)(50)\cos70°\]
\[XZ^2=3524-3200(0.3420)=2429.5\Rightarrow XZ=\sqrt{2429.5}=49.3\text{ km}.\]
(ii) Bearing of Z from X. By the sine rule,
\[\frac{\sin\angle YXZ}{ZY}=\frac{\sin70°}{XZ}\Rightarrow \sin\angle YXZ=\frac{50\sin70°}{49.3}=0.9532\]
\[\angle YXZ=72.4°.\]
From the diagram Z lies to the west of Y as seen from X, so the bearing of Z from X is measured anticlockwise from that of Y:
\[312°-72.4°=239.6°.\]
Tambaya 9 Rahoto
(a) A sector of a circle of radius 8cm subtends an angle of 90° at the centre of the circle. If the sector is folded without overlap to form the curved surface of a cone, find the :
(i) base radius ; (ii) height ; (iii) volume of the cone. [Take \(\pi = \frac{22}{7}\)].
(b) A map is drawn to a scale of 1 : 20,000. Use it to calculate the : (i) distance, in kilometres, represented by 4.5 cm on the map ;
(a) When the sector is folded into a cone, the arc length of the sector becomes the circumference of the base, and the sector radius becomes the slant height \(l = 8\text{ cm}\).
Arc length \(= \dfrac{90}{360}\times 2\pi(8) = \tfrac{1}{4}\times 16\pi = 4\pi\text{ cm}\).
(i) Base circumference \(2\pi r = 4\pi \ \Rightarrow\ \mathbf{r = 2\text{ cm}}\).
(ii) Height \(h = \sqrt{l^2 - r^2} = \sqrt{8^2 - 2^2} = \sqrt{60} = \mathbf{7.75\text{ cm}}\) (2 d.p.).
(iii) Volume \(= \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\times\tfrac{22}{7}\times 4 \times 7.746 = \mathbf{32.5\text{ cm}^3}\) (3 s.f.).
(b) Scale \(1 : 20{,}000\) means \(1\text{ cm}\) on the map \(= 20{,}000\text{ cm}\) on the ground.
(i) \(4.5\text{ cm}\) represents \(4.5\times 20{,}000 = 90{,}000\text{ cm} = 900\text{ m} = \mathbf{0.9\text{ km}}\).
Bayanin Amsa
(a) When the sector is folded into a cone, the arc length of the sector becomes the circumference of the base, and the sector radius becomes the slant height \(l = 8\text{ cm}\).
Arc length \(= \dfrac{90}{360}\times 2\pi(8) = \tfrac{1}{4}\times 16\pi = 4\pi\text{ cm}\).
(i) Base circumference \(2\pi r = 4\pi \ \Rightarrow\ \mathbf{r = 2\text{ cm}}\).
(ii) Height \(h = \sqrt{l^2 - r^2} = \sqrt{8^2 - 2^2} = \sqrt{60} = \mathbf{7.75\text{ cm}}\) (2 d.p.).
(iii) Volume \(= \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\times\tfrac{22}{7}\times 4 \times 7.746 = \mathbf{32.5\text{ cm}^3}\) (3 s.f.).
(b) Scale \(1 : 20{,}000\) means \(1\text{ cm}\) on the map \(= 20{,}000\text{ cm}\) on the ground.
(i) \(4.5\text{ cm}\) represents \(4.5\times 20{,}000 = 90{,}000\text{ cm} = 900\text{ m} = \mathbf{0.9\text{ km}}\).
Tambaya 10 Rahoto
(a) Solve for x and y in the following equations :
\(2x - y = \frac{9}{2}\)
\(x + 4y = 0\)
(b) 
In the diagram, TA is a tangent to the circle at A. If \(\stackrel\frown{BCA} = 40°\) and \(\stackrel\frown{DAT} = 52°\), find \(\stackrel\frown{BAD}\).
(a) Simultaneous equations.
\[2x-y=\frac{9}{2}\quad(1),\qquad x+4y=0\quad(2).\]
From (2), \(x=-4y\). Substitute into (1):
\[2(-4y)-y=\frac{9}{2}\]
\[-8y-y=\frac{9}{2}\]
\[-9y=\frac{9}{2}\ \Rightarrow\ y=-\frac{1}{2}.\]
Then \(x=-4y=-4\left(-\dfrac{1}{2}\right)=2\).
\[\boxed{x=2,\quad y=-\tfrac{1}{2}}.\]
(b) Tangent TA to the circle at A.
From the diagram, \(B, C, D, A\) lie on the circle, \(TA\) is the tangent at \(A\), \(\angle BCA=40^\circ\) (at \(C\)) and \(\angle DAT=52^\circ\) (between the tangent \(AT\) and chord \(AD\)).
Chord AB. By the alternate segment theorem, the tangent-chord angle between the tangent (on the side towards \(X\), opposite \(T\)) and chord \(AB\) equals the angle in the alternate segment \(\angle BCA\):
\[\angle BAX=\angle BCA=40^\circ.\]
Straight tangent line. The points \(X, A, T\) lie on the straight tangent line, so the three angles at \(A\) on one side sum to \(180^\circ\):
\[\angle BAX+\angle BAD+\angle DAT=180^\circ.\]
\[40^\circ+\angle BAD+52^\circ=180^\circ\]
\[\angle BAD=180^\circ-40^\circ-52^\circ=\boxed{88^\circ}.\]
Bayanin Amsa
(a) Simultaneous equations.
\[2x-y=\frac{9}{2}\quad(1),\qquad x+4y=0\quad(2).\]
From (2), \(x=-4y\). Substitute into (1):
\[2(-4y)-y=\frac{9}{2}\]
\[-8y-y=\frac{9}{2}\]
\[-9y=\frac{9}{2}\ \Rightarrow\ y=-\frac{1}{2}.\]
Then \(x=-4y=-4\left(-\dfrac{1}{2}\right)=2\).
\[\boxed{x=2,\quad y=-\tfrac{1}{2}}.\]
(b) Tangent TA to the circle at A.
From the diagram, \(B, C, D, A\) lie on the circle, \(TA\) is the tangent at \(A\), \(\angle BCA=40^\circ\) (at \(C\)) and \(\angle DAT=52^\circ\) (between the tangent \(AT\) and chord \(AD\)).
Chord AB. By the alternate segment theorem, the tangent-chord angle between the tangent (on the side towards \(X\), opposite \(T\)) and chord \(AB\) equals the angle in the alternate segment \(\angle BCA\):
\[\angle BAX=\angle BCA=40^\circ.\]
Straight tangent line. The points \(X, A, T\) lie on the straight tangent line, so the three angles at \(A\) on one side sum to \(180^\circ\):
\[\angle BAX+\angle BAD+\angle DAT=180^\circ.\]
\[40^\circ+\angle BAD+52^\circ=180^\circ\]
\[\angle BAD=180^\circ-40^\circ-52^\circ=\boxed{88^\circ}.\]
Tambaya 11 Rahoto
(a)(i) If \(4x < 2 + 3x\) and \(x - 8 < 3x\), what range of values of x satisfies both inequalities? ; (ii) Represent your result in (i) on the number line.
(b) A shop is sending out a bill for an amount less than £100. The accountant interchanges the two digits and so overcharges the customer by 45. Given that the sum of the two digits is 9, find how much the bill should be.
(a)(i) Solve both inequalities.
\(4x < 2 + 3x \ \Rightarrow\ x < 2\)
\(x - 8 < 3x \ \Rightarrow\ -8 < 2x \ \Rightarrow\ x > -4\)
Both hold when \(\mathbf{-4 < x < 2}\).
(ii) Number line: an open circle at \(-4\) and an open circle at \(2\), with the segment between them shaded (values strictly between, endpoints excluded).
(b) Let the correct bill be the two-digit number \(10a + b\) (\(a\) = tens digit, \(b\) = units digit). Interchanging gives \(10b + a\), which overcharges by 45:
\((10b + a) - (10a + b) = 45 \ \Rightarrow\ 9(b - a) = 45 \ \Rightarrow\ b - a = 5\)
Also \(a + b = 9\). Adding: \(2b = 14 \ \Rightarrow\ b = 7\), so \(a = 2\).
The bill should be \(10a + b = \mathbf{£27}\). (Check: \(72 - 27 = 45\).)
Bayanin Amsa
(a)(i) Solve both inequalities.
\(4x < 2 + 3x \ \Rightarrow\ x < 2\)
\(x - 8 < 3x \ \Rightarrow\ -8 < 2x \ \Rightarrow\ x > -4\)
Both hold when \(\mathbf{-4 < x < 2}\).
(ii) Number line: an open circle at \(-4\) and an open circle at \(2\), with the segment between them shaded (values strictly between, endpoints excluded).
(b) Let the correct bill be the two-digit number \(10a + b\) (\(a\) = tens digit, \(b\) = units digit). Interchanging gives \(10b + a\), which overcharges by 45:
\((10b + a) - (10a + b) = 45 \ \Rightarrow\ 9(b - a) = 45 \ \Rightarrow\ b - a = 5\)
Also \(a + b = 9\). Adding: \(2b = 14 \ \Rightarrow\ b = 7\), so \(a = 2\).
The bill should be \(10a + b = \mathbf{£27}\). (Check: \(72 - 27 = 45\).)
Tambaya 12 Rahoto
(a) In a class of 45 students, 32 offered Physics(P), 28 offered Government(G) and 12 did not offer any of the two subjects. (i) Draw the Venn diagram to represent the information ; (ii) How many students offered both subjects? (iii) What is \(n(P \cup G)\)?
(b) If \(p = \frac{2u}{1 - u}\) and \(q = \frac{1 + u}{1 - u}\) ; express \(\frac{p + q}{p - q}\) in terms of u.
(a)
Since 12 students offered neither subject, place 12 outside the two circles. Let the number in the overlap be \(x\).
\[ (32-x)+x+(28-x)+12=45 \]
\[72-x=45\]
\[x=27\]
Thus, \(P\) only \(=32-27=5\), and \(G\) only \(=28-27=1\).
(i) The required Venn diagram is:
(ii) \[\boxed{n(P\cap G)=27}\] Therefore, 27 students offered both Physics and Government.
(iii) \[n(P\cup G)=45-12=\boxed{33}\]
(b)
\[p+q=\frac{2u}{1-u}+\frac{1+u}{1-u}=\frac{2u+1+u}{1-u}=\frac{3u+1}{1-u}\]
\[p-q=\frac{2u}{1-u}-\frac{1+u}{1-u}=\frac{2u-1-u}{1-u}=\frac{u-1}{1-u}=-1\]
Hence,
\[\frac{p+q}{p-q}=\frac{\frac{3u+1}{1-u}}{-1}=-\frac{3u+1}{1-u}=\boxed{\frac{3u+1}{u-1}}\]
Bayanin Amsa
(a)
Since 12 students offered neither subject, place 12 outside the two circles. Let the number in the overlap be \(x\).
\[ (32-x)+x+(28-x)+12=45 \]
\[72-x=45\]
\[x=27\]
Thus, \(P\) only \(=32-27=5\), and \(G\) only \(=28-27=1\).
(i) The required Venn diagram is:
(ii) \[\boxed{n(P\cap G)=27}\] Therefore, 27 students offered both Physics and Government.
(iii) \[n(P\cup G)=45-12=\boxed{33}\]
(b)
\[p+q=\frac{2u}{1-u}+\frac{1+u}{1-u}=\frac{2u+1+u}{1-u}=\frac{3u+1}{1-u}\]
\[p-q=\frac{2u}{1-u}-\frac{1+u}{1-u}=\frac{2u-1-u}{1-u}=\frac{u-1}{1-u}=-1\]
Hence,
\[\frac{p+q}{p-q}=\frac{\frac{3u+1}{1-u}}{-1}=-\frac{3u+1}{1-u}=\boxed{\frac{3u+1}{u-1}}\]
Tambaya 13 Rahoto
In the diagram, ABCDEO is two- thirds of a circle centre O. The radius AO is 7cm and /AB/ = /BC/ = /CD/ = /DE/. Calculate, correct to the nearest whole number, the area of the shaded portion. [Take \(\pi = \frac{22}{7}\)].
Setting up. The figure \(ABCDEO\) is two-thirds of a circle, centre \(O\), radius \(AO=7\text{ cm}\). The sector angle is
\[\tfrac23\times360^\circ=240^\circ.\]The equal chords \(|AB|=|BC|=|CD|=|DE|\) divide the \(240^\circ\) arc into \(4\) equal arcs, each
\[\frac{240^\circ}{4}=60^\circ.\]So each of \(\triangle OAB,\ \triangle OBC,\ \triangle OCD,\ \triangle ODE\) has two sides equal to the radius \(7\) with a \(60^\circ\) angle between them, making each an equilateral triangle of side \(7\text{ cm}\).
Shaded portion = area of the \(240^\circ\) sector \(-\) area of the four triangles (it is the four segments between the chords and the outer arc).
Area of the \(240^\circ\) sector with \(\pi=\tfrac{22}{7}\):
\[\frac{240}{360}\times\frac{22}{7}\times7^2=\frac{2}{3}\times\frac{22}{7}\times49=\frac{2}{3}\times154=102.67\text{ cm}^2.\]Area of one equilateral triangle (side \(7\)):
\[\frac{\sqrt3}{4}\times7^2=\frac{\sqrt3}{4}\times49=21.22\text{ cm}^2.\]Four of them: \(4\times21.22=84.87\text{ cm}^2.\)
Shaded area:
\[102.67-84.87=17.8\approx \boxed{18\text{ cm}^2}.\](Equivalently, one \(60^\circ\) segment \(=\tfrac16(154)-21.22=25.67-21.22=4.45\text{ cm}^2\), and \(4\times4.45=17.8\text{ cm}^2.\))
Bayanin Amsa
Setting up. The figure \(ABCDEO\) is two-thirds of a circle, centre \(O\), radius \(AO=7\text{ cm}\). The sector angle is
\[\tfrac23\times360^\circ=240^\circ.\]The equal chords \(|AB|=|BC|=|CD|=|DE|\) divide the \(240^\circ\) arc into \(4\) equal arcs, each
\[\frac{240^\circ}{4}=60^\circ.\]So each of \(\triangle OAB,\ \triangle OBC,\ \triangle OCD,\ \triangle ODE\) has two sides equal to the radius \(7\) with a \(60^\circ\) angle between them, making each an equilateral triangle of side \(7\text{ cm}\).
Shaded portion = area of the \(240^\circ\) sector \(-\) area of the four triangles (it is the four segments between the chords and the outer arc).
Area of the \(240^\circ\) sector with \(\pi=\tfrac{22}{7}\):
\[\frac{240}{360}\times\frac{22}{7}\times7^2=\frac{2}{3}\times\frac{22}{7}\times49=\frac{2}{3}\times154=102.67\text{ cm}^2.\]Area of one equilateral triangle (side \(7\)):
\[\frac{\sqrt3}{4}\times7^2=\frac{\sqrt3}{4}\times49=21.22\text{ cm}^2.\]Four of them: \(4\times21.22=84.87\text{ cm}^2.\)
Shaded area:
\[102.67-84.87=17.8\approx \boxed{18\text{ cm}^2}.\](Equivalently, one \(60^\circ\) segment \(=\tfrac16(154)-21.22=25.67-21.22=4.45\text{ cm}^2\), and \(4\times4.45=17.8\text{ cm}^2.\))
