WAEC SSCE - General Mathematics - 2012 (Essay)
Tambaya 1 Rahoto
A point H is 20 m away from the foot of a tower on the same horizontal ground. From the point H, the angle of elevation of the point P on the tower and the top (T) of the tower are 30° and 50° respectively. Calculate, correct to 3 significant figures :
(a) /PT/; (b) the distance between H and the top of the tower
(c) The position of H if the angle of depression of H from the top of the tower is to be 40°.
\(H\) is 20 m from the foot of the tower. From \(H\), \(P\) has elevation \(30^\circ\) and the top \(T\) has elevation \(50^\circ\).
Heights above ground of the two points:
\[\text{height of }P = 20\tan 30^\circ = 20(0.5774) = 11.547\text{ m}\]
\[\text{height of }T = 20\tan 50^\circ = 20(1.1918) = 23.835\text{ m}\]
(a) \(|PT| = 23.835 - 11.547 = 12.288 \approx 12.3\text{ m (3 s.f.)}\).
(b) Distance \(|HT|\) (the slant line to the top):
\[|HT| = \frac{20}{\cos 50^\circ} = \frac{20}{0.6428} = 31.1\text{ m (3 s.f.)}\]
(c) For the angle of depression of \(H\) from the top to be \(40^\circ\), the angle of elevation of \(T\) from \(H\) must be \(40^\circ\). With the top still \(23.835\) m high, the required horizontal distance \(x\) satisfies:
\[\tan 40^\circ = \frac{23.835}{x} \;\Rightarrow\; x = \frac{23.835}{0.8391} = 28.4\text{ m (3 s.f.)}\]
So \(H\) must be \(28.4\) m from the foot of the tower.
Bayanin Amsa
\(H\) is 20 m from the foot of the tower. From \(H\), \(P\) has elevation \(30^\circ\) and the top \(T\) has elevation \(50^\circ\).
Heights above ground of the two points:
\[\text{height of }P = 20\tan 30^\circ = 20(0.5774) = 11.547\text{ m}\]
\[\text{height of }T = 20\tan 50^\circ = 20(1.1918) = 23.835\text{ m}\]
(a) \(|PT| = 23.835 - 11.547 = 12.288 \approx 12.3\text{ m (3 s.f.)}\).
(b) Distance \(|HT|\) (the slant line to the top):
\[|HT| = \frac{20}{\cos 50^\circ} = \frac{20}{0.6428} = 31.1\text{ m (3 s.f.)}\]
(c) For the angle of depression of \(H\) from the top to be \(40^\circ\), the angle of elevation of \(T\) from \(H\) must be \(40^\circ\). With the top still \(23.835\) m high, the required horizontal distance \(x\) satisfies:
\[\tan 40^\circ = \frac{23.835}{x} \;\Rightarrow\; x = \frac{23.835}{0.8391} = 28.4\text{ m (3 s.f.)}\]
So \(H\) must be \(28.4\) m from the foot of the tower.
Tambaya 2 Rahoto
Sonny is twice as old as Wale. Four years ago, he was four times as old as Wale. When will the sum of their ages be 66?
Let Wale's present age be \(w\). Since Sonny is twice as old, Sonny is \(2w\).
Four years ago Sonny was four times as old as Wale:
\[2w - 4 = 4(w - 4)\]
\[2w - 4 = 4w - 16 \;\Rightarrow\; 12 = 2w \;\Rightarrow\; w = 6\]
So presently Wale is \(6\) and Sonny is \(12\); their ages sum to \(18\).
Each year that passes adds \(2\) to the total (one year to each person). Let \(n\) be the number of years until the sum is \(66\):
\[18 + 2n = 66 \;\Rightarrow\; 2n = 48 \;\Rightarrow\; n = 24\]
The sum of their ages will be \(66\) in \(\mathbf{24}\) years' time (when Wale is 30 and Sonny is 36).
Bayanin Amsa
Let Wale's present age be \(w\). Since Sonny is twice as old, Sonny is \(2w\).
Four years ago Sonny was four times as old as Wale:
\[2w - 4 = 4(w - 4)\]
\[2w - 4 = 4w - 16 \;\Rightarrow\; 12 = 2w \;\Rightarrow\; w = 6\]
So presently Wale is \(6\) and Sonny is \(12\); their ages sum to \(18\).
Each year that passes adds \(2\) to the total (one year to each person). Let \(n\) be the number of years until the sum is \(66\):
\[18 + 2n = 66 \;\Rightarrow\; 2n = 48 \;\Rightarrow\; n = 24\]
The sum of their ages will be \(66\) in \(\mathbf{24}\) years' time (when Wale is 30 and Sonny is 36).
Tambaya 3 Rahoto
(a) 
In the diagram, TU is tangent to the circle. < RVU = 100° and < URS = 36°. Calculate the value of angle STU.
(b) In triangle XYZ, |XY| = 5 cm, |YZ| = 8 cm and |XZ| = 6 cm. P is a point on the side XY such that |XP| = 2 cm and the line through P, parallel to YZ meets XZ at Q. Calculate |QZ|.
(a) Reading the diagram. \(R, V, S, U\) lie on the circle (\(R\) at top, \(V\) on the left, \(S\) on the right, \(U\) at the bottom). The line \(TU\) is the tangent that touches the circle at \(U\), and the secant through \(R\) and \(S\) is produced to meet the tangent at the external point \(T\). We are given \(\angle RVU = 100^\circ\) and \(\angle URS = 36^\circ\).
Step 1: Find arc \(SU\). \(\angle URS = 36^\circ\) is an inscribed angle at \(R\) standing on arc \(SU\) (the arc not containing \(R\)):
\[\text{arc } SU = 2\times 36^\circ = 72^\circ.\]
Step 2: Find arc \(RU\) on the far side. \(\angle RVU = 100^\circ\) is an inscribed angle at \(V\) standing on the arc \(RU\) that does not contain \(V\) (the arc going \(R\to S\to U\)):
\[\text{arc } RSU = \text{arc } RS + \text{arc } SU = 2\times 100^\circ = 200^\circ.\]
The remaining arc from \(R\) to \(U\) through \(V\) is therefore
\[\text{arc } RVU = 360^\circ - 200^\circ = 160^\circ.\]
Step 3: Apply the tangent-secant angle rule at \(T\). The angle between a tangent and a secant drawn from an external point equals half the difference of the two intercepted arcs. Here the far arc (between tangent point \(U\) and far point \(R\), through \(V\)) is \(160^\circ\) and the near arc (between \(U\) and near point \(S\)) is \(72^\circ\):
\[\angle STU = \tfrac{1}{2}\left(\text{arc } RVU - \text{arc } SU\right) = \tfrac{1}{2}(160^\circ - 72^\circ) = \tfrac{1}{2}(88^\circ) = 44^\circ.\]
Answer (a): \(\angle STU = 44^\circ\).
(b) Triangle \(XYZ\): \(|XY| = 5\), \(|YZ| = 8\), \(|XZ| = 6\text{ cm}\), with \(P\) on \(XY\), \(|XP| = 2\text{ cm}\), and \(PQ \parallel YZ\) with \(Q\) on \(XZ\).
Since \(PQ \parallel YZ\), triangle \(XPQ\) is similar to triangle \(XYZ\), so corresponding sides are proportional:
\[\frac{XP}{XY} = \frac{XQ}{XZ} \;\Rightarrow\; \frac{2}{5} = \frac{XQ}{6}.\]
\[XQ = \frac{2}{5}\times 6 = 2.4\text{ cm}.\]
\[|QZ| = |XZ| - |XQ| = 6 - 2.4 = 3.6\text{ cm}.\]
Answer (b): \(|QZ| = 3.6\text{ cm}\).
Bayanin Amsa
(a) Reading the diagram. \(R, V, S, U\) lie on the circle (\(R\) at top, \(V\) on the left, \(S\) on the right, \(U\) at the bottom). The line \(TU\) is the tangent that touches the circle at \(U\), and the secant through \(R\) and \(S\) is produced to meet the tangent at the external point \(T\). We are given \(\angle RVU = 100^\circ\) and \(\angle URS = 36^\circ\).
Step 1: Find arc \(SU\). \(\angle URS = 36^\circ\) is an inscribed angle at \(R\) standing on arc \(SU\) (the arc not containing \(R\)):
\[\text{arc } SU = 2\times 36^\circ = 72^\circ.\]
Step 2: Find arc \(RU\) on the far side. \(\angle RVU = 100^\circ\) is an inscribed angle at \(V\) standing on the arc \(RU\) that does not contain \(V\) (the arc going \(R\to S\to U\)):
\[\text{arc } RSU = \text{arc } RS + \text{arc } SU = 2\times 100^\circ = 200^\circ.\]
The remaining arc from \(R\) to \(U\) through \(V\) is therefore
\[\text{arc } RVU = 360^\circ - 200^\circ = 160^\circ.\]
Step 3: Apply the tangent-secant angle rule at \(T\). The angle between a tangent and a secant drawn from an external point equals half the difference of the two intercepted arcs. Here the far arc (between tangent point \(U\) and far point \(R\), through \(V\)) is \(160^\circ\) and the near arc (between \(U\) and near point \(S\)) is \(72^\circ\):
\[\angle STU = \tfrac{1}{2}\left(\text{arc } RVU - \text{arc } SU\right) = \tfrac{1}{2}(160^\circ - 72^\circ) = \tfrac{1}{2}(88^\circ) = 44^\circ.\]
Answer (a): \(\angle STU = 44^\circ\).
(b) Triangle \(XYZ\): \(|XY| = 5\), \(|YZ| = 8\), \(|XZ| = 6\text{ cm}\), with \(P\) on \(XY\), \(|XP| = 2\text{ cm}\), and \(PQ \parallel YZ\) with \(Q\) on \(XZ\).
Since \(PQ \parallel YZ\), triangle \(XPQ\) is similar to triangle \(XYZ\), so corresponding sides are proportional:
\[\frac{XP}{XY} = \frac{XQ}{XZ} \;\Rightarrow\; \frac{2}{5} = \frac{XQ}{6}.\]
\[XQ = \frac{2}{5}\times 6 = 2.4\text{ cm}.\]
\[|QZ| = |XZ| - |XQ| = 6 - 2.4 = 3.6\text{ cm}.\]
Answer (b): \(|QZ| = 3.6\text{ cm}\).
Tambaya 4 Rahoto
In the diagram, O is the centre of the circleand XY is a chord. If the radius is 5 cm and /XY/ = 6 cm, calculate, correct to 2 decimal places, the :
(a) angle which XY subtends at the centre O ;
(b) area of the shaded portion.
From the diagram, \(O\) is the centre, radius \(OX = OY = 5\ \text{cm}\), chord \(|XY| = 6\ \text{cm}\), and the shaded portion is the minor segment cut off by the chord \(XY\) (the region between the chord and the minor arc on the right).
(a) Angle subtended at the centre, \(\angle XOY\):
Drop the perpendicular from \(O\) to the midpoint \(M\) of \(XY\). Then \(XM = 3\ \text{cm}\) and:
\[\sin\left(\frac{\angle XOY}{2}\right) = \frac{XM}{OX} = \frac{3}{5} = 0.6\]\[\frac{\angle XOY}{2} = \sin^{-1}(0.6) = 36.8699^\circ\]\[\angle XOY = 2 \times 36.8699^\circ = 73.7398^\circ\]\(\angle XOY \approx 73.74^\circ\)
(b) Area of the shaded (minor) segment \(=\) area of sector \(XOY\) \(-\) area of triangle \(XOY\).
Area of sector:
\[\frac{\theta}{360^\circ} \times \pi r^2 = \frac{73.7398}{360} \times \frac{22}{7} \times 5^2\]\[= 0.204833 \times 78.5714 = 16.09\ \text{cm}^2\]Area of triangle \(XOY\):
\[\frac{1}{2} r^2 \sin\theta = \frac{1}{2} \times 25 \times \sin 73.7398^\circ = \frac{1}{2} \times 25 \times 0.96 = 12.00\ \text{cm}^2\]Area of shaded segment:
\[16.09 - 12.00 = 4.09\ \text{cm}^2\]Shaded area \(\approx 4.09\ \text{cm}^2\) (to 2 decimal places).
Bayanin Amsa
From the diagram, \(O\) is the centre, radius \(OX = OY = 5\ \text{cm}\), chord \(|XY| = 6\ \text{cm}\), and the shaded portion is the minor segment cut off by the chord \(XY\) (the region between the chord and the minor arc on the right).
(a) Angle subtended at the centre, \(\angle XOY\):
Drop the perpendicular from \(O\) to the midpoint \(M\) of \(XY\). Then \(XM = 3\ \text{cm}\) and:
\[\sin\left(\frac{\angle XOY}{2}\right) = \frac{XM}{OX} = \frac{3}{5} = 0.6\]\[\frac{\angle XOY}{2} = \sin^{-1}(0.6) = 36.8699^\circ\]\[\angle XOY = 2 \times 36.8699^\circ = 73.7398^\circ\]\(\angle XOY \approx 73.74^\circ\)
(b) Area of the shaded (minor) segment \(=\) area of sector \(XOY\) \(-\) area of triangle \(XOY\).
Area of sector:
\[\frac{\theta}{360^\circ} \times \pi r^2 = \frac{73.7398}{360} \times \frac{22}{7} \times 5^2\]\[= 0.204833 \times 78.5714 = 16.09\ \text{cm}^2\]Area of triangle \(XOY\):
\[\frac{1}{2} r^2 \sin\theta = \frac{1}{2} \times 25 \times \sin 73.7398^\circ = \frac{1}{2} \times 25 \times 0.96 = 12.00\ \text{cm}^2\]Area of shaded segment:
\[16.09 - 12.00 = 4.09\ \text{cm}^2\]Shaded area \(\approx 4.09\ \text{cm}^2\) (to 2 decimal places).
Tambaya 5 Rahoto
(a) Simplify : \(\frac{1\frac{1}{4} + \frac{7}{9}}{1\frac{4}{9} - 2\frac{2}{3} \times \frac{9}{64}}\)
(b) Given that \(\sin x = \frac{2}{3}\), evaluate, leaving your answer in surd form and without using tables or calculator, \(\tan x - \cos x\).
(a) Convert to improper fractions.
Numerator: \(1\tfrac14 + \tfrac79 = \tfrac54 + \tfrac79 = \dfrac{45 + 28}{36} = \dfrac{73}{36}\).
Denominator: first \(2\tfrac23 \times \tfrac{9}{64} = \tfrac83 \times \tfrac{9}{64} = \dfrac{72}{192} = \dfrac38\). Then \(1\tfrac49 - \tfrac38 = \tfrac{13}{9} - \tfrac38 = \dfrac{104 - 27}{72} = \dfrac{77}{72}\).
\[\frac{73/36}{77/72} = \frac{73}{36}\times\frac{72}{77} = \frac{73 \times 2}{77} = \frac{146}{77} = 1\tfrac{69}{77}\]
(b) Given \(\sin x = \dfrac23\), the adjacent side is \(\sqrt{3^2 - 2^2} = \sqrt5\), so
\[\cos x = \frac{\sqrt5}{3}, \qquad \tan x = \frac{2}{\sqrt5} = \frac{2\sqrt5}{5}\]
\[\tan x - \cos x = \frac{2}{\sqrt5} - \frac{\sqrt5}{3} = \sqrt5\left(\frac{2}{5} - \frac{1}{3}\right) = \sqrt5\cdot\frac{6 - 5}{15} = \frac{\sqrt5}{15}\]
Bayanin Amsa
(a) Convert to improper fractions.
Numerator: \(1\tfrac14 + \tfrac79 = \tfrac54 + \tfrac79 = \dfrac{45 + 28}{36} = \dfrac{73}{36}\).
Denominator: first \(2\tfrac23 \times \tfrac{9}{64} = \tfrac83 \times \tfrac{9}{64} = \dfrac{72}{192} = \dfrac38\). Then \(1\tfrac49 - \tfrac38 = \tfrac{13}{9} - \tfrac38 = \dfrac{104 - 27}{72} = \dfrac{77}{72}\).
\[\frac{73/36}{77/72} = \frac{73}{36}\times\frac{72}{77} = \frac{73 \times 2}{77} = \frac{146}{77} = 1\tfrac{69}{77}\]
(b) Given \(\sin x = \dfrac23\), the adjacent side is \(\sqrt{3^2 - 2^2} = \sqrt5\), so
\[\cos x = \frac{\sqrt5}{3}, \qquad \tan x = \frac{2}{\sqrt5} = \frac{2\sqrt5}{5}\]
\[\tan x - \cos x = \frac{2}{\sqrt5} - \frac{\sqrt5}{3} = \sqrt5\left(\frac{2}{5} - \frac{1}{3}\right) = \sqrt5\cdot\frac{6 - 5}{15} = \frac{\sqrt5}{15}\]
Tambaya 6 Rahoto
(a) A box contains 40 identical discs which are either red or white. If the probability of picking a red disc is \(\frac{1}{4}\); Calculate the number of (i) white discs ; (ii) red discs that should be added such that the probability of picking a red disc will be \(\frac{1}{3}\).
(b) A salesman bought some plates at N50.00 each. If he sold all of them for N600.00 and made a profit of 20% on the transaction, how many plates did he buy?
(a) Total discs \(= 40\), and \(P(\text{red}) = \dfrac14\), so number of red \(= \dfrac14\times 40 = 10\).
(i) Number of white discs \(= 40 - 10 = 30\).
(ii) Let \(x\) red discs be added. New red count \(= 10 + x\); new total \(= 40 + x\). Require \(P(\text{red}) = \dfrac13\):
\[\frac{10 + x}{40 + x} = \frac13 \;\Rightarrow\; 3(10 + x) = 40 + x \;\Rightarrow\; 30 + 3x = 40 + x\]
\[2x = 10 \;\Rightarrow\; x = 5\]
So \(5\) red discs should be added.
(b) Cost price per plate \(=\) N50. A profit of \(20\%\) on total cost means
\[\text{Selling price} = 1.20 \times \text{Cost price}\]
\[600 = 1.20 \times \text{Cost price} \;\Rightarrow\; \text{Cost price} = \frac{600}{1.20} = \text{N}500\]
\[\text{Number of plates} = \frac{500}{50} = 10\]
Bayanin Amsa
(a) Total discs \(= 40\), and \(P(\text{red}) = \dfrac14\), so number of red \(= \dfrac14\times 40 = 10\).
(i) Number of white discs \(= 40 - 10 = 30\).
(ii) Let \(x\) red discs be added. New red count \(= 10 + x\); new total \(= 40 + x\). Require \(P(\text{red}) = \dfrac13\):
\[\frac{10 + x}{40 + x} = \frac13 \;\Rightarrow\; 3(10 + x) = 40 + x \;\Rightarrow\; 30 + 3x = 40 + x\]
\[2x = 10 \;\Rightarrow\; x = 5\]
So \(5\) red discs should be added.
(b) Cost price per plate \(=\) N50. A profit of \(20\%\) on total cost means
\[\text{Selling price} = 1.20 \times \text{Cost price}\]
\[600 = 1.20 \times \text{Cost price} \;\Rightarrow\; \text{Cost price} = \frac{600}{1.20} = \text{N}500\]
\[\text{Number of plates} = \frac{500}{50} = 10\]
Tambaya 7 Rahoto
(a) (i) Using a scale of 2 cm to 1 unit on both axes, on the same graph sheet, draw the graphs of \(y - \frac{3x}{4} = 3\) and \(y + 2x = 6\).
(ii) From your graph, find the coordinates of the point of intersection of the two graphs.
(iii) Show, on the graph sheet, the region satisfied by the inequality \(y - \frac{3}{4}x \geq 3\).
(b) Given that \(x^{2} + bx + 18\) is factorized as \((x + 2)(x + c)\). Find the values of c and b.
(a)(i) Rearrange the equations:
\[y-\frac{3x}{4}=3\quad\Rightarrow\quad y=\frac34x+3\]
\[y+2x=6\quad\Rightarrow\quad y=6-2x\]
| Line | Points used for plotting |
|---|---|
| \(y=\frac34x+3\) | \((-4,0),\ (0,3),\ (4,6)\) |
| \(y=6-2x\) | \((0,6),\ (1,4),\ (3,0)\) |
Using the scale of 2 cm to 1 unit on both axes, the required graph is:
The boundary \(y-\frac34x=3\) is drawn as a solid line since the inequality includes equality. Region \(R\), the side above this line, is shaded.
(ii) The two lines intersect at approximately
\[\boxed{(1.1,\ 3.8)}\]
This agrees with the exact solution:
\[\frac34x+3=6-2x\]
\[\frac{11}{4}x=3\quad\Rightarrow\quad x=\frac{12}{11}\]
\[y=6-2\left(\frac{12}{11}\right)=\frac{42}{11}\]
Thus the exact coordinates are \(\left(\frac{12}{11},\frac{42}{11}\right)\), which are approximately \((1.1,3.8)\).
(iii) \[y-\frac34x\geq3\quad\Rightarrow\quad y\geq\frac34x+3.\]
Hence, the required region is the half-plane on and above \(y=\frac34x+3\), labelled \(R\) on the graph.
(b)
\[(x+2)(x+c)=x^2+(c+2)x+2c.\]
Comparing with \(x^2+bx+18\):
\[2c=18\Rightarrow c=9,\]
\[b=c+2=9+2=11.\]
\[\boxed{c=9,\quad b=11}\]
Bayanin Amsa
(a)(i) Rearrange the equations:
\[y-\frac{3x}{4}=3\quad\Rightarrow\quad y=\frac34x+3\]
\[y+2x=6\quad\Rightarrow\quad y=6-2x\]
| Line | Points used for plotting |
|---|---|
| \(y=\frac34x+3\) | \((-4,0),\ (0,3),\ (4,6)\) |
| \(y=6-2x\) | \((0,6),\ (1,4),\ (3,0)\) |
Using the scale of 2 cm to 1 unit on both axes, the required graph is:
The boundary \(y-\frac34x=3\) is drawn as a solid line since the inequality includes equality. Region \(R\), the side above this line, is shaded.
(ii) The two lines intersect at approximately
\[\boxed{(1.1,\ 3.8)}\]
This agrees with the exact solution:
\[\frac34x+3=6-2x\]
\[\frac{11}{4}x=3\quad\Rightarrow\quad x=\frac{12}{11}\]
\[y=6-2\left(\frac{12}{11}\right)=\frac{42}{11}\]
Thus the exact coordinates are \(\left(\frac{12}{11},\frac{42}{11}\right)\), which are approximately \((1.1,3.8)\).
(iii) \[y-\frac34x\geq3\quad\Rightarrow\quad y\geq\frac34x+3.\]
Hence, the required region is the half-plane on and above \(y=\frac34x+3\), labelled \(R\) on the graph.
(b)
\[(x+2)(x+c)=x^2+(c+2)x+2c.\]
Comparing with \(x^2+bx+18\):
\[2c=18\Rightarrow c=9,\]
\[b=c+2=9+2=11.\]
\[\boxed{c=9,\quad b=11}\]
Tambaya 8 Rahoto
Three towns X, Y and Z are such that Y is 20 km from X and 22 km from Z. Town X is 18 km from Z. A health centre is to be built by the government to serve the three towns. The centre is to be located such that patients from X and Y travel equal distance to access the health centre while patients from Z will travel exactly 10 km to reach the Health centre.
(a) Using a scale of 1 cm to 2 km, find the construction, using a pair of compasses and ruler only, the possible positions the Health centre can be located.
(b) In how many possible locations can the Health centre be built?
(c) Measure and record the distances of the location from town X.
(d) Which of these locations would be convenient for all three towns?
(a) Construction using scale \(1\text{ cm}:2\text{ km}\)
- Draw \(XZ=18\) km, which is \(9\) cm on the scale.
- With centre \(X\), draw an arc of radius \(10\) cm. With centre \(Z\), draw an arc of radius \(11\) cm. Their intersection gives town \(Y\), so that \(XY=20\) km and \(YZ=22\) km.
- Construct the perpendicular bisector of \(XY\). Every point on this line is the same distance from \(X\) and \(Y\).
- With centre \(Z\), draw a circle of radius \(5\) cm, representing \(10\) km.
- The two points where this circle crosses the perpendicular bisector are the possible health-centre locations.
(b) There are two possible locations, because the perpendicular bisector of \(XY\) intersects the circle centred at \(Z\) at two points.
(c) Measuring from \(X\) on an accurate scale construction gives approximately:
- \(XH_1 \approx 12.7\) km;
- \(XH_2 \approx 28.0\) km.
Values such as \(12.4\) km and \(27.6\) km can result from ruler measurement on a less precise drawing, but the intended readings are approximately \(12\)–\(13\) km and \(28\) km respectively.
(d) The convenient location is \(H_1\), the site approximately \(12.7\) km from \(X\). It lies inside the triangle formed by the three towns, while still being \(10\) km from \(Z\) and equidistant from \(X\) and \(Y\). The other site lies well outside the triangle and is much farther from \(X\) and \(Y\).
Examination reminder: “Equal distance from two towns” means construct the perpendicular bisector of the line joining them. “A fixed distance from a town” means draw a circle centred on that town.
Bayanin Amsa
(a) Construction using scale \(1\text{ cm}:2\text{ km}\)
- Draw \(XZ=18\) km, which is \(9\) cm on the scale.
- With centre \(X\), draw an arc of radius \(10\) cm. With centre \(Z\), draw an arc of radius \(11\) cm. Their intersection gives town \(Y\), so that \(XY=20\) km and \(YZ=22\) km.
- Construct the perpendicular bisector of \(XY\). Every point on this line is the same distance from \(X\) and \(Y\).
- With centre \(Z\), draw a circle of radius \(5\) cm, representing \(10\) km.
- The two points where this circle crosses the perpendicular bisector are the possible health-centre locations.
(b) There are two possible locations, because the perpendicular bisector of \(XY\) intersects the circle centred at \(Z\) at two points.
(c) Measuring from \(X\) on an accurate scale construction gives approximately:
- \(XH_1 \approx 12.7\) km;
- \(XH_2 \approx 28.0\) km.
Values such as \(12.4\) km and \(27.6\) km can result from ruler measurement on a less precise drawing, but the intended readings are approximately \(12\)–\(13\) km and \(28\) km respectively.
(d) The convenient location is \(H_1\), the site approximately \(12.7\) km from \(X\). It lies inside the triangle formed by the three towns, while still being \(10\) km from \(Z\) and equidistant from \(X\) and \(Y\). The other site lies well outside the triangle and is much farther from \(X\) and \(Y\).
Examination reminder: “Equal distance from two towns” means construct the perpendicular bisector of the line joining them. “A fixed distance from a town” means draw a circle centred on that town.
Tambaya 9 Rahoto
(a) 
In the diagram, /PQ/ = 6 cm, /QR/ = 13 cm, /RS/ = 5 cm and < RSQ is a right- angled triangle. Calculate, correct to one decimal place, /PS/.
(b) 
The diagram show a wooden structure in the form of a cone mounted on a hemispherical base. The vertical height of the cone is 24 cm and the base radius 7 cm. Calculate, correct to 3 significant figures, the surface area of the structure. [Take \(\pi = \frac{22}{7}\)].
(a) Calculating /PS/
From the diagram, \(P, Q, R\) lie on one straight line with \(PQ = 6\text{ cm}\) and \(QR = 13\text{ cm}\). The triangle \(QSR\) is right-angled at \(S\), with \(RS = 5\text{ cm}\) and hypotenuse \(QR = 13\text{ cm}\).
Step 1: Find QS using Pythagoras in triangle QSR.
\[ QS^2 = QR^2 - RS^2 = 13^2 - 5^2 = 169 - 25 = 144 \]
\[ QS = 12\text{ cm} \]
Step 2: Find angle RQS.
\[ \cos(\angle RQS) = \frac{QS}{QR} = \frac{12}{13} \]
Step 3: Use triangle PQS. Since \(P, Q, R\) are collinear, \(\angle PQS\) and \(\angle RQS\) are supplementary, so \(\cos(\angle PQS) = -\dfrac{12}{13}\).
Applying the cosine rule in triangle \(PQS\) with \(PQ = 6\), \(QS = 12\):
\[ PS^2 = PQ^2 + QS^2 - 2\,(PQ)(QS)\cos(\angle PQS) \]
\[ PS^2 = 6^2 + 12^2 - 2(6)(12)\left(-\tfrac{12}{13}\right) \]
\[ PS^2 = 36 + 144 + \frac{1728}{13} = 180 + 132.92 = 312.92 \]
\[ PS = \sqrt{312.92} \approx 17.7\text{ cm} \]
/PS/ \(\approx\) 17.7 cm.
(b) Surface area of the structure
The structure is a cone (height \(24\text{ cm}\), base radius \(7\text{ cm}\)) mounted on a hemisphere of the same radius \(7\text{ cm}\). The exposed surface is the curved surface of the cone plus the curved surface of the hemisphere (the joining circle is hidden).
Slant height of cone:
\[ l = \sqrt{h^2 + r^2} = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25\text{ cm} \]
Curved surface area of cone:
\[ \pi r l = \frac{22}{7} \times 7 \times 25 = 550\text{ cm}^2 \]
Curved surface area of hemisphere:
\[ 2\pi r^2 = 2 \times \frac{22}{7} \times 7^2 = 2 \times 22 \times 7 = 308\text{ cm}^2 \]
Total surface area:
\[ 550 + 308 = 858\text{ cm}^2 \]
Surface area \(= 858\text{ cm}^2\) (to 3 significant figures).
Bayanin Amsa
(a) Calculating /PS/
From the diagram, \(P, Q, R\) lie on one straight line with \(PQ = 6\text{ cm}\) and \(QR = 13\text{ cm}\). The triangle \(QSR\) is right-angled at \(S\), with \(RS = 5\text{ cm}\) and hypotenuse \(QR = 13\text{ cm}\).
Step 1: Find QS using Pythagoras in triangle QSR.
\[ QS^2 = QR^2 - RS^2 = 13^2 - 5^2 = 169 - 25 = 144 \]
\[ QS = 12\text{ cm} \]
Step 2: Find angle RQS.
\[ \cos(\angle RQS) = \frac{QS}{QR} = \frac{12}{13} \]
Step 3: Use triangle PQS. Since \(P, Q, R\) are collinear, \(\angle PQS\) and \(\angle RQS\) are supplementary, so \(\cos(\angle PQS) = -\dfrac{12}{13}\).
Applying the cosine rule in triangle \(PQS\) with \(PQ = 6\), \(QS = 12\):
\[ PS^2 = PQ^2 + QS^2 - 2\,(PQ)(QS)\cos(\angle PQS) \]
\[ PS^2 = 6^2 + 12^2 - 2(6)(12)\left(-\tfrac{12}{13}\right) \]
\[ PS^2 = 36 + 144 + \frac{1728}{13} = 180 + 132.92 = 312.92 \]
\[ PS = \sqrt{312.92} \approx 17.7\text{ cm} \]
/PS/ \(\approx\) 17.7 cm.
(b) Surface area of the structure
The structure is a cone (height \(24\text{ cm}\), base radius \(7\text{ cm}\)) mounted on a hemisphere of the same radius \(7\text{ cm}\). The exposed surface is the curved surface of the cone plus the curved surface of the hemisphere (the joining circle is hidden).
Slant height of cone:
\[ l = \sqrt{h^2 + r^2} = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25\text{ cm} \]
Curved surface area of cone:
\[ \pi r l = \frac{22}{7} \times 7 \times 25 = 550\text{ cm}^2 \]
Curved surface area of hemisphere:
\[ 2\pi r^2 = 2 \times \frac{22}{7} \times 7^2 = 2 \times 22 \times 7 = 308\text{ cm}^2 \]
Total surface area:
\[ 550 + 308 = 858\text{ cm}^2 \]
Surface area \(= 858\text{ cm}^2\) (to 3 significant figures).
Tambaya 10 Rahoto
(a) Copy and complete the table of values for \(y = 1 - 4\cos x\).
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° | 210° | 240° | 270° | 300° |
| y | -3.0 | 1.0 | 4.5 | -1.0 |
(b) Using a scale of 2cm to 30° on the x- axis and 2cm to 1 unit on the y- axis, draw the graph of \(y = 1 - 4\cos x\) for \(0° \leq x \leq 360°\).
(c) Use the graph to : (i) solve the equation \(1 - 4\cos x = 0\) ; (ii) find the value of y when x = 105° ; (iii) find x when y = 1.5.
(a) For each value of x, calculate \(y=1-4\cos x\), correct to 1 decimal place.
| \(x\) | \(0^\circ\) | \(30^\circ\) | \(60^\circ\) | \(90^\circ\) | \(120^\circ\) | \(150^\circ\) | \(180^\circ\) | \(210^\circ\) | \(240^\circ\) | \(270^\circ\) | \(300^\circ\) |
|---|---|---|---|---|---|---|---|---|---|---|---|
| \(y\) | -3.0 | -2.5 | -1.0 | 1.0 | 3.0 | 4.5 | 5.0 | 4.5 | 3.0 | 1.0 | -1.0 |
(b) The completed graph of \(y=1-4\cos x\), including \(y=-2.5\) at \(330^\circ\) and \(y=-3.0\) at \(360^\circ\), is shown below.
(c)
(i) The points where the curve cuts the \(x\)-axis give
\[x\approx75^\circ\quad\text{or}\quad285^\circ.\]
(ii) At \(x=105^\circ\), the ordinate of the curve is
\[y\approx2.0.\]
(iii) Drawing the horizontal line \(y=1.5\) and reading the two intersections with the curve gives
\[x\approx96^\circ\quad\text{or}\quad264^\circ.\]
Bayanin Amsa
(a) For each value of x, calculate \(y=1-4\cos x\), correct to 1 decimal place.
| \(x\) | \(0^\circ\) | \(30^\circ\) | \(60^\circ\) | \(90^\circ\) | \(120^\circ\) | \(150^\circ\) | \(180^\circ\) | \(210^\circ\) | \(240^\circ\) | \(270^\circ\) | \(300^\circ\) |
|---|---|---|---|---|---|---|---|---|---|---|---|
| \(y\) | -3.0 | -2.5 | -1.0 | 1.0 | 3.0 | 4.5 | 5.0 | 4.5 | 3.0 | 1.0 | -1.0 |
(b) The completed graph of \(y=1-4\cos x\), including \(y=-2.5\) at \(330^\circ\) and \(y=-3.0\) at \(360^\circ\), is shown below.
(c)
(i) The points where the curve cuts the \(x\)-axis give
\[x\approx75^\circ\quad\text{or}\quad285^\circ.\]
(ii) At \(x=105^\circ\), the ordinate of the curve is
\[y\approx2.0.\]
(iii) Drawing the horizontal line \(y=1.5\) and reading the two intersections with the curve gives
\[x\approx96^\circ\quad\text{or}\quad264^\circ.\]
Tambaya 11 Rahoto
(a) A boy had M Dalasis (D). He spent D15 and shared the remainder equally with his sister. If the sister's share was equal to \(\frac{1}{3}\) of M, find the value of M.
(b) A number of tourists were interviewed on their choice of means of travel. Two- thirds said that they travelled by road, \(\frac{13}{30}\) by air and \(\frac{4}{15}\) by both air and road. If 20 tourists did not travel by either air or road ; (i) represent the information on a Venn diagram ; (ii) how many tourists (1) were interviewed ; (2) travelled by air only?
(a) The boy had \(M\) dalasis, spent D15, leaving \(M - 15\). This remainder is shared equally between him and his sister, so the sister's share is \(\dfrac{M - 15}{2}\). We are told this equals \(\dfrac13 M\):
\[\frac{M - 15}{2} = \frac{M}{3}\]
\[3(M - 15) = 2M \;\Rightarrow\; 3M - 45 = 2M \;\Rightarrow\; M = 45\]
So \(M = \text{D}45\).
(b) Let the total number of tourists be \(N\). Travelled by road \(= \tfrac23 N\), by air \(= \tfrac{13}{30}N\), by both \(= \tfrac{4}{15}N\).
(i) Venn diagram: two intersecting circles (Road, Air). Both region \(= \tfrac{4}{15}N\); road only \(= \tfrac23 N - \tfrac{4}{15}N = \tfrac{6}{15}N\); air only \(= \tfrac{13}{30}N - \tfrac{4}{15}N = \tfrac{5}{30}N\); outside both \(= 20\).
(ii)(1) Those using road or air:
\[\frac23 + \frac{13}{30} - \frac{4}{15} = \frac{20 + 13 - 8}{30} = \frac{25}{30} = \frac56\]
So the fraction using neither is \(1 - \tfrac56 = \tfrac16\), and \(\tfrac16 N = 20 \Rightarrow N = 120\). 120 tourists were interviewed.
(2) Air only \(= \tfrac{13}{30}N - \tfrac{4}{15}N = \tfrac{5}{30}\times 120 = 20\) tourists.
Bayanin Amsa
(a) The boy had \(M\) dalasis, spent D15, leaving \(M - 15\). This remainder is shared equally between him and his sister, so the sister's share is \(\dfrac{M - 15}{2}\). We are told this equals \(\dfrac13 M\):
\[\frac{M - 15}{2} = \frac{M}{3}\]
\[3(M - 15) = 2M \;\Rightarrow\; 3M - 45 = 2M \;\Rightarrow\; M = 45\]
So \(M = \text{D}45\).
(b) Let the total number of tourists be \(N\). Travelled by road \(= \tfrac23 N\), by air \(= \tfrac{13}{30}N\), by both \(= \tfrac{4}{15}N\).
(i) Venn diagram: two intersecting circles (Road, Air). Both region \(= \tfrac{4}{15}N\); road only \(= \tfrac23 N - \tfrac{4}{15}N = \tfrac{6}{15}N\); air only \(= \tfrac{13}{30}N - \tfrac{4}{15}N = \tfrac{5}{30}N\); outside both \(= 20\).
(ii)(1) Those using road or air:
\[\frac23 + \frac{13}{30} - \frac{4}{15} = \frac{20 + 13 - 8}{30} = \frac{25}{30} = \frac56\]
So the fraction using neither is \(1 - \tfrac56 = \tfrac16\), and \(\tfrac16 N = 20 \Rightarrow N = 120\). 120 tourists were interviewed.
(2) Air only \(= \tfrac{13}{30}N - \tfrac{4}{15}N = \tfrac{5}{30}\times 120 = 20\) tourists.
Tambaya 12 Rahoto
| Class Interval |
Frequency |
| 60 - 64 | 2 |
| 65 - 69 | 3 |
| 70 - 74 | 6 |
| 75 - 79 | 11 |
| 80 - 84 | 8 |
| 85 - 89 | 7 |
| 90 - 94 | 2 |
| 95 - 99 | 1 |
The table shows the distribution of marks scored by students in an examination. Calculate, correct to 2 decimal places, the
(a) mean ; (b) standard deviation of the distribution.
Working table (midpoint \(x\), \(N = 40\)).
| Class | x | f | fx | fx² |
|---|---|---|---|---|
| 60-64 | 62 | 2 | 124 | 7688 |
| 65-69 | 67 | 3 | 201 | 13467 |
| 70-74 | 72 | 6 | 432 | 31104 |
| 75-79 | 77 | 11 | 847 | 65219 |
| 80-84 | 82 | 8 | 656 | 53792 |
| 85-89 | 87 | 7 | 609 | 52983 |
| 90-94 | 92 | 2 | 184 | 16928 |
| 95-99 | 97 | 1 | 97 | 9409 |
| Total | 40 | 3150 | 250590 |
(a) Mean.
\[\bar{x} = \frac{\sum fx}{\sum f} = \frac{3150}{40} = \mathbf{78.75}\](b) Standard deviation.
\[\sigma = \sqrt{\frac{\sum fx^{2}}{N} - \bar{x}^{2}} = \sqrt{\frac{250590}{40} - 78.75^{2}}\] \[= \sqrt{6264.75 - 6201.5625} = \sqrt{63.1875} = \mathbf{7.95}\]Bayanin Amsa
Working table (midpoint \(x\), \(N = 40\)).
| Class | x | f | fx | fx² |
|---|---|---|---|---|
| 60-64 | 62 | 2 | 124 | 7688 |
| 65-69 | 67 | 3 | 201 | 13467 |
| 70-74 | 72 | 6 | 432 | 31104 |
| 75-79 | 77 | 11 | 847 | 65219 |
| 80-84 | 82 | 8 | 656 | 53792 |
| 85-89 | 87 | 7 | 609 | 52983 |
| 90-94 | 92 | 2 | 184 | 16928 |
| 95-99 | 97 | 1 | 97 | 9409 |
| Total | 40 | 3150 | 250590 |
(a) Mean.
\[\bar{x} = \frac{\sum fx}{\sum f} = \frac{3150}{40} = \mathbf{78.75}\](b) Standard deviation.
\[\sigma = \sqrt{\frac{\sum fx^{2}}{N} - \bar{x}^{2}} = \sqrt{\frac{250590}{40} - 78.75^{2}}\] \[= \sqrt{6264.75 - 6201.5625} = \sqrt{63.1875} = \mathbf{7.95}\]Tambaya 13 Rahoto
(a)
In the diagram, ABCD is a rectangular garden (3n - 1)m long and (2n + 1)m wide. A wire mesh 135m long is used to mark its boundary and to divide it into 8 equal plots. Find the value of n.
(b) A cylinder with base radius 14 cm has the same volume as a cube of side 22 cm. Calculate the ratio of the total surface area of the cylinder to that of the cube. [Take \(\pi = \frac{22}{7}\)]
(a) The diagram shows the rectangle divided into 4 columns and 2 rows, giving \(4 \times 2 = 8\) equal plots. Length \(AB = (3n-1)\ \text{m}\), width \(BC = (2n+1)\ \text{m}\).
The wire mesh forms the boundary and the internal dividing lines.
Horizontal lines (each of length \(3n-1\)): top, bottom and 1 internal line \(= 3\) lines.
Vertical lines (each of length \(2n+1\)): left, right and 3 internal lines \(= 5\) lines.
Total length of wire:
\[3(3n-1) + 5(2n+1) = 135\]\[9n - 3 + 10n + 5 = 135\]\[19n + 2 = 135\]\[19n = 133 \Rightarrow n = 7\]\(n = 7\)
(b) Take \(\pi = \frac{22}{7}\). Let the cylinder have radius \(r = 14\ \text{cm}\) and height \(h\).
Volume of cube (side \(22\ \text{cm}\)):
\[V = 22^3 = 10648\ \text{cm}^3\]Volume of cylinder equals this:
\[\pi r^2 h = \frac{22}{7} \times 14^2 \times h = 616h\]\[616h = 10648 \Rightarrow h = \frac{10648}{616} = \frac{121}{7}\ \text{cm} \;\left(=17\tfrac{2}{7}\right)\]Total surface area of cylinder:
\[2\pi r(r + h) = 2 \times \frac{22}{7} \times 14 \times \left(14 + \frac{121}{7}\right)\]\[= 88 \times \frac{98 + 121}{7} = 88 \times \frac{219}{7} = \frac{19272}{7}\ \text{cm}^2\]Total surface area of cube:
\[6 \times 22^2 = 6 \times 484 = 2904\ \text{cm}^2\]Ratio (cylinder : cube):
\[\frac{19272}{7} : 2904 = 19272 : 20328\]Dividing both by \(264\):
\[= 73 : 77\]Ratio \(= 73 : 77\)
Bayanin Amsa
(a) The diagram shows the rectangle divided into 4 columns and 2 rows, giving \(4 \times 2 = 8\) equal plots. Length \(AB = (3n-1)\ \text{m}\), width \(BC = (2n+1)\ \text{m}\).
The wire mesh forms the boundary and the internal dividing lines.
Horizontal lines (each of length \(3n-1\)): top, bottom and 1 internal line \(= 3\) lines.
Vertical lines (each of length \(2n+1\)): left, right and 3 internal lines \(= 5\) lines.
Total length of wire:
\[3(3n-1) + 5(2n+1) = 135\]\[9n - 3 + 10n + 5 = 135\]\[19n + 2 = 135\]\[19n = 133 \Rightarrow n = 7\]\(n = 7\)
(b) Take \(\pi = \frac{22}{7}\). Let the cylinder have radius \(r = 14\ \text{cm}\) and height \(h\).
Volume of cube (side \(22\ \text{cm}\)):
\[V = 22^3 = 10648\ \text{cm}^3\]Volume of cylinder equals this:
\[\pi r^2 h = \frac{22}{7} \times 14^2 \times h = 616h\]\[616h = 10648 \Rightarrow h = \frac{10648}{616} = \frac{121}{7}\ \text{cm} \;\left(=17\tfrac{2}{7}\right)\]Total surface area of cylinder:
\[2\pi r(r + h) = 2 \times \frac{22}{7} \times 14 \times \left(14 + \frac{121}{7}\right)\]\[= 88 \times \frac{98 + 121}{7} = 88 \times \frac{219}{7} = \frac{19272}{7}\ \text{cm}^2\]Total surface area of cube:
\[6 \times 22^2 = 6 \times 484 = 2904\ \text{cm}^2\]Ratio (cylinder : cube):
\[\frac{19272}{7} : 2904 = 19272 : 20328\]Dividing both by \(264\):
\[= 73 : 77\]Ratio \(= 73 : 77\)
