WAEC SSCE - General Mathematics - 2018 (Essay)
Question 1 Report
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:
\[a + (n - 1)d = 28 \Rightarrow 4 + 2(n - 1) = 28 \Rightarrow n - 1 = 12 \Rightarrow n = 13\]Answer Details
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:
\[a + (n - 1)d = 28 \Rightarrow 4 + 2(n - 1) = 28 \Rightarrow n - 1 = 12 \Rightarrow n = 13\]Question 2 Report
(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,
\[\sin\theta=\frac{h}{12}.\]
\[h=12(0.6)=7.2\text{ m}.\]
\[\boxed{\text{Ladder }M\text{ touches the wall }7.2\text{ m above the ground.}}\]
Answer Details
(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,
\[\sin\theta=\frac{h}{12}.\]
\[h=12(0.6)=7.2\text{ m}.\]
\[\boxed{\text{Ladder }M\text{ touches the wall }7.2\text{ m above the ground.}}\]
Question 3 Report
(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.
Answer Details
(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.
Question 4 Report
(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
\[\boxed{Q=(-3,5)}.\]
Answer Details
(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
\[\boxed{Q=(-3,5)}.\]
Question 5 Report
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 \(RQ = \tfrac{1}{2}TQ = \tfrac{1}{2}d\tan 28^{\circ}\).
\[\tan(\angle RPQ) = \frac{RQ}{PQ} = \tfrac{1}{2}\tan 28^{\circ} = \tfrac{1}{2}(0.5317) = 0.2659\] \[\angle RPQ = \tan^{-1}(0.2659) \approx 15^{\circ}\]Answer Details
(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 \(RQ = \tfrac{1}{2}TQ = \tfrac{1}{2}d\tan 28^{\circ}\).
\[\tan(\angle RPQ) = \frac{RQ}{PQ} = \tfrac{1}{2}\tan 28^{\circ} = \tfrac{1}{2}(0.5317) = 0.2659\] \[\angle RPQ = \tan^{-1}(0.2659) \approx 15^{\circ}\]Question 6 Report
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 \(30^\circ\):
\[\text{Area of } \triangle PSR = \tfrac{1}{2}(10)(15)\sin 30^\circ = \tfrac{1}{2}(150)(0.5) = 37.5\text{ cm}^2.\]
\[\text{Area of } \triangle PTQ = \tfrac{1}{2}(4)(6)\sin 30^\circ = \tfrac{1}{2}(24)(0.5) = 6\text{ cm}^2.\]
\[\text{Area of } TQRS = 37.5 - 6 = 31.5 \approx 32\text{ cm}^2.\]
\[\boxed{|SR| \approx 8\text{ cm},\qquad \text{Area of } TQRS \approx 32\text{ cm}^2.}\]
Answer Details
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 \(30^\circ\):
\[\text{Area of } \triangle PSR = \tfrac{1}{2}(10)(15)\sin 30^\circ = \tfrac{1}{2}(150)(0.5) = 37.5\text{ cm}^2.\]
\[\text{Area of } \triangle PTQ = \tfrac{1}{2}(4)(6)\sin 30^\circ = \tfrac{1}{2}(24)(0.5) = 6\text{ cm}^2.\]
\[\text{Area of } TQRS = 37.5 - 6 = 31.5 \approx 32\text{ cm}^2.\]
\[\boxed{|SR| \approx 8\text{ cm},\qquad \text{Area of } TQRS \approx 32\text{ cm}^2.}\]
Question 7 Report
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 \(= \dfrac{104.40}{180} \times 100\% = 58\%\).
Answer Details
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 \(= \dfrac{104.40}{180} \times 100\% = 58\%\).
Question 8 Report
(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:
\[\begin{aligned}W(2,3)&\longmapsto W_1(-3,2),\\X(4,-1)&\longmapsto X_1(1,4),\\Y(-3,-4)&\longmapsto Y_1(4,-3),\\Z(-3,2)&\longmapsto Z_1(-2,-3).\end{aligned}\]
Answer Details
(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:
\[\begin{aligned}W(2,3)&\longmapsto W_1(-3,2),\\X(4,-1)&\longmapsto X_1(1,4),\\Y(-3,-4)&\longmapsto Y_1(4,-3),\\Z(-3,2)&\longmapsto Z_1(-2,-3).\end{aligned}\]
Question 9 Report
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) \(\angle QRS \approx 55^\circ\).
Answer Details
(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) \(\angle QRS \approx 55^\circ\).
Question 10 Report
(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 \(= 50\% \times 500 = 250\).
Answer Details
(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 \(= 50\% \times 500 = 250\).
Question 11 Report
| 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
\[\boxed{44\text{ marks}}\]
Answer Details
(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
\[\boxed{44\text{ marks}}\]
Question 12 Report
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 \(= 4158 - 1232 = 2926 \text{ cm}^3\).
Answer Details
(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 \(= 4158 - 1232 = 2926 \text{ cm}^3\).
Question 13 Report
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.
Answer Details
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.
Question 14 Report
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,
\[P(\text{distinction})=\frac{10}{90}=\frac{1}{9}\approx0.111.\]
Answer Details
(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,
\[P(\text{distinction})=\frac{10}{90}=\frac{1}{9}\approx0.111.\]
Question 15 Report
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.
Answer Details
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.
Question 16 Report
(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
\[ 5\cos x - 4\tan x = 5\left(\frac{4}{5}\right) - 4\left(\frac{3}{4}\right) = 4 - 3 = 1. \]Answer Details
(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
\[ 5\cos x - 4\tan x = 5\left(\frac{4}{5}\right) - 4\left(\frac{3}{4}\right) = 4 - 3 = 1. \]Question 17 Report
(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, \[\boxed{x=-2\text{ or }x=2.5}.\]
Answer Details
(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, \[\boxed{x=-2\text{ or }x=2.5}.\]
Question 18 Report
(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?
Answer Details
None
Question 19 Report
(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 \(\frac{1 - x}{x} + \frac{x}{1 - x} = \frac{5}{2}\)
Answer Details
None
Question 20 Report
(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.
Question 21 Report
(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?
Answer Details
None
Question 22 Report
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.
Answer Details
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.
Question 23 Report
(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 \(2 < x < 6\).
Answer Details
(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 \(2 < x < 6\).
Question 24 Report
(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.
Answer Details
(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.
Question 25 Report
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 \(cm^3\)
Question 26 Report
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 \(4\).
Answer Details
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 \(4\).
Question 27 Report
(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.
Answer Details
(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.
Question 28 Report
(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):
\[\tfrac{1}{3} \times \tfrac{22}{7} \times r^2 \times 12 = 1848\] \[\tfrac{88}{7}r^2 = 1848 \Rightarrow r^2 = \frac{1848 \times 7}{88} = 147\] \[r = \sqrt{147} = 7\sqrt{3} \text{ cm}\]Answer Details
(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):
\[\tfrac{1}{3} \times \tfrac{22}{7} \times r^2 \times 12 = 1848\] \[\tfrac{88}{7}r^2 = 1848 \Rightarrow r^2 = \frac{1848 \times 7}{88} = 147\] \[r = \sqrt{147} = 7\sqrt{3} \text{ cm}\]Question 29 Report
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.
Answer Details
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.
Question 30 Report
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
\[ \angle QRS = 180^{\circ} - 62^{\circ} = 118^{\circ} \]
\(\angle QRS = 118^{\circ}\).
Answer Details
(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
\[ \angle QRS = 180^{\circ} - 62^{\circ} = 118^{\circ} \]
\(\angle QRS = 118^{\circ}\).
Question 31 Report
(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.
Question 32 Report
(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
\[\boxed{₦3,700,000.00}\]
Answer Details
(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
\[\boxed{₦3,700,000.00}\]
Question 33 Report
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 of values
| \(x\) | \(0^\circ\) | \(30^\circ\) | \(60^\circ\) | \(90^\circ\) | \(120^\circ\) | \(150^\circ\) | \(180^\circ\) | \(210^\circ\) | \(240^\circ\) | \(270^\circ\) | \(300^\circ\) |
|---|---|---|---|---|---|---|---|---|---|---|---|
| \(y=2\cos x-\sin x\) | 2.00 | 1.23 | 0.13 | −1.00 | −1.87 | −2.23 | −2.00 | −1.23 | −0.13 | 1.00 | 1.87 |
(b) Graph of \(y=2\cos x-\sin x\)
Using 2 cm to \(30^\circ\) on the \(x\)-axis and 2 cm to 1 unit on the \(y\)-axis, plot the points and join them with a smooth curve.
(c)
(i) From the intersections of the curve with \(y=1\),
\[x\approx 37^\circ\quad\text{or}\quad270^\circ.\]
(ii) \[2\cos x-\sin x=0\implies 2\cos x=\sin x\implies \tan x=2.\]
Thus, reading the \(x\)-intercepts of the graph,
\[x\approx 63^\circ\quad\text{or}\quad243^\circ.\]
Answer Details
(a) Completed table of values
| \(x\) | \(0^\circ\) | \(30^\circ\) | \(60^\circ\) | \(90^\circ\) | \(120^\circ\) | \(150^\circ\) | \(180^\circ\) | \(210^\circ\) | \(240^\circ\) | \(270^\circ\) | \(300^\circ\) |
|---|---|---|---|---|---|---|---|---|---|---|---|
| \(y=2\cos x-\sin x\) | 2.00 | 1.23 | 0.13 | −1.00 | −1.87 | −2.23 | −2.00 | −1.23 | −0.13 | 1.00 | 1.87 |
(b) Graph of \(y=2\cos x-\sin x\)
Using 2 cm to \(30^\circ\) on the \(x\)-axis and 2 cm to 1 unit on the \(y\)-axis, plot the points and join them with a smooth curve.
(c)
(i) From the intersections of the curve with \(y=1\),
\[x\approx 37^\circ\quad\text{or}\quad270^\circ.\]
(ii) \[2\cos x-\sin x=0\implies 2\cos x=\sin x\implies \tan x=2.\]
Thus, reading the \(x\)-intercepts of the graph,
\[x\approx 63^\circ\quad\text{or}\quad243^\circ.\]
Question 34 Report
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
\[1200\times\$560.00=\boxed{\$672,000.00}.\]
Answer Details
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
\[1200\times\$560.00=\boxed{\$672,000.00}.\]
Question 35 Report
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.
\[\text{angle west of north} = \tan^{-1}\frac{278.8}{128.5} = 65^{\circ} \Rightarrow \text{bearing} = 360^{\circ} - 65^{\circ} = 295^{\circ}\]Answer Details
(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.
\[\text{angle west of north} = \tan^{-1}\frac{278.8}{128.5} = 65^{\circ} \Rightarrow \text{bearing} = 360^{\circ} - 65^{\circ} = 295^{\circ}\]Question 36 Report
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
\[28+12+24=64\]
\[\boxed{64\text{ cars}}\]
Answer Details
(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
\[28+12+24=64\]
\[\boxed{64\text{ cars}}\]
Question 37 Report
(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.
Question 38 Report
(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.)
Answer Details
(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.)
