Détails de la réponse

Let \(P(A)=\tfrac23,\;P(K)=\tfrac58,\;P(F)=\tfrac34\), so the failure probabilities are \(\tfrac13,\tfrac38,\tfrac14\). The three events are independent, so multiply.

(a) All three pass: \[\frac23\times\frac58\times\frac34=\frac{30}{96}=\frac{5}{16}.\]

(b) None passes: \[\frac13\times\frac38\times\frac14=\frac{3}{96}=\frac{1}{32}.\]

(c) Ade and Kujo only (Fati fails): \[\frac23\times\frac58\times\frac14=\frac{10}{96}=\frac{5}{48}.\]

Détails de la réponse

The curve is a downward-opening parabola, so the coefficient of \(x^2\) is negative. Its roots are the \(x\)-coordinates where the curve crosses the horizontal axis.

(a) The roots are the \(x\)-intercepts:

\[x=-1.5 \quad \text{or} \quad x=2.\]

(b) The graph crosses the \(y\)-axis at \(6\), so \(c=6\). Therefore,

\[y=ax^2+bx+6.\]

Use each root, where \(y=0\).

When \(x=2\):

\[4a+2b+6=0 \quad\Rightarrow\quad 4a+2b=-6.\]

When \(x=-1.5\):

\[\frac{9}{4}a-\frac{3}{2}b+6=0 \quad\Rightarrow\quad 9a-6b=-24.\]

Multiplying \(4a+2b=-6\) by \(3\) gives \(12a+6b=-18\). Adding this to \(9a-6b=-24\):

\[21a=-42,\qquad a=-2.\]

Then:

\[4(-2)+2b=-6 \Rightarrow 2b=2 \Rightarrow b=1.\]

Hence,

\[a=-2,\qquad b=1,\qquad c=6,\]

and the relation is

\[\boxed{y=-2x^2+x+6}.\]

(c) The maximum is at the vertex. For \(y=ax^2+bx+c\), its \(x\)-coordinate is

\[x=\frac{-b}{2a}=\frac{-1}{2(-2)}=0.25.\]

\[y=-2(0.25)^2+0.25+6=6.125.\]

So the maximum value is

\[\boxed{6.125\text{ at }x=0.25}.\]

The graph may lead to approximate readings of \(x\approx0.2\) and \(y\approx6.2\), but \(0.25\) and \(6.125\) are the exact values from the equation.

(d) Set \(y=2\):

\[-2x^2+x+6=2.\]

\[2x^2-x-4=0.\]

\[x=\frac{1\pm\sqrt{(-1)^2-4(2)(-4)}}{2(2)} =\frac{1\pm\sqrt{33}}{4}.\]

\[\boxed{x\approx-1.2 \quad \text{or} \quad x\approx1.7}.\]

Examination reminder: Use the intercepts to find roots, the \(y\)-intercept to identify \(c\), and then substitute the known points into the quadratic to determine \(a\) and \(b\).

Détails de la réponse

```html

Geometrical Construction

Given: Construct triangle XYZ such that

• |XY| = 8 cm,
• ∠YXZ = 60°,
• ∠XYZ = 30°.

(a) Construction steps (ruler and compasses only).

1. Draw a line segment XY = 8 cm.
2. At X, construct an angle of 60° with XY. This may be done by constructing an equilateral triangle on XY, since every angle in an equilateral triangle is 60°.
3. At Y, construct an angle of 30° with YX. First construct a 60° angle at Y, then bisect it using the standard compass construction for an angle bisector.
4. Let the 60° ray from X and the 30° ray from Y meet at Z. This gives triangle XYZ.
5. Construct the perpendicular from Z to XY. Let it meet XY at T. Hence, ZT ⟂ XY.
6. The locus of points equidistant from the lines ZY and XY is the angle bisector of ∠ZYX. Construct the internal angle bisector of the 30° angle at Y and call it l₁.
7. Let l₁ meet ZT at S.

(b) Finding |ST|.

Since ∠XYZ = 30° and l<sub>1</sub> bisects this angle,

∠SYT = 15°.

Also, triangle ZXY is right-angled at Z, because

∠XZY = 180° − 60° − 30° = 90°.

In the 30°–60°–90° triangle XYZ, with hypotenuse XY = 8 cm:

XZ = 4 cm,    YZ = 4√3 cm.

The foot of the perpendicular is T, and

YT = 6 cm.

In right triangle YST:

tan 15° = ST / YT.

ST = 6 tan 15° = 6(2 − √3).

On a ruler-and-compass construction, the measured value should be approximately 1.6 cm.

```

Détails de la réponse

(a) AP with \(a=3,\;d=4,\;n=28\). \[S_{28}=\frac{n}{2}\bigl(2a+(n-1)d\bigr)=14\bigl(6+27\times4\bigr)=14\times114=1596.\]

(b) \(x=\dfrac{2m}{1-m^{2}}=\dfrac{2m}{(1-m)(1+m)}\) and \(y=\dfrac{2m}{1+m}\). \[2x-y=\frac{4m}{(1-m)(1+m)}-\frac{2m}{1+m}=\frac{4m-2m(1-m)}{(1-m)(1+m)}=\frac{2m+2m^{2}}{(1-m)(1+m)}=\frac{2m(1+m)}{(1-m)(1+m)}=\frac{2m}{1-m}.\]

(c) Interior angles of a pentagon sum to \(540^{\circ}\): \[x+2x+3x+2x+(3x-10)=540\Rightarrow 11x-10=540\Rightarrow 11x=550\Rightarrow x=50.\]

Détails de la réponse

From the diagram the solid is a slice (sector-prism) of a right circular cylinder: the sector angle at each centre is \(\angle ABC = \angle PQR = 120^\circ\), the radius is \(r = 7\text{ cm}\), and the height is \(h = 15\text{ cm}\). The solid is therefore \(\dfrac{120}{360} = \dfrac{1}{3}\) of the full cylinder.

(a) Volume

\[V = \frac{120}{360}\times \pi r^2 h = \frac{1}{3}\times \frac{22}{7}\times 7^2 \times 15\]

\[V = \frac{1}{3}\times \frac{22}{7}\times 49 \times 15 = \frac{1}{3}\times 22 \times 7 \times 15\]

\[V = \frac{1}{3}\times 2{,}310 = 770\text{ cm}^3\]

Correct to one decimal place, \(V = \mathbf{770.0\text{ cm}^3}\).

(b) Total surface area

The surface of the slice is made up of four parts:

1. Two sectoral faces (top \(ABC\) and bottom \(PQR\)), each a \(120^\circ\) sector of radius 7:

\[A_1 = 2\times \frac{120}{360}\pi r^2 = 2\times \frac{1}{3}\times \frac{22}{7}\times 49 = 2\times \frac{1}{3}\times 154 = 102.6667\text{ cm}^2\]

2. The curved outer face (a \(120^\circ\) strip of the cylinder wall):

\[A_2 = \frac{120}{360}\times 2\pi r h = \frac{1}{3}\times 2\times \frac{22}{7}\times 7 \times 15 = \frac{1}{3}\times 660 = 220\text{ cm}^2\]

3. The two flat rectangular faces formed by the straight cuts, each of dimensions \(r \times h = 7\times 15\):

\[A_3 = 2\times (7 \times 15) = 2\times 105 = 210\text{ cm}^2\]

Total surface area:

\[A = A_1 + A_2 + A_3 = 102.6667 + 220 + 210 = 532.6667\text{ cm}^2\]

Correct to one decimal place, \(A = \mathbf{532.7\text{ cm}^2}\).

Détails de la réponse

(a) Let width \(=w\). Then length \(=2w\) and depth \(=w-4\). An open tank has a base plus four sides: \[\text{Area}=\underbrace{(2w)(w)}_{\text{base}}+\underbrace{2(2w)(w-4)+2(w)(w-4)}_{\text{four sides}}=2w^{2}+6w(w-4).\] Set equal to \(1440\): \[2w^{2}+6w^{2}-24w=1440\Rightarrow 8w^{2}-24w-1440=0\Rightarrow w^{2}-3w-180=0.\] \[(w-15)(w+12)=0\Rightarrow w=15.\] Length \(=2w=\) 30 m.

(b) First arrangement: \(\dfrac{3000x}{100}+\dfrac{2000y}{100}=640\Rightarrow 3x+2y=64.\) Swapped: \(\dfrac{2000x}{100}+\dfrac{3000y}{100}=660\Rightarrow 2x+3y=66.\) Solving: from \(9x+6y=192\) and \(4x+6y=132\), subtract \(\Rightarrow 5x=60\Rightarrow x=12\); then \(3(12)+2y=64\Rightarrow y=14\). So x = 12, y = 14.

Détails de la réponse

(a) Simplify \((2a + b)^2 - (b - 2a)^2\). Note that \((b - 2a)^2 = (2a - b)^2\). This is a difference of two squares, \(P^2 - R^2 = (P + R)(P - R)\), with \(P = 2a + b\) and \(R = 2a - b\):

\[(2a + b)^2 - (2a - b)^2 = \big[(2a + b) + (2a - b)\big]\big[(2a + b) - (2a - b)\big].\]

\[= (4a)(2b) = 8ab.\]

(b) Given \(S = K\sqrt{m^2 + n^2}\).

(i) Make \(m\) the subject.

\[\frac{S}{K} = \sqrt{m^2 + n^2} \Rightarrow \left(\frac{S}{K}\right)^2 = m^2 + n^2.\]

\[m^2 = \left(\frac{S}{K}\right)^2 - n^2 \Rightarrow m = \sqrt{\left(\frac{S}{K}\right)^2 - n^2}.\]

(ii) With \(S = 12.2, K = 0.02, n = 1.1\):

\[\frac{S}{K} = \frac{12.2}{0.02} = 610, \qquad \left(\frac{S}{K}\right)^2 = 372100, \qquad n^2 = 1.21.\]

\[m = \sqrt{372100 - 1.21} = \sqrt{372098.79} = 609.999 \approx 610.\]

To the nearest whole number, the positive value of \(m = \mathbf{610}\).

Détails de la réponse

(a) Evaluate \(\dfrac{3}{\sqrt3}\left(\dfrac{2}{\sqrt3}-\dfrac{\sqrt{12}}{6}\right)\) without tables.

First simplify each surd. \(\dfrac{3}{\sqrt3}=\dfrac{3}{\sqrt3}\times\dfrac{\sqrt3}{\sqrt3}=\dfrac{3\sqrt3}{3}=\sqrt3.\)

Inside the bracket: \(\dfrac{2}{\sqrt3}=\dfrac{2\sqrt3}{3}\) and \(\dfrac{\sqrt{12}}{6}=\dfrac{2\sqrt3}{6}=\dfrac{\sqrt3}{3}.\)

\[\frac{2\sqrt3}{3}-\frac{\sqrt3}{3}=\frac{\sqrt3}{3}.\]

Therefore

\[\sqrt3\times\frac{\sqrt3}{3}=\frac{3}{3}=\boxed{1.}\]

(b) Circle, centre \(O\); \(AB\) produced to \(E\), \(\widehat{ACB}=49^\circ\), \(\widehat{CBE}=68^\circ\).

Since \(A,B,E\) are collinear, \(\widehat{ABC}\) and \(\widehat{CBE}\) are angles on a straight line:

\[\widehat{ABC}=180^\circ-68^\circ=112^\circ.\]

In \(\triangle ABC\):

\[\widehat{BAC}=180^\circ-\widehat{ABC}-\widehat{ACB}=180^\circ-112^\circ-49^\circ=19^\circ.\]

(i) Interior angle \(AOC\). \(\widehat{ABC}=112^\circ\) is the angle at the circumference standing on chord \(AC\); it subtends the major arc \(AC\), whose central angle (reflex \(AOC\)) is \(2\times112^\circ=224^\circ.\) Hence the interior (non-reflex) angle is

\[\widehat{AOC}=360^\circ-224^\circ=\boxed{136^\circ.}\]

(ii) Angle \(BOC\). \(\widehat{BAC}=19^\circ\) at the circumference subtends arc \(BC\); the angle at the centre on the same arc is twice as large:

\[\widehat{BOC}=2\times19^\circ=\boxed{38^\circ.}\]

(Check: \(\widehat{ACB}=49^\circ\Rightarrow\widehat{AOB}=98^\circ,\) and \(98^\circ+38^\circ+224^\circ=360^\circ.\))

Détails de la réponse

(a)(i) Marked price \(=\)N\(25{,}000\). Trade discount of \(20\%\): \[0.20\times 25000=\text{N}5000,\] so price becomes \(25000-5000=\)N\(20{,}000\). Prompt-payment discount of \(2\tfrac12\%\) on this: \[0.025\times 20000=\text{N}500.\] Total amount saved \(=5000+500=\) N5,500.

(a)(ii) After both discounts the distributor pays \(80\%\) then \(97.5\%\) of the mark: \[0.80\times 0.975\,x=0.78x.\] So \(0.78x=11700\Rightarrow x=\dfrac{11700}{0.78}=\) 15,000. The article is marked N15,000.

(b) Factorize \(6y^{2}-149y-102\). Product of ends \(=6\times(-102)=-612\); two numbers with product \(-612\) and sum \(-149\) are \(-153\) and \(+4\): \[6y^{2}-153y+4y-102=3y(2y-51)+2(2y-51)=(2y-51)(3y+2).\] Setting \((2y-51)(3y+2)=0\): \[y=\frac{51}{2}=25\tfrac12\quad\text{or}\quad y=-\frac{2}{3}.\]

Détails de la réponse

From the diagram: \(PQ=8\text{ m}\), \(QR=13\text{ m}\), the bearing of Q from P is \(050^\circ\), and the bearing of R from Q is \(130^\circ\).

(a)(i) Length PR. Find \(\angle PQR\). The bearing of P from Q is the back-bearing of \(050^\circ\):

\[050^\circ+180^\circ=230^\circ\]

The bearing of R from Q is \(130^\circ\), so

\[\angle PQR=230^\circ-130^\circ=100^\circ\]

Apply the cosine rule to triangle PQR:

\[PR^2=PQ^2+QR^2-2\,(PQ)(QR)\cos(\angle PQR)\]\[PR^2=8^2+13^2-2(8)(13)\cos100^\circ\]\[PR^2=64+169-208(-0.17365)=269.12\]\[PR=16.4\text{ m (3 s.f.)}\]

(a)(ii) Bearing of R from P. Use the sine rule to find \(\angle QPR\):

\[\frac{\sin\angle QPR}{QR}=\frac{\sin\angle PQR}{PR}\]\[\sin\angle QPR=\frac{13\sin100^\circ}{16.405}=\frac{12.803}{16.405}=0.78040\]\[\angle QPR=51.3^\circ\]

R lies clockwise of Q as seen from P, so the bearing of R from P is

\[050^\circ+51.3^\circ=101^\circ\ (\text{3 s.f.})\]

(b) Shortest distance from Q to PR, and the area. The area of triangle PQR is

\[\text{Area}=\tfrac12(PQ)(QR)\sin(\angle PQR)=\tfrac12(8)(13)\sin100^\circ\]\[\text{Area}=52(0.98481)=51.2\text{ m}^2\ (\text{3 s.f.})\]

The shortest distance from Q to PR is the perpendicular height h onto base PR. Using \(\text{Area}=\tfrac12(PR)(h)\):

\[h=\frac{2\times\text{Area}}{PR}=\frac{2(51.210)}{16.405}=6.24\text{ m (3 s.f.)}\]

Détails de la réponse

(a) Frequency table. Sorting the 30 lengths into the given class intervals and tallying gives:

Length (cm)	Tally	Frequency \(f\)	Midpoint \(x\)	\(fx\)
45 - 49	|	1	47	47
50 - 54	|||	3	52	156
55 - 59	|||| |	6	57	342
60 - 64	|||| ||	7	62	434
65 - 69	|||| |	6	67	402
70 - 74	||||	4	72	288
75 - 79	||	2	77	154
80 - 84	|	1	82	82
Total		30		1905

(b) Histogram of the distribution. The bars are drawn against the class boundaries \(44.5, 49.5, 54.5, 59.5, 64.5, 69.5, 74.5, 79.5, 84.5\) on the horizontal axis, with heights equal to the class frequencies. Since all class widths are equal, the bar heights are simply the frequencies.

(c) Mean of the distribution.

\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{1905}{30}=63.5\text{ cm}.\]

(d) Probability of selecting a rod in the modal class. The modal class is the one with the highest frequency, \(60 - 64\) (frequency \(7\)). The number of rods in this class is \(7\) out of \(30\), so

\[P(\text{modal class})=\frac{7}{30}.\]

Détails de la réponse

(a) Venn diagram

(b)(i) \(C \cap A'\) means the elements that are in \(C\) but not in \(A\).

From the diagram, these are the \(C\)-only region \(\{6,12\}\) and the \(B \cap C\) only region \(\{15\}\):

\[C \cap A'=\{6,12,15\}.\]

The supplied reference answer \(\{15\}\) is not consistent with the stated sets: \(6\) and \(12\) are both in \(C\) and neither is in \(A\), so they must be included.

(b)(ii) First identify all elements in \(B\) or \(C\):

\[B\cup C=\{2,3,5,6,7,9,11,12,15\}.\]

Intersecting with \(A'\) removes every element that is in \(A\), namely \(3,5,7,9,11\). Therefore:

\[A'\cap(B\cup C)=\{2,6,12,15\}.\]

Examination reminder: An intersection with \(A'\) keeps only elements outside \(A\). Check each region of the Venn diagram that lies outside the \(A\) circle.

Détails de la réponse

(a) \[\tfrac12\log_{10}25-2\log_{10}3+\log_{10}18=\log_{10}5-\log_{10}9+\log_{10}18=\log_{10}\!\left(\frac{5\times18}{9}\right)=\log_{10}10=1.\]

(b) \(123_{y}=1\cdot y^{2}+2\cdot y+3\). So \[y^{2}+2y+3=83\Rightarrow y^{2}+2y-80=0\Rightarrow(y+10)(y-8)=0.\] Since a base must be positive, \(y=8\).

(c) \[\frac{9^{2x-3}}{3^{x+3}}=1.\] Write \(9=3^{2}\): numerator \(=3^{2(2x-3)}=3^{4x-6}\). Then \[3^{(4x-6)-(x+3)}=3^{0}\Rightarrow 3x-9=0\Rightarrow x=3.\]