L.C. MATHS

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Past Papers

## Question 1

In a particular population $$15\%$$ of the people are left footed.
A soccer team of $$11$$ players, including $$1$$ goalkeeper, is picked at random from the population.

(a) Find the probability that there is exactly one left footed player on the team.
Give you answer correct to three decimal places.

$$0.325$$

Solution

\begin{align}P(\mbox{one left-footed})&={11\choose1}\times0.15^1\times0.85^{10}\\&\approx0.325\end{align}

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(b) Find the probability that less than three players on the team are left footed.
Give you answer correct to two decimal places.

$$0.78$$

Solution

\begin{align}P(\mbox{less than three are left-footed})&=P(\mbox{none are left-footed}) \mbox{ or } P(\mbox{one is left-footed}) \mbox{ or } P(\mbox{two are left-footed})\\&={11\choose0}\times0.85^{11}+{11\choose1}\times0.15^1\times0.85^{10}+{11\choose2}\times0.15^2\times0.85^{9}\\&\approx0.78\end{align}

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(c) The goalkeeper is left footed.
Find the probability that at least eight of the remainder of the team are right footed.
Give you answer correct to two decimal places.

$$0.82$$

Solution

\begin{align}P(\mbox{at least eight are right-footed})&=P(\mbox{less than three are left-footed})\\&=P(\mbox{none are left-footed})\mbox{ or }P(\mbox{one is left-footed})\mbox{ or }P(\mbox{two are left-footed})\\&={10\choose0}\times0.85^{10}+{10\choose1}\times0.15^1\times0.85^{9}+{10\choose2}\times0.15^2\times0.85^{8}\\&\approx0.82\end{align}

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## Question 2

(a) The line $$3x-6y+2=0$$ contains the point $$\left(k,\dfrac{2k+2}{3}\right)$$ where $$k\in\mathbb{R}$$.
Find the value of $$k$$.

$$k=-2$$

Solution

\begin{align}3x-6y+2=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3k-6\left(\frac{2k+2}{3}\right)+2=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3k-4k-4+2=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}-k-2=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}k=-2\end{align}

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(b) The point $$P(s,t)$$ is on the line $$x-2y-8=0$$.
The point $$P$$ is also a distance of $$1$$ unit from the line $$4x+3y+6=0$$.
Find a value of $$s$$ and the corresponding value of $$t$$.

$$s=2$$ and $$t=-3$$ or $$s=\dfrac{2}{11}$$ or $$t=-\dfrac{43}{11}$$

Solution

\begin{align}x-2y-8=0\end{align}

\begin{align}(x,y)=(s,t)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}s-2t-8=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}s=2t+8\end{align}

and

\begin{align}\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}=1\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{|4s+3t+6|}{\sqrt{4^2+3^2}}=1\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{|4(2t+8)+3t+6|}{5}=1\end{align}

\begin{align}\downarrow\end{align}

\begin{align}|8t+32+3t+6|=5\end{align}

\begin{align}\downarrow\end{align}

\begin{align}|11t+38|=5\end{align}

Choose positive option:

\begin{align}11t+38=5\end{align}

\begin{align}\downarrow\end{align}

\begin{align}t=-\frac{43}{11}\end{align}

and

\begin{align}s&=2t+8\\&=2\left(\frac{-43}{11}\right)+8\\&=\frac{2}{11}\end{align}

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(c) The points $$A(4,2)$$ and $$C(16,11)$$ are vertices of the triangle $$ABC$$ shown below.
$$D$$ and $$E$$ are points on $$[CA]$$ and $$[CB]$$ respectively.
The ratio $$|AD|:|DC|$$ is $$2:1$$.

(i) Find $$|AD|$$.

(ii) $$[AB]$$ and $$[DE]$$ are horizontal line segments.
$$|AB|=33$$ units.
Find the coordinates of $$B$$ and of $$E$$.

(i)Â $$10$$

(ii)Â $$(23,8)$$

Solution

(i)

\begin{align}|AC|&=\sqrt{(16-4)^2+(11-2)^2}\\&=\sqrt{12^2+9^2}\\&=\sqrt{225}\\&=15\end{align}

\begin{align}\downarrow\end{align}

(ii)

\begin{align}B&=(4+33,2)\\&=(37,2)\end{align}

and

\begin{align}E&=\left(16+\frac{1}{3}(21),11-\frac{1}{3}(9)\right)\\&=(23,8)\end{align}

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## Question 3

(a) The circle $$k$$ has centre $$C(1,-2)$$ and chord $$[AB]$$ where $$|AB|=4\sqrt{3}$$.
The point $$D(3,2)$$ is the midpoint of the chord $$[AB]$$, as shown below.
Find the radius of $$k$$. Give your answer in the form $$a\sqrt{b}$$, where $$a,b\in\mathbb{N}$$.

$$4\sqrt{2}$$

Solution

and

\begin{align}|CD|&=\sqrt{(3-1)^2+(2-(-2))^2}\\&=\sqrt{2^2+4^2}\\&=\sqrt{20}\end{align}

\begin{align}\downarrow\end{align}

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(b)

(i) Show that the circles $$c:x^2+y^2+4x-2y-95=0$$ and $$s:(x-7)^2+(y-13)^2=25$$ touch externally.Â

(ii) There are an infinite number of circles which touch circle $$c$$ externally at the same point that $$s$$ touches $$c$$.
Find the coordinates of the centre of one of these circles, apart from circle $$s$$.

(ii)Â e.g. $$(16,25)$$

Solution

(i)

Centre of $$c$$ is $$(-2,1)$$ and

\begin{align}r_c&=\sqrt{g^2+f^2-c}\\&=\sqrt{2^2+(-1)^2-(-95)}\\&=\sqrt{100}\\&=10\end{align}

and

Centre of $$s$$ is $$(7,13)$$ and $$r_s=5$$.

\begin{align}\downarrow\end{align}

\begin{align}r_c+r_s=15\end{align}

The distance $$d$$ between their centres is

\begin{align}d&=\sqrt{(-2-7)^2+(1-13)^2}\\&=\sqrt{81+144}\\&=\sqrt{225}\\&=15\end{align}

Therefore, they touch externally.

(ii)

\begin{align}(7+9,13+12)=(16,25)\end{align}

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## Question 4

(a)

(i) Prove that $$\cos 2A=\cos^2A-\sin^2A$$.

(ii) $$\sin\left(\dfrac{\theta}{2}\right)=\dfrac{1}{\sqrt{5}}$$, where $$0\leq\theta\leq\pi$$.
Use the formula $$\cos 2A=\cos^2A-\sin^2A$$ to find the value of $$\cos\theta$$.

(ii)Â $$\cos\theta=\dfrac{3}{5}$$

Solution

(i)

\begin{align}\cos(A+B)&=\cos A\cos B-\sin A\sin B\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\cos(A+A)&=\cos A\cos A-\sin A\sin A\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\cos(2A)&=\cos^2A-\sin^2A\end{align}

as required.

(ii)

\begin{align}\cos(2A)&=\cos^2A-\sin^2A\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\cos(\theta)&=\cos^2\left(\frac{\theta}{2}\right)-\sin^2\left(\frac{\theta}{2}\right)\\&=\left[1-\sin^2\left(\frac{\theta}{2}\right)\right]-\sin^2\left(\frac{\theta}{2}\right)\\&=1-2\sin^2\left(\frac{\theta}{2}\right)\\&=1-2\left(\frac{1}{\sqrt{5}}\right)^2\\&=1-\frac{2}{5}\\&=\frac{3}{5}\end{align}

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(b) Solve the equation:

\begin{align}\tan(B+150^{\circ})=-\sqrt{3}\end{align}

for $$0^{\circ}\leq B\leq360^{\circ}$$.

$$B=150^{\circ}$$ and $$B=330^{\circ}$$

Solution

\begin{align}\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\tan(B+150^{\circ})=\frac{\tan B+\tan 150^{\circ}}{1-\tan B\tan 150^{\circ}}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{\tan B+\tan 150^{\circ}}{1-\tan B\tan 150^{\circ}}=-\sqrt{3}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{\tan B+\left(-\frac{1}{\sqrt{3}}\right)}{1-\tan B\left(-\frac{1}{\sqrt{3}}\right)}=-\sqrt{3}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\tan B-\frac{1}{\sqrt{3}}=-\sqrt{3}-\tan B\end{align}

\begin{align}\downarrow\end{align}

\begin{align}2\tan B&=-\sqrt{3}+\frac{1}{\sqrt{3}}\\&=\frac{-3+1}{\sqrt{3}}\\&=-\frac{2}{\sqrt{3}}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\tan B=-\frac{1}{\sqrt{3}}\end{align}

For the reference angle $$0^{\circ}\leq A\leq 90^{\circ}$$:

\begin{align}\tan A=\frac{1}{\sqrt{3}}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}A=30^{\circ}\end{align}

Tangent is negative in both the second and fourth quadrants.

$\,$

\begin{align}B&=180^{\circ}-30^{\circ}\\&=150^{\circ}\end{align}

$\,$

\begin{align}B&=360^{\circ}-30^{\circ}\\&=330^{\circ}\end{align}

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## Question 5

(a) Two identical right-circular solid cones meet along their bases
and fit exactly inside a sphere, as shown in the diagram.

(i) Prove that the volume of the remaining space inside the sphere is exactly half the total volume of the sphere.

(ii) The combined volume of the two cones is $$\dfrac{686}{3}\pi\mbox{ cm}^3$$.
Find the radius of one of the cones.Â

(ii)Â $$7\mbox{ cm}$$

Solution

(i)

The volume $$V$$ of the remaining space is:

\begin{align}V&=\frac{4}{3}\pi r^3-2\left[\frac{1}{3}\pi r^2(r)\right]\\&=\frac{4}{3}\pi r^3-\frac{2}{3}\pi r^3\\&=\frac{2}{3}\pi r^3\\&=\frac{1}{2}\left(\frac{4}{3}\pi r^3\right)\end{align}

i.e. it is half of the volume of the sphere, as required.

(ii)

\begin{align}2\left[\frac{1}{3}\pi r^2(r)\right]=\frac{686}{\pi}\pi\end{align}

\begin{align}\downarrow\end{align}

\begin{align}2r^3=686\end{align}

\begin{align}\downarrow\end{align}

\begin{align}r&=\sqrt[3]{343}\\&=7\mbox{ cm}\end{align}

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(b) At 9.00 a.m. a delivery van leaves a factory.
It travels towards its destination at an average speed of $$60\mbox{ km/h}$$.
One hour and $$45$$ minutes later a second van leaves the factory on the same route.
It travels at an average speed of $$95\mbox{ km/h}$$.
Both vans arrive at their destination at the same time.
Find at what time they arrive.

$$13:45$$

Solution

\begin{align}d_1=d_2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}v_1t_1=v_2t_2\end{align}

\begin{align}\downarrow\end{align}

\begin{align}60t_1=95(t_1-1.75)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}60t_1=95t_1-166.25\end{align}

\begin{align}\downarrow\end{align}

\begin{align}t_1&=\frac{166.25}{95-60}\\&=4.75\mbox{ h} \end{align}

Therefore, the time that they arrive at is $$4.75$$ hours past $$9:00$$, i.e. it is $$13:45$$.

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## Question 6

(a) Prove that if two triangles $$\Delta ABC$$ and $$\Delta A’B’C’$$ are similar, then the lengths of their sides are proportional in order:

\begin{align}\frac{|AB|}{|A’B’|}=\frac{|BC|}{|B’C’|}=\frac{|CA|}{|C’A’|}\end{align}

Solution

Given

The similar triangles $$ABC$$ and $$A’B’C’$$.

$\,$

To Prove

\begin{align}\frac{|AB|}{|A’B’|}=\frac{|BC|}{|B’C’|}=\frac{|CA|}{|C’A’|}\end{align}

$\,$

Construction

• Mark $$B^*$$ on $$[AB]$$ such that $$[AB^*]=[A’B’]$$.
• Mark $$C^*$$ on $$[AC]$$ such that $$[AC^*]=[A’C’]$$.
• Connect $$B^*$$ to $$C^*$$.

$\,$

Proof

$$AB^{*}C^{*}$$ is congruent to $$AB’C’$$
Reason: SAS

\begin{align}\downarrow\end{align}

$$B^{*}C^{*}$$ is parallel to $$BC$$
Reason: Corresponding angles.

\begin{align}\downarrow\end{align}

\begin{align}\end{align}

\begin{align}\frac{|AB|}{|AB^{*}|}=\frac{|AC|}{|AC^{*}|}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{|AB|}{|A’B’|}=\frac{|AC|}{|A’C’|}\end{align}

as required. The above proof can also be repeated to obtain the second result.

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(b) In the diagram below, the lines $$PA$$, $$HK$$, and $$BR$$ are parallel.
Prove that $$|AH|\times |QB|=|AP|\times |HB|$$.
Give a reason for each geometrical statement you use.

Solution

$$|\angle HAP|$$ is equal to $$|\angle HBQ|$$ as they are alternate angles.

$$|\angle PHA|$$ is equal to $$|\angle QHB|$$ as they are vertically opposite each other.

Therefore, triangles $$APH$$ and $$HBQ$$ are similar as they contain the same three angles. Hence:

\begin{align}\frac{|AH|}{|HB|}=\frac{|AP|}{|QB|}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}|AH|\times|QB|=|AP|\times|HB|\end{align}

as required.

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## Question 7

The diagram (Triangle $$ABC$$) shows the $$3$$ sections of a level triathlon course.
In order to complete the triathlon, each contestant must swim $$4\mbox{ km}$$ from $$C$$ to $$B$$, cycle from $$B$$ to $$A$$, and then run $$28\mbox{ km}$$ from $$A$$ to $$C$$.
Mary can cycle at an average speed of $$25\mbox{ km/hour}$$.
It takes her $$1$$ hour and $$12$$ minutes to cycle from $$B$$ to $$A$$.

(a) Show that the total length of the course is $$62\mbox{ km}$$.

Solution

\begin{align}L&=|CB|+|BA|+|AC|\\&=4+(25\times1.2)+28\\&=62\mbox{ km}\end{align}

as required.

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(b) On average, Mary can run $$5.6$$ times as fast as she can swim.
It takes her $$4.8$$ hours to complete the course.
Find her average swimming speed in $$\mbox{ km/h}$$.

$$2.5\mbox{ km/h}$$

Solution

\begin{align}T_s+T_r&=4.8-1.2\\&=3.6\mbox{ h}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{d_s}{v_s}+\frac{d_r}{v_r}=3.6\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{4}{v_s}+\frac{28}{5.6v_s}=3.6\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{4(5.6)+28}{5.6v_s}=3.6\end{align}

\begin{align}\downarrow\end{align}

\begin{align}4(5.6)+28=(3.6)(5.6v_s)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}50.4=20.16v_s\end{align}

\begin{align}\downarrow\end{align}

\begin{align}v_s&=\frac{50.4}{20.16}\\&=2.5\mbox{ km/h}\end{align}

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(c) Show that $$|\angle ACB=116.5^{\circ}|$$, correct to $$1$$ decimal place.

Solution

\begin{align}|AB|^2=|AC|^2+|BC|^2-2|AC||BC|\cos|\angle ACB|\end{align}

\begin{align}\downarrow\end{align}

\begin{align}30^2=28^2+4^2-2(28)(4)\cos|\angle ACB|\end{align}

\begin{align}\downarrow\end{align}

\begin{align}|\angle ACB|&=\cos^{-1}\left(\frac{28^2+4^2-30^2}{2(28)(4)}\right)\\&\approx116.5^{\circ}\end{align}

as required.

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(d)Â To comply with safety regulations, the region inside the triangular course must be kept clear of people. Find the area of this region.
Give your answer, in $$\mbox{km}^2$$, correct to $$1$$ decimal place.Â

$$50.1\mbox{ km}^2$$

Solution

\begin{align}\mbox{Area}&=\frac{1}{2}|AC||BC|\sin |\angle ACB|\\&=\frac{1}{2}(28)(4)\sin116.5{^\circ}\\&\approx50.1\mbox{ km}^2\end{align}

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(e) Find the shortest distance from the point $$C$$ to the side $$AB$$.
Give your answer in $$\mbox{km}$$, correct to $$1$$ decimal place.

$$3.3\mbox{ km}$$

Solution

\begin{align}\mbox{Area}=\frac{1}{2}bh\end{align}

\begin{align}\downarrow\end{align}

\begin{align}50.1=\frac{1}{2}(30)h\end{align}

\begin{align}\downarrow\end{align}

\begin{align}h&=\frac{(2)(50.1)}{30}\\&\approx3.3\mbox{ km}\end{align}

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(f) The course is viewed from a camera tower which rises vertically from point $$A$$.
The top of the tower is point $$T$$. The angle of elevation of $$T$$ from $$B$$ is $$0.05^{\circ}$$.
Find $$|AT|$$, the vertical height of the tower.

$$26\mbox{ m}$$

Solution

\begin{align}\tan(0.05^{\circ})&=\frac{|AT|}{|AB|}\\&=\frac{|AT|}{30}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}|AT|&=30\tan(0.05^{\circ})\\&=0.02617…\mbox{ km}\\&\approx 26\mbox{ m}\end{align}

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## Question 8

(a) In a school all First Years sat a common maths exam.
The results, in integer values, were normally distributed with a mean of $$176$$ marks and a standard deviation of $$36$$ marks.
The top $$10\%$$ of students will go forward to a county maths competition.

(i) Find the minimum mark needed on the exam to progress to the county stage.

(ii) The school awarded a Certificate of Merit to any student who achieved between $$165$$ marks and $$210$$ marks.
Find the percentage of First Years who received the Certificate of Merit.Â

$$223$$

(ii) $$44.81\%$$

Solution

(i)

\begin{align}P(z<1.28)&=0.8997\end{align}

\begin{align}z=\frac{x-\mu}{\sigma}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}1.28=\frac{x-176}{36}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x&=(1.28)(36)+176\\&=222.08\end{align}

Therefore, the minimum mark is $$223$$.

(ii)

\begin{align}P\left(z<\frac{210-176}{36}\right)-P\left(z>\frac{165-176}{36}\right)&=P(z<0.94)-P(z>-0.31)\\&=0.8264-(1-0.6217)\\&=0.4481\end{align}

Therefore, $$44.81\%$$ of students got the Certificate of Merit.

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(b) A news report claimed that 6th year students in Ireland studied an average of $$21$$ hours per week, outside of class time. A Leaving Cert class surveyed $$60$$ students in 6th year, chosen at random, from different schools. It found that the average study time was $$19.8$$ hours and the standard deviation was $$5.2$$ hours.

(i) Find the test statistic (the $$z$$-score) of this sample mean.

(ii) Find the $$p$$-value of this test statistic. Comment on what can be concluded from its value, in a two-tailed hypothesis test at the $$5\%$$ level of significance, in relation to the news report claim.Â

(i)Â $$-1.787…$$

(ii)Â $$p=0.0734$$. As $$p>0.05$$, we do not have enough evidence to state that the claim is incorrect.

Solution

(i)

\begin{align}z&=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}\\&=\frac{19.8-21}{5.2/\sqrt{60}}\\&=-1.787…\end{align}

(ii)

\begin{align}p&=2[1-P(z<1.79)]\\&=2(1-0.9633)\\&=0.0734\end{align}

As $$0.0734>0.05$$, we do not have enough evidence to state that the claim is incorrect.

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(c) The school caretaker has a box with $$23$$ room keys in it.
$$12$$ of the keys are for general classrooms, $$6$$ for science labs and $$5$$ for offices.

(i) Four keys are drawn at random from the box.
What is the probability that the 4th key drawn is the first office key drawn?
Give your answer correct to $$4$$ decimal places.

(ii) All the keys are returned to the box. Then $$3$$ keys are drawn at random from the box one after the other, without replacement. What is the probability that one of them is for a general classroom, one is for a science lab and one is for an office?
Give your answer correct to $$4$$ decimal places.

(i)Â $$0.1152$$

(ii)Â $$0.2033$$

Solution

(i)

\begin{align}P&=\frac{18}{23}\times\frac{17}{22}\times\frac{16}{21}\times\frac{5}{20}\\&\approx0.1152\end{align}

(ii)

\begin{align}P&=3!\times\frac{12}{23}\times\frac{6}{22}\times\frac{5}{21}\\&\approx0.2033\end{align}

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## Question 9

(a) An aeroplane flies east from point $$A$$ for $$2$$ hours at a constant speed of $$420$$ km per hour until it reaches point $$B$$. It then changes direction by heading $$20^{\circ}$$ towards the south at the same speed until it reaches point $$C$$, as shown in the diagram below.
The direct distance from $$A$$ to $$C$$ is $$1450\mbox{ km}$$ and $$|\angle BAC|=8.57^{\circ}$$.

(i) Find how long it took to fly from $$B$$ to $$C$$.

(ii) The average fuel consumption of the plane is $$3.8$$ litres per second and the fuel capacity of the plane is $$100{,}000$$ litres.
Show that the plane will be able to complete the journey from $$A$$ to $$B$$ to $$C$$ and directly back to $$A$$ at a speed of $$420\mbox{ km/h}$$ without refuelling.

(i)Â $$1\mbox{ h }30\mbox{ min}$$

Solution

(i)

\begin{align}|AB|&=(2)(420)\\&=840\mbox{ km}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}|BC|^2=|AB|^2+|AC|^2-2|AB||AC|\cos |\angle BAC|\end{align}

\begin{align}\downarrow\end{align}

\begin{align}|BC|&=\sqrt{840^2+1450^2-2(840)(1450)(\cos8.57^{\circ}}\\&=631.90…\mbox{ km}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}t&=\frac{631.90…}{420}\\&=1.50…\mbox{ h}\\&\approx1\mbox{ h }30\mbox{ min}\end{align}

(ii)

\begin{align}T&=t_{AB}+t_{BC}+t_{CA}\\&=2+1.5+\frac{1450}{420}\\&=6.9523…\mbox{ h}\\&=25{,}028.57…\mbox{ s}\end{align}

As the fuel runs out in a longer time of $$\dfrac{100{,}000}{3.8}=26{,}315.7…\mbox{ s}$$, the plane can complete the journey.

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(b) The voltage, $$V(t)$$, (in Volts) of a certain alternating current is given by the function:

\begin{align}V(t)=110\sqrt{2}\sin(120\pi t)\end{align}

where $$t$$ is in seconds.

(i) Find the period and range of the function $$V(t)$$.

(ii) Sketch the function for $$0\leq t \leq p$$, where $$p$$ is the period of $$V(t)$$.
Indicate the period and range of the function on your graph.

(iii) Use $$V(t)$$ to find the voltage when $$t=6.67$$ seconds.

(iv) Find one value for $$t$$ where the voltage is $$110$$ Volts.
Give your answer in the form $$\dfrac{a}{b}$$ where $$a,b\in\mathbb{N}$$.

(v) Find the rate of change of the voltage when $$t=2$$ seconds.

(i) The period is $$\dfrac{1}{60}\mbox{ s}$$ and the range is $$[-110\sqrt{2},110\sqrt{2}]$$.

(ii)

(iii)Â $$147.95\mbox{ Volts}$$

(iv)Â $$t=\dfrac{1}{480}\mbox{ s}$$

(v)Â $$58{,}646\mbox{ Volts/sec}$$

Solution

(i)

\begin{align}\mbox{Period}&=\frac{2\pi}{\omega}\\&=\frac{2\pi}{120\pi}\\&=\frac{1}{60}\mbox{ s}\end{align}

and the range is $$[-110\sqrt{2},110\sqrt{2}]$$.

(ii)

(iii)

\begin{align}V(6.67)&=110\sqrt{2}\sin[120\pi(6.67)]\\&\approx147.95\mbox{ Volts}\end{align}

(iv)

\begin{align}110\sqrt{2}\sin(120\pi t)=110\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\sin(120\pi t)=\frac{1}{\sqrt{2}}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}120\pi t=\frac{\pi}{4}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}t=\frac{1}{480}\mbox{ s}\end{align}

(v)

\begin{align}V'(t)=(120\pi)[110\sqrt{2}\cos(120\pi t)]\end{align}

\begin{align}\downarrow\end{align}

\begin{align}V'(2)&=(120\pi)[110\sqrt{2}\cos(120\pi (2))]\\&\approx 58{,}646\mbox{ Volts/sec}\end{align}

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## Question 10

People with O-negative blood type are called “universal donors” because their blood can be given to anyone else. In Ireland approximately $$8\%$$ of the population have O-negative blood type (source: Blood Transfusion Service).

(a)

(i) At a blood donation clinic, ten donors give blood, one after the other.
Find the probability that the tenth person is the third O-negative donor.

(ii) At a blood donation clinic, five donors give blood.
What is the probability that at least one of the five donates O-negative blood?

(iii) Find the minimum number of blood donors required, so that the probability that at least one of them is type O-negative is greater than $$0.97$$.

(i)Â $$0.0103$$

(ii)Â $$0.3409$$

(iii)Â $$43$$

Solution

(i)

\begin{align}P(\mbox{tenth is third O neg})&=P(\mbox{exactly two from first nine are O neg})\times P(\mbox{tenth is O neg})\\&=\left[{9\choose2}\left(\frac{8}{100}\right)^2\left(\frac{92}{100}\right)^7\right]\times\frac{8}{100}\\&\approx0.0103\end{align}

(ii)

\begin{align}P(\mbox{at least one is O neg})&=1-P(\mbox{none are O neg})\\&=1-\left(\frac{92}{100}\right)^5\\&\approx0.3409\end{align}

(iii)

\begin{align}1-\left(\frac{92}{100}\right)^n=0.97\end{align}

\begin{align}\downarrow\end{align}

\begin{align}0.92^n=0.03\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\ln(0.92^n)=\ln(0.03)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}n\ln(0.92)=\ln(0.03)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}n&=\frac{\ln(0.03)}{\ln(0.92)}\\&42.05…\end{align}

Therefore, the minimum number of blood donors is $$43$$.

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(b) A homeowner has a problem with the heating system in her house. A plumber has identified the problem as a faulty part. The homeowner knows that in $$80\%$$ of cases a repair of the part will fix the problem and this repair will cost â‚¬$$70$$. If the repair does not work then a new part will have to be bought costing â‚¬$$150$$ and there will be an additional labour cost of â‚¬$$80$$ to replace the old part with the new.
Find the expected value of the cost of fixing this faulty system.

$$116\mbox{ euro}$$

Solution

\begin{align}E&=(0.8)(70)+(0.2)(70+15+80)\\&=116\mbox{ euro}\end{align}

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(c) A life insurance policy pays out â‚¬$$120{,}000$$ if the policy holder dies and â‚¬$$40{,}000$$ if the policy holder becomes disabled. The insurance company has calculated that in general, in any given year, the probability of death is $$0.0001$$ and the probability of disability is $$0.002$$.
The company has $$18{,}000$$ policy holders on its books at present who are all charged the same premium. The companyâ€™s goal is to make â‚¬$$900{,}000$$ profit in a particular year.
Find the annual premium it should charge its customers which in an average year would generate this level of profit.

$$142\mbox{ euro}$$

Solution

The average amount that will have to pay to each customer is:

\begin{align}(120{,}000)(0.0001)+(40{,}000)(0.002)=92\mbox{ euro}\end{align}

The amount of profit that they wish to make from each customer is instead

\begin{align}\frac{900{,}000}{18{,}000}=50\mbox{ euro}\end{align}

To make this profit, they there need to charge each person a premium of $$92+50=142\mbox{ euro}$$.Â

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