L.C. MATHS

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## Question 1

(a) The coordinates of three points are $$A(2,-6)$$, $$B(6,-12)$$ and $$C(-4,3)$$.
Find the perpendicular distance from $$A$$ to $$BC$$.
Based on your answer, what can you conclude about the relationship between the points $$A$$, $$B$$ and $$C$$?

As the distance is zero, all three points are collinear.

Solution

\begin{align}m_{BC}&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{3-(-12)}{-4-6}\\&=-\frac{15}{10}\\&=-\frac{3}{2}\end{align}

\begin{align}y-y_1=m(x-x_1)\end{align}

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

\begin{align}y-(-12)=-\frac{3}{2}(x-6)\end{align}

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

\begin{align}y+12=-\frac{3}{2}x+9\end{align}

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

\begin{align}2y+24=-3x+18\end{align}

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

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

The perpendicular is therefore

\begin{align}d&=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}\\&=\frac{3(2)+2(-6)+6}{\sqrt{3^2+2^2}}\\&=0\end{align}

and therefore all three points are collinear.

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(b) The diagram below shows two lines $$a$$ and $$b$$. The equation of $$a$$ is $$x-2y+1=0$$.
The acute angle between $$a$$ and $$b$$ is $$\theta$$. Line $$b$$ makes an angle of $$60^{\circ}$$ with the positive sense of the $$x$$-axis, as shown in the diagram.
Find the value of $$\theta$$, in degrees, correct to $$3$$ decimal places.

$$33.435^{\circ}$$

Solution

\begin{align}x-2y+1=0\end{align}

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

\begin{align}2y=x+1\end{align}

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

\begin{align}y=\frac{1}{2}x+1\end{align}

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

\begin{align}m_a=\frac{1}{2}\end{align}

and

\begin{align}m_b&=\tan60^{\circ}\\&=\sqrt{3}\end{align}

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

\begin{align}\tan\theta&=\pm\frac{m_a-m_b}{1+m_am_b}\\&=\pm\frac{\frac{1}{2}-\sqrt{3}}{1+\left(\frac{1}{2}\right)(\sqrt{3})}\\&=\pm\frac{1-2\sqrt{3}}{2+\sqrt{3}}\end{align}

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

\begin{align}\theta&=\tan^{-1}\left(\frac{1-2\sqrt{3}}{2+\sqrt{3}}\right)\\&\approx33.435^{\circ}\end{align}

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## Question 2

(a) The circle $$c$$ has equation $$x^2+y^2-4x+2y-4=0$$.
The point $$A$$ is the centre of the circle.
The line $$l$$ is a tangent to $$c$$ at the point $$T$$, as shown in the diagram.
The point $$B(5,8)$$ is on $$l$$.
Find $$|BT|$$.

$$|BT|=9$$

Solution

\begin{align}|AT|&=\sqrt{g^2+f^2-c}\\&=\sqrt{2^2+(-1)^2-(-4)}\\&=\sqrt{9}\\&=3\end{align}

and

\begin{align}|AB|&=\sqrt{(2-5)^2+(-1-8)^2}\\&=\sqrt{9+81}\\&=\sqrt{90}\end{align}

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

\begin{align}|BT|&=\sqrt{|AB|^2-|AT|^2}\\&=\sqrt{(\sqrt{90})^2-3^2}\\&=\sqrt{81}\\&=9\end{align}

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(b) Two circles, $$c_1$$ and $$c_2$$, have their centres on the $$x$$-axis. Each circle has a radius of $$5$$ units.
The point $$(1,4)$$ lies on each circle. Find the equation of $$c_1$$ and the equation of $$c_2$$.

$$(x-4)^2+y^2=25$$ and $$(x+2)^2+y^2=25$$

Solution

The centre of both circles is $$(-g,0)$$. Their radii therefore satisfy:

\begin{align}\sqrt{(-g-1)^2+(0-4)^2}=5\end{align}

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

\begin{align}\sqrt{g^2+2g+1+16}=5\end{align}

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

\begin{align}g^2+2g+17=25\end{align}

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

\begin{align}g^2+2g-8=0\end{align}

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

\begin{align}(g+4)(g-2)=0\end{align}

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

$$g=-4$$ and $$g=2$$

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

\begin{align}(x-4)^2+y^2=25\end{align}

and

\begin{align}(x+2)^2+y^2=25\end{align}

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## Question 3

(a) A flagpole $$[GH]$$, shown in the diagram, is vertical and the ground is inclined at an angle of $$5^{\circ}$$ to the horizontal between $$E$$ and $$G$$. The angles of elevation from $$E$$ and $$F$$ to the top of the pole are $$35^{\circ}$$ and $$52^{\circ}$$ respectively.
The distance from $$E$$ to $$F$$ along the incline is $$6\mbox{ m}$$.
Find how far $$F$$ is from the base of the pole ($$G$$) along the incline.

$$6.44\mbox{ m}$$

Solution

\begin{align}\frac{|HF|}{\sin35^{\circ}}=\frac{6}{\sin17^{\circ}}\end{align}

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

\begin{align}|HF|&=\frac{6\sin35^{\circ}}{\sin17^{\circ}}\\&=11.771…\mbox{ m}\end{align}

and

\begin{align}\frac{|FG|}{\sin33^{\circ}}=\frac{11.771…}{\sin95^{\circ}}\end{align}

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

\begin{align}|FG|&=\frac{(11.771…)(\sin33^{\circ})}{\sin95^{\circ}}\\&\approx6.44\mbox{ m}\end{align}

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(b) In the diagram the large circle $$s$$ has centre $$O$$ and the small circle $$c$$ has centre $$D$$.
The circle $$c$$ touches the circle $$s$$ at the point $$C$$.
$$OA$$ and $$OB$$ are tangents to $$c$$ as shown.
The radius of $$c$$ is $$r$$.
$$|\angle BOA|=60^{\circ}$$.
The ratio of the area of $$s$$ to the area of $$c$$ is $$k:1$$.
Find the value of $$k$$.Â

$$k=9$$

Solution

\begin{align}\sin30^{\circ}&=\frac{r}{|OD|}\end{align}

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

\begin{align}\frac{1}{2}&=\frac{r}{|OD|}\end{align}

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

\begin{align}|OD|=2r\end{align}

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

\begin{align}|OC|&=|OD|+r\\&=2r+3\\&=3r\end{align}

Therefore, the area of $$s$$ is $$k=3^2=9$$ times the area of $$c$$.

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## Question 4

(a)Â Find the two values of $$\theta$$ for which $$\tan \theta=-\dfrac{1}{\sqrt{3}}$$, where $$0\leq \theta \leq 4\pi$$.

$$\dfrac{5\pi}{3}$$ and $$\dfrac{11\pi}{3}$$

Solution

Let $$\alpha=\dfrac{\theta}{2}$$.

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

$\,$

Reference Angle

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

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

\begin{align}A=\frac{\pi}{6}\end{align}

Tangent is negative in both the second and fourth quadrants.

$\,$

\begin{align}\alpha&=\pi-\frac{\pi}{6}\\&=\frac{5\pi}{6}\end{align}

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

\begin{align}\frac{\theta}{2}=\frac{5\pi}{6}\end{align}

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

\begin{align}\theta=\frac{5\pi}{3}\end{align}

$\,$

\begin{align}\alpha&=2\pi-\frac{\pi}{6}\\&=\frac{11\pi}{6}\end{align}

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

\begin{align}\frac{\theta}{2}=\frac{11\pi}{6}\end{align}

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

\begin{align}\theta=\frac{11\pi}{3}\end{align}

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(b) The diagram shows $$OAB$$, a sector of a circle of radius $$7\mbox{ cm}$$ with centre $$O$$.
In the sector, $$|\angle BOA|=1.2$$ radians.
The area of the shaded region is $$21\mbox{ cm}^2$$.
Find $$|BC|$$.
Give your answer correct to $$1$$ decimal place.

$$4.4\mbox{ cm}$$

Solution

\begin{align}A_{ABO}=A_{ABC}+A_{ACO}\end{align}

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

\begin{align}\frac{1}{2}r^2\theta=21+\frac{1}{2}|CO||AO|\sin(|\angle AOC|)\end{align}

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

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

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

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

\begin{align}|BC|&=|BO|-|CO|\\&=7-2.57…\\&\approx4.4\mbox{ cm}\end{align}

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## Question 5

(a) Two events $$A$$ and $$B$$ are such that $$P(A)=\dfrac{3}{4}$$ and $$P(A\cap B)=\dfrac{1}{2}$$.

(i) Find $$P(B|A)$$. Give your answer as a fraction in its simplest form

(ii) $$P(A\cup B)=\dfrac{11}{12}$$. Investigate if the events $$A$$ and $$B$$ are independent.Â

(i)Â $$\dfrac{2}{3}$$

(ii) The events are independent.

Solution

(i)

\begin{align}P(B|A)&=\frac{P(A\cap B)}{P(A)}\\&=\frac{\frac{1}{2}}{\frac{3}{4}}\\&=\frac{2}{3}\end{align}

(ii)

\begin{align}P(A\cup B)=P(A)+P(B)-P(A\cap B)\end{align}

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

\begin{align}\frac{11}{12}=\frac{3}{4}+P(B)-\frac{1}{2}\end{align}

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

\begin{align}P(B)&=\frac{11}{12}-\frac{3}{4}+\frac{1}{2}\\&=\frac{2}{3}\end{align}

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

\begin{align}P(A)\times P(B)&=\frac{3}{4}\times\frac{2}{3}\\&=\frac{1}{2}\\&=P(A\cup B)\end{align}

Therefore, the events are independent.

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(b) A spinner consists of $$4$$ segments, as shown.
Each segment is equally likely to be landed on.
Liam, Sorcha and Lee play a game in which the spinner is spun twice and the numbers landed on are added together.
The result is divided by $$3$$ and the remainder is recorded.

If the remainder is $$0$$ then Liam wins the game.
If the remainder is $$1$$ then Sorcha wins the game.
If the remainder is $$2$$ then Lee wins the game.

Is this a fair game? (i.e. Are all $$3$$ participants equally likely to win?)

It is an unfair game.

Solution

There are $$3$$ players and $$4\times4=16$$ outcomes.

Therefore, all three people cannot have the same number of outcomes and it is therefore an unfair game.

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## Question 6

(a) A class group carried out a study of the makes and fuel types of cars in a large carpark.
It found that $$30\%$$ of the cars ran on diesel and $$70\%$$ of these diesel cars were Volkswagen.
It found that $$60\%$$ of the cars ran on petrol and $$25\%$$ of these petrol cars were Volkswagen.
It found that $$10\%$$ of the cars were hybrid/electric and $$9\%$$ of these cars were Volkswagen.
One car is selected at random from the car park.
Find the probability that it is a Volkswagen car.

$$0.369$$

Solution

\begin{align}P&=0.3\times0.7+0.6\times0.25+0.1\times0.09\\&=0.369\end{align}

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(b) The Road Safety Authority has data on driving test pass rates at all its test centres.

(i) In a particular Driving Test Centre the probability that a person taking the test for the first time will pass is $$\dfrac{1}{4}$$. All of the test results are independent.
In this centre on a particular day Joe, along with $$5$$ others, takes the test.
All six are taking the test for the first time.
Find the probability that Joe passes the test along with exactly $$2$$ others.

(ii) The overall pass rate for all drivers at another centre is $$\dfrac{1}{2}$$
(Whether it is their first attempt
or a subsequent attempt).
On a particular day, $$n$$ people take the test in this centre.
The probability that two people or less than two people pass the test can be written in the form

\begin{align}\frac{an^2+bn+c}{2^{n+1}}\end{align}

where $$a,b,c\in\mathbb{N}$$.

Find the value of $$a$$, the value of $$b$$, and the value of $$c$$.

(i)Â $$\dfrac{135}{2048}$$

(ii) $$a=1$$, $$b=1$$ and $$c=2$$

Solution

(i)

\begin{align}P&={5\choose 2}\left(\frac{1}{4}\right)^2\left(\frac{3}{4}\right)^3\left(\frac{1}{4}\right)\\&=\frac{135}{2048}\end{align}

(ii)

\begin{align}P(\mbox{at most two pass})&=P(\mbox{none pass})+P(\mbox{one passes})+P(\mbox{two pass})\\&=\left(\frac{1}{2}\right)^n+{n\choose1}\left[\left(\frac{1}{2}\right)^1\left(\frac{1}{2}\right)^{n-1}\right]+{n\choose2}\left[\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^{n-2}\right]\\&=\frac{1}{2^n}+\frac{n}{2^n}+\frac{n(n-1)}{2^{n+1}}\\&=\frac{2+2n+n^2-n}{2^{n+1}}\\&=\frac{n^2+n+2}{2^{n+1}}\end{align}

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

$$a=1$$, $$b=1$$ and $$c=2$$

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## Question 7

(a) A company makes biodegradable paper cups in the shape of a right circular cone. Each cup has radius of $$3.3\mbox{ cm}$$ and a slant height of $$9\mbox{ cm}$$, as shown.

(i) Show that the vertical height of the cup is $$8.37\mbox{ cm}$$, correct to $$2$$ decimal places.

(ii)Â Find the curved surface area of the cup correct to $$2$$ decimal places.

(iii) The diagram shows the net of the cup. Find, in degrees, the size of the angle $$\theta$$.

(ii)Â $$93.31\mbox{ cm}^2$$

(iii)Â $$132^{\circ}$$

Solution

(i)

\begin{align}9^2=h^2+3.3^2\end{align}

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

\begin{align}h&=\sqrt{9^2-3.3^2}\\&\approx8.37\mbox{ cm}\end{align}

as required.

(ii)

\begin{align}A&=\pi rl\\&=\pi(3.3)(9)\\&\approx93.31\mbox{ cm}^2\end{align}

(iii)

\begin{align}2\pi(3.3)&=[2\pi(9)]\left(\frac{\theta}{360^{\circ}}\right)\end{align}

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

\begin{align}\theta&=\frac{(3.3)(360^{\circ})}{9}\\&=132^{\circ}\end{align}

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(b) In order to avoid spillages each cup is marked with a dotted line at $$F$$ which is $$1\mbox{ cm}$$ vertically below the top of the cup, as shown.

Find the volume of water in the cup when it is filled as far as the dotted line.
Give your answer correct to $$1$$ decimal place.

$$65.2\mbox{ cm}^3$$

Solution

\begin{align}\frac{r}{7.37}=\frac{3.3}{8.37}\end{align}

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

\begin{align}r&=\frac{3.3(7.37)}{8.37}\\&=2.905…\mbox{ cm}\end{align}

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

\begin{align}V&=\frac{1}{3}\pi r^2h\\&=\frac{1}{3}\pi(2.905…)^2(7.37)\\&\approx65.2\mbox{ cm}^3\end{align}

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(c) Water flows into one of these cups through a cylindrical pipe of radius $$0.8\mbox{ cm}$$ at a flow rate of $$2.5\mbox{ cm/sec}$$. Find, to the nearest second, how long it will take to fill the cup to the line at $$F$$.

$$13\mbox{ s}$$

Solution

\begin{align}t&=\frac{65.2}{\pi(0.8^2)(2.5)}\\&\approx13\mbox{ s}\end{align}

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(d) The company decides to change the position of the line $$F$$ in order to limit the capacity of the cup to $$60\mbox{ cm}^3$$.
How far, vertically below the rim of the cup, should the line $$F$$ be drawn?
Give your answer, in $$\mbox{cm}$$, correct to $$1$$ decimal place.

$$1.2\mbox{ cm}$$

Solution

\begin{align}\frac{r}{h}=\frac{3.3}{8.37}\end{align}

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

\begin{align}r=\left(\frac{3.3}{8.37}\right)h\end{align}

and

\begin{align}\frac{1}{3}\pi r^2h=60\end{align}

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

\begin{align}\frac{1}{3}\pi \left(\frac{3.3}{8.37}\right)^2(h^2)h=60\end{align}

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

\begin{align}h&=\sqrt[3]{\frac{(60(8.37)^2(3)}{\pi(3.3)^2}}\\&=7.169…\mbox{ cm}\end{align}

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

\begin{align}x&=8.37-h\\&=8.37-7.169…\\&\approx1.2\mbox{ cm}\end{align}

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## Question 8

(a) An airline company Trans-sky Airways has designed an aptitude test for people applying for jobs as trainee pilots. The aptitude test is scored out of $$500$$ marks. The results are normally distributed with a mean score of $$280$$ and a standard deviation of $$90$$.

(i) The top $$25\%$$ of people taking the aptitude test are invited back for an interview.
Find the minimum mark needed on the test in order to be invited back for interview.

(ii) Anyone who scores above the $$40$$th percentile can re-sit the test later.
Eileen scored $$260$$ marks in the test.
Find out whether or not Eileen is eligible to re-sit the test.

(i)Â $$342$$

(ii) Eileen is eligible to re-sit the test.

Solution

(i)

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

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

\begin{align}0.68=\frac{x-280}{90}\end{align}

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

\begin{align}x&=280+(0.68)(90)\\&=341.2\end{align}

Therefore, the minimum mark is $$342$$.

(ii)

\begin{align}z&=\frac{260-280}{90}\\&=-0.222\end{align}

Since $$z=-0.25$$ corresponds to the 40th percentile, Eileen is eligible to re-sit the test.

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(b)

(i) Explain the relevance of the z-scores $$-1.96$$ and $$1.96$$ in the standard normal distribution.

(ii) Trans-sky Airways surveyed $$2500$$ of its passengers about a new service it proposed to introduce. The variable $$\hat{p}$$ is the proportion of respondents in the survey who said they would use the new service.
The radius of the $$95\%$$ confidence interval of the survey was $$0.01568$$.
Find the value of $$\hat{p}$$, where $$0.5<\hat{p}\leq1$$.

(i)Â $$95\%$$ of the curve is found in the interval $$-1.96\leq z\leq1.96$$.

(ii)Â $$\hat{p}=0.8$$

Solution

(i)Â $$95\%$$ of the curve is found in the interval $$-1.96\leq z\leq1.96$$

(ii)

\begin{align}1.96\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=r\end{align}

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

\begin{align}1.96\sqrt{\frac{\hat{p}(1-\hat{p})}{2500}}=0.01568\end{align}

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

\begin{align}\hat{p}(1-\hat{p})=2500\left(\frac{0.01568^2}{1.96^2}\right)\end{align}

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

\begin{align}\hat{p}^2-\hat{p}+0.16=0\end{align}

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

$$\hat{p}=0.2$$ or $$\hat{p}=0.8$$

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

$$\hat{p}=0.8$$

(since $$0.5<\hat{p}\leq1$$).

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(c) The weight of the Airline passengersâ€™ carry-on luggage is normally distributed with a mean of $$12\mbox{ kg}$$. The Airline has recently introduced a fee for non-carry-on luggage. After the fee was introduced, the Airline expected the mean weight of the carry-on luggage to change.
They selected a random sample of $$80$$ passengers and weighed their carry-on luggage.
The sample mean was $$13.1\mbox{ kg}$$ and the sample standard deviation was $$4.5\mbox{ kg}$$.

Test the hypothesis, at the $$5\%$$ level of significance, that the mean weight of the carry-on luggage has changed. State the null hypothesis and the alternative hypothesis.
Give your conclusion in the context of the question.

The mean weight of the carry-on luggage has changed.

Solution

Null hypothesis: Mean weight of the carry-on luggage has not changed.

Alternative hypothesis: Mean weight of the carry-on luggage has changed.

Calculations:

\begin{align}z&=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}\\&=\frac{13.1-12}{4.5/\sqrt{80}}\\&=2.186…\end{align}

Conclusion

As $$z>1.96$$, the mean weight of the carry-on luggage has changed.

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(d) The company bus can carry passengers up to a total maximum weight allowance of $$3000\mbox{ kg}$$.
The weight of passengers is normally distributed with a mean of $$73\mbox{ kg}$$ and a standard deviation of $$12\mbox{ kg}$$.
$$40$$ passengers board the bus.
Find the probability that the total passenger weight will be over the maximum weight allowance.
Give your answer as a percentage correct to $$2$$ decimal places.

$$14.69\%$$

Solution

The average weight of each person at the maximum wight allowance is

\begin{align}\mu&=\frac{3000}{4}\\&=75\mbox{ kg}\end{align}

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

\begin{align}z&=\frac{75-73}{12/\sqrt{40}}\\&=1.054\end{align}

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

\begin{align}P(\bar{x}<75)=0.8531\end{align}

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

\begin{align}P(\bar{x}>75)&=1-P(\bar{x}<75)\\&=1-0.8531…\\&=0.1469…\end{align}

Therefore, the probability that the total passenger weight will be over the maximum weight allowance is $$14.69\%$$.

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(e) A list consists of eight whole numbers. They are labelled from $$A$$ to $$H$$ as shown below.
The numbers are all greater than zero and are ordered from smallest to largest.
The difference between any two adjacent numbers is $$2$$ or more.
The median of the list is $$12.5$$.
The lower quartile (the median of the $$4$$ lowest numbers) of the list is $$7.5$$.
The interquartile range is $$12$$.
The second largest number is $$23$$, as shown.
The range of the list is $$21$$.
The mean of the list is $$13.5$$.

Find the numbers which satisfy all of the above conditions and write them into the boxes below.

$$(4,6,9,11,14,16,23,25)$$

Solution

\begin{align}D+E&=2(12.5)\\&=25\end{align}

and

\begin{align}B+C&=2(7.5)\\&=15\end{align}

and

\begin{align}F+G&=2(12+7.5)\\&=39\end{align}

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

\begin{align}F&=39-G\\&=39-23\\&=16\end{align}

and

\begin{align}H-A=21\end{align}

and

\begin{align}\frac{A+B+C+D+E+F+G+H}{8}=13.5\end{align}

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

\begin{align}A+B+C+D+E+F+G+H=108\end{align}

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

\begin{align}A+15+25+16+23+H=108\end{align}

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

\begin{align}A+H=29\end{align}

We therefore have the following two equations for two unknowns:

\begin{align}H-A=21\end{align}

\begin{align}A+H=29\end{align}

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

\begin{align}2H=50\end{align}

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

\begin{align}H=25\end{align}

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

\begin{align}A&=29-H\\&=29-25\\&=4\end{align}

Since $$D+E=25$$, the only possibilities are $$10+15$$, $$11+14$$ and $$12+13$$. The final option is ruled out as the difference between both numbers would be less than $$2$$. Likewise, the first option is ruled out as $$E$$ and $$F$$ would have the same problem.

Therefore, $$D=5$$ and $$E=25$$.

Likewise, since $$B+C=15$$, the only possibilities are $$5+10$$, $$6+9$$ and $$7+8$$. The final option is ruled out as the difference between both numbers would be less than $$2$$. Likewise, the first option is ruled out as $$A$$ and $$B$$ would have the same problem.

Therefore, $$B=6$$ and $$C=9$$.

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## Question 9

Two ships set sail at the same time, Ship $$A$$ from Port $$A$$ and Ship $$B$$ from Port $$B$$.
Port $$A$$ is $$90\mbox{ km}$$ due west of Port $$b$$, as shown below.
Ship $$A$$ is traveling due east at a speed of $$15\mbox{ km/h}$$.
Ship $$B$$ is travelling due south at a speed of $$30\mbox{ km/h}$$.

(a) Find the distance between the two ships $$30$$ minutes after they set sail.
Give your answer in $$\mbox{km}$$, correct to $$2$$ decimal places.

$$83.85\mbox{ km}$$

Solution

\begin{align}d&=\sqrt{\left(90-\frac{15}{2}\right)^2+\left(\frac{30}{2}\right)^2}\\&\approx83.85\mbox{ km}\end{align}

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##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(b) $$t$$ is the time in hours after the ships set sail.
Show that the distance between the ships at time $$t$$ can be given by the function

\begin{align}s(t)=(1125t^2-2700t+8100)^{\frac{1}{2}}\end{align}

where $$0\leq t\leq6$$.

Solution

\begin{align}s(t)&=[(90-15t)^2+(30t)^2]^{1/2}\\&=[8100-2700t+225t^2+900t^2]^{1/2}\\&=[1125t^2-2700t+8100]^{1/2}\end{align}

as required.

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##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(c) Use calculus to find the value of $$t$$ when the ships are closest to each other, and find the distance between the ships at your value of $$t$$.
Give the distance in $$\mbox{km}$$, correct to $$1$$ decimal place.

$$t=1.2\mbox{ hr}$$ and $$s=80.5\mbox{ km}$$

Solution

\begin{align}s'(t)=0\end{align}

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

\begin{align}\frac{1}{2}(1125t^2-2700t+8100)^{-1/2}(2250t-2700)=0\end{align}

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

\begin{align}2250t-2700=0\end{align}

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

\begin{align}t&=\frac{2700}{2250}\\&=1.2\mbox{ hr}\end{align}

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

\begin{align}s(1.2)&=[1125(1.2)^2-2700(1.2)+8100)^{1/2}\\&\approx80.5\mbox{ km}\end{align}

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