L.C. MATHS

Course Content
Higher Level (by year)
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Past Papers

## Question 1

Parts of the lines $$AC$$ and $$BC$$ are shown in the co-ordinate diagram below (not to scale).

(a)

(i) Find the slope of $$AC$$.

(ii) By using slopes, investigate if $$AC$$ is perpendicular to $$BC$$. Justify your answer.

(i) $$\dfrac{3}{2}$$

(ii) They are not perpendicular.

Solution

(i)

\begin{align}m_{AC}&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{3-0}{0-(-2)}\\&=\frac{3}{2}\end{align}

(ii)

\begin{align}m_{BC}&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{3-0}{0-5}\\&=-\frac{3}{5}\end{align}

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

\begin{align}m_{AC}\times m_{BC}&=\frac{3}{2}\times\left(-\frac{3}{5}\right)\\&=-\frac{9}{10}\\&\neq-1\end{align}

Therefore, they are not perpendicular.

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(b) The triangle $$LMN$$ is shown on the co-ordinate diagram below (not to scale).
The point $$M$$ has co-ordinates $$(9,1)$$
The triangle $$LMN$$ is symmetrical about the $$y$$-axis.

(i) Find the length $$|LM|$$.

(ii) Write down the equation of the horizontal line $$LM$$.

(iii) The line $$NM$$ has equation $$x+4y-13=0$$, where $$x,y\in\mathbb{R}$$.
Use this equation to find the co-ordinates of the point $$N$$.

(i) $$18$$

(ii) $$y=1$$

(iii) $$\left(0,\dfrac{13}{4}\right)$$

Solution

(i)

\begin{align}|LM|&=2\times9\\&=18\end{align}

(ii)

\begin{align}y=1\end{align}

(iii)

\begin{align}0+4y-13=0\end{align}

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

\begin{align}y=\frac{13}{4}\end{align}

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

\begin{align}N=\left(0,\frac{13}{4}\right)\end{align}

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## Question 2

(a) The circle $$k$$ has equation $$(x-4)^2+(y+2)^2=169$$.

(i) Write down the centre and radius of the circle $$k$$.

(ii) Is the point $$(11,10)$$ on the circle $$k$$, inside the circle $$k$$, or outside the circle $$k$$?

(i) Centre: $$(4,-2)$$ and radius: $$13$$

(ii) Outside

Solution

(i) Centre: $$(4,-2)$$ and radius: $$13$$

(ii)

\begin{align}(x-4)^2+(y+2)^2&=(11-4)^2+(10+2)^2\\&=49+144\\&=193\\&>169\end{align}

Therefore, this point is outside the circle $$k$$.

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(b) The diagram below shows two circles, $$s$$ and $$t$$. The circle $$s$$ has centre $$(22,13)$$.
The two circles touch at the point $$(12,11)$$.

(i) Find the co-ordinates of another point on the circle $$s$$, other than $$(12,11)$$.

(ii) The radius of the circle $$t$$ is half the radius of $$s$$.
Find the co-ordinates of the centre of the circle $$t$$.

(i) e.g. $$(32,15)$$

(ii) $$(7,10)$$

Solution

(i)

\begin{align}(12+10,11+2)\rightarrow(22+10,13+2)\rightarrow(32,15)\end{align}

(ii)

\begin{align}(22-10,13-2)\rightarrow(12-5,11-1)\rightarrow(7,10)\end{align}

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## Question 3

(a) A list of $$7$$ numbers is shown below.

\begin{align}17,8,9,8,14,11,28\end{align}

(i) Find the mean of these $$7$$ numbers, correct to $$1$$ decimal place.

(ii) Find the median of these $$7$$ numbers.

(iii) One more number is added to the list. The new median is $$10.5$$.
Find the number that was added to the list.

(i) $$13.6$$

(ii) $$11$$

(iii) $$10$$

Solution

(i)

\begin{align}\mbox{Mean}&=\frac{17+8+9+8+14+11+28}{7}\\&\approx13.6\end{align}

(ii)

\begin{align}8,8,9,11,14,17,28\end{align}

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

\begin{align}\mbox{Median}=11\end{align}

(iii)

\begin{align}\frac{x+11}{2}=10.5\end{align}

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

\begin{align}x+11=21\end{align}

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

\begin{align}x=10\end{align}

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(b) Ben has to choose three subjects to study.
He must pick one subject from each of these three groups:

Group A Group B Group C

French

Spanish

German

Biology

Physics

Chemistry

Art

Accounting

History

Geography

Home Economics

(i) How many different choices of three subjects can Ben make, picking one from each group?

(ii) The school is going to add one extra subject to one of these groups (A, B, or C).
Which group should the extra subject be added to, in order to make the number of different choices of three subjects that Ben can make as large as possible?

(i) $$60$$

(ii) Group A

Solution

(i)

\begin{align}3\times4\times5=60\end{align}

(ii)

\begin{align}\mathbf{A}:4\times4\times5=80\end{align}

or

\begin{align}\mathbf{B}:3\times5\times5=75\end{align}

or

\begin{align}\mathbf{C}:3\times4\times6=72\end{align}

It should therefore be added to Group A.

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## Question 4

(a) A group of students sat an exam. Each student was given a grade.
The following table shows how many students got each grade.

Grade Distinction High Merit Merit Achieved

Number of students

$$8$$

$$12$$

$$39$$

$$13$$

Complete the pie chart below to show this information.
Label each sector clearly. Show your working out.

The sector for Distinction is already given.
It has an angle of $$40^{\circ}$$.

Solution
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(b) A large group of people took a reading test.
The scores were normally distributed, with a mean of $$100$$ and a standard deviation of $$20$$.

Use the empirical rule to answer parts (b)(i) and (b)(ii).

(i) What percentage of the people had scores between $$80$$ and $$120$$?

(ii) The top $$2.5\%$$ of scores were given a grade of ‘Exceptional’.
What was the least score that was needed to get this grade?

(i) $$68\%$$

(ii) $$140$$

Solution

(i) $$68\%$$

(ii)

\begin{align}100+2\times20=140\end{align}

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(c) The scores of six people on this test were as follows:

\begin{align}104,82,94,113,98,105\end{align}

Find the range and the standard deviation of these numbers.
Give the standard deviation correct to $$1$$ decimal place.

Range: $$31$$

Standard deviation: $$9.8$$

Solution

Range is $$113-82=31$$ and standard deviation is $$\approx 9.8$$.

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## Question 5

(a) An equilateral triangle $$PQR$$ has sides of length $$8\mbox{ cm}$$.

(i) Write down the size of the angle $$\angle PQR$$.

(ii) Show that the area of the triangle $$PQR$$ is $$16\sqrt{3}\mbox{ cm}^2$$.

(iii) Hence, or otherwise, find the perpendicular height of the triangle $$PQR$$, taking $$PQ$$ as
the base. Give your answer in the form $$a\sqrt{b}\mbox{ cm}$$, where $$a,b\in\mathbb{N}$$.

(i) $$60^{\circ}$$

(iii) $$4\sqrt{3}\mbox{ cm}$$

Solution

(i) $$60^{\circ}$$

(ii)

\begin{align}\mbox{Area}&=\frac{1}{2}ab\sin C\\&=\frac{1}{2}(8)(8)\sin60^{\circ}\\&=32\left(\frac{\sqrt{3}}{2}\right)\\&=16\sqrt{3}\mbox{ cm}^2\end{align}

(iii)

\begin{align}16\sqrt{3}=\frac{1}{2}bh\end{align}

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

\begin{align}16\sqrt{3}=\frac{1}{2}(8)h\end{align}

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

\begin{align}h&=\frac{2(16\sqrt{3})}{8}\\&=4\sqrt{3}\mbox{ cm}\end{align}

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(b) $$GHK$$ is a right-angled triangle.
$$|\angle GHK|=90^{\circ}$$
$$|GH|=12\mbox{ cm}$$
and $$|GK|=30\mbox{ cm}$$.

Using the theorem of Pythagoras, find the distance $$|HK|$$.
Give your answer correct to $$1$$ decimal place.

$$27.5\mbox{ cm}$$

Solution

\begin{align}|GK|^2=|GH|^2+|HK|^2\end{align}

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

\begin{align}|HK|&=\sqrt{|GK|^2-|GH|^2}\\&=\sqrt{30^2-12^2}\\&\approx27.5\mbox{ cm}\end{align}

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## Question 6

(a) The circle $$c$$ is shown in the diagram below (not to scale). Its centre is at the point $$O$$.
The points $$A$$, $$B$$ and $$D$$ lie on the circle, and $$[AB]$$ is a diameter of the circle.

(i) Write down $$|\angle ADB|$$, the size of the total angle at the point $$D$$.

(ii) $$|\angle AOD|=130^{\circ}$$. Work out the size of the angle marked $$X$$ in the diagram.

(iii) The radius of the circle is $$18\mbox{ cm}$$. Find the length of the arc $$AD$$.
Give your answer in $$\mbox{cm}$$, in terms of $$\pi$$.

(i) $$90^{\circ}$$

(ii) $$65^{\circ}$$

(iii) $$13\pi\mbox{ cm}$$

Solution

(i) $$90^{\circ}$$

(ii)

\begin{align}2X=180^{\circ}-50^{\circ}\end{align}

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

\begin{align}X=65^{\circ}\end{align}

(iii)

\begin{align}\frac{130^{\circ}}{360^{\circ}}\times2\pi(18)=13\pi\mbox{ cm}\end{align}

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(b) Two statements, $$A$$ and $$B$$, are shown below.

State whether each statement below is true or false. Give a reason for your answer in each case.

(i) Statement A: If two triangles are similar, then they must be congruent.

(ii) Statement B: If two triangles are congruent, then they must be similar.

(i) False as two similar triangles may be of different sizes whereas two congruent triangles must be the same size.

(ii) True as two congruent triangles have the same angles which is also true of two similar triangles.

Solution

(i) False as two similar triangles may be of different sizes whereas two congruent triangles must be the same size.

(ii) True as two congruent triangles have the same angles which is also true of two similar triangles.

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## Question 7

An animal shelter takes cats and dogs.

(a) The staff record the weight of each animal in the shelter on two different days: the day they are taken in to the shelter (Day X) and a number of days later (Day Y).

The table below shows the weights of $$10$$ dogs (labelled A to J) on these two days.

Animal A B C D E F G H I J

Weight Day X (kg)

$$4.5$$

$$4.9$$

$$5.3$$

$$5.3$$

$$5.5$$

$$5.7$$

$$6.7$$

$$6.9$$

$$7.3$$

$$7.4$$

Weight Day Y (kg)

$$4.8$$

$$5.4$$

$$5.5$$

$$6.1$$

$$6.3$$

$$6.0$$

$$7.0$$

$$7.6$$

$$8.1$$

$$7.9$$

(i) Draw a back-to-back stem-and-leaf plot to show this information.

(ii) What does the stem-and-leaf plot show about the weights of the dogs on Day X and Day Y?

(iii) $$r$$ is the correlation coefficient between the weight on Day X and the weight on Day Y.
Based on the data in the table, which of the following is most likely value of $$r$$?

\begin{align}-0.9&&-0.2&&0.2&&0.9\end{align}

Give a reason for your answer, based on the data in the table.

(i)

(ii) The weight of the dogs has increased from Day X to Day Y.

(iii) $$0.9$$ as there is a strong, positive linear relationship.

Solution

(i)

(ii) The weight of the dogs has increased from Day X to Day Y.

(iii) $$0.9$$ as there is a strong, positive linear relationship.

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The table below shows the breakdown of the animals in the shelter, by type of animal (cat or dog) and by sex (male or female), on a particular Monday.

(b) Complete the table, by filling in the four missing values.

Male Female Total

Cats

$$5$$

$$9$$

$$\mathbf{14}$$

Dogs

$$11$$

Total

$$\mathbf{40}$$

Male Female Total

Cats

$$5$$

$$9$$

$$\mathbf{14}$$

Dogs

$$11$$

$$15$$

$$\mathbf{26}$$

Total

$$16$$

$$24$$

$$\mathbf{40}$$

Solution
Male Female Total

Cats

$$5$$

$$9$$

$$\mathbf{14}$$

Dogs

$$11$$

$$15$$

$$\mathbf{26}$$

Total

$$16$$

$$24$$

$$\mathbf{40}$$

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Three different animals were picked at random from the animals in the shelter on this Monday.

(c)

(i) Find the probability that the first animal picked was a cat.

(ii) Find the probability that all three animals picked were male dogs.
Give your answer correct to $$3$$ decimal places.

(i) $$\dfrac{7}{20}$$

(ii) $$0.017$$

Solution

(i)

\begin{align}P(\mbox{cat})&=\frac{14}{40}\\&=\frac{7}{20}\end{align}

(ii)

\begin{align}P(3\mbox{ male dogs})&=\frac{11}{40}\times\frac{10}{39}\times\frac{9}{38}\\&\approx0.017\end{align}

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(d) The $$9$$ female cats were put in $$9$$ separate pens.
Work out the number of ways in which this could have been done (that is, the number of different possible arrangements).

$$362{,}880$$

Solution

\begin{align}9!=362{,}880\end{align}

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(e) By the end of this week, $$10$$ of the animals had left the shelter, and no new animals had been taken in. If an animal was picked at random at the end of this week, the probability of picking a dog would be $$\dfrac{11}{15}$$.

Work out how many cats left the shelter during this week.

$$6$$

Solution

\begin{align}\mbox{Number of dogs}&=\frac{11}{15}\times(40-10)\\&=22\end{align}

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

\begin{align}\mbox{Number of cats}&=30-22\\&=8\end{align}

Therefore, the number of cats that left is $$14-8=6$$.

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## Question 8

The top of a particular lighthouse is in the shape of a hemisphere on top of a cylinder.
The hemisphere and the cylinder both have a radius of $$3\mbox{ m}$$.

(a)

(i) Find the volume of the hemisphere. Give your answer in $$\mbox{m}^3$$ in terms of $$\pi$$.

(ii) The volume of the cylinder is $$36\pi\mbox{ m}^3$$.

Work out $$h$$, the height of the cylinder.

(i) $$18\pi\mbox{ m}^3$$

(ii) $$4\mbox{ m}$$

Solution

(i)

\begin{align}V&=\frac{2}{3}\pi r^3\\&=\frac{2}{3}\pi(3^3)\\&=18\pi\mbox{ m}^3\end{align}

(ii)

\begin{align}\pi r^2h=36\pi\end{align}

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

\begin{align}(3^2)h=36\end{align}

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

\begin{align}h&=\frac{36}{3^2}\\&=4\mbox{ m}\end{align}

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(b) The diagram on the right below shows part of the base of the lighthouse (not to scale).
It is in the shape of a cone of radius $$7.5\mbox{ m}$$, from which the top part has been removed, leaving a horizontal circle of radius $$3\mbox{ m}$$.

(i) The height of the cone before the top part is removed is $$47\mbox{ m}$$.
Work out the size of the angle at the base of the cone, marked $$A$$ in the diagram above.

(ii) Find the distance marked $$k$$ on the diagram, the height after the top part is removed.

(i) $$81^{\circ}$$

(ii) $$28.2\mbox{ m}$$

Solution

(i)

\begin{align}A&=\tan^{-1}\left(\frac{47}{7.5}\right)\\&\approx81^{\circ}\end{align}

(ii)

\begin{align}\frac{x}{47}=\frac{3}{7.5}\end{align}

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

\begin{align}x&=\frac{(3)(47)}{7.5}\\&=18.8\end{align}

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

\begin{align}k&=47-x\\&=47-18.8\\&=28.2\mbox{ m}\end{align}

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(c) Assume that the Fastnet lighthouse can be seen from anywhere within a circle of radius $$50\mbox{ km}$$.

(i) Work out the area of the circle within which the Fastnet lighthouse can be seen.
Give your answer correct to the nearest $$\mbox{km}^2$$.

(ii) $$50\mbox{ km}=27\mbox{ nautical miles}$$.

Use this to work out how many $$\mbox{km}$$ are in $$1\mbox{ nautical mile}$$.

Give your answer correct to $$4$$ significant figures.

(i) $$7{,}854\mbox{ km}^2$$

(ii) $$1.852\mbox{ km}$$

Solution

(i)

\begin{align}A&=\pi r^2\\&=\pi(50^2)\\&\approx7{,}854\mbox{ km}^2\end{align}

(ii)

\begin{align}50\mbox{ km}=27\mbox{ nautical miles}\end{align}

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

\begin{align}\frac{50}{27}\mbox{ km}=1\mbox{ nautical mile}\end{align}

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

\begin{align}\approx1.852\mbox{ km}=1\mbox{ nautical mile}\end{align}

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(d) The top of the Fastnet lighthouse, $$F$$, is $$49\mbox{ m}$$ above sea level.

The angle of elevation of the top of the lighthouse from a ship $$S$$ is $$1.2^{\circ}$$, as shown in the
diagram below (not to scale).

Find the horizontal distance marked $$d$$ below, from the ship to the base of the lighthouse.
Give your answer in kilometres, correct to $$2$$ decimal places.

$$2.34\mbox{ km}$$

Solution

\begin{align}\tan1.2^{\circ}=\frac{49}{d}\end{align}

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

\begin{align}d&=\frac{49}{\tan1.2^{\circ}}\\&\approx2.34\mbox{ km}\end{align}

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## Question 9

Seán has built a shed. The diagrams below show the dimensions of Seán’s shed.
The shed is in the shape of a prism. Its front face is in the shape of a triangle on top of a rectangle.
Its highest point is directly above the centre of its base.

(a) State which of the following statements is most likely to be true, and write down a possible height of Seán that would support your answer.

• The shed at the highest point is $$3$$ times as high as Seán.
• The shed at the highest point is $$5$$ times as high as Seán.
• The shed at the highest point is $$8$$ times as high as Seán.

$$1.7\mbox{ m}$$

Solution

The shed at the highest point is $$5$$ times as high as Seán.

\begin{align}h&=\frac{8.5}{5}\\&=1.7\mbox{ m}\end{align}

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(b) Seán says that his shed has a capacity of over one million litres, where $$1\mbox{ m}^3=1{,}000\mbox{ litres}$$.
Work out the volume of Seán’s shed, to show that he is correct.

$$V=1{,}674{,}000\mbox{ litres}$$ which is over one million litres.

Solution

\begin{align}V&=\left[(7\times12)+\frac{1}{2}(12\times1.5)\right]\times 18\\&=1{,}674\mbox{ m}^3\\&=1{,}674{,}000\mbox{ litres}\end{align}

which is indeed over one million litres.

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(c) Use the theorem of Pythagoras to find the length of the distance marked $$d$$ in the diagram below, the slant length of the roof. Give your answer in metres, correct to $$1$$ decimal place.

$$6.2\mbox{ m}$$

Solution

\begin{align}d&=\sqrt{6^2+1.5^2}\\&\approx6.2\mbox{ m}\end{align}

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(d) A scale diagram of the front of the shed is shown below.
One point on the diagram is marked A.
Construct an enlargement of the diagram below, with centre A and a scale factor of $$3$$.
Show all of your construction lines clearly

Solution
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(e) The diagram below shows part of the roof of a smaller shed (not to scale).
Some measurements are marked on the diagram.

(i) Show that $$|BC|=4.65\mbox{ m}$$, correct to $$2$$ decimal places.

(ii) Find $$|\angle ACB|$$, the angle that the roof makes at the point $$C$$.
Remember that $$|BC|=4.65\mbox{ m}$$, correct to $$2$$ decimal places.

(ii) $$19^{\circ}$$

Solution

(i)

\begin{align}|BC|&=\sqrt{|AB|^2+|AC|^2-2|AB||AC|\cos|\angle BAC|}\\&=\sqrt{3^2+7^2-2(3)(7)\cos(30^{\circ}}\\&\approx4.65\mbox{ m}\end{align}

as required.

(ii)

\begin{align}|\angle ACB|&=\sin^{-1}\left(\frac{3\sin30^{\circ}}{4.65}\right)\\&\approx19^{\circ}\end{align}

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## Question 10

A survey was carried out to investigate the amount and type of exercise that adults in Ireland are taking in 2022. This survey was carried out on a sample of adults in Ireland.

(a) Write down one advantage and one disadvantage of carrying out a survey on a sample instead of a population.

Solution

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(b)

(i) A random sample of $$1500$$ people took part in the survey.
Show that the margin of error for the survey is $$2.6\%$$, correct to $$1$$ decimal place.

(ii) $$71\%$$ of the sample said that they walk for recreation.
Find the number of people from the sample who said that they walk for recreation.

(iii) Use the percentages given in (b)(i) and (b)(ii) to write down a $$95\%$$ confidence interval for the percentage of all adults in Ireland who walk for recreation, in 2022.

(iv) According to the 2019 Irish Sport Monitor Report, $$65\%$$ of the adults in Ireland walked
for recreation.

Carry out a Hypothesis Test, at the $$5\%$$ level of significance, to find out if this figure of $$65\%$$ has changed in 2022, based on the results of the above survey. Clearly state your conclusion and give a reason for your conclusion.

(ii) $$1065\mbox{ people}$$

(iii) $$68.4\leq p\leq73.6$$

(iv)

Null hypothesis: $$65%$$ of the adults in Ireland walk for recreation.

Alternative hypothesis: $$65%$$ of the adults in Ireland do not walk for recreation.

Calculations: As $$65\%$$ is not within the confidence interval of part (iii), we reject the null hypothesis.

Conclusion: $$65%$$ of the adults in Ireland do not walk for recreation.

Solution

(i)

\begin{align}\mbox{Margin of error}&=\frac{1}{\sqrt{n}}\\&=\frac{1}{\sqrt{1500}}\\&\approx2.6\%\end{align}

as required.

(ii)

\begin{align}N&=0.71\times1500\\&=1065\mbox{ people}\end{align}

(iii)

\begin{align}71-2.6\leq p\leq71+2.6\end{align}

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

\begin{align}68.4\leq p\leq73.6\end{align}

(iv)

Null hypothesis: $$65%$$ of the adults in Ireland walk for recreation.

Alternative hypothesis: $$65%$$ of the adults in Ireland do not walk for recreation.

Calculations: As $$65\%$$ is not within the confidence interval of part (iii), we reject the null hypothesis.

Conclusion: $$65%$$ of the adults in Ireland do not walk for recreation.

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(c) Assume that $$20\%$$ of adults in Ireland jog for recreation.
Three adults are picked at random.
Find the probability that exactly one of these adults jogs for recreation.

$$0.384$$

Solution

\begin{align}{3\choose1}\times0.8\times0.8\times0.2=0.384\end{align}

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Sinéad is joining a gym. She will be able to use the gym, and also go to classes.

(d) She could pay for individual classes, at €$$6$$ per class. Sinéad estimates that there is:

• a $$30\%$$ chance she will go to no classes in the year,
• a $$60\%$$ chance she will go to $$1$$ class a week, so $$52$$ classes in a year, and
• a $$10\%$$ chance she will go to $$2$$ classes a week, so $$104$$ classes a year.

Use these figures to work out the expected value of the cost of the classes for a year.

$$249.60\mbox{ euro}$$

Solution

\begin{align}E(x)&=0\times(6\times0.3)+52\times(6\times0.6)+104\times(6\times0.1)\\&=249.60\mbox{ euro}\end{align}

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(e) There are two price plans for the gym, Silver and Gold, as follows:

Silver Gold

€$$420$$ for the year.

Pay €$$6$$ for each class.

€$$670$$ for the year.

All classes are free.

Work out the least number of classes that Sinéad would have to go to in a year, so that the Gold plan would cost less than the Silver plan.

$$42$$

Solution

\begin{align}420+6n>670\end{align}

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

\begin{align}6n>250\end{align}

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

\begin{align}n>41.66…\end{align}

Therefore, the least number of classes is $$42$$.

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