Course Content
Higher Level (by year)
0/17
Higher Level (by topic)
0/13
Ordinary Level (by year)
0/17
Ordinary Level (by topic)
0/12
Past Papers

## Question 1

(a) An experiment consists of throwing two fair, standard, six-sided dice and recording the sum of the two numbers thrown. Some of the totals are shown in the table.

(a) An experiment consists of throwing two fair, standard, six-sided dice and recording the sum of the two numbers thrown. Some of the totals are shown in the table.

(i) Complete the table.

(ii) Find the probability of getting a total of $$7$$ or $$11$$.

(iii) Find the probability of getting a total which is a prime number.

(i)

(ii) $$\dfrac{2}{9}$$

(iii) $$\dfrac{5}{12}$$

Solution

(i)

(ii)

\begin{align}P&=\frac{6+2}{36}\\&=\frac{2}{9}\end{align}

(iii)

\begin{align}P&=\frac{1+2+4+6+2}{36}\\&=\frac{5}{12}\end{align}

Video Walkthrough
##### 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) A car distributor sells Ford cars and Renault cars.
It has $$30$$ cars for sale on a particular day; $$18$$ are Ford cars and $$12$$ are Renault cars.
$$7$$ of the Ford cars are red and $$4$$ of the Renault cars are red. One of the $$30$$ cars is chosen at random. What is the probability that the car chosen is a Ford car or a car which is not red?

$$\dfrac{13}{15}$$

Solution

\begin{align}P&=\frac{18}{30}+\frac{8}{30}\\&=\frac{13}{15}\end{align}

Video Walkthrough

## Question 2

The points $$P(7,10)$$, $$Q(1,2)$$ and $$R(11,4)$$ are the vertices of the triangle shown.
The point $$U(4,6)$$ is the midpoint of $$[PQ]$$ and the point $$V$$ is the midpoint of $$[PR]$$.

(a) Find the co-ordinates of $$V$$.

$$(9,7)$$

Solution

\begin{align}V&=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\&=\left(\frac{7+11}{2},\frac{10+4}{2}\right)\\&=(9,7)\end{align}

Video Walkthrough
##### 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) Show, by using slopes, that $$UV$$ is parallel to $$QR$$.

Solution

\begin{align}m_{UV}&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{7-6}{9-4}\\&=\frac{1}{5}\end{align}

and

\begin{align}m_{QR}&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{4-2}{11-1}\\&=\frac{1}{5}\end{align}

as required.

Video Walkthrough
##### 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) Find the area of the triangle $$PQR$$.

$$34\mbox{ units}^2$$

Solution

\begin{align}(7,10)&&(1,2)&&(11,4)\end{align}

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

\begin{align}(7-7,10-10)&&(1-7,2-10)&&(11-7,4-10)\end{align}

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

\begin{align}(0,0)&&(-6,-8)&&(4,-6)\end{align}

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

\begin{align}A&=\frac{1}{2}|x_1y_2-x_2y_1|\\&=\frac{1}{2}|(-6)(-6)-(4)(-8)|\\&=34\mbox{ units}^2\end{align}

Video Walkthrough
##### 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!

(d) The point $$S$$ is the image of the point $$Q$$ under the translation $$\vec{UV}$$.
Find the coordinates of $$S$$.

$$(6,3)$$

Solution

\begin{align}S&=(1+5,2+1)\\&=(6,3)\end{align}

Video Walkthrough

## Question 3

(a)

(i) Find the number of different arrangements that can be made using all the letters of the word RAINBOW. Each letter is used only once.

(ii) Find the number of different $$3$$-letter arrangements that can be made using the letters
of the word RAINBOW. Each letter is used at most once.

(i) $$5{,}040$$

(ii) $$210$$

Solution

(i)

\begin{align}7!=5{,}040\end{align}

(ii)

\begin{align}7\times6\times5=210\end{align}

Video Walkthrough
##### 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) A game, called Rainbow, uses an unbiased circular spinner.
The spinner has seven sectors coloured red ($$R$$), orange ($$O$$), yellow ($$Y$$), green ($$G$$), blue ($$B$$), indigo ($$I$$), and violet ($$V$$) as shown below.
The table below shows the angle of each sector.
It also shows the cash prize that a player wins if the spinner stops in that sector.

Colour Angle Probability Prize

Red

$$72^{\circ}$$

$$20\mbox{ euro}$$

Orange

$$30^{\circ}$$

$$60\mbox{ euro}$$

Yellow

$$45^{\circ}$$

$$\dfrac{1}{8}$$

$$24\mbox{ euro}$$

Green

$$90^{\circ}$$

$$8\mbox{ euro}$$

Blue

$$60^{\circ}$$

$$42\mbox{ euro}$$

Indigo

$$18^{\circ}$$

$$90\mbox{ euro}$$

Violet

$$45^{\circ}$$

$$48\mbox{ euro}$$

(i) Complete the “Probability” column of the table which shows the probability of the spinner coming to rest in each sector after one spin.

(ii) Find the expected value of the prize that a player wins if they play Rainbow.

(i)

$$t$$ $$0$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$8$$

$$\mathbf{h(t)}$$

$$42$$

(ii) $$31.50\mbox{ euro}$$

Solution

(i)

$$t$$ $$0$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$8$$

$$\mathbf{h(t)}$$

$$42$$

(ii)

\begin{align}E(X)&=20\left(\frac{1}{5}\right)+60\left(\frac{1}{12}\right)+24\left(\frac{1}{8}\right)+8\left(\frac{1}{4}\right)+42\left(\frac{1}{6}\right)+90\left(\frac{1}{20}\right)+48\left(\frac{1}{8}\right)\\&=31.50\mbox{ euro}\end{align}

Video Walkthrough

## Question 4

The points $$A(1,8)$$ and $$B(9,0)$$ are the end-points of a diameter of the circle $$w$$, as shown in the diagram.

(a) Find the co-ordinates of the centre of $$w$$.

$$(5,4)$$

Solution

\begin{align}\mbox{Centre}&=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\&=\left(\frac{1+9}{2},\frac{8+0}{2}\right)\\&=(5,4)\end{align}

Video Walkthrough
##### 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) Find the length of the radius of $$w$$. Give your answer in the form $$p\sqrt{q}$$, where $$p,q\in\mathbb{N}$$.

$$4\sqrt{2}\mbox{ units}$$

Solution

\begin{align}r&=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\&=\sqrt{(5-1)^2+(4-8)^2}\\&=\sqrt{32}\\&=\sqrt{16\times2}\\&=\sqrt{16}\times\sqrt{2}\\&=4\sqrt{2}\mbox{ units}\end{align}

Video Walkthrough
##### 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) Hence write down the equation of the circle $$w$$.

$$(x-5)^2+(y-4)^2=32$$

Solution

\begin{align}(x-5)^2+(y-4)^2=32\end{align}

Video Walkthrough
##### 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!

(d) Find the equation of the line that is a tangent to the circle $$w$$ at $$A$$.
Give your answer in the form $$ax+by+c=0$$, where $$a,b$$ and $$c\in\mathbb{Z}$$.

$$x-y+7=0$$

Solution

\begin{align}m_{AB}&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{0-8}{9-1}\\&=-1\end{align}

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

\begin{align}m_{\perp}=1\end{align}

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

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

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

\begin{align}y-8=1(x-1)\end{align}

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

\begin{align}y-8=x-1\end{align}

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

\begin{align}x-y+7=0\end{align}

Video Walkthrough

## Question 5

(a) The square $$ABCD$$ has sides of length $$7\mbox{ cm}$$.
The vertices of the square $$PQRS$$ lie on the perimeter of $$ABCD$$, as shown in the diagram, with $$|AQ|=5\mbox{ cm}$$.
Find the area of the square $$PQRS$$.

$$29\mbox{ cm}^2$$

Solution

\begin{align}|PQ|&=\sqrt{|AP|^2+|AQ|^2}\\&=\sqrt{2^2+5^2}\\&=\sqrt{29}\end{align}

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

\begin{align}A&=|PQ|^2\\&=(\sqrt{29})^2\\&=29\mbox{ cm}^2\end{align}

Video Walkthrough
##### 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) The circles $$u$$ and $$v$$ represent two wheels that are free to rotate about their centres, as shown.
The radius of $$u$$ is $$4\mbox{ cm}$$ and the radius of $$v$$ is $$6\mbox{ cm}$$.

(i) Find the length of the circumference of each circle.
Give your answers in $$\mbox{cm}$$ in terms of $$\pi$$.

(ii) The wheels $$u$$ and $$v$$ are in non-slip contact and therefore the rotation of one causes the other to rotate. Find the number of complete rotations wheel $$u$$ makes if wheel $$v$$ completes $$100$$ rotations.

(i) $$C_u=8\pi$$ and $$C_v=12\pi$$

(ii) $$150\mbox{ rotations}$$

Solution

(i)

\begin{align}C_u&=2\pi r_u\\&=2\pi(4)\\&=8\pi\mbox{ cm}\end{align}

and

\begin{align}C_v&=2\pi r_v\\&=2\pi(6)\\&=12\pi\mbox{ cm}\end{align}

(ii)

\begin{align}\frac{12\pi}{8\pi}\times100=150\mbox{ rotations}\end{align}

Video Walkthrough

## Question 6

(a)

(i) Construct the triangle $$ABC$$. where $$|AB|=10\mbox{ cm}$$, $$|\angle CAB|=60^{\circ}$$ and $$|\angle ABC|=40^{\circ}$$.
Label each vertex clearly.

(ii) Measure $$|BC|$$ and write your answer in $$\mbox{cm}$$, correct to $$1$$ decimal place.

(i)

(ii) $$8.8\mbox{ cm}$$

Solution

(i)

(ii)

\begin{align}|BC|=8.8\mbox{ cm}\end{align}

Video Walkthrough
##### 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) The diagram shows a parallelogram with vertices $$P$$, $$Q$$, $$R$$ and $$S$$.
$$|\angle SPQ|=115^{\circ}$$, $$|\angle QRS|=\alpha^{\circ}$$ and $$|\angle RSP|=\beta^{\circ}$$.

(i) Write down the value of $$\alpha$$ and the value of $$\beta$$.

(ii) Explain why the triangle $$PQR$$ is congruent to triangle $$RSP$$.
Give a reason for any statement you make in your explanation.

(i) $$\alpha=115^{\circ}$$ and $$\beta=65^{\circ}$$

Solution

(i)

\begin{align}\alpha=115^{\circ}\end{align}

and

\begin{align}\beta&=\frac{1}{2}(360^{\circ}-115^{\circ}-115^{\circ})\\&=65^{\circ}\end{align}

(ii)

$$|SR|=|PQ|$$ (opposite sides of parallelogram)

and

$$|SP|=|RQ|$$ (opposite sides of parallelogram)

and third side is shared. Therefore, according to SSS rule, they are congruent.

Video Walkthrough

## Question 7

The table below shows the total rainfall, in millimetres, and the total sunshine, in hours, at Valentia, County Kerry, during the month of June from $$2001$$ to $$2010$$.

Year $$2001$$ $$2002$$ $$2003$$ $$2004$$ $$2005$$ $$2006$$ $$2007$$ $$2008$$ $$2009$$ $$2010$$

Total Rainfall (mm)

$$72$$

$$133$$

$$155$$

$$101$$

$$94$$

$$47$$

$$149$$

$$134$$

$$94$$

$$84$$

Total Sunshine (hours)

$$169$$

$$124$$

$$180$$

$$173$$

$$173$$

$$239$$

$$159$$

$$168$$

$$228$$

$$205$$

###### Source: Met Éireann

(a) Based on the data in the table above write down:

(i) the range of the rainfall data

(ii) the year with the highest June rainfall

(iii) the year with the least June sunshine

(i) $$108\mbox{ mm}$$

(ii) $$2003$$

(ii) $$2002$$

Solution

(i) $$155-47=108\mbox{ mm}$$

(ii) $$2003$$

(ii) $$2002$$

Video Walkthrough
##### 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) Based on the data in the table, write down the year with the best June weather and give a reason for your answer.

$$2006$$ as it had both the least rainfall and the most sunshine.

Solution

$$2006$$ as it had both the least rainfall and the most sunshine.

Video Walkthrough
##### 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) Write the rainfall data in increasing order and hence find the median of the rainfall.

$$47,72,84,94,94,101,133,134,149,155$$ and $$\mbox{median}=97.5\mbox{ mm}$$

Solution

\begin{align}47&&72&&84&&94&&94&&101&&133&&134&&149&&155\end{align}

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

\begin{align}\mbox{Median}&=\frac{94+101}{2}\\&=97.5\mbox{ mm}\end{align}

Video Walkthrough
##### 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!

(d)

(i) Find the mean number of sunshine hours for June in Valentia between $$2001$$ and $$2010$$.

(ii) For what years was the sunshine data within $$5\%$$ of the mean number of sunshine hours in Valentia?

(i) $$181.8\mbox{ hours}$$

(ii) $$2003$$, $$2004$$ and $$2005$$

Solution

(i)

\begin{align}\mbox{Mean}&=\frac{169+124+180+173+173+239+159+168+228+205}{10}\\&=181.8\mbox{ hours}\end{align}

(ii)

\begin{align}0.95\times181<x<1.05\times181\end{align}

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

\begin{align}172.71<x<190.89\end{align}

The years within this range are $$2003$$, $$2004$$ and $$2005$$.

Video Walkthrough
##### 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!

(e) Find the standard deviation of the rainfall data, in $$\mbox{mm}$$, correct to $$1$$ decimal place.

$$33.5$$

Solution

\begin{align}\sigma\approx33.5\end{align}

Video Walkthrough
##### 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!

(f) Part of a scatterplot of the data in the table is shown below.
The first four data points are plotted.

(i) Complete the scatterplot.

(ii) One of the numbers in the table is the correlation coefficient for the data above, correct to $$1$$ decimal place.
Based on the scatterplot, select the number that you think most accurately reflects this data. Explain your choice.

Tick one box

$$0.6$$

$$0.1$$

$$-0.1$$

$$-0.6$$

(i)

(ii) $$-0.6$$ as the correlation is negative and moderate.

Solution

(i)

(ii) $$-0.6$$ as the correlation is negative and moderate.

Video Walkthrough

## Question 8

The diagram shows a section of a garden divided into three parts.
In the diagram: $$|Pr|=3.3\mbox{ m}$$, $$|PQ|=6.5\mbox{ m}$$, $$|QT|=|QS|=8\mbox{ m}$$, $$|\angle QRP|=90^{\circ}$$, $$|\angle PQR|=\alpha^{\circ}$$, and $$|\angle RQS|=\beta^{\circ}$$.

(a) Use the theorem of Pythagoras to find $$|RQ|$$.

$$5.6\mbox{ m}$$

Solution

\begin{align}|RQ|&=\sqrt{|PQ|^2-|PR|^2}\\&=\sqrt{6.5^2-3.3^2}\\&=5.6\mbox{ m}\end{align}

Video Walkthrough
##### 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) Show that $$\alpha=31^{\circ}$$, correct to the nearest degree.

Solution

\begin{align}\alpha&=\sin^{-1}\left(\frac{3.3}{6.5}\right)\\&\approx31^{\circ}\end{align}

as required.

Video Walkthrough
##### 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 the value of $$\alpha$$ given in part (b) to find the value of $$\beta$$.

$$149^{\circ}$$

Solution

\begin{align}\beta&=180^{\circ}-31^{\circ}\\&=149^{\circ}\end{align}

Video Walkthrough
##### 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!

(d) Use the Cosine Rule to find the length of $$[RS]$$.

$$13\mbox{ m}$$

Solution

\begin{align}|RS|&=\sqrt{|RQ|^2+|QS|^2-2|RQ||QS|\cos\beta}\\&=\sqrt{5.6^2+8^2-2(5.6)(8)\cos(149^{\circ})}\\&\approx13\mbox{ m}\end{align}

Video Walkthrough
##### 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!

(e) $$SQT$$ is a sector of a circle whose centre is $$Q$$.
Find the length of the arc $$TS$$.

$$4.3\mbox{ m}$$

Solution

\begin{align}l&=2\pi r\left(\frac{\theta}{360^{\circ}}\right)\\&=2\pi (8)\left(\frac{31^{\circ}}{360^{\circ}}\right)\\&\approx4.3\mbox{ m}\end{align}

Video Walkthrough
##### 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!

(f) Find the area of the sector $$SQT$$.

$$17.3\mbox{ m}^2$$

Solution

\begin{align}A&=\pi r^2\left(\frac{\theta}{360^{\circ}}\right)\\&2\pi (8)^2\left(\frac{31^{\circ}}{360^{\circ}}\right)\\&\approx17.3\mbox{ m}^2\end{align}

Video Walkthrough

## Question 9

(a)

(i) Find the volume of a solid sphere of radius $$0.3\mbox{ cm}$$.
Give your answer in $$\mbox{cm}^3$$, correct to $$3$$ decimal places.

(ii) The sphere is made of pure gold. Each $$\mbox{cm}^3$$ of pure gold weighs $$19.3$$ grams.
Find the number of grams of pure gold in the sphere.
Give your answer correct to $$2$$ decimal places.

(iii) It is known that there are approximately $$6.02\times 10^{23}$$ atoms in $$197$$ grams of pure gold.
Find the number of atoms of pure gold in the sphere.
Give your answer in the form $$a\times 10^n$$, where $$1\leq a<10$$, $$n\in\mathbb{N}$$, and where $$\alpha$$ is correct to $$2$$ significant figures.

(i) $$0.113\mbox{ cm}^3$$

(ii) $$2.18\mbox{ g}$$

(iii) $$6.7\times10^{21}\mbox{ atoms}$$

Solution

(i)

\begin{align}V&=\frac{4}{3}\pi r^3\\&=\frac{4}{3}\pi(0.3)^3\\&\approx0.113\mbox{ cm}^3\end{align}

(ii)

\begin{align}0.113\times19.3\approx2.18\mbox{ g}\end{align}

(iii)

\begin{align}2.18\times\frac{6.02\times10^{23}}{197}\approx6.7\times10^{21}\mbox{ atoms}\end{align}

Video Walkthrough
##### 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) A survey was carried out on behalf of a television station to investigate the popularity of a certain show.

(i) A random sample of $$1560$$ television viewers was surveyed. Find the margin of error of the survey. Give your answer as a percentage, correct to $$1$$ decimal place.

(ii) In the survey, $$546$$ of the $$1560$$ viewers surveyed said that they liked the show. Use your answer to part b(i) above to create a $$95%$$ confidence interval for the percentage of viewers who liked the show.

(iii) An executive for the television station had claimed that $$40\%$$ of viewers liked the show.
Use your answer to part b(ii) above to conduct a hypothesis test, at the $$5\%$$ level of significance, to test the executive’s claim. State your null hypothesis, your alternative hypothesis and give your conclusion in the context of the question.

(i) $$2.5\%$$

(ii) $$32.5\%<\hat{p}<37.5\%$$

(iii)

Null hypothesis: The percentage of viewers that liked the show was $$40\%$$.

Alternative hypothesis: The percentage of viewers that liked the show was not $$40\%$$.

Calculations: $$40%$$ is not within the confidence interval derived above.

Reasoning: The percentage of viewers that liked the show was not $$40\%$$.

Solution

(i)

\begin{align}\mbox{Margin of error}&=\frac{1}{\sqrt{n}}\\&=\frac{1}{\sqrt{1{,}560}}\\&=0.0253…\\&\approx2.5\%\end{align}

(ii)

\begin{align}\frac{546}{1{,}560}\times100-2.5<\hat{p}<\frac{546}{1{,}560}\times100+2.5\end{align}

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

\begin{align}32.5\%<\hat{p}<37.5\%\end{align}

(iii)

Null hypothesis: The percentage of viewers that liked the show was $$40\%$$.

Alternative hypothesis: The percentage of viewers that liked the show was not $$40\%$$.

Calculations: $$40%$$ is not within the confidence interval derived above.

Reasoning: The percentage of viewers that liked the show was not $$40\%$$.

Video Walkthrough