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2020 Ordinary Level - Paper Two

Section A

Question 1

(a) A restaurant is offering a three-course meal consisting of one starter, one main course and
one dessert. There are \(4\) different starters, \(6\) different main courses and \(8\) different desserts
to choose from.

(i) How many different three-course meal combinations are available?

(ii) When Jack visits the restaurant he discovers that the restaurant still has the \(4\) starters and the \(6\) main courses available but is sold out of some of the desserts.
Jack now has \(120\) different three-course meal combinations to choose from.
How many different desserts are still available to Jack? 

Answer

(i) \(192\)

(ii) \(5\)

Solution

(i)

\begin{align}4\times6\times8=192\end{align}

(ii)

\begin{align}4\times 6\times x=120\end{align}

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

\begin{align}x&=\frac{120}{4\times6}\\&=5\end{align}

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(b) In a large population \(1\) in \(8\) of the people play tennis.

(i) Four people are chosen at random from the population.
What is the probability that the fourth person chosen is the only one to play tennis?

(ii) Three people are chosen at random from the population.
What is the probability that exactly two of them play tennis?

Answer

(i) \(\dfrac{343}{4096}\)

(ii) \(\dfrac{21}{512}\)

Solution

(i)

\begin{align}\frac{7}{8}\times\frac{7}{8}\times\frac{7}{8}\times\frac{1}{8}=\frac{343}{4096}\end{align}

(ii)

\begin{align}{3\choose2}\times\left(\frac{1}{8}\right)^2\times\frac{7}{8}=\frac{21}{512}\end{align}

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Question 2

The points \(A(4,6)\), \(B(-2,2)\) and \(C(10,0)\) are the vertices of the triangle \(ABC\) shown below.

BAC

(a) Find \(|AB|\), the length of \([AB]\). Give your answer in the form \(a\sqrt{b}\) units, where \(a,b\in\mathbb{N}\).

Answer

\(2\sqrt{13}\)

Solution

\begin{align}|AB|&=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\&=\sqrt{(4-(-2)^2+(6-2)^2}\\&=\sqrt{36+16}\\&=\sqrt{52}\\&=\sqrt{4\times13}\\&=\sqrt{4}\times\sqrt{13}\\&=2\sqrt{13}\end{align}

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(b)

(i) Find the coordinates of \(D\), the midpoint of \([AB]\).

(ii) In the triangle \(ABC\), the point \(E(7,3)\) is the midpoint of \([AC]\).
Show that \(DE\) is parallel to \(BC\).

BDE(7, 3)AC
Answer

(i) \((1,4)\)

(ii) The answer is already in the question!

Solution

(i)

\begin{align}D&=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\&=\left(\frac{4+(-2)}{2},\frac{6+2}{2}\right)\\&=(1,4)\end{align}

(ii)

\begin{align}m_{DE}&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{3-4}{7-1}\\&=-\frac{1}{6}\end{align}

\begin{align}m_{BC}&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{0-2}{10-(-2)}\\&=-\frac{2}{12}\\&=-\frac{1}{6}\end{align}

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

\begin{align}m_{DE}=m_{BC}\end{align}

as required.

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(c) Find the area of the triangle \(ABC\).

Answer

\(30\mbox{ units}^2\)

Solution

\begin{align}(4,6)&&(-2,2)&&(10,0)\end{align}

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

\begin{align}(4-10,6)&&(-2-10,2)&&(10-10,0)\end{align}

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

\begin{align}(-6,6)&&(-12,2)&&(0,0)\end{align}

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

\begin{align}A&=\frac{1}{2}|x_1y_2-x_2y_1|\\&=\frac{1}{2}|(-6)(2)-(6)(-12)|\\&=\frac{1}{2}|60|\\&=30\mbox{ units}^2\end{align}

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Question 3

(a) In a survey, \(1000\) people are selected at random and asked some questions about online shopping.

(i) Find the margin of error of the survey.
Give your answer as a percentage correct to \(1\) decimal place.

(ii) Of those asked, \(762\) said they believe it is safe to give their credit card details when shopping online. Use your answer to Part a(i) above to create a \(95\%\) confidence interval for the percentage of people who believe it is safe to give their credit card details when shopping online.

(iii) An online media company claims that 80% of people believe it is safe to give their credit card details when shopping online.
Conduct a hypothesis test, at the 5% level of significance, to test the company’s claim.
Give your conclusion in the context of the question and give a reason for your conclusion.

Answer

(i) \(3.2\%\)

(ii) \(73\%<\hat{p}<79.4\%\)

(iii)

Null hypothesis: The percentage of people who believe it is safe to give their credit card details when shopping online is \(80\%\).

Alternative hypothesis: The percentage of people who believe it is safe to give their credit card details when shopping online is not \(80\%\).

Calculations: \(80%\) is not within the confidence interval above.

Conclusion: The percentage of people who believe it is safe to give their credit card details when shopping online is not \(80\%\).

Solution

(i)

\begin{align}\mbox{Margin of error}&=\frac{1}{\sqrt{n}}\\&=\frac{1}{\sqrt{1000}}\\&=0.03162…\\&\approx3.2\%\end{align}

(ii)

\begin{align}\hat{p}-\frac{1}{\sqrt{n}}<\hat{p}<\hat{p}+\frac{1}{\sqrt{n}}\end{align}

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

\begin{align}76.2\%-3.2\%<\hat{p}<76.2\%+3.2\%\end{align}

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

\begin{align}73\%<\hat{p}<79.4\%\end{align}

(iii)

Null hypothesis: The percentage of people who believe it is safe to give their credit card details when shopping online is \(80\%\).

Alternative hypothesis: The percentage of people who believe it is safe to give their credit card details when shopping online is not \(80\%\).

Calculations: \(80%\) is not within the confidence interval above.

Conclusion: The percentage of people who believe it is safe to give their credit card details when shopping online is not \(80\%\).

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(b)

(i) An aptitude test was taken by \(6500\) candidates.
The test scores were normally distributed.
The mean score was \(480\) and the standard deviation was \(90\).
On the distribution shown below, the shaded area represents all candidates who were within one standard deviation of the mean.
Write the value of A and the value of B into the boxes below.

AB480

(ii) Use the Empirical Rule to estimate the number of candidates in the shaded region.

Answer

(i) \(A=390\) and \(B=570\)

(ii) \(4{,}420\)

Solution

(i)

\begin{align}A&=480-90\\&=390\end{align}

and

\begin{align}B&=480+90\\&=570\end{align}

(ii)

\begin{align}0.68\times6{,}500=4{,}420\end{align}

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Question 4

(a)

(i) The circle \(c\) has equation \((x-1)^2+(y+4)^2=25\).
Find the centre and radius of \(c\).

(ii) The point \((1,k)\) is on \(c\). Find the two possible values of \(k\).

Answer

(i) Centre: \((1,04)\). Radius: \(5\).

(ii) \(k=-9\) or \(k=1\)

Solution

(i) Centre: \((1,04)\) and radius: \(\sqrt{25}=5\).

(ii)

\begin{align}(1-1)^2+(k+4)^2=25\end{align}

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

\begin{align}k^2+8k+16=25\end{align}

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

\begin{align}k^2+8k-9=0\end{align}

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

\begin{align}(k+9)(k-1)=0\end{align}

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

\(k=-9\) or \(k=1\)

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(b) The circle \(s\) has equation \(x^2+y^2=13\).
The point \(A(3,-2)\) is on \(s\).
Find the equation of \(t\), the tangent to the circle at the point \(A\).
Give your answer in the form \(ax+by+c=0\), where \(a,b,c\in\mathbb{Z}\).

yxstA(0, 0)
Answer

\(3x-2y-13=0\)

Solution

\begin{align}m_r&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{-2-0}{3-0}\\&=-\frac{2}{3}\end{align}

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

\begin{align}m_t=\frac{3}{2}\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-(-2)=\frac{3}{2}(x-3)\end{align}

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

\begin{align}2y+4=3x-9\end{align}

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

\begin{align}3x-2y-13=0\end{align}

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Question 5

(a) Two swimmers \(A\) and \(B\) stand at the same point \(X\), on one shore of a long, still rectangular shaped lake that is \(100\mbox{ m}\) wide, as shown below. (Diagram not to scale.)
Both swim to the opposite side of the lake.

θX55o100 m200 mSwimmer BSwimmer A
Answer

\(122\mbox{ m}\)

Solution

\begin{align}\sin55^{\circ}=\frac{100}{d}\end{align}

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

\begin{align}d&=\frac{100}{\sin55^{\circ}}\\&\approx122\mbox{ m}\end{align}

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(b) Swimmer \(B\) swims to the right and travels a distance of \(200\mbox{ m}\) to reach the other side, making an angle of \(\theta\) degrees with the bank on the other side of the lake, as shown.
Find the value of \(\theta\).

Answer

\(30^{\circ}\)

Solution

\begin{align}\sin \theta &=\frac{100}{200}\\&=\frac{1}{2}\end{align}

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

\begin{align}\theta &=\sin^{-1}\left(\frac{1}{2}\right)\\&=30^{\circ}\end{align}

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(c) The next day the swimmers again swim to the opposite side of the lake but in slightly different directions.
Swimmer \(A\) swims to the left, making an angle of \(45^{\circ}\) with the side of the lake and travels \(141.4\) metres as shown.
Swimmer \(B\) swims to the right, making an angle of \(40^{\circ}\) with the side of the lake and travels \(155.6\) metres as shown.
Find \(d\), the distance both swimmers are apart when they reach the opposite side of the lake.
Give you answer correct to the nearest metre.

Xd45o40o100 m155.6 m141.4 mSwimmer BSwimmer A
Answer

\(219\mbox{ m}\)

Solution

\begin{align}d&=\sqrt{141.4^2+155.6^2-2(141.4)(155.6)\cos95^{\circ}}\\&\approx219\mbox{ m}\end{align}

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Question 6

In the diagram below: \(|\angle CAB|=80^{\circ}\) and \(|\angle DCE|=60^{\circ}\).
\(|\angle ABC|=(x+y)^{\circ}\) and \(|\angle BCA|=(3x+y)^{\circ}\), where \(x,y\in\mathbb{N}\).

ABCDE(3x + y)o60o(x + y)o80o

(a) Find the value of \(x\) and the value of \(y\).

Answer

\(x=10^{\circ}\) and \(y=30^{\circ}\)

Solution

\begin{align}3x+y=60^{\circ}\end{align}

and

\begin{align}(x+y)+(3x+y)+80^{\circ}=180^{\circ}\end{align}

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

\begin{align}4x+2y=100^{\circ}\end{align}

We therefore have the following two equations with two unknowns:

\begin{align}3x+y=60^{\circ}\end{align}

\begin{align}4x+2y=100^{\circ}\end{align}

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

\begin{align}6x+2y=120^{\circ}\end{align}

\begin{align}4x+2y=100^{\circ}\end{align}

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

\begin{align}2x=20^{\circ}\end{align}

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

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

and

\begin{align}y&=60^{\circ}-3x\\&=60^{\circ}-3(10^{\circ})\\&=30^{\circ}\end{align}

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(b) In the diagram below, \(DE\) is parallel to \(FG\). \(|DH|=5\mbox{ cm}\). \(|HE|=12\mbox{ cm}\). \(|HG|=30\mbox{ cm}\).
The distance from \(D\) to \(F\) is \(x\mbox{ cm}\). Find the value of \(x\).

FDHE30 cmx cm12 cm5 cmG
Answer

\(7.5\mbox{ cm}\)

Solution

\begin{align}\frac{x}{5}=\frac{30-12}{12}\end{align}

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

\begin{align}x&=5\times\frac{18}{12}\\&=7.5\mbox{ cm}\end{align}

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Section B

Question 7

A vertical mobile phone mast, \([DC]\), of height \(h\mbox{ m}\), is secured with two cables: \([AC]\) of length \(x\mbox{ m}\),
and \([BC]\) of length \(y\mbox{ m}\), as shown in the diagram.
The angle of elevation to the top of the mast from \(A\) is \(30^{\circ}\) and from \(B\) is \(45^{\circ}\).

yxh30o45oCADB

(a)

(i) Explain why \(|\angle BCA|\) is \(105^{\circ}\).

(ii) The horizontal distance from \(A\) to \(B\) is \(100\mbox{ m}\).
Use the triangle \(ABC\) to find the length of \(y\).
Give your answer correct to one decimal place.

(iii) Using your answer to Part (a)(ii) or otherwise, find the value of \(h\) and the value of \(x\).
Give your answers correct to \(1\) decimal place. 

Answer

(i) The answer is already in the question!

(ii) \(y=51.8\mbox{ m}\)

(iii) \(h=36.6\mbox{ m}\) and \(x=73.2\mbox{ m}\)

Solution

(i)

\begin{align}|\angle BCA|&=|\angle ACD|+|\angle DCB|\\&=60^{\circ}+45^{\circ}\\&=10^{\circ}\end{align}

as required.

(ii)

\begin{align}\frac{y}{\sin30^{\circ}}=\frac{100}{\sin105^{\circ}}\end{align}

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

\begin{align}y&=\frac{100\sin30^{\circ}}{\sin105^{\circ}}\\&\approx51.8\mbox{ m}\end{align}

(iii)

\begin{align}\sin45^{\circ}&=\frac{h}{51.8}\end{align}

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

\begin{align}h&=51.8\sin45^{\circ}\\&\approx36.6\mbox{ m}\end{align}

and

\begin{align}\sin30^{\circ}=\frac{36.6}{x}\end{align}

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

\begin{align}x&=\frac{36.6}{\sin30^{\circ}}\\&\approx73.2\mbox{ m}\end{align}

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(b) The two cables to secure the mast costs €\(25\) per metre. The mast itself costs €\(580\) per metre. VAT at \(23\%\) is then added in each case.
Calculate the total cost of the cables and mast after VAT is included.

Answer

\(29{,}954.19\mbox{ euro}\)

Solution

\begin{align}[25(73.2+51.8)+580(36.6)]\times1.23=29{,}954.19\mbox{ euro}\end{align}

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(c)

(i) The mast can provide a strong signal for an area in the shape of a regular hexagon of side \(8\mbox{ km}\), as shown in the diagram.
Find the area of the hexagon.
Give your answer in \(\mbox{km}^2\), correct to \(2\) decimal places.

8 km8 km

(ii) A circle which touches all vertices of the hexagon can show areas where the signal is weak. One of these areas is shaded in the diagram.
Find this shaded area.
Give your answer in \(\mbox{km}^2\), correct to \(1\) decimal place.

Answer

(i) \(166.28\mbox{ km}^2\)

(ii) \(5.8\mbox{ km}^2\)

Solution

(i)

\begin{align}A&=6\times\left[\frac{1}{2}(8)(8)(\sin60^{\circ})\right]\\&\approx166.28\mbox{ km}^2\end{align}

(ii)

\begin{align}A&=\frac{\pi(8^2)-166.28}{6}\\&\approx5.8\mbox{ km}^2\end{align}

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Question 8

(a) The table below shows data (measured in kilotonnes) on how municipal waste was dealt with in Ireland from \(2005\) to \(2011\).

Year Total Waste Collected (Kilotonnes) Waste Sent to Landfill (Kilotonnes)

\(2005\)

\(2779\)

\(1833\)

\(2006\)

\(3100\)

\(1981\)

\(2007\)

\(3175\)

\(2015\)

\(2008\)

\(3104\)

\(1939\)

\(2009\)

\(2825\)

\(1724\)

\(2010\)

\(2580\)

\(1496\)

\(2011\)

\(2547\)

\(1344\)

Source: Environmental Protection Agency

(i) Find the difference between the percentage of the waste collected which was sent to landfill in \(2005\) and the percentage of waste sent to landfill in 2011.
Give your answer correct to \(1\) decimal place.

(ii) Find the mean amount of waste collected each year, from \(2005\) to \(2011\).
Give your answer correct to \(1\) decimal place.

(iii) The mean amount of waste sent to landfill in the \(4\) years from \(2009\) to \(2012\) was \(1398\) kilotonnes. Find the amount of waste sent to landfill in \(2012\). 

Answer

(i) \(13.2\%\)

(ii) \(2872.9\mbox{ kt}\)

(iii) \(1028\mbox{ kt}\)

Solution

(i)

\begin{align}\left(\frac{1833}{2779}-\frac{1344}{2547}\right)\times100\approx13.2\%\end{align}

(ii)

\begin{align}\frac{2779+3100+3175+3104+2825+2580+2547}{7}\approx2872.9\mbox{ kt}\end{align}

(iii)

\begin{align}\frac{1724+1496+1344+x}{4}=1398\end{align}

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

\begin{align}\frac{4564+x}{4}=1398\end{align}

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

\begin{align}x&=(4)(1398)-4564\\&=1028\mbox{ kt}\end{align}

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(b) The table below shows the percentages of all the different types of energy sources used in each of \(13\) European countries during \(2014\).

Country Renewable Oil Solid Fuel Gas Electricity Other

Austria

\(25\)

\(31\)

\(8\)

\(17\)

\(13\)

\(6\)

Bulgaria

\(8\)

\(19\)

\(28\)

\(11\)

\(11\)

\(23\)

Cyprus

\(5\)

\(82\)

\(0\)

\(0\)

\(13\)

\(0\)

Denmark

\(31\)

\(11\)

\(13\)

\(12\)

\(13\)

France

\(7\)

\(26\)

\(3\)

\(11\)

\(12\)

\(41\)

Germany

\(10\)

\(29\)

\(22\)

\(17\)

\(12\)

\(10\)

Greece

\(9\)

\(43\)

\(24\)

\(9\)

\(15\)

\(0\)

Ireland

\(6\)

\(42\)

\(13\)

\(24\)

\(15\)

\(0\)

Latvia

\(31\)

\(26\)

\(1\)

\(20\)

\(11\)

\(11\)

Luxembourg

\(4\)

\(61\)

\(1\)

\(19\)

\(12\)

\(3\)

Malta

\(2\)

\(82\)

\(0\)

\(0\)

\(16\)

\(0\)

Netherlands

\(4\)

\(38\)

\(10\)

\(33\)

\(10\)

\(5\)

Portugal

\(21\)

\(38\)

\(10\)

\(13\)

\(15\)

\(3\)

(i) What percentage of the energy used in Denmark in \(2014\) was renewable?

(ii) Name the four countries that used a lower percentage of renewable energy than Ireland.

(iii) Name a country which did not use gas as a fuel source.

(iv) The table below shows the percentage of the different types of energy that were used in Ireland during \(2014\).
It also shows one of the angles in a pie chart to represent this data.
The angle for renewable energy (\(6\%\)) is \(22^{\circ}\).
Complete the table to show the angle for each energy type.
Give each angle correct to the nearest degree. 

Type Percentage Angle

Renewable

\(6\)

\(22^{\circ}\)

Oil

\(42\)

Solid Fuel

\(13\)

Gas

\(24\)

Electricity

\(15\)

(v) Complete the pie chart below to show the energy types used in Ireland in \(2014\).
Label each section clearly. 

22oRenewable
Answer

(i) \(20\%\)

(ii) Cyprus, Luxembourg, Malta and Netherlands.

(iii) Cyprus or Malta

(iv)

Type Percentage Angle

Renewable

\(6\)

\(22^{\circ}\)

Oil

\(42\)

\(151^{\circ}\)

Solid Fuel

\(13\)

\(47^{\circ}\)

Gas

\(24\)

\(86^{\circ}\)

Electricity

\(15\)

\(54^{\circ}\)

(v)

22o151o47o86o54oRenewableOilSolid FuelGasElectricity
Solution

(i)

\begin{align}100-31-11-13-12-13=20\%\end{align}

(ii) Cyprus, Luxembourg, Malta and Netherlands.

(iii) Cyprus or Malta

(iv)

Type Percentage Angle

Renewable

\(6\)

\(22^{\circ}\)

Oil

\(42\)

\(151^{\circ}\)

Solid Fuel

\(13\)

\(47^{\circ}\)

Gas

\(24\)

\(86^{\circ}\)

Electricity

\(15\)

\(54^{\circ}\)

(v)

22o151o47o86o54oRenewableOilSolid FuelGasElectricity
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Question 9

A rectangular sheet of aluminium is used to make a cylindrical can of radius \(r\mbox{ cm}\) and height
\(10\mbox{ cm}\), as shown below. The aluminium does not overlap in the finished can.

16 cm10 cmr cmy cm

(a)

(i) Show that \(r\), the radius of the cylinder, is \(3\mbox{ cm}\).

(ii) Find the distance \(y\). Give your answer correct to the nearest centimetre.

(iii) Find the area, in \(\mbox{cm}^2\), of the waste aluminium after the top, bottom and side of the cylindrical can have been removed from the rectangular sheet.
Give your answer correct to the nearest \(\mbox{cm}^2\).

Answer

(i) The answer is already in the question!

(ii) \(19\mbox{ cm}\)

(iii) \(57\mbox{ cm}^2\)

Solution

(i)

\begin{align}r&=\frac{16-10}{2}\\&=3\mbox{ cm}\end{align}

as required.

(ii)

\begin{align}y&=2\pi r\\&=2\pi(3)\\&\approx19\mbox{ cm}\end{align}

(iii)

\begin{align}A&=(16)(19)-(10)(19)-2\pi(3^2)\\&\approx57\mbox{ cm}^2\end{align}

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(b)

(i) Find the volume of a spherical ice cube of radius \(1.5\mbox{ cm}\).
Give your answer in terms of \(\pi\).

h cm

(ii) Three of the spherical ice cubes of radius \(1.5\mbox {cm}\) are added to a cylinder of internal radius \(3.5zmbox{ cm}\) which is partially filled with water.
All of the ice cubes are completely submerged in the water and the water does not overflow.
Find the rise, \(h\mbox{ cm}\), in the water level. Give your answer correct to \(1\) decimal place.

Answer

(i) \(4.5\pi\mbox{ cm}^3\)

(ii) \(1.1\mbox{ cm}\)

Solution

(i)

\begin{align}V&=\frac{4}{3}\pi r^3\\&=\frac{4}{3}\pi(1.5^3)\\&=4.5\pi\mbox{ cm}^3\end{align}

(ii)

\begin{align}3\times 4.5\pi=\pi(3.5^2)h\end{align}

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

\begin{align}h&=\frac{3\times4.5}{3.5^2}\\&\approx1.1\mbox{ cm}\end{align}

Video Walkthrough
Our Video Walkthroughs are in the final stages of editing and will go live during Summer 2023.
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!
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