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2017 Ordinary Level - Paper One
Section A
Question 1
(a) A new machine is bought for €\(30000\). Its value depreciates by \(15\%\) each year for five years.
Find the value of the machine at the end of the five years.
\(13{,}311.16\mbox{ euro}\)
\begin{align}30{,}000\times0.85^5=13{,}311.16\mbox{ euro}\end{align}
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(b) A sum of money was invested for two years at \(3\%\) compound interest per year.
At the end of the two years it amounted to €\(30000\). Find the sum invested.
\(28{,}277.38\mbox{ euro}\)
\begin{align}P\times1.03^2=30{,}000\end{align}
\begin{align}\downarrow\end{align}
\begin{align}P&=\frac{30{,}000}{1.03^2}\\&=28{,}277.38\mbox{ euro}\end{align}
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(c) A company invested €\(25000\) for three years at a fixed rate of compound interest.
At the end of the three years it amounted to €\(26530.20\). Find the rate of interest.
\(2\%\)
\begin{align}25{,}000(1+i)^3=26{,}530.20\end{align}
\begin{align}\downarrow\end{align}
\begin{align}1+i&=\sqrt[3]{\frac{26{,}530.20}{25{,}000}}\\&=1.02\end{align}
\begin{align}\downarrow\end{align}
\begin{align}i&=0.02\\&=2\%\end{align}
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Question 2
(a) The complex number \(z_1=a+bi\), where \(i^2=-1\), is shown on the Argand Diagram below.
(i) Write down the value of \(a\) and the value of \(b\).
(ii) \(z_2=-1+2i\). Plot \(z_2\) on the Argand Diagram.
(iii) \(z_3=\dfrac{z_1}{z_2}\). Write \(z_3\) in the form \(x+yi\), where \(x,y\in\mathbb{R}\).
(i) \(a=3\) and \(b=-1\)
(ii)
(iii) \(-1-i\)
(i) \(a=3\) and \(b=-1\)
(ii)
(iii)
\begin{align}z_3&=\frac{z_1}{z_2}\\&=\frac{3-i}{-1+2i}\\&=\frac{3-i}{-1+2i}\times\frac{-1-2i}{-1-2i}\\&=\frac{-3-6i+i-2}{1+2i-2i+4}\\&=\frac{-5-5i}{5}\\&=-1-i\end{align}
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(b) Solve for \(z\)
\begin{align}2z-6(4-6i)=(-1+i)(4-2i)\end{align}
\(z=11-15i\)
\begin{align}2z-6(4-6i)=(-1+i)(4-2i)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2z-24+36i=-4+2i+4i+2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2z&=-4+2i+4i+2+24-36i\\&=22-30i\end{align}
\begin{align}\downarrow\end{align}
\begin{align}z=11-15i\end{align}
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Question 3
(a) Find the two values of \(x\) for which \(3x^2-6x-8=0\).
Give each answer correct to \(1\) decimal place.
\(x=-0.9\) or \(x=2.9\)
\begin{align}a=3&&b=-6&&c=-8\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\&=\frac{-(-6)\pm\sqrt{(-6)^2-4(3)(-8)}}{2(3)}\\&=\frac{6\pm\sqrt{132}}{6}\end{align}
\begin{align}\downarrow\end{align}
\(x\approx-0.9\) or \(x\approx2.9\)
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(b) Find the co-ordinates of the minimum point of the function \(f(x)=3x^2-6x-8\), where \(x\in\mathbb{R}\).
\((1,-11)\)
\begin{align}f'(x)=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}6x-6=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x=1\end{align}
\begin{align}\downarrow\end{align}
\begin{align}f(1)&=3(1^2)-6(1)-8\\&=-11\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(1,-11)\end{align}
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Question 4
(a) Solve for \(x\):
\begin{align}11x-5(2x-1)=3(6-x)+3\end{align}
\(x=4\)
\begin{align}11x-5(2x-1)=3(6-x)+3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}11x-10x+5=18-3x+3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x+5=21-3x\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3x+x=21-5\end{align}
\begin{align}\downarrow\end{align}
\begin{align}4x=16\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x&=\frac{16}{4}\\&=4\end{align}
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(b) Solve the simultaneous equations:
\begin{align}y+5&=2x\\x^2+y^2&=25\end{align}
\(x=0\) and \(y=-5\) or \(x=4\) and \(y=3\)
\begin{align}y+5&=2x\\x^2+y^2&=25\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y=2x-5\end{align}
\begin{align}x^2+y^2=25\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x^2+(2x-5)^2=25\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x^2+4x^2-20x+25=25\end{align}
\begin{align}\downarrow\end{align}
\begin{align}5x^2-20x-0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}5x(x-4)=0\end{align}
\begin{align}\downarrow\end{align}
\(x=0\) or \(x=4\)
\begin{align}\downarrow\end{align}
\begin{align}y&=2x-5\\&=2(0)-5\\&=-5\end{align}
or
\begin{align}y&=2(4)-5\\&=2(4)-5\\&=3\end{align}
\[\,\]
Solution 1
\(x=0\) and \(y=-5\)
\[\,\]
Solution 2
\(x=4\) and \(y=3\)
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Question 5
A field is divided into eight sections as shown below. The width of each section is \(3\) metres.
The height, in metres, of each section is given in the diagram.
Use the Trapezoidal rule to estimate the area of the field.
\(145.8\mbox{ m}^2\)
\begin{align}A&=\frac{3}{2}[4+3+2(5.8+7+6.5+6+4.8+6+6.5)]\\&=145.8\mbox{ m}^2\end{align}
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(b) The area of the same field was re-estimated by applying the Trapezoidal rule again.
This time, a different section width (\(4\mbox{ m}\)) and a different set of section heights were used, as
shown below. The area was found to be \(145.6\mbox{ m}^2\).
Use this information to find the value of the height marked \(x\) on the diagram.
\(x=4.9\mbox{ m}\)
\begin{align}\frac{4}{2}[4+8+2(6.4+6.9+6+x+6.2)]=145.6\end{align}
\begin{align}\downarrow\end{align}
\begin{align}4+8+2(6.4+6.9+6+x+6.2)=72.8\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2(6.4+6.9+6+x+6.2)=60.8\end{align}
\begin{align}\downarrow\end{align}
\begin{align}12.8+13.8+12+2x+12.4=60.8\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2x=9.8\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x=4.9\mbox{ m}\end{align}
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Question 6
(a) A salesman earns a basic salary of €\(150\) per week. In addition, he gets commission of \(20\%\) on sales up to the value of €\(1000\) in the week and \(30\%\) commission on any sales above this.
Find his total income for a week when his total sales amount to €\(3000\).
\(950\mbox{ euro}\)
\begin{align}150+0.2\times1{,}000+0.3\times(3{,}000-1{,}000)=950\mbox{ euro}\end{align}
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(b) On a different week his total income is €\(1160\). Find his total sales for this week.
\(3{,}700\mbox{ euro}\)
\begin{align}\frac{1{,}160-150-0.2\times1{,}000}{0.3}+1{,}000=3{,}700\mbox{ euro}\end{align}
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Section B
Question 7
The first three patterns in a sequence of patterns of tiles are shown in the diagram below.
(a) Draw the next pattern of tiles.
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(b) Based on the patterns shown, complete the table below.
| Pattern Number | Number of Tiles |
|---|---|
|
\(1\) |
\(5\) |
|
\(2\) |
|
|
\(3\) |
|
|
\(4\) |
|
|
\(5\) |
|
| Pattern Number | Number of Tiles |
|---|---|
|
\(1\) |
\(5\) |
|
\(2\) |
\(8\) |
|
\(3\) |
\(11\) |
|
\(4\) |
\(14\) |
|
\(5\) |
\(17\) |
| Pattern Number | Number of Tiles |
|---|---|
|
\(1\) |
\(5\) |
|
\(2\) |
\(8\) |
|
\(3\) |
\(11\) |
|
\(4\) |
\(14\) |
|
\(5\) |
\(17\) |
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(c)
(i) Assuming the pattern continues, the number of tiles in the \(n\)th pattern of the sequence is given by the formula \(T_n=pn+q\), where \(p\) and \(q\in\mathbb{N}\).
Find the value of \(p\) and the value of \(q\).
(ii) How many tiles are in the \(20\)th pattern?
(iii) Find which pattern has exactly \(290\) tiles.
(i) \(p=3\) and \(q=2\)
(ii) \(62\)
(iii) \(96\mbox{th}\)
(i)
\begin{align}a=5&&d=3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}T_n&=a+(n-1)d\\&=5+(n-1)3\\&=5+3n-3\\&=3n+2\end{align}
\begin{align}\downarrow\end{align}
\(p=3\) and \(q=2\)
(ii)
\begin{align}T_{20}&=3(20)+2\\&=62\end{align}
(iii)
\begin{align}T_n=290\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3n+2=290\end{align}
\begin{align}\downarrow\end{align}
\begin{align}n&=\frac{290-2}{3}\\&=96\end{align}
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(d)
(i) Show that \(S_n=\dfrac{3n^2+7n}{2}\) is a formula for the total number of tiles needed to build the first \(n\) patterns.
(ii) Find the total number of tiles needed to build the first \(30\) patterns.
(i) The answer is already in the question!
(ii) \(1{,}455\)
(i)
\begin{align}a=5&&d=3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}S_n&=\frac{n}{2}[2a+(n-1)d]\\&=\frac{n}{2}[2(5)+(n-1)3]\\&=\frac{n}{2}(10+3n-3)\\&=\frac{n}{2}(3n+7)\\&=\frac{3n^2+7n}{2}\end{align}
as required.
(ii)
\begin{align}S_{30}&=\frac{3(30^2)+7(30)}{2}\\&=1{,}455\end{align}
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Question 8
A company makes and sells fibre optic cable. It can sell, at most, \(200\) kilometres of cable in a week. For a certain range of its production the company has found that profit can be modelled using the function:
\(P(x)=275x-x^2-2000\), where \(x\leq 200\)
In the function, \(x\) is the number of kilometres of fibre optic cable sold and \(P(x)\) is the profit in euro.
(a) Use the profit function, \(P(x)\), to find how much money the company loses if it does not sell any cable.
\(2{,}000\mbox{ euro}\)
\begin{align}P(0)&=275(0)-0^2-2{,}000\\&=-2{,}000\end{align}
i.e. the company will lose \(2{,}000\mbox{ euro}\).
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(b) In a particular week the company made a profit of €\(8350\).
Find the number of kilometres of cable it sold that week.
\(45\mbox{ km}\)
\begin{align}P(x)=8{,}350\end{align}
\begin{align}\downarrow\end{align}
\begin{align}275-x^2-2{,}000=8{,}350\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x^2-275x+10{,}350=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(x-45)(x-230)=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x=45\mbox{ km}\end{align}
(as \(x\leq200\))
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(c)
(i) The table below shows some of the data representing the profit on sales.
Use the profit function, \(P(x)\), to complete the table.
| Number of km of cable sold (\(x\)) | \(50\) | \(60\) | \(70\) | \(80\) | \(90\) | \(100\) |
|---|---|---|---|---|---|---|
|
Profit (€) |
|
|
\(12{,}350\) |
|
|
|
(ii) Use the data in the table to draw the graph of the profit function on the axes below for \(50\leq x\leq 100\), \(x\in\mathbb{R}\).
(iii) Use your graph to estimate the lower and upper range of sales (in \(\mbox{km}\) of cable) in order to make a profit of between €\(10{,}000\) and €\(14{,}000\) in a particular week.
Show your work on the graph above.
(i)
| Number of km of cable sold (\(x\)) | \(50\) | \(60\) | \(70\) | \(80\) | \(90\) | \(100\) |
|---|---|---|---|---|---|---|
|
Profit (€) |
\(9{,}250\) |
\(10{,}900\) |
\(12{,}350\) |
\(13{,}600\) |
\(14{,}650\) |
\(15{,}500\) |
(ii)
(iii) Lower: \(55\mbox{ km}\). Upper: \(83\mbox{ km}\).
(i)
| Number of km of cable sold (\(x\)) | \(50\) | \(60\) | \(70\) | \(80\) | \(90\) | \(100\) |
|---|---|---|---|---|---|---|
|
Profit (€) |
\(9{,}250\) |
\(10{,}900\) |
\(12{,}350\) |
\(13{,}600\) |
\(14{,}650\) |
\(15{,}500\) |
(ii)
(iii) Lower: \(55\mbox{ km}\). Upper: \(83\mbox{ km}\).
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(d) Use calculus to find the number of kilometres of cable sold when the profit is increasing at a rate of €\(105\) per \(\mbox{km}\).
\(85\mbox{ km}\)
\begin{align}P'(x)=105\end{align}
\begin{align}\downarrow\end{align}
\begin{align}275-2x=105\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x&=\frac{275-105}{2}\\&=85\mbox{ km}\end{align}
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Question 9
Forensic scientists can estimate the height of a person from the lengths of their bones.
One method uses a function which relates the length of the femur bone, \(x\) ,to the height of the person. Using this method the heights of males and females are estimated using the following functions:
Male: \(m(x)=2.3x+65.53\),
where \(m(x)\) is the height and \(x\) is the length of the femur, in \(\mbox{cm}\).
Female: \(f(x)=2.5x+54.13\),
where \(f(x)\) is the height and \(x\) is the length of the femur, in \(\mbox{cm}\).
(a) Use the functions above to estimate the height of a male and the height of a female each of whose femur is \(47.54\mbox{ cm}\) in length. Give both answers correct to \(2\) decimal places.
\(174.87\mbox{ cm}\) and \(172.98\mbox{ cm}\) respectively
\begin{align}m(47.54)&=2.3(47.54)+65.53\\&=174.87\mbox{ cm}\end{align}
and
\begin{align}f(47.54)&=2.5(47.54)+54.13\\&=172.98\mbox{ cm}\end{align}
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(b) Use \(m(x)\) to estimate the femur length of a male whose height is \(184\mbox{ cm}\).
Give your answer correct to \(2\) decimal places.
\(51.51\mbox{ cm}\)
\begin{align}m(x)=184\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2.3x+65.53=184\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x&=\frac{184-65.53}{2.3}\\&\approx51.51\mbox{ cm}\end{align}
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(c) Conor’s femur length is \(44.2\mbox{ cm}\). His height is \(171\mbox{P cm}\).
Find the percentage error in using \(m(x)\) to estimate his height.
Give your answer correct to \(2\) decimal places.
\(2.23\%\)
\begin{align}m(44.2)&=2.3(44.2)+65.53\\&=167.19\mbox{ cm}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\%\mbox{error}&=\frac{171-167.19}{171}\times100\\&\approx2.23\%\end{align}
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(d) Find the length of a femur for which the estimated height of a male and the estimated height of a female are the same and find this estimated height.
\(\mbox{Length of femur}=57\mbox{ cm}\) and \(\mbox{ height}=196.63\mbox{ cm}\).
\begin{align}2.3x+65.53=2.5x+54.13\end{align}
\begin{align}\downarrow\end{align}
\begin{align}0.2x=11.4\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x&=\frac{11.4}{0.2}\\&=57\mbox{ cm}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}m(57)&=2.3(57)+65.53\\&=196.63\mbox{ cm}\end{align}
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(e) The Ponderal Index is a number which relates a person’s height to their weight.
The formula for the Ponderal Index, \(P\), is
\begin{align}P=\frac{M}{h^3}\end{align}
where \(M\) is weight in \(\mbox{kg}\) and \(h\) is height in metres.
(i) Find the Ponderal Index for a person who is \(1.60\mbox{ m}\) tall and weighs \(72.5\mbox {kg}\).
Give your answer correct to \(1\) decimal place.
(ii) Rearrange the formula \(P=\dfrac{M}{h^3}\) to give a formula which will give the height of a person in terms of their weight and Ponderal index.
(iii) Mary has a Ponderal Index of \(13\) and a weight of \(67.5\mbox{ kg}\). Find her height.
Give your answer in metres, correct to \(2\) decimal places.
(i) \(17.7\)
(ii) \(h=\sqrt[3]{\dfrac{M}{P}}\)
(iii) \(1.73\mbox{ m}\)
(i)
\begin{align}P&=\frac{72.5}{1.6^3}\\&\approx17.7\end{align}
(ii)
\begin{align}P=\frac{M}{h^3}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}h^3=\frac{M}{P}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}h=\sqrt[3]{\frac{M}{P}}\end{align}
(iii)
\begin{align}h&=\sqrt[3]{\frac{67.5}{13}}\\&\approx1.73\mbox{ m}\end{align}
