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2022 Ordinary Level - Paper One
Section A
Question 1
The complex number \(z_1\) is shown on the Argand diagram below.
(a) Using the Argand diagram:
(i) write down the values of \(z_1\) and \(\bar{z_1}\), where \(\bar{z_1}\) is the complex conjugate of \(z_1\).
(ii) plot and label \(\bar{z_1}\) on the Argand diagram above.
(i) \(z_1=-2-3i\) and \(\bar{z_1}=-2+3i\)
(ii)
(i)
\begin{align}z_1=-2-3i && \bar{z_1}=-2+3i\end{align}
(ii)
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\(z_2\) and \(z_3\) are two other complex numbers.
\(z_2=-5+3i\) and \(z_3=4-2i\), where \(i^2=-1\).
(b) Plot and label \(z_2\) and \(z_3\) on the Argand diagram from part (a).
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(c) Write \(z_2-z_3\) in the form \(a+bi\), where \(a,b,\in\mathbb{R}\), \(i^2=-1\), and hence find \(|z_2-z_3|\).
\(z_2-z_3=-9+5i\) and \(|z_2-z_3|=\sqrt{106}\)
\begin{align}z_2-z_3&=(-5+3i)-(4-2i)\\&=-5+3i-4+2i\\&=-9+5i\end{align}
\begin{align}\downarrow\end{align}
\begin{align}|z_2-z_3|&=|-9+5i|\\&=\sqrt{(-9)^2+5^2}\\&=\sqrt{106}\end{align}
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(d) Investigate if \(z_3=4-2i\) is a solution of the equation \(z^2+2iz-7i=0\).
\(z_3\) is not a solution.
\begin{align}z^2+2iz-7i&=(4-2i)^2+2i(4-2i)-7i\\&=16-16i-4+8i+4-7i\\&=16-15i\\&\neq0\end{align}
Therefore, \(z_3\) is not a solution.
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Question 2
(a) Solve the following equation in \(x\):
\begin{align}2(3x-5)+8=4x-5\end{align}
\(x=-\dfrac{3}{2}\)
\begin{align}2(3x-5)+8=4x-5\end{align}
\begin{align}\downarrow\end{align}
\begin{align}6x-10+8=4x-5\end{align}
\begin{align}\downarrow\end{align}
\begin{align}6x-4x=-5+10-8\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2x=-3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x=-\frac{3}{2}\end{align}
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(b) Write \(\dfrac{(3^4)^5}{3^6}\) in the form \(3^k\), whee \(k\in\mathbb{R}\).
\(3^{14}\)
\begin{align}\frac{(3^4)^5}{3^6}&=\frac{3^{20}}{6}\\&=3^{14}\end{align}
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(c) Solve the simultaneous equations below to find the value of \(x\) and the value of \(y\).
\begin{align}3x+2y&=1\\7x+5y&=-2\end{align}
\(x=9\) and \(y=-13\)
\begin{align}3x+2y&=1\\7x+5y&=-2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}21x+14y&=7\\21x+15y&=-6\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-y=7-(-6)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-y=13\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y=-13\end{align}
and
\begin{align}3x+2y&=1\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x&=\frac{1-2y}{3}\\&=\frac{1-2(-13)}{3}\\&=9\end{align}
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Question 3
Joe, Émile, and Wei are all PAYE workers.
Each of them has an annual tax credit of €3300.
Their tax rates and bands are shown in the table below.
Assume that no other deductions are made from their income.
| Annual Income | Tax Rate |
|---|---|
|
First €\(35{,}300\) |
\(20\%\) |
|
Balance |
\(40\%\) |
(a) Joe’s gross annual income is €\(27{,}500\). Joe only pays tax at the lower rate.
Work out Joe’s net annual income.
\(25{,}300\mbox{ euro}\)
Income Tax
\begin{align}27{,}500\times0.2=5{,}500\mbox{ euro}\end{align}
\[\,\]
Tax Due
\begin{align}5{,}500-3{,}300=2{,}200\mbox{ euro}\end{align}
\[\,\]
Net Income
\begin{align}27{,}500-2{,}200=25{,}300\mbox{ euro}\end{align}
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(b) Émile’s gross annual income is €\(43{,}450\).
Work out Émile’s net annual income.
\(36{,}430\mbox{ euro}\)
Income Tax
\begin{align}35{,}300\times0.2+(43{,}450-35{,}300)\times0.4=10{,}320\mbox{ euro}\end{align}
\[\,\]
Tax Due
\begin{align}10{,}320-3{,}300=7{,}020\mbox{ euro}\end{align}
\[\,\]
Net Income
\begin{align}43{,}450-7{,}020=36{,}430\mbox{ euro}\end{align}
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(c) Wei’s gross annual income is over €\(35{,}300\), so she pays tax at both rates.
Wei is looking for a pay rise.
She wants her net income to increase by €\(80\) each month.
Work out how much her gross annual income will need to increase by, in order for this to happen.
\(1{,}600\mbox{ euro}\)
Net Annual Increase
\begin{align}80\times12=960\mbox{ euro}\end{align}
\[\,\]
Gross Annual Increase
\begin{align}\frac{960}{60}\times100=1{,}600\mbox{ euro}\end{align}
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Question 4
(a) \(g(x)=x^3-7x^2+x-12\), where \(x\in\mathbb{R}\).
(i) Work out the value of \(g(5)\).
(ii) Find \(g'(x)\), the derivative of \(g(x)\).
(iii) \(g'(5)=6\).
Use this to find the equation of the tangent to the curve \(y=g(x)\) when \(x=5\).
Give your answer in the form \(ax+by+c=0\), where \(a,b,c\in\mathbb{R}\).
(i) \(g(5)=-57\)
(ii) \(g'(x)=3x^2-14x+1\)
(iii) \(6x-y-87=0\)
(i)
\begin{align}g(5)&=5^3-7(5^2)+5-12\\&=125-175+5-12\\&=-57\end{align}
(ii)
\begin{align}g'(x)=3x^2-14x+1\end{align}
(iii)
\begin{align}(x_1,y_1)=(5,-57) && \mbox{ and } && m=6\end{align}
\begin{align}y-y_1=m(x-x_1)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y-(-57)=6(x-5)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y+57=6x-30\end{align}
\begin{align}\downarrow\end{align}
\begin{align}6x-y-87=0\end{align}
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(b) The graph of the function \(y=u(x)\) is shown below, for \(0\leq x\leq10\), \(x\in\mathbb{R}\).
\(u'(x)\) is the derivative of \(u(x)\).
(i) Using the graph, write down a value of \(x\) for which \(u'(x)\) is negative.
(ii) On the diagram above, draw the tangent to \(u(x)\) at the point \((4,2)\) and use the tangent that you draw to work out an estimate for the value of \(u'(4)\).
(i) e.g. \(x=8\)
(ii)
\(u'(4)=\dfrac{1}{2}\)
(i) \(x=8\)
(ii)
\begin{align}(x_1,y_1)=(2,1) \mbox{ and }(x_2,y_2)=(6,3)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}u'(4)&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{3-1}{6-2}\\&=\frac{1}{2}\end{align}
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Question 5
(a) Write each of the following values in the form \(a\times10^n\) where \(1\leq a<10\) and \(n\in\mathbb{Z}\).
(i) \(1200\)
(ii) \(0.27\)
(i) \(1.2\times 10^3\)
(ii) \(2.7\times10^{-1}\)
(i) \(1.2\times 10^3\)
(ii) \(2.7\times10^{-1}\)
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(b) A falcon can dive at a speed of up to \(120\) miles per hour.
\(1\) mile is approximately \(1.6\) kilometres.
Use this to work out how long it would take the falcon to travel \(100\) metres, when diving at this speed. Give your answer in seconds, correct to one decimal place.
\(1.9\mbox{ seconds}\)
\begin{align}120\mbox{ miles in }1\mbox{ hour}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}120\mbox{ miles in }60\times60=3{,}600\mbox{ seconds}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}120\times1.6=192\mbox{ kilometres in }3{,}600\mbox{ seconds}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}192{,}000\mbox{ metres in }3{,}600\mbox{ seconds}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}1\mbox{ metre in }\frac{3{,}600}{192{,}000}=0.01875\mbox{ seconds}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}100\mbox{ metres in }100\times0.01875\approx1.9\mbox{ seconds}\end{align}
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(c) The diagram below shows the graphs of the functions \(k(x)\) and \(m(x)\), for \(0\leq x\leq 5\), \(x\in\mathbb{R}\).
Use the graphs to estimate each of the following, for \(0\leq x\leq 5\):
(i) the two values of \(x\) for which \(m(x)=0\)
(ii) the range of values of \(x\) for which \(k(x)\) is less than \(m(x)\).
(i) \(x=1\) and \(x=4.5\)
(ii) \(2<x<3.5\)
(i) \(x=1\) and \(x=4.5\)
(ii) \(2<x<3.5\)
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Question 6
The \(n\)th term of an arithmetic sequence is given by the following expression, for \(n\in\mathbb{N}\):
\begin{align}T_n=-254+(n-1)(4)\end{align}
(a)
(i) Find the value of \(T_1\) , the first term of this sequence.
(ii) Find the value of the common difference for this sequence (that is, \(T_2-T_1\)).
(i) \(-254\)
(ii) \(4\)
(i)
\begin{align}T_1&=-254+(1-1)(4)\\&=-254\end{align}
(ii)
\begin{align}T_2&=-254+(2-1)(4)\\&=-250\end{align}
\begin{align}\downarrow\end{align}
\begin{align}d&=T_2-T_1\\&=-250-(-254)\\&=4\end{align}
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(b) Find the smallest value of \(n\in\mathbb{N}\) for which
\begin{align}-254+(n-1)(4)>0\end{align}
\(65\)
\begin{align}-254+(n-1)(4)>0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-254+4n-4>0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}4n>258\end{align}
\begin{align}\downarrow\end{align}
\begin{align}n>64.5\end{align}
\begin{align}\downarrow\end{align}
\begin{align}n=65\end{align}
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(c) The sum of the first \(n\) terms of this sequence is given by \(S_n=\dfrac{n}{2}[2(-254)+4n-4]\).
Solve the following equation for \(n\in\mathbb{N}\). Note that \(n\neq0\).
\begin{align}\frac{n}{2}[2(-254)+4n-4]=0\end{align}
\(128\)
\begin{align}\frac{n}{2}[2(-254)+4n-4]=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2(-254)+4n-4=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-508+4n-4=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}4n=512\end{align}
\begin{align}\downarrow\end{align}
\begin{align}n&=\frac{512}{4}\\&=128\end{align}
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Section B
Question 7
Joseph is doing a training session. During the session, his heart-rate, \(h(x)\), is measured in beats
per minute (BPM). For part of the session, \(h(x)\) can be modelled using the following function:
\begin{align}h(x)=-0.38x^3+2.6x^2-0.13x+158\end{align}
where \(x\) is the time, in minutes, from the start of the session, and \(0\leq x \leq 6\), \(x\in\mathbb{R}\).
(a)
(i) Complete the table below to show the values of \(h(x)\) for the given values of \(x\).
Give each value of \(h(x)\) correct to the nearest whole number.
| Time (minutes) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
|
Heart-rate (BPM) |
|
\(160\) |
|
\(171\) |
|
|
\(169\) |
(ii) Draw the graph of \(y=h(x)\) on the axes below, for \(0\leq x\leq 6\), \(x\in\mathbb{R}\).
Note that the point A \((6,169)\) is on the graph.
(i)
| Time (minutes) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
|
Heart-rate (BPM) |
\(158\) |
\(160\) |
\(165\) |
\(171\) |
\(175\) |
\(175\) |
\(169\) |
(ii)
(i)
| Time (minutes) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
|
Heart-rate (BPM) |
\(158\) |
\(160\) |
\(165\) |
\(171\) |
\(175\) |
\(175\) |
\(169\) |
(ii)
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(b) Explain what the co-ordinates of the point A \((6,169)\) represent, in the context of Joseph’s heart-rate.
\(6\) minutes from the start of the session, Joseph’s heart-rate is \(169\) BPM.
\(6\) minutes from the start of the session, Joseph’s heart-rate is \(169\) BPM.
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(c) Using the same axes and scales, continue your graph on the previous page to show the following information. From the point represented by A \((6,169)\), Joseph’s heart-rate:
- stays at the same level for the next \(2\) minutes, and then
- decreases at a steady rate of \(10\) BPM per minute for \(2\) minutes.
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(d) During his training session, the number of calories per minute that Joseph is burning after \(x\) minutes can be modelled by \(c(x)\) as follows, where \(h(x)\) is Joseph’s heart-rate at that time:
\begin{align}c(x)=0.1h(x)-7\end{align}
Using the information in the table or graph, work out \(c(6)\), the number of calories per minute that Joseph is burning \(6\) minutes after the start of the session.
\(9.9\mbox{ calories per minute}\)
\begin{align}c(6)&=0.1h(6)-7\\&=0.1(169)-7\\&=9.9\mbox{ calories per minute}\end{align}
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(e) Joseph has a smart watch that beeps every \(15\) seconds during the session.
It beeps for the first time at exactly 2: 55 p.m., as Joseph starts his session.
It beeps for the last time at exactly 3: 23 p.m., as Joseph finishes his session.
Work out how many times, in total, the smart watch beeps during the session, including the first and last beep.
\(113\) times
\begin{align}&3\colon23\\\underline{-}&\underline{2\colon55}\\&0\colon28\end{align}
The watch will beep \((4\times28)+1=113\) times.
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(f) Solve the equation
\begin{align}h'(x)=-1.14x^2+5.2x-0.13=0\end{align}
to find how long after the start of the session Joseph’s heart-rate is at a maximum, for \(0\leq x\leq6\), \(x\in\mathbb{R}\). Give your answer in minutes, correct to \(2\) decimal places.
\(4.54\) minutes
\begin{align}a=-1.14&&b=5.2&&c=-0.13\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\&=\frac{-5.2\pm\sqrt{5.2^2-4(-1.14)(-0.13)}}{2(-1.14)}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x\approx0.025&&x\approx4.536\end{align}
According to the graph, his heart-rate is therefore a maximum after \(4.536\approx 4.54\) minutes.
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Question 8
(a) Jessica is a scientist. Jessica is making up a solution of acid.
She has two different bottles, each with the following concentration of the acid:
| Bottle A | Bottle B |
|---|---|
|
Concentration: \(12\%\) |
Concentration: \(5\%\) |
This means that, for example, in every \(100\mbox{ ml}\) of liquid in Bottle A, there are \(12\mbox{ ml}\) of acid.
(i) Work out how many \(\mbox{ml}\) of acid are in \(200\mbox{ ml}\) of liquid from Bottle A.
(ii) Jessica mixes \(200\mbox{ ml}\) of liquid from Bottle A with \(300\mbox{ ml}\) of liquid from Bottle B.
Work out the overall concentration of the acid in Jessica’s mixture.
Give your answer as a percentage.
(iii) Explain why Jessica could not make a solution with a \(4\%\) concentration of acid by mixing liquid from Bottle A and Bottle B.
(iv) When she is making another mixture, Jessica makes a mistake in measuring.
She wants to measure out \(250\mbox{ ml}\) but she measures out \(260\mbox{ ml}\) instead.
Work out the percentage error in this measurement.
(i) \(24\mbox{ ml}\)
(ii) \(7.8\%\)
(iii) The solution can only be between \(5\%\) and \(12\%\).
(iv) \(4\%\)
(i)
\begin{align}200\times\frac{12}{100}=24\mbox{ ml}\end{align}
(ii)
\begin{align}\frac{24+300\times0.05}{200+300}\times 100=7.8\%\end{align}
(iii) The solution can only be between \(5\%\) and \(12\%\).
(iv)
\begin{align}\frac{260-260}{250}\times100=4\%\end{align}
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(b) If a solid is made up of faces with straight edges, then the following identity is often true:
\begin{align}C-E+F=2\end{align}
where:
- \(C\) is the number of corners
- \(E\) is the number of edges
- \(F\) is the number of faces
(i) The value of \(E\) for a cube is \(12\). Write down the values of \(C\) and \(F\) for a cube, and
show that \(C-E+F=2\) for these values.
(ii) Each of the faces of a different solid is in the shape of a triangle of area \(5\mbox{ cm}^2\).
This solid has \(12\) corners (\(C\)) and \(30\) edges (\(E\)), and \(C-E+F=2\) for this solid.
Work out the surface area of this solid, in \(\mbox{cm}^2\).
(iii) The surface of a third solid is made up of \(h\) hexagons and \(p\) pentagons, where \(h,p\in\mathbb{N}\). For this solid:
\begin{align}\frac{6h+5p}{3}-\frac{6h+5p}{2}+h+p=2\end{align}
Use this equation to find the number of pentagons in the surface of this solid (that is, the value of \(p\)).
(i) \(C=8\) and \(F=6\)
(ii) \(100\mbox{ cm}^2\)
(iii) \(p=12\)
(i)
\begin{align}C=8 && F=6\end{align}
\begin{align}\downarrow\end{align}
\begin{align}C-E+F&=8-12+6\\&=2\end{align}
as required.
(ii)
\begin{align}C-E+F=2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}12-30+F=2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}F&=2-12+30\\&=20\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\mbox{Surface area}&=20\times5\\&=100\mbox{ cm}^2\end{align}
(iii)
\begin{align}\frac{6h+5p}{3}-\frac{6h+5p}{2}+h+p=2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2(6h+5p)-3(6h+5p)+6h+6p=12\end{align}
\begin{align}\downarrow\end{align}
\begin{align}12h+10p-18h-15p+6h+6p=12\end{align}
\begin{align}\downarrow\end{align}
\begin{align}p=12\end{align}
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Question 9
Brian buys a new car.
The graph below represents a model that can be used to predict the value of this car, \(V\), for the next
number of years. This model assumes that the value of the car reduces (depreciates) by a fixed percentage each year.
(a)
(i) Use the graph to write down \(V(0)\), the initial value of Brian’s car, and \(V(1)\), the value of Brian’s car after \(1\) year.
(ii) Show that the value of the car will reduce by \(20\%\) in its first year, according to this model.
(i) \(V(0)=30{,}000\mbox{ euro}\) and \(V(1)=24{,}000\mbox{ euro}\)
(ii) The answer is already in the question!
(i)
\begin{align}V(0)=30{,}000\mbox{ euro} && V(1)=24{,}000\mbox{ euro}\end{align}
(ii)
\begin{align}\frac{30{,}000-24{,}000}{30{,}000}\times 100=20\%\end{align}
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(b)
(i) Based on this model, write a formula for \(V(t)\), the value of Brian’s car after \(t\) years, in terms of the age of the car (\(t\)).
Use the fact that the value decreases by \(20\%\) each year.
(ii) Hence, or otherwise, work out the value of Brian’s car after \(4\) years, according to this model. Show your working out.
(i) \(V(t)=30{,}000(0.8)^t\)
(ii) \(12{,}288\mbox{ euro}\)
(i)
\begin{align}V(t)=30{,}000(0.8)^t\end{align}
(ii)
\begin{align}V(4)&=30{,}000(0.8)^4\\&=12{,}288\mbox{ euro}\end{align}
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(c) The graph from part (a) is shown again below.
A different (linear) model assumes that the value of the car reduces (depreciates) by a fixed amount each year. The value of the car will also reduce by \(20\%\) in its first year, according to this model.
(i) Draw a line on the diagram above, passing through the first two points on the graph
with whole-number values of \(t\) (\(t=0\) and \(t=1\)). Continue your line until it reaches the horizontal axis.
(ii) Hence, or otherwise, estimate \(T\), the age of Brian’s car when its value would be €\(0\), according to this new model.
(i)
(ii) \(T=5\mbox{ years}\)
(i)
(ii) \(T=5\mbox{ years}\)
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(d) Eva buys a new car that has a price of €\(19{,}445\).
She pays \(30\%\) of this price as a deposit and makes repayments of €\(206.97\) each month for the following 3 years. At the end of the \(3\) years, she pays an additional lump sum of €\(7{,}389\).
Work out the total cost of the car for Eva.
\(20{,}673.42\mbox{ euro}\)
\begin{align}0.3\times19{,}445+36\times206.97+7{,}389=20{,}673.42\mbox{ euro}\end{align}
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(e) Eva drives her car home from the garage, a distance of \(12\mbox{ km}\).
Eva usually drives this journey at an average speed of \(60\mbox{ km/hr}\).
On this day, there are roadworks, so her average speed is only \(40\mbox{ km/hr}\) for the journey.
Work out the percentage increase in the time it takes Eva to drive home, because of the roadworks.
\(50\%\)
\begin{align}\frac{60-40}{40}\times100=50\%\end{align}
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Question 10
Keith plays hurling.
(a) During a match, Keith hits the ball with his hurl.
The height of the ball could be modelled by the following quadratic function:
\begin{align}h=-2t^2+5t+1.2\end{align}
where \(h\) is the height of the ball, in metres, \(t\) seconds after being hit, and \(t\in\mathbb{R}\).
(i) How high, in metres, was the ball when it was hit (when \(t=0\))?
(ii) The ball was caught after \(2.4\) seconds.
How high, in metres, was the ball when it was caught?
(iii) When the ball passed over the halfway line, it was at a height of \(3.2\) metres and its height was decreasing.
How many seconds after it was hit did the ball pass over the halfway line?
Remember that \(h=-2t^2+5t+1.2\).
(iv) Find \(\dfrac{dh}{dt}\) and hence find how long it took the ball to reach its greatest height.
Give your answer in seconds.
(i) \(1.2\mbox{ m}\)
(ii) \(1.68\mbox{ m}\)
(iii) \(2\mbox{ s}\)
(iv) \(\dfrac{5}{4}\mbox{ s}\)
(i)
\begin{align}h(0)&=-2(0^2)+5(0)+1.2\\&=1.2\mbox{ m}\end{align}
(ii)
\begin{align}h(2.4)&=-2(2.4^2)+5(2.4)+1.2\\&=1.68\mbox{ m}\end{align}
(iii)
\begin{align}h(t)&=3.2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-2t^2+5t+1.2=3.2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2t^2-5t+2=0\end{align}
\begin{align}\downarrow\end{align}
\(t=\dfrac{1}{2}\) or \(t=2\)
As the height was decreasing, it must be the second time, i.e. \(2\) seconds.
(iv)
\begin{align}\frac{dh}{dt}=-4t+5\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-4t+5=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t=\frac{5}{4}\mbox{ s}\end{align}
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(b) Later in the game, Keith hit the ball again. This time, the height of the ball \(t\) seconds after it was hit could be modelled by a different quadratic function, \(y=k(t)\), where \(k\) is in metres.
This time, the ball was \(1\) metre high when Keith hit it.
Its greatest height was \(5\) metres, which it reached after \(2\) seconds.
It hit the ground without being caught.
Using the information above, write down the co-ordinates of three points that must be on the graph of \(y=k(t)\), and draw the graph of \(y=k(t)\) on the axes below, from when the ball is hit until it hits the ground.
\begin{align}(0,1)&&(2,5)&&(4,1)\end{align}
\begin{align}(0,1)&&(2,5)&&(4,1)\end{align}
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(c)
(i) Keith buys a new hurl. It usually costs €\(33\).
Keith gets a student discount of \(15\%\).
Work out the price Keith pays for the hurl.
(ii) Keith also buys a jersey. This costs €\(49.50\), including VAT at \(23\%\).
Work out the VAT on this jersey. Give your answer correct to the nearest cent.
(i) \(28.05\mbox{ euro}\)
(ii) \(9.26\mbox{ euro}\)
(i)
\begin{align}33\times0.85=28.05\mbox{ euro}\end{align}
(ii)
\begin{align}\frac{49.50}{123}\times23\approx9.26\mbox{ euro}\end{align}