L.C. MATHS

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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)

Solution

(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).

Solution
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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 $$|z_2-z_3|=\sqrt{106}$$

Solution

\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.

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}

$$x=-\dfrac{3}{2}$$

Solution

\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 $$k\in\mathbb{R}$$.

$$3^{14}$$

Solution

\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 $$y=-13$$

Solution

\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.

$$25{,}300\mbox{ euro}$$

Solution

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.

$$36{,}430\mbox{ euro}$$

Solution

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.

$$1{,}600\mbox{ euro}$$

Solution

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) $$6x-y-87=0$$

Solution

(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)

$$u'(4)=\dfrac{1}{2}$$

Solution

(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) $$2.7\times10^{-1}$$

Solution

(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.

$$1.9\mbox{ seconds}$$

Solution

\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) $$2<x<3.5$$

Solution

(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) $$4$$

Solution

(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}

$$65$$

Solution

\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}

$$128$$

Solution

\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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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)

Solution

(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.

Solution

$$6$$ 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.
Solution
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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.

$$9.9\mbox{ calories per minute}$$

Solution

\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

Solution

\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

Solution

\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.

(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) $$4\%$$

Solution

(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) $$p=12$$

Solution

(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

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 $$V(1)=24{,}000\mbox{ euro}$$

Solution

(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) $$12{,}288\mbox{ euro}$$

Solution

(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) $$T=5\mbox{ years}$$

Solution

(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.

$$20{,}673.42\mbox{ euro}$$

Solution

\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.

$$50\%$$

Solution

\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.

(i) $$1.2\mbox{ m}$$

(ii) $$1.68\mbox{ m}$$

(iii) $$2\mbox{ s}$$

(iv) $$\dfrac{5}{4}\mbox{ s}$$

Solution

(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}

Solution

\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) $$9.26\mbox{ euro}$$

Solution

(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}

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