L.C. MATHS

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## Question 1

When Sean joined a sales company he was offered a choice of two different salary contracts.
The details of the contracts are outlined in the table below.

$$t$$ Salary End of year commision on total sales

Contract A

$$35{,}000$$ euro

$$2\%$$

Contract B

$$30{,}000$$ euro

$$3\%$$

(a) Find how much Sean would earn under each contract in a year where his total sales were €$$400{,}000$$.

Contract A: $$43{,}000\mbox{ euro}$$

Contract B: $$42{,}000\mbox{ euro}$$

Solution

Contract A

\begin{align}35{,}000+400{,}000\times0.02=43{,}000\mbox{ euro}\end{align}

$\,$

Contract B

\begin{align}30{,}000+400{,}000\times0.03=42{,}000\mbox{ euro}\end{align}

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(b) Another employee, Mary, earned €$$50{,}000$$ in a particular year. She is on Contract $$A$$.
Find her total sales for that year.

$$750{,}000\mbox{ euro}$$

Solution

\begin{align}\frac{50{,}000-35{,}000}{0.02}=750{,}000\mbox{ euro}\end{align}

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(c) Find the total sales for which a salesperson would earn the same amount of money under each contract.

$$500{,}000\mbox{ euro}$$

Solution

\begin{align}35{,}000+0.02x=30{,}000+0.03x\end{align}

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

\begin{align}0.01x=5{,}000\end{align}

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

\begin{align}x&=\frac{5{,}000}{0.01}\\&=500{,}000\mbox{ euro}\end{align}

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## Question 2

$$z_1=-2+3i$$ and $$z_2=-3-2i$$, where $$i^2=-1$$.
$$z_3=z_1-z_2$$.

(a) Plot $$z_1$$, $$z_2$$, and $$z_3$$ on the Argand Diagram.
Label each point clearly.

Solution

\begin{align}z_3&=z_1-z_2\\&=(-2+3i)-(-3-2i)\\&=-2+3i+3+2i\\&=1+5i\end{align}

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(b) Investigate if $$|z_3|=|z_1|+|z_2$$.

$$|z_3|\neq|z_1|+|z_2|$$

Solution

\begin{align}|z_3|&=\sqrt{1^2+5^2}\\&=\sqrt{26}\end{align}

and

\begin{align}|z_1|+|z_2|&=\sqrt{(-2)^2+3^2}+\sqrt{(-3)^2+(-2)^2}\\&=\sqrt{13}+\sqrt{13}\\&=2\sqrt{13}\end{align}

Therefore, $$|z_3|\neq|z_1|+|z_2|$$.

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(c) $$z_4=\dfrac{z_1}{z_2}$$. Write $$z_4$$ in the form $$x+yi$$,݅ where $$x,y,\in\mathbb{R}$$.

\begin{align}z_4=-i\end{align}

Solution

\begin{align}z_4&=\frac{-2+3i}{-3-2i}\\&=\frac{-2+3i}{-3-2i}\times\frac{-3+2i}{-3+2i}\\&=\frac{6-4i-9i-6}{9-6i+6i+4}\\&=\frac{-13i}{13}\\&=-i\end{align}

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## Question 3

(a) Solve the equation $$2x^2-7x-3=0$$. Give each answer correct to $$2$$ decimal places.

$$x=-0.39$$ or $$x=3.89$$

Solution

\begin{align}a=2&&b=-7&&c=-3\end{align}

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

\begin{align}x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\&=\frac{-(-7)\pm\sqrt{(-7)^2-4(2)(-3)}}{2(2)}\\&=\frac{7\pm\sqrt{73}}{4}\end{align}

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

$$x\approx-0.39$$ or $$x\approx3.89$$

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(b) Solve the simultaneous equations below to find the value of $$a$$ and the value of $$b$$.

\begin{align}2a+3b&=15\\5a+b&=-8\end{align}

$$a=-3$$ and $$b=7$$

Solution

\begin{align}2a+3b&=15\\5a+b&=-8\end{align}

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

\begin{align}2a+3b&=15\\15a+3b&=-24\end{align}

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

\begin{align}13a=-39\end{align}

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

\begin{align}a=-3\end{align}

and

\begin{align}b&=-8-5a\\&=-8-5(-3)\\&=7\end{align}

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## Question 4

The first three stages in a pattern of grey and white tiles are shown in the diagram below.

(a) Draw the next stage of tiles (Stage 4) onto the diagram above

Solution
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(b) Based on the pattern shown, complete the table below.

Stage ($$n$$) Number of Grey Tiles Number of White Tiles Total

$$1$$

$$4$$

$$6$$

$$2$$

$$3$$

$$4$$

$$5$$

Stage ($$n$$) Number of Grey Tiles Number of White Tiles Total

$$1$$

$$4$$

$$2$$

$$6$$

$$2$$

$$9$$

$$2$$

$$11$$

$$3$$

$$16$$

$$2$$

$$18$$

$$4$$

$$25$$

$$2$$

$$18$$

$$5$$

$$36$$

$$2$$

$$38$$

Solution
Stage ($$n$$) Number of Grey Tiles Number of White Tiles Total

$$1$$

$$4$$

$$2$$

$$6$$

$$2$$

$$9$$

$$2$$

$$11$$

$$3$$

$$16$$

$$2$$

$$18$$

$$4$$

$$25$$

$$2$$

$$18$$

$$5$$

$$36$$

$$2$$

$$38$$

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(c) Assuming the pattern continues, the total number of tiles in stage ݊$$n$$ $$T_n$$ is given by the formula $$T_n=n^2+bn+c$$, where $$b$$ and $$c\in\mathbb{N}$$.
Find the value of $$b$$ and the value of $$c$$.

$$b=2$$ and $$c=3$$

Solution

\begin{align}T_1=6\end{align}

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

\begin{align}1^2+b(1)+c=6\end{align}

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

\begin{align}b+c=5\end{align}

and

\begin{align}T_2=11\end{align}

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

\begin{align}2^2+b(2)+c=11\end{align}

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

\begin{align}2b+c=7\end{align}

We therefore have the following two equations with two unknowns:

\begin{align}b+c=5\end{align}

\begin{align}2b+c=7\end{align}

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

\begin{align}b=2\end{align}

and

\begin{align}c&=5-b\\&=5-2\\&=3\end{align}

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(d) Find the number of the stage which has $$443$$ tiles in total.

$$n=20$$

Solution

\begin{align}T_n=443\end{align}

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

\begin{align}n^2+2n+3=443\end{align}

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

\begin{align}n^2+2n-440=0\end{align}

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

\begin{align}(n+22)(n-2)=0\end{align}

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

\begin{align}n=20\end{align}
(as $$n>0$$)

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## Question 5

The diagram shows the graph of a quadratic function $$f$$.

(a) Write down the co-ordinates of $$A$$, $$B$$ and $$C$$.

$$A=(0,6)$$, $$B=(-2,0)$$ and $$C=(3,0)$$

Solution

\begin{align}A=(0,6)&&B=(-2,0)&&C=(3,0)\end{align}

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(b) Show that the function can be written as $$f(x)=-x^2+x+6$$.

Solution

\begin{align}f(x)&=-(x+2)(x-3)\\&=-x^2+x+6\end{align}

as required.

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(c) Show, using calculus, that the maximum point of $$f(x)$$ is $$(0.5,6.25)$$.

Solution

\begin{align}f'(x)=0\end{align}

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

\begin{align}-2x+1=0\end{align}

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

\begin{align}x=0.5\end{align}

and

\begin{align}f(0.5)&=-(0.5^2)+0.5+6\\&=6.25\end{align}

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## Question 6

(a) Solve for $$x$$.

\begin{align}(x+5)(3x-4)-3(x^2+2)+4=0\end{align}

$$x=2$$

Solution

\begin{align}(x+5)(3x-4)-3(x^2+2)+4=0\end{align}

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

\begin{align}3x^2-4x+15x-20-3x^2-6+4=0\end{align}

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

\begin{align}11x-22=0\end{align}

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

\begin{align}x&=\frac{22}{11}\\&=2\end{align}

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(b) Find the solutions of

$$\dfrac{5}{x+3}-\dfrac{1}{x}=\dfrac{1}{2}$$, where $$x\neq-3,0,x\in\mathbb{R}$$.

$$x=2$$ or $$x=3$$

Solution

\begin{align}\frac{5x-1(x+3)}{(x+3)(x)}=\frac{1}{2}\end{align}

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

\begin{align}\frac{4x-3}{x^2+3x}=\frac{1}{2}\end{align}

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

\begin{align}2(4x-3)=1(x^2+3x)\end{align}

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

\begin{align}8x-6=x^2+3x\end{align}

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

\begin{align}x^2-5x+6=0\end{align}

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

\begin{align}(x-2)(x-3)=0\end{align}

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

$$x=2$$ or $$x=3$$

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## Question 7

Part of the seating arrangement in a theatre is shown in the diagram below. The seats are arranged in rows. Row $$1$$ is nearest the stage and has $$28$$ seats. Each subsequent row behind that contains one more seat. i.e. row $$2$$ has $$29$$ seats, row $$3$$ has $$30$$ seats, and so on.

(a) Find the number of seats in row $$10$$.

$$37$$

Solution

\begin{align}28+(10-1)(1)=37\end{align}

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(b) There are $$50$$ seats in the last row. How many rows of seats are there in the theatre?

$$23$$

Solution

\begin{align}28+(n-1)(1)=50\end{align}

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

\begin{align}n&=50-28+1\\&=23\end{align}

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(c) Find the total number of seats in the theatre.

$$897$$

Solution

\begin{align}S_{23}&=\frac{23}{2}[2(28)+(23-1)(1)]\\&=897\end{align}

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(d) On a particular night $$600$$ people are at a show in the theatre. Assume that people are seated only in the rows closest to the stage, i.e. they have filled the first ݊ rows and there are some people seated in the next row.
Find the value of $$n$$ and find the number of people seated in the next row.

$$n=16$$ and the number of people in the next row is $$32$$.

Solution

\begin{align}\frac{n}{2}[2(28)+(n-1)(1)]=600\end{align}

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

\begin{align}n^2+55n-1200=0\end{align}

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

\begin{align}n&=\frac{-55\pm\sqrt{55^2-4(1)(-1200)}}{2(1)}\end{align}

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

\begin{align}n\approx16.7\end{align}
($$n>0$$)

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

\begin{align}n=16\end{align}
($$n\in\mathbb{N}$$)

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

\begin{align}S_{16}&=\frac{16}{2}[2(28)+(16-1)(1)]\\&=568\end{align}

The number of people in the next row is therefore $$600-568=32$$.

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(e) For a particular show, adult tickets cost €$$25$$ each and children’s tickets cost €$$12$$ each.
Find the total income from ticket sales if $$276$$ adult tickets and $$212$$ children’s tickets were sold.

$$9{,}444\mbox{ euro}$$

Solution

\begin{align}276\times25+212\times12=9{,}444\mbox{ euro}\end{align}

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(f)

(i) For a different show $$752$$ tickets were sold.
The ratio of adult tickets sold to children’s tickets sold is $$3:1$$.
Find how many adult tickets and how many children’s tickets were sold for that show.

(ii) For the show referred to in part f(i) an adult ticket cost $$2\dfrac{1}{2}$$ times as much as a children’s ticket. The income was €$$17{,}578$$. Find the cost of a children’s ticket for this show.

(i) $$\mbox{Number of children’s tickets}=188$$ and $$\mbox{Number of adult tickets}=564$$

(ii) $$11\mbox{ euro}$$

Solution

(i)

\begin{align}\mbox{Number of children’s tickets}&=\frac{752}{1+3}\\&=188\end{align}

(ii)

\begin{align}188x+564(2.5x)=17{,}758\end{align}

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

\begin{align}1{,}598x=17{,}578\end{align}

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

\begin{align}x&=\frac{17{,}758}{1{,}598}\\&=11\mbox{ euro}\end{align}

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## Question 8

The amount, in appropriate units, of a certain medicinal drug in the bloodstream $$t$$ hours after it has been taken can be estimated by the function:

$$C(t)=-t^3+4.5t^2+54t$$, where $$0\leq t\leq9$$, $$t\in\mathbb{R}$$

(a) Use the drug amount function, $$C(t)$$, to show that the amount of the drug in the bloodstream $$4$$ hours after the drug has been taken is $$224$$ units.

Solution

\begin{align}C(4)&=-4^3+4.5(4^2)+54(4)\\&=224\mbox{ units}\end{align}

as required.

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(b) Use the function $$C(t)$$ to complete the table below.

$$t$$ (Hours) $$0$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$8$$ $$9$$

$$C(t)$$ (Units)

$$0$$

$$57.5$$

$$224$$

$$t$$ (Hours) $$0$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$8$$ $$9$$

$$C(t)$$ (Units)

$$0$$

$$57.5$$

$$118$$

$$175.5$$

$$224$$

$$257.5$$

$$270$$

$$255.5$$

$$208$$

$$121.5$$

Solution
$$t$$ (Hours) $$0$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$8$$ $$9$$

$$C(t)$$ (Units)

$$0$$

$$57.5$$

$$118$$

$$175.5$$

$$224$$

$$257.5$$

$$270$$

$$255.5$$

$$208$$

$$121.5$$

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(c) Draw the graph of the function $$C(t)$$ for $$0\leq t\leq9$$ where $$t\in\mathbb{R}$$.

Solution
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(d) Use your graph to estimate each of the following values.
In each case show your work on the graph above.

(i) The amount of the drug in the bloodstream after $$3\dfrac{1}{2}$$ hours.

(ii) How long after taking the drug will the amount
of the drug in the bloodstream be $$100$$ units?

(i) $$205\mbox{ units}$$

(ii) $$1.7\mbox{ hours}$$

Solution

(i)

\begin{align}205\mbox{ units}\end{align}

(ii)

\begin{align}1.7\mbox{ hours}\end{align}

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(e)

(i) Use the drug amount function $$C(t)=-t^3+4.5t^2+54t$$ to find, in terms of $$t$$, the rate at which the drug amount is changing after $$4$$ hours.

(ii) Use your answer to part e(i) to find the rate at which the drug amount is changing after $$4$$ hours.

(iii) Use your answer to part e(i) to find the maximum amount of the drug in the bloodstream over the first $$9$$ hours.

(iv) Use your answer to part e(i) to show that the drug amount in the bloodstream is
decreasing $$7$$ hours after the drug has been taken. Explain your reasoning.

(i) $$C'(t)=-3t^2+9t+54$$

(ii) $$42\mbox{ units/hr}$$

(iii) $$270\mbox{ units}$$

(iv) $$C'(7)=-30\mbox{ units/hr}$$. Therefore, the amount is decreasing as the rate of change is negative.

Solution

(i)

\begin{align}C'(t)&=-3t^2+9t+54\end{align}

(ii)

\begin{align}C'(4)&=-3(4^2)+9(4)+54\\&=42\mbox{ units/hr}\end{align}

(iii)

\begin{align}C'(t)=0\end{align}

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

\begin{align}-3t^2+9t+54=0\end{align}

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

\begin{align}t^2-3t-18=0\end{align}

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

\begin{align}(t+3)(t-6)=0\end{align}

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

\begin{align}t=6\mbox{ s}\end{align}
(as $$t>0$$)

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

\begin{align}C(6)&=-6^3+4.5(6^2)+54(6)\\&=270\mbox{ units}\end{align}

(iv)

\begin{align}C'(7)&=-3(7^2)+9(7)+54\\&=-30\mbox{ units/hr}\end{align}

Therefore, the amount is decreasing as the rate of change is negative.

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## Question 9

When earthquakes occur under the sea, they can cause large tidal waves called tsunamis.
Scientists can estimate the arrival times of tsunamis to nearby countries.
The average speed at which a tsunami travels is given by the formula:

\begin{align}s=\sqrt{g\times d}\end{align}

where $$s$$ is the speed of the tsunami (in metres per second),
$$d$$ is the depth of the ocean (in metres) at the location where the earthquake occurred,
and $$g=9.8$$ metres per second$$^2$$.

(a)

(i) Find $$s$$, the speed of a tsunami when the an earthquake occurs at a depth of $$2000\mbox{ m}$$.

(ii) A tsunami has been identified as beginning $$400\mbox{ km}$$ from land.
The depth of the ocean at that point is $$2000\mbox{ m}$$.
Find how long it will take this tsunami to reach land.

(i) $$140\mbox{ m/s}$$

(ii) $$48\mbox{ min}$$

Solution

(i)

\begin{align}s(2{,}000)&=\sqrt{g\times2{,}000}\\&=\sqrt{9.8\times2{,}000}\\&=140\mbox{ m/s}\end{align}

(ii)

\begin{align}t&=\frac{d}{s}\\&=\frac{400\times1{,}000}{140}\\&=2{,}857.14…\mbox{ s}\\&\approx48\mbox{ min}\end{align}

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(b)

(i) Rearrange the formula to give $$d$$ in terms of $$g$$ and $$s$$.

(ii) Hence, or otherwise, find the depth of the ocean at the place where an earthquake occurred, if the resulting tsunami has a speed of $$55$$ metres per second.

(i) $$d=\dfrac{s^2}{g}$$

(ii) $$309\mbox{ m}$$

Solution

(i)

\begin{align}s=\sqrt{g\times d}\end{align}

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

\begin{align}s^2=g\times d\end{align}

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

\begin{align}d=\frac{s^2}{g}\end{align}

(ii)

\begin{align}d&=\frac{s^2}{g}\\&=\frac{55^2}{9.8}\\&\approx309\mbox{ m}\end{align}

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