One-to-One Grinds
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2018 Ordinary Level - Paper One
Section A
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: \(42{,}000\mbox{ euro}\)
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.
\(750{,}000\mbox{ euro}\)
\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.
\(500{,}000\mbox{ euro}\)
\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.
\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 \(|z_3|=|z_1|+|z_2\).
\(|z_3|\neq|z_1|+|z_2|\)
\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}
\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 \(x=3.89\)
\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 \(b=7\)
\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
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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 |
|---|---|---|---|
|
\(1\) |
\(4\) |
\(2\) |
\(6\) |
|
\(2\) |
\(9\) |
\(2\) |
\(11\) |
|
\(3\) |
\(16\) |
\(2\) |
\(18\) |
|
\(4\) |
\(25\) |
\(2\) |
\(18\) |
|
\(5\) |
\(36\) |
\(2\) |
\(38\) |
| 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 \(c=3\)
\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.
\(n=20\)
\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 \(C=(3,0)\)
\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 \(f(x)=-x^2+x+6\).
The answer is already in the question!
\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 \((0.5,6.25)\).
The answer is already in the question!
\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}
\(x=2\)
\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 \(x=3\)
\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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Section B
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 \(10\).
\(37\)
\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?
\(23\)
\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.
\(897\)
\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 \(32\).
\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.
\(9{,}444\mbox{ euro}\)
\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) \(11\mbox{ euro}\)
(i)
\begin{align}\mbox{Number of children’s tickets}&=\frac{752}{1+3}\\&=188\end{align}
\begin{align}\mbox{Number of adult tickets}&=3\times188\\&=564\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.
The answer is already in the question!
\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) |
\(0\) |
\(57.5\) |
\(118\) |
\(175.5\) |
\(224\) |
\(257.5\) |
\(270\) |
\(255.5\) |
\(208\) |
\(121.5\) |
| \(t\) (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 \(t\in\mathbb{R}\).
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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) \(1.7\mbox{ hours}\)
(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.
(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 \(2000\mbox{ m}\).
Give your answer in metres per second.
(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.
Give your answer correct to the nearest minute.
(i) \(140\mbox{ m/s}\)
(ii) \(48\mbox{ min}\)
(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.
Give your answer correct to the nearest metre.
(i) \(d=\dfrac{s^2}{g}\)
(ii) \(309\mbox{ m}\)
(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}
