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2021 Ordinary Level - Paper One
Section A
Question 1
(a) A television costs €\(380\) before VAT at \(21\%\) has been added.
Find the cost of the television after VAT has been added.
\(459.80\mbox{ euro}\)
\begin{align}380\times1.21=459.80\mbox{ euro}\end{align}
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(b) When VAT at \(21\%\) is included, the price of a laptop increases by €\(130.20\).
Find the total cost of the laptop including VAT.
\(750.20\mbox{ euro}\)
\begin{align}\frac{130.20}{0.21}+130.20=750.20\mbox{ euro}\end{align}
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(c) A printer is priced at €\(290.40\) including VAT at \(21\%\).
Find how much VAT is included in the price of this printer.
\(50.40\mbox{ euro}\)
\begin{align}290.40\times\frac{21}{121}=50.40\mbox{ euro}\end{align}
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(d) On September 1, 2020 the Standard Rate of VAT in Ireland was reduced from \(23\%\) to \(21\%\).
A company bought \(30\) computers in September, all at the same price.
The company calculated that it saved €\(336\) due to the reduction in the VAT rate.
Find the price of one computer before VAT had been added.
\(560\mbox{ euro}\)
\begin{align}\frac{100}{23-21}\times\frac{336}{30}=560\mbox{ euro}\end{align}
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Question 2
\(z_1=-3+4i\) and \(z_2=4+3i\), where \(i^2=-1\).
(a) Plot and label \(z_1\), \(z_2\) and \(z_1+z_2\) on the Argand Diagram.
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(b) \(z_3=\dfrac{z_1}{z_2}\). Find \(z_3\) in the form \(a+bi\), where \(a,b\in\mathbb{Z}\).
\(z_3=i\)
\begin{align}z_3&=\frac{z_1}{z_2}\\&=\frac{-3+4i}{4+3i}\\&=\frac{-3+4i}{4+3i}\times\frac{4-3i}{4-3i}\\&=\frac{-12+9i+16i+12}{16-12i+12i+9}\\&=\frac{25i}{25}\\&=i\end{align}
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(c) Find \(|\bar{z_1}-z_2|\), where \(\bar z_1\) is the complex conjugate of \(z_1\).
Give your answer in the form \(p\sqrt{q}\), where \(p\) and \(q\in\mathbb{N}\)
\(7\sqrt{2}\)
\begin{align}|\bar{z_1}-z_2|&=|(-3-4i)-(4+3i)|\\&=|-7-7i|\\&=\sqrt{(-7)^2+(-7)^2}\\&=\sqrt{49\times2}\\&=\sqrt{49}\sqrt{2}\\&=7\sqrt{2}\end{align}
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Question 3
(a) Show that \(x=4\) is a solution of the equation \(x^2-2x-8=0\).
The answer is already in the question!
\begin{align}4^2-2(4)-8&=16-8-8\\&=0\end{align}
as required.
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(b) The equation \(x^2+ax+b=0\), where \(a,b,\in\mathbb{Z}\), has solutions \(x=5\) and \(x=-2\).
Find the value of \(a\) and the value of \(b\).
\(a=-3\) and \(b=-10\)
\begin{align}5^2+a(5)+b=0\end{align}
and
\begin{align}(-2)^2+a(-2)+b=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}25+5a+b=0\end{align}
\begin{align}4-2a+b=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}21+7a=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}a=-3\end{align}
and
\begin{align}b&=2a-4\\&=2(-3)-4\\&=-10\end{align}
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(c) Find the solutions of the equation \(5x^2-2x-9=0\), where \(x\in\mathbb{R}\).
Give each answer correct to \(2\) decimal places.
\(x=-1.16\) or \(x=1.56\)
\begin{align}a=5&&b=-2&&c=-9\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\&=\frac{-(-2)\pm\sqrt{(-2)^2-4(5)(-9)}}{2(5)}\\&=\frac{2\pm\sqrt{184}}{10}\end{align}
\begin{align}\downarrow\end{align}
\(x\approx-1.16\) or \(x\approx1.56\)
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Question 4
(a) Solve the equation:
\(4(2x+3)-7=3(x-5)\), where \(x\in\mathbb{R}\).
\(x=-4\)
\begin{align}4(2x+3)-7=3(x-5)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}8x+12-7=3x-15\end{align}
\begin{align}\downarrow\end{align}
\begin{align}8x-3x=-15-12+7\end{align}
\begin{align}\downarrow\end{align}
\begin{align}5x=-20\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x&=-\frac{20}{5}\\&=-4\end{align}
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(b) Solve the simultaneous equations:
\begin{align}2x-y&=7\\x^2+y^2&=49\end{align}
\(x=0\) and \(y=-7\) or \(x=\dfrac{28}{5}\) and \(y=\dfrac{21}{5}\)
\begin{align}2x-y&=7\\x^2+y^2&=49\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y=2x-7\\x^2+y^2=49\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x^2+(2x-7)^2=49\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x^2+4x^2-28x+49=49\end{align}
\begin{align}\downarrow\end{align}
\begin{align}5x^2-28x=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x(5x-28)=0\end{align}
\begin{align}\downarrow\end{align}
\(x=0\) or \(x=\dfrac{28}{5}\)
and
\begin{align}y&=2x-7\\&=2(0)-7\\&=-7\end{align}
or
\begin{align}y&=2x-7\\&=2\left(\frac{28}{5}\right)-7\\&=\frac{21}{5}\end{align}
\[\,\]
Solutions
\(x=0\) and \(y=-7\)
or
\(x=\dfrac{28}{5}\) and \(y=\dfrac{21}{5}\)
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Question 5
The function, \(f\) is defined as \(f(x)=3x^2-6x+7\), where \(x\in\mathbb{R}\).
(a) Find \(f(0.67)\), correct to \(2\) decimal places.
\(4.33\)
\begin{align}f(0.67)&=3(0.67)^2-6(0.67)+7\\&\approx4.33\end{align}
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(b) Find the value of \(x\) when \(f(x)=4\).
\(x=1\)
\begin{align}f(x)=4\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3x^2-6x+7=4\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3x^2-6x+3=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x^2-2x+1=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(x-1)^2=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x=1\end{align}
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(c) Use calculus to find the co-ordinates of the local minimum point of \(f\).
\((1,4)\)
\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}
and
\begin{align}f(1)&=3(1^2)-6(1)+7\\&=4\end{align}
\begin{align}\downarrow\end{align}
Minimum: \((1,4)\)
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Question 6
(a) The first three terms of an arithmetic sequence are \(-5,k,1\).
(i) Find \(k\) and hence or otherwise show that the common difference is \(3\).
(ii) Find \(T_{10}\), the \(10\)th term in the sequence.
(iii) Find which term in the sequence has a value of \(247\).
(i) \(k=-2\)
(ii) \(T_{10}=22\)
(iii) \(n=85\)
(i)
\begin{align}T_2-T_1=T_3-T_2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}k-(-5)=1-k\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2k=-4\end{align}
\begin{align}\downarrow\end{align}
\begin{align}k=-2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}T_2-T_1&=k-(-5)\\&=-2+5\\&=3\end{align}
as required.
(ii)
\begin{align}a=-5&&d=3\end{align}
\begin{align}T_n=a+(n-1)d\end{align}
\begin{align}\downarrow\end{align}
\begin{align}T_{10}&=-5+(10-1)(3)\\&=22\end{align}
(iii)
\begin{align}T_n=247\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-5+(n-1)(3)=247\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-5+3n-3=247\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3n=255\end{align}
\begin{align}\downarrow\end{align}
\begin{align}n=85\end{align}
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(b) The first three terms of a different arithmetic sequence are \(4,9,14\).
Find \(S_{50}\), the sum of the first \(50\) terms of the sequence.
\(S_{50}=6{,}325\)
\begin{align}a=4&&d=5\end{align}
\begin{align}S_n=\frac{n}{2}[2a+(n-1)d]\end{align}
\begin{align}\downarrow\end{align}
\begin{align}S_{50}&=\frac{50}{2}[2(4)+(50-1)(5)]\\&=6{,}325\end{align}
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Section B
Question 7
The rates and thresholds of the Universal Social Charge (USC) in Ireland (excluding the top rate) during \(2020\) are given in the table below.
| Annual Income | Rate |
|---|---|
|
First €\(12{,}012\) |
\(0.5\%\) |
|
Next €\(8{,}472\) |
\(2\%\) |
|
Next €\(49{,}560\) |
\(5%\) |
|
Balance |
(Top Rate) |
(a) At what level of annual income does a worker start paying the top rate of USC?
\(70{,}044\mbox{ euro}\)
\begin{align}12{,}012+8{,}472+49{,}560=70{,}044\mbox{ euro}\end{align}
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(b) How much USC will a worker have paid in total if they pay the maximum amount due at each of the first three rates?
\(2{,}459.70\mbox{ euro}\)
\begin{align}\frac{0.05}{100}\times12{,}012+\frac{2}{100}\times8{,}472+\frac{4.5}{100}\times49{,}560=2{,}459.70\mbox{ euro}\end{align}
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(c) John’s annual income is €\(54{,}800\). Find the amount of USC he will pay.
\(1{,}773.72\mbox{ euro}\)
\begin{align}\frac{0.05}{100}\times12{,}012+\frac{2}{100}\times8{,}472+\frac{4.5}{100}\times(54{,}800-12{,}012-8{,}472)=1{,}773.72\mbox{ euro}\end{align}
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(d) Mary pays €\(1602.72\) in USC in \(2020\). Find her annual income.
\(51{,}000\mbox{ euro}\)
\begin{align}1{,}602.72-\frac{0.05}{100}\times12{,}012-\frac{2}{100}\times8{,}472=1{,}373.22\mbox{ euro}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\mbox{Annual Income}&=\frac{1373.22\times100}{4.5}+12{,}012+8{,}472\\&=51{,}000\mbox{ euro}\end{align}
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(e) Patrick has an annual income of €\(83{,}000\) and pays a total of €\(3496.18\) in USC.
Find the top rate of USC.
\(8\%\)
\begin{align}\frac{(3{,}496.18-2{,}459.70)\times100}{83{,}000-70{,}044}=8\%\end{align}
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Question 8
A square sheet of cardboard, of side \(10\) units, is used to make an open box.
Squares of side \(x\) units, where \(x\in\mathbb{R}\), are removed from each corner of the cardboard and it is then folded along the dotted lines, as shown in the diagram below, in order to create the box.
(a) The length (\(l\)), breadth (\(b\)) and height (\(h\)) of the box are shown in the diagram above.
Write \(l\), \(b\), and \(h\) in terms of \(x\).
\(l=10-2x\), \(b=10-2x\) and \(h=x\)
\begin{align}l=10-2x\end{align}
\begin{align}b=10-2x\end{align}
\begin{align}h=x\end{align}
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(b) Show that the volume of the box can be written as
\begin{align}V(x)=4x^3-40x^2+100x\end{align}
The answer is already in the question!
\begin{align}V&=lbh\\&=(10-2x)(10-2x)(x)\\&=(10-2x)(10x-2x^2)\\&=100x-20x^2-20x^2+4x^3\\&=4x^3-40x^2+100x\end{align}
as required.
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(c) Explain why a box of height \(6\) units cannot be made from the sheet of cardboard.
The length and breadth would be negative, which is not possible.
The length and breadth would be negative, which is not possible.
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(d) Complete the table below to show the values of \(V(x)=4x^3-40x^2+100x\), where \(x\in\mathbb{R}\),
for the given values of \(x\) in the domain \(0\leq c\leq5\).
| \(x\mbox{ m}\) | \(0\) | \(0.5\) | \(1\) | \(1.5\) | \(2\) | \(2.5\) | \(3\) | \(3.5\) | \(4\) | \(4.5\) | \(5\) |
|---|---|---|---|---|---|---|---|---|---|---|---|
|
\(V(x)\mbox{ m}^3\) |
|
\(40.5\) |
|
|
|
|
|
|
|
\(4.5\) |
|
| \(x\mbox{ m}\) | \(0\) | \(0.5\) | \(1\) | \(1.5\) | \(2\) | \(2.5\) | \(3\) | \(3.5\) | \(4\) | \(4.5\) | \(5\) |
|---|---|---|---|---|---|---|---|---|---|---|---|
|
\(V(x)\mbox{ m}^3\) |
\(0\) |
\(40.5\) |
\(64\) |
\(73.5\) |
\(72\) |
\(62.5\) |
\(48\) |
\(31.5\) |
\(16\) |
\(4.5\) |
\(0\) |
| \(x\mbox{ m}\) | \(0\) | \(0.5\) | \(1\) | \(1.5\) | \(2\) | \(2.5\) | \(3\) | \(3.5\) | \(4\) | \(4.5\) | \(5\) |
|---|---|---|---|---|---|---|---|---|---|---|---|
|
\(V(x)\mbox{ m}^3\) |
\(0\) |
\(40.5\) |
\(64\) |
\(73.5\) |
\(72\) |
\(62.5\) |
\(48\) |
\(31.5\) |
\(16\) |
\(4.5\) |
\(0\) |
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(e) Draw the graph of the function \(V(x)\) on the grid below.
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(f) Use your graph to estimate each of the following values.
In each case show your work on the graph above.
(i) The maximum volume of the box.
(ii) The values of \(x\) which will create a box which has a volume of \(30\) units cubed.
(iii) The volume of the box when \(x\) is \(2.8\) units.
(i) \(74\)
(ii) \(x=0.35\) and \(x=3.55\)
(iii) \(54\)
(i) \(74\)
(ii) \(x=0.35\) and \(x=3.55\)
(iii) \(54\)
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Question 9
(a) The first three patterns in a sequence of patterns containing dots and crosses are shown below.
(i) Draw the fourth pattern in the sequence into the box below.
(ii) Find a formula, in \(n\), for the number of dots in pattern \(n\) of the sequence \(T_n\)).
(iii) Find the total number of dots in the first 20 patterns of the sequence.
(iv) The table shows the number of crosses for the first two patterns. Complete the table, and hence find a formula for the number of crosses in pattern \(n\) of the sequence.
| Pattern | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) |
|---|---|---|---|---|---|---|
|
Number of crosses |
\(1\) |
\(4\) |
|
|
|
|
(v) Find the number of crosses in pattern \(20\) of the sequence.
(vi) Find the number of shapes (dots and crosses) in pattern \(10\) of the sequence.
(i)
(ii) \(T_n=n\)
(iii) \(210\)
(iv)
| Pattern | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) |
|---|---|---|---|---|---|---|
|
Number of crosses |
\(1\) |
\(4\) |
\(9\) |
\(16\) |
\(25\) |
\(36\) |
\begin{align}T_n=n^2\end{align}
(v) \(144\)
(vi) \(110\)
(i)
(ii)
\begin{align}a=1&&d=1\end{align}
\begin{align}\downarrow\end{align}
\begin{align}T_n&=a+(n-1)d\\&=1+(n-1)(1)\\&=1+n-1\\&=n\end{align}
(iii)
\begin{align}a=1&&d=1\end{align}
\begin{align}S_n=\frac{n}{2}[2a+(n-1)d]\end{align}
\begin{align}\downarrow\end{align}
\begin{align}S_{20}&=\frac{20}{2}[2(1)+(20-1)(1)]\\&=210\end{align}
(iv)
| Pattern | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) |
|---|---|---|---|---|---|---|
|
Number of crosses |
\(1\) |
\(4\) |
\(9\) |
\(16\) |
\(25\) |
\(36\) |
\begin{align}T_n=n^2\end{align}
(v)
\begin{align}T_{12}&=12^2\\&=144\end{align}
(vi)
\begin{align}10^2+10=110\end{align}
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(b) The first two patterns in a sequence of patterns of crosses are shown below.
The number of crosses in pattern \(n\) is \(T_n\).
The general term describing \(T_n\) can be written in the form:
\(T_n=\dfrac{n^2}{2}+bn+c\), where \(b,c\in\mathbb{R}\).
(i) Use substitution for \(n\) to write \(T_1\) and \(T_2\) in terms of \(b\), \(c\) and a number.
(ii) Hence, or otherwise, find the value of \(b\) and the value of \(c\).
(i) \(T_1=b+c+\dfrac{1}{2}\) and \(T_2=2b+c+2\)
(ii) \(b=\dfrac{1}{2}\) and \(c=0\)
(i)
\begin{align}T_1&=\frac{1^2}{2}+b(1)+c\\&=b+c+\frac{1}{2}\end{align}
and
\begin{align}T_2&=\frac{2^2}{2}+b(2)+c\\&=2b+c+2\end{align}
(ii)
\begin{align}T_1=1\end{align}
\begin{align}\downarrow\end{align}
\begin{align}b+c+\frac{1}{2}=1\end{align}
\begin{align}\downarrow\end{align}
\begin{align}b+c=\frac{1}{2}\end{align}
and
\begin{align}T_2=3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2b+c+2=3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2b+c=1\end{align}
We therefore have the following two equations with two unknowns:
\begin{align}b+c=\frac{1}{2}\end{align}
\begin{align}2b+c=1\end{align}
\begin{align}\downarrow\end{align}
\begin{align}b=\frac{1}{2}\end{align}
and
\begin{align}c&=\frac{1}{2}-b\\&=\frac{1}{2}-\frac{1}{2}=0\end{align}
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Question 10
(a) The water behind a dam is normally released at a rate of \(250{,}000\) litres per second.
(i) Find how long it takes to release \(1\) million cubic metres (\(1{,}000{,}000\mbox{ m}^3\)) of water.
Note \(1\mbox{ m}^3=1000\mbox{ litres}\).
Give your answer correct to the nearest minute.
(ii) Due to heavy rainfall, the operators of the dam decide to increase the flow by \(10\%\) for \(24\) hours. Find how many \(\mbox{m}^3\) of water were released in that \(24\) hour period.
Give your answer in the form \(a\times10^n\), where \(1\leq a<10\), and \(n\in\mathbb{N}\).
Give the value of \(a\) correct to \(3\) significant figures.
(i) \(67\mbox{ min}\)
(ii) \(2.38\times10^7\mbox{ m}^3\)
(i)
\begin{align}t&=\frac{1{,}000{,}000\times100}{250{,}000}=440\mbox{ s}\\&\approx67\mbox{ min}\end{align}
(ii)
\begin{align}\frac{250{,}000}{1000}\times1.1\times60\times60\times24\approx2.38\times10^7\mbox{ m}^3\end{align}
Our Video Walkthroughs are in the final stages of editing and will go live during Summer 2023.
In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!
(b)
(i) John walks around a circular trail of radius \(0.5\mbox{ km}\) at a steady speed of \(6\mbox{ km/h}\).
How long will it take him to complete \(3\) full circuits of the trail?
Give your answer correct to the nearest minute.
(ii) Mary decides to walk every day over a \(5\) day period.
She walks a distance of \(3\mbox{ km}\) on day one.
She increases the length of her walk by \(15\%\) each day for the next four days.
Her average speed on day \(5\) is \(4\mbox{ km/h}\).
Find how long it will take her to complete her walk on day \(5\).
Give your answer in minutes, correct to the nearest minute.
(iii) One day, during John’s walk he meets Mary at point \(P\) on the trail.
Mary is walking in the opposite direction at a steady speed of \(4\mbox{ km/h}\).
John continues walking at \(6\mbox{ km/h}\).
How far will he travel until he meets Mary again?
Give your answer correct to the nearest metre.
(i) \(94\mbox{ min}\)
(ii) \(79\mbox{ min}\)
(iii) \(1{,}885\mbox{ m}\)
(i)
\begin{align}t&=3\times\frac{2\pi r}{s}\\&=3\times\frac{2\pi(0.5)}{6}\\&=1.5707..\mbox{ hr}\\&\approx94\mbox{ min}\end{align}
(ii)
\begin{align}t&=\frac{d}{s}\\&=\frac{3(1.15^4)}{4}\\&=1.2117…\mbox{ hr}\\&\approx79\mbox{ min}\end{align}
(iii)
\begin{align}d&=\frac{6}{10}\times(2\pi r)\\&=\frac{6}{10}\times[2\pi(0.5)]\\&\approx1{,}885\mbox{ m}\end{align}
