OL Mock Exam A Paper 1

Section A

Question 1

VAT is charged at \(23\%\) on all video-game related products.

**(a) **A video game console costs \(220\mbox{ euro}\) *before* VAT is added on.

Find the selling price of the games console.

##
Answer

\(270.60\mbox{ euro}\)

##
Solution

\begin{align}220\times1.23=270.60\mbox{ euro}\end{align}

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**(b) **A video game controller costs \(40\mbox{ euro}\) *including* VAT.

Calculate how much VAT is included in the price of the controller, correct to the nearest cent.

##
Answer

\(7.48\mbox{ euro}\)

##
Solution

\begin{align}40\div1.23&\approx32.52\mbox{ euro}\end{align}

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

\begin{align}\mbox{VAT}&=40-32.52\\&=7.48\mbox{ euro}\end{align}

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**(c) **The VAT for a particular video game is \(11.22\mbox{ euro}\).

Calculate the selling price of the game, correct to the nearest cent.

##
Answer

\(60\mbox{ euro}\)

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Solution

\begin{align}\frac{11.22}{0.23} +11.22=60\mbox{ euro}\end{align}

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**(d) **The console from part **(a)** costs \(160\mbox{ euro}\) to manufacture.

Calculate the percentage profit that the manufacturer makes for each console sold.

##
Answer

\(37.5\%\)

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Solution

\begin{align}\frac{220-160}{160}\times100=37.5\%\end{align}

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Question 2

Consider the complex number \(z_1=4-2i\), where \(i^2=-1\).

**(a) **

**(i)** Calculate \(|z_1|\). Write your answer in surd form.

**(ii)** Write \(z_2=-\dfrac{1}{2}z_1\) in the form \(a+bi\), where \(a,b,\in\mathbb{Z}\).

**(iii)** Plot \(z_1\) and \(z_2\) on the following Argand diagram.

##
Answer

**(i) **\(\sqrt{20}\)

**(ii)** \(-2+i\)

**(iii)**

##
Solution

**(i)**

\begin{align}|z_1|&=\sqrt{4^2+(-2)^2}\\&=\sqrt{20}\end{align}

**(ii)**

\begin{align}z_2&=-\frac{1}{2}z_1\\&=-\frac{1}{2}(4-2i)\\&=-2+i\end{align}

**(iii)**

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**(b) **Show that \(\bar{z_1}\) is a solution to the equation \(z^2-8z+20=0\).

##
Answer

\(\pm\left(\dfrac{\sqrt{6}}{2}+\dfrac{1}{\sqrt{2}}i\right)\)

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Solution

\begin{align}z^2-8z+20=0\end{align}

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

\begin{align}(4+2i)^2-8(4+2i)+20=0\end{align}

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

\begin{align}16+16i-4-32-16i+20=0\end{align}

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

\begin{align}0=0\end{align}

Therefore, \(\bar{z_1}\) is a solution, as required.

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Question 3

**(a)** Solve the following equation:

\begin{align}\frac{2x+1}{3}+\frac{x-4}{5}=10\end{align}

##
Answer

\(x=7\)

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Solution

\begin{align}\frac{2x+1}{3}+\frac{x-4}{5}=\frac{28}{5}\end{align}

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

\begin{align}5(2x+1)+3(x-4)=84\end{align}

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

\begin{align}10x+5+3x-12=84\end{align}

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

\begin{align}13x-7=84\end{align}

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

\begin{align}13x=91\end{align}

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

\begin{align}x&=\frac{91}{13}\\&=7\end{align}

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**(b)** Solve the following simultaneous equations:

\begin{align}x+2y=8\end{align}

\begin{align}x^2-2x-y+3=0\end{align}

##
Answer

\(x=2\) and \(y=3\) or \(x=-\dfrac{1}{2}\) and \(y=\dfrac{17}{4}\)

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Solution

\begin{align}x+2y=8\end{align}

\begin{align}x^2-2x-y+3=0\end{align}

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

\begin{align}x=8-2y\end{align}

\begin{align}x^2-2x-y+3=0\end{align}

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

\begin{align}(8-2y)^2-2(8-2y)-y+3=0\end{align}

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

\begin{align}64-32y+4y^2-16+4y-y+3=0\end{align}

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

\begin{align}4y^2-29y+51=0\end{align}

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

\begin{align}(y-3)(4y-17)=0\end{align}

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

\(y=3\) and \(x=8-2(3)=2\)

or

\(y=\dfrac{17}{4}\) and \(x=8-2\left(\dfrac{17}{4}\right)=-\dfrac{1}{2}\)

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Question 4

**(a)** Consider the following equation:

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

**(i)** Show that \(x=2\) is a solution to this equation.

**(ii)** Find the other solution.

##
Answer

**(i)** The answer is already in the question!

**(ii)** \(x=\dfrac{1}{2}\)

##
Solution

**(i)**

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

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

\begin{align}8-10+2=0\end{align}

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

\begin{align}0=0\end{align}

Therefore, \(x=2\) is a solution to the equation, as required.

**(ii)**

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

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

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

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

\(x=\dfrac{1}{2}\) or \(x=2\)

Therefore, the other solution is \(x=\dfrac{1}{2}\).

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**(b)** Solve the following inequality for \(x\in\mathbb{R}\):

\begin{align}3(5-x)>9\end{align}

##
Answer

\(x<2\)

##
Solution

\begin{align}3(5-x)>9\end{align}

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

\begin{align}15-3x>9\end{align}

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

\begin{align}-3x>-6\end{align}

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

\begin{align}3x<6\end{align}

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

\begin{align}x<2\end{align}

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**(c)** Solve the following equation for \(x\):

\begin{align}9^{5x-1}=27\end{align}

##
Answer

\(x=\dfrac{1}{2}\)

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Solution

\begin{align}9^{5x-1}=27\end{align}

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

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

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

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

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

\begin{align}10x-2=3\end{align}

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

\begin{align}10x=5\end{align}

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

\begin{align}x=\frac{1}{2}\end{align}

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Question 5

**(a) **The second and third terms of an arithmetic sequence are \(10\) and \(14\).

**(i)** Find the first and fourth terms.

**(ii)** Write an expression for \(T_n\), the \(n\)th term of the sequence.

**(iii)** For what value of \(n\) is the term in the sequence \(82\)? Show your workings.

##
Answer

**(i) **\(T_1=6\) and \(T_4=18\)

**(ii)** \(T_n=4n+2\)

**(iii)** \(n=20\)

##
Solution

**(i)**

\begin{align}d&=14-10\\&=4\end{align}

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

\begin{align}T_1&=T_2-d\\&=10-4\\&=6\end{align}

and

\begin{align}T_4&=T_3+d\\&=14+4\\&=18\end{align}

**(ii)**

\begin{align}T_n&=a+(n-1)d\\&=6+(n-1)(4)\\&=6+4n-4\\&=4n+2\end{align}

**(iii)**

\begin{align}T_n=82\end{align}

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

\begin{align}4n+2=82\end{align}

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

\begin{align}4n=80\end{align}

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

\begin{align}n=20\end{align}

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**(b) **The sum of the first \(n\) terms of a sequence is given by:

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

Find \(T_2\), the second term of the sequence.

##
Answer

\(9\)

##
Solution

\begin{align}S_1&=3(1^2)+2\\&=5\end{align}

and

\begin{align}S_2&=3(2^2)+2\\&=14\end{align}

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

\begin{align}T_2&=S_2-S_1\\&=14-5\\&=9\end{align}

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Question 6

Consider the following quadratic function:

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

**(a)** Calculate the derivative of \(f(x)\) with respect to \(x\).

##
Answer

\(-2x+2\)

##
Solution

\begin{align}\frac{df}{dx}=-2x+2\end{align}

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**(b)** Using part **(a)**, or otherwise, find the maximum point of \(f(x)\).

##
Answer

\((1,4)\)

##
Solution

\begin{align}\frac{df}{dx}=0\end{align}

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

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

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

\begin{align}x=1\end{align}

and

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

Therefore, the maximum point is \((1,4)\).

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**(c)** Find the equation of the tangent to \(f(x)\) at the maximum point.

##
Answer

\(y=4\)

##
Solution

\begin{align}y-y_1=m(x-x_1)\end{align}

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

\begin{align}y-4=0(x-1)\end{align}

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

\begin{align}y=4\end{align}

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Section B

Question 7

Jeremy wishes to invest his \(12{,}000\mbox{ euro}\) savings with a bank for five years.

He is weighing up the different options from three different banks.

**(a) **Bank \(A\) offers *simple* interest at an interest rate of \(4\%\) per year.

**(i)** Calculate how interest Jeremy will have at the end of the second year if he chooses this option.

**(ii)** Calculate how much money Jeremy will have at the end of the five years if he chooses this option.

##
Answer

**(i)** \(960\mbox{ euro}\)

**(ii)** \(14{,}400\mbox{ euro}\)

##
Solution

**(i)**

\begin{align}2\times12{,}000\times0.04=960\mbox{ euro}\end{align}

**(ii)**

\begin{align}12{,}000+5\times12{,}000\times0.04=14{,}400\mbox{ euro}\end{align}

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**(b) **Bank \(B\) offers a \(15\%\) return at the end of his investment.

Calculate how much money Jeremy will have at the end of the five years if he chooses this option.

##
Answer

\(13{,}800\mbox{ euro}\)

##
Solution

\begin{align}12{,}000\times1.15=13{,}800\mbox{ euro}\end{align}

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**(c) **Finally, bank \(C\) offers *compound* interest at an interest rate of \(0.3\%\) per *month*.

**(i)** Find the corresponding AER (annual equivalent rate) for this investment, correct to three significant figures.

**(ii)** Calculate how much money interest will have made at the end of the five years if he chooses this option, correct to the nearest cent.

##
Answer

**(i) **\(3.66\%\)

**(ii)** \(14{,}362.74\mbox{ euro}\)

##
Solution

**(i)**

\begin{align}(1+0.003)^{12}=1+i\end{align}

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

\begin{align}i&=(1+0.003)^{12}-1\\&=0.0366…\\&=3.66\%\end{align}

**(ii)**

\begin{align}F&=P(1+i)^t\\&=(12{,}000)(1+0.0366)^5\\&\approx14{,}362.74\mbox{ euro}\end{align}

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**(d) **Complete the following graph representing the amount \(A\) of the investment after \(t\) years in all three cases for \(0\leq t\leq3\).

##
Answer

**(i)** The Conjugate Roots theorem cannot be used as the coefficients of the equation are not all real.

**(ii)** \(z_2=-i\)

##
Solution

**(i)** The Conjugate Roots Theorem only applies if the coefficients are real. In this case, this equation is of the form \(az^2+bz+c=0\), but \(c\) is not real.

**(ii)**

\begin{align}[z-(2+i)][z-(a+bi)]=z^2-2z+1-2i=0\end{align}

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

\begin{align}z[z-(a+bi)]-(2+i)[z-(a+bi)]=z^2-2z+1-2i\end{align}

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

\begin{align}z^2-(a+bi)z-2z+2(a+bi)-iz+i(a+bi)=z^2-2z+1-2i\end{align}

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

\begin{align}z^2-az-biz-2z+2a+2bi-iz+ai-b=z^2-2z+1-2i\end{align}

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

\begin{align}z^2+(-a-bi-2-i)z+(2a+2bi+ai-b)=z^2-2z+1-2i\end{align}

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

\begin{align}z^2+[(-a-2)+(-b-1)i]z+[(2a-b)+(2b+a)i]=z^2-2z+1-2i\end{align}

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

\begin{align}2a-b&=1\\2b+a&=-2\end{align}

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

\begin{align}b&=2a-1\\2b+a&=-2\end{align}

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

\begin{align}2(2a-1)+a=-2\end{align}

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

\begin{align}4a-2+a=-2\end{align}

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

\begin{align}5a=0\end{align}

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

\begin{align}a=0\end{align}

and

\begin{align}b&=2a-1\\&=2(0)-1\\&=-1\end{align}

Therefore, the other root is \(z_2=-i\).

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Question 8

A car is slowing down as it approaches a red light.

The distance \(D\) (in metres) that the car is from the red light at time \(t\) (in seconds) is given by:

\begin{align}D(t)=-4t^2+25t+30\end{align}

where \(t=0\) corresponds to when the driver started braking.

**(a)** How far was the car from the red light when the driver started braking?

##
Answer

\(30\mbox{ m}\)

##
Solution

\begin{align}D(0)&=-4(0^2)+25(0)+30\\&=30\mbox{ m}\end{align}

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**(b)** It can be shown that the derivative of \(D\) with respect to time is the car’s *speed* \(v\) at time \(t\), i.e.

\begin{align}v=\frac{dD}{dt}\end{align}

**(i)** Write an expression for \(v\) as a function of \(t\).

**(ii)** Using part **(i)**, or otherwise, find the car’s speed at the moment the driver started braking.

**(iii)** How long did it take the car to come to a complete stop?

**(iv) **How far had the car travelled in total while braking?

##
Answer

**(i)** \(v(t)=-8t+25\)

**(ii)** \(25\mbox{ m/s}\)

**(iii)** \(3.125\mbox{ s}\)

**(iv)** \(69.0625\mbox{ m}\)

##
Solution

**(i)**

\begin{align}v&=\frac{dD}{dt}\\&=-8t+25\end{align}

**(ii)**

\begin{align}v(0)&=-8(0)+25\\&=25\mbox{ m/s}\end{align}

**(iii)**

\begin{align}v(t)=0\end{align}

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

\begin{align}-8t+25=0\end{align}

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

\begin{align}t&=\frac{25}{8}\\&=3.125\mbox{ s}\end{align}

**(iv)**

\begin{align}D(3.125)&=-4(3.125)^2+25(3.125)+30\\&=69.0625\mbox{ m}\end{align}

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**(c)** It can also be shown that the derivative of \(v\) with respect to time is the car’s *acceleration *\(a\) at time \(t\), i.e.

\begin{align}a=\frac{dv}{dt}\end{align}

**(i)** Show that the car’s acceleration during the braking process was a constant value of \(-k\mbox{ m/s}^2\), where \(k\in\mathbb{N}\).

**(ii)** Explain the significance of the negative sign in the acceleration from part **(i)**.

##
Answer

**(i)** The answer is already in the question!

**(ii)** The car’s speed is decreasing with time.

##
Solution

**(i)**

\begin{align}a&=\frac{dv}{dt}\\&=-8\mbox{ m/s}^2\end{align}

as required.

**(ii) **The car’s speed is decreasing with time (i.e. it is decelerating.)

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Question 9

A sphere, a cone and a cylinder all have a diameter of \(1.5\mbox{ m}\) and share equal heights when positioned as shown below.

**(a)** Find the volume of the sphere in terms of \(\pi\).

##
Answer

\(4.5\pi\mbox{ m}^3\)

##
Solution

\begin{align}V_{sp}&=\frac{4}{3}\pi r^3\\&=\frac{4}{3}\pi (1.5^3)\\&=4.5\pi\mbox{ m}^3\end{align}

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**(b)** Find the *total* surface area of the cylinder in terms of \(\pi\).

##
Answer

\(13.5\pi\mbox{ m}^2\)

##
Solution

\begin{align}A_{cy}&=2\pi rh+2\pi r^2\\&=2\pi(1.5)(3)+2\pi(1.5^2)\\&=9\pi+4.5\pi\\&=13.5\pi\mbox{ m}^2\end{align}

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**(c)** Calculate the ratio of the total surface area of the sphere and cylinder in its simplest form \(a:b\), where \(a,b\in\mathbb{N}\) and \(a>b\).

##
Answer

\(27:18\)

##
Solution

\begin{align}A_{cy}=13.5\pi\mbox{ m}^2\end{align}

and

\begin{align}A_{sp}&=4\pi r^2\\&=4\pi(1.5^2)\\&=9\pi\mbox{ m}^2\end{align}

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

\begin{align}\mbox{Ratio}&=13.5\pi:9\pi\\&=27:18\end{align}

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**(d)** Calculate the ratio of the volume of all three objects in its simplest form \(a:b:c\), where \(a,b,c\in\mathbb{N}\) and \(a>b>c\).

##
Answer

\(27:18:11\)

##
Solution

\begin{align}V_{sp}=4.5\pi\mbox{m^3} \end{align}

and

\begin{align}V_{cy}&=\pi r^2h\\&=\pi(1.5^2)(3)\\&=6.75\pi\mbox{ m^3}\end{align}

and

\begin{align}V_{co}&=\frac{1}{3}\pi r^2h\\&=\frac{1}{3}\pi(1.5^2)(3)\\&=2.25\pi\mbox{ m^3}\end{align}

\begin{align}\mbox{Ratio}&=6.75\pi:4.5\pi:2.25\pi\\&=27:18:11\end{align}

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