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Mock Exams

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.

$$270.60\mbox{ euro}$$

\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.

$$7.48\mbox{ euro}$$

\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.

$$60\mbox{ euro}$$

\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.

$$37.5\%$$

\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.

(i) $$\sqrt{20}$$

(ii) $$-2+i$$

(iii)

(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 $$z^2-8z+20=0$$.

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

\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}

$$x=7$$

\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}

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

\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.

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

(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}

$$x<2$$

\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}

$$x=\dfrac{1}{2}$$

\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.

(i) $$T_1=6$$ and $$T_4=18$$

(ii) $$T_n=4n+2$$

(iii) $$n=20$$

(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.

$$9$$

\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

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

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

$$-2x+2$$

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

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

$$(1,4)$$

\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.

$$y=4$$

\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.

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

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

(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.

$$13{,}800\mbox{ euro}$$

\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.

(i) $$3.66\%$$

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

(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 $$0\leq t\leq3$$.

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

(ii) $$z_2=-i$$

(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?

$$30\mbox{ m}$$

\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?

(i) $$v(t)=-8t+25$$

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

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

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

(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).

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

(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 $$\pi$$.

$$4.5\pi\mbox{ m}^3$$

\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 $$\pi$$.

$$13.5\pi\mbox{ m}^2$$

\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 $$a>b$$.

$$27:18$$

\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 $$a>b>c$$.

$$27:18:11$$

\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}