HL Mock Exam A

Question 1

Consider the graph of the lines \(l_1\) and \(l_2\) below.

**(a)** State whether each line has a positive or negative slope.

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Answer

\(l_1\) has a positive slope and \(l_2\) has a negative slope.

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Solution

\(l_1\) has a positive slope and \(l_2\) has a negative slope.

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**(b) **The equation of the line \(l_1\) is \(y=2x+1\).

**(i)** Write down the equation of *any* line that is parallel to \(l_1\).

\(l_1\) passes through the point \((-3,k)\).

**(ii)** Find the value of \(k\).

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Answer

**(i)** \(y=2x+3\)

**(ii)** \(k=-5\)

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Solution

**(i)** \(y=2x+3\)

**(ii)**

\begin{align}k&=2(-3)+1\\&=-5\end{align}

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**(c) **The line \(l_2\) contains the points \(A(3,1)\) and \(B(-4,0)\).

Find \(|AB|\). Write your answer in surd form.

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Answer

\(\sqrt{50}\mbox{ units}\)

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Solution

\begin{align}|AB|&=\sqrt{(x_2-x_1)^2-(y_2-y_1)^2}\\&=\sqrt{(-4-3)^2+(0-1)^2}\\&=\sqrt{49+1}\\&=\sqrt{50}\mbox{ units}\end{align}

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**(d) **The equation of the line \(l_2\) is \(y=-4x+3\).

Find the intersection point of \(l_1\) and \(l_2\).

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Answer

\(\left(\dfrac{1}{3},\dfrac{5}{3}\right)\)

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Solution

\begin{align}y=2x+1\end{align}

\begin{align}y=-4x+3\end{align}

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

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

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

\begin{align}x&=\frac{2}{6}\\&=\frac{1}{3}\end{align}

and

\begin{align}y&=2x+1\\&=2\left(\frac{1}{3}\right)+1\\&=\frac{5}{3}\end{align}

Therefore, the intersection point is \(\left(\dfrac{1}{3},\dfrac{5}{3}\right)\).

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Question 2

A class \(U\) of \(25\) students were asked if they own a laptop \(L\) and/or a tablet \(T\).

\(4\) students said that own had a laptop *only*.

**(a)** Which of the letters \(A\), \(B\) and \(C\) is therefore equal to \(4\)?

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Answer

\(B\)

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Solution

\(B\)

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**(b) **We also know that \(60\%\) of the students own a laptop.

Find the number of students who own both a laptop and a tablet.

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Answer

\(11\)

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Solution

\begin{align}25\times0.6-4=11\end{align}

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**(c) ***“\(\#(L’)\) is equal to the number of students who own a tablet only.”*

Is this statement correct? Explain why.

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Answer

No as \(\#(L’)\) is equal to the number of students who own a tablet only plus the number of students who own neither.

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Solution

No as \(\#(L’)\) is equal to the number of students who own a tablet only plus the number of students who own neither.

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Question 3

Consider the following inequality:

\begin{align}3-x<1\end{align}

where \(x\in\mathbb{N}\).

**(a)** State any value of \(x\) which satisfies this inequality.

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Answer

\(x=3\)

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Solution

\(x=3\)

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**(b)** Solve the inequality.

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Answer

\(x>2\)

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Solution

\begin{align}3-x<1\end{align}

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

\begin{align}-x<1-3\end{align}

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

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

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

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

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**(c)** *“If we allowed \(x\) to be integers rather than just natural numbers, the solution to the inequality would still be the same.”*

Explain why this statement is correct.

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Answer

Since \(x>2\), only positive integers, i.e. natural numbers, are possible values of \(x\).

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Solution

Since \(x>2\), only positive integers, i.e. natural numbers, are possible values of \(x\).

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Question 4

Jacob wishes to travel from \(A\) to \(B\) by car.

However, there is no direct route from \(A\) to \(B\).

Instead, he must travel \(1.2 \mbox{ km}\) from \(A\) to \(C\), turn \(90^{\circ}\) left at \(C\) and then travel \(700\mbox{ m}\) from \(C\) to \(B\).

**(a)** Find the distance Jacob will travel in his car. Write your answer in kilometers.

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Answer

\(1.9\mbox{ km}\)

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Solution

\begin{align}d&=1.2+0.7\\&=1.9\mbox{ km}\end{align}

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**(b)** Find the straight-line distance between \(A\) and \(B\). Write your answer in metres correct to two decimal places.

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Answer

\(1{,}389.24\mbox{ m}\)

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Solution

\begin{align}|AB|&=\sqrt{1.2^2+0.7^2}\\&=1.389244…\mbox{ km}\\&\approx1{,}389.24\mbox{ m}\end{align}

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**(c) **Calculate the angle \(\theta\) in the diagram above. Give your answer in degrees, correct to the nearest degree.

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Answer

\(60^{\circ}\)

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Solution

\begin{align}\tan\theta=\frac{1.2}{0.7}\end{align}

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

\begin{align}\theta&=\tan^{-1}\left(\frac{1.2}{0.7}\right)\\&\approx60^{\circ}\end{align}

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**(d) **Jacob’s journey ended up taking a total of \(1.2\) minutes.

Find Jacob’s average speed for the journey. Write your answer in kilometres per hour.

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Answer

\(95\mbox{ km/hr}\)

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Solution

\begin{align}s&=\frac{d}{t}\\&=\frac{1.9}{1.2/60}\\&=95\mbox{ km/hr}\end{align}

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Question 5

Consider the following shape.

**(a)** Find an expression for the perimeter of this shape in terms of \(x\). Write this expression in the form \((a+\sqrt{b})x\), where \(a,b,\in\mathbb{N}\).

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Answer

\((4+\sqrt{2})x\)

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Solution

\begin{align}P&=2x+x+x+\sqrt{x^2+x^2}\\&=4x+\sqrt{2}x\\&=(4+\sqrt{2})x\end{align}

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**(b) **The area of this shape is \(14\mbox{ units}^2\).

Find the value of \(x\). Write your answer correct to two decimal places.

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Answer

\(3.06\mbox{ units}\)

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Solution

\begin{align}A=14\end{align}

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

\begin{align}x^2+\frac{1}{2}x^2=14\end{align}

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

\begin{align}\frac{3}{2}x^2=14\end{align}

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

\begin{align}x&=\sqrt{\frac{28}{3}}\\&\approx3.06\mbox{ units}\end{align}

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Question 6

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

\begin{align}8x^2-18y^2\end{align}

**(i)** If \(x=-3\) and \(y=1\), work out the value of the value of the expression.

**(ii)** Fully factorise the expression.

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Answer

**(i)Â **\(54\)

**(ii)** \(2(2x+3y)(2x-3y)\)

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Solution

**(i)**

\begin{align}8x^2-18y^2&=8(-3)^2-18(1^2)\\&=54\end{align}

**(ii)**

\begin{align}8x^2-18y^2&=2(4x^2-9y^2)\\&=2[(2x)^2-(3y)^2]\\&=2(2x+3y)(2x-3y)\end{align}

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

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

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Answer

\(x=\dfrac{1}{2}\) or \(x=-3\)

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Solution

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

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

\begin{align}(2x-1)(x+3)=0\end{align}

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

\(2x-1=0\) or \(x+3=0\)

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

\(x=\dfrac{1}{2}\) or \(x=-3\)

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Question 7

Brandon purchased a video games console for \(300\mbox{ euro}\).

**(a)** This selling price included VAT at \(23\%\).

Find the VAT that Brandon paid. Write your answer correct to the nearest cent.

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Answer

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

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

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Solution

\begin{align}\mbox{VAT}&=300\frac{300}{1.23}\\&=56.10\mbox{ euro}\end{align}

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**(b)** \(1\mbox{ euro}=160\mbox{ yen}\).

Find the selling price of the console in yen.

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Answer

\(48{,}000\mbox{ yen}\)

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Solution

\begin{align}300\times160=48{,}000\mbox{ yen}\end{align}

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**(c)** The console depreciates in value by \(15\%\) each year.

**(i)** Find the value of the console \(5\) years after Brandon purchased it, correct to the nearest cent.

**(ii)** Find the percentage loss Brandon would make if he sold it at this depreciated value at the end of the 5th year, correct to the nearest percent.

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Answer

**(i)** \(133.11\mbox{ euro}\)

**(ii)** \(56\%\)

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Solution

**(i)**

\begin{align}300(0.85)^5\approx133.11\mbox{ euro}\end{align}

**(ii)**

\begin{align}\frac{300-133.11}{300}\times100\approx56\%\end{align}

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Question 8

Below are the ages (in years) of everyone in a small classroom (including the teacher!):

\begin{align}12&&11&&12&&10&&12&&10&&38\end{align}

**(a)** Find the range of these ages.

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Answer

\(28\mbox{ years}\)

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Solution

\begin{align}38-10=28\mbox{ years}\end{align}

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**(b)** Which of the seven values would be considered an *outlier*?

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Answer

\(38\)

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Solution

\(38\)

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**(c)** Find the mean of these ages.

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Answer

\(15\mbox{ years old}\)

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Solution

\begin{align}\frac{12+11+12+10+12+10+38}{7}=15\mbox{ years old}\end{align}

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**(d)** Explain why the mean would be a bad choice to measure the average in this case.

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Answer

Six of the seven data values are less than the mean.

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Solution

Six of the seven data values are less than the mean.

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Question 9

The first three terms of a sequence are as follows:

\begin{align}11,15,19\end{align}

**(a)** Is this sequence linear, quadratic or exponential?

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Answer

Linear

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Solution

Linear

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**(b)** What is the next term in the sequence?

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Answer

\(23\)

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Solution

\(23\)

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**(c)** \(T_n\) refers to the \(n\)th term of a sequence.

**(i)** Write an expression for \(T_n\) for this sequence.

**(ii)** Using part (i), or otherwise, find the value of \(n\) for the term \(103\) in the sequence.

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Answer

**(i)** \(T_n=4n+7\)

**(ii)** \(24\)

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Solution

**(i)** \begin{align}T_n&=11+4n-4\\&=4n+7\end{align}

**(ii)**

\begin{align}T_n=103\end{align}

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

\begin{align}4n+7=103\end{align}

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

\begin{align}n&=\frac{103-7}{4}\\&=24\end{align}

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Question 10

Three tennis balls of radius \(7\mbox{ cm}\) are placed inside their cylindrical container as shown.

**(a)** Find the surface area of one of the tennis balls. Write your answer in terms of \(\pi\).

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Answer

\(196\pi\mbox{ cm}^2\)

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Solution

\begin{align}A&=4\pi r^2\\&=4\pi(7^2)\\&=196\pi\mbox{ cm}^2\end{align}

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No gap is left between the top of the highest tennis ball and the top of the container.

**(b)** Find the height of the container.

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Answer

\(42\mbox{ cm}\)

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Solution

\begin{align}h&=6r\\&=6(7)\\&=42\mbox{ cm}\end{align}

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**(c)** Find the volume of the container. Write your answer correct to \(1\) decimal place.

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Answer

\(6465.4\mbox{ cm}^3\)

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Solution

\begin{align}V&=\pi r^2h\\&=\pi(7^2)(42)\\&\approx6465.4\mbox{ cm}^3\end{align}

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**(d)** Find the volume of air in the container. Give your answer correct to one deicmal place.

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Answer

\(2155.1\mbox{ m}^3\)

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Solution

\begin{align}V&=6465.4-3\left(\frac{4}{3}\pi r^3\right)\\&=6465.4-4\pi(7^3)\\&\approx2155.1\mbox{ m}^3\end{align}