L.C. MATHS

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Past Papers

## Question 1

(a) The table below gives some details on the number of different types of student in a
university. There are $$22{,}714$$ students in the university in total.

Aged 23 or younger Aged 24 or older Total

$$12{,}785$$

$$2{,}922$$

$$15{,}707$$

$$1{,}353$$

Total

$$8{,}576$$

$$22{,}714$$

(i) Fill in the three missing values to complete the table above.

(ii) One student is picked at random from the students in the university.
Let $$O$$ be the event that the student is $$24$$ years old, or older.
Let $$U$$ be the event that the student is an undergraduate.
Are the events $$O$$ and $$U$$ independent? Justify your answer.

(i)

Aged 23 or younger Aged 24 or older Total

$$12{,}785$$

$$2{,}922$$

$$15{,}707$$

$$1{,}353$$

$$5{,}654$$

$$7{,}007$$

Total

$$14{,}138$$

$$8{,}576$$

$$22{,}714$$

(ii) Since $$P(O)\times P(U)\neq P(O\cap U)$$, they are not independent.

Solution

(i)

Aged 23 or younger Aged 24 or older Total

$$12{,}785$$

$$2{,}922$$

$$15{,}707$$

$$1{,}353$$

$$5{,}654$$

$$7{,}007$$

Total

$$14{,}138$$

$$8{,}576$$

$$22{,}714$$

(ii)

\begin{align}P(O)=\frac{8{,}576}{22{,}714}\end{align}

and

\begin{align}P(U)=\frac{15{,}707}{22{,}714}\end{align}

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

\begin{align}P(O)\times P(U)&=\frac{8{,}576}{22{,}714}\times\frac{15{,}707}{22{,}714}\\&\approx0.261\end{align}

and

\begin{align}P(O\cap U)&=\frac{2{,}922}{22{,}714}\\&\approx0.129\end{align}

Since $$P(O)\times P(U)\neq P(O\cap U)$$, they are not independent.

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(b) Three people are picked at random from a class.
Find the probability that all three were born on the same day of the week.
Assume that the probability of being born on each day is the same.

$$\dfrac{1}{49}$$

Solution

\begin{align}7\times\left(\frac{1}{7}\times\frac{1}{7}\times\frac{1}{7}\right)=\frac{1}{49}\end{align}

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(c) There are $$b$$ boys and $$g$$ girls in a class, where $$b,g\in\mathbb{N}$$.
$$\dfrac{3}{5}$$ of the students in the class are girls.

$$4$$ boys and $$4$$ girls join the class.
One student is then picked at random from the whole class.
The probability that this student is a girl is now $$\dfrac{4}{7}$$.

Find the value of $$b$$ and the value of $$g$$.

$$b=8$$ and $$g=12$$

Solution

\begin{align}\frac{g}{b+g}=\frac{3}{5}\end{align}

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

\begin{align}3(b+g)=5g\end{align}

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

\begin{align}3b-2g=0\end{align}

and

\begin{align}\frac{g+4}{(b+4)+(g+4)}=\frac{4}{7}\end{align}

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

\begin{align}4(b+g+8)=7(g+4)\end{align}

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

\begin{align}4b-3g=-4\end{align}

We therefore have the following two equations for two unknowns:

\begin{align}3b-2g=0\end{align}

\begin{align}4b-3g=-4\end{align}

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

\begin{align}12b-8g=0\end{align}

\begin{align}12b-9g=-12\end{align}

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

\begin{align}-g=-12\end{align}

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

\begin{align}g=12\end{align}

and

\begin{align}3b-2g=0\end{align}

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

\begin{align}b&=\frac{2g}{3}\\&=\frac{2(12)}{3}\\&=8\end{align}

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## Question 2

(a) The points $$A (8,-4)$$ and $$B(-1,3)$$ are the endpoints of the line segment $$[AB]$$.

Find the coordinates of the point $$C$$, which divides $$[AB]$$ internally in the ratio $$4:1$$.

$$\left(\dfrac{4}{5},\dfrac{8}{5}\right)$$

Solution

\begin{align}C&=\left(\frac{bx_1+ax_2}{b+a},\frac{by_1+ay_2}{b+a}\right)\\&=\left(\frac{(1)(8)+(4)(-1)}{4+1},\frac{(1)(-4)+(4)(3)}{4+1}\right)\\&=\left(\frac{4}{5},\frac{8}{5}\right)\end{align}

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(b) The line $$l$$ has a slope of $$m$$ and contains the point $$(q,r)$$, where $$m,q,r\in\mathbb{R}$$ are all positive.
Find the co-ordinates of the point where $$l$$ cuts the $$y$$-axis, in terms of $$m$$, $$q$$ and $$r$$.

$$(0,r-mq)$$

Solution

\begin{align}y=mx+c\end{align}

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

\begin{align}r=mq+c\end{align}

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

\begin{align}c=r-mq\end{align}

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

\begin{align}(0,r-mq)\end{align}

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(c) The line $$k$$ has a slope of $$-2$$.
The line $$j$$ makes an angle of $$30^{\circ}$$ with $$k$$.

Find one possible value of the slope of the line $$j$$.
Give your answer in the form $$d+e\sqrt{f}$$, where $$d,e,f\in\mathbb{Z}$$.

$$8+5\sqrt{3}$$

Solution

\begin{align}\tan\theta=\pm\frac{m_1-m_2}{1+m_1m_2}\end{align}

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

\begin{align}\tan30^{\circ}=\frac{-2-m_2}{1+(-2)m_2}\end{align}

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

\begin{align}\frac{1}{\sqrt{3}}=\frac{-2-m_2}{1-2m_2}\end{align}

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

\begin{align}1-2m_2=-2\sqrt{3}-\sqrt{3}m_2\end{align}

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

\begin{align}m_2&=\frac{1+2\sqrt{3}}{2-\sqrt{3}}\\&=\frac{1+2\sqrt{3}}{2-\sqrt{3}}\times\frac{2+\sqrt{3}}{2+\sqrt{3}}\\&=\frac{2+\sqrt{3}+4\sqrt{3}+6}{4+2\sqrt{3}-2\sqrt{3}-3}\\&=8+5\sqrt{3}\end{align}

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## Question 3

(a) The circle $$c$$ has equation $$x^2+y^2-2x+8y+k=0$$, where $$k\in\mathbb{R}$$.
The radius of $$c$$ is $$5\sqrt{3}$$.

Find the value of $$k$$.

$$k=-58$$

Solution

\begin{align}\sqrt{g^2+f^2-c}=5\sqrt{3}\end{align}

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

\begin{align}\sqrt{(-1)^2+4^2-k}=5\sqrt{3}\end{align}

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

\begin{align}1+16-k=75\end{align}

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

\begin{align}k=-58\end{align}

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(b) The circle $$(x-5)^2+(y+2)^2=20$$ has a tangent at the point $$(9,-4)$$.
Find the slope of this tangent.

$$2$$

Solution

The circle has a centre $$(5,-2)$$. The slope of the line going through this point and $$(9,-4)$$ is

\begin{align}m&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{-4-(-2)}{9-5}\\&=-\frac{1}{2}\end{align}

The tangent therefore has a slope of $$2$$.

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(c) Two circles each have both the $$x$$-axis and the $$y$$-axis as tangents, and each contains the point $$(1,-8)$$, as shown in the diagram.

Find the equation of each of these circles.

$$(x-5)^2+(y+5)^2=25$$ and $$(x-13)^2+(y+13)^2=169$$

Solution

\begin{align}(x-h)^2+(y-k)^2=r^2\end{align}

Centre: $$(h,k)=(r,-r)$$

Point on circumference: $$(x,y)=(1,-8)$$

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

\begin{align}(1-r)^2+(-8-(-r))^2=r^2\end{align}

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

\begin{align}1-2r+r^2+64-16r+r^2=r^2\end{align}

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

\begin{align}r^2-18r+65=0\end{align}

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

\begin{align}(r-5)(r-13)=0\end{align}

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

$$r=5$$ or $$r=13$$

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

\begin{align}(x-5)^2+(y+5)^2=25\end{align}

or

\begin{align}(x-13)^2+(y+13)^2=169\end{align}

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## Question 4

(a)

(i) Prove that $$\tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A \tan B}$$.

(ii) Write $$\tan 15^{\circ}$$ in the form $$\dfrac{\sqrt{a}-1}{\sqrt{a}+1}$$, where $$a\in\mathbb{N}$$.

(ii) $$\dfrac{\sqrt{3}-1}{\sqrt{3}+1}$$

Solution

(i)

\begin{align}\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\end{align}

Replacing $$B$$ with $$-B$$:

\begin{align}\tan(A+(-B))=\frac{\tan A+\tan (-B)}{1-\tan A\tan (-B)}\end{align}

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

\begin{align}\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}\end{align}

where we have used $$\tan(-B)=-\tan B$$.

(ii)

\begin{align}\tan15^{\circ}&=\tan(60^{\circ}-45^{\circ})\\&=\frac{\tan 60^{\circ}-\tan 45^{\circ}}{1+(\tan 60^{\circ})(\tan 45^{\circ})}\\&=\frac{\sqrt{3}-1}{1+(\sqrt{3})(1)}\\&=\frac{\sqrt{3}-1}{\sqrt{3}+1}\end{align}

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(b) The triangle $$ABC$$ is shown in the diagram below.

$$|AC|=|BC|$$ and $$|\angle ACB|=45^{\circ}$$.

$$|AB|=10\sqrt{2-\sqrt{2}}$$, as shown.

Find the length $$|AC|$$.

$$10$$

Solution

As the triangle is isosceles, the other two angles are

\begin{align}\frac{180^{\circ}-45^{\circ}}{2}=67.5^{\circ}\end{align}

$\,$

Sine Rule

\begin{align}\frac{|AC|}{\sin67.5^{\circ}}=\frac{10\sqrt{2-\sqrt{2}}}{\sin45^{\circ}}\end{align}

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

\begin{align}|AC|&=\frac{10\sqrt{2-\sqrt{2}}(\sin67.5^{\circ})}{\sin45^{\circ}}\\&=10\end{align}

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## Question 5

(a) A survey on remote learning was carried out on a random sample of $$400$$ students.
$$135$$ of the students preferred remote learning over in-person learning.
For parts (a)(i), (a)(ii), and (a)(iii), give all solutions as decimals, correct to $$4$$ decimal places.

(i) Work out the proportion of the sample that preferred remote learning.

(ii) Use the margin of error $$\left(\frac{1}{\sqrt{n}}\right)$$ to create a $$95\%$$ confidence interval for the proportion of the population that preferred remote learning.

(iii) Using the proportion from part (a)(i), create a $$95\%$$ confidence interval for this population proportion that is more accurate than the $$95\%$$ confidence interval based on the margin of error.

(i) $$0.3375$$

(ii) $$[0.2875,0.3875]$$

(iii) $$[0.2912,0.3838]$$

Solution

(i)

\begin{align}\hat{p}&=\frac{135}{400}\\&=0.3375\end{align}

(ii)

\begin{align}\hat{p}\pm\mbox{error }&=0.3375\pm\frac{1}{\sqrt{400}}\\&=0.3375\pm0.05\end{align}

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

Confidence Interval: $$[0.2875,0.3875]$$

(iii)

\begin{align}\hat{p}\pm 1.96\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}&=0.3375\pm1.96\sqrt{\frac{(0.3375)(1-0.3375)}{400}}\\&=0.3375\pm0.04633…\end{align}

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

Confidence Interval: $$[0.2912,0.3838]$$

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(b) In $$2019$$, people with a pre-pay mobile phone plan spent an average (mean) of €$$20.79$$ on
their mobile phone each month (source: www.comreg.ie).

In $$2021$$, some students carried out a survey to see if this figure had changed.
They surveyed a random sample of $$500$$ people with pre-pay mobile phone plans.
For this sample, the mean amount spent per month was €$$22.16$$ and the standard deviation was €$$8.12$$.

Carry out a hypothesis test at the $$5\%$$ level of significance to see if this shows a change in the mean monthly spend on mobile phones for people with a pre-pay plan.

The average amount has changed.

Solution

Null Hypothesis: Average amount has not changed.

Alternative Hypothesis: Average amount has changed.

$\,$

Calculations:

\begin{align}\hat{p}\pm 1.96\frac{\sigma}{\sqrt{n}}&=22.16\pm1.96\frac{8.12}{\sqrt{500}}\\&=22.16\pm0.7117…\end{align}

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

Confidence Interval: $$[21.44…,22.87…]$$

$\,$

Conclusion: As $$20.79$$ lies outside of this interval, the average amount has changed.

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## Question 6

(a) Construct the circumcentre of the triangle $$XYZ$$ shown below, using only a compass and straight edge. Label the circumcentre $$C$$. Show your construction lines clearly.

Solution
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(b) The points $$A$$, $$B$$, $$C$$ and $$D$$ lie on a circle, as shown in the diagram (not to scale).

$$[AB]$$ is a diameter of the circle.
$$|\angle DAC|=40^{\circ}$$, as shown.
The triangle $$ABD$$ is isosceles.
Find $$|\angle ADC|$$.

$$95^{\circ}$$

Solution

Since $$|AB|$$ is a diameter, $$\angle ADB$$ is a right angle.

Therefore, since we’re told that $$ABD$$ is an isosceles triangle, both $$|\angle DAB|$$ and $$|\angle ABD|$$ are $$45^{\circ}$$.

Since $$|\angle ABD|$$ and $$|\angle ACD|$$ are on the same arc, $$|\angle ACD|$$ must also be $$45^{\circ}$$ and therefore

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The diagram shows the triangle $$PQR$$ (not to scale).

The angle at $$R$$ is greater than $$90^{\circ}$$.

$$k$$ is the circumcircle of $$PQR$$, as shown,
and $$O$$ is the circumcentre.
$$O$$ is not shown in the diagram.

(c) Prove that $$O$$ cannot be inside the triangle $$PQR$$.

If you are proving this by contradiction, your first line should be:

“Assume that $$O$$ is inside the triangle $$PQR$$.”

Solution

Assume that $$O$$ is inside the triangle $$PQR$$.

As $$|PS|$$ is a diameter, $$|\angle PRS|$$ is a right angle.

According to the diagram, $$|\angle PRQ|<|\angle PRS|$$, i.e. $$|\angle PRQ|$$ is less than $$90^{\circ}$$.

However, the question states that $$|\angle PRQ|$$ is greater than $$90^{\circ}$$ – a contradiction.

Therefore, $$O$$ cannot be inside the triangle $$PQR$$.

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## Question 7

A company makes and sells candles of different shapes and sizes.

(a) A candle in the shape of a cylinder has a diameter of $$10\mbox{ cm}$$ and a volume of $$450\pi\mbox{ cm}^3$$.
Work out the height of this candle.

$$18\mbox{ cm}$$

Solution

\begin{align}\pi r^2h=450\pi\end{align}

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

\begin{align}\pi (5^2)h=450\pi\end{align}

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

\begin{align}h&=\frac{450}{5^2}\\&=18\mbox{ cm}\end{align}

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(b) A small candle in the shape of a cone has a volume of $$12\pi\mbox{ cm}^3$$.
A large candle, also in the shape of a cone, has a volume of $$150\pi\mbox{ cm}^3$$.
The height of the small candle is $$h$$, and the height of the large candle is $$2h$$.
The large candle has a radius that is $$k$$ times that of the small candle, where $$k\in\mathbb{R}$$.

Work out the value of $$k$$.

$$2.5$$

Solution

Small Candle

\begin{align}\frac{1}{3}\pi r^2 h=12\pi\end{align}

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

\begin{align}r^2h=36\end{align}

$\,$

Large Candle

\begin{align}\frac{1}{3}\pi (kr)^2 (2h)=150\pi\end{align}

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

\begin{align}k^2(r^2h)=225\end{align}

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

\begin{align}k^2(36)=225\end{align}

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

\begin{align}k&=\sqrt{\frac{225}{36}}\\&=2.5\end{align}

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(c) A third conical candle has its curved surface area covered in cloth.

The net of the cloth is shown in the diagram below (not to scale).
This net covers the cone perfectly, with no overlapping material.

In the diagram, $$|\angle BOA|=216^{\circ}$$, as shown. $$|OA|=8\mbox{ cm}$$.

Find the length of the arc from $$B$$ to $$A$$, in terms of $$\pi$$, and hence find the radius of the cone.

$$l=9.6\pi\mbox{ cm}$$ and $$r=4.8\mbox{ cm}$$

Solution

\begin{align}l&=2\pi r\left(\frac{\theta}{360^{\circ}}\right)\\&=2\pi (8)\left(\frac{216^{\circ}}{360^{\circ}}\right)\\&=9.6\pi\mbox{ cm}\end{align}

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

\begin{align}2\pi r=9.6\pi\end{align}

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

\begin{align}r&=\frac{9.6}{2}\\&=4.8\mbox{ cm}\end{align}

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(d)

(i) A spherical ball of wax is used as a candle. The radius of the sphere is $$2.7\mbox{ cm}$$.
Find the volume of the sphere, correct to $$3$$ decimal places.

(ii) A horizontal slice is cut off this sphere so that the candle will balance on level surfaces.
The area of the circular base of this candle is $$5.4\mbox{ cm}^2$$.

Find the value of $$l$$, the vertical distance from the top of the candle to where the cut is made.
As shown in the diagram on the right, $$l>2.7\mbox{ cm}$$.

Give your answer in $$\mbox{cm}$$, correct to the nearest $$\mbox{mm}$$.

(i) $$82.448\mbox{ cm}^3$$

(ii) $$5.1\mbox{ cm}$$

Solution

(i)

\begin{align}V&=\frac{4}{3}\pi r^3\\&=\frac{4}{3}\pi(2.7^3)\\&\approx82.448\mbox{ cm}^3\end{align}

(ii)

\begin{align}\pi r^2=5.4\end{align}

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

\begin{align}r&=\sqrt{\frac{5.4}{\pi}}\\&=1.311…\end{align}

Let $$x$$ be vertical distance from the sphere’s centre to the horizontal slice.

\begin{align}x^2+(1.311…)^2=2.7^2\end{align}

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

\begin{align}x&=\sqrt{2.7^2-(1.311…)^2}\\&=2.36…\end{align}

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

\begin{align}l&=2.7+x\\&=2.7+2.36…\\&\approx 5.1\mbox{ cm}\end{align}

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(e) Part of the logo of the company is shown below.
$$ABCD$$ is a square, with sides of length $$30\mbox{ cm}$$.
The points $$E$$ and $$F$$ are the midpoints of $$[AB]$$ and $$[AD]$$, respectively.
The lines $$EC$$ and $$FB$$ are perpendicular, and meet at the point $$O$$.

Using similar triangles or trigonometry, find the length $$|EO|$$.
Give your answer in the form $$a\sqrt{b}\mbox{ cm}$$, where $$a,b\in\mathbb{N}$$.

$$3\sqrt{5}\mbox{ cm}$$

Solution

\begin{align}|AF|=|FD|=|AE|=|EB|=15\mbox{ cm}\end{align}

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

\begin{align}|EC|&=\sqrt{30^2+15^2}\\&=15\sqrt{5}\mbox{ cm}\end{align}

Since $$|\angle EOB|=|\angle EBC|$$ and since triangles $$EOB$$ and $$EBC$$ share $$|\angle BEC|$$, the other angle in both triangles must also be the same.

Therefore, triangles $$EOB$$ and $$EBC$$ are similar.

\begin{align}\frac{|EO|}{|EB|}=\frac{|EB|}{|EC|}\end{align}

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

\begin{align}|EO|&=\frac{|EB|^2}{|EC|}\\&=\frac{15^2}{15\sqrt{5}}\\&=\frac{15}{\sqrt{5}}\\&=\frac{3\times(\sqrt{5})^2}{\sqrt{5}}\\&=3\sqrt{5}\mbox{ cm}\end{align}

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## Question 8

(a) Jena is researching fuel consumption in cars. She finds the following data for the number of miles per gallon ($$\mbox{m/g}$$) for eight different cars, labelled A to H, when driving in the city and on the motorway:

Car City ($$\mbox{m/g}$$) Motorway ($$\mbox{m/g}$$)

A

$$22$$

$$34$$

B

$$27$$

$$38$$

C

$$24$$

$$34$$

D

$$16$$

$$27$$

E

$$15$$

$$24$$

F

$$21$$

$$30$$

G

$$\mathbf{30}$$

$$\mathbf{40}$$

H

$$\mathbf{17}$$

$$\mathbf{30}$$

(i) The scatterplot below shows this data for cars A to F.

Using the data in the table above, plot and label points to represent cars G and H on the scatterplot below.

(ii) On the scatterplot, draw the line of best fit for the data, by eye.

(iii) Two other cars, K and L, have the miles per gallon values given in the following table.

Use your line of best fit on the scatterplot to fill in an estimate for each of the two missing values in the table below. Show your work on the scatterplot.

Car City ($$\mbox{m/g}$$) Motorway ($$\mbox{m/g}$$)

K

$$20$$

L

$$60$$

(iv) Based on the data given, would you be more confident in the value you estimated for K or for L? Give a reason for your answer.

(v) Find the value of $$r$$, the correlation coefficient between city and motorway miles per gallon. Use only the values for the $$8$$ cars A to H in the table on the previous page.
Give your answer correct to $$3$$ decimal places.

(i)

(ii)

(iii)

Car City ($$\mbox{m/g}$$) Motorway ($$\mbox{m/g}$$)

K

$$20$$

$$32$$

L

$$56$$

$$60$$

(iv) $$K$$ as it is in the neighbourhood of the other points.

(v) $$r\approx0.966$$

Solution

(i)

(ii)

(iii)

Car City ($$\mbox{m/g}$$) Motorway ($$\mbox{m/g}$$)

K

$$20$$

$$32$$

L

$$56$$

$$60$$

(iv) $$K$$ as it is in the neighbourhood of the other points.

(v) Calculator: $$r\approx0.966$$

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(b) The scatterplot shows some values of fuel consumption ($$F$$) for the given values of engine speed ($$S$$), for a particular car.
For the points in this scatterplot, $$F$$ can be closely approximated by a quadratic function of $$S$$.

$$r_{FS}$$ is the correlation coefficient between $$F$$ and $$S$$, based on the points in this scatterplot.
Give a reason why you might think that $$r_{FS}$$ is very close to $$0$$.

The line of best fit is close to being horizontal.

Solution

The line of best fit is close to being horizontal.

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(c) $$13$$ customers rated their experience in a garage, by giving a whole-number score out of $$100$$.
The mean score was $$52$$. The median score was $$54$$. No two scores were the same.

The data below shows the score for each of the $$13$$ customers (in no particular order).
Stephen gave a score of S and Mary gave a score of M, where S, M $$\in\mathbb{N}$$.
Find the least value and the greatest value that S could be.

\begin{align}46&&68&&24&&74&&42&&30&&61&&54&&28&&50&&57&&\mbox{S}&&\mbox{M}\end{align}

Least Value: $$55$$

Greatest Value: $$87$$

Solution

Least Value

The $$11$$ know numbers listed in order are

\begin{align}24&&28&&30&&42&&46&&50&&54&&57&&61&&68&&74\end{align}

The median of these $$11$$ numbers is therefore $$50$$.

The only way the median can increase to $$54$$ is if both $$S$$ and $$M$$ are larger than $$54$$.

Therefore, the smallest possible value of $$S$$ is $$55$$.

$\,$

Greatest Value

\begin{align}\frac{46+68+24+74+42+30+61+54+28+50+57+S+M}{13}=52\end{align}

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

\begin{align}\frac{534+S+M}{13}=52\end{align}

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

\begin{align}534+S+M=676\end{align}

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

\begin{align}S=142-M\end{align}

The smallest possible value of $$M$$ is $$55$$. Therefore, the largest possible value of $$S$$ is

\begin{align}S&=142-M\\&=142-55\\&=87\end{align}

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(d) John bought a car a number of years ago. The table below gives an estimate of the probability that each of the following three events happens to John’s car in the next year.

Event Probability

$$0.095$$

Timing belt goes

$$0.041$$

Air filters break

$$0.073$$

(i) If the head gasket blows, John will have to replace his car, at an estimated cost of €$$20{,}000$$. If the head gasket is replaced now, it will cost €$$1{,}450$$, and the probability that it blows in the next year will be reduced to $$0.005$$.

Based on these figures, use expected values to work out if it is worth replacing the head gasket now, or not.

(ii) Work out the probability that at least one of the events in the table above happens to John’s car this year, taking these events to be independent. Give your answer correct to $$3$$ decimal places.

(i) It is worth replacing it now.

(ii) $$0.195$$

Solution

(i)

Doesn’t replace

\begin{align}E(X)&=(0.095)(20{,}000)\\&=1{,}900\mbox{ euro}\end{align}

$\,$

Replaces

\begin{align}E(X)&=(0.005)(20{,}000)+1{,}450\\&=1{,}550\mbox{ euro}\end{align}

It is therefore worth replacing it now.

(ii)

\begin{align}P(\mbox{at least one})&=1-P(\mbox{none}
)\\&=1-(1-0.095)\times(1-0.041)\times(1-0.073)\\&\approx0.195\end{align}

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## Question 9

Oscar is taking some measurements and is using trigonometry to work out some angles, distances, and areas.

First, Oscar takes measurements of two adjacent triangular fields, Field 1 ($$ABC$$) and Field 2 ($$BDC$$), as shown in the diagram below (not to scale).
$$B$$ lies on the line $$AD$$.
$$|AB|=30\mbox{ m}$$.
$$|BD|=10\mbox{ m}$$.
$$|AC|=35\mbox{ m}$$.
$$|\angle CAD|=50^{\circ}$$.

Note: the angle $$ABC$$ is not a right angle.

(a) Find the area of Field 1 and, hence, find the area of Field 2.
Give each answer correct to the nearest $$\mbox{m}^2$$.

$$A_1=402\mbox{ m}^2$$ and $$A_2=134\mbox{ m}^2$$

Solution

\begin{align}A_1&=\frac{1}{2}|AC||AB|\sin A\\&=\frac{1}{2}(35)(30)(\sin 50^{\circ})\\&=402.1…\\&\approx402\mbox{ m}^2\end{align}

and

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

\begin{align}A_2&=A_{\mbox{total}}-A_1\\&=536.2…-402.1…\\&\approx134\mbox{ m}^2\end{align}

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(b) Find the length of the perimeter of Field 1.

$$93\mbox{ m}$$

Solution

\begin{align}|CB|&=\sqrt{|CA|^2+|AB|^2-2|CA|AB|\cos A}\\&=\sqrt{35^2+30^2-2(35)(3)(\cos50^{\circ})}\\&\approx 28\mbox{ m}\end{align}

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

\begin{align}\mbox{Perimeter}&=35+30+28\\&=93\mbox{ m}\end{align}

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Oscar is also watching an airplane, P, fly directly over his head. He is standing at the point O in the diagrams below. The $$x$$-axis is the horizontal ground and the $$y$$-axis runs vertically up from Oscar.
The airplane P is flying at a constant height of $$10\mbox{ km}$$ above the ground, as shown in the diagrams.

(c) Sound travels at a speed of roughly $$343$$ metres per second in air.

(i) As the airplane flies, its engine makes noise. It takes some time for this sound to reach Oscar. Use the information in Diagram 1 to show that it takes $$41$$ seconds for the sound the airplane makes at $$P_1$$ to reach Oscar, correct to the nearest second.

(ii) The airplane P is flying at a constant speed of $$255$$ metres per second. By the time Oscar hears the sound the airplane made at the point $$P_1$$ , the airplane has flown on to the point $$P_2$$, as shown in Diagram 2 above.

Work out the size of the angle marked $$\theta$$ in Diagram 2, correct to $$1$$ decimal place.

(ii) $$2.6^{\circ}$$

Solution

(i)

\begin{align}\cos 45^{\circ}&=\frac{10}{|P_1O|}\end{align}

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

\begin{align}|P_1O|&=\frac{10}{\cos 45^{\circ}}\\&=14.142…\mbox{ km}\\&\approx14{,}142\mbox{ m}\end{align}

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

\begin{align}t&=\frac{d}{v}\\&=\frac{14{,}142}{343}\\&\approx 41\mbox{ s}\end{align}

(ii)

\begin{align}|P_1P_2|&=vt\\&=(255)(41)\\&=10{,}455\mbox{ m}\end{align}

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

\begin{align}\theta&=\tan^{-1}\left(\frac{10{,}455-10{,}000}{10{,}000}\right)\\&\approx2.6^{\circ}\end{align}

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(d) $$P_3$$ and $$P_4$$ are two other points on the flightpath of airplane P.

By the time Oscar hears the sound the airplane made at the point $$P_3$$, the airplane has flown on to the point $$P_4$$, as shown in the diagram below (not to scale).
$$P_3$$ and $$P_4$$ are both a distance of $$d \mbox{ km}$$ from the $$y$$-axis.

(i) Explain briefly why the following equation holds:

\begin{align}\frac{\sqrt{100+d^2}}{0.343}=\frac{2d}{0.255}\end{align}

(ii) Solve the equation above to find the value of $$d$$, correct to $$1$$ decimal place.

(i) The left hand side is the time taken for the sound to go from $$P_3$$ to $$O$$. The right hand side is the time taken for the plane to go from from $$P_3$$ to $$P_4$$. These should be the same.

(ii) $$4.0\mbox{ km}$$

Solution

(i) The left hand side is the time taken for the sound to go from $$P_3$$ to $$O$$. The right hand side is the time taken for the plane to go from from $$P_3$$ to $$P_4$$. These should be the same.

(ii)

\begin{align}\frac{\sqrt{100+d^2}}{0.343}=\frac{2d}{0.255}\end{align}

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

\begin{align}\sqrt{100+d^2}&=\frac{2d(0.343)}{0.255}\\&=(2.69…)d\end{align}

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

\begin{align}100+d^2=(7.23…)d^2\end{align}

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

\begin{align}(6.23…)d^2=100\end{align}

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

\begin{align}d&=\sqrt{\frac{100}{6.23…}}\\&\approx4.0\mbox{ km}\end{align}

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## Question 10

(a) In an athletics competition, there were a number of heats of the $$1500\mbox{ m}$$ race.
In the heats, the times that it took the runners to complete the $$1500\mbox{ m}$$ were approximately normally distributed, with a mean time of $$225$$ seconds and a standard deviation of $$12$$ seconds.

(i) Find the percentage of runners in these heats who took more than $$240$$ seconds to run the $$1500\mbox{ m}$$.

(ii) The $$20\%$$ of runners with the fastest times qualified for the final.
Assuming the race times were normally distributed as described above, work out the time needed to qualify for the final, correct to the nearest second.

(i) $$10.56\%$$

(ii) $$215\mbox{ s}$$

Solution

(i)

\begin{align}z&=\frac{x-\mu}{\sigma}\\&=\frac{240-225}{12}\\&=1.25\end{align}

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

\begin{align}P(x>240)&=P(z>1.25)\\&=1-P(z<1.25)\\&=1-0.8944\\&=0.1056\end{align}

Therefore, the percentage of runners in these heats who took more than $$240$$ seconds to run the $$1500\mbox{ m}$$ is $$10.56\%$$.

(ii) When $$P=0.8$$, $$z\approx0.84$$ and therefore

\begin{align}0.84=\frac{225-x}{12}\end{align}

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

\begin{align}x&=225-12(0.84)\\&\approx215\mbox{ s}\end{align}

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(b) Sally takes part in a number of different races in the competition.
The probability that she makes a false start in any given race is $$5\%$$.
Find the probability that she makes her first false start in her fourth race.
Give your answer correct to $$4$$ decimal places.

$$0.0429$$

Solution

\begin{align}P&=0.95\times0.95\times0.95\times(1-0.95)\\&\approx0.0429\end{align}

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(c) $$20$$ relay teams took part in the competition. For any particular team, the probability that they drop the baton at some point during the competition is $$0.1$$.

Find the probability that at most 𝟐 teams drop the baton during the competition.
Give your answer correct to $$4$$ decimal places.

$$0.6769$$

Solution

\begin{align}P(\mbox{at most 2})&=P(\mbox{0 or 1 or 2)}\\&=P(0)\mbox{ or }P(1)\mbox{ or }P(2)\\&={20\choose0}0.9^{20}+{20\choose1}0.1^10.9^{19}+{20\choose2}0.1^20.9^{18}\\&\approx0.6769\end{align}

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(d) $$300$$ runners take part in a road race.
Each runner has a number, from $$1$$ to $$300$$ inclusive. No two runners have the same number.

Two runners are picked at random from the runners in this race.
Work out the probability that the sum of their numbers is $$101$$.

$$\dfrac{1}{897}$$

Solution

\begin{align}P&=\frac{100}{300}\times\frac{1}{299}\\&=\frac{1}{897}\end{align}

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(e) Sorcha ran two different marathons: the Windy Marathon and the Sunny Marathon.
The table below gives some details on the finishing times for the two marathons.
For each marathon, the finishing times of the runners were approximately normally distributed.

Sorcha’s position in each marathon was based on her finishing time.
Sorcha came $$5265$$th in the Windy Marathon, so exactly $$5264$$ of the $$6000$$ runners had a finishing time that was less than Sorcha’s.

Sorcha’s finishing time for both marathons was the same.

Use this fact, and the details in the table, to estimate Sorcha’s position in the Sunny Marathon. Show all of your working out.

Mean finishing time (minutes) Standard deviation of finishing times (minutes) Number of runners Sorcha's position

Windy Marathon

$$254$$

$$38$$

$$6000$$

$$5265$$th

Sunny Marathon

$$247$$

$$29$$

$$2000$$

$$1{,}923$$

Solution

Windy Marathon

\begin{align}P=\frac{5{,}265}{6{,}000}=0.8775\end{align}

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

\begin{align}z\approx1.16\end{align}

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

\begin{align}1.16&=\frac{x-\mu}{\sigma}\\&=\frac{t-254}{38}\end{align}

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

\begin{align}t&=1.16(38)+254\\&=298.08\mbox{ min}\end{align}

$\,$

Sunny Marathon

\begin{align}z&=\frac{x-\mu}{\sigma}\\&=\frac{298.08-247}{29}\\&\approx1.76\end{align}

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

\begin{align}P=0.9608\end{align}

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

\begin{align}\frac{\mbox{Position}}{2{,}000}=0.9608\end{align}

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

\begin{align}\mbox{Position}&=(0.9608)(2{,}000)\\&\approx1{,}923\end{align}

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