Course Content
Higher Level (by year)
0/17
Higher Level (by topic)
0/13
Ordinary Level (by year)
0/17
Ordinary Level (by topic)
0/12
Past Papers

## Question 1

(a) A new machine is bought for €$$30000$$. Its value depreciates by $$15\%$$ each year for five years.
Find the value of the machine at the end of the five years.

$$13{,}311.16\mbox{ euro}$$

Solution

\begin{align}30{,}000\times0.85^5=13{,}311.16\mbox{ euro}\end{align}

Video Walkthrough
##### 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) A sum of money was invested for two years at $$3\%$$ compound interest per year.
At the end of the two years it amounted to €$$30000$$. Find the sum invested.

$$28{,}277.38\mbox{ euro}$$

Solution

\begin{align}P\times1.03^2=30{,}000\end{align}

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

\begin{align}P&=\frac{30{,}000}{1.03^2}\\&=28{,}277.38\mbox{ euro}\end{align}

Video Walkthrough
##### 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!

(c) A company invested €$$25000$$ for three years at a fixed rate of compound interest.
At the end of the three years it amounted to €$$26530.20$$. Find the rate of interest.

$$2\%$$

Solution

\begin{align}25{,}000(1+i)^3=26{,}530.20\end{align}

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

\begin{align}1+i&=\sqrt{\frac{26{,}530.20}{25{,}000}}\\&=1.02\end{align}

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

\begin{align}i&=0.02\\&=2\%\end{align}

Video Walkthrough

## Question 2

(a) The complex number $$z_1=a+bi$$, where $$i^2=-1$$, is shown on the Argand Diagram below.

(i) Write down the value of $$a$$ and the value of $$b$$.

(ii) $$z_2=-1+2i$$. Plot $$z_2$$ on the Argand Diagram.

(iii) $$z_3=\dfrac{z_1}{z_2}$$. Write $$z_3$$ in the form $$x+yi$$, where $$x,y\in\mathbb{R}$$.

(i) $$a=3$$ and $$b=-1$$

(ii)

(iii) $$-1-i$$

Solution

(i) $$a=3$$ and $$b=-1$$

(ii)

(iii)

\begin{align}z_3&=\frac{z_1}{z_2}\\&=\frac{3-i}{-1+2i}\\&=\frac{3-i}{-1+2i}\times\frac{-1-2i}{-1-2i}\\&=\frac{-3-6i+i-2}{1+2i-2i+4}\\&=\frac{-5-5i}{5}\\&=-1-i\end{align}

Video Walkthrough
##### 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) Solve for $$z$$

\begin{align}2z-6(4-6i)=(-1+i)(4-2i)\end{align}

$$z=11-15i$$

Solution

\begin{align}2z-6(4-6i)=(-1+i)(4-2i)\end{align}

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

\begin{align}2z-24+36i=-4+2i+4i+2\end{align}

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

\begin{align}2z&=-4+2i+4i+2+24-36i\\&=22-30i\end{align}

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

\begin{align}z=11-15i\end{align}

Video Walkthrough

## Question 3

(a) Find the two values of $$x$$ for which $$3x^2-6x-8=0$$.
Give each answer correct to $$1$$ decimal place.

$$x=-0.9$$ or $$x=2.9$$

Solution

\begin{align}a=3&&b=-6&&c=-8\end{align}

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

\begin{align}x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\&=\frac{-(-6)\pm\sqrt{(-6)^2-4(3)(-8)}}{2(3)}\\&=\frac{6\pm\sqrt{132}}{6}\end{align}

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

$$x\approx-0.9$$ or $$x\approx2.9$$

Video Walkthrough
##### 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) Find the co-ordinates of the minimum point of the function $$f(x)=3x^2-6x-8$$, where $$x\in\mathbb{R}$$.

$$(1,-11)$$

Solution

\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}

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

\begin{align}f(1)&=3(1^2)-6(1)-8\\&=-11\end{align}

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

\begin{align}(1,-11)\end{align}

Video Walkthrough

## Question 4

(a) Solve for $$x$$:

\begin{align}11x-5(2x-1)=3(6-x)+3\end{align}

$$x=4$$

Solution

\begin{align}11x-5(2x-1)=3(6-x)+3\end{align}

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

\begin{align}11x-10x+5=18-3x+3\end{align}

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

\begin{align}x+5=21-3x\end{align}

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

\begin{align}3x+x=21-5\end{align}

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

\begin{align}4x=16\end{align}

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

\begin{align}x&=\frac{16}{4}\\&=4\end{align}

Video Walkthrough
##### 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) Solve the simultaneous equations:

\begin{align}y+5&=2x\\x^2+y^2&=25\end{align}

$$x=0$$ and $$y=-5$$ or $$x=4$$ and $$y=3$$

Solution

\begin{align}y+5&=2x\\x^2+y^2&=25\end{align}

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

\begin{align}y=2x-5\end{align}

\begin{align}x^2+y^2=25\end{align}

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

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

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

\begin{align}x^2+4x^2-20x+25=25\end{align}

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

\begin{align}5x^2-20x-0\end{align}

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

\begin{align}5x(x-4)=0\end{align}

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

$$x=0$$ or $$x=4$$

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

\begin{align}y&=2x-5\\&=2(0)-5\\&=-5\end{align}

or

\begin{align}y&=2(4)-5\\&=2(4)-5\\&=3\end{align}

$\,$

Solution 1

$$x=0$$ and $$y=-5$$

$\,$

Solution 2

$$x=4$$ and $$y=3$$

Video Walkthrough

## Question 5

A field is divided into eight sections as shown below. The width of each section is $$3$$ metres.
The height, in metres, of each section is given in the diagram.
Use the Trapezoidal rule to estimate the area of the field.

$$145.8\mbox{ m}^2$$

Solution

\begin{align}A&=\frac{3}{2}[4+3+2(5.8+7+6.5+6+4.8+6+6.5)]\\&=145.8\mbox{ m}^2\end{align}

Video Walkthrough
##### 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) The area of the same field was re-estimated by applying the Trapezoidal rule again.
This time, a different section width ($$4\mbox{ m}$$) and a different set of section heights were used, as
shown below. The area was found to be $$145.6\mbox{ m}^2$$.
Use this information to find the value of the height marked $$x$$ on the diagram.

$$x=4.9\mbox{ m}$$

Solution

\begin{align}\frac{4}{2}[4+8+2(6.4+6.9+6+x+6.2)]=145.6\end{align}

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

\begin{align}4+8+2(6.4+6.9+6+x+6.2)=72.8\end{align}

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

\begin{align}2(6.4+6.9+6+x+6.2)=60.8\end{align}

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

\begin{align}12.8+13.8+12+2x+12.4=60.8\end{align}

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

\begin{align}2x=9.8\end{align}

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

\begin{align}x=4.9\mbox{ m}\end{align}

Video Walkthrough

## Question 6

(a) A salesman earns a basic salary of €$$150$$ per week. In addition, he gets commission of $$20\%$$ on sales up to the value of €$$1000$$ in the week and $$30\%$$ commission on any sales above this.
Find his total income for a week when his total sales amount to €$$3000$$.

$$950\mbox{ euro}$$

Solution

\begin{align}150+0.2\times1{,}000+0.3\times(3{,}000-1{,}000)=950\mbox{ euro}\end{align}

Video Walkthrough
##### 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) On a different week his total income is €$$1160$$. Find his total sales for this week.

$$3{,}700\mbox{ euro}$$

Solution

\begin{align}\frac{1{,}160-150-0.2\times1{,}000}{0.3}+1{,}000=3{,}700\mbox{ euro}\end{align}

Video Walkthrough

## Question 7

The first three patterns in a sequence of patterns of tiles are shown in the diagram below.

(a) Draw the next pattern of tiles.

Solution
Video Walkthrough
##### 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) Based on the patterns shown, complete the table below.

Pattern Number Number of Tiles

$$1$$

$$5$$

$$2$$

$$3$$

$$4$$

$$5$$

Pattern Number Number of Tiles

$$1$$

$$5$$

$$2$$

$$8$$

$$3$$

$$11$$

$$4$$

$$14$$

$$5$$

$$17$$

Solution
Pattern Number Number of Tiles

$$1$$

$$5$$

$$2$$

$$8$$

$$3$$

$$11$$

$$4$$

$$14$$

$$5$$

$$17$$

Video Walkthrough
##### 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!

(c)

(i) Assuming the pattern continues, the number of tiles in the $$n$$th pattern of the sequence is given by the formula $$T_n=pn+q$$, where $$p$$ and $$q\in\mathbb{N}$$.
Find the value of $$p$$ and the value of $$q$$.

(ii) How many tiles are in the $$20$$th pattern?

(iii) Find which pattern has exactly $$290$$ tiles.

(i) $$p=3$$ and $$q=2$$

(ii) $$62$$

(iii) $$96\mbox{th}$$

Solution

(i)

\begin{align}a=5&&d=3\end{align}

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

\begin{align}T_n&=a+(n-1)d\\&=5+(n-1)3\\&=5+3n-3\\&=3n+2\end{align}

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

$$p=3$$ and $$q=2$$

(ii)

\begin{align}T_{20}&=3(20)+2\\&=62\end{align}

(iii)

\begin{align}T_n=290\end{align}

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

\begin{align}3n+2=290\end{align}

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

\begin{align}n&=\frac{290-2}{3}\\&=96\end{align}

Video Walkthrough
##### 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!

(d)

(i) Show that $$S_n=\dfrac{3n^2+7n}{2}$$ is a formula for the total number of tiles needed to build the first $$n$$ patterns.

(ii) Find the total number of tiles needed to build the first $$30$$ patterns.

(ii) $$1{,}455$$

Solution

(i)

\begin{align}a=5&&d=3\end{align}

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

\begin{align}S_n&=\frac{n}{2}[2a+(n-1)d]\\&=\frac{n}{2}[2(5)+(n-1)3]\\&=\frac{n}{2}(10+3n-3)\\&=\frac{n}{2}(3n+7)\\&=\frac{3n^2+7n}{2}\end{align}

as required.

(ii)

\begin{align}S_{30}&=\frac{3(30^2)+7(30)}{2}\\&=1{,}455\end{align}

Video Walkthrough

## Question 8

A company makes and sells fibre optic cable. It can sell, at most, $$200$$ kilometres of cable in a week. For a certain range of its production the company has found that profit can be modelled using the function:

$$P(x)=275x-x^2-2000$$, where $$x\leq 200$$

In the function, $$x$$ is the number of kilometres of fibre optic cable sold and $$P(x)$$ is the profit in euro.

(a) Use the profit function, $$P(x)$$, to find how much money the company loses if it does not sell any cable.

$$2{,}000\mbox{ euro}$$

Solution

\begin{align}P(0)&=275(0)-0^2-2{,}000\\&=-2{,}000\end{align}

i.e. the company will lose $$2{,}000\mbox{ euro}$$.

Video Walkthrough
##### 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) In a particular week the company made a profit of €$$8350$$.
Find the number of kilometres of cable it sold that week.

$$45\mbox{ km}$$

Solution

\begin{align}P(x)=8{,}350\end{align}

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

\begin{align}275-x^2-2{,}000=8{,}350\end{align}

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

\begin{align}x^2-275x+10{,}350=0\end{align}

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

\begin{align}(x-45)(x-230)=0\end{align}

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

\begin{align}x=45\mbox{ km}\end{align}
(as $$x\leq200$$)

Video Walkthrough
##### 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!

(c)

(i) The table below shows some of the data representing the profit on sales.
Use the profit function, $$P(x)$$, to complete the table.

Number of km of cable sold ($$x$$) $$50$$ $$60$$ $$70$$ $$80$$ $$90$$ $$100$$

Profit (€)

$$12{,}350$$

(ii) Use the data in the table to draw the graph of the profit function on the axes below for $$50\leq x\leq 100$$, $$x\in\mathbb{R}$$.

(iii) Use your graph to estimate the lower and upper range of sales (in $$\mbox{km}$$ of cable) in order to make a profit of between €$$10{,}000$$ and €$$14{,}000$$ in a particular week.
Show your work on the graph above.

(i)

Number of km of cable sold ($$x$$) $$50$$ $$60$$ $$70$$ $$80$$ $$90$$ $$100$$

Profit (€)

$$9{,}250$$

$$10{,}900$$

$$12{,}350$$

$$13{,}600$$

$$14{,}650$$

$$15{,}500$$

(ii)

(iii) Lower: $$55\mbox{ km}$$. Upper: $$83\mbox{ km}$$.

Solution

(i)

Number of km of cable sold ($$x$$) $$50$$ $$60$$ $$70$$ $$80$$ $$90$$ $$100$$

Profit (€)

$$9{,}250$$

$$10{,}900$$

$$12{,}350$$

$$13{,}600$$

$$14{,}650$$

$$15{,}500$$

(ii)

(iii) Lower: $$55\mbox{ km}$$. Upper: $$83\mbox{ km}$$.

Video Walkthrough
##### 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!

(d) Use calculus to find the number of kilometres of cable sold when the profit is increasing at a rate of €$$105$$ per $$\mbox{km}$$.

$$85\mbox{ km}$$

Solution

\begin{align}P'(x)=105\end{align}

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

\begin{align}275-2x=105\end{align}

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

\begin{align}x&=\frac{275-105}{2}\\&=85\mbox{ km}\end{align}

Video Walkthrough

## Question 9

Forensic scientists can estimate the height of a person from the lengths of their bones.
One method uses a function which relates the length of the femur bone, $$x$$ ,to the height of the person. Using this method the heights of males and females are estimated using the following functions:

Male: $$m(x)=2.3x+65.53$$,

where $$m(x)$$ is the height and $$x$$ is the length of the femur, in $$\mbox{cm}$$.

Female: $$f(x)=2.5x+54.13$$,

where $$f(x)$$ is the height and $$x$$ is the length of the femur, in $$\mbox{cm}$$.

(a) Use the functions above to estimate the height of a male and the height of a female each of whose femur is $$47.54\mbox{ cm}$$ in length. Give both answers correct to $$2$$ decimal places.

$$174.87\mbox{ cm}$$ and $$172.98\mbox{ cm}$$ respectively

Solution

\begin{align}m(47.54)&=2.3(47.54)+65.53\\&=174.87\mbox{ cm}\end{align}

and

\begin{align}f(47.54)&=2.5(47.54)+54.13\\&=172.98\mbox{ cm}\end{align}

Video Walkthrough
##### 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) Use $$m(x)$$ to estimate the femur length of a male whose height is $$184\mbox{ cm}$$.
Give your answer correct to $$2$$ decimal places.

$$51.51\mbox{ cm}$$

Solution

\begin{align}m(x)=184\end{align}

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

\begin{align}2.3x+65.53=184\end{align}

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

\begin{align}x&=\frac{184-65.53}{2.3}\\&\approx51.51\mbox{ cm}\end{align}

Video Walkthrough
##### 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!

(c) Conor’s femur length is $$44.2\mbox{ cm}$$. His height is $$171\mbox{P cm}$$.
Find the percentage error in using $$m(x)$$ to estimate his height.
Give your answer correct to $$2$$ decimal places.

$$2.23\%$$

Solution

\begin{align}m(44.2)&=2.3(44.2)+65.53\\&=167.19\mbox{ cm}\end{align}

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

\begin{align}\%\mbox{error}&=\frac{171-167.19}{171}\times100\\&\approx2.23\%\end{align}

Video Walkthrough
##### 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!

(d) Find the length of a femur for which the estimated height of a male and the estimated height of a female are the same and find this estimated height.

$$\mbox{Length of femur}=57\mbox{ cm}$$ and $$\mbox{ height}=196.63\mbox{ cm}$$.

Solution

\begin{align}2.3x+65.53=2.5x+54.13\end{align}

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

\begin{align}0.2x=11.4\end{align}

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

\begin{align}x&=\frac{11.4}{0.2}\\&=57\mbox{ cm}\end{align}

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

\begin{align}m(57)&=2.3(57)+65.53\\&=196.63\mbox{ cm}\end{align}

Video Walkthrough
##### 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!

(e) The Ponderal Index is a number which relates a person’s height to their weight.
The formula for the Ponderal Index, $$P$$, is

\begin{align}P=\frac{M}{h^3}\end{align}

where $$M$$ is weight in $$\mbox{kg}$$ and $$h$$ is height in metres.

(i) Find the Ponderal Index for a person who is $$1.60\mbox{ m}$$ tall and weighs $$72.5\mbox {kg}$$.
Give your answer correct to $$1$$ decimal place.

(ii) Rearrange the formula $$P=\dfrac{M}{h^3}$$ to give a formula which will give the height of a person in terms of their weight and Ponderal index.

(iii) Mary has a Ponderal Index of $$13$$ and a weight of $$67.5\mbox{ kg}$$. Find her height.
Give your answer in metres, correct to $$2$$ decimal places.

(i) $$17.7$$

(ii) $$h=\sqrt{\dfrac{M}{P}}$$

(iii) $$1.73\mbox{ m}$$

Solution

(i)

\begin{align}P&=\frac{72.5}{1.6^3}\\&\approx17.7\end{align}

(ii)

\begin{align}P=\frac{M}{h^3}\end{align}

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

\begin{align}h^3=\frac{M}{P}\end{align}

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

\begin{align}h=\sqrt{\frac{M}{P}}\end{align}

(iii)

\begin{align}h&=\sqrt{\frac{67.5}{13}}\\&\approx1.73\mbox{ m}\end{align}

Video Walkthrough