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Past Papers

## Algebra

NOTE: Clicking an entry in the right column will take you directly to that question!

(BLUE = Paper 1, RED = Paper 2)

Since 2015, the following subtopic... Has explicitly appeared in...

Factorisation

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Linear Equations

Simultaneous Equations

Indices & Surds

Inequalities

Distance, Speed & Time

Percentages

## 2022 Paper 1 Question 2

(a) Solve the following equation in $$x$$:

\begin{align}2(3x-5)+8=4x-5\end{align}

$$x=-\dfrac{3}{2}$$

Solution

\begin{align}2(3x-5)+8=4x-5\end{align}

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

\begin{align}6x-10+8=4x-5\end{align}

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

\begin{align}6x-4x=-5+10-8\end{align}

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

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

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

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

Video Walkthrough
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(b) Write $$\dfrac{(3^4)^5}{3^6}$$ in the form $$3^k$$, whee $$k\in\mathbb{R}$$.

$$3^{14}$$

Solution

\begin{align}\frac{(3^4)^5}{3^6}&=\frac{3^{20}}{6}\\&=3^{14}\end{align}

Video Walkthrough
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(c) Solve the simultaneous equations below to find the value of $$x$$ and the value of $$y$$.

\begin{align}3x+2y&=1\\7x+5y&=-2\end{align}

$$x=9$$ and $$y=-13$$

Solution

\begin{align}3x+2y&=1\\7x+5y&=-2\end{align}

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

\begin{align}21x+14y&=7\\21x+15y&=-6\end{align}

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

\begin{align}-y=7-(-6)\end{align}

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

\begin{align}-y=13\end{align}

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

\begin{align}y=-13\end{align}

and

\begin{align}3x+2y&=1\end{align}

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

\begin{align}x&=\frac{1-2y}{3}\\&=\frac{1-2(-13)}{3}\\&=9\end{align}

Video Walkthrough

## 2022 Paper 1 Question 5(a) & (b)

(a) Write each of the following values in the form $$a\times10^n$$ where $$1\leq a<10$$ and $$n\in\mathbb{Z}$$.

(i) $$1200$$

(ii) $$0.27$$

(i) $$1.2\times 10^3$$

(ii) $$2.7\times10^{-1}$$

Solution

(i) $$1.2\times 10^3$$

(ii) $$2.7\times10^{-1}$$

Video Walkthrough
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(b) A falcon can dive at a speed of up to $$120$$ miles per hour.
$$1$$ mile is approximately $$1.6$$ kilometres.

Use this to work out how long it would take the falcon to travel $$100$$ metres, when diving at this speed. Give your answer in seconds, correct to one decimal place.

$$1.9\mbox{ seconds}$$

Solution

\begin{align}120\mbox{ miles in }1\mbox{ hour}\end{align}

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

\begin{align}120\mbox{ miles in }60\times60=3{,}600\mbox{ seconds}\end{align}

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

\begin{align}120\times1.6=192\mbox{ kilometres in }3{,}600\mbox{ seconds}\end{align}

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

\begin{align}192{,}000\mbox{ metres in }3{,}600\mbox{ seconds}\end{align}

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

\begin{align}1\mbox{ metre in }\frac{3{,}600}{192{,}000}=0.01875\mbox{ seconds}\end{align}

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

\begin{align}100\mbox{ metres in }100\times0.01875\approx1.9\mbox{ seconds}\end{align}

Video Walkthrough

## 2022 Paper 1 Question 6(b) & (c)

(b) Find the smallest value of $$n\in\mathbb{N}$$ for which

\begin{align}-254+(n-1)(4)>0\end{align}

$$65$$

Solution

\begin{align}-254+(n-1)(4)>0\end{align}

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

\begin{align}-254+4n-4>0\end{align}

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

\begin{align}4n>258\end{align}

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

\begin{align}n>64.5\end{align}

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

\begin{align}n=65\end{align}

Video Walkthrough
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(c) The sum of the first $$n$$ terms of this sequence is given by $$S_n=\dfrac{n}{2}[2(-254)+4n-4]$$.

Solve the following equation for $$n\in\mathbb{N}$$. Note that $$n\neq0$$.

\begin{align}\frac{n}{2}[2(-254)+4n-4]=0\end{align}

$$128$$

Solution

\begin{align}\frac{n}{2}[2(-254)+4n-4]=0\end{align}

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

\begin{align}2(-254)+4n-4=0\end{align}

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

\begin{align}-508+4n-4=0\end{align}

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

\begin{align}4n=512\end{align}

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

\begin{align}n&=\frac{512}{4}\\&=128\end{align}

Video Walkthrough

## 2022 Paper 1 Question 7(e) - (f)

(e) Joseph has a smart watch that beeps every $$15$$ seconds during the session.
It beeps for the first time at exactly 2: 55 p.m., as Joseph starts his session.
It beeps for the last time at exactly 3: 23 p.m., as Joseph finishes his session.

Work out how many times, in total, the smart watch beeps during the session, including the first and last beep.

$$113$$ times

Solution

\begin{align}&3\colon23\\\underline{-}&\underline{2\colon55}\\&0\colon28\end{align}

The watch will beep $$(4\times28)+1=113$$ times.

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(f) Solve the equation

\begin{align}h'(x)=-1.14x^2+5.2x-0.13=0\end{align}

to find how long after the start of the session Joseph’s heart-rate is at a maximum, for $$0\leq x\leq6$$, $$x\in\mathbb{R}$$. Give your answer in minutes, correct to $$2$$ decimal places.

$$4.54$$ minutes

Solution

\begin{align}a=-1.14&&b=5.2&&c=-0.13\end{align}

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

\begin{align}x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\&=\frac{-5.2\pm\sqrt{5.2^2-4(-1.14)(-0.13)}}{2(-1.14)}\end{align}

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

\begin{align}x\approx0.025&&x\approx4.536\end{align}

According to the graph, his heart-rate is therefore a maximum after $$4.536\approx 4.54$$ minutes.

Video Walkthrough

## 2022 Paper 1 Question 8

(a) Jessica is a scientist. Jessica is making up a solution of acid.
She has two different bottles, each with the following concentration of the acid:

Bottle A Bottle B

Concentration: $$12\%$$

Concentration: $$5\%$$

This means that, for example, in every $$100\mbox{ ml}$$ of liquid in Bottle A, there are $$12\mbox{ ml}$$ of acid.

(i) Work out how many $$\mbox{ml}$$ of acid are in $$200\mbox{ ml}$$ of liquid from Bottle A.

(ii) Jessica mixes $$200\mbox{ ml}$$ of liquid from Bottle A with $$300\mbox{ ml}$$ of liquid from Bottle B.
Work out the overall concentration of the acid in Jessica’s mixture.

(iii) Explain why Jessica could not make a solution with a $$4\%$$ concentration of acid by mixing liquid from Bottle A and Bottle B.

(iv) When she is making another mixture, Jessica makes a mistake in measuring.
She wants to measure out $$250\mbox{ ml}$$ but she measures out $$260\mbox{ ml}$$ instead.

Work out the percentage error in this measurement.

(i) $$24\mbox{ ml}$$

(ii) $$7.8\%$$

(iii) The solution can only be between $$5\%$$ and $$12\%$$.

(iv) $$4\%$$

Solution

(i)

\begin{align}200\times\frac{12}{100}=24\mbox{ ml}\end{align}

(ii)

\begin{align}\frac{24+300\times0.05}{200+300}\times 100=7.8\%\end{align}

(iii) The solution can only be between $$5\%$$ and $$12\%$$.

(iv)

\begin{align}\frac{260-260}{250}\times100=4\%\end{align}

Video Walkthrough
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(b) If a solid is made up of faces with straight edges, then the following identity is often true:

\begin{align}C-E+F=2\end{align}

where:

• $$C$$ is the number of corners
• $$E$$ is the number of edges
• $$F$$ is the number of faces

(i) The value of $$E$$ for a cube is $$12$$. Write down the values of $$C$$ and $$F$$ for a cube, and
show that $$C-E+F=2$$ for these values.

(ii) Each of the faces of a different solid is in the shape of a triangle of area $$5\mbox{ cm}^2$$.
This solid has $$12$$ corners ($$C$$) and $$30$$ edges ($$E$$), and $$C-E+F=2$$ for this solid.
Work out the surface area of this solid, in $$\mbox{cm}^2$$.

(iii) The surface of a third solid is made up of $$h$$ hexagons and $$p$$ pentagons, where $$h,p\in\mathbb{N}$$. For this solid:

\begin{align}\frac{6h+5p}{3}-\frac{6h+5p}{2}+h+p=2\end{align}

Use this equation to find the number of pentagons in the surface of this solid (that is, the value of $$p$$).

(i) $$C=8$$ and $$F=6$$

(ii) $$100\mbox{ cm}^2$$

(iii) $$p=12$$

Solution

(i)

\begin{align}C=8 && F=6\end{align}

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

\begin{align}C-E+F&=8-12+6\\&=2\end{align}

as required.

(ii)

\begin{align}C-E+F=2\end{align}

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

\begin{align}12-30+F=2\end{align}

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

\begin{align}F&=2-12+30\\&=20\end{align}

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

\begin{align}\mbox{Surface area}&=20\times5\\&=100\mbox{ cm}^2\end{align}

(iii)

\begin{align}\frac{6h+5p}{3}-\frac{6h+5p}{2}+h+p=2\end{align}

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

\begin{align}2(6h+5p)-3(6h+5p)+6h+6p=12\end{align}

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

\begin{align}12h+10p-18h-15p+6h+6p=12\end{align}

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

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

Video Walkthrough

## 2022 Paper 1 Question 9(e)

(e) Eva drives her car home from the garage, a distance of $$12\mbox{ km}$$.

Eva usually drives this journey at an average speed of $$60\mbox{ km/hr}$$.
On this day, there are roadworks, so her average speed is only $$40\mbox{ km/hr}$$ for the journey.

Work out the percentage increase in the time it takes Eva to drive home, because of the roadworks.

$$50\%$$

Solution

\begin{align}\frac{60-40}{40}\times100=50\%\end{align}

Video Walkthrough

## 2022 Paper 1 Question 10(a)

Keith plays hurling.

(a) During a match, Keith hits the ball with his hurl.
The height of the ball could be modelled by the following quadratic function:

\begin{align}h=-2t^2+5t+1.2\end{align}

where $$h$$ is the height of the ball, in metres, $$t$$ seconds after being hit, and $$t\in\mathbb{R}$$.

(i) How high, in metres, was the ball when it was hit (when $$t=0$$)?

(ii) The ball was caught after $$2.4$$ seconds.
How high, in metres, was the ball when it was caught?

(iii) When the ball passed over the halfway line, it was at a height of $$3.2$$ metres and its height was decreasing.

How many seconds after it was hit did the ball pass over the halfway line?
Remember that $$h=-2t^2+5t+1.2$$.

(iv) Find $$\dfrac{dh}{dt}$$ and hence find how long it took the ball to reach its greatest height.

(i) $$1.2\mbox{ m}$$

(ii) $$1.68\mbox{ m}$$

(iii) $$2\mbox{ s}$$

(iv) $$\dfrac{5}{4}\mbox{ s}$$

Solution

(i)

\begin{align}h(0)&=-2(0^2)+5(0)+1.2\\&=1.2\mbox{ m}\end{align}

(ii)

\begin{align}h(2.4)&=-2(2.4^2)+5(2.4)+1.2\\&=1.68\mbox{ m}\end{align}

(iii)

\begin{align}h(t)&=3.2\end{align}

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

\begin{align}-2t^2+5t+1.2=3.2\end{align}

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

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

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

$$t=\dfrac{1}{2}$$ or $$t=2$$

As the height was decreasing, it must be the second time, i.e. $$2$$ seconds.

(iv)

\begin{align}\frac{dh}{dt}=-4t+5\end{align}

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

\begin{align}-4t+5=0\end{align}

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

\begin{align}t=\frac{5}{4}\mbox{ s}\end{align}

Video Walkthrough

## 2021 Paper 1 Question 3

(a) Show that $$x=4$$ is a solution of the equation $$x^2-2x-8=0$$.

Solution

\begin{align}4^2-2(4)-8&=16-8-8\\&=0\end{align}

as required.

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(b) The equation $$x^2+ax+b=0$$, where $$a,b,\in\mathbb{Z}$$, has solutions $$x=5$$ and $$x=-2$$.
Find the value of $$a$$ and the value of $$b$$.

$$a=-3$$ and $$b=-10$$

Solution

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

and

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

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

\begin{align}25+5a+b=0\end{align}

\begin{align}4-2a+b=0\end{align}

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

\begin{align}21+7a=0\end{align}

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

\begin{align}a=-3\end{align}

and

\begin{align}b&=2a-4\\&=2(-3)-4\\&=-10\end{align}

Video Walkthrough
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(c) Find the solutions of the equation $$5x^2-2x-9=0$$, where $$x\in\mathbb{R}$$.
Give each answer correct to $$2$$ decimal places.

$$x=-1.16$$ or $$x=1.56$$

Solution

\begin{align}a=5&&b=-2&&c=-9\end{align}

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

\begin{align}x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\&=\frac{-(-2)\pm\sqrt{(-2)^2-4(5)(-9)}}{2(5)}\\&=\frac{2\pm\sqrt{184}}{10}\end{align}

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

$$x\approx-1.16$$ or $$x\approx1.56$$

Video Walkthrough

## 2021 Paper 1 Question 4

(a) Solve the equation:

$$4(2x+3)-7=3(x-5)$$, where $$x\in\mathbb{R}$$.

$$x=-4$$

Solution

\begin{align}4(2x+3)-7=3(x-5)\end{align}

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

\begin{align}8x+12-7=3x-15\end{align}

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

\begin{align}8x-3x=-15-12+7\end{align}

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

\begin{align}5x=-20\end{align}

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

\begin{align}x&=-\frac{20}{5}\\&=-4\end{align}

Video Walkthrough
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(b) Solve the simultaneous equations:

\begin{align}2x-y&=7\\x^2+y^2&=49\end{align}

$$x=0$$ and $$y=-7$$ or $$x=\dfrac{28}{5}$$ and $$y=\dfrac{21}{5}$$

Solution

\begin{align}2x-y&=7\\x^2+y^2&=49\end{align}

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

\begin{align}y=2x-7\\x^2+y^2=49\end{align}

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

\begin{align}x^2+(2x-7)^2=49\end{align}

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

\begin{align}x^2+4x^2-28x+49=49\end{align}

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

\begin{align}5x^2-28x=0\end{align}

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

\begin{align}x(5x-28)=0\end{align}

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

$$x=0$$ or $$x=\dfrac{28}{5}$$

and

\begin{align}y&=2x-7\\&=2(0)-7\\&=-7\end{align}

or

\begin{align}y&=2x-7\\&=2\left(\frac{28}{5}\right)-7\\&=\frac{21}{5}\end{align}

$\,$

Solutions

$$x=0$$ and $$y=-7$$

or

$$x=\dfrac{28}{5}$$ and $$y=\dfrac{21}{5}$$

Video Walkthrough

## 2021 Paper 1 Question 10

(a) The water behind a dam is normally released at a rate of $$250{,}000$$ litres per second.

(i) Find how long it takes to release $$1$$ million cubic metres ($$1{,}000{,}000\mbox{ m}^3$$) of water.
Note $$1\mbox{ m}^3=1000\mbox{ litres}$$.

(ii) Due to heavy rainfall, the operators of the dam decide to increase the flow by $$10\%$$ for $$24$$ hours. Find how many $$\mbox{m}^3$$ of water were released in that $$24$$ hour period.
Give your answer in the form $$a\times10^n$$, where $$1\leq a<10$$, and $$n\in\mathbb{N}$$.
Give the value of $$a$$ correct to $$3$$ significant figures.

(i) $$67\mbox{ min}$$

(ii) $$2.38\times10^7\mbox{ m}^3$$

Solution

(i)

\begin{align}t&=\frac{1{,}000{,}000\times100}{250{,}000}=440\mbox{ s}\\&\approx67\mbox{ min}\end{align}

(ii)

\begin{align}\frac{250{,}000}{1000}\times1.1\times60\times60\times24\approx2.38\times10^7\mbox{ m}^3\end{align}

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(b)

(i) John walks around a circular trail of radius $$0.5\mbox{ km}$$ at a steady speed of $$6\mbox{ km/h}$$.
How long will it take him to complete $$3$$ full circuits of the trail?

(ii) Mary decides to walk every day over a $$5$$ day period.
She walks a distance of $$3\mbox{ km}$$ on day one.
She increases the length of her walk by $$15\%$$ each day for the next four days.
Her average speed on day $$5$$ is $$4\mbox{ km/h}$$.
Find how long it will take her to complete her walk on day $$5$$.

(iii) One day, during John’s walk he meets Mary at point $$P$$ on the trail.
Mary is walking in the opposite direction at a steady speed of $$4\mbox{ km/h}$$.
John continues walking at $$6\mbox{ km/h}$$.
How far will he travel until he meets Mary again?

(i) $$94\mbox{ min}$$

(ii) $$79\mbox{ min}$$

(iii) $$1{,}885\mbox{ m}$$

Solution

(i)

\begin{align}t&=3\times\frac{2\pi r}{s}\\&=3\times\frac{2\pi(0.5)}{6}\\&=1.5707..\mbox{ hr}\\&\approx94\mbox{ min}\end{align}

(ii)

\begin{align}t&=\frac{d}{s}\\&=\frac{3(1.15^4)}{4}\\&=1.2117…\mbox{ hr}\\&\approx79\mbox{ min}\end{align}

(iii)

\begin{align}d&=\frac{6}{10}\times(2\pi r)\\&=\frac{6}{10}\times[2\pi(0.5)]\\&\approx1{,}885\mbox{ m}\end{align}

Video Walkthrough

## 2020 Paper 1 Question 2

(a) Solve the equation:

\begin{align}\frac{9x-6}{2}=\frac{3x-14}{3}+\frac{9x}{4}\end{align}

$$x=-\dfrac{4}{3}$$

Solution

\begin{align}\frac{9x-6}{2}=\frac{3x-14}{3}+\frac{9x}{4}\end{align}

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

\begin{align}\frac{6(9x-6)}{12}=\frac{4(3x-14)+3(9x)}{12}\end{align}

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

\begin{align}6(9x-6)=4(3x-14)+3(9x)\end{align}

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

\begin{align}54x-36=12x-56+27x\end{align}

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

\begin{align}54x-12x-27x=36-56\end{align}

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

\begin{align}15x=-20\end{align}

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

\begin{align}x&=-\frac{20}{15}\\&=-\frac{4}{3}\end{align}

Video Walkthrough
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(b) Solve the simultaneous equations:

\begin{align}3x-y&=4\\4x^2-3xy&=4\end{align}

$$x=\dfrac{2}{5}$$ and $$y=-\dfrac{14}{5}$$ or $$x=2$$ and $$y=2$$

Solution

\begin{align}3x-y&=4\\4x^2-3xy&=4\end{align}

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

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

\begin{align}4x^2-3xy=4\end{align}

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

\begin{align}4x^2-3x(3x-4)=4\end{align}

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

\begin{align}4x^2-9x^2+12x=4\end{align}

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

\begin{align}5x^2-12x+4=0\end{align}

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

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

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

$$x=\dfrac{2}{5}$$ or $$x=2$$

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

\begin{align}y&=3x-4\\&=3\left(\frac{2}{5}\right)-4\\&=-\frac{14}{5}\end{align}

or

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

$\,$

First Solution

$$x=\dfrac{2}{5}$$ and $$y=-\dfrac{14}{5}$$

$\,$

Second Solution

$$x=2$$ and $$y=2$$

Video Walkthrough

## 2020 Paper 1 Question 4

(a) The Golden Gate Bridge in San Francisco is constantly being repainted. It is estimated that the surface area of exposed steel
that needs to be painted is approximately $$10$$ million square feet.

(i) Given that $$1$$ metre is equal to $$3.28$$ feet, convert the surface area of the bridge into square metres, giving your answer in the form $$a\times 10^n$$, where $$1\leq a<10$$ and $$n\in\mathbb{N}$$.
Give the value of $$a$$ correct to two significant figures.

(ii) A litre of the paint used on the Golden Gate Bridge will cover approximately $$5$$ square
metres. This paint comes in $$25$$ litre tins. Find the minimum number of tins of paint that will be needed to paint the entire bridge.

(i) $$9.3\times10^5\mbox{ square metres}$$

(ii) $$7{,}440\mbox{ tins}$$

Solution

(i)

\begin{align}10{,}000{,}000\mbox{ square feet}&=\frac{10{,}000{,}000}{3.28^2}\mbox{ square metres}\\&\approx929{,}506\mbox{ square metres}\\&\approx9.3\times10^5\mbox{ square metres}\end{align}

(ii)

\begin{align}\frac{\frac{930{,}000}{5}}{25}=7{,}440\mbox{ tins}\end{align}

Video Walkthrough
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(b) Solve the equation $$2^{9x-1}=8^{2x}$$.

$$x=\dfrac{1}{3}$$

Solution

\begin{align}2^{9x-1}=8^{2x}\end{align}

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

\begin{align}2^{9x-1}=2^{3(2x)}\end{align}

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

\begin{align}9x-1=2(3x)\end{align}

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

\begin{align}9x-1=6x\end{align}

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

\begin{align}3x=1\end{align}

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

\begin{align}x=\frac{1}{3}\end{align}

Video Walkthrough

## 2020 Paper 1 Question 5(a)

(a) Solve the equation $$x^2-3x-4=0$$.

$$x=-1$$ or $$x=4$$

Solution

\begin{align}x^2-3x-4=0\end{align}

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

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

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

$$x=-1$$ or $$x=4$$

Video Walkthrough

## 2020 Paper 1 Question 8(d)

(d) In the $$2016$$ Summer Olympics, Michael Phelps, won the $$200\mbox{ m}$$ Butterfly final in a time of $$1$$ minute and $$53.36$$ seconds ( $$1:53.36$$). The time it took to swim each of the four $$50\mbox{ m}$$ sections of the race (split time) is given in the table below. The percentage increase in the time taken to swim one section is also given. Thus, it took the swimmer $$14.7\%$$ longer to swim the second $$50$$ metres compared to the first $$50$$ metres.

Race Section Split time (secs) Percentage Increase

$$0$$ - $$50\mbox{ m}$$

$$24.85$$

\

$$50$$ - $$100\mbox{ m}$$

$$28.5$$

$$14.7\%$$

$$100$$ - $$150\mbox{ m}$$

$$29.33$$

$$150$$ - $$200\mbox{ m}$$

$$30.68$$

Total

$$1\colon53.36$$

\

(i) Complete the table by finding the percentage increase in split time by comparing each
split time with the previous one.

(ii) Another swimmer in the race completed the first $$50\mbox{ m}$$ in a time of $$25.01$$ seconds.
His subsequent $$50\mbox{ m}$$ split times increased at the same rates as each of Michael Phelps’ times. Using the table below, or otherwise, find the difference between his finishing time and that of Michael Phelps.
Give your answers in seconds correct to $$2$$ decimal places.

Race Section Split time (secs) Percentage Increase

$$0$$ - $$50\mbox{ m}$$

$$25.01$$

\

$$50$$ - $$100\mbox{ m}$$

$$14.7\%$$

$$100$$ - $$150\mbox{ m}$$

$$150$$ - $$200\mbox{ m}$$

Total

\

(i)

Race Section Split time (secs) Percentage Increase

$$0$$ - $$50\mbox{ m}$$

$$24.85$$

\

$$50$$ - $$100\mbox{ m}$$

$$28.5$$

$$14.7\%$$

$$100$$ - $$150\mbox{ m}$$

$$29.33$$

$$2.9\%$$

$$150$$ - $$200\mbox{ m}$$

$$30.68$$

$$4.6\%$$

Total

$$1\colon53.36$$

\

(ii)

Race Section Split time (secs) Percentage Increase

$$0$$ - $$50\mbox{ m}$$

$$25.01$$

\

$$50$$ - $$100\mbox{ m}$$

$$28.686$$

$$14.7\%$$

$$100$$ - $$150\mbox{ m}$$

$$29.518$$

$$14.7\%$$

$$150$$ - $$200\mbox{ m}$$

$$30.876$$

$$4.6\%$$

Total

$$1\colon54\colon09$$

\

\begin{align}0.73\mbox{ s}\end{align}

Solution

(i)

Race Section Split time (secs) Percentage Increase

$$0$$ - $$50\mbox{ m}$$

$$24.85$$

\

$$50$$ - $$100\mbox{ m}$$

$$28.5$$

$$14.7\%$$

$$100$$ - $$150\mbox{ m}$$

$$29.33$$

$$2.9\%$$

$$150$$ - $$200\mbox{ m}$$

$$30.68$$

$$4.6\%$$

Total

$$1\colon53.36$$

\

(ii)

Race Section Split time (secs) Percentage Increase

$$0$$ - $$50\mbox{ m}$$

$$25.01$$

\

$$50$$ - $$100\mbox{ m}$$

$$28.686$$

$$14.7\%$$

$$100$$ - $$150\mbox{ m}$$

$$29.518$$

$$14.7\%$$

$$150$$ - $$200\mbox{ m}$$

$$30.876$$

$$4.6\%$$

Total

$$1\colon54\colon09$$

\

\begin{align}1\colon54\colon09-1\colon53\colon36&\approx0.73\mbox{ s}\end{align}

Video Walkthrough

## 2019 Paper 1 Question 4

(a) Solve for $$x$$:

\begin{align}\frac{3x+1}{5}+\frac{x-2}{2}=\frac{47}{10}\end{align}

$$x=5$$

Solution

\begin{align}\frac{2(3x+1)}{10}+\frac{5(x-2)}{10}=\frac{47}{10}\end{align}

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

\begin{align}2(3x+1)+5(x-2)=47\end{align}

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

\begin{align}6x+2+5x-10=47\end{align}

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

\begin{align}11x=55\end{align}

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

\begin{align}x&=\frac{55}{11}\\&=5\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}x-5y&=-13\\x^2+y^2&=13\end{align}

$$x=-3$$ and $$y=2$$ or $$x=2$$ and $$y=3$$

Solution

\begin{align}x-5y&=-13\\x^2+y^2&=13\end{align}

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

\begin{align}x=5y-13\end{align}

\begin{align}x^2+y^2=13\end{align}

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

\begin{align}(5y-13)^2+y^2=13\end{align}

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

\begin{align}25y^2-130y+169+y^2=13\end{align}

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

\begin{align}26y^2-130y+156=0\end{align}

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

\begin{align}y^2-5y+6=0\end{align}

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

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

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

$$y=2$$ or $$y=3$$

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

\begin{align}x&=5(2)-13\\&=-3\end{align}

or

\begin{align}x&=5(3)-13\\&=2\end{align}

$\,$

Solutions

$$x=-3$$ and $$y=2$$

or

$$x=2$$ and $$y=3$$

Video Walkthrough

## 2019 Paper 1 Question 5(b)

(b) Nuala can walk at a speed of $$1.6$$ metres per second.
Write this speed in kilometres per hour.

$$5.76\mbox{ km/hr}$$

Solution

\begin{align}1.6\frac{\mbox{metres}}{\mbox{second}}&=1.6\frac{\frac{1}{1{,}000}\mbox{kilometres}}{\frac{1}{60\times60}\mbox{hour}}\\&=\frac{1.6(60\times60)}{1{,}000}\frac{\mbox{kilometres}}{\mbox{hour}}\\&=5.76\frac{\mbox{kilometres}}{\mbox{hour}}\end{align}

Video Walkthrough

## 2019 Paper 1 Question 6

(a) Solve the following inequality for $$x\in\mathbb{R}$$ and show your solution on the numberline below:

\begin{align}2(3-x)<8\end{align}

$$x>-1$$

Solution

\begin{align}2(3-x)<8\end{align}

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

\begin{align}6-2x<8\end{align}

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

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

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

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

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

\begin{align}x>-1\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 $$x$$:

\begin{align}2^{2x-1}=64\end{align}

$$x=\dfrac{7}{2}$$

Solution

\begin{align}2^{2x-1}=64\end{align}

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

\begin{align}2^{2x-1}=2^6\end{align}

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

\begin{align}2x-1=6\end{align}

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

\begin{align}2x=7\end{align}

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

\begin{align}x=\frac{7}{2}\end{align}

Video Walkthrough

## 2019 Paper 1 Question 8(a)

(a) The power ($$P$$) of an engine is measured in horsepower using the formula:

\begin{align}P=\frac{R\times T}{5252}\end{align}

where $$R$$ is the engine speed measured in revolutions per minute ($$\mbox{RPM}$$) and $$T$$ is the torque measured in appropriate units.

(i) Find the power of an engine that generates $$480$$ units of torque at $$2500\mbox{ RPM}$$.

(ii) Rearrange the formula to write $$R$$ in terms of $$P$$ and $$T$$.

(i) $$228\mbox{ horsepower}$$

(ii) $$R=\dfrac{5{,}252P}{T}$$

Solution

(i)

\begin{align}P&=\frac{2,{}500\times480}{5{,}252}\\&\approx228\mbox{ horsepower}\end{align}

(ii)

\begin{align}R=\frac{5{,}252P}{T}\end{align}

Video Walkthrough

## 2018 Paper 1 Question 3

(a) Solve the equation $$2x^2-7x-3=0$$. Give each answer correct to $$2$$ decimal places.

$$x=-0.39$$ or $$x=3.89$$

Solution

\begin{align}a=2&&b=-7&&c=-3\end{align}

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

\begin{align}x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\&=\frac{-(-7)\pm\sqrt{(-7)^2-4(2)(-3)}}{2(2)}\\&=\frac{7\pm\sqrt{73}}{4}\end{align}

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

$$x\approx-0.39$$ or $$x\approx3.89$$

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 below to find the value of $$a$$ and the value of $$b$$.

\begin{align}2a+3b&=15\\5a+b&=-8\end{align}

$$a=-3$$ and $$b=7$$

Solution

\begin{align}2a+3b&=15\\5a+b&=-8\end{align}

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

\begin{align}2a+3b&=15\\15a+3b&=-24\end{align}

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

\begin{align}13a=-39\end{align}

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

\begin{align}a=-3\end{align}

and

\begin{align}b&=-8-5a\\&=-8-5(-3)\\&=7\end{align}

Video Walkthrough

## 2018 Paper 1 Question 6

(a) Solve for $$x$$.

\begin{align}(x+5)(3x-4)-3(x^2+2)+4=0\end{align}

$$x=2$$

Solution

\begin{align}(x+5)(3x-4)-3(x^2+2)+4=0\end{align}

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

\begin{align}3x^2-4x+15x-20-3x^2-6+4=0\end{align}

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

\begin{align}11x-22=0\end{align}

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

\begin{align}x&=\frac{22}{11}\\&=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) Find the solutions of

$$\dfrac{5}{x+3}-\dfrac{1}{x}=\dfrac{1}{2}$$, where $$x\neq-3,0,x\in\mathbb{R}$$.

$$x=2$$ or $$x=3$$

Solution

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

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

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

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

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

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

\begin{align}8x-6=x^2+3x\end{align}

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

\begin{align}x^2-5x+6=0\end{align}

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

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

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

$$x=2$$ or $$x=3$$

Video Walkthrough

## 2018 Paper 1 Question 9

When earthquakes occur under the sea, they can cause large tidal waves called tsunamis.
Scientists can estimate the arrival times of tsunamis to nearby countries.
The average speed at which a tsunami travels is given by the formula:

\begin{align}s=\sqrt{g\times d}\end{align}

where $$s$$ is the speed of the tsunami (in metres per second),
$$d$$ is the depth of the ocean (in metres) at the location where the earthquake occurred,
and $$g=9.8$$ metres per second$$^2$$.

(a)

(i) Find $$s$$, the speed of a tsunami when the an earthquake occurs at a depth of $$2000\mbox{ m}$$.

(ii) A tsunami has been identified as beginning $$400\mbox{ km}$$ from land.
The depth of the ocean at that point is $$2000\mbox{ m}$$.
Find how long it will take this tsunami to reach land.

(i) $$140\mbox{ m/s}$$

(ii) $$48\mbox{ min}$$

Solution

(i)

\begin{align}s(2{,}000)&=\sqrt{g\times2{,}000}\\&=\sqrt{9.8\times2{,}000}\\&=140\mbox{ m/s}\end{align}

(ii)

\begin{align}t&=\frac{d}{s}\\&=\frac{400\times1{,}000}{140}\\&=2{,}857.14…\mbox{ s}\\&\approx48\mbox{ min}\end{align}

Video Walkthrough