Below, you will find features that will be included with a membership on Junior Maths from the 2023/24 school year onwards.

Note that these are already live on A.M. Online, our Leaving Cert Applied Maths website. You can check out the free lessons on A.M. Online to see what the style of the lessons will be like here also!

Great question!

We started working full time creating that content across all three of our websites in the summer of 2017 with the hopes of finishing it by 2020.

We instead finished in January 2022.

However, yes, you read that correctly. We actually do have all of the content completed for this website as well!

However, there was still one important step that we were yet to complete on any of our three websites - quality checks. We were yet to quadruple check our lessons for grammar mistakes, quadruple check our solutions for maths errors etc.

We don't want our first students to be the ones finding those mistakes for us!

When we started those checks in January, we assumed that it would take approximately 2 months to perform those checks on each website, thereby allowing us to have everything completed by the summer.

However, even with some colleagues helping us, we found that this quality check process was taking longer than we expected. A lot longer in fact.

This is primarily due to the fact that our original plan was to make the content on each website as long as a typical schoolbook. Instead, the content on each website was almost ten times longer!

We therefore decided to prioritise getting A.M. Online finished first for two reasons:

• Many applied maths students study the subject from home and therefore a website like A.M. Online is of particular benefit to them.
• The applied maths syllabus is changing for the 2023 Leaving Cert onwards. Therefore, there is a lack of online resources and support for all applied maths students for the new syllabus.

By prioritising A.M. Online, we completed these quality checks in August 2022, taking over half a year for just one website.

We are therefore spending the 2022/23 school year repeating that very same process for our other two websites with the goal of having them ready to go for the 2023/24 school year!

## INTERACTIVE LESSONS

Junior Maths is not simply a collection of teacher notes and recorded videos.

Instead, this one-of-a-kind website is composed of professional, interactive lessons that make learning Junior Cycle Maths an enjoyable experience for students.

Schoolbooks take a “fill in the gaps” approach, whereby a teacher is expected to expand on the core concepts of the book.

As we are not limited by what can fit into a schoolbook, our lessons instead describe every concept within Junior Cycle Maths with as much detail as possible.

Throughout each lesson, we also highlight three important aspects:

• Key Points – the most important parts of the lesson.
• Causes of Confusion – common mistakes that students make.
• Exam Tips – tips to give you that extra edge on the exam!

### Key Point

A linear equation is any equation of the form

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

### Cause of Confusion

A line of zero slope corresponds to a horizontal line, not a vertical line.

### Exam Tip

When choosing two points to draw a line, you can always choose $$(0,c)$$ as one of those points.

## ANIMATIONS

Animations make for a much more helpful and beneficial learning experience compared to the static images of a schoolbook.

We therefore provide rich, unique animations, together with thousands of illustrations, throughout the lessons in order to provide students with what we believe are the best learning resources available for Junior Cycle Maths.

Note: Animations, equations and indeed many other advanced features of this website may not render correctly on browser versions released way back in 2017 or earlier. In the unlikely event that you are still using such an outdated version, updating the browser will get everything working smoothly!

## FULL SOLUTIONS

Schoolbooks typically only provide the final answers to practice questions in the back pages rather than the full details of how those answers are obtained.

Wouldn’t it again be awesome if your book instead provided those full details?

As you might have guessed, Junior Maths has got that covered!

We provide detailed, written solutions to EVERY exercise question, and we have just as many questions to practice with as you’d find in a schoolbook!

Can you imagine having a detailed written solution for every single exercise question in your schoolbook?! Wow!

In fact, we go even further! We provide a total of four different methods for students to check their understanding of every question:

• First, check to see if you got the right Answer. If you didn’t…
• Read the Solution, i.e. what you should write down when answering the question. If you don’t understand the solution…
• Read the Walkthrough, i.e. what a teacher would say if they were explaining the solution to you. If you don’t understand the walkthrough…
• Ask Mr. Kenny for further guidance on Teacher Chat!

These “stepping stones” greatly reduce the likelihood of a student “giving up” if they don’t understand the solution to a question.

Note: The equations on this page may load slowly (or not at all) on certain mobile/tablet devices. Don’t worry! This page is very resource-heavy compared to a typical lesson page that you’ll find on this website. As long as everything loads correctly on e.g. this preview lesson page on A.M. Online, that device will work fine!

## Practice Question

Solve the following equation:

\begin{align}\frac{x+2}{4}+\frac{5x-2}{3}=\frac{91}{6}\end{align}

$$x=8$$

\begin{align}\frac{x+2}{4}+\frac{5x-2}{3}=\frac{91}{6}\end{align}

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

\begin{align}\frac{3(x+2)}{3(4)}+\frac{4(5x-2)}{4(3)}=\frac{91}{6}\end{align}

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

\begin{align}\frac{3x+6}{12}+\frac{20x-8}{12}=\frac{91}{6}\end{align}

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

\begin{align}\frac{(3x+6)+(20x-8)}{12}=\frac{91}{6}\end{align}

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

\begin{align}\frac{23x-2}{12}=\frac{91}{6}\end{align}

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

\begin{align}23x-2=12\times \frac{91}{6}\end{align}

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

\begin{align}23x-2=182\end{align}

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

\begin{align}23x=184\end{align}

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

\begin{align}x&=\frac{184}{23}\\&=8\end{align}

First, we need to add the two algebraic fractions on the left together. This can be done only when both fractions have the same denominator.

As always, one way of achieving this is by writing the denominator of both fractions as $$4\times 3=12$$. Such a denominator is obtained by:

• multiplying the first fraction above and below by $$3$$
• multiplying the second fraction above and below by $$4$$

\begin{align}\frac{3(x+2)}{3(4)}+\frac{4(5x-2)}{4(3)}=\frac{91}{6}\end{align}

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

\begin{align}\frac{3x+6}{12}+\frac{20x-8}{12}=\frac{91}{6}\end{align}

Now that both fractions on the left hand side have the same denominator, they can be added together by simply adding the numerators together.

\begin{align}\frac{(3x+6)+(20x-8)}{12}=\frac{91}{6}\end{align}

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

\begin{align}\frac{23x-2}{12}=\frac{91}{6}\end{align}

We can remove the denominator on the left hand side by multiplying both sides by $$12$$.

\begin{align}23x-2=12\times \frac{91}{6}\end{align}

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

\begin{align}23x-2=182\end{align}

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

\begin{align}23x=184\end{align}

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

\begin{align}x&=\frac{184}{23}\\&=8\end{align}

Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.

In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat!

## PAST PAPER SOLUTIONS

Official past papers, and their marking schemes, are freely available online.

However, these marking schemes are often hard to follow as they do not fully explain how to go from one step to the next.

As such, we provide our own unique, detailed solutions to past examinations papers, both at Higher and Ordinary Level.

In fact, we provide the same four “stepping stones” as we provide for our practice questions, including solutions and walkthroughs!

We don’t just provide these solutions for past papers of the last few years. Instead, we provide our “stepping stones” for every single past paper question of the current syllabus.

For example, the question shown is Higher Level Paper 1 Question 1 from all the way back in 2015!

We organise these past paper questions both 1) by year so that students can practice with full papers and 2) by topic so that students can improve their understanding in specific areas.

## Past Paper Question

The sets $$A$$, $$B$$ and $$C$$ are as follows:

\begin{align}A = \{1,2,3,5,6,7\} && B=\{2,3,4,5,8,9\} && C=\{1,4,5,10\}\end{align}

(a) Complete the Venn diagram below.

(b) List the elements of each of the following sets:

(i) $$A\cup B$$

(ii) $$A\setminus C$$

(iii) $$A\cup (B\cap C)$$

(c) Complete the following identity:

\begin{align}A\cup (B\cap C)=(A\cup B)\cap (\rule{2cm}{1pt})\end{align}

(a)

(b)

(i) $$\{1,2,3,4,5,6,7,8,9\}$$

(ii) $$\{2,3,6,7\}$$

(iii) $$\{1,2,3,4,5,6,7\}$$

(c) $$A\cup (B\cap C)=(A\cup B)\cap (A\cup C)$$

(a)

(b)

(i)

\begin{align}A\cup B = \{1,2,3,4,5,6,7,8,9\}\end{align}

(ii)

\begin{align}A \setminus C =\{2,3,6,7\} \end{align}

(iii)

\begin{align}B \cap C=\{4,5\}\end{align}

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

\begin{align}A \cup(B \cap C)=\{1,2,3,4,5,6,7\}\end{align}

(c)

\begin{align}A\cup (B\cap C)=(A\cup B)\cap (A\cup C)\end{align}

(a) Each of the numbers from $$1$$ to $$10$$ are included in at least one of the the three sets. Therefore, let’s go through each of these elements one by one.

• The element $$1$$ is in sets $$A$$ and $$C$$ but not $$B$$.
• The element $$2$$ is in sets $$A$$ and $$B$$ but not in $$C$$.
• The element $$3$$ is in sets $$A$$ and $$B$$ but not in $$C$$.
• The element $$4$$ is in sets $$B$$ and $$C$$ but not in $$A$$.
• The element $$5$$ is in sets $$A$$, $$B$$ and $$C$$.
• The element $$6$$ is in set $$A$$ but not in sets $$B$$ and $$C$$.
• The element $$7$$ is in set $$A$$ but not in sets $$B$$ and $$C$$.
• The element $$8$$ is in set $$B$$ but not in sets $$A$$ and $$C$$.
• The element $$9$$ is in set $$B$$ but not in sets $$A$$ and $$C$$.
• The element $$10$$ is in set $$C$$ but not in sets $$A$$ and $$B$$.

Therefore, the Venn diagram looks as follows:

(b)

(i) $$A\cup B$$ corresponds to all of the elements in sets $$A$$ and $$B$$, including those that are in both sets.

\begin{align}A\cup B = \{1,2,3,4,5,6,7,8,9\}\end{align}

(ii) $$A\setminus C$$ corresponds to all of the elements in set $$A$$ that are not in set $$C$$.

\begin{align}A \setminus C =\{2,3,6,7\} \end{align}

(iii) As always, we determine what’s in brackets first.

$$B \cap C$$ corresponds to all of the elements that are in both $$B$$ and $$C$$.

\begin{align}B \cap C=\{4,5\}\end{align}

$$A\cup(B \cap C)$$ then corresponds to these elements in addition to all of the elements in set $$A$$.

\begin{align}A \cup(B \cap C)=\{1,2,3,4,5,6,7\}\end{align}

(c) Finally, we are merely being asked to state one of the two set distributive laws that we discussed in the last lesson.

\begin{align}A\cup (B\cap C)=(A\cup B)\cap (A\cup C)\end{align}

Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.

In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat!

## Sample Paper Question

Consider the following four points on the $$x$$-$$y$$ plane.

(a) State the coordinates of each of these four points.

(b) State the slopes of the line segments $$AB$$, $$BC$$, $$CD$$ and $$DA$$.

(c) Using part (b), show that the quadrilateral $$ABCD$$ is a parallelogram.

(d) Is $$ABCD$$ also a rectangle? Explain why.

(a) $$A = (1,2)$$, $$B = (3,5)$$, $$C = (5,4)$$, $$D = (3,1)$$

(b) $$m_{AB} = 1.5$$, $$m_{BC} = -0.5$$, $$m_{CD} = 1.5$$, $$m_{DA} = -0.5$$

(c) As all opposite sides have the same slope, the quadrilateral $$ABCD$$ is a parallelogram.

(d) As all adjacent sides are not perpendicular to each other, $$ABCD$$ is not a rectangle.

(a)

\begin{align}A = (1,2) && B=(3,5)\end{align}

\begin{align}C = (5,4) && D=(3,1)\end{align}

(b)

$$\underline{\mathbf{AB}}$$

\begin{align}m&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{5-2}{3-1}\\&=\frac{3}{2}\\&=1.5\end{align}

$\,$

$$\underline{\mathbf{BC}}$$

\begin{align}m&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{4-5}{5-3}\\&=\frac{-1}{2}\\&=-0.5\end{align}

$\,$

$$\underline{\mathbf{CD}}$$

\begin{align}m&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{1-4}{3-5}\\&=\frac{-3}{-2}\\&=1.5\end{align}

$\,$

$$\underline{\mathbf{DA}}$$

\begin{align}m&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{2-1}{1-3}\\&=\frac{1}{-2}\\&=-0.5\end{align}

(c) Both $$AB$$ and $$CD$$ have the same slope of $$1.5$$.

Both $$BC$$ and $$DA$$ have the same slope of $$-0.5$$.

Therefore, as all opposite sides have the same slope, this quadrilateral is indeed a parallelogram.

(d) For $$AB$$ and $$BC$$:

\begin{align}m_1\times m_2 &=1.5\times -0.5\\&=-0.75\end{align}

As $$m_1 \times m_2 \neq -1$$, this parallelogram is not a rectangle.

(a) The coordinates $$(x,y)$$ of each of the four points are found by looking the corresponding grid lines.

\begin{align}A = (1,2) && B=(3,5)\end{align}

\begin{align}C = (5,4) && D=(3,1)\end{align}

(b) As we know the coordinates of two points $$(x_1,y_1)$$ and $$(x_2,y_2)$$ on each of these lines segments, we can use our usual slope formula in each case.

\begin{align}m=\frac{y_2-y_1}{x_2-x_1}\end{align}

For the line segment $$AB$$ with endpoints $$(1,2)$$ and $$(3,5)$$, this formula gives

\begin{align}m&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{5-2}{3-1}\\&=\frac{3}{2}\\&=1.5\end{align}

For the line segment $$BC$$ with endpoints $$(3,5)$$ and $$(5,4)$$, we obtain

\begin{align}m&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{4-5}{5-3}\\&=\frac{-1}{2}\\&=-0.5\end{align}

For the line segment $$CD$$ with endpoints $$(5,4)$$ and $$(3,1)$$, we obtain

\begin{align}m&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{1-4}{3-5}\\&=\frac{-3}{-2}\\&=1.5\end{align}

Finally, for the line segment $$DA$$ with endpoints $$(3,1)$$ and $$(1,2)$$, we obtain

\begin{align}m&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{2-1}{1-3}\\&=\frac{1}{-2}\\&=-0.5\end{align}

(c) $$ABCD$$ is a parallelogram only if all pairs of opposite sides are parallel to each other, i.e. if all opposite sides have the same slope.

Both $$AB$$ and $$CD$$ have the same slope of $$1.5$$.

Both $$BC$$ and $$DA$$ have the same slope of $$-0.5$$.

Therefore, as all opposite sides have the same slope, this quadrilateral is indeed a parallelogram.

(d) $$ABCD$$ is also a rectangle if all adjacent sides are perpendicular to each other.

To determine if this is the case, we can use the fact that two lines of slopes $$m_1$$ and $$m_2$$ respectively are perpendicular to each other only if they satisfy the following condition:

\begin{align}m_1\times m_2=-1\end{align}

For example, the product of the slopes of the adjacent line segments $$AB$$ and $$BC$$ is

\begin{align}m_1\times m_2 &=1.5\times -0.5\\&=-0.75\end{align}

As $$m_1 \times m_2 \neq -1$$, we don’t have to check this condition for the other adjacent sides as we can already say that this parallelogram is not a rectangle.

Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.

In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat!

## SAMPLE PAPERS

Just in case past paper solutions aren’t enough, we also provide an additional TEN sample examination papers so that both Higher and Ordinary Level students don’t have to worry about not having enough exam questions to practice with!

These papers can only be found on Junior Maths and, as you may have guessed by now, also come with the same four “stepping stones” for every question!

As with past papers, these sample paper questions are also organised both by year and by topic.

## KNOWLEDGE CHECKS

Before jumping directly into answering full exercise questions, we first test each student’s basic understanding of the lesson content with quick-fire questions in the form of Knowledge Checks.

These questions give each student immediate feedback as to whether a lesson “clicked” or not, thereby indicating whether they are ready to answer the corresponding exercise set or whether they need to do some quick revision first.

$\,$Knowledge Check 18

$\,$A quick check on your understanding of:

Linear Sequences

Exponential Sequences

$\,$

An arithmetic sequence is also known as a linear sequence.

1 / 10

The sequence $$10,15,20,25,...$$ is arithmetic.

An exponential sequence has a general term of the form $$T_n = a^n$$.

2 / 10

$$T_n = n^2$$ and $$T_n = 2^n$$ are both general terms of quadratic sequences.

What is the value of $$T_1$$ for each sequence?

3 / 10

What is the general term of the sequence $$10,15,20,25,...$$?

What is the value of $$T_1$$ for each sequence?

4 / 10

What is the general term of the sequence $$12,17,22,27,...$$?

What is the value of $$T_1$$ for each sequence?

5 / 10

What is the general term of the sequence $$7,12,19,28,...$$?

For this sequence, the difference between successive terms is constant.

6 / 10

For a particular sequence, we find that $$T_2-T_1=3$$, $$T_3-T_2=3$$, $$T_4-T_3=3$$ etc. This sequence is said to be

For a linear sequence, the difference between successive terms is constant.

7 / 10

The first two terms of a sequence are $$3$$ and $$5$$. If this sequence is linear, then the next term will be

For a quadratic sequence, the "difference of the differences" is constant.

8 / 10

For a particular sequence, we find that $$T_{12}-T_{11} = 21$$ and $$T_{13}-T_{12} = 26$$. If this sequence is quadratic, then what is $$T_{14}-T_{13}$$?

For a geometric sequence, the ratio of successive terms is a constant.

9 / 10

Which of the following is an exponential sequence?

If it is not quadratic, what kind of sequence would it be?

10 / 10

A sequence of the form $$T_n = an^2+c$$ is not quadratic as there is no "$$bn$$" term.

0%

## Membership

#### ONLINE COURSE:

Interactive Lessons

Practice Questions

Animations

Sample Papers

Walkthrough Solutions

and more!

Next year, students can get access to all of the above until June 30th 2024 for the same regular membership price of €39.

## CHEAP AND CHEERFUL

Similar websites typically charge hundreds of euro per year for an online course that contains significantly less features. These are businesses whose primary goal is to make a profit.

Junior Maths, on the other hand, was created by a teacher whose sole ambition is to give all students access to the best junior cycle maths resources, all in one place.

As such, even though our online course is more feature-rich than any other online course that you can find, our weekly grinds and our separate Teacher Chat feature will still also be included with a membership!

For the same price as a typical grind or two, we will therefore provide everything needed to ace the exam for an entire school year, all on one website.

Alternative Junior Maths

Ask a teacher for help only during class.

Ask a teacher for help throughout the day, every day.

Only answers at the back of the schoolbook.

Full, detailed solutions to every question in addition to detailed  walkthroughs.

Learn using dull, static images.

Learn using vibrant animations.

Attempt to fill the gaps left out from marking schemes.

Solutions and walkthroughs included both for past papers and sample papers.