Lesson video

Well done for taking that first step and choosing to load this video to learn today.

My name is Ms. Davies and I'm gonna be helping you as you work your way through this lesson.

Lots of skills that we're gonna be using, so feel free to pause the video and give things a go.

Give yourself time to think about what you are doing and then I'll help you out by adding any hints, any tips as we work our way through.

Let's get started then.

Welcome to today's very exciting lesson where we're gonna be problem solving with advanced algebraic fractions.

We're gonna have a look at some real-life applications of algebraic fractions, and then we're gonna have a look at some more exciting problems that we can now solve and how we can manipulate these.

There are a few keywords that we do need to be confident with.

So we're gonna talk about compound interest.

Now compound interest is the interest calculated on the original amount and the interest accumulated over the previous period.

The second part of the lesson, we're gonna have a look at tree diagrams. Tree diagrams are a representation used to model statistical or probability questions.

The branches represent different possible events or outcomes, specifically a probability tree shows all the possible outcomes from an event or from a stage of a trial along with the probability of that outcome happening.

And I will talk you through these as we go.

So we're gonna start by looking at loan calculations.

Jacob is borrowing 10 pound from Lucas.

Lucas wants to know how much Jacob will pay him back.

So Jacob says, That's easy.

I borrowed 10 pound, so I'll pay you back 10 pound." Lucas has a problem with this, "No, sorry, because I am gonna charge you interest.

If you pay me back after one day, I will charge you 10% interest." 10% interest is crazy.

Let's see what happens to the amount Jacob owes.

How much will Jacob owe Lucas after one day? So the 10 pound he owed him plus 10%, so it'll be 11 pound.

Jacob's very understanding, "But if I don't get pocket money for three more days, how about if I pay you back then?" Lucas is happy with that, "Absolutely, but I will charge you an extra 10% interest each day until you pay me back." "So that would be 13?" says Jacob.

Nope, 13 pound 31 actually.

Right, Lucas seems to be being a little bit mean.

However, how has he calculated this amount? Can you see where that value's come from? Have a think.

So this comes from compound interest.

If it's 10% each day, when that value's been increased by 10%, we then calculate 10% of the new amount.

The quickest way to do this is to use multipliers and then to use exponents.

So to increase by 10%, you can multiply by 1.

1.

We're gonna do that three times.

So you can multiply by 1.

1, then by 1.

1, then by 1.

1 again, or 1.

1 cubed, and that's where the 13 pound 31 comes from.

You might wanna try that with your calculator if you haven't done so already.

So a quick recap on compound interest.

A quick way to calculate a value after compound interest is using multipliers for percentage increase.

So if we had a 5% increase, then the multiplier would be 1.

05.

That's a hundred percent plus 5%, which is 105%.

Then that can be written as 1.

05.

If it's been increased by 5% three times using compound interest, the multiplier would be 1.

05 cubed.

So we can write a formula for borrowing money with compound interest as the amount owed equals the amount borrowed times percentage increase as a multiplier to the power of n.

This is similar to a compound interest formula you may have seen before.

N is the number of time periods.

So if the percentage is 5% a week and then you're borrowing it for four weeks, n would be four.

To make that easier, we're gonna write this as A equals C times P to the power of n where A is the amount owed, C is the amount borrowed, P is the percentage increase as a multiplier, and n is the number of time periods.

Quick check then.

So there's the formula as we saw it before.

Which of these would be the formula for the amount owed when borrowing money with 2% compound interest? What do you think? Well done if you said it's C, times 1.

02 to the power of n.

The multiplier to increase by 2% is 1.

02.

Right, I'd like you now to make C the subject of that formula.

What can we do? Lovely, if we divide both sides by p to the power of n, we get C equals A over p to the power of n.

Okay, so we're gonna use that formula, and I've just defined what C, A, P and n are.

Jacob says, "How much could I borrow if I paid you back 15 pounds after seven days?" So maybe he knows that in seven days' time he's gonna have 15 pounds.

Let's see if we can use the formula to work that out.

Lucas is being a bit nicer this time and says, I'll charge you 5% interest compounded every day.

That's still a high interest rate.

5% is high anyway.

The fact that it's compounded every day is quite harsh.

Let's work it out though.

So substituting into our new formula, we've got C equals 15 over 1.

05 to the power of 7, 10 pound 66.

So Jacob says, "I can only borrow 10 pound 66 and still have to pay you 15 pounds." Well, like we just said, that's a really high compound interest rate.

"Okay," says Jacob, "how about I pay you five pound after one day and the rest after three days? How much will it cost to borrow 10 pound then?" Lucas says, "If you start making multiple repayments, the calculations are trickier, but there's a formula we can use." So let's have a look at what happens if we're making two instalments.

So of course we've still got compound interest, but Jacob's gonna pay off some of what he owes after some amount of time, but then his loan is still gonna gather interest until he pays off the final amount.

So the formula ends up looking like this.

So C is the amount you can borrow.

A1 is the first repayment.

A2 is then the final repayment.

P is still the multiplier to increase.

The n1 is the time periods since borrowing until the first repayment is made.

So if it's two days until he makes the first repayment and the percentage interest is applied every day, then that will be two.

N is the time periods elapsed since borrowing until the final repayment is made.

Let's put that into practise then.

So we've got 10 pounds, so that's the amount he wants to borrow.

The first repayment is gonna be five pounds.

Then our denominator for the first fraction is 1.

05 to the power of 1 because he's gonna pay it back after one day and the interest is daily.

The second fraction, we don't know what that final repayment's gonna be.

We're gonna try and work that out.

And then 1.

05 to the power of 3, because he is gonna pay the whole lot back three days after he borrowed it.

So we're gonna rearrange this.

So if we subtract that whole fraction from the left-hand side and then multiply it by the denominator.

Because we're using a calculator, I don't need to rearrange that anymore.

I can type that in.

And we get six pounds and six pence.

So in total, the first repayment was five pounds and the second repayment was six pounds and six pence.

"That's not quite as bad," says Jacob.

So obviously paying some off early does reduce the amount that you then owe.

Right, quick check with this formula then.

So I've written the formula for you at the top.

I'd like you to fill in the missing explanations for the variables.

Give that a go and then we'll run through it together.

Right, well done if you remembered that C is the amount borrowed.

A1 is the first repayment, so A2 is the final repayment if we've got two repayments.

If you said the second repayment, it means the same thing in this example.

P is the percentage increase as a multiplier, so it needs to be a decimal multiplier for the percentage increase.

N is the number of time periods since borrowing.

So n1 is the number of time periods until the first payment is made.

Okay, Jacob's thinking about how we can get this to work, "Say I pay two equal instalments, one after two days and one after four days, how much would it cost to borrow 10 pound at Lucas's rather mean rate of 5% interest?" Now the important thing here is he wants the payments to be equal, so that's gonna require a little bit more work.

The formula for two equal repayments can be written as shown.

This time, A is standing for the repayment amount and they're both gonna be the same repayment amount.

Let's start by making A the subject.

We're gonna practise our algebraic fraction skills.

There's a couple of ways to do this.

I'm gonna look at making a common denominator to start with.

So the common denominator would be p to the n1 times p to the n.

That means the first fraction, we need to multiply the numerator and the denominator by p to the n and the second fraction, multiply the numerator and the denominator by p to the n1.

Then we can add, so we've got C equals Ap to the n plus Ap to the n1 all over p to the n1 p to the n.

Now we can multiply both sides by the denominator.

It might look a bit complicated, but there are only variables.

So we've got C, P to the n1, p to the n equals Ap to the n plus Ap to the n1.

Now we've got the variable that we want to make the subject in two different terms, so we need to factorise.

Then you can divide by the bracketed expression, so that's our final equation.

So now we can substitute the values Jacob wanted.

So, C, remember, is the amount he borrowed.

P is the increase as a multiplier, so 1.

05.

N1 is the time until he pays back the first instalment, so that'll be 2.

N is the time until he pays back the second instalment, so that'll be four.

So we've substituted in all our values and now all we need to do is type it into our calculator.

You should get 5.

78 to two decimal places.

So if each repayment is 5 pound 78, in total, he's gonna pay back 11 pound 56.

Lucas says, "I think it would've been easier to substitute the values in first without making A the subject." Now this can be the case.

You might have done lots of practise at making variables the subject, but sometimes that's actually adding extra work for yourself and it can be just as easy to substitute the values in first and then rearrange.

It's entirely up to you, but it's worth knowing that there's two methods.

So if we substitute in the values, now remember I said that there was a couple of ways to make A the subject.

This time I'm gonna show you a slightly different way.

We can actually factorise out A now.

So if we factorised out an A, that's A multiplied by 1 over 1.

05 squared plus 1 over 1.

05 to the 4.

Then because we're using a calculator, we can just type that in as normal 10 divided by, and then that whole expression.

Right, quick check, which of those are equivalent to C equals A over p1 plus A over p2? Notice that I've gone for the powers of one and two for this one.

Give that a go.

Well done if you saw the two different methods we looked at.

One where you factorised out an A and one when you make a common denominator.

Either is gonna help us make A the subject.

Right, time for a practise.

For this task, you're gonna be using these two formula that we've talked about.

So you might want to refer back to those as you work your way through or write those down so that you can use them.

Have a read at this question.

Think about what values you'll know and what values you're working out.

When you're happy with your answer, we'll have a look at the next bit together.

And then question two, what I want you to notice in this question is that we're charging 1% compound interest in both the first time the 1% is being applied every week, second time every day.

Think about how that's gonna change your equation.

I've put the formula in the top right-hand corner to help you.

Off you go.

And question four, so we're using the formula with two repayments this time.

I've helped you by telling you what the denominators are gonna be.

I'd like you to make A2 the subject and then use that to answer the second question.

Give it a go.

So Lucas's uncle borrows 50 pounds and pays back 30 pound the first month.

Lucas says, "I have done this calculation without first making A2 the subject." So can you finish off Lucas's working? So you need to substitute in the right values and fill in his answers at each stage.

Give it a go.

Perfect, so time for you to have a go yourself this time.

Work out the answer to A and then for B, you can see that I've changed the formula a little bit because both instalments have to be the same, so I made the numerator the same variable.

Think about how you're then gonna work out what A is.

Give that one a go.

Come back to the last question.

And finally, I'm showing you the formula for paying back a loan in three equal monthly repayments.

You'll notice the similarities with the previous question.

So C is the amount borrowed.

A is the repayment, and p is the monthly interest as a multiplier as normal.

I'd like you to rearrange to show that A equals Cp cubed over p squared plus p plus 1 and then you could use that to work out what each repayment would be if 100 pound was borrowed at an interest rate of 2% each month.

When you're happy with your answer, we'll look at those together.

Well done, there's so much more that you can explore with loan repayments.

Play around with the numbers and see what happens as you increase the interest rate.

So if that's something you're interested in, it's definitely something you can go ahead and have a look at in more detail.

So the first one, C equals A over 1.

08 to the power of 5.

For B, we want the repayment, so the first formula is actually gonna be easier to use, and we get 29 pound 39.

For C, we want to know how much was borrowed, so we can use our second formula, substitute in our values, and we get 13.

61.

That's 13 pounds and 61 in context.

So per week, so we're gonna have a denominator of 1.

01 to the power of 4, so 1% each week for four weeks.

That's 9 pounds 61.

If I'm borrowing it for four weeks, but the interest being charged every day, that's gonna happen 28 times, isn't it? 'Cause what I'm actually doing is I'm borrowing it for 28 days at 1% each day.

So that means I can only borrow 7 pound 57 and then pay back 10 pound.

It's worth knowing when the interest is being charged.

It makes a big difference.

For four, there's different ways you could've written that.

I've given you two examples there.

Have a look and see if you've got something that's equivalent to either of those two.

I'm gonna use that second form, because it's a lot easier to substitute into.

So for B, I can substitute my values in and I get A2 as 28 pounds 88.

So to fill in Lucas' working, we've got 50 is the amount he borrowed and 30 is the amount he paid back after one month.

Then I can do 50 subtract that whole fraction and then I'm gonna need to multiply that by 1.

05 squared.

I can use my calculator to help me and it's 23 pound 63, that second repayment.

All I need to do for 5 a is substitute in the values that I know and you've got 13 pound 86.

For b, I've got equal repayments, so I need to factorise out an A.

Then I can do 100 divided by and then I've got the sum of those two fractions.

I'm using a calculator so there's no need to additively combine those fractions.

So each instalment is 51 pound 50, but don't forget the overall cost is the two instalments paid in total, so 103 pounds and a penny.

And finally I'd like you to pause the video and look through my steps to make A the subject.

Your two main options would be to factorise out A and write that as A lots of 1 over p to the 1 plus 1 over p squared plus 1 over p cubed.

But what I decided would be easier, because of the way that the denominators are written, is to actually use a common denominator.

So if I use the common denominator p cubed, p to the 1 times p squared is p cubed.

So if I multiply the numerator and the denominator by p squared, I get Ap squared over p cubed.

Similarly, if I multiply the numerator and denominator at the second fraction by p, I get Ap over p cubed and then I can additively combine them.

Read through the rest of that now.

And that's really helpful to me because all I need to do now is substitute in the values and I get A is 34 pounds 68.

Fantastic effort with that one, guys.

Right, we're now gonna bring our algebraic fraction skills together with our probability skills.

So we can use probability trees to represent the outcome of two-stage trials.

So Izzy is spinning a spinner and flipping this fair coin.

So a quick recap, we can draw one stage of a probability tree.

So if we do the spinner first, we've got two outcomes, A or B.

A has a probability of 5/8 and B 3/8.

We're gonna spin the spinner and flip the coin.

The coin can land on heads or tails.

It's a fair coin, so the probabilities are both 1/2.

These branches are representing the outcome of the coin flip if the spinner lands on A.

So how can we complete the diagram? What more do we need to do? Perfect, we need to do the branches when the spinner lands on B.

It's gonna be exactly the same half and a half for heads and tails.

Izzy's remembered something.

To get the probability of one thing happening, then another, you can multiply the probabilities.

So probabilities of getting an A on the spinner and then heads is 5/8 times 1/2, which is 5/16.

We can do the same for tails.

Then B in heads, then B in tails.

If we've done it correctly, these should sum to one.

You could pause the video and check if you need to.

Right, so now we've had a recap of that, we can apply it to problems with unknowns.

Jun's little cousin has a bag containing 10 coloured blocks.

They are either red or yellow.

They take a block out at random, put it back in the bag and then take another block out at random.

So what's the probability of getting red block the first time? Now we don't know how many red blocks there are, but we know that there are 10 in total.

So let's say the number of red blocks is A, the probability of picking a red block will be A over 10.

How could we write the probability of picking a yellow block? Can you work this one out? So if there's a red blocks, there are 10 minus a yellow blocks.

So the probability is 10 minus a over 10.

We can display that now on our probability tree.

So how could we express the probability of getting a red, then another red.

What would it be? Perfect, it's gonna be a squared over 100.

Right, I'd like you to calculate the expressions for the other three outcomes.

Give it a go.

Right, we should have 10 a subtract a squared over 100.

And then the same for yellow, then red.

And we need the product of two binomials for the bottom one.

So we've got 10 multiplied by 10 is 10O.

10 multiplied by negative a is negative 10a.

Negative a times 10 negative 10a.

Negative a times negative a positive a squared.

So your four partial products, that simplifies to 100 minus 20a plus a squared all over 100.

Well done if you've got all of those.

So the probability of getting a red and a yellow is 8/25.

Let's see how many of each block there are.

So the probability of getting a red and a yellow would be getting a red, then a yellow, or a yellow, then a red.

So if we wanted the probability of either of those two options, we could add the probabilities together.

So we could add those two fractions and put it equal to 8/25.

Now because those two fractions are actually the same, there's a slightly easier way.

One of those probabilities must be 4/25.

Now we just need to apply our algebraic fraction skills.

I'm gonna put them over a common denominator.

Because the denominators are the same, we can put the numerators equal to each other, rearrange to equals 0, and then we've got a quadratic, so we can factorise.

And we have two solutions, a is 8 or a is 2.

Now don't forget, we need to answer the question.

The question was how many of each block are there? Well, there's either eight red and two yellow or two red and eight yellow.

We can do a quick check and it works for both options.

The probability of getting one red and one yellow then will be the combination of those two probabilities.

So 8/25, which is what we started with.

Slightly different question.

There are n number of pieces of paper in a bag, five of them have win, and the rest they lose.

Can you complete the probability tree? Give this one a go.

Right, well if there's five out of n that say win, we've got n subtract 5 over n is the probability of getting a lose, then we can fill in the remaining parts.

Then we need to do a little bit of multiplying of our probabilities.

So you get 5n minus 25 over n squared.

Same for the next probability.

And then n squared minus 10n plus 25 over n squared for the last one.

Right, so this time, the probability of getting a win, then a lose is 2/9.

How many pieces of paper are in the bag? Well, let's try it.

So win, then a lose is 5n minus 25 over n squared.

If we multiply both sides by n squared and both sides by 9, rearrange to equal 0.

And then this is a little bit trickier to factorise, but if we know our square numbers, 225 is 15 squared.

So negative 15 and negative 15 might work.

Then you just need to try it out and, yes, we do get negative 45n.

So n is 15 or n is 15 over 2.

But it was pieces of paper in a bag, so it has to be an integer, so there must be 15 pieces of paper.

Right, you've got all the skills you need now then to crack some of these problems. So to start with, I'd like you to read this scenario and complete the probability tree.

I'd then like an expression for the probability of getting two milk chocolates.

Give that one a go.

So have a read of question two and then your final question is to work out how many of each type of coin there are.

See if you can apply what we did in the examples to this practise.

Give it a go.

Well done, so this time we don't know the number of counters in a bag, but we know that three are blue and the rest are pink.

Can you use the information to work out how many pink counters there are if the probability of getting two pinks is 4 over 9? Give this one a go.

So this time, we know the probability of getting a red, then a yellow is 2/9.

Can you use that to work out how many yellow blocks there are? Try this one out.

And finally, there are 12 counters in a bag that are either green or purple.

The probability of picking two counters of the same colour is 5/9.

How many of each colour are there? See if you can do this one.

Well done, there were some tricky questions there at the end.

Right, I'd like you to pause and just check.

You've got the correct probability tree and then the probability of getting two milk chocolates is 49 over a squared.

For question two, we're looking for the probability of getting a bronze, then a silver.

So if we make a the number of bronze coins and the probability of getting a bronze and a silver is 8a minus a squared over 64.

So if we now use the probability of 3/16, the easiest way is to put them over a common denominator of 64.

Rearrange, that one factor rises nicely.

So A is 6 or a is 2.

So six bronze and two silver, or two bronze and six silver.

Well done if you got that one.

Right, the probability of getting two pink counters is n squared minus six n plus 9 over n squared.

Now there's a bit of a trick with this second one.

You could've put that equation equal to 4/9.

However, there's a more efficient way.

Because we've got the probability of two things happening and they're the same thing, so the same probability, we can actually square root both sides of our equation.

So we know that n minus three over n must be the square root of 4/9.

which is 2/3.

And this makes our question a lot easier.

So you get n is nine, but don't forget to answer the question, how many pink counters there are.

Well, it's 9 minus 3, which is 6.

You could always check that back and see if it works.

Then we're getting onto the really challenging ones.

So a red than a yellow would be 8n minus 64 over n squared and that's equal to 2/9.

I've made my life easier here by dividing both sides by two.

Then I can multiply both sides by nine and both sides by n squared.

Now this is a horrible one to factorise, so well done if you managed to spot this.

There's no reason why you couldn't have used the function on your calculator to help you if you have a calculator with an equation solving function.

It is n minus 24 and n minus 12.

Well done if you got that.

Therefore n is 24 or 12.

So there could be 16 yellow blocks or four yellow blocks.

You might wanna just check that, make sure that works.

So if there was 24 in total, you'd have 8/24 times 16 over 24 and that does give you 2/9.

There's 12 in total, you'd have 8 over 12, and 4 over 12, and that also gives you 2/9.

So we're reassured that we've got the right answer.

And then finally is the probability of getting two of the same colour.

So let's have the number of green counters as x.

Two green and two purple.

So probability of getting two of the same is if we add those two probabilities together.

So we get 2x squared minus 24x plus 144 all over 144.

You might want to just pause and check you are happy with that so far.

(mouse clicks) I've simplified by dividing the numerator and denominator by two and then I've put it equal to 5/9.

So now the easiest thing to do is get a common denominator, so 72 if we multiply the numerator and denominator by 8.

Then we can make the numerators equal and actually we end up with a nice quadratic to factorise.

We get x minus 8 and x minus 4.

Our solutions then is eight green and four purple, or four green and eight purple.

It might have looked complicated to start that one, but it ended up with a really nice quadratic and a nice way of getting our two solutions and as always, we could check that they work.

Well done.

We've covered loads in that lesson.

I hope you really enjoyed playing around with the idea of loans and you can think a little bit more about what is a sensible interest rate.

Probably not if you're borrowing from Lucas.

Well done with that second part of the lesson too, bringing those probability skills together.

Really got to practise loads of our algebraic manipulation.

Thank you for joining me for that lesson.

I really enjoyed it and I really hope to see you again.