Hi, it's Jenni with the How and Why of Mathematics and this week I want to go
through my favourite questions from the past
Australian Mathematics Competition papers. So you can buy these booklets...
this is what you would have gotten back sort of 30 years ago. This is what they
look like today. And I'll put a link in the description
for where you can buy these, but they have great questions in them, really
challenging, tricky, really fun ones that need a lot of creativity to solve. And
this week I just wanna be going through questions from the Junior
Division which is for students who are about 13 years old. Okay, first question:
so we've got a shed that's 9 metres by 7 metres and there's a goat tethered
to the corner of the shed. I'll try to draw a little goat here.
He's tied onto the corner there by a rope that's ten metres long and the
question is: how much grass can the goat graze on, like what's the area of the
grass that it can reach? This was from the 1989 competition and only 5% of
students got this one right. So pause the video, give that a go and
then I'll go through the solution. All right so the first thing we want to do
is try to visualize the area that this goat can reach. So he can go here
anywhere within ten metres so I'm sort of tracing out this circle that's ten
metres and then we'll figure out what happens there in a minute. So I'm
gonna go ten meters in every direction. Okay, and then what happens when he
reaches here? So he's used seven metres of his rope to get to there and then
he's got three metres left. So you could imagine him up there but
then he can also reach out here anywhere within three meters of there he can
reach cuz he's still got three metres of rope. So that's gonna make a little
quarter circle with a radius of three metres.
And similar sort of thing over here where... so he's used nine metres of
his rope to get to here and then he's got one metre left so he can go around in a
little quarter circle with a radius of one metre. Okay so, well... we know the area
of a circle is pi times the radius squared. If we've just got a fraction of
a circle like a quarter of a circle it's just going to be a quarter of that and
this bit is three-quarters of a circle with a radius of 10 metres. So the whole
circle with a radius of 10 metres would be pi times 10 squared but then we just
want three-quarters of that so that is this whole bit here. All right let's do
this bit: that's gonna be pi times - radius is three - squared but it's just a quarter
of that so I'll divide it by four cuz it's just a quarter circle. And then for
this one the radius is one metre so we've got pi times one squared and again
on four cuz it's just a quarter circle. So we just need to add all these
up. That one is gonna make 300 pi on four because 10 squared is 100 and then you
multiply that by the 3. We could simplify the 300 on 4 but because we're going to
be adding it to these ones which have a 4 in the bottom as well I'll just
leave it like that for now. So that one is... 3 squared is 9... 9pi/4 and that one's
just pi on 4. Now if you haven't done much algebra you might be a bit confused
about how to add all these with the pi in them. What you could just think of it
as is 300 on 4 so that much pi and then 9 on 4 pi and then another one on 4 pi.
So we'll work out what that is and then just... that's how much pi, basically. So
if we add all those up... you can just add the numerators so that'd make 310/4 and
then we can simplify that because these are both even, so let's divide both of
them by 2. That one divided by 2... I would work it out like 150 is half of 300 and then 5 is
half of 10 so it's 155.
And then we just need to multiply by pi so the answer would be 155 pi over 2.
Ok next question: we've got two jars containing a mixture of water and
vinegar. Now the jars are identical except this
one contains a mixture of water and vinegar in the ratio two to one and for
this second jar the water and vinegar is in the ratio of three to one but they
both have the same amount of liquid in them. And the question is: if we poured
both these jars into another container, what would be the ratio of water to
vinegar in that container? And this is from the 1988 competition and only two
percent of students got this one right. Okay let's have a look at it...
so with this one it's much easier to work with fractions
rather than ratios, like let's try to work out what fraction of this first jar
is water and what fraction is vinegar. So if it's two parts water to one part
vinegar then there's three parts in total so two out of three of it is
water and one out of three is vinegar. So it's like we've got two thirds of a
jar of water and one-third of a jar of vinegar. In the second jar we have four
parts in total: three parts of water and one part vinegar makes four
parts. So 3/4 of the jar is water and 1/4 is vinegar. So you can actually kind of
think of this as being the amounts of water and vinegar measured in units of
one jarsworth. So there's two thirds of a jarsworth, so you could imagine that the
jar is a litre or a gallon or whatever and that this is 2/3 of a
litre or whatever, if you want. So let's look at how much water and vinegar is in
the final container. So the amount of water... we've got two thirds of a jar plus
3/4 of a jar so we just need to add these two fractions. We need to make a
common denominator so we can do just 3 times 4 is 12 and I know this sometimes
confuses people about what to multiply by what. You just remember that to make
an equivalent fraction you multiply the top and bottom by the same thing.
So we've multiplied the 3 by 4 to get the 12; that means we have to multiply
the top by 4 as well so that's 8. And then when we do the second fraction
we've multiplied the bottom by 3 to make 12 so we've got to multiply the top by
3 as well. And then we just add these up, so 17 over 12 or 17 twelfths of a
jar of water is in the final container. All right what about the vinegar? We've
got the third of a jar plus 1/4 of a jar. Make the common denominator again
multiply that one by 4 and multiply that one by 3, so we get 7 over 12.
So the ratio... remember the question was what is the ratio of water to vinegar in the
final mixture. So we've got 17 over 12 to 7 over 12 and then usually with ratios
what you want to do is simplify that to a whole number. So you can just multiply
both of them by 12, just like with fractions you can multiply top and bottom by 12
and get a similar– an equivalent fraction, so multiply both of
those by 12: we get 17 to 7 is the ratio of water to vinegar in the container.
Okay, next question: in this one we've got four points on a straight line
and they're all equally spaced; they each have one meter between them. And the
question is: how do we get from P to S in the shortest way possible? But we have to
stay at least one metre away from Q and R, so we're not allowed to walk straight
through here, we have to go sort of around there somehow, but we want to know
the shortest path from P to S. And this one was from the 1988 competition as
well and only 5% of students got this one right. Okay so let's start by
figuring out what is our path going to look like? And what I would do is imagine
the area that you're not allowed inside is actually like a circle around Q,
because if we're not allowed within one metre of Q then we're not allowed inside
here, and same with not being allowed within a metre of R, that's like this.
So we have to stay outside of this region. So if we just walked like along the
edge of there, that's good. And we'll have to finish up by... by walking along that
edge there. But in between what do we do? We want to go sort of from here to
here in the shortest way possible. Well we can just go straight there – a straight
line is the shortest path between two points. And that is just directly above
Q and that's just directly above R despite my bad drawing. So that means
that this will just be a quarter of a circle. So we know that circumference of
a circle is 2 pI r. The radius here is 1 so that's just 2 pi but if we just want
to go a quarter of the way around the circle then it's going to be this
divided by 4 so 2 pi r over 4 which is just pi r over 2 if you simplify that.
And then since r is 1, it's just pi on two is that distance there. And then
this bit... remember they're all just one metre apart so there's a metre from Q
to R, this is gonna be the same, so that's one metre. And then this will be the same
as the pi on 2. So we've got that distance plus that distance plus that
distance that will make one plus pi. Okay, last question. So in this one they said that
you've got this big rectangle and there's all these squares inside it and
the smallest square has as an area of two by two, so its sides are each two. And
we want to know the side length of the next biggest square. Well with this
one you had to read the question quite carefully because it's sort of drawn
inaccurately and they do that on purpose so that you can't just eyeball what... what
it looks like the next square is. Because what they said was that this was
actually a rectangle, even though it's drawn as a square, so it isn't
necessarily square, like that's not necessarily the same length as that. And
they said that all of these are squares so even though you know this one is very
much not drawn as a square they're saying that that length is the same as
that length. Okay so I'm just going to write the two on this one so that we
know that the side length is 2 and then because we're trying to work out this
side length I might just call that x because we don't know what it is yet
But if... if each of the sides of this is x so
that's x and that's x, that's x. This... let's have a look at this square because
we've got this side here, we can work that out, because we know that that one's
2 and that one's x so that's going to be 2 plus x and then you've got this
square here – remember these are all squares – so if we know that that's 2 plus
x then this also has to be 2 + x. And this side's 2, so we've got a 2 and a 2
plus x so that length there is 4 + x, and that also means this length here is
4 + x even though it doesn't look the same as that one. Okay what else can we
work out? We could look at this square, so we've got that length is 2 + x and
that one's 4 + x so together that's gonna make 6 because the 2 plus 4 is 6,
and then you've got two x's. All right, now we could look at this one. We've got
a 6 + 2x and a 4 + x so 6 plus 4 is 10 and then we've got a 2x and an x;
that makes 3x. We could also do this one. We've got that length is 6 + 2x
that one's 2 + x and that one's x so let's add all those up. The 6 and the 2
make 8 and then you've got a 2x and x and another x so that'll be 4x. Okay, we
could do this one. We've got the 8 + 4x and the x that'll make 8 + 5x and
then we could also do this one from all of these lengths added up. So a 2 and a
4 + x and the 10 + 3x. Two plus 4 plus 10 makes 16,
and then we've got an x and a 3x is 4x. Okay,
well now what we could look at is this length along here, so we could measure it
as being the x plus 8 + 5x or we could also work it out from being 2
and 16 + 4x. Those have to be the same thing, so let's write that down
as an equation. We've got the x plus the 8 + 5x and that has to equal
2 plus the 16 + 4x, so that'll give us an equation that we can solve
for x. Let's simplify this. We've got an x and 5x is 6x. The 2 plus the 16 make 18.
Now what we could do is move the 4x over to the left, so subtract 4x from both
sides of the equation. That'll give us 2x + 8 = 18 and then we could
subtract the 8 from both sides to move it across.
That'll give us 2x = 10 and then we can divide by 2.
That'll give us x = 5, so there's the answer. All right, let me know in the
comments what your favourite maths question is and if you liked this video
then subscribe because I'm planning to make more videos like this in the future.