GCSE Maths Foundation "Best Guess" Predicted Paper 1 | 14th May 2026 | TGMT

Added: 2026-05-12

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[00:02:24]So, write these numbers in order of size. Start with the smallest number.

[00:02:27]So, 67 is our smallest number here as that's in the tens. We're then looking at the hundreds digits. So we have two that are in the 100s. We have 112. And then we have 151. And then we have a number in the 200s 245. And our final number in the 300s 390.

[00:02:49]When we are writing a decimal as a percentage. We will there are two ways that we can go about this. We can either write this as a percentage straight away or we could actually write this as a fraction if it helps. Seven is in the hundreds position. So it is 7 hundreds.

[00:03:05]And then any number as a percentage is the number over 100. So that would just be 7%.

[00:03:11]Just being careful there because a lot of people accidentally write 70% potentially. So just being really careful when you are converting your decimal to a percentage. We're solving a one-step equation. Remember to think logically. This does just mean a number plus 9 is equal to 8. And we're just trying to figure out what that number is. Now to solve this algebraically, we would subtract 9 from both sides and we get n is equal to and 8 takeway 9 is -1.

[00:03:41]And there we go. Solving a one-step equation. Writing a number or an improper fraction as a mixed number. We want to know how many times 8 goes into 35. To make sure we get this right, we can just write down sum of the eight times table. So 8 16 24 32. So it's going to fit in four times and there's a remainder of three there up to 35. So we have three out of eight left over and that's converting it into a mixed number.

[00:04:11]There are a few main angle types that we need to know. This one here is less than 90. So we call this an acute angle.

[00:04:18]Obviously on the 90° we call that a right angle. Then we have an obtuse angle up to 180. And then we have a reflex angle over 180. And there we go.

[00:04:29]This one is an acute angle.

[00:04:33]This one says, "A cafe is creating meal deals. A customer can choose one item from menu A and one item from menu B."

[00:04:40]It wants us to write down all the possible combinations a customer can choose. So when we're doing this, we want to pick the first option, which is the sandwich, and we're going to pair that with each of the three options in menu B. So we'll have SJ, S M, and S C. And what we're going to do is replicate that with all the other options from menu A. So we'll have WJ, W M, and W C.

[00:05:11]And then onto the next one, we have A for salad. So A, J, A, M, A, C, and then P for pasta. P J, P M, and then P C. And there we go. That is systematic listing. Just listing those different options and combining them in a nice logical order.

[00:05:36]Okay. So, we have a list of numbers here. And it says from the list, write down the odd number. So, the odd number is over here. It's 27. The multiple of seven. So, numbers in the seven times table. So, 7 14. So, there we go. 14 is the multiple of 7. The square number, well, I wouldn't there's no 1 * 1. 2 * 2 is 4. 3 * 3 is 9. We don't have 16. We don't have 25, but we do have 36 at the end there. So, the square number is 36.

[00:06:06]The prime number over here is two. Two is the only even prime number. And all of these other numbers divide by two except for 27, which divides by 3 and 9.

[00:06:16]So, that's not prime. So, there we go.

[00:06:18]two is our prime number and then two numbers with a sum of 41. So we want to get one of the larger numbers in order to get to 41. Hopefully we can spot it there. It is actually 27 and 14. 27 add 4 is 31 plus the 10 is 41. So the two numbers would be 14 and 27. There we go. That's identifying some number types.

[00:06:44]Here we are collecting like terms. So it says to simplify 7 a + 5 b - 4 a and - 6 b. So I like to pair them up. So we have a 7 a and we also have a - 4 a. So 7 take away 4 gives us 3 a. For the next one we also have a 5 b minus 6 b. That's one more than five there when we're thinking about the size of five and six.

[00:07:10]So that's going to go down to minus 1 b.

[00:07:13]So we can just write minus b for that.

[00:07:14]We don't need to put the one in front, although it wouldn't be wrong to do so.

[00:07:17]And there we go. That is collecting like terms.

[00:07:21]So, we've got a lot more words in this question. We are constructing a two-way table. Now, it says here there are 60 students in total. So, we're going to go down and put 60 in our total box. 40 of the students are boys. Now, we can just keep going through ticking these off and filling the numbers in. But, if one pops up where we can find a missing number, typically I'll go and fill that in. So, we can work this out. That is going to be 20. That gives us that total there of 60. We're then told 12 of the girls play the piano. So 12 here in the girls that play piano. 15 of the students who play the guitar are boys. So 15 in the boys guitar.

[00:08:01]And then eight of the 20 students who play the drums are girls. So it tells us eight are girls out of the 20 students that play drums. So that gives us two bits of information there. So watching out for one of those lines. And we can also figure out there for the boys that play drums. They've got to add up to 20.

[00:08:18]So that would be 12.

[00:08:21]Each student plays exactly one instrument, piano, guitar, or drums. Use information to complete the two-way table. So we've got most of the information in there. We now just need to complete the rest using the numbers.

[00:08:33]So looking at the top row, that has to add up to 40. At the moment, we have 15 and we have 12. That adds up to 27 which gives us 13 to go over here in the boys's piano. We can now work out the total of that row as well or that column there. So 13 + 12 which is 25. Probably a little bit nicer than working out the row. The row when we're going across adds up to 20. We have girls that play the piano as 12 and girls that play the drums as eight. That already adds up to 20. So there are no girls here that play the guitar and that leaves us 15 as a total for that column. Says one of the students is chosen at random. Write down the probability that the student is a girl who plays the piano. So identifying the amount of girls that play the piano.

[00:09:19]We have 12 girls that play the piano.

[00:09:22]And that is going to be then 12 out of the 60 students. Just watch out for the language in a question like that. This one does say one of the students is chosen. It could say something like one of the girls is chosen which would change our denominator of 60 here to the total amount of girls. So it' be 12 over 20. But it does say one of the students.

[00:09:40]So it is out of 60 there. And that is writing it as a probability.

[00:09:45]And looking at some coordinates we have the diagram shows two points B and C on a grid. Write down the coordinates of point C. So we want to find that x coordinate first. So line up with the x axis. We have a coordinate of one there.

[00:10:00]So the first number in our coordinate will be 1. And then we want to look down the y- axis, line it up, and it's down here at minus2. So 1 and minus2 is our coordinate for point c. Says the coordinates of point a are -3, -2. On the grid, mark with a cross the position of a and label the cross a. So -3 will go left along the x-axis to -3 here, down by two, which gets us to here. So we'll put a cross there, and we'll label that as a. And that is labeling a coordinate.

[00:10:32]Now it says to find the coordinates of the midpoint of BC. Now the midpoint of a line we want to have a look down the line here which is going diagonally and we want to find where halfway is. Now hopefully we can see that visually. We can see here there are two diagonals getting me to that point and another two diagonals getting me to this point. So that indicates where the middle is just there. And hopefully we can kind of see it visually most of the time. But that is our midpoint. Now being careful because we again we want to give our x coordinate first. The x coordinate is on -1 so -1 but it is on the x axis. So the ycoordinate there is zero and there is the midpoint of bc.

[00:11:13]Final part here says on the grid draw the line with equation x= 3. Now the x axis is just here. If we highlight the x so when we are drawing the line x= 3 we want to actually find three on the x axis. Now, here is three. And we're just going to draw a straight line with a ruler going down through three. There we go. Make sure that's nice and straight.

[00:11:35]And there is the line x= 3. Now, you don't have to, but you can label your line as well. There we go. There's x= 3 going straight down through where x is equal to three. A nice little question here. If you know your polygon names, it says, write down the mathematical name for a sixsided polygon. Now, a six-sided polygon is called a hexagon.

[00:11:58]So, you just need to know some of those polygon names. You've got your pentagon, which is five sides, hexagon, six sides.

[00:12:05]One another one to definitely know is an octagon, which has got eight sides.

[00:12:08]Beyond that, we do tend to call them 12-sided polygons, you know, and 20sided polygons rather than necessarily using the name, but there are some that you're definitely expected to know, and hexagon is one of them. When it comes to unit conversions, there are quite a few different ones, but this one here very important to know that we there are 100 cm in a meter. So in the 4 m, that would be 400 cm. But this is 4.8 that 8 there is less than 100 that is 80 cm. So in total we have 480 cm and that is our unit conversion here converting meters into centm.

[00:12:51]There are lots of different types when it comes to using a ratio. This one here says that there are only blue pens and black pens in a box and blue to black is 2:5. Says there are 24 blue pens in the box. Work out the total number of pens in the box. So here we have 2 to five and blue is the number on the left. So it's telling us that the 2 is equal to 24. So we need to figure out, well, how do we get from 2 to 24? And that is multiplying by 12. What we have to do is do exactly the same on the other side.

[00:13:22]So we're going to do 5 * 12. Taking your time to work that out. 5 * 12 is 60. And then we can answer any question here. We could say that there are 60 black pens.

[00:13:32]This specific question says the total number. So the total number here is the combined total. and that is 84 pens. So 60 + 24 gives us 84 in total.

[00:13:45]So we told here this graph can be used to change between US dollars and British pounds. Change 15 British pounds into US dollars. Now 15 British pounds is on the axis. It's just down here. So what we need to do is trace up really carefully to our line and then go across. And you can see there that lands on $20.

[00:14:09]So our answer would be 20. And obviously we'll include the symbol for dollars unless it's on our dotted line. The next part says change 52 into British pounds. So getting us to properly use the actual scale on our axes. So between 50 to 60 up here there are five squares meaning each square represents two. So 52 will be one square up from 50. So again, we'll use a ruler to go across. We'll hit our line, bounce it down, and it actually lands perfectly on £40. And there we go. £40 is our answer. Now, the final part here says Steven needs $30 to pay for a hotel stay. He has £250 British pounds. Show that Steven has enough money to pay for his hotel stay. Now 250 British pounds is not anywhere on this particular graph. It only goes up to 50. So we need to find a factor of 250 which will nicely then scale up to 250. And there are lots of different options here. Now of course we're doing this without a calculator. So we want to find one that's relatively easy. For example, going for five and then trying to scale that up is not going to be as nice as either picking 25 or even 50. Now 25 I think is quite a nice one because we just need to multiply that by 10 to get to 250. So I would go down to here at 25. We'll go up across. Now we're going to read this as best as we can because it is ever so slightly above that line there, but that comes out around 32. So, if £25 is equal to £32, then to get to £250, which is how much Steven has, we need to multiply that by 10. And when we times that by 10, we get £320, and therefore he has enough. He needed 300. So, 320 is going to cover the bill for his hotel stay. Here, we're looking at a scale model. It says here is a prism and we have a diagram of the prism. It says male makes a model of the shape. He uses a length of wire to make each edge of the model. Each edge of the model is 5 cm long. Male has 90 cm of wire. What length of wire does he have left after making the model? So we are here looking at the edges of the shape.

[00:16:39]So if I was to highlight one of the edges, this here is an edge.

[00:16:45]Now on this shape, we have five edges on this front face. So that's five in total for that bit. We also have some at the back. We've got one, two, three, four, five again. So there are five on the front, five on the back, and then other than that, we just have these edges here that are connecting the front and the back. And again, there are five. So in total we have 15 edges on this particular prism. Each edge is 5 cm long. So the calculation that we need to do is 15 edges multiplied by the 5 cm. Now we're doing this without a calculator. So we want to make sure we get that correct. 5 * 5 25.

[00:17:36]5 * 1 is 5. Add the two 7. So 75 cm.

[00:17:43]Now it says here that male has 90 cm of wire. So he has 90 cm.

[00:17:51]He's using 75. And that leaves us with 15 cm of wire that's left over. And there is our final answer. That is the length of wire that's left after making the model.

[00:18:04]So when we are looking at finding midpoints, this particular question says find the number that's exactly halfway between 8 over 25 and 0.88.

[00:18:14]Now when we are looking at midpoints, we definitely want these in the same form.

[00:18:19]So we have two options really. We could either convert them both to fractions or we can convert them both to decimals.

[00:18:27]Now it doesn't actually really matter either way. Both is technically the exact same process. It just depends which one you prefer looking at. Now, as a fraction, 8 over 25, for starters, we want to make that over 100 so that we could maybe write it as a decimal. Or I could write 0.88 straight away as a fraction over 100. Either way, we need to get that over 100. So, I'm going to times the top and bottom by four, which gives me 32 over 100 or 0.32, 0.32.

[00:19:01]So when I compare that now to 0.88, I can actually think about a process for finding the midpoint. Now I don't need to worry too much about the fact that it's a 0 point number. What I'm concerned about is the 88 or 88 and 32.

[00:19:18]If we think about these as whole numbers rather than decimals, it can be a little bit nicer just to find the midpoint because 32 and 88 to find the middle you just have to add them together and divide it by two. Now when you add that together you get 120. So it's 120 / two which is 60.

[00:19:42]Now of course we need to obviously re relay that back into the fact that it was a decimal. So the midpoint of 0.88 and 0.32 would be 0.60 or you could just write 0.6.

[00:19:57]Now of course there are different ways of writing that. If you did that as a fraction it would be 60 over 100. That is the midpoint if we keep it over 100.

[00:20:07]Of course you could simplify it 6 over 10 or 3 over 5. It doesn't really matter. It doesn't state here that the answer has to be given in a particular form. So any of those would be absolutely fine. But there we go. 0.6 is the one that I've got written down. So when we're looking at laws of indices, there are only three major ones that we are going to be using quite a lot. We have the division law here. As long as we have the same base, which we do, a and a. We can just subtract powers when dividing. So 12 subtract 4 is 8. So my answer is a to the power of 8.

[00:20:45]For part B, we are multiplying and we've got various different parts here. Now, numbers are always treated as numbers.

[00:20:53]So, what I mean by that is the three and the five, they just multiply to make 15.

[00:20:58]It's only the powers that we are now going to apply power rules. Now, when multiplying, which is the inverse of dividing. So, instead of subtracting the powers, we add the powers and we look at these letters individually. So we have a b to the^ of 4 multiplied by a b to the^ of two. Adding those powers gives us b to the power of six. And the final piece here is the c. And you got to be really careful because this c over here doesn't have a power written with it. Which means it is a what I call a secret power of one. So those powers there are going to add not forgetting the one. So c to the five add one is c to the power of six.

[00:21:42]So in that question there, we had the division and multiplication rule. The only one we didn't have is where there are powers outside of brackets. When there are powers outside of brackets, you multiply the powers. It's the only other major one that sits alongside these. But there we go. Division and multiplication using laws of indices.

[00:21:59]Now when we are looking at solving equations, of course, we have two main types where there is an unknown on one side and like this one where there is an unknown on both sides of the equal sign.

[00:22:10]Now, in this question, we do have a bracket on the right hand side there.

[00:22:13]And typically, whenever I see a bracket in a question, I'd always expand it. So, before I even try and start solving this, I'm going to expand this bracket, multiplying both of those by 3. So, on the left, we still have 5x - 6. And on the right, we now have 3 * x, which is 3 x, 3 * 1, which is 3. and keeping the symbol the same.

[00:22:38]Now we need to have a look for the smallest coefficient of x, meaning the smallest number in front of an x value.

[00:22:46]So on the right we have a 3x. On the left it's a 5x. So 3 is the smaller number there. Watch out cuz that can sometimes be negative which means you have to add it to the other side. This is a positive 3x though. So we need to subtract 3x's from both sides of the equation. That gives me 2x on the left.

[00:23:05]Still - 6 equals. And here's where we got to be really careful because that is a -3 there. And that negative doesn't disappear just because we've got rid of the 3x. So I need to keep -3 on the right hand side.

[00:23:20]I now want to isolate that x. So I don't I just want it to say 2x equals. So to get to that point, I have to add six to both sides. Thankfully that does make the right hand side positive now. So we get 2x is equal to -3 add 6 is pos3.

[00:23:38]I can now figure out the value of 1x. So we have 2 x's on the left. So to get to 1 x I'm going to need to divide by two.

[00:23:46]And on the right there three doesn't divide perfectly by two meaning it doesn't give us a whole number. So I can just leave that as a fraction 3 / two.

[00:23:57]Now you might know cuz 3 / two is quite a nice one. Or you could write your answer as 1.5 or 1 and a half. Any of those would be absolutely fine. But I would just leave my answer as a fraction unless the question ever said give your answer as a decimal. So there we go.

[00:24:12]Solving an equation with unknowns both sides and brackets.

[00:24:17]So in a reflection and here we are looking at a reflection. Hopefully you can see that from the diagram. It says describe fully the single transformation that maps triangle B onto triangle C. So we need to state the type of transformation. That's the first word we should write down. It's either a reflection, a rotation, a translation, which is to move it, and an enlargement.

[00:24:40]So here we want to say it's a reflection, and that's the very first word that we are going to write down.

[00:24:48]Now, every type of transformation, you have to describe how it's been done. So in this case, a reflection is reflected in a what I guess we could refer to as a mirror line or a reflection line. Now the reflection line in this case is quite a nice one. It's straight in the middle of these two going down the y- axis. So we would just say a reflection in the yaxis.

[00:25:15]There is another way of describing the y- axis. I'd never use it, but the y- axis can also be described as x= 0. cuz it's the line that goes through x= 0.

[00:25:25]But this one here, when it's on the axis, I just keep it nice and easy and just write the y- axis or of course if it was the other way around, the x-axis.

[00:25:33]There we go. That is describing a reflection. So in this question, we are told the nth term of a sequence is 8 - 6n. Is 58 a term of this sequence? You must show how you get your answer. Well, a minus 6n does mean that the sequence is going down. So there is a very good possibility that -58 is in the sequence.

[00:25:56]Now to generate a sequence we substitute the position in place of n. So the first number in the sequence n will equal 1.

[00:26:06]And when we do that the first number would give us 8 takeway 6 * 1. Got to be careful there. We do the multiplication first. 8 takeway 6 is equal to two. So the first number in the sequence is two. Now what we could do if we understand what that code means it means it's going now down in sixes. So if I wanted to just keep writing the sequence out I could go 2 takeway 6 which is -4 takeway 6 which is - 10 and so on and so on. So that is my fail safe method. I got to be really careful there that I don't make a mistake cuz I am counting in negatives. Little tip there is just to count up in sixes from 10 and stick the negative with it. So 10 add 6 is 16. So the next one's -6. 16 add 6 is 22. So - 22 if that helps. But we could try and jump a little bit closer to 58.

[00:27:00]I know that 8 take away let's say 60 should get me pretty close. Now you might argue that actually I'm just going to count down in sixes. But if I do 8 take away 60 where n equals 10 that gets me pretty close. 8 take away 60 which is 52.

[00:27:24]So I'm really really close to 58. And actually if I take away another six I actually get 58 which is 8 take away 66.

[00:27:36]So actually it doesn't really matter which method you use here. But the important part is is 58 a term of the sequence? I would say yes. And whichever method I have used obviously it's important that I show that. So the method on the left I could show that I have subtracted six each time to get there. Or the method on the right I have shown that I have substituted in particular values and actually have found that it's 58. Now if it wasn't in the sequence so for example maybe the question says is minus 56 in the sequence that method on the right there or on the left would show that nicely too as you would get 52 and then jump past 56 and we'd hit 58. So either one of those methods is fine whichever you choose whichever you prefer. Okay so looking at some fraction calculations this one here we have a multiplication of fractions. So when we are dealing with fractions we want to make sure that these are written as improper fractions.

[00:28:30]So to do that we'll take the large number in this case 1 multiply that by our denominator which is four. So that's that one represents four quarters plus the extra three which is 74 or 7 over 4.

[00:28:45]The same for the right we have 1 * 3 which is 3. Add the one which is 4/3 or 4 over 3. Now from here because we are only multiplying we can actually just multiply the top and multiply the bottom. When dividing you have to times by the reciprocal. So we have to flip that second fraction over when dividing and do a multiplication. But with a multiplication straight away we can just times the top 7 * 4 28 and 4 * 3 12. So 28 over 12 is our answer as an improper fraction. Now this question does say give your answer as a mixed number. So to write it as a mixed number we want to write how many times 12 fits into 28.

[00:29:27]Now it fits in twice up to 24. So we get a large two and from 24 to 28 there is a remainder of 4. So 4 over 12.

[00:29:37]Now this is an interesting question because it only actually says give your answer as a mixed number and normally it has this other language along with it that says in its simplest form. Now this one doesn't say that. So 2 and 4 12ths would be absolutely fine, but 4 over 12 does simplify. And I think it's always good practice to do so if it doesn't say to do it in an exam question. You don't need to, but in this case, I think it's a nice one to simplify. They both divide by 4, which would give you a 1 over three. So 2 and a3 would be your simplified version of that. You might have simplified it in the first place.

[00:30:08]You might multiply fractions by cross-ancelling, which is another method people can use as well. But there we go.

[00:30:15]There is our final answer. 2 and 1/3 for this fraction calculation.

[00:30:22]Now, a quick little question here. When you know what a chord of a circle is, there are certain circle pieces that you need to make sure that you know, particularly radius and diameter, which you're going to use when working out the area or circumference of a circle. Also, a tangent, which touches the circle at one point on the outside.

[00:30:42]But a chord is just a straight line that passes from one point on the circumference to another point on the circumference. So we're going to pick any point, join it up to any other point, and there we go. There is a chord. Now, a chord could also go through the center, but that does have the special name, the diameter. However, it is a chord. But there we go. That is our chord drawn for us. A nice quick little question on circle parts.

[00:31:09]Okay, so a larger problem here. And you can kind of spot when a question is going to be a little bit more complicated, particularly when we have a diagram. When there's two shapes involved, there's going to be more to think about. Says here is a triangle and a rectangle. The height of the triangle is x cm, which we can see on the diagram. The area of the triangle is 20 cm squared greater than the area of the rectangle. Work out the height of the triangle. So this concept is about reverse area. Essentially, we can work out what the area of the triangle is because it's 20 cm greater than the area of the rectangle. And the rectangle, we have the length and the width. So, we can start here by working out the area of the rectangle. The area of the rectangle is length time width. So, 5 * 6, which is equal to 30. So, that's the area of the rectangle. It says that the triangle is 20 cm squared greater. So if we add 20 to 30, that gives us an area of 50 for our triangle. Now's the part where we have to really know about area of a triangle because we're going to have to work in reverse here. Now, there's two ways that you could think about this, but ultimately to work out the area of a triangle, we do base time height. So 8 * x you could write as 8 x / 2 for a triangle equ= 50.

[00:32:36]So we kind of create this little equation. Now the little equation isn't actually too bad. If we get rid of that time sign, it's 8x / 2 equals 50. And we just need to solve that. Now to solve an equation like this, we need to first do the inverse of dividing by two. So getting rid of this divide by two and we are left with 8x equ= 100.

[00:33:03]Here's where the slightly complex part comes in for a non-cal question because 8 doesn't fit perfectly into 100. Now it is one of those ones which when you've seen it enough times you might just remember how many times eight goes into 100. But if you don't, particularly if you're in an exam situation and this pops up, we can do bus stop division. So 8 into 100, 8 fits into 10 once, remainder two fits into 20 twice, up to 16, remainder four. So I need to put a zero in, carry the four over, keep my decimal point in the same place. And 8 fits into 40 five times. So x equ= 12.5 and that would be my answer 12.5 cm. So that 100 divided by 8 1/8 might be one that's nice that you can remember. So there we go. 12.5 was my final answer for that one. Don't let things like that throw you. If you need to use bus stop division in a non-calculated paper, just use it straight away. Don't hesitate.

[00:34:13]Okay. Okay, so looking at some estimations, it says a charity sells tickets for a raffle. It receives 91 p for every 10 tickets sold. A total of 513 tickets are sold. The charity gives 5% of the money raised to a local animal shelter. Work out an estimate for the amount of money donated to the local animal shelter. So for this question, it says estimate. So we are not going to do 513 tickets times 91 p. We are going to round those values ideally to one significant figure. Now because 91 has been written in terms of pounds rounding that to one significant figure would give us 90 p.

[00:35:00]So 90 p which you could also just write in pence if you want 90p and the tickets 513 to one significant figures is going to be 500 tickets.

[00:35:16]So the charity it says it receives 91 p for every 10 tickets sold and we sell 500 tickets in our estimation. So for starters, we want to know well how many 91 PS or in our case 90 PS are we actually going to get. So 500 divided by the 10 tickets is going to give us 50. So that's 50 lots of 90p that we are going to get. We can now do 50 multiplied by the 90p. You could either do time 90 or time 0.9. Might actually be easier because we're doing estimations, which is what makes these calculations nice and easy just to keep it as 90p.

[00:36:02]5 * 9 is 45 and two zeros. Just be careful cuz that is still in pence. So when we put that into pounds or divide by 100 for pounds, £45.

[00:36:19]So we have £45 from this particular set of tickets or this raffle.

[00:36:26]It says the charity gives5% of the money raised to a local animal shelter. So we need to work out 5% of £45.

[00:36:35]So £45 we can work out 10%.

[00:36:40]Divide that by 10 that would be £450.

[00:36:46]to get to 5% from 10% divide by two. So we can divide that by two quite nicely. Again we could use bus stop division but 4 / 2 is 2. 50p / 2 is 25p. So I end up with £225.

[00:37:08]There we go. There's our answer. £225.

[00:37:12]says in part B, is your answer to part A an underestimate or an overestimate?

[00:37:18]Give a reason for your answer.

[00:37:20]Well, look back at our numbers because this all depends on how we have changed our numbers in terms of our calculation.

[00:37:27]We changed 513 tickets to 500 tickets.

[00:37:31]So that's actually 13 less tickets.

[00:37:34]We also changed the amount of money from 91p to 90p. So we in our calculation have sold less tickets for less money.

[00:37:44]So ours is an underestimate and our reason for that is because we have rounded both the tickets down and the cost of the ticket down. So you would need to write down an explanation for that. I'm not going to write one down here because you could come up with your own words. But it's an underestimate because both the tickets and the amount of money were both rounded down. So we would have got a larger value.

[00:38:05]Okay. So finding the highest common factor of 72 and 90. Now typically, especially when we have some slightly harder numbers like this, I'm going to want to look for the easier number to start finding factors of. And to be honest, neither of these are necessarily easier. But 90 actually has quite a lot of factors. We have one and 90, 2 and 45, 3 and 30. Four doesn't go in. Five does.

[00:38:35]Five is always one that people tend to miss. So remember, if a number divides by 10, especially, think about how many times 10 fits in. 10 fits in nine times.

[00:38:44]So five fits in twice as many times as that. 18 times. Just a little trick for dividing by five, which I quite like.

[00:38:52]Six. Does six go in? That's one of the ones that I think we'd have to just probably go to the side and maybe do some bus stop division, which is absolutely fine to do. But rather people do this to the side and make sure they don't miss a factor than potentially making a guess and losing a mark. So six goes into nine once remainder three and then five times. So it does go in 15 times. Another one that I think a lot of people there would miss.

[00:39:16]Seven doesn't, eight doesn't, nine does, 10 times. Now those numbers are joined up. I know I've got the others right, but I know I've got all the factors there. Now, I wouldn't always write down all the factors, but this question here, a lot of people, I think, would write the answer nine. And actually, when you start to go from the top of the list here, 90 doesn't fit into 72. Neither does 45. Neither does 30. 18. Actually, it does. 18. 36. Not the nicest one to actually count up in here. 54 72. fits in four times. So actually the highest common factor is 18. I do think that's the type of question where a lot of people are going to think ah I know that it's nine and just write down nine.

[00:40:04]So always check work your way through at least one of them whether that's 72 or 90. Write down all the factors you can possibly find and then just double check from the top. Have I definitely not missed one? So there we go. Our answer highest common factor here 18.

[00:40:20]Okay. So dividing decimals definitely one that I would use bus stop division for. There are other methods. Feel free to use which whichever you prefer. I just think taking the decimals out. So writing 63,6,372 divided by 12 and just working my way through using bus stop division. I might want to write down some of the 12 times table just to make sure I get this absolutely perfect. So 36 48 60. I'm going to stop there because I'm only going into 63 here. So that's five times remainder three. My list still works for that one. 36 three times remainder one.

[00:41:01]And then 12 once. So actually it worked out quite nicely. I only have to write out a few of the 12 times table. So we get the answer 531.

[00:41:12]Now I think in this sort of process in this type of question the best thing to do just to get the answer nice and quick here is just to estimate the answer. So here if I was going to estimate it 60 / 1 is equal to 60. So I know that my answer should be close to 60. So this would be 53.1.

[00:41:40]And there we go. And that would be Manta 53.1.

[00:41:44]And there we go. Dividing decimals. Nice and simple, nice and quick. Not stressing out at this particular point here. Just getting your bus stop division done. Estimate the answer to see what size it should be and putting your decimal in the correct place. Okay.

[00:41:58]One of these worded questions where we have a mixture of fractions, percentages, and ratio. It says, "A warehouse stores 520 boxes of goods. The boxes contain electronics, clothing, toys, or books.

[00:42:12]3/8s of the 520 books contain electronics.

[00:42:16]We could just start working that out straight away because sometimes when you go through this whole question, you can kind of get lost. But I think reading through it twice is probably the best idea. So 38 electronics. We'll come back to that. 30% contain clothing. And then toys to books is in the ratio 6 to 7.

[00:42:34]work out the number of bo boxes containing books.

[00:42:38]So first of all we will work out the 3/8s of 520.

[00:42:44]So 1/8 is found by dividing by 8. So 520 divide by 8. We have a lot of bus stop division going on. 8 again take your time. Just count your way up. Write your eight times table down to the side if you need to. 8 * 6 is 48. So it goes in six times remainder four. and then five times into 40. So 1/8 is equal to 65.

[00:43:07]Now we need 38s. So we want to multiply this by three. So 65 * 3, take your time with that. 60 * 3 is 180. 5 * 3 is 15.

[00:43:21]So add those together, 195.

[00:43:25]30% of the 520 box uh boxes contain clothing. So for this one, we'll work out 10%.

[00:43:32]I always say our most important percentage and that is equal to 52. We want 30%. So similar to the fraction there, we're actually going to times this by 3. 52 * 3, 50 * 3 is 150. 2 * 3 is 6. So 156.

[00:43:52]So we have 195 electronics, 156 clothing.

[00:43:58]We need to work out what's in the remaining boxes. So where it says the number of boxes containing toys to the number of boxes containing books is in the ratio 6 to 7, that's not the original 520. That's only in the boxes that are toys and books. So we do need to work out the total of the electronics and the clothing and then figure out what's left over to split into that ratio. So add these together.

[00:44:26]15 351 and we want to take that away from 520.

[00:44:33]So 520 subtract 351. Take your time.

[00:44:37]This question is all about written methods. So we're going to have to borrow. There we go. 10 take away 1 is 9. Going to have to borrow again.

[00:44:47]11 take away 5 is 6. 4 takeway 3 is 1.

[00:44:51]And we have 169 left over which is split into the ratio 6 to 7.

[00:44:59]Now that ratio there is boxes to books six 7. We need to figure out what one part is worth. So in total there there are 13 parts.

[00:45:18]169 / 13. It's actually a square number.

[00:45:23]It's actually 13 squared. But if you're not sure of that, 13 into 16 goes once, 13 into 39 goes 3 is 13. 13 * 13 is 169.

[00:45:34]So, actually, these add up to 13, but we also need to times them by 13. So, again, you want to take your time working these out. 6 * 13 6 * 10 is 60.

[00:45:46]6 * 3 is 18. So that becomes 78. 7 * 10 is 70. 7 * 3 is 21.

[00:45:57]So 91. And there we go. So 91 is for the books. It says, work out the number of boxes containing books. Well, we found that 91.

[00:46:08]And there we go. There is our final answer. So a long question. It involves a lot of steps. It's a long question because it's it's all the written methods involved. Addition, subtraction, multiplication, division, everything is within that as well as knowing about fractions, ratios, and percentages. They are answered very badly these types of questions. So the really important thing is keep everything super organized.

[00:46:30]Okay, so we've got quite a complex linear graph here. If you look, it says that the equation of this linear graph is 3x + 2 y = 6. We have to draw the graph from the x values -2 to 4. Now typically and if the equation was just you know y = 3x + 2. Quite nice. We just set up a table. We do x and y times each x value by three and add two and we draw it in. Really nice. So this is a much harder equation. We've got two options to do this. Number one is we can either rearrange the equation. So making it y equals I don't still don't think it's particularly nice but if we minus 3x we get 2 y = - 3x + 6. Then you can divide by two. So it says y = which is - 3 over 2x + 3. You could go through tsing each x value by - 1.5 and adding three. That will work absolutely fine. It's not the nicest of equations though. And I think there's a really nice trick that you can apply to a question like this. So if I write that equation out, look 3x + 2 y = 6. There are two values which we can find really easily. Those two values are the value that sits on the x axis or and the value that sits on the y- axis. If we can find two values, as long as we're nice and neat with a ruler, we can extend that line. Now the way I do that is I say okay well on the x- axis y equ= 0. If this equals 0 essentially this is gone. So we've got 3x = 6. Well in that case when y equals 0 x = 2. 3 * 2 is 6. That gives us a point. When y is 0 x is 2 right there.

[00:48:30]So that gives us one coordinate straight away and quite a nice one. We can apply the same logic for the other side. I say okay well what about when x is zero?

[00:48:41]That gets rid of this side or does y equal 2 y = 6? So y is equal to 3 and that goes just there.

[00:48:52]So applying that bit of logic and just thinking well one when one is zero what must the other be actually gives us these two coordinates really nicely.

[00:49:02]If you're super neat with a ruler you could now extend that line. Or what you could do is just have a look now at the pattern between the points. So when I say pattern I mean it goes 1 2 3 down and then in terms of the squares 1 2 3 four across. That's actually just two across because of the way the axes are scaled. But in terms of the squares, it's four across. If I do the same, 1 2 3 1 2 3 4. That gets me here. And I can do the same going backwards. 1 2 3 4 1 2 3 Right at the very top there.

[00:49:44]And there it is. I can just now get my ruler, join it up, and there we go. There is my straight line drawn. So, a little bit of a trick there. Not you don't have to use that method. Of course, you can always draw a table, which I'd advise for most types of linear graphs. Just draw a table, get it all nice and confidently drawn in, but just like this one where there is a something a little bit horrible in there. That little method might just give you a little trick if there's a nasty question that comes up.

[00:50:14]Okay. So looking at inequalities says here is an inequality in x shown on a number line. So it says here it's in x.

[00:50:21]If it doesn't say that just check the inequality there on the number line. We are using the letter x here. Write down the inequality.

[00:50:30]So this inequality here is pointing to the right. It's pointing to the bigger numbers bigger than -1. And that is the exact inequality we're going to write down.

[00:50:40]You don't have to write the letter first, but I always write the letter first. Then I draw my inequality pointing in the same direction as the arrow, which means greater than.

[00:50:51]Literally the same direction as the arrow is pointing. And then we write the number that it's above. So x is greater than one.

[00:51:02]Now, watch out that circle. If the circle, which this one isn't, but if it was colored in, I refer to it as having extra ink in the circle. We think of it as pen being drawn, then there's an extra line on the inequality. Okay, this one doesn't have that. It just had a nice open circle. So, there is our final answer. The next part here says -12 is less than or equal to 3 y and then less than six. Says y is an integer, meaning a whole number. find all the possible values of y. So you need to watch out here because in this inequality there's a 3 y in the middle. Now we only want there to be one letter in the middle, one y. And in order to get one y in the middle, we would have to divide both of these numbers by three. You've probably seen lots of these questions and typically you'll just write down the numbers from -12 to 6, not including one if it doesn't have that equal to symbol. So, it's trying to catch you out a little bit, hoping maybe that you'll miss the 3 y. Okay, we're not going to miss that 3 y or whatever's in the middle there. You solve it like an equation. It's just that you do the same to both sides of the inequality.

[00:52:15]So, dividing them both by 3, 6 becomes 2, -12 becomes -4.

[00:52:22]Now, we can find all the possible values of y.

[00:52:27]-4, -3, -2, -1, 0, and 1. And I'm not going to write down two because this inequality here doesn't have the equal to symbol on it. So, it has to be less than two. So, they are the whole numbers that are greater than or equal to -4, but also less than two.

[00:52:51]Okay. Factoriize fully. That word fully there being really key gives you a massive hint. It's just not one thing coming out of the bracket. There's a lot of powers, a lot of letters here. This does go into a single bracket thankfully. So the first thing we want to do is have a look at the numbers within the expression 15 and 21. We are factorizing fully. So I need to spot the highest common factor of 15 and 21.

[00:53:21]3 goes into 15 and 21. That is also the highest common factor. And there's a little test we can do because if we divide them both by three, we get five and seven. And there's nothing that goes into five and seven. So three goes on the outside.

[00:53:36]Both parts of the expression also have an a in. They don't have a b in. So we are finished once we've got the a figured out. We've definitely got an a on the outside.

[00:53:47]We want to have a look now at which is the smaller power of a. So there's an a cubed on the left and an a squ on the right. A squ is the smaller one. So we can take a squ out of the bracket. We can't take a b out. So we'll open up the bracket. So it's 3 a 2 that would go on the outside.

[00:54:08]Now focusing on the numbers. 15 / 3 is five.

[00:54:13]A cubed. To get to a cubed, we'd need another a here. Otherwise when we expand this bracket, it would stay as a squ. We would need it to turn into an a cubed.

[00:54:22]We also have the b ^ 2 that needs to go in there. There's no b's on the outside, so it's just replicating what's in the question. Keep the symbol the same. To get from 3 to 21, we'd have to times by 7. And that's already an a 2 a squ there. So, we can just close the bracket because when we do 3 a 2 * 7, we'll get 21 a 2. So, there we go. That is factorizing that. And hopefully you can just see there. I really took my time with that just to make sure I got every single little piece correct. I've factorized it fully. There's nothing left in the bracket that can divide out five and seven. They don't divide by anything. There's not an A in both.

[00:54:56]There's not a b in both. It is fully factorized. And there we go. Factorizing a slightly harder expression. Okay. So, looking at reverse mean says the mean mass of five parcels is 3.6 kg. One of the parcels has a mass of 4.8 kg. Work out the mean mass of the other four parcels.

[00:55:16]So when we have a reverse mean or we are told a mean, we can take that mean value 3.6 and multiply it by how many things it's related to. In this case, five parcels. That will tell us the total weight of those five parcels. So in this case, we can do 3.6 * 5. So whatever method of multiplication you use for multiplying decimals is up to you. 5 * 6 is 30. 5 * 3 is 15 plus the 3 is 18. And there we go. We can hop the decimal back in. 18.0.

[00:55:51]So 18.0 is the total weight of the five parcels.

[00:55:57]It says one of the parcels has a mass of 4.8. So to work out the total mass of those other four parcels that we're going to need to work out the mean of, we need to subtract 4.8 from the total weight.

[00:56:12]Now, when we do that, again, we've got some written methods. I'm going to need to do some borrowing. There we go. So, 10 - 8 is 2.

[00:56:22]7 take away 4 is three. And then we're left with the one. So, I have 13.2 as the total weight of the other four parcels.

[00:56:32]So, the other four parcels, which have a total weight of 13.2, will need to be divided by four. That's how we work out the mean. We take the total and divide by how many it's related to. In this case, four parcels. So four goes into 13 three times up to 12, which leaves us with a remainder of one. And then it fits into 12 three times. And therefore our mean for the other four parcels is 3.3 kg.

[00:57:02]And there we go. That is reverse mean there. So working out the total using the mean, taking away in this case that one parcel that it told us the mass of and then working out the mean of the remaining four parcels.

[00:57:16]And onto our final question here, which is the subject of a formula. Now this type of formula where there is a fraction separated from a number always tends to catch people out. Most people are pretty happy with the fact that we're going to need to multiply by five to get rid of this divide by five. Now, when we do multiply by five, if and we normally do this first, if we multiply by five first because the two is separate from that fraction to the side, the two needs to be multiplied by five as well. Now what I would advise is when the divide by 5 isn't locking everything in. For example, here would be an alternative m + 2 all divided by five. In this case, everything is locked in by that divide by five. And when we times by five, it doesn't actually affect the visual look of the two. The two stays as a two when we write the next step down. However, in this case, it does affect it. So in this particular situation, what I would actually do is subtract 2 first from both sides, which leaves you with p minus 2 is equal to m over 5.

[00:58:29]Now when we divide by five times by five the inverse of dividing by five normally people don't now forget that we need to multiply both pieces on the left by five which makes it either five brackets p minus 2 or you can actually times them both by five. So 5 p min - 10 is equal to m and we get m as the subject. Now, had I have multiplied by five in the first step, of course, the plus two would have turned into + 10 and then we'd have subtracted 10. We'd got the same answer. If you do tend to forget that step there, it might be worth from here to here just writing it as five brackets P minus 2 equals M. You've still made M the subject. And it would be absolutely fine to leave your answer like that. So, there we go. Just watching out for changing the subject or any sort of solving or rearranging process where there is a fraction separate from a little number like that is really really key. But there we go.

[00:59:31]That is changing the subject of a formula. And where we had the little fraction separate to the