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Okay dokey so just turn on. Yeah I think it would be a good idea for us to have a go at the rest of the $11 plus paper today. I've already got noted down that we've got sort of slightly more complex decimal conversion, straight line graphs and transformations to do as kind of topic intensive classes later down the line. But we'll Carry on with this paper. Let's respond to what we got up to last. Bus. It's been very slow today. It's just sort of warming up still 1s. Okay, this paper. And we did all of this. We did all of that. We did that. We did that. Did that, did this. We did that. I think we were just starting to do these, as I remember. I think you were pretty good on the notes. So we're not going to bother with those and we're going to go straight to question 16 now. So the mean average of five numbers is 28. The mean of a different set of twelve numbers is eleven. Calculate the mean of all of the numbers together. So at the mean of the five, number 28. The five of numbers that together will be. Hundred. If twelve, number eleven be eleven times twelve is 132, add them together is a 272. Divide by divide. Them by 17 years. Well, do me think. Great. What's my that? 1616, very ry, very nice, fantastic, absolutely brilliantly done, really quick, really competent answering. Well done, Leo. Okay, question 17, working out the value of x from the diagram below. Oh. Okay. Sorry. Why are you doing the right thing? I'm not really sure you're doing it perfectly. So that I mean the thing to sort here, I think you just said that this top angle here is 110 which is absolutely right. 180 degrees in a straight line. Just look here that we've actually got another yes. Oh yes, I saw the triangle. Okay, yes. So that be 70 divided by 2:35. Good. So we know that both of these are 30 fine. On the top one, the next 70, 35, absolutely it is well spotted. And then on 80 minus five is 75. Wait a minute, I can just work out with 180-70, 710. 110 is. Very good, okay? The shape below is made up of equal actual triangles. The side length of the largest triangle is 16 cm. Work out the total length of all of the lines used to draw the shape. Bit of a nightmare, but you can chip away at it. Okay, so all the lines. So okay, so first do the biggest lines. So that will be the big triangles, which is 16, 16, 16. Okay, yes, 16. And then at. The from this. 16, one, two, three, four, five, six, seven, eight, so move the beadivithe eight parts, so third, so the triangles. Oh, wait, wait. I挺住的。So. Oh, no, Yeah, not really good. Oh, my God. So in the square, in the big triangle, there's four of this, like this big gle, not really big, but like bigger triangle. And as you calculate, sir. So. That will be four times four. Yeah that's a four times plus 14 plus 4:14 plus 14 times four. Yeah so can I just just to you might be doing the right thing, Leo. I'm just slightly lost in what you're doing now. So if we if we label this really clearly, so we've got the the red part there we agree is three times 16 oops, sorry, the red part running around the outside. And then I think what you were going for was this kind of medium size triangle in the middle here. Am I wrong in thinking, Oh yes, I forgot that like all of the lines are covered. So that will be just like 16 plus 16 plus 16 plus 14 plus 14 plus 14 which definitely is definitely Oh, I keep forgetting as forit's forceful. So how how long so you know how long is one of these sides of the small triangle just from there to there? I've not drawn that very well, but so that's two. That's two. So what what therefore must each side of the blue triangle be equal to? Eight. Good. So we can say that this here is just going to be three times eight. Relatively clear, straightforward. Yep. And we've got these yellow triangles that I'm gonna draw here. Now let's figure out yellow, what the matths are, smaller ones. So that's four times four. That's Yeah so each each side is is 4 cm and there were three sides. Yep. Three times, four times. And how many of them are there? Good. Three times, four times. Three. Absolutely right. And then we've got these kind of purple triangles here. What are the maths going to be for these guys? So that is the smaller, smaller triangle. And there is. One, two, three, four, five, six, 78, nine, nine of them. So thatbe, three times, two times nine, three times, two times nine. Absolutely right. And we're finished. Now I go, there's nothing. And then add them together. So three times 16 is 48 plus 24, plus 36, plus 54 equals. Now you just add them together like 48 plus 24 is 72, plus 36. It is 108 plus 54 is 152. 152. Yes, very, very good. 72 plus 100. Is it 100? Did you say 152? 100 so Yeah, 152I'm not like, Yeah, Yeah. So we've got I mean, we've got 48 plus 24, as you said, that's 72. Yeah, okay, I know, I know. It's 162. 62, very, very good. 36 plus 54, it's 90. That's a nice one and a little W, see, 162. Good. So as long as you don't panic, can you just chip away at that question? I think it's relatively straightforward. Really weird. Yeah. Okay, question 19, John has 30% more money than Miguel and together they have 46 pounds. Calculate how much money John has, okay, 30% more than Miguel. So that means Miguel is 100% and Joan is 130%. And Yeah gathers 230% and 46. Oh, so if it's 23 therebe two pounds per percent and so will be it's 20 pence per percent. So how much money John has usually just a 20 times 130. Yeah, I can just do two pounds times 13 honny. Next wait, no, Yeah, 26. So join us to advance. Nice. Yeah Yeah, that's absolutely right. Very, very good. Really like the way that you did that actually very, very accomplished. Yeah. We've just got sort of 230% of something and remember what you're were really good on. Actually, what's very easy, it's very easy to get confused in a question like this because we took John as our reference sorry, we took Miguel as our reference point. But John is actually the guy that they're asking about. So he said if Miguel is 100%, John is 130%. So 230% of Miguel, the money that Miguel has is 46. So we've got 2.3, but something six to 46. So we can say Miguel has 20. And of course, if John has those percent more, yes, time six. Really, really nicely done. 20. A group of children are playing a game where they can either score two, three or five points on each turn. The children have had eight turns each and recorded their scores in a table, but some of the entries have been deleted. Fill in the missing numbers to give a complete set of scores. Okay. So. Eight turns. Okay. So if there's four turns, four of two points and two of three points as eight plus six is 14. 14 and then plus there's two turns left. So two times five is 1014 plus ten is 24. So the show to school is 24. So that will be 24 plus ten. Yeah total school is 20 53 points of. Three points of three is nine points, 25 divide minus nine is 16. Well if it's 16 there's only one combination which is 25 should that be two zero five and it will be six minutes that should that will be three turns on I just I think you did say it right. I was just a little bit confused. Yeah. So I see what you mean. You mean he scored ten points but that's 25 points right here. Oh yes. What did I write? You wrote ten. Oh Yeah, I wrote ten. What it amounts to? Yeah, no, it's fine. It's perfect. Good. So Yeah p. Yeah the first taste not that hard, but the but last two like Yeah a little harder. He's doing really well understanding standing the question very, very well indeed. So imagine we had 15 so therebe 30 left. Others like. Yeah Yeah. So if we did 15, 13 thatbe 27. So it can't be 25.23 thatbe six is 24. So that could be it. So 24 will be twelve. So that can't be because therebe like 15 goes one, five, three, three is a 21. Note one, five, two, three plus twelve plus five is 17. Yeah seven. Oh, I could just 13 minus twelve, 30 minus twelve is 18. That could be, I'll be six, two, three W, no, because that's eleven goes one, five, five. I'll be thatbe sweet times five times three is 15. And 15 plus five, 21. No, be, there will be 15 left. One. Five, six degrees I'll be 80 plus 75 30 itbe, twelve flat and six days no, because I'll be like 30 noand. One, five, seven, three es three times seven is 21, 21 plus 5:26. So Crobe 15. I feel like this is going to take like ages. I know what you mean. I mean, I think the fact that we've got a relatively quite big number here, I think it's probably a clue that we're going to be maxing out on fives and probably the twos and the threes. We might have a few but not very many. And likewise the fact that we have a very small number here, it's another clue as to sort of where to distribute most of the points. The 18, most of the points are probably going to be two or three, right? Yeah, but that's two times 9:16. So think of the smallest way that you can possibly make a five using 23 like one, two of one, right? Okay, let's give that a go. So one, two, one, three, what are we left with? You could go six. Okay, I get it now. No. Wait, can they like score zero zero things or like zero three es? Yeah, you can absolutely. Someone might not have spoilled any. I can make it like I can make the numbers correct, but I kept to use like seven goes. This is like. It's got to be eight goes again. Think about how small a number 18 is. We really. It probably means we're going to be easing out of one or of the five points. What happens if Charlotte scores no five times? Can you get to 18? Yeah Yeah can go two and threes each three turns. We'll make three, five, which ches six goes and we'll be 50 and plus a three is 18, but it only needs seven coso. What about if you use more two points and fewer three points? Because we're still we're going over, we're getting to 18 and 70 or two points and two esday points. Yes, costattherebe eight plus six is. 14. And that's six gsts m plus weighty. So 23 es and 62. Good. Absolutely right. Six, two os, two, three s and way back. Good though. Really nice six marfor that one. So I guess they they assume it's going to take some time. Well done that. Really nice 21. Write each of the numbers 80, 81, 82, 83, 84 and 85 in the spaces below, using each number only once to make all of the statements. Okay, so 81a square number very good. 85 to multiple 45. Yep. Like I skipped out the prime number one as well. That's a nice one to come back to. 84 is a multiple three. 84 is a multiple of three. Well spotted. See. So let's check, is it prime number now is 80? Double prime number divided by two, 14, nine. So 80 must be a product of two prime numbers. Well, did you? Absolutely. Which prime numbers would those be? I mean, they could be a lot of them, but I think. 39 and 41. So product to remember is multiplication, not addition. Oh yes, product means so that can be like. Well, I'm not really sure. Yeah I mean look, there's to make if we think about most prime numbers are odd, aren't they? So in fact, two is the only even prime number. And when you times two odd numbers together, you times 35 together, do you get an odd number or do you get an even number? Oh, you got like an odd number. Yeah, an odd number. So the only way to make 80 using if one of the numbers are prime, would be to use two. But if if we do use two, we've got two times 40 and 40 is not a prime number. So that should be cause for you to I like how you deduced it, how you kind of whittled it down. But now might be a time to have a rethink. Okay. Because there are no if we use any other two prime numbers, say seven times eleven, we're going to ramp against the same problem again and again, which is that they they always give us an odd number we can't get even unless we use two. Oh, Yeah. So can't be 80. Okay so I must got like something wrong then right we all know that so Oh it could be 80 that's multiple five good and this is absolutely the question is designed to make you rethink there's nothing wrong in to the 85 productive do grand numbers five and eleven wait no no no no no not five kind eleven like Yeah. We go on five and what? It's. 17Yeah and that does work, does it? Those are two prime numbers. Fantastic. Yeah, let me just grab. Now what's the double of a prime number? There's only 82 and 83 left. So 82 by by two, 41, which is a prime number. And 8383 is a prime number. Fantastic well done. K 22, James cycles 4000m at 10m per second. Tom takes 100s more de cycle the same distance. At what speed does Tom cycle? So 10m per seconds, that's quite fast. So that will be 400s Yeah so 500s we'll start. 500 to thatbe 8m per second. Yes 8s to absolutely 4000 divided by 500. Eight eigha second very very nicely built. K 23, the instruction x spade y means double x and add y. So they've made this up. It's not really a thing, but I've seen if you can take the instructions and run with them. So for example, five spade three means two times five plus three, which is 13. So using that same logic, what would 14 spade 55 be? Okay. So thatbe double 14 plus 55. Thatbe 28 plus 55 will be 70, 83 plus four. Yeah, 83. Well done. And what is the value of a if nine spade a is equal to 31? Nice, fit well. So is double nine is 18. Yep, and 31-18 is 13. So 13 absolutely is 13. And I'll make this a bit smaller so we don't forget what the instructions are. What is the value of b if b spade ten is equal to 16 spade b? So b spade ten equals 16 spade b. Okay. Okay, so double 16 is 3232 plus something. Equals 22 times someday plus ten. Okay. So. Okay, I get it. So if two times this plus ten, there's something plus that, 20, 32 plus that. If I go like 32 minus ten, it will be two b equals to 22 plus b very good. That means 22 is b very nice. I try. That will be 22. So that be 44 plus ten is 52 and 32. Wait, what? Oh Yeah. So therebe 44Yeah 44 plus ten is 54 and that will be 32 plus 20 to 54. Yeah very, very nice. I love the way that you formed and solved an equation there. Absolutely the right way to do it, by my book. Very, very good. So the numbers c and d are such that c spade d is equal to d spade c work out. Then the value of d minus C A lot. So like it could be the same number like. Yeah that could be the same number, like 44. Four times two plus four is three times four is twelve. And two times four plus four is also like, wow, right? Yeah, absolutely. And you just see minacy is zero, zero, very, very good. And you know we could also prove that algebraically, couldn't we? We could say two c plus d is equal to 2D plus c. And if we rearrange that, you know, we get the c over to the other side. So minus c from both sides. That leaves us one c over there, minus d from both sides. As you said, c is equal to d, so d minus c or c minus d for that matter, always going to be zero. Really, really good. Well done, Leo. The cards below have letters on one side and numbers one, two, three, four, five, six, seven on the other side, but not necessarily in this order on the cards. So one, two, three, four, five, six, seven, eight. Oh, no eight. Well, there's no eight. Yeah. So one, two, three, four, five, six, seven, Yeah, I think I forgot to 77, didn't that? So we're told that alphabet adds up to 35 and we've got to now try and figure out the value of a. Can I look at the next question because I'm pretty sure it says like something adds up to 20. Ais that alpha wait, I can change, I can scroll it. I just realized Yeah so Alva is add up to 20 so that means spat is 15. Yeah and that is the interesting what the combinations it could be one, two and twelve now it could be one, three and well, I would just be that you might be right that I'm having a slight doubt about whether we have can we spell alphabet? We of course we can spell alphabet using these letters but we have some letters which are repeated. Yeah exactly. A so Oh, I see what you mean alfa. Yeah got you, got you, got you. I'm with you now. So very good. So bet is 35. Yeah so Yeah, that beis 15 as you just said, yes, 15. And I don't think you really need to work it out because just feel like. So there will be 20 left. If there's 20 left divided by five, so the average will be full. Yeah Yeah. So the average is full. For example, let's say. And two a's will be eight, that will be 20 minus eight is twelve therebe three number left and they have to be like about four. Well, divide by 43. So that's about right, let's say. That's takes. Three, five and no, that can't be three, five and 94. Maybe three. Three, six thatbe nine and another 39. What about 12 therewill be nine left. So no one, three, one, three kg, 48 left. No, Yeah. So Yeah, and conyou. So 15, say 15, and there's seven left. Who's 157? So a could be four. Because. Yeah because so it was if it was four. All of the other letters get used at least once. So we've got A P that gets we've got an H, we've got A B, we've got an l, we've got an e. We've got and now you'll get used once. So it doesn't really matter the order. We're just going to use every other number except for so one, two, three, four. We already have five, six, seven. I don't know if they're in that order, but does that they are going to be they would have have to be somewhere those numbers does that add up to 35? So one plus two plus three plus four plus 56 plus seven that's three plus three, six. So will be ten, 50, 21, 28 plus four, 32. Okay. So that's a real big Yeah. So that's a clue. That's a real clue. Oh, so I just have to see. I just have to add. I just have to use all of the numbers once, which there's one plus two, three, three plus three, six, six plus four is ten, ten plus five, 15 plus six is 21. So thatbe 28 and what's left? So that be a seventh, so a seven. Absolutely right. Very, very okay. You are also told that alpha adds up to 20. There go. It doesn't really matter. We just need to know bet, so I already knows that beis 15. So 15 for seven, 22. Very nicely done. Brilliant. Really, really like the way you did that. Okay we've got more weird problem solving going on here. So along any line drawn the numbers in the two circles at the end of the line I'll tell you what I'm going to make this question a little bit bigger because it's very we on the page drag this little bit okay so any on any line you're on the numbers in the two circles at the end of a line should add together to make the number in the square between the two circles. So you can see 20 is 9:11 added together twelve is 933 is 14 is three and eleven. Fill in the remaining circles with the numbers obeying this rule. Okay. Okay. So the three numbers like in the circle makes up the square between them. Okay. So that's just like what that like a really typical question in China. Yeah it's like a add another one will be like a like triangle and there's nine and there's one. So lay all of the squares and here's also circle. You have to fill it up like right, one, two, three, four, five, six and now have to add up to like 15. Now let me do this one. That is actually a big number. So. 80 and 55. So 80 is bigger than 47, 55 bigger than 47. So there must be a very big number in the circle between 18 and 55. Let's say it's started to you. Let's say that's starting. And the other ones have to be 50, so can't be. Yeah, Yeah. No, that's Yeah. I feel like 40 is good one, so the other one will be 40. So can, like any two numbers, be the same? Yeah doesn't say anything about them not being the same numbers. So theybe seven, 40 plus. Okay, I get it. So. The gap between 47 and 55, it's eight now. The number on the left side of 80 and then on the right side of the 80 have have to have a different of eight. So that could be 36 and 44. Thatbe way up, sorry. So this is 36 oops, and that's 44. And the top one could be eleven. And Yeah, it's Yeah, it's 1136. Checout, very nicely done. And same thing for this one here. So it's 30, survival 2:15 and the gap between 20.5 and 25.5, it's also five. So the needs like a different spy. Let's say 13.5. So out of 0.5 will be no copy. Yeah 12.5. So 12.5 plus 12.5 is 25 and plus five. Yeah. So 12.5 scoso, okay, that's 12.5 and the other must be 17.5 at that's eight and 17 plus eight is 25. So that's good. Very, very nice. Fantastic. Good. And we are on to the very last question, 26. So each of the diagrams bshow a pair of triangles drawn on a grid of squares, side length 1 cm, a smaller shaded triangle where these overlap. Each corner of each of the larger triangles lies exactly at the corner of a square on the grid. Work out the area of the shade triangle in each case. Il? A. I don't really understand the question. Okay, so that's it's a these are really hard actually may have a big memory of doing this before. That's not easy at all. So let's let's just imagine a much easier question. They're basically asking you to find the area of the black shaded triangle. So if you had a much easier question would be something like this where you've got a triangle that looks like that and that's just copy and paste. Like I just don't understand what the question means. Like the each of the diagram below shows a pair of triangles drawn on a grid of squares of silence once and reasons are small. So let's imagine it was a much, much easier question. And it said, find the area of this black shaded triangle. Now we can see here really clearly I haven't. It's not brilliantly drawn, but we can see that the height of the triangle is one actually this isn't much easier question, I'd say, and not that much easier. So our job now to try and figure out the area would be to try and figure out the width. That's where these questions are difficult. And because the example that I gave is just as hard as the actual question itself. Didn't intend it to be, but it was. We're just going to ditch that and we're gonna go straight into the question. So I think it's more easy work out, but you going work out the areas of the two big fples. Try. So the area of the problem one is four times. Four divide by two. That's eight. And for the bottom, one, one, two, three, four, five, six, six times four is 24, twelve, so twelve. So that's twelve and that's eight. Yeah, okay. Now there's a bit of lax. I can't I know the width of that little triangle is one, but what's like the height? What's the height? That's the confusing part. And what you were about to do is is great logic for when you have a shape within a shape. So if you had a you know like a big triangle like this from a slightly smaller triangle within it, and you were being asked, what's the area of this shaded section here? Absolutely. It's a good idea to work out the areas of both triangles. You minus the area of the smaller triangle from the area of the bigger triangle, and you're left with the shaded section. Now, unfortunately, that doesn't really apply to this question, because even if we work out the areas of both triangles, it says nothing about where they overlap. You know I could drag this triangle further up and there would be a bigger area of overlap. There or there Yeah or there. You know so eight and twelve isn't actually telling us much about what is the area of overlap. So you're absolutely right to kind of course correct there and say, I need to work out the height. Yes, that's the hard part of this question, isn't it? We know this is one that's nice and easy because we can just see it, but what the hell is this? What fraction of one is this? Well, I just realized the black bed looks exactly same as this bit, right? Good. I like that you're on the right track. No. And I know the black ackbeplus oak oops. Right, wait, go on. The black bed t plus. Oh, my God, this bit is a square. I just Oh, you're thinking along the right lines, Leo, to be confident. This is really hard. What is this? Okay, let's get let's get a grid up because I think you're pretty much there and I don't really want to give you the answer because I really do think you can figure this out. Let's get some graph paper up and I'm gonna to try to more successfully design an easier question. So here is my easier question which I think you will be able to figure out. So let's do a triangle that looks like this triangle it really accurately this time since it's gonna to be four therebe eight like that and we've got the line going at that. And let's draw another sort of smaller triangle, something which looks maybe. Snwith myself here. Okay, so we've got that triangle there and just going to drag this so that it looks that would be a good place. Yeah. Okay. Yeah. Actually do you know what? I haven't I haven't joined this as well as I thought I had. Sorry. Let me just correct this slightly meant triangle like that. Again, this is just so that the question is a little bit easier. Let's drag this triangle down. It looks like. That more or less. So imagine that this time they asked you, what is the area of this section here, this black section? Well, I don't know. That's why I tried to look out the last question. Yeah. So ignoring my not very well, do you know what? Let me let me try one more time. And I'm going to try and be as accurate as I possibly can in class. And sometimes it's really hard to draw with absolute accuracy. So we've got a triangle like this. Four, eight, let. Me try and be as pinpoint precise as I possibly can be. There we go. There we go. And let's have another a nice little triangle like. Like that. And I'm going to this time, I'm going to ask you, what is the area of this little blue section here? Well, two times one derived by two. One Yeah two times one diviby two is one. So this is an easier question because we can see it clearly, 21. But one of the things to notice, the reason why, if we're being really kind of pedantic about it, the reason why we know the height is one is actually by looking in detail at the gradients of this diagonal line here, we can track the fact that for every two squares it goes across, it's going one up, two across, one up, two across, one up, two across, one. And what that means is when we have a slightly more tricky question, I'm slightly hurrying along that. So I realize we've gone over time, when we have a trickier question like this, we can do the same thing. We can go, okay, so this is going, what, three along, one up, three along. Let's sorry, two up. Actually I one up. So that's the gradient. And if we divide that by three. Because we we want to figure out not what this distance here is, which is two, but rather, as you said, this distance is. So if this is two. Yeah the two divided by 42 divided by four is actually three got Yeah exactly. So it's going to be wait, let me write it down. Two thoughts. Oh, Oh, Oh. And that looks about right, doesn't it? Yeah. Two, three by three and prolike 0.6666 recurring. Yes, it is. Absolutely. And we've got one here. And to work out the area of the triangle, what's that now going to be? Thatbe one times two thoughts. Do we need to do anything else? It's a triangle. Sowar war so so two thirds divided by two. Two third by by two equals. Equals so two to earth, two over one. And if you want to divide it, you just have to flip it. So itbe, one over two. So two times one from 26. So that's one over six. Good. Is it one over six or is it one over three? I mean, it's 22 over three times one over two, which is two over six, which is one over three. Yeah, okay, one. So we got a third centimesquare nasty, really, really hard. We're gonna we're gonna stop there because we've gone over time. Now I don't want to keep you too long, but I'm going to send you the next question for prep. See if you can figure it out. It's exactly the same working out the area of the shaded section a little bit trickier because we're actually unsure of the width as well as the hedes. But it's the same idea in that can you you can figure it out using the gradients. This is hard. This is really, really hard what we're doing now. But I think you can do at great stuff today. As ever, have a wonderful rest of your Sunday and I'll see you next time. Take care, Leo. Yeah.
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{
    "header_icon": "fas fa-crown",
    "course_title_en": "Language Course Summary",
    "course_title_cn": "语言课程总结",
    "course_subtitle_en": "1v1 English Lesson - Topic Name",
    "course_subtitle_cn": "1v1 英语课程 - 主题名称",
    "course_name_en": "CAT4 Test Preparation",
    "course_name_cn": "CAT4 考试准备",
    "course_topic_en": "11+ Paper Practice (Cont.) - Quantitative Reasoning",
    "course_topic_cn": "11+ 试卷练习(续)- 数理逻辑",
    "course_date_en": "Current Date (Implied)",
    "course_date_cn": "当前日期(推断)",
    "student_name": "Leo",
    "teaching_focus_en": "Continuation of the 11+ paper, focusing on complex problem-solving in Data & Space sections (Mean, Geometry, Logic Puzzles, Coordinates).",
    "teaching_focus_cn": "继续完成11+试卷,重点关注数据与空间部分(平均数、几何、逻辑谜题、坐标)的复杂问题解决。",
    "teaching_objectives": [
        {
            "en": "Complete the remaining questions (Q16-Q26) from the assigned 11+ paper.",
            "cn": "完成分配的11+试卷中剩余的题目(Q16-Q26)。"
        },
        {
            "en": "Reinforce understanding and application of mean\/average calculations.",
            "cn": "加强对平均数计算的理解和应用。"
        },
        {
            "en": "Develop strategies for complex spatial reasoning and pattern recognition problems (e.g., fractal geometry area calculation).",
            "cn": "培养解决复杂空间推理和模式识别问题的策略(例如,分形几何面积计算)。"
        }
    ],
    "timeline_activities": [
        {
            "time": "Start - Q16",
            "title_en": "Mean Average Calculation",
            "title_cn": "平均数计算",
            "description_en": "Calculating the combined mean of two sets of numbers given their individual means.",
            "description_cn": "根据两个数集的个体平均数计算其总平均数。"
        },
        {
            "time": "Q17",
            "title_en": "Angle\/Geometry Problem Solving",
            "title_cn": "角度\/几何问题求解",
            "description_en": "Working out an unknown angle 'x' using properties of straight lines and triangles.",
            "description_cn": "利用直线和三角形的性质求解未知角'x'。"
        },
        {
            "time": "Q18",
            "title_en": "Complex Spatial Reasoning - Perimeter\/Length Calculation",
            "title_cn": "复杂空间推理 - 周长\/长度计算",
            "description_en": "Calculating the total length of lines in a complex fractal-like shape composed of equal triangles.",
            "description_cn": "计算由等边三角形组成的复杂分形结构中所有线条的总长度。"
        },
        {
            "time": "Q19",
            "title_en": "Percentage and Ratio Problem Solving",
            "title_cn": "百分比和比例问题求解",
            "description_en": "Solving a word problem involving relative amounts (John has 30% more than Miguel).",
            "description_cn": "解决涉及相对数量(John比Miguel多30%)的应用题。"
        },
        {
            "time": "Q20",
            "title_en": "Number Combinations\/Constraints",
            "title_cn": "数字组合\/约束条件",
            "description_en": "Finding the correct distribution of scores (2, 3, or 5 points over 8 turns) to reach a specific total (53 or 18 points).",
            "description_cn": "找出在8轮中得分(2、3或5分)的正确分配方式,以达到特定总分(53分或18分)。"
        },
        {
            "time": "Q21",
            "title_en": "Number Properties\/Categorization",
            "title_cn": "数字性质\/分类",
            "description_en": "Assigning numbers (80-85) to statements based on properties like prime, square, multiple.",
            "description_cn": "根据质数、平方数、倍数等性质将数字(80-85)分配给相应的陈述。"
        },
        {
            "time": "Q22",
            "title_en": "Speed, Distance, Time Calculation",
            "title_cn": "速度、距离、时间计算",
            "description_en": "Calculating the speed of the second person given the distance and time difference.",
            "description_cn": "给定距离和时间差后计算第二个人的速度。"
        },
        {
            "time": "Q23",
            "title_en": "Abstract Rule Interpretation & Equation Solving",
            "title_cn": "抽象规则解释与方程求解",
            "description_en": "Applying a custom binary operation (x spade y = 2x + y) and solving resulting algebraic equations.",
            "description_cn": "应用自定义的二元运算(x spade y = 2x + y)并求解由此产生的代数方程。"
        },
        {
            "time": "Q24",
            "title_en": "Letter\/Number Constraints Puzzle",
            "title_cn": "字母\/数字约束谜题",
            "description_en": "Using the sum of letter values (ALPHABET = 35) and subset sums (BETA = 15) to find the value of 'A', utilizing the constraint that all numbers 1-7 must be used exactly once.",
            "description_cn": "利用字母值的总和(ALPHABET = 35)和子集总和(BETA = 15)来找到'A'的值,利用数字1-7必须恰好使用一次的约束。"
        },
        {
            "time": "Q25",
            "title_en": "Number Line Addition\/Subtraction Logic",
            "title_cn": "数轴加减逻辑",
            "description_en": "Filling in circles such that the sum of endpoints equals the middle square value along a line segment.",
            "description_cn": "填充圆圈,使沿线段的端点之和等于中间方块的值。"
        },
        {
            "time": "Q26",
            "title_en": "Area of Overlap (Hard Geometry)",
            "title_cn": "重叠面积(困难几何)",
            "description_en": "Calculating the area of an overlapping shaded triangle on a grid by analyzing gradients\/slopes.",
            "description_cn": "通过分析梯度\/斜率来计算网格上重叠阴影三角形的面积。"
        }
    ],
    "vocabulary_en": "Mean, conversion, transformations, decimal, triangle, diagram, angle, isosceles, side length, total length, percentage, calculate, score, turn, set, prime number, square number, multiple, product, speed, cycle, distance, instruction, operation, gradient, area, overlap.",
    "vocabulary_cn": "平均数,转换,变换,小数,三角形,图表,角度,等腰,边长,总长度,百分比,计算,得分,轮次,集合,质数,平方数,倍数,乘积,速度,骑行,距离,指令,运算,梯度,面积,重叠。",
    "concepts_en": "Mean calculation (weighted average), geometric angle properties, area calculation via decomposition\/subtraction (initial attempt), ratio translation (percentage increase), number theory constraints, abstract algebraic definitions, spatial reasoning using gradients (slope analysis for area).",
    "concepts_cn": "平均数计算(加权平均),几何角度性质,通过分解\/相减计算面积(初步尝试),比例换算(百分比增长),数论约束,抽象代数定义,利用梯度(斜率分析求面积)进行空间推理。",
    "skills_practiced_en": "Mental arithmetic (especially for mean calculation), deductive reasoning, precise calculation under time pressure, interpreting novel mathematical instructions, using graphical analysis (gradients) for exact area measurement.",
    "skills_practiced_cn": "心算(特别是平均数计算),演绎推理,在时间压力下的精确计算,解释新颖的数学指令,使用图形分析(梯度)进行精确面积测量。",
    "teaching_resources": [
        {
            "en": "Assigned 11+ Paper (Quantitative Reasoning section)",
            "cn": "分配的11+试卷(数理逻辑部分)"
        },
        {
            "en": "Graph paper\/Grid visualization for Q26",
            "cn": "用于Q26的方格纸\/网格可视化"
        }
    ],
    "participation_assessment": [
        {
            "en": "Excellent engagement, especially on complex problems like Q23 and Q26. Leo was vocal about his thought process, even when confused.",
            "cn": "参与度极佳,特别是在Q23和Q26等复杂问题上。Leo清晰地表达了他的思考过程,即使感到困惑。"
        }
    ],
    "comprehension_assessment": [
        {
            "en": "Very strong comprehension of calculation-based questions (Q16, Q19, Q22). Struggled initially with the abstraction of Q26 but demonstrated capacity to follow the teacher's guidance on gradient analysis.",
            "cn": "对计算类问题(Q16, Q19, Q22)的理解非常扎实。最初在Q26的抽象性上有些困难,但展示了跟随老师关于梯度分析指导的能力。"
        }
    ],
    "oral_assessment": [
        {
            "en": "High level of oral fluency when explaining methods, showing competency in setting up equations (Q23) and simplifying complex steps (Q18).",
            "cn": "在解释方法时口语流利度很高,在建立方程(Q23)和简化复杂步骤(Q18)方面表现出能力。"
        }
    ],
    "written_assessment_en": "Not explicitly assessed, but performance implied high accuracy on calculation steps.",
    "written_assessment_cn": "未明确评估,但表现暗示计算步骤的准确性很高。",
    "student_strengths": [
        {
            "en": "Speed and accuracy on standard calculation problems (e.g., Q16, Q19).",
            "cn": "在标准计算问题(如Q16, Q19)上速度和准确性很高。"
        },
        {
            "en": "Strong algebraic formulation when required (Q23).",
            "cn": "在需要时能建立强大的代数公式(Q23)。"
        },
        {
            "en": "Ability to 'chip away' at large, daunting problems like Q18 by breaking them into manageable parts.",
            "cn": "有能力通过将大型、艰巨的问题(如Q18)分解成可管理的部分来逐步解决。"
        }
    ],
    "improvement_areas": [
        {
            "en": "Initial confusion\/hesitation on highly abstract spatial problems (Q26), needing teacher scaffolding to apply gradient analysis.",
            "cn": "在高度抽象的空间问题(Q26)上最初感到困惑\/犹豫,需要老师的引导来应用梯度分析。"
        },
        {
            "en": "Risk of getting lost in complex scenarios where multiple constraints overlap (e.g., Q18 decomposition, Q20 constraint testing).",
            "cn": "在多个约束重叠的复杂场景中(如Q18分解,Q20约束测试)有迷失的风险。"
        }
    ],
    "teaching_effectiveness": [
        {
            "en": "The teacher managed the pace well, allowing the student time to struggle productively on Q18 and Q26.",
            "cn": "教师很好地控制了节奏,允许学生在Q18和Q26上进行有成效的挣扎。"
        },
        {
            "en": "Effective scaffolding was provided for the most difficult questions (Q26), shifting from area subtraction to gradient analysis.",
            "cn": "为最难的问题(Q26)提供了有效的脚手架支持,从面积相减转向梯度分析。"
        }
    ],
    "pace_management": [
        {
            "en": "The pace was generally fast, but slowed appropriately for the highly complex reasoning questions at the end of the paper.",
            "cn": "节奏总体较快,但在试卷末尾遇到高度复杂的推理题时,节奏放慢得当。"
        }
    ],
    "classroom_atmosphere_en": "Positive, encouraging, and intellectually challenging, with the teacher acknowledging the high difficulty of the later problems.",
    "classroom_atmosphere_cn": "积极、鼓励和具有智力挑战性,老师承认试卷后半部分问题的难度很高。",
    "objective_achievement": [
        {
            "en": "Most objectives were met, completing the paper. The understanding of advanced topics (gradients\/area) was initiated but needs reinforcement.",
            "cn": "大多数目标已达成,完成了试卷。对高级主题(梯度\/面积)的理解已经开始,但需要加强。"
        }
    ],
    "teaching_strengths": {
        "identified_strengths": [
            {
                "en": "Exceptional scaffolding in Q26, allowing Leo to reach the correct analytical path (gradient analysis) despite initial confusion.",
                "cn": "在Q26中提供了出色的脚手架支持,使Leo尽管最初感到困惑,仍能达到正确的分析路径(梯度分析)。"
            },
            {
                "en": "Strong positive reinforcement, especially noting Leo's accomplishment in forming and solving the equation in Q23.",
                "cn": "强有力的积极强化,特别提到了Leo在Q23中建立和求解方程的成就。"
            }
        ],
        "effective_methods": [
            {
                "en": "When Leo was confused in Q18, the teacher immediately simplified the problem context by focusing on measurable segments (side lengths of smaller triangles).",
                "cn": "当Leo在Q18感到困惑时,老师通过关注可测量的段(更小三角形的边长)立即简化了问题背景。"
            },
            {
                "en": "Breaking down Q21's logic by contrasting the properties of odd\/even numbers regarding prime multiplication.",
                "cn": "通过对比质数乘法中奇数\/偶数的性质来分解Q21的逻辑。"
            }
        ],
        "positive_feedback": [
            {
                "en": "Feedback was specific and detailed, highlighting 'very nice,' 'absolutely right,' and 'brilliantly done' for high-accuracy answers.",
                "cn": "反馈具体而详细,对高准确度的答案使用了'very nice'、'absolutely right'和'brilliantly done'等评价。"
            }
        ]
    },
    "specific_suggestions": [
        {
            "icon": "fas fa-ruler-combined",
            "category_en": "Geometry & Spatial Reasoning",
            "category_cn": "几何与空间推理",
            "suggestions": [
                {
                    "en": "Review gradient calculation (rise\/run) and its direct application to finding dimensions within complex shapes on a grid, especially for area calculation (Q26 style problems).",
                    "cn": "复习梯度计算(垂直变化\/水平变化)及其在网格上复杂图形中确定尺寸的直接应用,特别是用于面积计算(Q26类型的题目)。"
                },
                {
                    "en": "Practice dissecting complex figures (like Q18) by systematically calculating the perimeter\/length contributions from each level of repetition.",
                    "cn": "练习解剖复杂图形(如Q18),通过系统地计算每一层重复结构对周长\/长度的贡献。"
                }
            ]
        },
        {
            "icon": "fas fa-calculator",
            "category_en": "Quantitative Reasoning & Logic",
            "category_cn": "数理逻辑与推理",
            "suggestions": [
                {
                    "en": "For constraint problems (Q20, Q21), practice listing possible combinations systematically when dealing with small totals to avoid getting lost in trial and error.",
                    "cn": "对于约束问题(Q20, Q21),练习在处理小总数时系统地列出可能的组合,以避免陷入试错的泥潭。"
                },
                {
                    "en": "Solidify the method for problems involving finding the value of a single component when given the total and relative proportions (Q19).",
                    "cn": "巩固在已知总数和相对比例时求解单个组成部分值的方法(Q19)。"
                }
            ]
        }
    ],
    "next_focus": [
        {
            "en": "Focus on the preparation topics noted earlier: More complex decimal conversion, Straight Line Graphs, and Transformations.",
            "cn": "重点关注先前记录的准备主题:更复杂的十进制转换、直线图和几何变换。"
        },
        {
            "en": "Revisiting complex area\/volume questions that rely on coordinate geometry or gradient analysis.",
            "cn": "重新审视依赖坐标几何或梯度分析的复杂面积\/体积问题。"
        }
    ],
    "homework_resources": [
        {
            "en": "Complete Question 26 using the gradient method independently to confirm understanding.",
            "cn": "独立完成问题26,使用梯度方法来确认理解。"
        },
        {
            "en": "Review topic-intensive class notes for Decimal Conversion\/Transformations as scheduled.",
            "cn": "按计划复习小数转换\/变换的主题强化课程笔记。"
        }
    ]
}
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