大家好 欢迎回来
Hey there, welcome back!
今天 我将带领大家看另一门非常古老的考试
Today, I’m going to look at another very old exam.
这就是始于1869年的麻省理工入学考试
This one is an MIT entrance exam from 1869.
我在麻省理工档案馆的网站上发现了这个
And I found this on the MIT archive’s website.
所以 我将在说明中添加上链接
So, I’ll put the link to that in the description.
或许它值得你们亲自去核实查证
And it might be worth checking it out yourself,
因为他们实际上还保有许多古老的考试
because they actually have a lot of their really old exams
和一些答案
and some solutions there as well.
言归正传 MIT入学考试貌似是从1869年被引进的
Now, the MIT entrance exam seems like it was introduced in 1869,
尽管学校最初开始授课是在1865年
although the school first had classes starting in 1865.
因此 实际上有几年是不需要入学考试的
So, there was actually a few years there where no entrance exam was required,
学生们达到合适的教育标准就允许入学
and students were just expected to have a proper education to get in.
我发现这个考试的内容覆盖英语 代数 几何和算数
This exam that I found covers English, Algebra, Geometry and Arithmetic.
为了尽量保持视频时间简短
But I’m just going to look at the Algebra section,
我将只着眼于代数部分
in an effort to keep this video kind of short.
并且我已经在上一个视频里讲解了一些老式算术题
And also because I already looked at some old fashioned Arithmetic questions in my last video.
网站上有一张摄于1873年的班级合影
There was a photo on the website of a class from 1873.
很有可能这张照片上的人
And it’s likely that the people in this photo
曾经参加过我将要带领大家了解的这个考试
would have been the ones to sit this exam that I’m going to look through.
所以 那将是一件十分有意思的事情
So, that’s kind of interesting to know.
在说这句话的时候我会十分的谨慎
I will say and I do take a bit of caution in saying this,
但我不得不说代数部分实在是太简单了
but I sort of think this Algebra section is surprisingly easy.
我认为它达不到一项真正竞争性考试的试题标准
I think it’s less to do with, you know, being a really competitive exam,
反倒是可以用来检测你在进入MIT之前
and more to do with just showing that you do have some sort of basic education
在代数学方面是否接受过最基础的教育
in something like Algebra, and you’re ready to go off to MIT.
我想有一点是需要记在心里的
I think it’s useful to keep in mind that
时间退回到十九世纪
back in the 19th century,
很少有人读书到八年级
it was not that common for people to be getting pass say 8th grade,
或者达到类似水平
or something like that,
亦或是接受了完全合适的教育
or to be getting full proper educations.
所以像MIT这样的学校
So for schools like MIT,
它们的学生生源是有限的
they had a very limited pool of students
它们需要吸引学生来让学校变得成功
that they needed to attract to their college for them to be successful.
因此 我认为这个主意对于我们这代人来说是比较新鲜的
So, I think this idea that getting into one of these top schools
即招收大量合格的学生提高竞争力 从而成为这些顶尖高校的一员
being highly competitive for many qualified students is a bit new to sort of our generation,
然而 在19世纪却并不是这么回事
and looking back at the 19th century, that probably wasn’t the case.
接下来 我们来看看考试试题吧
So, let’s just have a look at it.
这是代数部分
So, this is the Algebra section here.
我会通读全部题目 但仅仅是粗略地看一遍
And I’ll go through the questions, but just have a glance at them here,
如果你有数学基础或背景 一看到这些题目
you probably might recognise how to do most of them already,
你可能会发现大部分题目你都会做
if you do have a math education or a math background.
我们需要做的就是化简表达式
We’re just doing things like simplifying an expression,
进行相乘计算求值
multiplying something out,
将表达式化成最简形式
reducing something to its simplest form,
化简 代数运算
simplifying, solving algebra
这些都是各种最基本的方程式
And these are all kind of basic equations.
今天你看到的试卷内容主要就是这些了
I think these are the sorts of things you’ll probably see today
有点像8年级的课程内容
in something like 8th grade,
或者 确切的说属于高中数学范畴
or, yeah, definitely within high school math education.
了解那时候的书写语言也是很有趣的
It’s also interesting to know that the language too,
直到今天它仍然适用
still holds true today.
所以 这个考试已经近乎成熟了
So, this exam has aged pretty well.
不需要我过多地讲解上下文
I think it’s because I don’t go too much into context,
也不用尝试解释任何深层次的东西
and try to explain anything too deeply.
但是你知道的 诸如倍数 化简到最简形式
But, you know, words like multiple, reduce to its lowest terms,
化简 都已经十分成熟了
simplify, you know, that can’t really age.
这些仅仅只是数学的基础部分
These are just the basics of math,
直至今日 我们仍然通过这种方式书写数学问题
and these are still how we would write math questions today.
所以 我能够理解这些问题在问我什么
So, I understand what I’m being asked for in all of these questions.
第一部分 你得到这个表达式后会发现里面有一串E
Part one, you got this expression and it’s got a bunch of Es in it.
我起初被欺骗了去思考E是什么
Now, I was initially fooled into thinking that this was,
就像E等于2.7 诸如此类的
you know, like E as in, like 2.7 something something,
但它仅仅是一个变量 我已经说过E等于8
but it’s just a variable. I’ve said E is equal to 8.
我不知道这是否是最差的工程师所求出来的E的近似值
I don’t know if that’s the worst engineer to do proximation for E ever,
但是它实际上让这个问题变得简单许多
but it actually makes this question kind of simple.
你仅需进行替换 你知道的 把8替换到表达式中
You just need to sub, you know, 8 into here.
所以 你将用8减去8加1的和开平方根 计算结果为9
So, you’ve got like 8 minus the square root of 8 plus 1, that’s 9.
8减去8开立方根 计算结果等于2
8 minus the cube root of 8, which is gonna be 2,
接下来是4开平方根 结果将会是
and then square root of 4, that’s gonna be.
我们先把这些计算出来
Let’s just work some of these out,
然后再对它们做整体运算
and I’ll do them over here.
8减去括号3加2 加上8减2
8 minus 3 plus 2 plus 8 minus 2.
实际结果等于6
Well, actually that’s just gonna be 6.
乘以4开平方根 那将会是乘以2
Times the squared of 4, that’s going to be times 2.
所以 这个部分将会是加上12
So, this part here will be adding 12.
8减5 所以这里等于3加12
8 minus 5. So, that’s gonna be 3 plus 12.
我们计算出的答案是15
We’re gonna have 15 as our answer there.
我猜我已经有点暴露自己的算术和代数能力了
I guess I’m kind of exposing my own abilities with some Arithmetic and Algebra here.
希望我在完成这份试卷的时候做得不是太差
Hopefully I’m not doing too bad of a job.
问题二 简化以下表达式 去除括号和合并项
Question two, simplify the following expression, remove the brackets and collect like terms.
我们开始吧 一层一层的来理清和解答
So let’s do it, sort out layer by layer.
这里的大括号将会是我们解决的第一层
Our first layer will be these curly brackets in here.
所以 我们将去除这些小括号
So, we’ll get rid of those to have’em here,
B加上2A减B 然后减去A再加B
B plus 2A minus B minus A plus B.
接下来 我们将去除下一层的大括号
Then, we’ll get rid of our next layer of brackets.
我们将得到3A减B减2A加B加A减B
We’ll have 3A minus B minus 2A plus B plus A minus B.
将它简化 然后把这个A和前面3A相加
Let’s simplify this, okay, then plus A, so put that there.
所以 我想我们在这里就结束了 结果是2A减B
So, I think what we would end up with here in the end will be 2A minus B.
问题三 用这一项乘以这个表达式
Question three: Multiple this expression here by this one,
然后用结果除以A加B
and divide the product by A plus B.
解决这个问题的一种糟糕方法是先做乘法
Now, a bad way to approach this question would be to do the complete multiplication first
它将会让表达式扩大到十分复杂
and to completely expand, you know,
如果你将这个表达式括起来 再与前面的表达式相乘
what you’re gonna get if you multiple this and brackets by this expression here,
然后用A加B来除那个答案
and then trying to divide that answer by A plus B.
那可不是件令人愉快的事情
That’s not gonna be that pleasant.
我推荐的方法是分解出一个A加B
Instead, what I would recommend is to try to factor out an A plus B
从这两个表达式中的一个
from either of these two expressions,
这就意味着你不必在后面做除法
which will mean you don’t have to divide by afterwards.
所以让我们尝试从这个表达式里面得出一个A加B
So, let’s try and take an A plus B out of this one here.
我们将分解出一个A加B
We’ll factor out an A plus B.
因此 如果可能的话
So, if it’s possible,
我们将在这个括号里面填入一个A
what we’re going to have in this bracket would be a A
我们就得到一个A²
that gives us our A squared.
我们需要得到一个B² 但是也想有一个负数3
We’re gonna have a B squared , but we want to have a minus 3.
所以 这样试一下
So, try this, and yep,
就可以得到A²减3AB加AB减3B²
that’ll give us A squared minus 3AB plus AB minus 3B squared.
这就是一个正确的因式分解
That’s a correct factorization there.
现在 我们仅需化简这个表达式
Now, we just want to take this,
用3A²加AB减B²乘以它
multiple it by 3A squared plus AB minus B squared,
就是我们需要的表达式
which is our expression there,
不要忘记我们想要的
and don’t forget we want,
我们需要除去这个A加B
we’re dividing out this A plus B.
这样就把它分解了 我们可以将它删掉
So, we factored it, so that we can get rid of it,
这就意味着我们仅需将两个括号里的表达式相乘就能得到我们的答案
which means we’re just section-multiplying this bracket by this bracket here to get our answer.
这样我们得到3A³
That’s going to give us 3A cubed plus. Okay,
接下来我将计算这一项 A²B 减去B²A
I’ll do this term next, A squared B minus B squared A,
再加上这一项
add in this term.
那样我们可以得到3A³
That would give us 3A cubed.
这里我们有一个A²B
We’ve got A squared B here, and here,
我们就可以得到－8A²B
that’s gonna give us minus 8A squared B.
我将先把这些删除掉
I’m gonna get rid of those.
我们有一个－B²A和一个－3B²A
Our B squared As, we’re gonna have minus 1 and minus 3.
最后 我们得到一个3B³
And lastly, our 3B cubed.
因此 这就是我们第三题的答案
So, that’s our answer to three right there.
问题四：将下面的分式化简到最简形式
Question four: Reduce the following fraction to its lowest terms.
仅仅是想让我们简化这个分数形式
That just want us to simplify this.
好吧 我们试一下 提取任何我们可以从这提取的东西
Okay. So, let’s try and extract whatever we can out of here.
我们先把这块锁住
If we just lock off this.
在顶行中 我可以在这两项中各得到一个X
I’ve got an X in both terms on the top line,
那是X的六次方加上A²X³Y
That’s X to the six plus A squared X to the cubed Y.
所以 我可以提取一个X³出来
So, what I could do is take an X cubed out,
然后我们将会得到另一个X³加上A²Y
and then we’d have another X cubed plus A squared Y,
这样我们就得到一个全新的顶行
that would be our new top line.
那就是我想要得到的顶行分子的精简形式
And that’s I think as condensed as we can make the top line, the numerator.
所以 如果我们能从底行 也就是分母里 提取出X³加上A²Y
So, what would be great is if we could take this X cubed plus A squared Y
那将会是非常棒的
also out of the bottom line, out of the denominator.
让我们尝试着做一下吧
So, let’s try and do that.
所以 我将尝试着从第二行中提取一下
So, I’m trying take that out of the second line.
我将会剩下什么呢？
And what would I be left with?
实际上我这里有一个X³减去
And actually I’d have an X cubed minus.
这就是两个二次方之间的差别
This is gonna be the difference of two squares,
因为我们仅仅想要得到两项
because we only want to get two terms.
另外一个A²Y
Another A squared Y.
你看 这个就是我们的分母
Alright, so that should give us our denominator,
意味着我们可以删除这两个式子
meaning that we can cancel out these two,
最终 我们的最简答案将是X³除以X³减去A²Y
and our final most simplified answer would be X cubed over X cubed minus A squared Y.
好 把我们的目光转到第五题
Alright, let’s move on to question five.
我们想要简化这个表达式
We want to simplify this expression here.
因此 我们要用前面这组括号除以后面这组括号
So, we have these brackets divided by this set of brackets here.
并且这两组括号里的内容是一样的
And these two sets of brackets are the same,
除了这组括号表达式的中间是加号 而这边是减号
except that this one is plus in the middle and this one is minus.
我们能够在这里增加或者提取一个分数
To be able to add fractions or to subtract fractions,
我们想让它们有共同的分母
we want them to have common denominators.
所以 我必须用A加B乘以这边
So, I’m going to essentially take this side times it by A plus B,
而在这边用A减B来乘以它
take this side and times it by A minus B,
这样就可以将他们转换成拥有共同的分母
and that would convert them to having the same denominators,
以便于我们将这两部分相加
so that we can actually add them.
就像是这样：A加B
It’ll look like this: A plus B
我们要用前半部分除以后半部分
We want to divide it by this next thing,
所以也要把后半部分转换成一样的分母
and I’m gonna do the same to that.
但是我们要记住 除以一个分数
But keep in mind that dividing by a fraction
就等于乘以那个分数的倒数
is the same as multiplying by that fraction’s reciprocal,
就意味着将分子和分母交换位置
which means swapping the numerator and denominator.
所以 我将乘以后面这个分数的倒数
So I’m going to multiply it by the reciprocal of this,
但要先将它转换成具有相同的分母
but convert it to have the same denominators.
所以它会变成这个样子
So it’s gonna look like this.
现在我们可以做一些消除
We can do some cancelling now.
我们可以把这一整行和这一行消除掉
We can cancel this whole line with this one here.
我们答案是这一部分在这部分的上面
Our answer would just be this on top of this,
但是这个表达式需要我们进行扩展
but they probably want us to expand that out.
因此 我们快速对其进行扩展
So we’ll do that kind of quickly.
现在我们将消除一些相同项
Okay. Now we wanna cancel some terms.
我们可以把最终的答案写成
And we can write out final answer as being
2A²加上2B²再除以4AB
2 A squared plus 2 B squared over 4AB.
实际上 你可以将每一项除以一个2
And in fact, you can take 2 out of all of these.
因此 我们在这里去掉一个2 在这里也去掉一个2
So we’d get rid of 2, get rid of 2,
这里分母的系数会变成2 而不是4 这就是最后的答案
this one here would become a 2, instead of a 4. There’s your answer.
第六题 我们将解答这个表达式
So question six, we wanna solve for this expression here.
我想做的是先消除这些分母
And what I wanna do is get rid of these things on the bottom.
那样这个表达式将变得简单许多
That’s gonna make it a lot easier.
所以 我们开始一步一步来做吧
So, let’s just do it one by one.
首先 我们将在底部消除一个2
First of all, we’re gonna get rid of the 2 on the bottom.
所以我将把每一项都乘以2
So I’m gonna times each term by 2,
这样就会把每一项的分母都除以2
which will get rid of that.
我们将把一个2放在这一项的上面
We will get a 2 on the top here,
实际上 分母将会变成4
which will actually be like changing this into a 4,
同样的 这里将会变成8
and likewise, this will change into an 8.
下一步 我们想要消掉4
Next step, we want to get rid of 4.
把每一项都乘以4 就会把这项中的分母4消除
Times every term by 4, that would get rid of this.
在这一排乘以4 第三项的分母将会变成2
Would put a 4 up on this line, and it would change this to a 2. Alright.
最后 我们想要消掉这个2
Lastly, we want to get rid of this 2.
用每一项都乘以2 这样就可以删掉
Times every term by 2. It’s gonna rid of that.
这个4将会变成8 希望不要太复杂
That 4 is gonna change to an 8. Hope this isn’t too messy.
我们需要把这一项的每个数都乘以2
We’re gonna get a times 2 for everything there.
现在 让我们在纸上将刚写下的表达式进行扩展
Now on paper, let’s just write that out expanded.
8乘以3X减4 24X减去4乘以8 32减去
So 8 times 3X minus 4, 24X minus, 4 times 8, 32 minus,
2乘以6 等于12X
2 times 6, 12X.
它在这是个双负号 我们要加10 等于3X减1
It’s a double negative in here, we’re gonna add 10, is equal to 3X minus 1.
把相同项进行合并 24X减去12X 合并后等于12X
Okay, collect like terms, 24X minus 12X, it’s gonna be 12X,
减去22 等于3X减1
minus 22, equal to 3X minus 1.
把我们带X的项放到等式左边 把常数放到右边
Get our Xs onto this side and our constants over there,
我们将得到12减3 等式为9X等于21
we’re going to get 12 minus 3, 9X is equal to 21.
X就等于9分之21
X is gonna be equal to 21 over 9.
或者写成3分之7 这样就对了
Or that can be written as 7 over 3. There you go.
我们的最后一题 题目七
And our last one is question seven.
这是两个未知的双变量 并且有两个等式
This is two variables, two unknowns with two equations.
不管它叫什么 联立方程或其它名称
So it’s like, whatever it’s called, simultaneous equations or something.
我们将要挑选其中一个进行重新排列
So what we’re gonna do is select one of them and rearrange it
来得到只有一个变量的表达式
to get an expression for one of the variables.
我们从这个等式开始写吧
So let’s just take this one here.
我将会为X进行重新排列等式
I’m gonna rearrange this for X.
所以 X等于11加3Y除以4
So X would be equal to 11 plus 3Y over 4.
我将把X的值代入到这个等式里面
I’m gonna sub that value of X into this equation here.
我知道有好几种方法来解答这个问题
I know there are several types, several ways of solving problems like this,
这可能是个糟糕的方法 但是我还不确定
this might be a bad way, but I don’t know,
现在对我来说 它看起来是一个简单的解题方法
it just seems like an easy way for me to do right now,
考虑到我已经有段时间没有做过这些问题了
considering I haven’t done these problems for a while.
我将用4来乘以等式中的每一项
I’m taking that 4 and multiplying every term by it,
因为我认为 相比用每项都除以4 这样会更简单
cause I think that’s easier than dividing these numbers by 4.
不错 减去这个数 结果为19
Okay. Subtract this. That’s going to be 19.
计算出来Y等于19 太酷了
And that would be Y equals 19. Cool.
现在 我们仅需把Y替换回去来求X的值
Now we just need to sub that back in for X,
求出X等于多少
and find out what X would be.
X将等于11加3乘以19
It would be 11 plus 3 times 19.
这里汇总求和等于68 然后除以4
That would bring the total up there to 68, divide by 4.
X将会等于17
X would be equal to 17.
好吧 就这样 这门考试就已经为大家解答完毕了
Alright. Well, this exam is solved for you,
可能并不需要我为大家进行解答
but maybe you don’t need me to solve it for you,
就像我之前说的 这门考试超级简单
because, I don’t know, I sort of think this exam is surprisingly easy, like I said before.
它有点儿像8年级的代数
It is kind of 8th grade Algebra.
所以 也许在19世纪考入MIT比现在通过8年级考试要简单的多
So maybe getting into MIT in the 19th century was easier than passing 8th grade today.
我不知道 也许是时代发生了一点改变
I don’t know. Maybe times have changed a little bit.
我认为 你知道的 务必要记住
I would say, you know, do keep in mind that
也许这个考试并不是为了达到和如今入学考试同样的目的
maybe this exam didn’t serve the same purpose that entrance exams do now,
或许只是为了缩小申请人的数量
which have to like really narrow down the amount of people applying.
也许这能反映出一个更具代表性的事实
Maybe this was more representative of the fact that
这些大学是供上流社会学习的机构
these colleges were really, you know, upper crust institutions.
并且这些大学招收的学生数量有限 很难将申请入学者摆在首位
And they had a very limited pool of students from which, you know, to get applicants in the first place.
所以 这个阶段拒绝大量的学生
So they probably, you know, wasn’t in their best interest to be turning away
可能并不是出于他们的最大利益考虑
a lot of students at this stage.
感谢观看 大家有任何想法都可以留在评论区
So thanks for watching, and leave any thoughts in the comments.
我想尝试一下其他部分的内容
I will likely attempt some of the other sections,
也许在下一个视频中给大家带来几何学
maybe the geometry section in a later video.
感谢brilliant.org赞助我制作这个视频
Thanks to brilliant.org for sponsoring me to make this video.
如果你想复习一下自己的代数技巧
If you want to brush up on your own algebra skills,
你可以通过brilliant网站上的课程来学习
then you can work through brilliant’s course,
不需要死记硬背的前提下 探索代数和逻辑的基础
exploring the foundations of algebra and logic without any rote memorisation.
你可以点击brilliant.org/tibees进行免费注册
You can hit to brilliant.org/tibees and sign up for free.
前200名注册者享受20%的年度优质订阅费用减免
And the first 200 people to so, get 20% off in annual premium subscription.
如果你仍在寻找一份假期礼物的话
If you’re still looking for a gift this holiday season,
也许可以考虑送给他一份优质订阅哦
then maybe consider gifting someone a premium subscription.
也要感谢那些在Patreon平台上支持我的小伙伴
Thanks also go to those who support me on Patreon.