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Harvard's entrance exam from 1869...
"The New York Times remembers back to when 'college was a buyer's bazaar' and digs up 19th-century classified ads from Columbia, Harvard, Yale, and others. In competitive efforts to attract students from the limited pool of qualified candidates, applications were taken as late as September for an October freshman class. Vassar offered lush room accommodations. The expectations were high: Latin, Greek, Virgil, Caesar's Commentaries; Harvard's entrance exam from 1869 is posted (PDF). Could any of us pass the exam today?"
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Re: Harvard's entrance exam from 1869...
Holy shit, Greek Composition, Greek Grammar, Latin Grammar, and Latin translations?
:lol
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Re: Harvard's entrance exam from 1869...
Eh, there's a reason not many of us write in Latin now. It's great to understand origins of words, and word backbones, but back then it was a sign that you were erudite.
The math section is still relevant, obviously.
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Re: Harvard's entrance exam from 1869...
I can see the teachers pay hasn't changed any.
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Re: Harvard's entrance exam from 1869...
How about the math...
The 5th root question. I know how to calculate square roots longhand, but never learned 3rd roots plus.
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Re: Harvard's entrance exam from 1869...
Quote:
Originally Posted by
mouse
I can see the teachers pay hasn't changed any.
At Harvard?
Really?
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Re: Harvard's entrance exam from 1869...
Quote:
Originally Posted by
Wild Cobra
How about the math...
The 5th root question. I know how to calculate square roots longhand, but never learned 3rd roots plus.
I'd just start guessing at that point. "Let's try .5 times .5 times .5, and see what that gives us... ok, how about .3 cubed..." :lol
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Re: Harvard's entrance exam from 1869...
Quote:
Originally Posted by
LnGrrrR
I'd just start guessing at that point. "Let's try .5 times .5 times .5, and see what that gives us... ok, how about .3 cubed..." :lol
There are ways of doing it that are not trial and error. I just never learned you to solve that one without trial and error.
I can do square roots though!
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Re: Harvard's entrance exam from 1869...
Quote:
Originally Posted by
Wild Cobra
There are ways of doing it that are not trial and error. I just never learned you to solve that one without trial and error.
I can do square roots though!
Yeah, neither have I. Whenever I get stuck on a math problem, I try the "educated guess" method. :lol
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Re: Harvard's entrance exam from 1869...
Quote:
Originally Posted by
ElNono
"The New York Times remembers back to when '
college was a buyer's bazaar' and digs up 19th-century classified ads from Columbia, Harvard, Yale, and others. In competitive efforts to attract students from the limited pool of qualified candidates, applications were taken as late as September for an October freshman class. Vassar offered lush room accommodations. The expectations were high: Latin, Greek, Virgil, Caesar's Commentaries;
Harvard's entrance exam from 1869 is posted (PDF). Could any of us pass the exam today?"
Quote:
Originally Posted by question 7 from the algebra section
"A man bought a watch, a chain, and a locket that together cost $216.
The watch and locket together cost three times as much as the chain, and the chain and locket together cost half as much as the watch. What was the price of each?"
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Re: Harvard's entrance exam from 1869...
Took me a lot of scribbling, but I got it. whoot.
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Re: Harvard's entrance exam from 1869...
Quote:
Originally Posted by
RandomGuy
Originally Posted by question 7 from the algebra section
"A man bought a watch, a chain, and a locket that together cost $216.
The watch and locket together cost three times as much as the chain, and the chain and locket together cost half as much as the watch. What was the price of each?"
If you want someone to solve that simple problem, maybe you should put it in this thread:
48÷2(9+3) = ????
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Re: Harvard's entrance exam from 1869...
That question 7 looks like it's going to be easy, but is it a bit of a pain.
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Re: Harvard's entrance exam from 1869...
I was looking for instructions to tell the test taker to read all the questions carefully, then return to the first page and begin the test. Then the last paragraph would say to write your name at the top and then submit the exam.
I was disappointed.
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Re: Harvard's entrance exam from 1869...
Quote:
Originally Posted by
Wild Cobra
How about the math...
The 5th root question. I know how to calculate square roots longhand, but never learned 3rd roots plus.
The Newton-Raphson method applied to the polynomial f(x) = x^n - y can be used to find the nth root of y (assuming y > 0), and it converges very quickly. The algorithm from Heron of Alexandria for finding square roots is a special case of this method.
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Re: Harvard's entrance exam from 1869...
Quote:
Originally Posted by
Wild Cobra
If you want someone to solve that simple problem, maybe you should put it in this thread:
48÷2(9+3) = ????
Erk.... missed that one. Spaced out the "left to right" rule.
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Re: Harvard's entrance exam from 1869...
Quote:
Originally Posted by
baseline bum
The Newton-Raphson method applied to the polynomial f(x) = x^n - y can be used to find the nth root of y (assuming y > 0), and it converges very quickly. The algorithm from Heron of Alexandria for finding square roots is a special case of this method.
Thanx, I'll look that up.
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Re: Harvard's entrance exam from 1869...
Quote:
Originally Posted by
LnGrrrR
That question 7 looks like it's going to be easy, but is it a bit of a pain.
oops...
When I wrote the equation, I originally solved for:
the chain and locket together cost twice as much as the watch.
not:
the chain and locket together cost half as much as the watch.
Changes the answer dramatically, except the chain. It remained the same price!
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Re: Harvard's entrance exam from 1869...
For those curious about the number 7, answers/work below! (If I'm taking the time to answer it, I'm bragging. :p)
SPOILERS BELOW, HIGHLIGHT FOR ANSWER
The question gives three different equations.
a) (W+L) = 3C
b) 2(L+C) = W
c) W+C+L = 216
Since we know that 2(L+C) = W, we can insert that into the first equation. This gives us (2L + 2C + L) = 3C, which ultimately boils down to 3L=C.
Now that we know that 3L=C, we also know that W= 2(L+C) or W = 2(L+3L), or W=8L.
So C = 3L, and W=8L. Time to substitute those into equation C.
8L + 3L + L = 216. From this we get that L equals 18. And due to the info above, that means that C = 54, and W = 144. Doublechecking by plugging those numbers into the equations above, we see that it works. Voila!
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Re: Harvard's entrance exam from 1869...
Quote:
Originally Posted by
Wild Cobra
Hint...
Chain squared plus Watch squared equals Locket squared.
Yeah, just posted the answer. I was hoping I could guesstimate the answer quick, but I kept guessing wrong, so I actually had to do the work. :lol
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Re: Harvard's entrance exam from 1869...
Quote:
Originally Posted by
Wild Cobra
Thanx, I'll look that up.
It's a pretty interesting algorithm useful for finding roots of algebraic equations in many cases, but the one Heron of Alexandria came up with for square roots is essentially just smoothing out an oscillating signal by averaging.
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Re: Harvard's entrance exam from 1869...
Quote:
Originally Posted by
LnGrrrR
Yeah, just posted the answer. I was hoping I could guesstimate the answer quick, but I kept guessing wrong, so I actually had to do the work. :lol
You caught that before I caught my mistake and removed it...
What I posted before editing is true with the "twice" rather than "half."
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Re: Harvard's entrance exam from 1869...
Believe it or not, the Ancient Greek part wasn't as scary as you'd think.
I only took 3 semesters of it in college (over 15 years ago) (fuck), getting mediocre to bad grades on the reg, and I was still able to get a fair amount of it with stuff I learned in 101.
I think LnGrrr's right, though: the exam was really just a formality that any kid of aristocratic breeding back then could have breezed through on a ripping laudanum high... or whatever shit they were into then.
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Re: Harvard's entrance exam from 1869...
Quote:
Originally Posted by
baseline bum
It's a pretty interesting algorithm useful for finding roots of algebraic equations in many cases, but the one Heron of Alexandria came up with for square roots is essentially just smoothing out an oscillating signal by averaging.
Well, I was hoping for a straightforward method like there is for solving square roots, with no guessing.
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Re: Harvard's entrance exam from 1869...
Quote:
Originally Posted by
admiralsnackbar
Believe it or not, the Ancient Greek part wasn't as scary as you'd think.
I only took 3 semesters of it in college (over 15 years ago) (fuck), getting mediocre to bad grades on the reg, and I was still able to get a fair amount of it with stuff I learned in 101.
I think LnGrrr's right, though: the exam was really just a formality that any kid of aristocratic breeding back then could have breezed through on a ripping laudanum high... or whatever shit they were into then.
I wish I took Greek as a language. Sure would like to read revelations in the original tongue.