/Jeo/ - Jeopardy

Hey, it's been a while Veeky Forums.
While I'm updating my computer, let's play a game.

A while back I downloaded a quarter million clues from a Jeopardy archive. Give a category and difficulty

category | approx number of questions
science 3964
economics 2352
physics 1048
chemistry 424
biology 471
astronomy 599
math 401
engineering 281
geology 256
inventions 1537
psychology 521

I have math puzzles/problems, but they aren't jeopardy questions

post a few top difficulty math ones

Or did you mean the puzzles?

What is the additive identity

>.>
correct.
category, difficulty?

Science 800 idk

...

What is plastic?

correct, category?

Math

Hardest (idk the number)

...

What is calculus

correct, category?

Chemistry hardest

...

What is chaos?

Correct, category?

Not guy who asked question but what is chaos.

Engineering 600

Math second hardest.

...

Who was Eiffel?

Correct

google

Correct, category?

Biology 200

...

What is a Dove?

If correct: Astronomy 600

...

What is squab

Correct, sorry for the wait

What is solar wind?

Correct, category?

Chemistry 2nd hardest

...

Tungsten

Correct, category?

math nigga, hard.

didn't know they call wolfram tungsten, thx

...

What is numbers

What is number?

what is multiply?

More math plx

tau/2

correct

forgot my image, sorry for the wait

(four and twenty) / three

correct, category?

math

During a baseball game in Mudville, Case was Mudville's lead-off batter. There were no substitutions or changes in the batting order of the nine Mudville men throughout the nine-inning game. It tured out that Casey came to bat in every inning. What is the least number of runs Mudville could have scored?

user, the one above you asked for a math question, not a sport one

13?

nope, there's a smaller answer.
I also require a short proof for these.
Here have another:
With a 7-minute hourglass and an 11-minute hourglass, what is the quickest way to time the boiling of an egg for 15 minutes?

i'm still working on the baseball one but for the hourglass one you need to wait 7+4+11=22 minutes?

boil the egg for 15 minutes

>7+4+11=22 minutes?
That's a solution, but not the optimal one.
Also show how!

set the timer to 7,5 twice

9 runs? 3 runs for every third inning?

How do you set an 11 minute hourglass to 7.5?
> 9 runs?
nope. fewer

you eye ball it

Flip the 11-minute and 7-minute at the same time. When the seven-minute is finished, flip the eleven-minute and boil the egg. When the 11-minute is finished, flip it. When it is finished, 4+11=15 minutes will have passed, and the egg is done.

Wait, no: you can do it in 15 minutes, the minimum. Boil the egg and set both hourglasses. When the seven is finished, flip the eleven, and when that is finished you are done. No solution can be faster because it would take less time than it takes the egg to boil.

Math, hardest.

Start both hourglasses at the same time. When the 7 minute one runs out, you'll have 4 minutes left in the 11 minute one. Start your egg. When the 11 minute one runs out, turn it over. When the 11 minute one is out, the egg has been cooking for 15 minutes.

That's 14 min. Pretty sure
And by math do you want puzzles or more jeopardy questions?

fuck I never thought to delay starting the egg

Set both timers and boil the egg. After the 7 runs out reset it. After the 11 runs out flip the 7. When the 7 runs out 15 minutes should have passed. Pretty sure that's the optimal.

It's fifteen. And, math puzzles sound nice.

MAKE WITH THE JEOPARDY-STYLE MATH QUESTIONS OP.

Correct.
Give me a min for more clues

>Here have another:
>With a 7-minute hourglass and an 11-minute hourglass, what is the quickest way to time the boiling of an egg for 15 minutes?

What's a general form for solving these types of questions?

I don't think there is one. That's the beauty.
Here's another one for whatever algo you want to develope:
How do you measure 9 minutes with a 4min and 7min hourglass?

and for the puzzle, how about a classic:
Three missionaries and three cannibals must cross a river using a canoe, which can only carry two people. But if there are ever more cannibals than missionaries on either shore, the cannibals will overpower the poor missionaries and eat them. There's a strong current so the boat cannot be sent across the river without any people rowing.

Start 4 and 7.
When 4 is out flip both.
When 7 is out flip 4
Let 4 run out.

Set both timers. When the 4 finishes reset it (4 mins). When the 7 finishes (7 mins) flip the 7. When the 4 finishes (8 mins) flip the 7 again and wait till the minute's worth if sand runs out.

youre getting nice digits this evening

and the jeopardy seems really weird, is their something apart from a protractor that measures angles?

Geology for 500 points

Sextant? The etymology seems wrong.

I'm going to post "octant" before checking.

What is an octant?

MAKE WITH ANOTHER MATH QUESTION. ENGAGE WITH YOUR THREAD.

Octant nigs

I'll agree that's right for 1/8.th. Never knew it was a measuring device

correct

Altitude.
More math or linguistics

In diagram form
Left side | Right Side | Boat direction
mmmccc | | ->
mmcc | mc |
mmm | ccc |
mc | mmcc |
cc | mmmc |
c | mmmcc |
| mmmccc |

Also did anyone get the 9 minute candle right?

What is

22 is absolutely the optimal solution. The only periods of time when a timer can have finished before 22 minutes are minute 0, 7, 11, 14, 18, 21, and 22. Only 22-7 allows you to measure out 15 minutes.

The method to cook the egg is to run the eleven minute timer twice and the seven minute timer once. Start boiling the egg after the seven minute timer finishes.

It wasn't asked, but burn one candle from one end and the other candle from both ends. When the second candle finishes, burn the other end of the first candle.

See

I meant the hourglass but fucked up.

oh

candle problems are different from hourglass problems

you can understand my confusion

Seems valid

>> 9 runs?
>nope. fewer
That doesn't seem possible. If the first batter pitches in every inning it means you need 9*8+1 or 73 batters at least. Each inning 3 of these batters will get an out, and contribute nothing to the run count, so we can subtract off another 3*9=27. Leaving us with 56 batters. Essentially every 4 batters, a run is scored, which should make the minimum 14, yet I already found a solution for 13 so that can't be right. If you want to get it so that there's less than a run per inning you'd have to arrange it in a very clever way. This kinda reminds me of the 8 queens puzzle.

Seems they don't need a run in the last inning for Casey to bat if he bats first.

So apparently this was a real game.
The solution is 0 runs.
I'll type it out later because theres lots of color.

Thisll prob be it for today. My comp is back to full power

Please do I'm really curious about this.

Actually I just looked it up because I couldn't wait. I knew I was missing some sneaky baseball rule. It turns out if you're at bat while another player gets out stealing a base you will get first at bat next inning. It was the extra piece I needed to actually solve this. Feel free to type out your colorful solution though.

Biology, 304

Can you query this backwards. Can I ask a question and you give an asinine jeopardy answer.

That was indeed an amusing problem concerning Casey and the Mudville nine--amusing, that is, to all save lovers of Mudville. For on the unfortunate day described in your problem, Mudville scored not a single run. This is what happened:

In the first inning, Casey and two of his confreres reached base, but batters four through six struck, flied, or otherwise made out. No runs.

In the second inning, batters seven and eight struck out, let us say, but the Mudville pitcher, to the surprise of all, reached base on a bobbled infield roller. Casey came up to bat, frowning mightily. With the count two and two, the perfidious rival pitcher, ignoring the best interests of poetry, baseball mythology and Mudville, whirled toward first and picked off his opposite number, who, dreaming of Cooperstown and the Hall of Fame, had strolled too far from the bag. THe crowd sighed, Casey glowered, and the inning was over: no runs.

Now as you know, if an inning ends with a pick-off play at any base, the batter who was in the box at the time becomes the first batter next inning. So it was with Casey; once again Mudville loaded the bases; but once again three outs were made with no runs scoring, so that the inning ended with batter si making the last out.

Life may be linear but fate is cyclic: innings four, six, and eight followed precisely the same pattern as inning two (though you may be sure that after his second miscue in the fourth, the Mudville pitcher was lifted and his relief was responsible for only the flood of runs the opponents scored, but for similar cloud-gazing base-paths). And of course, Case led off the fifth, seventh, and ninth innings as he had the third--and again, Mudville would load the bases, but could not deliver (if I remember correctly, the gentlemen responsible for this orgy of weak hitting were Cooney, Burrows, Blake and Flynn). Grand total for Mudville: a goose egg.

Sorry I was unaware of that trick too. Becomes a baseball question.

To make up for it:

Each end of a 10-foot length of rope is tied securely to a man's ankles. Without cutting or untying the rope, is it possible to remove his trousers, turn them inside out on the rope, and put them back on correctly?

Most of these puzzles were taken from Martin Gardner's Colossal Book of Short Puzzles and Problems.

Yes, take one leg of the trousers and move them up through the other leg of the trousers. If you pull all the way through the pants will be turned inside out and can then be worn.