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Favre4Ever

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I love these... Feel free to post your own if you have some!

 

-----------------------------

 

Two grasshoppers are hanging around, and each is boasting that he is faster than the other. To settle the argument, they decide to have a race.

 

The larger of the grasshoppers, Throckmorton -- or Throckie, as he's known -- can jump 10 inches at a single bound. The other grasshopper, Rocky, can jump only six inches at a shot. So the larger grasshopper says, "We're going to set up a racecourse that's 24 feet long: 12 feet out and 12 feet back."

 

They're each at the starting point. Vinnie Goombatz-Hopper shoots the gun, and they take off.

 

Now, even though Throckie, the bigger guy, can jump 10 inches at a shot, the little guy, Rocky, jumps more often. So when they get to the five-foot mark -- which is 60 inches -- the big guy has jumped six times and the little guy has jumped 10 times, but they're dead even...neck and neck...antennae to antennae.

 

The question is: Which one wins the race, and why?

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So basically what this is saying is that the Grasshoppers are tied every 5 feet. So essentially this is a 4 foot race. If we say it takes a minute for these grasshoppers to make it 5 feet (because I need a time reference), then we can say the small grass hopper is going 6 inches at each 6 second interval, while Throckie is going 10 inches at each 10 second interval.

 

So it takes Rocky 48 seconds to reach the four feet mark (48 inches)

While it takes Throckie an extra 2 seconds to reach 50 inches, because at 48 seconds, he's still at 40 inches.

 

Rocky wins. Do you have any more of these? They are fun.

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You might get another if you get this one right first. :p

 

They jump at the same speed.

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I love these... Feel free to post your own if you have some!

 

-----------------------------

 

Two grasshoppers are hanging around, and each is boasting that he is faster than the other. To settle the argument, they decide to have a race.

 

The larger of the grasshoppers, Throckmorton -- or Throckie, as he's known -- can jump 10 inches at a single bound. The other grasshopper, Rocky, can jump only six inches at a shot. So the larger grasshopper says, "We're going to set up a racecourse that's 24 feet long: 12 feet out and 12 feet back."

 

They're each at the starting point. Vinnie Goombatz-Hopper shoots the gun, and they take off.

 

Now, even though Throckie, the bigger guy, can jump 10 inches at a shot, the little guy, Rocky, jumps more often. So when they get to the five-foot mark -- which is 60 inches -- the big guy has jumped six times and the little guy has jumped 10 times, but they're dead even...neck and neck...antennae to antennae.

 

The question is: Which one wins the race, and why?

 

You don't need a time reference.

 

Here's how you do it. First, we figure out how many inches long the racecourse is- 144 in each direction, 288 total.

 

I think the trick is you can't have a partial jump.

 

So Throck would take 144/10, 14.4, or 15 jumps in one direction. He would then be 6 inches over the line, and so the reverse course would be, for him, 12.5 feet long, or 150 inches, which would take him exactly 15 jumps, for a total of 30.

 

Rock jumps 6 inches each time he jumps, (60 inches at the five foot mark, divided by 10 jumps= 6 inches), and he will be exactly at the line both ways, since its exactly 12 feet each direction. He will take 48 total jumps in order to finish the course.

 

So, since we know that if Rock jumps 10 times and Throck jumps 6 times, they would be dead even, that means the question is, is 48/30 > 10/6? If it is, then Rock wins. If its not, then Throck wins. 48/30 = 1.6, and 10/6 = 1.667.

 

So Throck wins the race.

Edited by Thanatos19

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You don't need a time reference.

 

Here's how you do it. First, we figure out how many inches long the racecourse is- 144 in each direction, 288 total.

 

I think the trick is you can't have a partial jump.

 

So Throck would take 144/10, 14.4, or 15 jumps in one direction. He would then be 6 inches over the line, and so the reverse course would be, for him, 12.5 feet long, or 150 inches, which would take him exactly 15 jumps, for a total of 30.

 

Rock jumps 6 inches each time he jumps, (60 inches at the five foot mark, divided by 10 jumps= 6 inches), and he will be exactly at the line both ways, since its exactly 12 feet each direction. He will take 48 total jumps in order to finish the course.

 

So, since we know that if Rock jumps 10 times and Throck jumps 6 times, they would be dead even, that means the question is, is 48/30 > 10/6? If it is, then Rock wins. If its not, then Throck wins. 48/30 = 1.6, and 10/6 = 1.667.

 

So Throck wins the race.

 

You were doing really good, then you lost it. Your train of thought was better than Razor's.. then the ending kinda ruined it.

Edited by Favre4Ever

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Ah, I think I see.

 

At five feet, they are tied. At ten feet, they are tied.

 

Now we're getting close to the first line. They need to go 24 inches. Rocky takes 4 jumps and hits it perfect, and turns around to continue the race. Throck takes 3 jumps, and is now 6 inches behind the line.

 

Now they head back. When Rocky is at the 7 foot mark, Throcky is at the 7 foot-6 inch mark.

 

When Rocky is at the 2 foot mark, Throcky is at the 2 foot-6 inch mark.

 

Now we are approaching the finish line. Rocky needs 24 inches to get there. Rocky takes 4 jumps and lands right on it. Throcky needs 30 inches to get there, which he does in 3 jumps.

 

If Rocky can take 10 jumps to Throcky's six, then he can clearly take 4 jumps to Throcky's 3.

 

So Rocky wins.

 

That was the original ending of the post, but the analysis is not quite correct, although the conclusion is the same. It actually ends up being each one needing to take 3 jumps. The problem with this explanation is that we don't end up with the correct ratio of jumps at the halfway area. Rocky jumped 4 times to get there, and Throcky jumped 3, but the ratio is 10 jumps of Rocky's to 6 jumps of Throcky's. So the correct entire thought process is below, (I think. Screw you for making me do math at 2 in the morning).

 

Now we have an anomaly. Rocky takes six more jumps in the same time it takes Throcky to do 3 more jumps, (keeping the 10 to 6 ratio, since they had a 4/3 ratio to get to the halfway point), which leaves Rocky sitting at 108 inches to go, (6 jumps at 6 inches each=36 inches, 144-36=108).

 

Throcky must go 150 inches to get to the finish line, and 3 more jumps puts him at 120 inches at the same time Rocky is at 108 inches. They then each travel 60 inches, and hold the same distance, (they go the same distance every five feet), which leaves Rocky with 48 inches, or exactly 4 feet to go, and Throcky with exactly 60 inches, or five feet to go.

 

Our ratio was 10 jumps to 6 jumps, but we can drop that down further, 5 jumps of Rocky's, for every 3 jumps of Throcky's, (10/6=5/3). So they're also tied every 30 inches. So each goes another 30 inches, which leaves Rocky at 18 inches, and Throcky at 30 inches. Each will reach the finish line in exactly 3 jumps. Since Rocky can easily jump 3 times faster than Throcky can jump 3 times, Rocky wins the race.

Edited by Thanatos19

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Bingo.

 

That extra jump the big guy needs at the halfway point puts him behind and being that the two jump at the same speed, he can never catch up and pass Rocky.

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I'm posting this before reading any of the replies so I can see if I get it right.

 

Rocky would win because the length of the track is divisible by 6 and not 10 like Throckie's jump. Throckie jumps 6 inches over the turning point line to 150 inches. Rocky lands directly at 144 inches, turns around, and wins the race because Throckie can't make up the difference in inches.

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Take yourself back in time to California, the Gold Rush, 1849. You're prospecting for gold. You've had a pretty good run of luck.

 

So you decide it's time to clean up and go into the big city to celebrate.

 

You stumble out of one of the saloons, having spent most of your money on women and wine -- and you're about to squander the rest -- when you hear someone call out to you.

 

From the inky shadows emerges a well-dressed gentleman who proposes a game of chance. He says, "I have this little silk bag. In it are three cards. One of them is green on both sides. Another one is red on both sides. And the third is red on one side and green on the other.

 

"I'm going to allow you to inspect the bag and put the cards inside. Without looking, I will let you pull one of the cards and place it on this little table in front of me without revealing what's on the bottom of the card."

 

You reach into the bag, deftly pull out one card, and put it on the table. You see a red face.

 

The con man says, "I'll bet you even money that the other side of the card is also red."

 

Should you take the bet?

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No, you shouldn't. The trick to this is that you're picking a card's side out of the bag, not a card. 2/3 of the sides with red on them will have red on the other side. Unless you've got really good chi going on, don't take that bet.

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Razor redeems himself.

 

-------------------------------------

 

When my kids were in school, they, like all the other kids I guess, had to learn their numbers. So each day for homework, they would bring home a list of numbers on a piece of paper, and they were asked to write out the letters that spelled that number, right next to each of them. So the number seven would be there, there'd be a blank space, the kids would have to write S - E - V - E - N. And of course they were also asked which numbers were spelled out by the various combinations of letters, so they'd see S - I - X - T - Y and write Sixty, etc.

 

One day, son number two presented me with a list of numbers and he said, "These numbers are different. There's something special about them." Here are the numbers:

 

Four, Six, Twelve, Thirty, Thirty Three, Thirty Six, Forty, Forty Five, Fifty, Fifty Four, Fifty Six, Sixty, Seventy, Eighty One, Eighty Eight, Ninety, and a Hundred.

 

Now there are no other numbers between one and one hundred inclusive that share this same characteristic. There's something unusual about these numbers that son number two figured out. And I'll give you an additional hint that order does not matter. The best hint is that he determined that these numbers should be on the list perhaps from his homework assignment.

 

What is special about this list of numbers?

Edited by Favre4Ever

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Each number is divisible by the number of letters in their name. Except that it should be "One Hundred" and not "a hundred".

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I am vastly impressed.

 

-----------------------------------------------------------------

A landscaper returns from work and is sitting at the kitchen table with his kids. The kids ask, "Did you work hard today, Daddy?"

 

Dad says, "I did. I planted five rows of four trees each." His little third grader, wanting to show off her new found skills with the multiplication table, says, "You planted 20 trees, Daddy!"

 

He says, "No, I'm sorry, you little twerp. That's wrong. I planted 10 trees." She responds, "That's impossible!"

 

How does the landscaper plant five rows of four and end up with ten total trees?

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Satanpaintskills.png

 

Behold my master artisitic skills. The black dots are trees and the red are the rows it goes by. You didn't tell me the lumberjack was a satanist though... FOUR ROWS OF FIVE!

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Nice one razor...

 

In recognition of the upcoming baseball season, here is one dedicated to America's Past time

----------------------------------------------

 

An avid baseball fan came home from work one day and realized this his favorite team had played earlier that afternoon. He was wondering how his favorite pitcher did, so he brought up the box score of the game.

 

When he looked at the box score, it said...

 

Innings pitched, 1. Hits, 0. So he pitched an inning and allowed no hits. Runs, 1. Earned runs, 1. Bases on balls, 0.

 

So, in other words, he pitched one inning, and in doing so, he recorded 3 outs, gave up no hits or walks, and was still charged with an earned run. He also struck out the first batter he faced.

 

How was this possible?

Edited by Favre4Ever

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Nice one razor...

 

In recognition of the upcoming baseball season, here is one dedicated to America's Past time

----------------------------------------------

 

An avid baseball fan came home from work one day and realized this his favorite team had played earlier that afternoon. He was wondering how his favorite pitcher did, so he brought up the box score of the game.

 

When he looked at the box score, it said...

 

Innings pitched, 1. Hits, 0. So he pitched an inning and allowed no hits. Runs, 1. Earned runs, 1. Strikeouts, 1. Bases on balls, 0.

 

So, in other words, he pitched one inning, and in doing so, he recorded 3 outs, gave up no hits or walks, and was still charged with an earned run.

 

How was this possible?

I don't know if an HBP counts as a walk, but my best guess would be 2 HBP's and two Sacrifice flies that advance base runners. So:

 

HBP, HBP, SAC, SAC, Runner scores.

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Wells... there's a first. Technically.. You would be correct.. But not what I was looking for. Therefore, I changed the puzzle to avoid loop holes like that. ;D

 

I propose you get 3 points for that one.

Edited by Favre4Ever

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Wells... there's a first. Technically.. You would be correct.. But not what I was looking for. Therefore, I changed the puzzle to avoid loop holes like that. ;D

 

I propose you get 3 points for that one.

Hmmm...How about a pitching change after an HBP? So:

Batter gets hit by pitch with zero outs, the offending pitcher is pulled and reliever comes into the game. That puts the reliever in a bind to start with, of course. The reliever strikes out the next batter. From there (I think), my prior scenario may work. 2 HBP, 2 Sacrifice flies.

 

Man on base. Strike out, HBP, Sacrifice fly, Sacrifice Fly.

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Hmmm...How about a pitching change after an HBP? So:

Batter gets hit by pitch with zero outs, the offending pitcher is pulled and reliever comes into the game. That puts the reliever in a bind to start with, of course. The reliever strikes out the next batter. From there (I think), my prior scenario may work. 2 HBP, 2 Sacrifice flies.

 

Man on base. Strike out, HBP, Sacrifice fly, Sacrifice Fly.

 

The earned run would go to the first pitcher and inning would be over on outs before he accumulated an ER. One pitcher accumulates the stats you see above.

 

On Base

Out

On Base

On Base

Out 2, ER to pitcher 1

Out 3, no runs

 

Although you are on the right track. Just... kind of in the wrong order.

Edited by Favre4Ever

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The earned run would go to the first pitcher and inning would be over on outs before he accumulated an ER. One pitcher accumulates the stats you see above.

 

On Base

Out

On Base

On Base

Out 2, ER to pitcher 1

Out 3, no runs

 

Although you are on the right track. Just... kind of in the wrong order.

Okay. My brain is tired (it is 4am, after all), so I will try and work it out the rest of the way on the morrow.

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No, you shouldn't. The trick to this is that you're picking a card's side out of the bag, not a card. 2/3 of the sides with red on them will have red on the other side. Unless you've got really good chi going on, don't take that bet.

 

I don't get this one.

 

If the card you pulled out has red on one side, assuming he's telling the truth, only 1 of the 3 cards in the bags has red on both sides. There's a green/green card, a green/red card, and a red/red card. Obviously the green/green card is not the one you pulled, so its the either the red/green card or the red/red card.

 

Why is not then a 50-50 shot?

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I don't get this one.

 

If the card you pulled out has red on one side, assuming he's telling the truth, only 1 of the 3 cards in the bags has red on both sides. There's a green/green card, a green/red card, and a red/red card. Obviously the green/green card is not the one you pulled, so its the either the red/green card or the red/red card.

 

Why is not then a 50-50 shot?

 

Like Razor said... You need to think about as sides, not necessarily a probability at first.

 

Possible options with all three cards...

 

Red Red Red Green Green Green

 

You take out one card and its red...

 

Red Red Green Green Green

 

You know the card can't be double green...

 

Red Red Green

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When he strikes out the batter, he does so on a wild pitch, which still makes the runner count as an earned run. The runner reaches base, so the strikeout does not result in an out.

 

Then there are multiple ways to get home, he either steals all the bases, the pitcher throws 3 more wild pitches and he scores, or 2 sac flies happen and then either he steals home or a wild pitch lets him take home, or any combination of these things.

 

No hits, 1 strikeout, no walks, and yet 1 ER.

Edited by Thanatos19

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This question has far too many loopholes... lol.

 

The pitcher is a relief pitcher. He gets his first out, and finishes his inning without allowing a walk, hit, or run. He begins the next inning, but beans the first batter. The manager pulls him and the new pitcher gives up a homerun. Pitcher #1 is responsible for the batter that scored after reaching on a HBP.

 

 

Not gonna do any other baseball ones... >_>

 

 

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This is short and sweet. I’m going to give you a series of numbers and ask you what the next number in the series is. Here are the numbers:

 

4, 6, 12, 18, 30, 42, 60, X, 102

 

What is X and why?

Edited by Favre4Ever

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This question has far too many loopholes... lol.

 

The pitcher is a relief pitcher. He gets his first out, and finishes his inning without allowing a walk, hit, or run. He begins the next inning, but beans the first batter. The manager pulls him and the new pitcher gives up a homerun. Pitcher #1 is responsible for the batter that scored after reaching on a HBP.

 

 

Not gonna do any other baseball ones... >_>

 

 

----------------------------------------------

This is short and sweet. I’m going to give you a series of numbers and ask you what the next number in the series is. Here are the numbers:

 

4, 6, 12, 18, 30, 42, 60, X, 102

 

What is X and why?

 

Ahha, I was thinking too hard.

 

Well here's the pattern. The answer is 78, I think, this is too much to be a coincidence.

 

4+2=6, 6+6=12, 12+6=18, 18+12=30, 30+12=42, 42+18=60, 60+18=78, 78+24=102

 

It's adding 6 to 6 and 12, then 6x2 to 18 and 30, then 6x3 to 42 and 60, then 6x4 to 78. The next number in the sequence would be 126.

 

This doesn't account for the first number of the series, however. That 4... not sure why its there.

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