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post #1 of 34 (permalink) Old 05-24-2005, 05:51 PM Thread Starter
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How good are you in MATH??

i got this of some other forum so here it goes...

3 guys get a motel room.. They all share the same room. The cost of the room is $30 that = how much each? $10.The manager comes in and says to the lady at the front desk that you over charged the guys by $5. The front desk lady sends the bell boy to give the guys there $5 back. He is thinking how am I going to do this, I have to split $5 three ways. He said I will make it easy and pocket 2 of the 5 dollars.

Now the guys are paying $9 each for the room That = $27 now Then the bellboy has the $2.Where did that dollar go?

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post #2 of 34 (permalink) Old 05-24-2005, 05:58 PM
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post #3 of 34 (permalink) Old 05-24-2005, 06:07 PM Thread Starter
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smarta$$ lol where man i didn't see it

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post #4 of 34 (permalink) Old 05-24-2005, 06:11 PM
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hahah too lazy to go looking for it
muwhahah
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post #5 of 34 (permalink) Old 05-24-2005, 06:13 PM
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They aren't paying $9 for the room each. If the total bill for the room was 25 bucks, the equal share of that is 8.33 each (with one guy getting screwed paying 8.34 ). Add that extra dollar to each person's share and you are at 9.33 each. Multiply by 3, add the 2 bucks that the bellboy stole and you have 30 bucks.

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post #6 of 34 (permalink) Old 05-24-2005, 06:13 PM
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There is no missing dollar. It's phrased wrong.

Amount Initially Paid: $30

Amount After $1x3 Refund: $27

Amount Tipped to Bellboy: $2

Amount of Room: $25

Now that should be enough for you to figure out the solution.

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post #7 of 34 (permalink) Old 05-24-2005, 06:15 PM
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If people are still arguing it when I get back on later, I'll explain a bit further.

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post #8 of 34 (permalink) Old 05-24-2005, 06:24 PM
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For both palmguy and bill, congratulations for both being able to say 'i'm correct, i win'.

And it's still a REPOST!

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post #9 of 34 (permalink) Old 05-25-2005, 01:44 AM
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lol...when i read the title of this thread, i thought i was gonna have to bust out the partial differential equations book :p
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post #10 of 34 (permalink) Old 05-25-2005, 04:00 PM
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I have the dollar. Problem solved. Next?

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post #11 of 34 (permalink) Old 05-25-2005, 11:10 PM
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That was disappointing. I've heard that question too many times. Here's a better one.

Imagine a 55 gallon barrel filled with water. Would it drain faster with a 6-inch hole or two 3-inch holes in the bottom of the barrel? Assume the rest of the barrel is enclosed.

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post #12 of 34 (permalink) Old 05-26-2005, 12:20 AM
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Quote:
Originally Posted by kenshi
That was disappointing. I've heard that question too many times. Here's a better one.

Imagine a 55 gallon barrel filled with water. Would it drain faster with a 6-inch hole or two 3-inch holes in the bottom of the barrel? Assume the rest of the barrel is enclosed.
hmmm.... im leaning towards a 6'' hole... theoriticaly I imagine it should be the same... but im leaning towards the gravity of the water pushing harder on one single bigger hole.... hang on im gonna go poke some holes in my water jugs

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post #13 of 34 (permalink) Old 05-26-2005, 01:07 AM
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Quote:
Originally Posted by awdRocks
hmmm.... im leaning towards a 6'' hole... theoriticaly I imagine it should be the same... but im leaning towards the gravity of the water pushing harder on one single bigger hole.... hang on im gonna go poke some holes in my water jugs
I won't say whether you're right or not about it being a 6" hole since that would kind of give it away, but your reasoning is incorrect.

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post #14 of 34 (permalink) Old 05-26-2005, 07:55 AM
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The two three inch holes because they are in the bottom of the barrel and can hence drain it completely, the 6 inch hole could be anywhere since it wasn't specified in the word problem and might not be able to get all the water out.

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post #15 of 34 (permalink) Old 05-26-2005, 08:09 AM
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Here's one to get you thinking a little. Take 4 steel balls of exactly 1" in diameter. Place three of them on a flat table top, touching each other so they form a triangle. Now place the fourth on top of them, nesting it in the center. Can everyone see this? What is the distance to the center of the ball on top from the table surface?


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post #16 of 34 (permalink) Old 05-26-2005, 08:11 AM
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1/2" after the three on the bottom roll out and it falls.

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post #17 of 34 (permalink) Old 05-26-2005, 08:21 AM
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Quote:
Originally Posted by kenshi
Imagine a 55 gallon barrel filled with water. Would it drain faster with a 6-inch hole or two 3-inch holes in the bottom of the barrel? Assume the rest of the barrel is enclosed.

FIrst you didn't say whether they are square holes or round holes, but it doesn't really matter.

If they are round holes, the area of 2 - 3" holes is about 14 square inches, while the area of 1 - 6" hole is about 28". If they are square holes, the area of the 3" holes is 18 square inches, while the 6" hole is 36 square inches.

In either case, the one 6" hole will drain faster, due to its larger opening.


I just noticed the part about the 'rest of the barrel is enclosed'. Now I don't know, but if both 3" holes are in the bottom, I don't think it changes much. If one of the 3" holes were in the top and one in the bottom, then that would definitely be the faster answer.

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post #18 of 34 (permalink) Old 05-26-2005, 08:37 AM
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Quote:
Originally Posted by ShadowDragon
1/2" after the three on the bottom roll out and it falls.
Alright, smarta$$!! I intentionally left that part out, just because I wanted to see who the first person to say that was. You win! That was actually what I said too, right before I calculated it. Ok, so now assume the balls won't roll apart...they're captured, retained, glued, whatever. Carry-on.

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post #19 of 34 (permalink) Old 05-26-2005, 09:16 AM
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As far as the barrel goes, I'm gonna figure 2- 3" holes. One draining and one replenishing air into the drum.

The balls.....I couldn't tell you exactly. I would have to figure somewhere around 1.25" to center???

Just guessing like everyone else.

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post #20 of 34 (permalink) Old 05-26-2005, 09:21 AM
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Approximately 1.32" if my calculations are correct.

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post #21 of 34 (permalink) Old 05-26-2005, 09:25 AM
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Ding ding ding ding....PalmGuy's the winner. Good job. Now, did you calculate that, or use CAD to figure it out?

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post #22 of 34 (permalink) Old 05-26-2005, 09:27 AM
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Alright, smarta$$!! I intentionally left that part out, just because I wanted to see who the first person to say that was. You win!
That's the difference between real-life math and theoretical math.

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post #23 of 34 (permalink) Old 05-26-2005, 09:32 AM
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Quote:
Originally Posted by Toombs
Here's one to get you thinking a little. Take 4 steel balls of exactly 1" in diameter. Place three of them on a flat table top, touching each other so they form a triangle. Now place the fourth on top of them, nesting it in the center. Can everyone see this? What is the distance to the center of the ball on top from the table surface?

Mike

Well, even with seeing an explanation, that problem is just too tough for me.

But here is a bunch of geeks discussing it. They don't seem to come to an agreement, so I don't know if any of them are right either.

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post #24 of 34 (permalink) Old 05-26-2005, 09:43 AM
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Hmmm, what a life.
Sitting around wondering what the height of 4 ping pong balls could possibly be. I must be doing something wrong.

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post #25 of 34 (permalink) Old 05-26-2005, 09:44 AM
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Interesting find. Actually, that Dook character is right. Well, on the right track anyway. He just left out accounting for the balls themselves. He figured the vertical distance from the center of one of the lower balls to the center of the top ball. For overall height (what they were discussing) he would have to add 2R to his answer. For what I asked, you only add R. But either way, Dook had the right idea. That other guy likes to hear himself talk, uses what he considers to be his extensive vocabulary to confuse the issue with mathematical jargon and sound smarter than he really is. Gosh, I remember how much I hate math for the sake of math now. The complex calculations that apply to design I find fun in, but math just for math, i.e. in math class....sucks! Anyway, I've gone off on a tangent.

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post #26 of 34 (permalink) Old 05-26-2005, 09:47 AM
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uh...actually I guess we are doing the same thing eh??LOL!

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post #27 of 34 (permalink) Old 05-26-2005, 10:30 AM
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Ding ding ding ding....PalmGuy's the winner. Good job. Now, did you calculate that, or use CAD to figure it out?

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a completely random, unrelated question, but . . . Is that an AutoCAD or Solid Edge rendering in that one picture?

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post #28 of 34 (permalink) Old 05-26-2005, 10:57 AM
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None of the above....it's SolidWorks. Por que?

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post #29 of 34 (permalink) Old 05-26-2005, 11:03 AM
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the two 3" inch holes placed on oppisited sides of the bottom of the barrel, then place the barrel on its side, the air would enter the top hole, while fluid pores out the bottom whole...

oh, is that why there is two 3" holes on most barrels?

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post #30 of 34 (permalink) Old 05-26-2005, 11:05 AM
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Sure its not closer to 1 1/3 " on the balls?

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