Logic problem

[Buddy_Christ]

i thought this was pretty interesting.

you have 2 large coffee cans, both are filled with beads. one can has red beads, the other has blue. both cans have the exact same number of beads in them.

someone takes a scooper and dips into the can of blue beads. they count exactly how many blue beads were removed, then place those blue beads into the can filled with red beads. the can with the red beads is stirred so the blue beads are mixed in with the red.

then, the same scooper is dipped into the red can. the amount of beads scooped out is counted so it is exactly the same as how many were removed from the blue can. then these beads are dumped into the can with the blue beads.

does the amount of blue beads in the blue can still equal the number of red beads in the red can?

you are not able to see the scooping or mixing process, nor are you allowed to see into the cans once the process has been started.

equal or not equal? those are the only choices.

 

[BileStew]

Who's asking, you or the derranged alter?

 

[Buddy_Christ]

me. the derranged alter is upstairs screwing my girlfriend. i don't think she's ever had this many orgasims in her life.

hell, i didn't even know she could have orgasims until recently. i thought the female orgasim was just a big fat lie.

 

[mylife]

Seemingly too easy but I'll play along.

Unequal

 

[MAX 300]

The information given in this problem is not enough to come to a definite conclusion.

Suppose one can had 3 red beads, the other 3 blue.

2 blue beads are scooped out and mixed with the red.

2 beads are scooped from the red can, they could be all blue, all red, or one blue and one red.

Those 2 beads are placed in the blue can.

The only definite is that both cans will contain the same number of beads.

 

[Buddy_Christ]

good point max. i didn't think of what'd happen with an odd number.

ok, the number of beads in each can is an even number.

no more clues.

 

[Buddy_Christ]

nope, it can still be done with an odd number.

you have plenty of information.

 

[BradG]

Well, I thought they would not be equal, but according to math if you assume the scoop to be 1/8 the size of the container, and you take the blue first and put it into the red, the you have a bucket composed of 8/8 red and 1/8 blue. Then if you assume the mixture to be homogeneous (perfectly mixed) then you take a scoop from it that 1/8 scoop will be composed of 8/9 red and 1/9 blue (0.11111 red and 0.014 blue). So, the original red bucket red count goes from 8/8 (1) to 0.89 (1-0.11=0.89). The blue content of that same bucket (originally all red) goes from 1/8 blue (0.125) to 0.111 blue (0.125-0.014=0.111). The sum of these is 1.001, pretty much one. Then the blue bucket has 7/8 blue and the scoop is added. So, now the blue gives 7/8 (0.875) plus the 0.0147 blue from the scoop, which yeilds 0.889 blue (0.875+0.0147=0.889). The amount of red in the blue bucket is 0.0 plus 0.111 from the scoop so 0.111 (0+0.111=0.111). The sum ov the red and blue is 0.889 blue + 0.111 red = 1.0. So, now we see the red percentage in each is equal (0.111 vs 0.111) and the blue percentage is equal (0.89 vs. 0.89), and each bucket has a full 8/8ths in it. So, yes they will be equal given perfect mixing.

I must hedge my bet. I am very close to finishing a liter of alcohol within a couple of hours. So my logic may be off and the detailed nature of my response is, I am sure due to the alcohol.

Really, I am not a dork. I am actually a pretty cool guy. I just understand mat to a degree. (Agian if I fucked up it is because I am drunk )

 

[MAX 300]

I don't see it.

I think your logic problem assumes that after the the blue beads are mixed with the red, the final scoop will include an even number of red beads and blue beads in that scoop - or that the distribution of red and blue beads in the final scoop will match the distribution of red and blue beads in the red can.

However your problem does NOT say that.

 

[MAX 300]

bradg, you made two assumptions which were not stated in the original problem.

Given those assumptions the mix will be equal.

But I maintain: "The only definite is that both cans will contain the same number of beads."

 

[BradG]

Yes, I did make two assumptions, but those assuptions were ones that had to be decided upon to solve the problem. Otherwise, we can't come up with an answer other than "It depends".

 

[MAX 300]

Exactly

Crak, your problem sucks man! It can't be solved with the info you gave.

 

[BradG]

To be honest, only one assumption that I made had any effect on the answer. The scoop size is pretty much irrelevant, but the "perfect mixing assumption is crutial.

 

[Buddy_Christ]

the problem can be solved with the info given.

the question is - after the scooping/mixing/scooping process is all done, will the number of blue beads in the blue can be equal or not equal to the number of red beads in the red can?

one of my professors gave this problem to 3rd grade kids and to college students. the 3rd grade kids got it right, the college students got it wrong.

 

[MAX 300]

Ok, I'll stick my head out.

The number of red beads in in the red can will be equal to the number of blue beads in the blue can.

I can't think of a scenario right now where they won't.

 

[BradG]

Ok, YES. The number of blue beads in the blue can can be equal to the number of red beads in the red can. It can also be greater or lower depending on the mixing.

 

[Buddy_Christ]

there's only one answer. bradg, you seem to be straddling 2.

 

[jackangel]

they will be equal.

max300, have you actually looked at that one example you mention earlier? whatever the outcome, the # of blue beads in one can will equal the # of read beads in the other. remember that along with the blue beads going back in, are likely to be some red beads. but it doesn't matter, even if there aren't.

there are 3 general cases here:

the 2nd scoop going back into the blue can is made up of:

1)all blue beads.
2)some red beads and some blue beads.
3)all red beads.

whatever combination, the subtraction and addition of red and blue beads on each side will always balance out so that the # of blue beads in the blue can equals the # of red beads in the red can.

algebraically:

x = # blue beads and # red beads
y = # blue scooped out in the first scoop
z = # blue scooped out in 2nd scoop
i = # red scooped out in 2nd scoop
h = # red beads in red can after
l = # of blue beads in blue can after

y= z + i
h = x - i
l = x - y + z

-> l = x - (z + i) + z
-> l = x - z - i + z
-> l = x - i

thus, l = h

 

[BradG]

Jack, I tend to agree, but I will see in the morning when I am sober

 

[Deleted member 33117]

Equal

 

[mylife]

Yes, the number of blue beads in blue can will be equal or not equal to the # of red beads in the red can.

 

[BradG]

Yes, equal.

 

[Buddy_Christ]

mylife - you can't pick both answers.

 

[Beachbum1546]

It is equal. Let me explain.

Ok imagine each cup has 100 beads. 100 red, 100 blue. Okay, you scoop the blue cup first picking up 20 blue beads. That leaves 80 blue beads in the blue cup.

You place the 20 blue scooped beads into the 100 red bead cup. Now that cup is 100 red, 20 blue. Okay, now you scoop the red/blue mixed cup. You sccop and get 15red and 5blue.

That 15r/5b scoop is put back into the blue cup. Now lets add it all up.

The 5 blue beads put back into the blue cup equals 85 (remember it was 80 because we first scooped 20 beads out of it first).
The 15 red we scooped out of the red/blue mix can minus the 100 original in there equals 85red.
85 blue in blue cup, 85 red in red cup.

in the red cup (85 red, 15 blue). in the blue cup (85blue, 15 red).

tada, equal.

Now give me something for writing all that out.

 

[Buddy_Christ]

beachbum1546 put this problem into the simplest terms out of anyone. everyone that said equal is correct.

the mixing of the beads is irrelevant. the more the mixing is emphasized, the more confusion sets in with some people.

those of you that made the long responses, i'm curious as to how much time you put into this.

this problem was presented to us in class yesterday (adolescent psych). a few of us began to argue with the professor about the answer until she shut us up with an answer very similar to what beachbum gave. some of the math majors were doing all sorts of complicated stuff to try to mathmatically prove her wrong. they failed.

anyway, thought some of you would get a kick out of this.

 

[Deleted member 33117]

beachbum1546 put this problem into the simplest terms out of anyone. everyone that said equal is correct.

the mixing of the beads is irrelevant. the more the mixing is emphasized, the more confusion sets in with some people.

those of you that made the long responses, i'm curious as to how much time you put into this.

this problem was presented to us in class yesterday (adolescent psych). a few of us began to argue with the professor about the answer until she shut us up with an answer very similar to what beachbum gave. some of the math majors were doing all sorts of complicated stuff to try to mathmatically prove her wrong. they failed.

anyway, thought some of you would get a kick out of this.

Right, they all over analyzed this. I damn near did the same thing, but then realized that it whatever answer I gave would have to apply regardless of how many beads we were dealing with and what mixing method was used. Beachbum's answer was simple and correct. You just owned alot of peeps orb.

 

[Beachbum1546]

i like jackangel's answer the best. very 12th grade thinking. my answer is a little more elementary.

 

[mylife]

Lmao, that's a good one.

Stumped me so much I brought out two glasses and a jar full of beads.

 

[sh4dowf4lcon]

equal.


















lol

 

[Buddy_Christ]

the simplest answer is actually the best. the reason 3rd grade kids get this right and college students get it wrong is because college students use too much of their formal operational thinking skills and overanalyze problems like this.

still impressive some of the things you guys did to solve this.

and bbf - i don't think i "owned" anyone. i actually hate that term almost as much as i hate "orb."

 

[Buddy_Christ]

Lmao, that's a good one.

Stumped me so much I brought out two glasses and a jar full of beads.

you could've done it with as little as 6 coins, 3 of each type. that's what i did when i was challeneged on the odd number thing.

 

[Deleted member 33117]

the simplest answer is actually the best. the reason 3rd grade kids get this right and college students get it wrong is because college students use too much of their formal operational thinking skills and overanalyze problems like this.

still impressive some of the things you guys did to solve this.

and bbf - i don't think i "owned" anyone. i actually hate that term almost as much as i hate "orb."

Look here, orb, maybe you didn't own them, but you still got the best of them, and it was amusing.

 

[Buddy_Christ]

i exercised some brains, definetly. that was the intent. it was definetly interesting to see who the thinkers are around here and to see how many people didn't even venture a response...the number of views/replies is disproportionate, showing that some people didn't even want to take the challenge.

 

[jackangel]

for the record, no one owned me.

simplicity is good, but a complete answer involves more than giving one specific example which proves your point. it's the difference between proving it and not proving it. this is all math, so a mathematically sound explanation which generalizes to any possibility is the best. that algebra is on a 9th grade level, if that. not overly complicated, or involving too much analysis (maybe more effort than an EF thread is worth, though).

but i did spend too much time on it. i'm slow, and i was stoned to boot.

 

[Buddy_Christ]

you had a very good analysis of the whole thing though jackangel. your answer was pretty impressive. much better than i think i could've ever done when tokin

 

[jackangel]

worth trying, just to make sure.

for what it's worth, i had the wrong answer at first, because i wasn't paying attention.

 

[Buddy_Christ]

this is something you could do to amaze and wow your friends when drunk. just use 2 jars and coins.