# Games Question: Four Flasks, Oct. 1994

#### Lawyerr

##### Games Question: Four Flasks, Oct. 1994
« on: July 20, 2005, 08:12:52 PM »
Hey folks, Im new here - been lurking for quite some time. I usually find games to be a breeze, but this one stopped me dead in my tracks. Specifically, the second iteration questions. The rules dont specify as to, when mixing, say, flask 1 and 3 together, which flask the new  mixture is contained in. Consequently, the rules can't be applied to future mixtures because the rules are based off the flask numbers, not the color of their contents. Of course, one could map out all of the possibilities, but, based on my experience with other games, the incredibly small amount of empty space available on the test page, and the insane scale of this test, I can only describe that as completely fking unreasonable. What am I missing?

#### tjking82

• 302
##### Re: Games Question: Four Flasks, Oct. 1994
« Reply #1 on: July 20, 2005, 09:09:11 PM »
Please post a summary of the game if you expect responses - thanks!

#### Lawyerr

##### Re: Games Question: Four Flasks, Oct. 1994
« Reply #2 on: July 20, 2005, 09:53:26 PM »
All apologies -

"A science student has exactly four flasks - 1,2,3, and 4 - originally containing a red, a blue, a green, and an orange chemical, respectively. An experiment consists of mixing exactly two of these chemicals together by completely emptying the contents of one of the flasks into another of the flasks. The following conditions apply -

The product of an experiment cannot be used in further experiments.
Mixing the contents of 1 and 2 produces a red chemical.
Mixing the contents of 2 and 3 produces an orange chemical.
Mixing the contents of 3 with the contents of either 1 or 4 makes blue.
Mixing the contents of 4 with the contents of either 1 or 2 makes green."

There are a few questions that ask "If one experiment is performed..." - I got these correct. But the other questions...

"19. If the student performs exactly two experiments, which one of the following could be the colors of the chemicals in the resulting two nonempty flasks?

a. blue, blue
b. blue, orange
c. blue, red
d. green, red
e. orange, orange

22. If the student will perform exactly two experiments and after the first exactly one of the resulting three nonempty flasks contains an orange chemical, then in the second experiment the student could mix together the contents of flasks...

a.1 and 2
b.1 and 3
c.1 and 4
d.2 and 3
e.3 and 4

24. If the student performs exactly two experiments and exactly one of the resulting two nonempty flasks contains an orange chemical, then it must be true that the contents of the other nonempty flask is...

a. obtained by mixing flasks 1 and 2
b. obtained by mixing flasks 2 and 4
c. blue
d. green
e. red."

Dr Yamata was a breeze, but I don't get this at all. Many thanks to anyone willing to take the time to help me out on this.

#### chambers

##### Re: Games Question: Four Flasks, Oct. 1994
« Reply #3 on: July 20, 2005, 09:57:11 PM »
R B G O

1 + 2 = R

2 + 3 = O

3 + 1 = B

3 + 4 = B

4 + 1 = G

4 + 2 = G

After 1st Experiment

RGO (1+2)

ROO (2+3)

BBO (3+1)

RBB (3+4)

BGG (4+1)

RGG (4+2)

After 2nd Experiment  (Keep in mind, you can't mix the same bottles once they've been used.  When you think about it, there's only two ways we can do the 2nd experiment.)

RB [(1+2) + (3+4)]

GO [(1+4) + (2+3)]

Now the entire game has been uncovered.  Enjoy.

#### Atlas429

##### Re: Games Question: Four Flasks, Oct. 1994
« Reply #4 on: July 21, 2005, 05:47:52 AM »
Let's look at 19.

It's important to bear in mind the rule "The product of an experiment cannot be used in further experiments". So, if you mix 1 and 2 together, neither can be used in the next experiment. You have to use 3 and 4 in the next one.

The way I approached this question was to look at the "combination" rules and see if two experiments using all four flasks could produce the colors in question.

No combination of 1,2,3,4 will give you blue, blue.

The two experiments that would make blue and orange both involve 3. This can't be done according to the rules.

You can do blue and red. 1+2 = red and 3+4 = blue.

Green and red is wrong for the same reason blue and orange are, in that both experiments involve either 1 or 2.

No unique combination of 1,2,3,4 would give you orange and orange.

Hence C) is correct.

#### Atlas429

##### Re: Games Question: Four Flasks, Oct. 1994
« Reply #5 on: July 21, 2005, 05:57:38 AM »
22)

Key point ".....and after the first exactly one of the resulting three nonempty flasks contains an orange chemical...".

2 + 3 = orange and 4 is already orange. This means you cannot mix 2+3 the first time and 4 must not be mixed the first time. Notice, to mix 4 at first would leave no orange chemical and 2+3 would mean exactly TWO orange chemicals.

So, we cant mix 4 at first and we cant mix 2+3 at first.

So what are our options?
1+2  3   4
1+3  2   4
Those are the only two possible scenarios given the question.

Now, its easy. 1+2 is wrong since 1 will definitely be used. Same thing for 1+3 and 1+4.With 2+3 one or the other is getting used the first time, so its wrong too.

Hence E), 3+4, is the only possible experiment.

#### Atlas429

##### Re: Games Question: Four Flasks, Oct. 1994
« Reply #6 on: July 21, 2005, 06:02:04 AM »
24)

Exactly one is orange and two experiments are done. If two experiments are done, all 4 flasks must be used.

One of those experiments is definitely 2+3, since 2+3  is the only mixture that makes orange.

1 and 4 are left. 1+4 = green.

So D) is right.

#### Lawyerr

##### Re: Games Question: Four Flasks, Oct. 1994
« Reply #7 on: July 21, 2005, 12:04:58 PM »
Awesome. Somehow or another, the non-use rule got lost in the mix. Thanks ya'll.

#### Atlas429

##### Re: Games Question: Four Flasks, Oct. 1994
« Reply #8 on: July 21, 2005, 01:54:31 PM »
Awesome. Somehow or another, the non-use rule got lost in the mix. Thanks ya'll.

Yeah that's the key, in my opinion.