Fun and (mostly) simple logic questions

[Vort]

I like lemonade. I like Hawaiian Punch®. But most of all, I like an equal mixture of lemonade and Hawaiian Punch®. But it has to be exactly half and half -- otherwise, no go.

I have a two-liter container that holds some fresh lemonade. Let us assume that the container has gradation markings sufficient to measure the amount of liquid it contains to an acceptable degree.

I also have a gallon jug (let's say four liters) containing some Hawaiian Punch. The jug is not marked for amount, but through the magical magic of, I don't know, magic, I guess, I know how much Hawaiian Punch there is in the jug.

These are the only two containers I am allowed to use.

In every case, my goal is to make a 50/50 mixture of lemonade and Hawaiian Punch in the two-liter bottle. Your job, should you choose to accept it: Describe the method by which I can reach my goal. I will try to make cases that proceed roughly from easiest to hardest. (Note: I haven't figured all of these out yet. I am just assuming that they are solvable. The first few are pretty easy, at least.)

For the purposes of this thread, I propose that all answers be given in a white font, so that you have to select them with the cursor to see them. Like this: --> Hi! <-- To write in a white font, write your answer, select it, then click on the A dropdown at the top right of the box where you type and select the bottom right-hand color, which is white.

Case I

I have one liter of lemonade and two liters of Hawaiian Punch.

Case II

I have half a liter of lemonade and two liters of Hawaiian Punch.

Case III

I have one liter of lemonade and 3/4 liter of Hawaiian Punch.

Case IV

I have 3/4 liter of lemonade and 3/4 liter of Hawaiian Punch.

Case V

I have 1 1/2 liters of lemonade and 2 liters of Hawaiian Punch.

Case VI

I have 2 liters of lemonade and 3 liters of Hawaiian Punch.

Case VII

I have n liters of lemonade (n <= 2) and m liters of Hawaiian Punch (m >= n).

[Vort]

The first four are quite easy.

Case I

I have one liter of lemonade and two liters of Hawaiian Punch.

Answer: Fill the half-full two-liter bottle with Hawaiian Punch.

Case II

I have half a liter of lemonade and two liters of Hawaiian Punch.

Answer: Put Hawaiian Punch in the two-liter bottle until it shows a liter total.

Case III

I have one liter of lemonade and 3/4 liter of Hawaiian Punch.

Answer: Can't be done. If you ever start out with more lemonade than Hawaiian Punch, the problem is unsolvable. Note that I can create a 50/50 mixture by pouring 3/4 liter of lemonade into the Hawaiian Punch jug, but then my 50/50 mixture is in the jug, not in the two-liter bottle as required.

Case IV

I have 3/4 liter of lemonade and 3/4 liter of Hawaiian Punch.

Answer: Pour the lemonade into the Hawaiian Punch jug, making a 50/50 mixture. Then refill the two-liter bottle with that mixture.

Edited April 15, 2016 by Vort

[zil]

Hmm. Did someone spin the planet backward far enough that I've ended up back in grade school?

[mordorbund]

You should probably mention that you don't want to poor any down the sink. Otherwise, they all reduce to Case II.

[Vort]

Case V

I have 1 1/2 liters of lemonade and 2 liters of Hawaiian Punch.

Answer: This one should be a pretty straightforward algebra problem. (zil, I took algebra in junior high, not grade school, so you're way ahead of me.)

I start out with 1.5 liters of lemonade, and pour x liters into the Hawaiian Punch, leaving me (1.5 - x) liters of lemonade and a 2 / (2 + x) Hawaiian Punch concentration.

Let's think about this a bit. If I were to pour half a liter of lemonade into the Hawaiian Punch jug, I would have exactly a liter of lemonade. If I could then fill that up with pure Hawaiian Punch, I'd have it made. Alas, I have only the Hawaiian Punch/lemonade mixture, and adding that to my liter of lemonade will give me too much lemonade.

What if I added all of the lemonade into the Hawaiian Punch jug, except a very small amount -- say, a teaspoon? Then I would have a teaspoon of pure lemonade and a jug full of an (almost exactly) 4/3 mixture of Hawaiian Punch to lemonade (2 liters of Hawaiian Punch to 1.5 liters, minus a teaspoon, of lemonade). So if I add in seven teaspoons of that mixture, that will give me 7(4/7) t = 4 t of Hawaiian Punch and 1 t + 7(3/7) t = 4 t of lemonade -- perfect! Well, not "perfect", because I'm a teaspoon shy of having a perfect 4/3 mixture, but the point is, I know I can create at least small quantities of a perfect mixture, if I get my math right.

But of course, I want more than a mere sip of my elixir. Eight teaspoons? That's, like, a swallow. I want the maximum amount possible!

Did someone say "maximum"? AHA! I know what that means! Calculus! Take the derivative and set it equal to zero!

...Well, yes, but I don't think we actually need calculus. I think this is solvable purely with algebra. I suspect calculus would unnecessarily complicate things.

So after pouring x liter(s) of lemonade into my Hawaiian Punch jug, I want to pour the mixture back into my (1.5 - x) liter(s) of lemonade until I have the same amount of lemonade and Hawaiian Punch. And I want the MAXIMUM amount, which would be two liters.

My Hawaiian Punch jug now contains 2 + x liters of drink, of which 2 liters are Hawaiian Punch and x liter(s) is/are lemonade. So in any given amount of that mixture, 2 / (2 + x) of it will be Hawaiian Punch and  x / (2 + x) of it will be lemonade.

Since I have (1.5 - x) liter(s) of lemonade in my 2-liter bottle, and I want to fill the bottle up, I need to add 2 - (1.5 - x) = 0.5 + x liter(s) of mixture back into my lemonade. At that point, I will have a PERFECT mixture: Half lemonade (1 liter) and half Hawaiian Punch (1 liter).

Total lemonade:

(1.5 - x) + [x / (2 + x)] (0.5 + x) liter

= 1.5 - x + [(0.5x + x²) / (2 + x)] liter

= (3 + 1.5x - 2x - x² + 0.5x + x²) / (2 + x) liter

= (x² - x² + 1.5x + 0.5x - 2x + 3) / (2 + x) liter

= 3 / (2 + x) liter

= 1 liter

Total Hawaiian Punch:

[2 / (2 + x)] (0.5 + x) liter

= (1 + 2x) / (2 + x) liter

= 1 liter

Since they are both equal to 1 liter, we can set them equal to each other:

3 / (2 + x) liter = (1 + 2x) / (2 + x) liter

3 liter = (1 + 2x) liter

x = 1 liter

So here's what you do: Pour a liter of lemonade into the Hawaiian Punch, leaving a half liter in the bottle and making a 2:1 mixture of Hawaiian Punch to lemonade. Then add 1.5 liters of this mixture (2/3 of which is Hawaiian Punch, or 1 liter, and 1/3 of which is lemonade, or 1/2 liter) back into the half liter of lemonade, giving you two liters of bliss, consisting of a liter of Hawaiian Punch and a liter of lemonade.

No calculus involved!

Edited April 16, 2016 by Vort

[Vort]

24 minutes ago, mordorbund said:

You should probably mention that you don't want to poor any down the sink. Otherwise, they all reduce to Case II.

Ooh. You're right. But how could I have thought to mention that it is criminal to pour any of this divine substance down the drain? It's like having to explicitly forbid murder or Mountain Dew.

[Guest]

Case VI: See solution to Case VII.  That solution solves all other cases.

Case VII: This is not a single answer.  We have two methods based on punch quantity or lemonade quantity.

Condition 1: n <= 1 L: Just pour an equal amount of punch into the lemonade jug.

Condition 2: n > 1 L: The constants change based on m.  But the equations are essentially the same.

n1 = initial lemonade. n2 = lemonade poured into punch.  q = amount of mixture poured back into graduated jug.

****m=2 L

n2 = n1/(n1-1) * (n1-1)

q, then is whatever is needed to bring the balance to 50/50.

But we need limits on this as well.  When m and n are great enough that the required n2 will not fit in the remaining volume of punch, then it is impossible.

I hope I got all the notations correct.  Can someone check my equations?

Edited April 16, 2016 by Guest

[Vort]

5 hours ago, Carborendum said:

When m and n are great enough that the required n2 will not fit in the remaining volume of punch, then it is impossible.

As long as [lemonade in bottle < Hawaiian Punch in jug < 4 liters], the problem is solvable. But as lemonade in bottle approaches 2 liters and Hawaiian Punch in jug approaches 4 liters, it gets into a ridiculous number of iterations. (You are allowed to make as many pours as you like.)

[mordorbund]

On 4/15/2016 at 4:40 PM, Vort said:

Ooh. You're right. But how could I have thought to mention that it is criminal to pour any of this divine substance down the drain? It's like having to explicitly forbid murder or Mountain Dew.

After you solve case V and VI you have a jug of a sub-optimal mixture. Can you weight the following for me:

• Lemonade
• Hawaiian Punch
• 50/50 mixture
• HP > L mixture (this can be a ratio, like L/2HP where ideal mixture is 1)
• L > HP mixture (this can also be a ratio)

I'll run my min-max (or mix-man) on it and tell you if it's even worth attempting to mix.  

[Vort]

24 minutes ago, mordorbund said:

After you solve case V and VI you have a jug of a sub-optimal mixture. Can you weight the following for me:

• Lemonade
• Hawaiian Punch
• 50/50 mixture
• HP > L mixture (this can be a ratio, like L/2HP where ideal mixture is 1)
• L > HP mixture (this can also be a ratio)

I'll run my min-max (or mix-man) on it and tell you if it's even worth attempting to mix.  

Case V: 2 liters of 50/50 in the bottle, 1 1/2 liters of 2:1 Hawaiian Punch-to-lemonade ratio in the jug.

I did Case VI, but I messed up the math somewhere and haven't had the chance to go figure out where my mistake was. (This actually turns out not to be much of a logic problem, but really just a picky algebra problem, which to me is not nearly as fun.)

[mordorbund]

Assigning the following values:

• Lemonade - .25 (like)
• Hawaiian Punch - .25 (like)
• 50/50 - 1 (Really Like)
• other mix - 0 (no go)

The best solution for Case V turns a .875 delight to a 1.5, so if "Really Like" is 4 times as delicious as a "like" (as above) it's worth it. If, on the other hand, "Really like" is only 2 1/3 times as good (or less), you're better off leaving it alone and just drinking the two separately.

You're best case scenario for Case VI (2L of 50/50 and 3L of nogo) turns a 1.25 delight into a 1.5 (using those dummy weights). If "Really Like" is anything less than 3x as good (3 1/3) then you're better off leaving it alone.

And so long as any other mixture is "no go" then you'll save yourself a lot of calculating just pouring all but 1L lemonade down the drain. At least that boosts your 1.5 experience with the additional HP.

[Vort]

[Blackmarch]

On 4/15/2016 at 1:02 PM, Vort said:

I like lemonade. I like Hawaiian Punch®. But most of all, I like an equal mixture of lemonade and Hawaiian Punch®. But it has to be exactly half and half -- otherwise, no go.

I have a two-liter container that holds some fresh lemonade. Let us assume that the container has gradation markings sufficient to measure the amount of liquid it contains to an acceptable degree.

I also have a gallon jug (let's say four liters) containing some Hawaiian Punch. The jug is not marked for amount, but through the magical magic of, I don't know, magic, I guess, I know how much Hawaiian Punch there is in the jug.

These are the only two containers I am allowed to use.

In every case, my goal is to make a 50/50 mixture of lemonade and Hawaiian Punch in the two-liter bottle. Your job, should you choose to accept it: Describe the method by which I can reach my goal. I will try to make cases that proceed roughly from easiest to hardest. (Note: I haven't figured all of these out yet. I am just assuming that they are solvable. The first few are pretty easy, at least.)

For the purposes of this thread, I propose that all answers be given in a white font, so that you have to select them with the cursor to see them. Like this: --> Hi! <-- To write in a white font, write your answer, select it, then click on the A dropdown at the top right of the box where you type and select the bottom right-hand color, which is white.

Case I

I have one liter of lemonade and two liters of Hawaiian Punch.

Case II

I have half a liter of lemonade and two liters of Hawaiian Punch.

Case III

I have one liter of lemonade and 3/4 liter of Hawaiian Punch.

Case IV

I have 3/4 liter of lemonade and 3/4 liter of Hawaiian Punch.

Case V

I have 1 1/2 liters of lemonade and 2 liters of Hawaiian Punch.

Case VI

I have 2 liters of lemonade and 3 liters of Hawaiian Punch.

Case VII

I have n liters of lemonade (n <= 2) and m liters of Hawaiian Punch (m >= n).

Answer
Since it's the 2 liter bottle that is marked, it is then the bottle that is used for everything-
IF the two liter bottle is over half full, then it has to be emptied till it is at half and then punch poured in to filling.
OR if the bottle is so designed for it, measure how much lemonade there is, freeze and remove it, then pour in punch equivalent to the lemonade and dump the rest of the punch and then put everything back in the bigger jug.

IF the two liter bottle is less than half then your pour in punch only to the amount that it takes to double what was originally in the two liter bottle.


 

[Guest]

On 4/17/2016 at 8:26 PM, Vort said:

This actually turns out not to be much of a logic problem, but really just a picky algebra problem, which to me is not nearly as fun.

So the real logic question is: Why would anyone want a Lemonade/Fruit Punch mixture when orange Gatorade can be had?