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# NeoPets Riddles (Lenny Conundrums) and Answers Here

stapel

1
Fri 15 Jul, 2005 04:18 pm
Quote:
...the desert heat uniformly melts the surface of the ice at a rate of 1.5mm of depth per minute....hexagonal bath measuring 1.5 meters on a side, and 0.8 meters deep.

Quote:
...This time, they returned with three ice spheres: one was 0.9 meters in diameter; one was 1.5 meters in diameter; and the last was 1.8 meters in diameter. They all started melting at the same time. Also, the water in the bath evaporated at a rate of 0.1mm of depth per minute.

Otherwise assuming the same conditions as last week's Lenny Conundrum, how many minutes will elapse before the bath is completely full? Please round up to the nearest minute.

[All the following measurements are in millimeters, square millimeters, or cubic millimeters.]

======================================

The formula for the area A of an hexagon with side-length s is:

__A = (3/2)(s^2)(cot(30°))

For the tub, s = 1500 mm, so the cross-sectional area is given by:

__A = (3/2)(1 500^2)(cot(30°)) = (1.5)(2 250 000)(sqrt(3)) = 3 375 000(sqrt(3))

...or about 5845671.5 mm^2. This is the surface area of the water that will be evaporating. Since 0.1 mm of depth evaporates every minute, we have an evaporative volume of:

__3 375 000(sqrt(3))(0.1) = (337 500)(sqrt(3))

...or about 584 567.15 mm^3 water lost to evaporation every minute. Then, after t minutes, the total volume evaporated will be:

__337 500(sqrt(3))t

Since the height of the tub is 800 mm, then the total volume to be filled is:

__3 375 000(sqrt(3))(800) = 2 700 000 000(sqrt(3))

...or about 4 676 537 180 mm^3.

The radii of the three spheres are 450 mm, 750 mm, and 900 mm. The volume SV of a sphere with radius r is given by:

__SV = (4/3)(pi)(r^3)

The radii are decreasing by 1.5 mm each minute. The total water added to the tub due to ice-melt after t minutes will be the total original volume of the ice, less the volume remaining after t minutes, multiplied by 0.92, to account for the fact that water expands upon freezing (that is, a given mass of ice takes up a greater volume than the same mass of water).

The original volume of ice is:

__(4/3)(pi)[450^3 + 750^3 + 900^3] = 1 656 000 000(pi)

The radii, after t minutes, are 450 - 1.5t mm, 750 - 1.5t mm, and 900 - 1.5t mm, respectively. Then the ice volume remaining after t minutes is:

__(4/3)(pi)[(450 - 1.5t)^3 + (750 - 1.5t)^3 + (900 - 1.5t)^3]

The volume of water due to ice-melt after t minutes is then:

__(0.92)[1 656 000 000(pi) - (4/3)(pi)[(450 - 1.5t)^3 + (750 - 1.5t)^3 + (900 - 1.5t)^3]]

The volume of water in the tub after t minutes is the added ice-melt less the evaporative loss, or:

__(0.92)[1 656 000 000(pi) - (4/3)(pi)[(450 - 1.5t)^3 + (750 - 1.5t)^3 + (900 - 1.5t)^3]] - 337 500(sqrt(3))t

Setting this volume equal to the total volume of the tub, we get:

__(0.92)[1 656 000 000(pi) - (4/3)(pi)[(450 - 1.5t)^3 + (750 - 1.5t)^3 + (900 - 1.5t)^3]] - 337 500(sqrt(3))t = 2 700 000 000(sqrt(3))

Using a solver utility, the numerical solution is at t = 554.09050490076. Rounding up, this is 555 minutes.

======================================

I don't know if you're supposed to take the density-volume issue into consideration.

Eliz.
0 Replies

lem

1
Fri 15 Jul, 2005 05:09 pm
stapel, there are two problems with your solution:

1. In the problem it is specifically stated that volume does not change between solid and liquid states (I guess the terror mountain water has some physical properties we are not familiar with :/)

2. I did not check every detail in your workout, but there is a major problem with volume of first sphere [ (450 - 1.5t)^3 ] part -- the thing you need to realize, is that when t is greater than 300 it becomes negative. Or in outer words, after 300 minutes the first sphere will completely disappear. However your formula considers it to be of a negative volume, and effectively sucking the water out of the bath as it expands negativly over time! This is a big source of error. And in your case, I think after 500 minutes the second sphere disappears as well.

That is why your formula (even if take out 0.92 factor) will yield greater time than in fact needed.
0 Replies

stapel

1
Fri 15 Jul, 2005 05:39 pm
Thanks for the feedback. I was out of town for a couple of weeks, so I'd missed the density-volume proviso. And you're quite right about the first sphere. So, recalculating:

The first sphere's radius shrinks according to:

__r = 450 - 1.5t

...where t is in minutes and the radius r is in millimeters. Then the radius will be zero for:

__450 - 1.5t = 0
__450 = 1.5t
__300 = t

That is, in 300 minutes, the first sphere will be gone.

The volume of ice-melt at this point ("t = 300") will be:

__1 656 000 000(pi) - (4/3)(pi)[(450 - 1.5t)^3 + (750 - 1.5t)^3 + (900 - 1.5t)^3]
__1 656 000 000(pi) - (4/3)(pi)[(0)^3 + (300)^3 + (450)^3]
__1 498 500 000(pi)

The volume lost due to evaporation at this point ("t = 300") will be:

__337 500(sqrt(3))t = 101 250 000(sqrt(3))

The volume of water remaining in the tub after evaporative losses then will be:

__1 498 500 000(pi) - 101 250 000(sqrt(3))

The total volume of the tub is 2 700 000 000(sqrt(3)), so this leaves 2 700 000 000(sqrt(3)) - [1 498 500 000(pi) - 101 250 000(sqrt(3))], or about 144 230 733 cubic millimeters (about 3% of the tub) yet to be filled.

At t = 300, the remaining volume of ice is:

__(4/3)(pi)[300^3 + 450^3] = 157 500 000(pi)

The ice-melt at time t > 300 will be:

__157 500 000(pi) - (4/3)(pi)[(300 - 1.5(t - 300))^3 + (450 - 1.5(t - 300))^3]

...and the evaporative losses will be:

__337 500(sqrt(3))(t - 300)

Then the net amount of water added in t > 300 minutes is given by:

__157 500 000(pi) - (4/3)(pi)[(300 - 1.5(t - 300))^3 + (450 - 1.5(t - 300))^3] - 337 500(sqrt(3))(t - 300)

Set this equal to the remaining volume to be filled:

__157 500 000(pi) - (4/3)(pi)[(300 - 1.5(t - 300))^3 + (450 - 1.5(t - 300))^3] - 337 500(sqrt(3))(t - 300)
_____ = 2 700 000 000(sqrt(3)) - [1 498 500 000(pi) - 101 250 000(sqrt(3))]

A solver utility gives a solution of t = 334.01387430002, which rounds up to 335 minutes, a close match to previous solutions.

Eliz.
0 Replies

lem

1
Fri 15 Jul, 2005 06:08 pm
Indeed, it is the answer. I did not calculate the exact number of minutes it will take to fill the bath, but rather used a "brute force" excel tactic by setting up formulas for the condition of a tub at any given minute, then stretched it out until it hit the mark. So All I know is that at the end of 334th minute it was almost full, but off by just a fraction and at 335 minutes it would've spilled over.

Because we were asked to round up to the nearest minute this is all we needed to know. Your solution is purely mathematical - probably the way it was meant to be solved (and where the catch with negative size of first sphere is not as apparent - so to catch more ppl unprepared), hence you got the exact number of minutes until it's full.

By the way markr used a similar approach and even posted his excel formulas into this thread if you look back...
0 Replies

markr

1
Fri 15 Jul, 2005 06:54 pm
lem wrote:

Except for the solver. :wink:
0 Replies

lem

1
Fri 15 Jul, 2005 07:49 pm
markr wrote:
lem wrote:

Except for the solver. :wink:

Well, yes, but solver solved the equasion with 1 unknown. That's elementary algebra. The most tricky part was to come up with one big equasion to fit all variables (and not to forget what happens after 300 minutes)

I didn't bother to, I just set up cells for all three volumes at the beginning of first minute, then calculated volumes at the end, then added difference to volume in bath from previous row and subtracted water lost to evaporation, and then dragged it down until saw figures large enough to fill the tub

Even more brute force approach than yours - you actually calculated the volume of water less evaporation at the end of minute "t" for each row, but I just added up to previous rows sequentially...
0 Replies

markr

1
Fri 15 Jul, 2005 10:42 pm
Without knowing a priori that the solution is less than 500 minutes, you'd probably want to set up three equations assuming:
t <= 300
300 < t <= 500
500 < t <= 700

The max function in Excel eliminated the need for that in my solution.

Mine is actually one big equation with one unknown spread across several cells to make it easier to comprehend. Solver could have been set up to vary the time until the volume matched the tub.
0 Replies

lem

1
Fri 15 Jul, 2005 11:39 pm
markr wrote:
Without knowing a priori that the solution is less than 500 minutes, you'd probably want to set up three equations assuming:
t <= 300
300 < t <= 500
500 < t <= 700

Not necesserily. You need to set up the first one and solve for t. If t>300, only then you'd have to set up the second equasion and solve it. And if you still get t>500 only then you'd need the third one (which you don't for this problem)... The main thing is to realize that you need to check for these conditions.

Quote:
Mine is actually one big equation with one unknown spread across several cells to make it easier to comprehend. Solver could have been set up to vary the time until the volume matched the tub.

That's why I said that my solution was more of a brute force than yours. Well, actually mine was also a wide spread equasion, but it was not set-up to solve for t=xxx, it was set-up to calculate a series of values v(t) (volume over time):

v(t)=v(t-1)+v1(t-1)+v2(t-1)+v3(t-1)-v1(t)-v2(t)-v3(t)-evaporation

Where v1, v2, v3 are series that define the volume of each sphere over time (and I did use a similar excel trick to eliminate negative volumes). So yeah - I can call my solution mathematical as well. Only I stopped there, and instead of coming up with an equasion to calculate the n'th member of the series I used the references to previous row in excel for v(t-1) and friends, then calculated every member individually until it hit the mark.

Basically your solution takes it one step further by coming up with that equasion for v(t), with added bonus, that you actually can stick t=334.01387430002 into your spreadsheet and see that it will in fact give you the exact match. Though calculating it to the 10th decimal point by substitution would've probably taken you some time :wink:

That's why your solution is more mathematical than mine. We were going the same way, but I gave up the algebra and used the power of my CPU for brute force solution one step earlier than you!
0 Replies

markr

1
Fri 15 Jul, 2005 11:48 pm
Are you a neopet member, or do you just like solving the math problems like me (and I assume stapel)?
0 Replies

lem

1
Sat 16 Jul, 2005 12:13 am
markr wrote:
Are you a neopet member, or do you just like solving the math problems like me (and I assume stapel)?

Both

--- I am NP member mostly to play some of their games (this being one of them) and sometimes when time permits do some other things...

--- I like solving problems, especially tricky ones like this. I'm not much of a fun of solving straight-forward out of the textbooks problems, just for the sake of solving something... Like many posted in maths forum, where kids are looking for someone to do their homework :wink:
0 Replies

x

1
Wed 20 Jul, 2005 08:09 am
markr wrote:
Except for the solver. :wink:

Well, I think we can all agree that the cubic formula's out for this one...
0 Replies

monetangel

1
Thu 21 Jul, 2005 10:18 am
WOOHOO!
Congratulations! You have guessed correctly in the Lenny Conundrum game (Round 125). We have given you a Bath Tub, an Avatar, and 5421 NP!
0 Replies

stapel

1
Thu 21 Jul, 2005 10:25 am
Re: WOOHOO!
monetangel wrote:
You have guessed correctly....

What value did you submit? Thanks.

Eliz.
0 Replies

monetangel

1
Thu 21 Jul, 2005 10:26 am
I submitted the 335 number
0 Replies

stapel

1
Thu 21 Jul, 2005 10:27 am
Cool!

Thanks.

Eliz.
0 Replies

lem

1
Thu 21 Jul, 2005 11:40 am
Re: WOOHOO!
monetangel wrote:
Congratulations! You have guessed correctly in the Lenny Conundrum game (Round 125). We have given you a Bath Tub, an Avatar, and 5421 NP!

Wow. That means that only 369 correct answers were submitted! Impressive. Especially since for the last round, where the question was tricky, but less mathematically challenging there were something like 1200 correct entries!

That's probably how many people were fooled by the first sphere completely disappearing prematurely (or who submitted 334 by rounding to the nearest integer, rather than rounding up to that nearest integer)...

(for those who are not NP members -- the figures come from the fact that for every round there is a 2,000,000 neopoints prize pool, which is evenly divided between everybody who submitted the correct answer. So divide 2Mil by your winnings, and you'll get a number of correct entries)
0 Replies

stapel

1
Thu 21 Jul, 2005 11:55 am
Re: WOOHOO!
lem wrote:
for those who are not NP members...

Thank you for that info!

Eliz.
0 Replies

volleyball babe04

1
Thu 21 Jul, 2005 06:30 pm
new
A new one is up!

There are 32 letters on the wheel. I'm guessing you pair up the letters across from each other and keep moving around it, but I don't know where to start from.
0 Replies

sobec 13

1
Thu 21 Jul, 2005 06:32 pm
new lc
The new Lenny C is up. It's a circle of letters all mixed up. From the top going right they are:
FRSDTHIAWHTAIAPSENRHUSEENUTMJSON

I have no idea what this is. I'm thinking it's a question we have to answer. And possibly...to begin unscrambling the sentence we start with the letter W, the only one in the circle because most questions begin with a W. What? Where? Who? Why?
By doing this I was able to get the word WHAT from the letters after the W.
But that's all I could think of.
0 Replies

volleyball babe04

1
Thu 21 Jul, 2005 06:36 pm
image
Here's a link to the picture for non-neopet users:

http://images.neopets.com/games/conundrum/126_circle.gif
0 Replies

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