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NeoPets Riddles (Lenny Conundrums) and Answers Here

 
 
Siku
 
  1  
Reply Thu 17 Nov, 2005 03:04 pm
lem wrote:
Not sure why, but I'm off by 3 servings...

My calculation up to the frosting are same as yours... but the frosting ammount differs in my version:

1: frosting on top of the cake
pi * 9^2 * 0.375 * 470 = 14276.25 * pi

2: frosting around cake. Height this time is 3.5 + 3.5 + 0.25 + 0.375 = 7.625in, ourter radius is 9.375, inner radius is 9:

outer: pi*9.375^2 * 7.625 * 470 = 314978 *pi
inner: pi*9^2 * 7.625 * 470 = 290283.8 *pi

so total frosting callories are (314978-290283.8+14276.25) *pi=38970.53*pi= 122429.5

With two other figures being the same as in your calculation my total calorie count was 392166.7, which makes 490.21 servings or 490 complete servings...


I put your answer just to guess...

But truth is I would solve this one but I am used to metrics system so I didn't even bother.
0 Replies
 
lem
 
  1  
Reply Thu 17 Nov, 2005 03:15 pm
Siku wrote:

But truth is I would solve this one but I am used to metrics system so I didn't even bother.


I am too used to metric system -- gotta love it 10mm=1cm, 100cm=1m, 1000m=1km, etc... no 12's 16's, etc Very Happy

But.... for this problem... it makes no difference at all! Since TNT was nice enough to use only one type of unit (inch), it doestn't really matter. If everything was centimeters (e.g. cake diameter was 18cm, etc) and you were given callories per cm^2, the answer would have been exactly the same (cake would've been a lot smaller, but much more rich in callories though)
0 Replies
 
lem
 
  1  
Reply Thu 17 Nov, 2005 03:20 pm
geddoe wrote:
so is it 490 or 493?


A good question. We'll have to wait for somebody else to go through this and confirm one or another (or challenge both) answer(s)... I'm too tired right now to go through stapel's math, but if nobody volunteers to cross-check, I may as well do it later tonight.
0 Replies
 
Isfan
 
  1  
Reply Thu 17 Nov, 2005 03:30 pm
stapel wrote:
The adjusted calorie total for the cake then is:

. . . . .79380(pi) cal + 6480(pi) cal + 39780.17578(pi) cal
. . . . . . . .= 125670.1758(pi) calories


I found an error right here.

Look at the 10's digit for each value:
80.0 + 80.0 + 80.----- = 70.----- ?

One of these numbers is wrong.

79380 + 6480 + 39780.17578 =/= 125670.1758
79380 + 6480 + 39780.17578 = 125640.17578

I just finished AP calc homework, so I'm too mentally exhausted to go through and figure out what numbers are right or not, but i'd just like to bring up this point. See if anything changes.
0 Replies
 
Isfan
 
  1  
Reply Thu 17 Nov, 2005 03:40 pm
lem wrote:
Not sure why, but I'm off by 3 servings...

My calculation up to the frosting are same as yours... but the frosting ammount differs in my version:

1: frosting on top of the cake
pi * 9^2 * 0.375 * 470 = 14276.25 * pi

2: frosting around cake. Height this time is 3.5 + 3.5 + 0.25 + 0.375 = 7.625in, ourter radius is 9.375, inner radius is 9:

outer: pi*9.375^2 * 7.625 * 470 = 314978 *pi
inner: pi*9^2 * 7.625 * 470 = 290283.8 *pi

so total frosting callories are (314978-290283.8+14276.25) *pi=38970.53*pi= 122429.5

With two other figures being the same as in your calculation my total calorie count was 392166.7, which makes 490.21 servings or 490 complete servings...


I get this answer too. (122429.5224 to be exact)

What I realize Eliz did was that when she miscalculated the values with the chocolate layer being twice the height it was suppost to be, it affected ALL of the other values as well. She only recalculated the VC and the chocolate layer, but she forgot that the height value of the Butter Cream on the sides was also affected by the hieght of the two cake layers plus the icing. By giving the Butter Cream on the side an extra .25 inchs of hieght could easily add three more servings to the cake.

I'm going for 490, since 493 was calculated with too much Butter cream.
0 Replies
 
TheLatterRain
 
  1  
Reply Thu 17 Nov, 2005 06:08 pm
4ni74 wrote:
It says a quarter of inch by the chocolate layer indeed Wink

But the rest of the work is the same as mine Smile

The only thing I'm not sure about is that the 3/8" buttercream layer could be seen as a layer on both sides of the cake. So that r would be 9.75. But since nobody seems to see it that way, it's probably just me reading the wrong thing Wink


That doesn't matter because you're not finding the volume of the cake as a whole, you're finding the volume of the individual parts of the cake.
0 Replies
 
TheLatterRain
 
  1  
Reply Thu 17 Nov, 2005 06:21 pm
I worked everything out and got the following:

Cake--
V = (pi)(81)(3.5) = 890.6415(2) = 1781.283 cubic inches
1781.283 cubic inches(140 cal.) = 249379.6248 calories

Choc. Icing--
V = (pi)(81)(0.25) = 63.617 cubic inches
63.617 cubic inches(320 cal.) = 20357.5204 calories

Buttercream Icing--
V = [(56.549)(7.25)(0.375)]+[(pi)(81)(0.375)] = 249.167 cubic inches
249.167 cubic inches(470 calories) = 117108.432 calories
[56.549= the circumference of the circle(2*pi*r)]

TOTAL CALORIES = 386945.5772 calories
Divide this by 800 and you get: 483.5569715, or 483 complete servings
0 Replies
 
fmorel
 
  1  
Reply Thu 17 Nov, 2005 07:24 pm
http://bellsouthpwp.net/f/m/fmorel90/book1.xls

there's how i did it...and i got 490
0 Replies
 
lem
 
  1  
Reply Thu 17 Nov, 2005 07:58 pm
TheLatterRain wrote:
I worked everything out and got the following:

Cake--
V = (pi)(81)(3.5) = 890.6415(2) = 1781.283 cubic inches
1781.283 cubic inches(140 cal.) = 249379.6248 calories

Choc. Icing--
V = (pi)(81)(0.25) = 63.617 cubic inches
63.617 cubic inches(320 cal.) = 20357.5204 calories

Buttercream Icing--
V = [(56.549)(7.25)(0.375)]+[(pi)(81)(0.375)] = 249.167 cubic inches
249.167 cubic inches(470 calories) = 117108.432 calories
[56.549= the circumference of the circle(2*pi*r)]

TOTAL CALORIES = 386945.5772 calories
Divide this by 800 and you get: 483.5569715, or 483 complete servings


Ok, we've got candidate #3...
I see 2 problems here.

1. Ok, this is less of a problem, but more of a conflict of assumptions. Your frosting on the sides of a cake does not go up the height of the cake. It ends where the top of the upper vanilla pie ends (7.25 = 3.5+3.5+0.25). You don't account for the height of frosting on top. It may just be that you are right and everyone else is wrong, but I would stick to the common assumption that frosting on sides goes all the way up making the cake a perfect cylinder.

2. your formula for outer frosting cylinder is not correct. You are multiplying the circumference inside the frosting belt (so to say) by the depth. If you imagine it in real life. Let's say you take a layer of frosting that is .375in deep and exactly as long and as inner circumference and lay it on the table. Now try wrapping it around the cake. While the inside part will touch precisely, the outer part will not, leaving a gap, because outer radius is greater, and therefore outer circumference is supposed to be bigger. So your solution leads to cutting the number of servings a bit short, even if you used 7.625 for height of the frosting belt... I think the best way to get this volume is as folks used here: calculate the volume of cylinders of outer and inner radius, then subtract one from another.
0 Replies
 
markr
 
  1  
Reply Thu 17 Nov, 2005 10:49 pm
The answer is 490.
0 Replies
 
stapel
 
  1  
Reply Thu 17 Nov, 2005 11:13 pm
Okay, that "h = 0.5 versus h = 0.25" error is turning out to be more persistant than I'd realized, so I'll recalculate everything....

By the way, when computing the volume of buttercream along the sides, you can't use "(circumference) times (3/8" width)". Think about it: Take a sheet of paper the same length as the circumference and the same height as the layers. (In other words, this paper could "wrap" the sides of the cake.) Slather the paper with 3/8" of frosting. Now curl the paper into a circle. Either the frosting is on the outside of your paper loop and is now (due to stretching) less than 3/8" thick, or else is on the inside of your lopp and is now (due to compression) more than 3/8" thick. This is why you have to take the outer volume (cake plus frosting) and subtract the inner volume (cake only) to find the volume of frosting along the sides.

Note: In my calculations, I included all decimal places displayed by the calculator, carried what I could in the calculator, and waited to multiply through by "pi" until as late as possible, so as to minimize round-off error.

====================================
==========PROBLEM STATEMENT===========
====================================


A large birthday cake was baked to celebrate the 6th birthday of Neopets. It was a round cake, made of two layers of vanilla cake; each layer was 18 inches in diameter and 3.5 inches tall. Between the two layers was a quarter inch of chocolate frosting. And covering the top and side of the cake was 3/8" of buttercream frosting.

If the vanilla cake has 140 calories per cubic inch, and the chocolate frosting has 320 calories per cubic inch, and the buttercream frosting has 470 calories per cubic inch, how many 800-calorie servings are in the cake? Please round down to the nearest serving.

====================================
===========PROBLEM SOLUTION===========
====================================


Calories due to cake layers

height of one layer: 3.5 inches
height of both layers: 3.5" + 3.5" = 7 inches

diameter of layers: 18 inches
radius of layers: 9 inches

volume of right circular cylinder: V = (pi)(r^2)(h)

volume of cake layers: (pi)(9" ^2)(7") = 567(pi) cubic inches

calories per cubic inch of cake: 140

total calories in cake layers: (567(pi) cu. in.)(140 cal/cu. in.) = 79380(pi) calories

Calories due to chocolate frosting between layers

thickness of frosting: 1/4" or 0.25 inches

radius of frosting: 9 inches

volume of frosting: (pi)(9" ^2)(0.25") = 20.25(pi) cubic inches

calories per cubic inch of chocolate frosting: 320

total calories in chocolate frosting: (20.25(pi) cu. in.)(320 cal/cu. in.) = 6480(pi) calories

Calories due to buttercream frosting on top

The buttercream has to lap over the edge by 3/8", so the radius of the top "layer" is increased to 9" + 0.375" = 9.375".

height of layer: 0.375"
radius of layer: 9.375"

volume of layer: (pi)(9.375" ^2)(0.375") = 32.958984375(pi) cubic inches

calories per cubic inch of buttercream frosting: 470

total calories in buttercream across top of cake: (32.958984375(pi) cu. in.)(470 cal/cu. in.) = 15490.7226563(pi) calories

Calories due to buttercream frosting around the sides

The frosting on the sides does not include the frosting on the top. The extra 3/8" was added to that radius to account for the "side" frosting that was also "top" frosting. So what follows includes only that frosting along the sides from the plate to the height of the top of the second layer.

height: 3.5" + 0.25" + 3.5" = 7.25 inches

This includes the two 3.5-inch cake layers and the 1/4-inch frosting layer.

inner radius: 9 inches
outer radius: 9.375 inches

To find the volume of frosting, subtract the inner volume (of cake) from the total volume (of cake plus frosting).

volume: (pi)(9.375" ^2)(7.25") - (pi)(9" ^2)(7.25") = 49.95703125(pi) cubic inches

calories per cubic inch of buttercream frosting: 470

total calories in buttercream along the sides: (49.95703125(pi) cu. in.)(470 cal/cu. in.) = 23479.8046875(pi) calories

Combined total calories

(cake layers) + (chocolate between layers) + (buttercream across top) + (buttercream along sides)

(79380(pi) cal) + (6480(pi) cal) + (15490.7226563(pi) cal) + (23479.8046875(pi) cal)

= 124830.527344(pi) calories

= 392166.667647 calories

Number of servings

(392166.667647 calories) ÷ (800 calories / serving) = 490.208334559 servings

...or 490 servings (after rounding down to the nearest entire serving).

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


Please verify. Thank you.

Eliz.
0 Replies
 
geddoe
 
  1  
Reply Sat 19 Nov, 2005 03:28 pm
ooook im goin with 490
0 Replies
 
catchprj
 
  1  
Reply Wed 23 Nov, 2005 06:37 pm
You won the Lenny Conundrum!
Dear Ryan Catchpole,

Congratulations! You have guessed correctly in the Lenny Conundrum game (round 143). You have won 1213 NP!

Yours Sincerely,
The Neopets Team!

I think I put 490 (can't really remember!)
0 Replies
 
fmorel
 
  1  
Reply Wed 23 Nov, 2005 06:48 pm
i got the money and put in 490 as my answer
0 Replies
 
sianhuk
 
  1  
Reply Thu 24 Nov, 2005 10:47 am
new lenny culdrum.
a angelpuss?????? Confused
0 Replies
 
sianhuk
 
  1  
Reply Thu 24 Nov, 2005 10:48 am
lenny....
Larry, a shy Kacheek shepherd from Meri Acres, was well known for taking great care of his Petpets. He was so devoted to them, in fact, that he had no prolem caring for over the fourteen he had.

Recently, he heard there was a sale on Whinies at Ye Olde Petpets. Always looking for ways to increase his menagerie, Larry really wanted to go. However, the backyard gate was being repaired today, so he didn't dare leave his flock in the yard for fear they would liberate themselves and eradicate his vegetable patch, AGAIN. He would simply have to bring them along.

Larry, being a resourceful Kacheek, decided the best way to handle the situation would be to organize the Petpets in a line. He tied a string to the lead Mazzew's collar, and tied another string from his collar to that of the Angelpuss behind him, and so forth down the line. This way, none of them could become seprated from the others and become irretrevably lost.

Proud of his ingenious solution, Larry held his head with pride as he led his colorful band of fourteen Petpets down the lane. Past a large pile of potatoes, around a giant marrow, over the snoring Turmaculus, and even a playful jump over the Mysterious Symol Hole. Finally, he and his whimsical line of Petpets approached Ye Olde Petpet. Nancy, the blue Ixi that ran the shop, welcomed him.

"Ahh, Larry! A pleasant surprise as always," she said, smiling. "And look at that! Thirteen Petpets in tow with you!"

"Thirteen?!?" he shouted, and whirled around. Larry gasped in horror as he saw that the last bit of string was simply dragging in the dust.

What was the species of the missing Petpet?
0 Replies
 
sianhuk
 
  1  
Reply Thu 24 Nov, 2005 11:16 am
ello??
0 Replies
 
stapel
 
  1  
Reply Thu 24 Nov, 2005 11:47 am
I guess you're supposed to figure out, somehow, what the other twelve pets were, and how they were organized. But I have no idea how one would do that, just from the information provided...?

Eliz.
0 Replies
 
lunababy107
 
  1  
Reply Thu 24 Nov, 2005 11:48 am
answer
I kinda tend to believe that the answer is a Symol because in the riddle it said:
Quote:
Proud of his ingenious solution, Larry held his head with pride as he led his colorful band of fourteen Petpets down the lane. Past a large pile of potatoes, around a giant marrow, over the snoring Turmaculus, and even a playful jump over the Mysterious Symol Hole

logically, as these petpets tend to do, the Symol could've jumped right in
0 Replies
 
pinkmustangxxx
 
  1  
Reply Thu 24 Nov, 2005 01:33 pm
I did a search on neopets for the description of a symol.. and this is what it says :
"Symols have a habit of burrowing into places they shouldnt be!"

So I believe the answer would be a symol as well.. there is no other logical answer.. omg my brain hurts
0 Replies
 
 

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