Date Rewarded: May 15th, 2008.
What's the next word in this sequence?
Cream
Radish
And
______
Submit only a one-word answer.
Prize
Radish Meringue
Radish Meringue
Click to show/hide solution
Thursday, May 8, 2008
Round 261
Date Released: May 8th, 2008.
Date Rewarded: May 15th, 2008.
What's the next word in this sequence?
Cream
Radish
And
______
Click to show/hide solution
Date Rewarded: May 15th, 2008.
What's the next word in this sequence?
Cream
Radish
And
______
Click to show/hide solution
Thursday, May 1, 2008
Round 260
Date Released: May 1st, 2008.
Date Rewarded: May 8th, 2008.
A Chomby was walking down a path one day when Jhudora appeared in front of it. "Solve this puzzle, and you may pass," she said.
Suppose you have a very long steel bar, with a regular pentagonal cross-section where each side of the pentagon is 2 cm in length. This bar will be suspended vertically over a very deep chasm. At the bottom end of the steel bar, there is a spherical steel weight with a diameter of 50 cm. Going up the bar, at every 3 metre increment, there is a spherical steel weight with a diameter of 30cm.
If the density of the steel is exactly 7.8 g/cm3, the steel can withstand a tensile stress of 400 MPa, and gravity is 9.81m/s2, what is the maximum possible length, in metres, that the bar can extend without breaking? Round down to the nearest metre, submit only a number (with no additional or extraneous information), and disregard any deformation of the bar or any other forces acting on the bar.
Click to show/hide solution
Using the stress equation [1]:
Now, we can calculate the force that the steel bar can withstand:
Date Rewarded: May 8th, 2008.
A Chomby was walking down a path one day when Jhudora appeared in front of it. "Solve this puzzle, and you may pass," she said.
Suppose you have a very long steel bar, with a regular pentagonal cross-section where each side of the pentagon is 2 cm in length. This bar will be suspended vertically over a very deep chasm. At the bottom end of the steel bar, there is a spherical steel weight with a diameter of 50 cm. Going up the bar, at every 3 metre increment, there is a spherical steel weight with a diameter of 30cm.
If the density of the steel is exactly 7.8 g/cm3, the steel can withstand a tensile stress of 400 MPa, and gravity is 9.81m/s2, what is the maximum possible length, in metres, that the bar can extend without breaking? Round down to the nearest metre, submit only a number (with no additional or extraneous information), and disregard any deformation of the bar or any other forces acting on the bar.
Prize
Steel Negg
Steel Negg
Click to show/hide solution
Using the stress equation [1]:
rho = F/A
where t = 2 cm = 0.02 m is the side of the pentagon.where rho = 400 MPa = 400,000,000 kg/m·s2 is the tensile stress that steel can withstand, F is the force that the steel has to withstand and A is the crossed area of the steel bar.
The crossed area is a regular pentagon can be calculated by the following equation [2]:
A = 5/4 t2 tan(54°) = 0.000688190960236 m2
Now, we can calculate the force that the steel bar can withstand:
F = rho * A = rho * 5/4 t2 tan(54°)
Using Newton's gravity law [3], we have:
F = m * g
Therefore, the mass that the steel bar can widthstand is:
where g = 9.81 m/s2 is the gravitational acceleration.
Don't round anything yet. The mass is contributed by 3 components: the big sphere m1, small spheres m2, and the steel bar m3. Since all of them are steel, they have the same density. We have:
where h is the length of the steel bar, int is the floor function (every 3 meter, we have one small sphere), d = 7.8 g/cm3 = 7,800 kg/m3 is the density of steel, V1, V2 are the volumes of big and small spheres. The volume of the steel bar is the product of the crossed area with the length of the bar.
We can calculate V1, V2 by the sphere volume formula [4]:
where the radius r1 = d1/2 = 25 cm = 0.25 m, r2 = d2/2 = 15 cm = 0.15 m. Don't round both volume either. We already know A, so the equation now should be:
We know the right hand side of the equation but the left hand side is a little bit tricky. So, we can do a method of trial-and-error to test. Solve h by the following equation:
m = F /g = rho * 5/4 t2 tan(54°) / g = 28060.793485650833 kg
where g = 9.81 m/s2 is the gravitational acceleration.
Don't round anything yet. The mass is contributed by 3 components: the big sphere m1, small spheres m2, and the steel bar m3. Since all of them are steel, they have the same density. We have:
m1 + int(h/3) * m2 + m3 = m
<=>[V1 + int(h/3) * V2 + A * h] * d = m
<=>[V1 + int(h/3) * V2 + A * h] * d = m
where h is the length of the steel bar, int is the floor function (every 3 meter, we have one small sphere), d = 7.8 g/cm3 = 7,800 kg/m3 is the density of steel, V1, V2 are the volumes of big and small spheres. The volume of the steel bar is the product of the crossed area with the length of the bar.
We can calculate V1, V2 by the sphere volume formula [4]:
V1 = 4/3 * pi * (r1)3 = 0.065449846949787 m3
V2 = 4/3 * pi * (r2)3 = 0.014137166941154 m3
V2 = 4/3 * pi * (r2)3 = 0.014137166941154 m3
where the radius r1 = d1/2 = 25 cm = 0.25 m, r2 = d2/2 = 15 cm = 0.15 m. Don't round both volume either. We already know A, so the equation now should be:
V1 + int(h/3) * V2 + A * h = m/d
<=>int(h/3) * V2 + A * h = m/d - V1
<=>int(h/3) * V2 + A * h = m/d - V1
We know the right hand side of the equation but the left hand side is a little bit tricky. So, we can do a method of trial-and-error to test. Solve h by the following equation:
h/3 * V2 + A * h = m/d - V1
<=>(V2/3 + A) * h = m/d - V1
<=>h = (m/d - V1)/(V2/3 + A) = 654.0200901109209 m
<=>h = (m/d - V1)/(V2/3 + A) = 654.0200901109209 m
Now round h down to an integer. We then have h = 654 m. Next, test h into the equation int(h/3)* V2 + A * h = m/d - V1. If the left hand side is smaller than the right hand side, the steel bar can withstand the gravitational forces. The largest h that is satisfied that the left hand side is smaller than the right hand side will be the answer.
Using h = 654 m, the left hand side: int(h/3)* V2 + A * h = 3.5320. The right hand side: m/d - V1 = 3.5321 (good). Now increase h by 1 (h = 655 m) to see if the steel bar can still withstand. The left hand side: int(h/3)* V2 + A * h = 3.5327. The right hand side: m/d - V1 = 3.5321. (not good). Therefore, the answer is h = 654 m.
Results: 986 people guessed the correct answer earning themselves 2029 NP each.
References:
[1] Stress
[2] Regular pentagon area
[3] Newton's gravity law
[4] Sphere volume
Using h = 654 m, the left hand side: int(h/3)* V2 + A * h = 3.5320. The right hand side: m/d - V1 = 3.5321 (good). Now increase h by 1 (h = 655 m) to see if the steel bar can still withstand. The left hand side: int(h/3)* V2 + A * h = 3.5327. The right hand side: m/d - V1 = 3.5321. (not good). Therefore, the answer is h = 654 m.
Results: 986 people guessed the correct answer earning themselves 2029 NP each.
References:
[1] Stress
[2] Regular pentagon area
[3] Newton's gravity law
[4] Sphere volume
Thursday, April 24, 2008
Round 259
Date Released: April 24th, 2008.
Date Rewarded: May 1st, 2008.
A Chomby was walking down a path one day when Jhudora appeared in front of it. "Unscramble these tiles and solve the puzzle, and you may pass," she said.
Click to show/hide solution
Same as round 257 except this is a trickier one. The question is: How many neopoints does a game of tyranu evavu cost? The answer is 30.
Results: 4335 people guessed the correct answer earning themselves 462 NP each.
Date Rewarded: May 1st, 2008.
A Chomby was walking down a path one day when Jhudora appeared in front of it. "Unscramble these tiles and solve the puzzle, and you may pass," she said.
Prize
Speak Tyrannian
Speak Tyrannian
Click to show/hide solution
Same as round 257 except this is a trickier one. The question is: How many neopoints does a game of tyranu evavu cost? The answer is 30.
Results: 4335 people guessed the correct answer earning themselves 462 NP each.
Thursday, April 17, 2008
Round 258
Date Released: April 17th, 2008.
Date Rewarded: April 24th, 2008.
A Chomby was walking down a path one day when Jhudora appeared in front of it. "Answer this question, and you may pass," she said.
Suppose you have a perfectly spherical water tank with an inside diameter of 8.6 metres. If the drain at the bottom of the tank can't handle a hydrostatic pressure of more than 50 kilopascals, what is the maximum volume of water, in litres, that can be contained in the tank? Assume that gravitational acceleration is exactly 9.81 m/s2. Please round to the nearest 10 litre increment, and please submit only a number for your answer. (For example, if you calculate the answer to be 16277 litres, submit 16280 as your answer)
Click to show/hide solution
First off, we need to find the maximum height of the water tank that the drain can handle using hydrostatic pressure equation [1]:
where P = 50k Pa = 50,000 kg/m·s2 is the maximum hydrostatic pressure, rho = 1000 kg/m3 is density of water, g = 9.81 m/s2 is the gravitational acceleration. Remember that do not round h yet.
Then, we can start to calculate the partial volume of the sphere by the following formula (explanation of the formula is at the end of the solution):
Date Rewarded: April 24th, 2008.
A Chomby was walking down a path one day when Jhudora appeared in front of it. "Answer this question, and you may pass," she said.
Suppose you have a perfectly spherical water tank with an inside diameter of 8.6 metres. If the drain at the bottom of the tank can't handle a hydrostatic pressure of more than 50 kilopascals, what is the maximum volume of water, in litres, that can be contained in the tank? Assume that gravitational acceleration is exactly 9.81 m/s2. Please round to the nearest 10 litre increment, and please submit only a number for your answer. (For example, if you calculate the answer to be 16277 litres, submit 16280 as your answer)
Prize
Underwater Tour
Underwater Tour
Click to show/hide solution
First off, we need to find the maximum height of the water tank that the drain can handle using hydrostatic pressure equation [1]:
P = rho * g * h <=> h = P / (rho * g) = 5.096839959225281 m
where P = 50k Pa = 50,000 kg/m·s2 is the maximum hydrostatic pressure, rho = 1000 kg/m3 is density of water, g = 9.81 m/s2 is the gravitational acceleration. Remember that do not round h yet.
Then, we can start to calculate the partial volume of the sphere by the following formula (explanation of the formula is at the end of the solution):
V = 1000 * pi * h2 * (r - h/3) = 212276.11
where r = d/2 = 4.3 m, is the radius of the spherical water tank, h is the above maximum height.
We know that: 1 m3 = 1000 dm3 = 1000 liters. We have a 1000 factor in the above formula because the answer should be in liters while we calculated in cubic meters. Now perform rounding as the requirements and we get V = 212,280 liters.
Results: 462 people guessed the correct answer earning themselves 4330 NP each.
If you are interested to see how the partial volume of a sphere is derived, you can continue to read. Otherwise stop here.
We can consider the sphere with center at origin (above picture). We cut the partial sphere into small slices. Each slice is a circle, perpendicular to the vertical direction. Using Pythagorean theorem [2], we can calculate area of each circle at a certain height x:
Then we do integration on the whole vertical direction to add all area of the slices to get the volume. Since we choose the origin at the center of the sphere, the limit of the integration should be from -r to h-r
References:
[1] Hydrostatic pressure
[2] Pythagorean theorem
where r = d/2 = 4.3 m, is the radius of the spherical water tank, h is the above maximum height.
We know that: 1 m3 = 1000 dm3 = 1000 liters. We have a 1000 factor in the above formula because the answer should be in liters while we calculated in cubic meters. Now perform rounding as the requirements and we get V = 212,280 liters.
Results: 462 people guessed the correct answer earning themselves 4330 NP each.
If you are interested to see how the partial volume of a sphere is derived, you can continue to read. Otherwise stop here.
We can consider the sphere with center at origin (above picture). We cut the partial sphere into small slices. Each slice is a circle, perpendicular to the vertical direction. Using Pythagorean theorem [2], we can calculate area of each circle at a certain height x:
Then we do integration on the whole vertical direction to add all area of the slices to get the volume. Since we choose the origin at the center of the sphere, the limit of the integration should be from -r to h-rReferences:
[1] Hydrostatic pressure
[2] Pythagorean theorem
Thursday, April 10, 2008
Round 257
Date Released: April 10th, 2008.
Date Rewarded: April 17th, 2008.
Figure out how to unscramble these tiles to discover this week's Lenny Conundrum
Prize
Zeenana Crepe
Click to show/hide solution
One of they way to solve this problem is to print out the picture, use scissors to cut and try to unscramble the letters. The question is: What neopet species is last in alphabetical order? The answer is Zafara.
Results: 2272 people guessed the correct answer earning themselves 881 NP each.
Date Rewarded: April 17th, 2008.
Figure out how to unscramble these tiles to discover this week's Lenny Conundrum
Prize
Zeenana Crepe
Click to show/hide solution
Results: 2272 people guessed the correct answer earning themselves 881 NP each.
Tuesday, April 1, 2008
About us
Every week, The Neopets Team (TNT) releases a new Lenny Conundrum puzzle. We will try to post the hints to solve the current Lenny Conundrum puzzle while we don't want to give out the answer directly. Sometimes, solution is given out by us. Other times, the puzzle is solved by others. In any case, we always put the solution so you have chance to guess the answer. Answer will be posted when the contest is over. So, please don't include a specific number/answer in the comments. We will also try to post all the previous solutions of the Lenny Conundrum puzzles.
We hope you enjoy our blog and if you have any feedback to improve our blogs, feel free to drop us a comment. Thanks.
Notice: There was a busy time that we thought we had to shut down this blog. Fortunately, we didn't. This is our plan until further notice.
Every Wednesday: we create an entry for those who are interested in discussion. We may or may not be here to put the solution up. If we are here, we will update the solution.
Every weekend: we will guarantee to put the solutions up by weekends, could be earlier than that. You can check back any time. You can use the following feature or feed feature for easier notice.
We hope you enjoy our blog and if you have any feedback to improve our blogs, feel free to drop us a comment. Thanks.
Notice: There was a busy time that we thought we had to shut down this blog. Fortunately, we didn't. This is our plan until further notice.
Every Wednesday: we create an entry for those who are interested in discussion. We may or may not be here to put the solution up. If we are here, we will update the solution.
Every weekend: we will guarantee to put the solutions up by weekends, could be earlier than that. You can check back any time. You can use the following feature or feed feature for easier notice.
Thursday, February 1, 2007
Round 201
Date Released: February 1st, 2007.
Date Rewarded: February 8th, 2007.
Click to show/hide solution
This is another guessing game. If you are lucky, you guess it right. The answer is 35000.
Results: 95 people guessed the correct answer earning themselves 21053 NP each.
Date Rewarded: February 8th, 2007.
How many people answered last week's Lenny Conundrum incorrectly? Please round to the nearest 1000.
Prize
Guess the Card Game Table
Guess the Card Game Table
Click to show/hide solution
This is another guessing game. If you are lucky, you guess it right. The answer is 35000.
Results: 95 people guessed the correct answer earning themselves 21053 NP each.
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The 1st item in Appertisers: Maraquan Cream Broth.
The 2nd item in Main Courses: Spicy Radish Salad.
The 3rd item in Desserts: Peachpa and Stramberry Shell.
We observe that the words in the sequence are the second word in the 1st, 2nd and 3rd items on the 1st, 2nd and 3rd menu. Therefore, the next word should be the second word in the 4th items on the 4th menu (Cocktails): Phear Juice Tonic.
Results: 3477 people guessed the correct answer earning themselves 576 NP each.