Jump to content


Photo

Pony Mathematics test, word problems


  • Please log in to reply
22 replies to this topic

#1 Trotsky

Trotsky

    The last Brohoof admin

  • Banned
  • 512 posts
  • minecraft:
    HellSteedTrotsky

Posted 22 February 2016 - 12:44 AM

Princess Cadence has 34 bits. Princess Luna has ten times as many bits as Princess Cadence. Princess Celestia has the same number of bits Luna has, multiplied by the number of bits Luna has. How many bits do they have in total?

 

Show your answer with all your work and then write the next question.



#2 Aurora

Aurora

    The final hour is upon us.

  • Pony
  • 50 posts
  • minecraft:
    stevsumm

Posted 22 February 2016 - 12:51 AM

Together, the Princesses have 3,774 Bits.

 

With the number of Bits above, the three Princesses go shopping!  But first they have to pay for airship tickets to Saddle Arabia, which costs 1,039 Bits per ticket.  How many Bits will they have left after purchasing the airship tickets?



#3 jimme neutron corndog

jimme neutron corndog

    Is a woman

  • Builder
  • 1822 posts
  • minecraft:
    Sierif

Posted 22 February 2016 - 01:05 AM

657

Now Luna and cadance goes on one train, whilst celestia goes on the other.

two trains, Train A and Train B,which simultaneously depart Station A and Station B. Station A and Station B are 252.5 miles apart from each other. Train A is moving at 124.7mph towards Station B, and Train B is moving at 253.5mph towards Station A. If both trains departed at 10:00AM and it is now 10:08, how much longer until both trains pass each other?

#4 Trotsky

Trotsky

    The last Brohoof admin

  • Banned
  • 512 posts
  • minecraft:
    HellSteedTrotsky

Posted 22 February 2016 - 01:09 AM

Your answer is incorrect the actual number is 34+34*10+340^2 = 115974 bits

 

Using 3774 bits because that's what your problem is stated as the solution is:

 

3774 - 3*1039 = 657

 

Having bought tickets to Saddle Arabia, the three Princesses realize they'll need more bits to return to Equestria, or face a long journey home. Work is available for all three in Saddle Arabia paying 35 bits per hour doing admin work. The journey home would take 45 hours of flying and walking to get back to Canterlot, which choice gets the Princesses home faster? Continue the assumption that tickets cost 1039 bits.


Me and Sierif posted at the same time so please solve both problems and then just continue with a new problem.


Edited by Trotsky, 22 February 2016 - 02:07 AM.


#5 Hepolite

Hepolite

    Questionably questionable

  • Builder
  • 143 posts
  • minecraft:
    Hepolite

Posted 22 February 2016 - 02:14 AM

the actual number is 34+34*10+34^2 = 1530 bits

Cadence has 34 bits. Luna has ten times that, 340 bits. Celestia has the same amount as bits as Luna does, squared ("same number of bits Luna has, multiplied by the number of bits Luna has"). 340^2 = 115600. Total bit count is then 115974. But then again, Celestia has probably had a lot, lot more time to gather all these bits...

 

two trains, Train A and Train B,which simultaneously depart Station A and Station B. Station A and Station B are 252.5 miles apart from each other. Train A is moving at 124.7mph towards Station B, and Train B is moving at 253.5mph towards Station A. If both trains departed at 10:00AM and it is now 10:08, how much longer until both trains pass each other?

d = 406.3km, v_a = 200.6km/h, v_b = 407,9km/h.

t = d / (v_a + v_ b) = 0.668h, or approximately 40 minutes. This then leaves 32 minutes of travel time before they meet.

 

Having bought tickets to Saddle Arabia, the three Princesses realize they'll need more bits to return to Equestria, or face a long journey home. Work is available for all three in Saddle Arabia paying 35 bits per hour doing admin work. The journey home would take 45 hours of flying and walking to get back to Canterlot, which choice gets the Princesses home faster? Continue the assumption that tickets cost 1039 bits.

In order to earn enough to get home, they would need 3117 bits in total. Assuming initial money to be 657, since the actual sum would be significantly more, based on the first question. This means they need 2460, which implies that they have to work for 23.4 hours each. Without information on how fast the airship is, this question can't be fully answered. If the airship completes the trip in less than 21.6 hours, it would be faster to work, otherwise manual flight is faster.

 

...

 

Later, when got back to Canterlot they wanted to play a game where they could chose a sequence of three numbers from a pool of five. Getting all three numbers correct would give them a reward. Assuming they chose a sequence of numbers each (total three unique sequences), what are the odds that at least one of them won something?


Edited by Hepolite, 22 February 2016 - 02:29 AM.


#6 JoyJoy

JoyJoy

    I don't listen to rap music

  • Builder
  • 2172 posts
  • minecraft:
    JoyJoy

Posted 22 February 2016 - 06:48 AM

x: cadence

y: luna

z: celestia

 

x = 34

y = 10x

z = y2

 

y = 340

z = 3402

 

z = 115600

 

x + y + z = 115974

 

 

 



#7 Luna Lens

Luna Lens

    Hiya

  • Administrator
  • 2617 posts
  • minecraft:
    Princess_Luna

Posted 23 February 2016 - 05:45 AM

Later, when got back to Canterlot they wanted to play a game where they could chose a sequence of three numbers from a pool of five. Getting all three numbers correct would give them a reward. Assuming they chose a sequence of numbers each (total three unique sequences), what are the odds that at least one of them won something?

 
Hopefully I've done everything right. I calculated the probability based on the total number of combinations
 
Total number of combinations.
Where,
n = number of objects from which the combination is formed.
r = number of objects used to form the combination.
C(n,r) = Total combinations of n using r.
C(n,r) = n! / ( r! (n - r)! )
C(5,3) = 5! / ( 3! (5 - 3)! )
C(5,3) = 120 / ( 6 x 2!)
C(5,3) = 120 / 12
C(5,3) = 10

where, 
n(A) - number of event occurs in A
n(S) - number of possible outcomes
n(B) - number of event occurs in B
n(C) - number of event occurs in C
Each princess can choose 1 out of 10 possible combinations.

Probability that event A occurs P(A) = n(A) / n(S).
1 / 10  = 0.1
Probability that event B occurs P(B) = n(B) / n(S).
1 / 10 = 0.1
Probability that event C occurs P(C) = n(C) / n(S).
1 / 10= 0.1

Probability that all events occur P(A ∩ B ∩ C) = P(A) x P(B) x P(C).
0.1 X 0.1 X 0.1 = 0.001

Probability that either of event occurs P(A ∪ B ∪ C) = P(A) + P(B) + P(C) - P(A ∩ B ∩ C).
0.1 + 0.1 + 0.1 - 0.001 = 0.299
So the probability that one of the three princesses draws the correct numbers is apparently 0.299.

#8 Trotsky

Trotsky

    The last Brohoof admin

  • Banned
  • 512 posts
  • minecraft:
    HellSteedTrotsky

Posted 23 February 2016 - 03:58 PM

so that would be 30% roughly...

 

New Problem:

Twilight Sparkle is sorting her books, and she decides to order them by title and the scientific notation of how many title shes has starting with that letter. Considering she has 1.36 x 10^3 total books. She wants to arrange them such that no bookshelf contains more than 3.9 x 10^2 books per shelf. She has two dozen shelves. How many additional books can she have or how many does she need to get rid of to perfectly fill her shelves?



#9 Aurora

Aurora

    The final hour is upon us.

  • Pony
  • 50 posts
  • minecraft:
    stevsumm

Posted 23 February 2016 - 04:13 PM

If my math is right (probably not). Twilight has room for 8,000 more books. Ya, that doesn't sound right to me. Could someone let me know where I went wrong?

1.36 x 10^3 = 1,360 total books
3.9 x 10^2 = 390 books per shelf
390 x 24 = 9,360 # of books it would take to fill every shelf
9,360 - 1,360 = 8,000 # of books she has room for

Edited by Aurora, 23 February 2016 - 04:19 PM.


#10 Trotsky

Trotsky

    The last Brohoof admin

  • Banned
  • 512 posts
  • minecraft:
    HellSteedTrotsky

Posted 23 February 2016 - 04:17 PM

two dozen = 24



#11 Aurora

Aurora

    The final hour is upon us.

  • Pony
  • 50 posts
  • minecraft:
    stevsumm

Posted 23 February 2016 - 04:19 PM

That was a typo, the answer is for 24, I just put down 12

#12 Trotsky

Trotsky

    The last Brohoof admin

  • Banned
  • 512 posts
  • minecraft:
    HellSteedTrotsky

Posted 23 February 2016 - 04:21 PM

I made the problem up on the fly so I don't know what the answer is, could have come out to be some nice pretty round number, don't know.



#13 Hepolite

Hepolite

    Questionably questionable

  • Builder
  • 143 posts
  • minecraft:
    Hepolite

Posted 23 February 2016 - 07:24 PM

 Hopefully I've done everything right. I calculated the probability based on the total number of combinations
 [Snip]

So the probability that one of the three princesses draws the correct numbers is apparently 0.299.

Very close, but off by just a tiny fraction. When calculating P(A B C), P(A B C) isn't needed here. P( A ), P( B ) and P( C ) are all separate in the probability space with no overlap (thus no P(x | y) terms), so they can be simply added together for the final solution. The correct solution is then 0.30, or 30% chance of at least one of them winning.

Stupid emoticons mess up the posts with math... ._.


Edited by Hepolite, 23 February 2016 - 07:26 PM.


#14 Super Ultra Uber C00l Dood

Super Ultra Uber C00l Dood

    ▒▒▒▒▒██ ██▒▒██▒▒██▒▒██

  • Pony
  • 955 posts
  • minecraft:
    StateFarmJoe

Posted 23 February 2016 - 10:07 PM

what's 2 + 2



#15 Trotsky

Trotsky

    The last Brohoof admin

  • Banned
  • 512 posts
  • minecraft:
    HellSteedTrotsky

Posted 24 February 2016 - 03:13 AM

5



#16 Luna Lens

Luna Lens

    Hiya

  • Administrator
  • 2617 posts
  • minecraft:
    Princess_Luna

Posted 24 February 2016 - 03:49 AM

what's 2 + 2


10

#17 Trotsky

Trotsky

    The last Brohoof admin

  • Banned
  • 512 posts
  • minecraft:
    HellSteedTrotsky

Posted 24 February 2016 - 04:45 AM

Problem:

 

A pony train engineer is trying to stop his train to precisely at the correct stopping point. He approximates that he is 1000 ft from the stopping point. He is going at 20 mph, and he knows from experience that the brakes decelerate the train at a constant 5ft/s. In how many seconds should he apply the brake to stop at the correct point, or is he already going to pass the mark? (Don't worry he won't crash, plenty of open track ahead.)



#18 Trotsky

Trotsky

    The last Brohoof admin

  • Banned
  • 512 posts
  • minecraft:
    HellSteedTrotsky

Posted 02 March 2016 - 05:58 AM

Here's a more practical problem, So given a ponies total height from hoof to top of mane I can get a rough estimate of their total mass using the assumption that ponies are the same density as water (which is roughly true for most mammals) So if someone can give me a total height for one of the mane six or Princess Luna or Celestia, I can give you an approximation of their mass based on that measurement (I can't do weight because I don't know Equestria's gravitational value, I can only compute mass. If you want the weight on Earth just multiply by 9.81 m/s^2 for SI units and 32.2 ft/s^2 for English units. The results being in Newtons and Pounds respectively from multiplying in those constants.)

 

If you give me heights in English units I will give you output in Slugs, If in SI units I will give you Kilograms.


Edited by Trotsky, 02 March 2016 - 05:59 AM.


#19 Luna Lens

Luna Lens

    Hiya

  • Administrator
  • 2617 posts
  • minecraft:
    Princess_Luna

Posted 02 March 2016 - 10:13 PM

my-little-pony-%D1%84%D1%8D%D0%BD%D0%B4%

#20 Trotsky

Trotsky

    The last Brohoof admin

  • Banned
  • 512 posts
  • minecraft:
    HellSteedTrotsky

Posted 03 March 2016 - 05:09 AM

Luna is roughly 6ft so my calculator did the analysis and said that Luna is 20,160.9 in^3 multiplying by the factor of 0.036127Ibm/in^3, I get 728 pounds mass. which on earth would be the same weight of 728 pounds force. This is a reasonable answer given that a normal horse weighs around 1000-1400 Ib. Therefore a pony of Luna's size could be expected to be around half the mass of a horse. This equation also means that Luna's wings must support at least this force in level flight. GIven that the cross section of Luna's wing bones appears to be 1 inch underneath the feathers. Making the assumption that Luna's center of lift is 2ft from her body, and that the end of Luna's wing is held up by a simple pinned bone bearing and and an applied muscle force applied 4 inches into her body. Then the maximum shear stress in the bone would be solving for x in the equation 728 Ib *24 in - x*4 in =0 by taking moments about the muscle force to determine the maximum force acting on the bone. We get that at the bone support the bone is holding a load of 4368 Ib over the area of that cross section of pi * 1 in ^2/4 gives a stress of 5561.5 psi . Given that an experiment showed that bone starts to fail in shear at 51.6 MPa. We need to convert psi to Pa. Using an online converter this turned out to be 38.35 MPa meaning that Luna's wing bone has a factor of safety of 1.34 the maximum upward acceleration Luna can achieve before being at severe risk of wing failure is about 0.304 ft/s^2. A more reasonable 0.15 ft/s^2 might be actually achieved.

 

All in a days work for a young mechanical engineering pony...



#21 Trotsky

Trotsky

    The last Brohoof admin

  • Banned
  • 512 posts
  • minecraft:
    HellSteedTrotsky

Posted 03 March 2016 - 02:15 PM

Realized in the above problem I derped and used Luna's full wing load on only one wing. I'd have to repeat with half the initial load to make it work right again, but it was just for fun anyway. The mass calculation should be fine.



#22 Tesla

Tesla

    CAPCOM (Mission Control)

  • Pony
  • 139 posts
  • minecraft:
    __Tesla__

Posted 06 March 2016 - 06:43 AM

New Question:

Princess Celestia wants to fly to Manehattan. To do so, she leaves Canterlot with a bearing of N 58 degrees W and flies for 3 hours and 43 minutes at a constant speed of 70mph. After she arrives, she receives news of an urgent request for aid from the city of Baltimare. To get there, she flies for 2 hours and 11 minutes at a constant speed of 100mph. After she helps calm the riots, what direction (bearing) does Celestia need to head to return to Canterlot on the most direct path? How long will it take Celestia to get to Canterlot at a speed of 70mph? What bearing did she face to get from Manehattan to Baltimare?

Assume Baltimare and Canterlot are at the same latitude.

Edited by Tesla, 06 March 2016 - 07:16 AM.


#23 mewtwotwotwo2

mewtwotwotwo2

    Background Pony

  • Pony
  • 8 posts
  • minecraft:
    mewtwotwotwo2

Posted 22 March 2016 - 02:16 AM

my-little-pony-%D1%84%D1%8D%D0%BD%D0%B4%

does big mac have a horn?.... XD




0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users