G'day possums, here are yesterdays results as voted for in a nationwide poll.
Mark:
before:
300 men, 276 women
after:
351 men, 325 women
Q. The AtwoK dating service has m men and w women, such that m>w. Introductions are made by randomly drawing any two names (without replacement) out of the total membership. If one man and one woman is drawn then a blind date is arranged. The probability that a draw will result in one of each gender is exactly 50%. After an advertising campaign to attract women to the club, 100 more members join the service. After the increase in membership, both properties still hold true: m>w and a 50% probability of a successful drawing.
After the increase, how many men and how many women are in the service?
First let's define some terms:
m = men before increase
w = women before increase
m' = men after increase
w' = women after increase
d = m-w
d' = m'-w'
The fact that a random drawing will result in a date 50% of the time tells us that 2*(m/(m+w))*(w/(m+w-1)) = 0.5
(m/(m+w))*(w/(m+w-1)) = .25
m/(m+w) = (m+w-1)/(4w)
4mw = (m+w)*(m+w-1) 4mw = m^2 + mw -m + mw + w^2 - w
4mw = m^2 + w^2 -2mw - m - w
(m-w)^2 = m + w, or d^2 = m+w
By the same reasoning:
(m'-w')^2 = m' + w', or d'^2 = m'+w'
We are also given:
m'+w'-m-w=100
d'^2 - d^2 = 100
(d'+d)*(d'-d) = 100
We know that d and d' must be integers so the only possible values for d'+d and d'-d are: (100,1), (50,2), (25,4), (20,5), (10,10).
Let's consider the first possibility of (100,1). If this were the solution then d'+d=100 and d'-d=1. Adding the two equations together yields 2d'=101 --> d'=101/2.
However since m' and w' are integers then d' is also an integer and cannot be 101/2. For the same reason (25,4) and (20,5) are not possible because the sum of the two factor is odd.
Let's consider (10,10). Then d'+d=10, d'-d = 10 --> d=0 --> m-w=0. However the problem stated m>w, so (10,10) doesn't work.
That leaves only (d'+d,d'-d)=(50,2). Adding the two equations gives us d'=m'-w'=26.
Remember, d'2 = m'+w', so m'+w'=262 = 676.
Remember also that m'-w'=26
Adding the two equations give us 2m'=702 --> m'=351 --> w'=325.
So after in increase there are 325 women and 351 men. From here, it can also be easily found that before the increase there were 300 men and 276 women.
Many thanks to Shari for her help in formulating this problem.
Highway overpass
foriegn language
Hey! That's not my avatar. I'll have you know, that's a family portrait.
Mark, who has now been removed as the head of my re-election committee, posted the following on the FBI board; "Message from stalker Try to Shari" (edited)
Playing card riddle:
Much I marvelled to hear discourse so plainly,
Though its answer little meaning - little relevancy bore;
Then this vision beguiling my sad fancy into smiling,
By the grave and stern decorum of the countenance she wore -
Shari just missed the Big prize with, "7 of clubs, jack of diamonds and ace of spades."
It is clear to me that Mark picked up on the clues and came up with the answer:
SHARI'S ONE HOT CHICK
KS, QS, QH
Let's label the clues as follows:
1. To the right of the King there's a Queen or a two.
2. To the left of a Queen there's a Queen or a two.
3. To the left of a Heart there's a Spade or a two.
4. To the right of a Spade there's a Spade or a two.
From clues 1 and 2 we can infer that a king is on the left and a queen on the right and the middle card is either a queen or a 2.
First consider the possibility that a 2 is in the middle. If there are two consecutive spades (from clue 4) then then three cards must be the king of spades, 2 of spades, queen of hearts. If to the right of a spade there is a 2 (again from rule 4) then the three cards are the kings of spades, a two of unknown suit, and the queen of hearts. So if we assume a 2 in the middle it is not certain what the suit of the 2 is.
Next consider the possibility that a queen is in the middle. Clue 4 tells us there must be two consecutive spades, since there is no 2. Both queens can not be spades since they are from the same deck, thus the left and center cards must be spades. Clue 3 tells us the right queen is a heart. So the three cards would be the king of spades, queen of spades, and queen of hearts.
So a 2 in the middle does not lead to a definitive answer, but a queen in the middle does. The question implies there is just one possible answer. Therefore, there is a queen in the middle, leading to the final answer of:
King of spades, queen of spades, and queen of hearts.
Credit for this problem goes to Martin Gardner who published it in the November 1965 Scientific American.
"Maybe that's why I need glasses"
They say, girls who wear glasses, have the cutest
Arrh say's the same thing.
Rap:
Train speed
Slow train speed is s (mph)
Fast train speed is 2s (mph)
t is time to meet (hr)
d is distance between stations
2st+st=d (mi)
3st=d (mi)
meeting (mo) point is d/3 (mi) if both trains leave at same time
if slow train is 5 minutes late
fast train covers 2s/12 (mi) and the trains are d-2s/12 (mi) apart
the meeting point (m1) is (d-2s/12)/3mi and |m0-m1|=2
|d/3-(d-2s/12)/3|=2 mi
or 2s/36=2 and s=36 mph
if fast train is 5 minutes late
slow train covers s/12 (mi) and the trains are d-s/12 (,i) apart
the meeting point (m1) is s/12+(d-s/12)/3(mi)=(d+2s/12)/s mi and |m0-m1|=2
|d/3-(d-2s/12)/3|=2 mi
or 2s/36=2mi and s=36mph
So slow train speed is 36mph
Mark agrees, "I did it graphically. I ended up with the sum of the times it takes each train to go 2 miles is 5 minutes. Therefore, the slow train goes 36mph."
He also finds time to come up with:
many thanks
safety in numbers
under cover
makeup
Dadpad I am reliably informed, is still walking the track
Welcome Xtwitch, your arrival has been long awaited. If you have high ambitions and an enquiring mind dedicated to eradicating world hunger and poverty, then you are probably in the wrong place. If you just wanna have fun, git reading and jump right in.
FLUBADENCE
OONWET
ELSEVEVEN
STEFRANKIN
A, B, and C are three towns, each connected by a network of roads.
There are 82 ways to get from A to B, including those routes that pass through C.
There are 62 ways to get from B to C, including those routes that pass through A.
The number of ways to get from A to C, including those ways passing through B, is less than 300.
How many ways are there to get from A to C
